THEORY OF STRUCTURES
AND
STRENGTH OF MATERIALS.
WITH
DIAGRAMS, ILLUSTRATIONS, AND EXAMPLES.
BY
HENRY T. BOVEY, M.A., D.C.L., F.R.S.C.,
PROFESSOR OF CIVIL ENGINEERING AND APPLIED MECHANICS, M*GILL UNIVERSITY, MONTREAL.
MEMBER OF THE INSTITUTION OF CIVIL ENGINEERS; MEMBER OF THE INSTITU-
TION OF MECHANICAL ENGINEERS; LATE FELLOW OP
QUEENS' COLLEGE, CAMBRIDGE (ENG.).
FIRST THOUSAND.
NEW YORK:
JOHN WILEY & SONS,
53 EAST TENTH STREET.
1893.
COPYRIGH^, 1893,
BV
HENRY T. BOVEY.
ROBERT DRTJMMOND,
886 Pearl Street,
New York.
444 & 446 Pearl Street,
New York.
DEDICATED
WHOSE BENEFACTIONS TO M*GILL UNIVERSITY
HAVE DONE SO MUCH TO ADVANCE THE CAUSE OF
SCIENTIFIC EDUCATION.
PREFACE.
THE present work treats of that portion of Applied
Mechanics which has to do with the Design of Structures.
Free reference has been made to the works of other
authors, yet a considerable amount of new matter has been
introduced, as, for example, the Articles on " Surface Loading"
by Carus-Wilson, " The Flexure of Columns " by Findlay, and
" The Efficiency of Riveted Joints " by Nicolson ; also my
own Articles on " Maximum Shearing Forces and Bending
Moments," " The Flexure of Long Columns," " The Theorem
of Three Moments," etc.
I am much indebted to Messrs. C. F. Findlay and W. B.
Dawson for valuable information respecting the treatment of
Cantilever Bridges, Arched Ribs, and the Live Loads on Bridges.
To Messrs. J. M. Wilson, P. A. Peterson, C. Macdonald,
and others, many thanks are due for data respecting the Dead
Weights of Bridges.
I am under deep obligation to my friend Prof. Chandler,
who has kindly revised the proof-sheets, and who has made
many important suggestions.
I have endeavored so to arrange the matter that the
student may omit the advanced portions and obtain a com-
plete elementary course in natural sequence.
At the end of each chapter, a number of Examples,
VI PREFACE.
selected for the most part from my own experience, are
arranged with a view to illustrating the subject-matter an
important feature, as it is admitted that the student who care-
fully works out examples obtains a mastery of the subject
which is otherwise impossible.
The various Tables in the volume have been prepared from
the most recent and reliable results.
A few years ago I published a work on " Applied Me-
chanics," consisting mainly of a collection of notes intended
for the use of my own students. The present volume may be
considered as a second edition of that work, but the subject-
matter has been so much added to and rearranged as to make
it almost a new book. I venture to hope that this volume
may prove acceptable not only to students, but to the profes-
sion at large.
HENRY T. BOVEY.
McGiLL COLLEGE, MONTREAL,
November, 1892.
CONTENTS.
CHAPTER I.
FRAMED STRUCTURES.
PAGE
Definitions I
Frames of Two or More Members 2
Funicular Polygon 3
Polygon of Forces 4
Line of Loads 5
Mansard Roof 6
Non-closing Polygons ... 7
Funicular Curve 10
Centre of Gravity 1 1
Moment of Inertia 12
Cranes, Jib 13
" Derrick 16
" Composite 31
Shear Legs ^ \~f
Bridge Trusses 17
Roof Trusses i >
King-post Truss 21
Incomplete Frames 27
Queen-post Truss 31
Composite Frames 32
Roof-weights 37
Wind-pressure 38
Distribution of Loads 39
Examples of Roof Trusses 41
Examples of Bridge Trusses (Fink, Bollman, Howe, Bowstring, Single-
intersection, etc) 52
Method of Sections 62
Piers 65
vii
VI il CONTENTS.
Tables of Roof -weights and Wind-pressures ............................ 67
Examples ............................................................. 7
CHAPTER II.
SHEARING FORCES AND BENDING MOMENTS.
Equilibrium of Beams ................................................ 93
Shearing Force .................................. ...................... 95
Bending Moment ................... ................................ 96
Examples of Shearing Force and Bending Moment ...................... 97
Relation between Shearing Force and Bending Moment .................. 108
Effect of Live (or Rolling) Load ........................................ in
Graphical Representation of Moment of Forces with Respect to a Point.. . 116
Relation between Bending Moments and Funicular Polygon .............. 118
Maximum Shear and Maximum Bending Moment at any Point of an Arbi-
trarily Loaded Girder ............................................. 121
Hinged Girders ....................................................... 126
Examples ........... ' ................................................ 131
CHAPTER III.
GENERAL PRINCIPLES, ETC.
Definitions. . . ....................................... ................ 140
Stress, Simple ........................................................ 140
" Compound .................................................... 140
Hooke's Law ........................................................ 141
Coefficient of Elasticity ................................... ............ 141
Poisson's Ratio ........................ . .............................. 142
Effect of Temperature ................................................. 142
Specific Weight ....................................................... 143
Limit of Elasticity ................................................... 143
Breaking Stress ....... ............................................... 147
Dead and Live Loads ................................................. 143
Repeated Stress Effect ................................................ 145
Wohler's Experiments ................................................ 145
Testing of Metals .................................................... 147
Launhardt's Formula .................................................. 159
Wey ranch's Formula ............... ................................... 153
Unwin's Formula ......................................... . ......... 159
Flow of Solids ....................................................... 162
Work, Internal and External ........................................... 168
Energy, Kinetic and Potential ......................................... 167
Oblique Resistance .................................................. 169
Values of k .......................................................... 174
Momentum. Impulse .................. . ............................ 176
Angular Momentum .................................................. 177
CONTENTS. . IX
PAGE
Useful Work. Waste Work 178
Centrifugal Force 181
I m pact 1 84
Extension of a Prismatic Bar 189
Oscillatory Motion of a Weight at the End of a Vertical Elastic Rod 190
Inertia 198
Balancing 198
Curves of Piston Velocity 205
Linear Diagrams of Velocity 206
Curves of Crank-effort. . . 207
Curves of Energy 207
Fluctuation of Energy 207
Tables of Strengths, Elasticities, and Weights of Materials 210
Tables of the Breaking Weights and Coefficients of Bending Strengths of
Beams 213
Table of the Weights and Crushing Weights of Rocks, etc 214
Table of Expansions of Solids 215
Examples 216
CHAPTER IV.
STRESSES, STRAINS, EARTHWORK, AND RETAINING WALLS.
Internal Stresses 235
Simple Strain 235
Compound Strain 236
Principal Stresses 240
Curves of Maximum Shear and Normal Intensity 240
Combined Bending and Twisting Stresses 244
Combined Longitudinal and Twisting Stresses 247
Conjugate Stresses 247
Relation between Principal and Conjugate Stresses 247
Ratio of Conjugate Stresses 250
Relation between Stress and Strain 251
Rankine's Earthwork Theory 255
Pressure against a Vertical Plane 257
Earth Foundations 258
Retaining Walls 260
Retaining Walls. Conditions of Equilibrium 260
Rankine's Earthwork Theory applied to Retaining Wall? 264
Line of Rupture 265
Practical Rules respecting Retaining Walls 267
Reservoir Walls 271
General Case of Reservoir Walls 275
General Equations of Stress 276
Ellipsoid of Stress 281
Stress-strain Equations 281
X CONTENTS.
PAGE
Isotropic Bodies 283
Relation between A , A, and G 285
Traction 287
Torsion 288
Work done in the Small Strain of a Body (Clapeyron) 292
Examples 294
CHAPTER V.
FRICTION.
Friction 300
Laws of Friction 300
Inclined Plane 301
Wedge , 302
Screws . 306
Endless Screw 309
Rolling Friction 310
Journal Friction 312
Pivot 316
Cylindrical Pivot 316
Wear 318
Conical Pivot 319
Schiele's Pivot {anti-friction) 320
Belts and Ropes 321
Brakes 323
Effective Tension of a Belt 324
Effect of High Speed 325
Slip of Belts 326
Prony's Dynamometer 327
Stiffness of Belts and Ropes 327
Wheel and Axle 329
Toothed Gearing 331
Bevel- wheels 335
Efficiency of Mechanisms 335
Table of Coefficients of Journal Friction 336
Examples 337
CHAPTER VI.
t
TRANSVERSE STRENGTH OF BEAMS.
Elastic Moment 340
Moment of Resistance 340
Neutral Axis 340
Transverse Deformation 344
Coefficient of Bending Strength 344
CONTENTS. XI
PAGE
Equalization of Stress 349
Surface Loading 350
Effect of Bending Moment in a Plane which is not a Principal Plane 354
Springs 355
Beams of Uniform Strength 358
Flanged Girders 365
Classification of Flanged Girders 365
Equilibrium of Flanged Girders , 366
Moments of Inertia of I and other Sections 371
Design of a Girder of I-section 381
Deflection of Girders 384
Camber 387
Stiffness 389
Distribution of Shearing Stress 391
Beam acted upon by Forces Oblique to its Direction 396
Similar Girders 401
Allowance to be made for Weight of Beam 405
Examples 407
CHAPTER VII.
TRANSVERSE STRENGTH OF BEAMS {Continued.}
General Equations 428
Interpretation of the General Equations 432
Examples Cantilever 435
Girder upon Two Supports 439
" fixed at One End and resting upon Support at the
Other 442
" " fixed at Both Ends 445
" " upon Two Supports not in the same Horizontal Plane. 446
" Neutral Axis of Arbitrarily Loaded Girders 448
" Cantilever with Varying Section 455
" Girder Encastre at the Ends 458
Springs 456
Work done in bending a Beam 460
Transverse Vibrations of a Beam supported at Both Ends , 461
Imperfect Fixture 461
Continuous Girders 463
Theorem of Three Moments 463
Swing-bridges 470
Maximum Bending Moment at the Points of Support of a Continuous
Girder of n Spans 475
General Theorem of Three Moments 484
Comparative Merits of Continuous Girders 486
Examples 490
Xll CONTENTS.
CHAPTER VIII.
PILLARS.
PAGE
Classification of Pillars 513
Form " " 514
Failure " " 515
Uniform Stress 516
Uniformly Varying Stress 517
Hodgkinson's Formulae 520
Gordon's Formula 522
Values of the Coefficients (a and f) in Gordon's Formula 523
Graphical Representation of Strength of Pillars 524
Rankine's Modification of Gordon's Formula 526
Formula for Safe Working Stress 526
Value of " Radius of Gyration " for Different Sections 526
American Iron Columns 532
Long Thin Pillars 534
Long Columns of Uniform Section (Euler's Theory) 538
Resistance of Columns to Buckling (Weyrauch's Theory) 550
Baker's Formulae .... 549
Flexure of Columns , 554, 557
Examples 563
CHAPTER IX.
TORSION.
Definition 568
Coulomb's Laws 568
Torsional Strength of Shafting 569
St. Venant's Results 572
Torsional Rupture 572
Resilience of Shafting 574
Effect of Combined Bending and Twisting 574
Distance between Bearings for Shafting 575
Efficiency of Shafting 577
Spiral Springs 577
Figures illustrating the Distortion produced by twisting Round, Square,
and Rectangular Iron Bars 5850, 585^
Examples 580
CHAPTER X.
CYLINDRICAL AND SPHERICAL BOILERS.
Cylinders 586
Efficiency of Riveted Joints in Boilers 587
CONTENTS. X1U
PAGE
Thick Hollow Cylinder 588
Spherical Shells 591
Practical Formulae 592
Examples 594
CHAPTER XL
BRIDGES.
Classification 597
Curved and Horizontal Flanges 597
Depth of Girders (or Trusses) 597
Position of Platform 598
Comparative Advantages of Two, Three, and Four Main Girders 600
Dead Load 600
Live Load 600
Trellis (or Lattice) Girder 600
Warren Girder 603
Howe Truss 61 1
Single-intersection Truss 616
Double-intersection Truss 616
Whipple Truss 618
Linvillc Truss 618
Post Truss 618
Quadrangular Truss 618
Bowstring Truss 618
Bowstring Truss with Isosceles Bracing 624
Bowstring Suspension-bridge (Lenticular Truss) 626
Cantilever Trusses 627
Curve of Cantilever Boom 634
Deflection of Cantilevers 638
Rollers 639
Wind-pressure 651
Regulations respecting Wind-pressure 653
Lateral Bracing 654
Chords 655
Stringers 656
Maximum Allowable Working Stresses 657
Camber 659
Rivet Connections between Flange and Web 660
Eye-bars, Pins, and Rivets 661
Steel Eyebars . 665
Rivets 666
Dimensions of Rivets 667
Strength of Punched and Drilled Plates 668
Riveted Joints 668
Theory of Riveted Joints 671
XIV CONTENTS.
Covers 675
Efficiency of Riveted Joints 676
Tables of Weights of Actual Bridges 682-687
Table of Loads for Highway Bridges 688
Examples 689-702
CHAPTER XII.
SUSPENSION-BRIDGES.
Cables 703
Anchorage 704
Suspenders . . 706
Curve of Cable (catenary) 706
Link " Cable" 709
Length of Cable 712
Weight of Cable 713
Deflection due to Change of Length 714
Pressure upon Piers 718
Stiffening Truss 719
Stiffening Truss hinged at the Centre 725
Suspension-bridge Loads 730
Modifications of the Suspension-bridge proper 731
Examples 734~739
CHAPTER XIII.
ARCHED RIBS.
Definitions 740
Equilibrated Polygon (Line of Resistance) 741
Polygon of Pressures 743
Linear Arch 743
Conditions of Equilibrium 745
Joint of Rupture 747
Minimum Thickness of Abutment 749
Empirical Formulae 750
Linear Arch in Form of a Parabola 750
" " " " " Transformed Catenary 750
" " " " " Circular Arc 753
" " " " " Elliptic Arc 753
Hydrostatic Arch 757
Geostatic Arch , 759
General Arch Theory 760
Arched Ribs 762
Bending Moment and Thrust at any Point of an Arched Rib 763
Rib with Hinged Ends 764
CONTENTS. XV
PAGE
Semicircular Rib with Hinged Ends ' 765
Graphical Determination of the Thrust at any Point of an Arched Rib 767
Rib in Form of a Circular Arc 769
Rib with Fixed Ends * 771
" " " " in Form of a Circular Arc 773
" " " " " ' " Semicircle 775
" " " " " " " Parabola 775
Effect of a Change of Temperature 777, 786
Deflection of an Arched Rib 780, 802
Elementary Deformation of an Arched Rib 781
Rib of Uniform Stiffness 788
Parabolic Rib of Uniform Depth and Stiffness 789, 795, 800
Arched Rib of Uniform Stiffness with Fixed Ends 804
Stresses in Spandril Posts and Diagonals 804
Maxwell's Method of Determining the Stresses in a Framed Arch 806
Examples 809-812
THEORY OF STRUCTURES.
CHAPTER I.
FRAMES LOADED AT THE JOINTS.
I. Definitions. Frames are rigid structures composed oi
straight struts and ties, jointed together by means of bolts,
straps, mortises and tenons, etc. Struts are members in com-
pression, ties members in tension, and the term brace is applied;
to either.
The external forces upon a frame are the loads and the
reactions at the points of support, from which may be found
the resultant forces at the joints. The greatest care should be
exercised in the design of the joints. The resultant forces
should severally coincide in direction with the axes of the
members upon which they act, and should intersect the joints
in their centres of gravity. Owing to a want of homogeneity
in the material, errors of workmanship, etc., this coincidence is
not always practicable, but it should be remembered that the
smallest deviation introduces a bending action. Such an
action will also be caused by joint friction when the frame is
insufficiently braced. The points in which the lines of action
of the resultants intersect the joints are also called the centres
of resistance, and the figure formed by joining the centres of
resistance in order is usually a polygon, which is designated the
line of resistance of the frame.
The position of the centres should on no account be allowed
to vary. It is assumed, and is practically true, that the joints
of a frame are flexible, and that the frame under a given load
2 THEORY OF STRUCTURES.
i
does not sensibly change in form. Thus an individual mem.
her is merely stretched or compressed in the direction of its
length, i.e., along its line of resistance, while the frame as a
whole may be subjected to a bending action.
The term truss is often applied to a frame supporting a
weight,
2. Frame of Two Members. OA, OB are two bars
jointed at O and supported at the ends A, B. The frame in
FIG.
FIG. 2.
FIG. 3.
Fig. I consists of two ties, in Fig. 3 of two struts, and in Fig.
2 of a strut and a tie.
Let P be the resultant force at the joint, and let it act in
the direction OC. Take OC equal to P in magnitude, and
draw CD parallel to OB. OD is the stress along OA, and CD
is that along OB.
Let the angle AOB a, and the angle COD = /3.
Let S, , 5 2 be the stresses along OA, O, respectively.
S, _ OD _ sin(<x ft)
~P ~~OC~ sin a~
, S, _ CD _ sin ft
P ~ OC ~ sin a
3. Frame of Three or More Members- Let A.A^A^ . . .
be a polygonal frame jointed at A l9 A tt A 91 . . . Let P l ,
/*,, P 3 , . . . be the resultant forces at the joints A lt A 9 , A 3 ,
. . . , respectively. Let S lt S a , 5 3 , . . . be the forces along
A t A 9t A^A Z , . . . , respectively.
Consider the joint A^.
The lines of action of three forces, P lt S l , and S 9 , intersect
in this joint, and the forces, being in equilibrium, may be
represented in direction and magnitude by the sides of the
FRAMES LOADED AT THE JOINTS. 3
triangle Os,s 6 , in which v 6 is parallel to P lt Os J to S lt and Os 9
to 5..
Similarly,/^, S t , 5 2 maybe represented by the sides of
the triangle Os^., which has one side, Os iy common to the
triangle Os^^, and so on.
Thus every joint furnishes a triangle having a side common
to each of the two adjacent triangles, and all the triangles to-
gether form a closed polygon s^^. . . The sides of this
polygon represent in magnitude and direction the resultant
FIG. 4.
forces at the joints, and the radii from the pole O to the angles
jjVs .' represent in magnitude, direction and character,
the forces along the several sides of the frame A^A Z A^, . .
The polygon A^A^A 3 ... is the line of resistance of the
frame, and is called the funicular polygon of the forces P lt P 39
/*,... with respect to the pole O.
The two polygons are said to be reciprocal, and, in general,
two figures in graphical statics are said to be reciprocal when
the sides in the one figure are parallel or perpendicular to cor-
responding sides in the other.
A triangle or polygon is also said to be the reciprocal of a
point when its sides are parallel or perpendicular to correspond-
ing lines radiating from the point. Thus the triangle Os^ is
4 THEORY OF STRUCTURES.
the reciprocal of the point A lt and the polygon
is the reciprocal of the point O.
If more than two members meet at a joint, or if the joint is
subjected to more than one load, the resulting force diagram
will be a quadrilateral, pentagon, hexagon, . . . according as
the number of members is 3, 4, 5, ... or the number of loads
2* 3 4, *
In practice it is usually required to determine the stresses
in a number of members radiating from a joint in a framed
structure. If the reciprocal of the joint can be drawn, its
sides will represent in direction and magnitude the stresses in
the corresponding members.
Corollary. The converse of the preceding is evidently true.
For if a system of forces is in equilibrium, the polygon of
forces -V 2 .y 3 . . . must close, and therefore the polygon which
has its sides respectively parallel to the radii from a pole O to
the angles s 1 , s 2 , s s , . . . and which has its angles upon the
lines of action of the forces, must also close.
EXAMPLE I. Let O be a joint in a framed structure, and
let Os 1} Os^, Os s , ... be the axes of the members radiating
from it. The polygon A^A^A^ ... is the reciprocal of O, the
side A^A^ representing the stress along Os lt the side A^A 3 that
along Os z , etc.
Ex. 2. Let the resultant forces at the joints be paral-
lel. The polygon of forces becomes the straight line v 5 ,
FIG. 6.
FIG. 7.
FRAMES LOADED AT THE JOINTS. 5
which is often termed the line of loads. Thus, the forces P s ,
/>..../>, are represented by the sides s,s^ s y s 3 , . . . v*> which
are in one straight line closed by s^ t and s & s t , representing the
remaining forces P l and P t , and the triangles Os t s tt Os^ 3 , . . .
are the reciprocals of the points A t , A,. . . . Draw OH per-
pendicular to s t s t . The projection of each of the lines Os lt
Oft, Os 3 , . . . perpendicular to s^ is the same and equal to
OH, which therefore represents in magnitude and direction
the stress which is the same for each member of the frame.
Let a lf # 2 , a 3 , . . . be the inclinations of the members
A,A^ t A^AS, . . . respectively, to the line of loads. Then
OH Hs l tan a l = Hs & tan a b ;
.-. Off (cot a ; -f cot a 6 ) = ffs 1 + ffs, s.s,
and OH, in direction and magnitude, is equal to the stress
common to each member. Also, the stress in any member,
e.g., A^AS Os, Offcosec <x t .
Corollary. Let the resultant forces at the joints A lt A 6
be inclined to the common direction of the remaining forces,
and act in the directions shown by the dotted lines. Let P/,
P t ' be the magnitudes of the new forces ; draw s^' parallel to
the direction of P t ' so as to meet Os 6 in s e ' ; join s^s^. Since
there is equilibrium, s 6 's 6 must be parallel to the line Sj
of action of P t ' . Thus, s^s^ is the force polygon.
Ex. 3. The forces, or loads, />, P t , . . . P 6 are
generally vertical, while P lt P 6 are the vertical re-
actions of the two supports.
Suppose, e.g., that A t A t . . . A 6 is a rope or chain
suspended from the points A lt A 6 , in a horizontal
plane and loaded at A^A S . . . with weights P 9 , P tt . . .
The chain will hang in a form dependent upon the
magnitude of these weights. The points H and S 6
will coincide, and Off will represent the horizontal
riG. 8.
tension of the chain.
Let the polygon A^A^ . . . A t be inverted, and let the rope
be replaced by rigid" bars, A.A^ A^A Z . . . The diagram of
O THEORY OF STRUCTURES.
forces will remain the same, and the frame will be in
equilibrium under fat given loads. The equilibrium, however,
is unstable as the chain, and consequently the inverted frame
will change form if the weights vary. Braces must then be
introduced to prevent distortion.
-H
FIG. 9.
FIG. 10.
Tal^e the case of a frame DCBA . . . symmetrical with
respect to a vertical through A, and let the weights at A, B, C,
. . . be W lt W^ W z , . . . , respectively.
Drawing the stress diagram in the usual manner, OH rep-
resents the horizontal thrust of the frame.
The portions s^^, s^s 3 , ... of the line of loads give a
definite relation between the weights for which the truss will
be stable. The result may be expressed analytically, as
follows :
Let arj, or a , of,, . . . be the inclinations of AB, BC, CD, . . . ,
respectively, to the horizontal.
Let the horizontal thrust OH = H. Then
It ^ = IV,= IV 3 = . ..
cot af 1 = 3 cot or 2 = 5 cot of a = . . .
If there are two bars only, viz., AB, BC, on each side of the
vertical centre line, the frame will have a double slope, and in
this form is employed to support a Mansard roof.
FRAMES LOADED A T THE JOINTS. ?
4. Non-closing Polygons. Let a number of forces P 19
P 9 , P 3 , . . . act upon a structure, and let these forces, taken in
order, be represented in direction and magnitude by the sides
of the unclosed figure MNPQ . . . This figure is the unclosed
polygon of forces, and its closing line TM represents in direction
and magnitude the resultant of the forces P l , P 2 , P 3 , . . .
For PM is the resultant of P^ and P^ and may replace
them ; QM may replace PM and P 3 , i.e., P lt .P 9 , and />,; and
so on.
Take any point O and join OM, ON, OP, . . .
Draw a line AB parallel to OM and intersecting the line of
action of P l in any point B. Through B draw BC parallel to
ON and cutting the line of action of P^ in C. Similarly, draw
CD parallel to OP, DE to OQ, EF to OR, . . . The figure
ABCD ... is called the funicular polygon of the given forces
with respect to the pole O. The position of the pole O is arbi-
trary, and therefore an infinite number of funicular polygons
may be drawn with different poles.
Also the position of the point B in the line of action of P l
is arbitrary, and hence an infinite number of funicular polygons
with their corresponding sides parallel, i.e., an infinite number
of similar funicular polygons, may be drawn with the same pole.
8
THEORY OF STRUCJ^URES.
5. To show that the Intersection of the First and Last
Sides of the Funicular Polygon (i.e., the Point G) is a Point
on the Actual Resultant of the System of Forces P t , P,,
P 3 , . . . First consider two points P t , P 3 , MNP being the
force and ABCD the funicular polygon.
Let AB, DC, the first and last sides of the latter, be pro-
FIG. 12.
duced to meet in g l \ also let DC produced meet the line of
action of P l in H.
Produce OP and MN to meet in K.
Let the lines of action of P t and P a meet in L.
By similar triangles,
KP _HC KN _ HB_ KO
KN ~ Hi ; KO ~~~~ HC ; KM ~ HB
KPKNKO __ HC HB Hg,
Hence
or
KP
KM^~ HL>
and therefore, since the angle H is equal to the angle K, the
line PM \s parallel to the line Lg^.
FRAMES LOADED A T THE JOINTS.
But PM represents in magnitude the resultant of the forces
P lt Py, and is parallel to it in direction.
Therefore Lg^ is also parallel to the direction of the re-
sultant.
But L is evidently a point on the actual resultant of P l , P^ .
Hence ^ must be a point on this resultant.
Next, let there be three forces, P, , P t , P a .
Replace P l , P a by their resultant X acting in the direction
Lg r The force and funicular polygons for the forces X and
PI are evidently MPQ and AgfiE, respectively; ana^, the
point of intersection of Ag l and ED produced, is, as already
proved, a point on the actual resultant of X and P 9 , i.e., of
P lt P 9 , and P 3 .
Hence the first and last sides, AB, ED, of the funicular
polygon ABODE of the forces P^, P^, PI, with respect to the
pole O, intersect in a point which is on the actual resultant of
the given forces.
The proof may be similarly extended to four, five, and any
number of forces.
If the forces are all parallel, the force polygon of the two
forces P l , P 9 becomes a straight line, MNQ. Draw the funicular
M
?
FIG. 13.
polygon ABCD as before, and through g v , the intersection of the
first and last sides, draw^-, Y parallel to MQ, and cutting BC in Y.
By similar triangles,
ON
MN
ON
BY
and
P, QN
ON ON : CY '
' ' P.
CY
BY'
10
THEORY OF STRUCTURES.
Hence Yg lt which is parallel to the direction of the forces P t9
P y , divides the distance between their lines of action into seg-
ments which are inversely proportional to the forces, and must
therefore be the line of action of their resultant. The proof
may be extended to any number of forces, as in the preceding.
Funicular Curve. Let the weights upon a beam AB become
infinite in number, and let the distances between the weights
diminish indefinitely.
The load then becomes continuous, and the funicular poly-
gon is a curve, called the funicular curve.
The equation to this curve may be found as follows :
Let the tangents at two consecutive points Pand Q meet
in R. This point is on the vertical through the centre of
gravity of the load upon the portion MN of the beam.
FIG. 14. FIG. 15.
Let in be the line of loads, and let OS, OT be the radial
lines from O, the pole, parallel to the tangents at P and Q.
Take A as the origin.
Let 6 be the inclination of the tangent at P to the beam,
and let the polar distance OV = p.
wdx = the load upon the portion MN. Then
wdx = ST= SV TV p tan p tan (0 -f dff)
= pdti, approximately.
d
since u =
Integrating twice,
:, and a being constants of integration.
CENTRES OF GRAVITY.
II
If the intensity, w, of the load is constant,
wx
and the curve is a parabola.
6. Centres of Gravity. Let it be required to determine
the centre of gravity of any plane area symmetrical with re-
spect to an axis XX. Divide the area into suitable elementary
areas A lt A 9t A t , . . . having known centres of gravity.
T
FIG. 16.
FIG. 17.
Draw the force (the line in) and funicular polygons corre-
sponding to these areas, and let g be the point in which the first
and last sides of the funicular polygon meet. The line drawn
through g parallel to in must pass through the centre of gravity
of all the elementary areas and, therefore, of the whole area.
Hence it is the point G in which this line intersects the axis XX.
Rail and similar sections may be divided into elementary
areas by drawing a number of parallel lines at right angles to
the axis of symmetry, and at such distances
apart that each elementary figure may, with-
out sensible error, be considered a rectangle
of an area equal to the product of its breadth
by its mean height.
In the case of a very irregular section, an
accurate template of the section may be cut
out of cardboard or thin metal. If the tem-
plate is then suspended from a pin through a point near the
FIG. 1 8.
12 THEORY OF STRUCTURES.
edge, the centre of gravity of the section will lie in the vertical
through the pin. By changing the point of suspension, a new
line in which the centre of gravity lies may be found. The
intersection of the two lines must, therefore, be the centre of
gravity required. Another method of finding the centre of
gravity is to carefully balance the template upon a needle-point.
The area of such a section may be determined either by
means of a planimeter or by balancing the template against a
rectangle cut out of the same material, the area of the rectangle
being evidently the same as that of the section.
7. Moment of Inertia of a Plane Area. Let any two
consecutive sides, C^ , C 3 67 4 , of the funicular polygon meet
line gG in the points m 3 , n s .
Let ^ , .r a , ;r 3 , . . . be the lengths of the perpendiculars from
the centres of gravity of A v , A t , A z , . . . , respectively, upon gG.
Draw the line OH perpendicular to the line of loads, and
let OH=p.
By the similar triangles C 9 m z n z and 6^34,
34
--,-. -, or ,,=;-;
<7 3 X* X^
.-. = wj = area of triangle Cjnji, .
But the total area A bounded by the funicular polygon
i . . . and the lines gC lt gk is the sum of all the triangular
areas C l gm l1 C 9 mji 9 , C 4 mji t , . . ., described in the same manner
as C 3 m 9 n 3 .
5*' *,*'
~~ p 2 " H 7T"
The sum 2(ax*) is the moment of inertia, 7, of the plane
area with respect to gG. Hence,
MOMENT OF INERTIA. 13
The moment of inertia I y of the area, with respect to a
parallel axis at distance ^ from gG, is given by the equation
where S=A 1 + A 9 + ...
Let the new axis intersect C^g and kg in the points q and r.
The triangles qgr and Oin are similar.
qr in S
'^ = 7~ = / ;
and, therefore, the area A' of the triangle qgr
2 l 2p
Hence
T 72 y4
Note. If p be made = = ,
f=A* and
and
The angle lOn is also evidently a right angle.
8. Cranes. (a). Jib-crane. Fig. 19 is a skeleton diagram of
an ordinary jib-crane. OA is the post fixed in the ground at
O ; OB is the jib ; AB is the tie. The jib, tie, and gearing are
suspended from the top of the post by a cross-head, which
admits of a free rotation round the axis of the post.
Let the crane lift a weight W.
THEORY OF STRUCTURES.
Three forces in equilibrium meet at B ; viz., W, the tension
T in the tie, and the thrust C along the jib.
m
FIG. 3o.
Draw the reciprocal figure SS& of B, 5,5 2 representing W.
T SS n AB
and
C_ SS, _ BO
AO'
The load is not suspended directly from B, but is carried by
a chain passing over pulleys to a chain-barrel usually fixed to
the crane-post. The stress 5 in the chain depends upon the
w
system of pulleys, and is, e.g., , if n is the number of falls of
chain from B and if friction is neglected. In order to obtain
the true values of T and C this tension 5 must be compounded
with W.
Draw S^k parallel to the direction in which the chain passes
from B to the chain-barrel, and take S^k to represent 5 in
CRANES. 1 5
magnitude. The line Sfi evidently represents the resultant
force at B due to W and 5.
Draw kt parallel to AB.
The tension in the tie and the thrust in the jib are now
evidently represented by tk, tS^ , respectively.
Generally the effect of chain-tension is to diminish the ten-
sion of the tie and to increase the thrust on the jib.
The vertical component of T, viz., T-^-g W-^Q, is trans-
mitted through the post.
The total resultant pressure along the post at O
-- = W.
The pull upon the tie tends to upset the crane, and its
moment with respect to O is
= WAD = WOF y
OF being the horizontal projection of AB.
OF is often called the radius or throw of the crane.
If the post revolves about its axis (as in ///-cranes), the jib
and gearing are bolted to it, and the whole turns on a pivot at
the toe G. In this case, the frame, as a whole, is kept in
equilibrium by the weight W, the horizontal reaction H of the
web-plate at O, and the reaction R at G. The first two forces
meet in F and, therefore, the reaction at G must also pass
through F.
Hence, since OFG may be taken to represent the triangle
of forces,
and R=
In a portable crane the tendency to upset is counteracted
by means of a weight Q placed upon a horizontal platform
OL attached to the post and supported by the tie AL.
The horizontal projection tm of tk represents the horizontal
i6
THEORY OF STRUCTURES.
FIG. 21.
pull at A, and if tn be drawn parallel to AL, the intercept ww
cut off on the vertical through m by the lines tm and tn repre-
sents the counter-weight required at L.
(U) Derrick- crane. The figure shows a combination of a der-
B rick and crane, called a derrick-
crane. It is distinguished from
the jib-crane by having two
back-stays, AD, AE. One end
of the jib is hinged at or near
the foot of the post, and the
other is held by a chain which
passes over pulleys to a winch
on the post, so that the jib may
be raised or lowered as required.
The derrick-crane is gener-
ally of wood, is simple in con-
struction, is easily erected, has
a vertical as well as a lateral
motion, and a range equal to a circle of from 10 to 60 feet
radius. It is therefore useful for temporary works, setting
masonry, etc.
The stresses in the jib and tie are calculated as in the jib-
crane, and those in the back-stays and post maybe obtained as
follows :
Let the plane of the tie and jib intersect the plane DAE of
the two back-stays in the line AF, and suppose the back-stays
replaced by a single tie AF. Take OF to represent the hori-
zontal pull at A. -The pull on the " imaginary" stay AF is then
represented by AF and is evidently the resultant pull on the
two back-stays. Completing the parallelogram FGAH, AH
will represent the pull on the back-stay AF, and AG that upon
AD, their horizontal components being OK, OL, respectively.
The figure OKFL is also a parallelogram.
If the back-stays lie in planes at right angles to each other,
OL = OF cos 8 = T sin a cos 6, and is a max. when 6 = o,
and
OK = OF sin = T sin a sin 6, and is a max. when 6 = 90,
BRIDGE AND ROOF TRUSSES. I/
6 being the angle FOL, and a the inclination of the tie to the
vertical.
Hence the stress in a back-stay is a maximum when the"
plane of the back-stay and post coincides with that of the jib
and tie.
Again, let ft be the inclination of the back-stays to the ver-
tical. The vertical components of the back-stay stresses are
Ts'm a cos cot ft and T sin a sin 6 cot ft;
and, therefore, the corresponding stress along the post is
T sin a cot ft (cos -\- sin 0),
which is a maximum when = 45.
9. Shear Legs (or Shears) and Tripods (or Gins) are
FIG. 22.
often employed when heavy weights are to be lifted. The
former consists of two struts, AD, AE, united at A and sup-
ported by a tie A C, which may be made adjustable so as to
admit of being lengthened or shortened. The weight is sus-
pended from A, and the legs are capable of revolving around
DE as an axis. Let the plane of the tie and weight intersect
the plane of the legs in AF, and suppose the two legs replaced
by a single strut AF. The thrust along AF. can now be
easily obtained, and hence its components along the two legs.
In tripods one of the three legs is usually longer than the
others. They are united at the top, to which point the tackle
is also attached.
10. Bridge and Roof Trusses of Small Span. A single
girder is the simplest kind of bridge, but is only suitable for
IS
THEORY OF STRUCTURES.
very short spans. When the spans are wider, the centre of the
girder may be supported 'by struts OC, OD, through which a
portion of the weight is transmitted to the abutments.
FIG.
23.
FIG. 24.
Take the vertical line 5,5,, to represent P, the weight at O.
Draw 55 t parallel to OC, and SS^ to OD.
Draw the horizontal SH, and let the angle AOC '= a.
The thrust along OC = S^ = S^ff cosec a = cosec a.
The tension along OA = SH = S^H cot a = cot of.
The horizontal and vertical thrusts upon the masonry at C
P P
(or D) are cot a and , respectively.
If the girder is uniformly loaded, P is one half of the whole
load.
II. In the figure a straining rill, EF, is introduced, and the
girder is supported at two intermediate points.
A B
FIG. a$.
FIG. 26.
Let Pbe the weight at each of the points E and F.
Draw the reciprocal SS^oi the point E, ^//representing P.
BRIDGE AND ROOF TRUSSES.
= P. and the
The thrust in EC (or FD) = 55, =
horizontal thrust in the straining piece = SH = P = P
S,H 'AC
If a load is uniformly distributed over AB, it may be
assumed that each strut carries one half of the load upon AF
(or BE), and that each abutment carries one half of the load
upon AE (or BF).
By means of straining cills the girders may be supported at
several points, I, 2, . . . , and the
weight concentrated at each may
be assumed to be one half of the
load between the two adjacent
points of support. The calcula-
tions for the stresses in the struts,
etc., are made precisely as above.
If the struts are very long they are liable to bend, and
counterbraces, AM, BN, are added to counteract this tendency.
12. The triangle is the only geometrical figure of which
the form cannot be changed without varying the lengths of the
sides. For this reason, all compound trusses for bridges, roofs,
etc., are made up of triangular frames.
Fig. 28 represents the simplest form of roof-truss. AC,
BC are rafters of equal length inclined to the horizontal at an
angle a> and each carries a uniformly distributed load W.
FIG. 27.
Y
W
FIG. 28.
The rafters react horizontally upon each other at C, and
their feet are kept in position by the tie-beam AB. Consider
the rafter A C.
The resultant of the load upon AC, i.e., W, acts through
the middle point D.
2O THEORY OF STRUCTURES.
Let it meet the horizontal thrust // of BC upon AC in F.
For equilibrium, the resultant thrust at A must also act
through F.
The sides of the triangle AFE evidently represent the
three forces. Hence
AE WAE W
- = - = cot a\
2 DE 2
f = w^J-.
The thrust R produces a tension H in the tie-beam, and a
vertical pressure W upon the support.
Also, if y 1S tne an gl e FAE,
EF DE
If the rafters AC, BC are unequal, let <*,, <* a be their in-
clinations to A, B, respectively.
Let W, be the uniformly distributed load upon AC, W 9
that upon BC.
Let the direction of the mutual thrust P at C make an
angle ft with the vertical, so that if CO is drawn perpendicular
ROOF TRUSSES.
21
to FC, the angle COB = ft\ the angle ACF=yo AGO
= 90 - (ft - aj.
Draw AM perpendicular to the direction of P, and consider
the rafter A C. As before, the thrust R v at A, the resultant
weight W^ at the middle point of AC, and the thrust P at C
meet in the point F.
Take moments about A. Then
But
and
P.AM=
sin ACM = AC cos (/? tf x ),
- cos a l .
W 7 ! cos or,
2 cos(/? or,)'
Similarly, by considering the rafter BC y
^
P
cos
cos
Hence
W l cos ar, _ W^ cos tf 2
2 cos(/? or,) " 2 cos(/J + 2 ) '
and therefore
**/=
W l tan a'., ^ 9 tan a l '
The horizontal thrust of each rafter = P sin /?.
The vertical thrust upon the support A = W l P cos ft.
The vertical thrust .upon the support B = W t + P cos /?.
13. King-post Truss. The simple triangular truss may
be modified by introducing a
king-post CO, which carries a
portion of the weight of the
beam AB, and transfers it
through the rafters so as to act ^
upon the tie in the form of a tensile stress.
FIG. 30.
22
THEORY OF STRUCTURES.
Let P be the weight borne by the king-post ; represent it
by CO.
Draw OD parallel to BC, and DE parallel to AB.
CE P
DC = -cosec a is the thrust in CA due to P. ana
sin a 2
is of course equal to DO, i.e., the thrust along CB.
P
DE = CE cot a = cot <* is the horizontal thrust on
each rafter, and is also the tension in the tie due to P.
Let Wbe the uniformly distributed load upon each rafter.
The total horizontal thrust upon each rafter = ( W-\- P) .
p
The total vertical pressure upon each support = W -\ .
If the apex C is not vertically over the centre of the tie-
beam take CO, as before, to represent the weight /'borne by
r the king-post ; draw OD parallel
to BC, and DE parallel to AB.
The weight P produces a
thrust CD along CA, DO along
CB, and a horizontal thrust DE
upon each rafter.
FIG. 31.
CE is the portion of P supported at A, and EO that sup-
ported at B.
DE, and therefore the tension in the tie AB, diminishes
with AO, being zero when AC
is vertical.
Sometimes it is expedient
to support the centre of the tie-
beam upon a column or wall,
the king-post being a pillar
against which the heads of the
rafters rest.
Consider the rafter AC.
The normal reaction R' of
CO upon AC, the resultant
FIG. 32. weight W at the middle point
D, and the thrust R at A meet in the point K
ROOF TRUSSES.
Take moments about A. Then
W
R'AC = W. AE, or R' = cos
Thus the total thrust transmitted through CO to the sup-
W
port at O is 2 cos a . cos or Fcos 2 #.
The horizontal thrust upon each rafter
W
= cos
2
W
sin a = sin 2ct.
4
14. If the rafters are inconveniently long, or if they are in
danger of bending or breaking transversely, the centres may
be supported by struts OD t OE. A portion of the weight upon
FIG. 33-
FIG.
the rafters is then transmitted through the struts to the
vertical tie (king-post or rod) CO, which again transmits it
through the rafters to act partly as a vertical pressure upon
the supports, and partly as a tension on the tie-beam. The
main duty, indeed, of struts and ties is to transform transverse
into longitudinal stresses.
This king-post truss is the simplest and most economical
frame for spans of less than thirty feet. In larger spans two
or more suspenders may be introduced, or the truss otherwise
modified.
24 THEORY OF STRUCTURES.
Let there be a load 2W uniformly distributed over the
rafters AC, BC> and assume it to be concentrated at the joints
W W W W W
A, D, C, E, B, in the proportion , , , , - .
42224
Also, let the load (including a portion of the weight upon
the tie-beam AB, and the weights of the members OD, OE,
OC) borne directly at O be P.
p
The total reaction at each support is W-\- -, and acts in
W
an opposite direction to the weight there concentrated.
4
Hence the resultant reaction at a support is
Thus, the weights at the points of support A and B are
taken up by the abutments, and need not be considered in de-
termining the stresses in the several members of the frame.
Draw the reciprocal SflS^ of A. Then
^W P
= 1 ; ^5., = tension in AO\
4 2
5 a 5 1 = compression in AD.
Draw the reciprocal 5,5,535, of D. Then
5,5, = compression in OD\ 5 3 5 4 = compression in DC;
.,
\ - 5 4 5, = = weight at D.
Draw the reciprocal 5 4 5 3 5 5 5 8 5 4 of C. Then
W
= tension in CO = + />; 55 8 = compression in CE ;
2
W
5 8 5 4 = = weight at C.
ROOF TRUSSES.
Draw the reciprocal 5 8 S 5 5 6 5 7 5 8 of E. Then
,5 6 = compression in OE ; S 6 S 7 = compression in BE ;
" 7 5 8 = = weight at E.
Draw S 6 K horizontally. Then
,S 6 is evidently the reciprocal of B ; KS, =
being uie reaction at B, and 5 6 ATthe tension in the tie BO. The
reciprocal of O is also the figure S^HKS^S^S^ , and HK = P.
15. Collar-beams (DE), queen-posts (DF, G), braces, etc.,
may be employed to prevent the deflection of the rafters.
The complexity of the truss necessarily increases with the span
and with the weight to be borne.
FIG. 35,
FIG. 36.
With a single collar-beam and a uniformly distributed load,
StffSs is the reciprocal of A, and S l S^S 3 S t S l the reciprocal of
D ; S^H being the reaction at A, and 6^5, the weight at D.
FIG. 37.
With a collar-beam DE, two king-posts DF, EG, and a
uniformly distributed load, the stresses at the joints D and E
26
THEORY OF STRUCTURES.
become indeterminate. To render them determinate it is
sometimes assumed that the components of the weights at D
and E, normal to the rafters, are taken up by the collar-beam
and corresponding king-post. Thus S^HS^ is the reciprocal of
A, and S 1 S^S 3 S 4 S & the reciprocal of D, Sfl being the reaction at
A, SlS t the weight at D\ S 4 S 2 is the normal component of the
weight, and the components of S 4 S 2 , viz., S t S 3 horizontal and
S 3 >S 2 vertical, represent the stresses borne by DE and DF, re-
spectively.
This frame belongs to the incomplete (Art. 1 8) class, and if
it has to support an unequally distributed load, braces must be
introduced from D to G and from E to F.
16. The truss ABC, Fig. 40, having the rafters supported
at two intermediate points, may be employed for spans of from
30 to 50 feet. Suppose that these intermediate points of sup-
port trisect the rafters, and let each rafter carry a uniformly dis-
tributed load W.
FIG. 39. FIG. 40.
Then a weight may be considered as concentrated at each
of the joints H, D, C, E, K. This weight = .
Let P be the weight directly supported at each of the
joints F, G.
The resultant reaction at A = *W-}-P.
^ is the reciprocal of A, 6V/ representing %W + P.
, is the reciprocal of H.
,S Z is the reciprocal of F, HK representing P, the
weight directly borne at F.
S 3 S 6 5 6 S 7 S 4 S 3 is the reciprocal of D, 5 7 5 4 representing the
weight at D, S<S, the thrust along HD, S 3 S 5 the
tension in DF, S,S, the thrust along ED, and
5 C 5 7 the thrust alon- CD.
INCOMPLETE FRAMES. 2/
As in the preceding case, this truss will be found incomplete
if the load is unevenly distributed, and the reciprocals of D and
E will not close. In practice, however, the friction at the joints,
the stiffness of the several members, and the mode of construc-
tion render the truss sufficiently strong to meet the ordinary
variations of load.
17. General Remarks. In the trusses described in Arts.
13 and 14 the vertical members are ties, i.e., are in tension, and
the inclined members are struts, i.e., are in compression. By
inverting the respective figures another type of truss is obtained
in which the verticals are struts while the inclined members are
ties. Both systems are widely used, and the method of calcu-
lating the stresses is precisely the same in each.
In designing any particular member, allowance must be
made for every kind of stress to which it may be subjected.
The collar-beam DE, for example, must be treated as a pillar
subjected to a thrust in the direction of its length at each end ;
if it carry a transverse load, its strength as a beam, supported
at the points D and E, must also be determined. Similarly,
the rafters AC, BC, etc., must be designed to carry transverse
loads and to act as pillars. But it must be remembered that
struts and queen-posts provide additional points of support
over which the rafters are continuous, and it is practically suf-
ficient to assume thai the rafters are divided into a number of
short lengths, each of which carries one half of the load between
the two adjacent supports.
When a tie-beam is so long as to require to be spliced,
allowance must be made for the weakening effect of the splice.
18. Incomplete Frames. The frames discussed in the
preceding articles (excepting those referred to in Art. 15) will
support, without change of form, any load consistent with
strength, and the stresses in the several members can be found
in terms of the load. It sometimes happens, however, that a
frame is incomplete, so that it tends to change form under every
distribution of load. An example of this class is the simple
trapezoidal truss, consisting of the two horizontal members AB,
DE, and the two equal inclined members AD, BE, Fig. 41.
First, let there be a weight W at each of 'the points D, E.
THEORY OF STRUCTURES.
The triangles of forces for the joints D and E, viz., SSfl and
fl, can be drawn, and hence it follows that there must be
r
FIG. 41.
FIG.
equilibrium. This is also evident from the symmetrical char-
acter of the loading.
The same triangles represent the forces at the points of
support A, B.
.'. reaction at A = 5,11= W '= S,ff = reaction at B.
Next, let there be a weight W l at D and a weight
<< W,) at E.
FIG. 43.
FIG.
It will now be found that the diagram of forces will not
close, so that there cannot be equilibrium. The joint D will be
pushed in and the frame distorted. The distortion may be
prevented by introducing a brace from A to E or from -B to D.
In the latter case S.mSS^S, represents the stress-diagram, the
triangle S^ffS being the reciprocal of the joint E, and the quad-
rilateral 5w5,//that of the joint D. Drawing the horizontal
mn, the triangle mnS, and the quadrilateral mSSji are evidently
the reciprocals of A and B, respectively.
.-. nS l = reaction at A and nS^ = reaction at B.
INCOMPLETE FRAMES.
In practice the loads are usually transmitted to D and E by
means of two vertical queen-posts (queen-rods or queens] DF, EG.
If there are no diagonal braces DG, EF, the distortion of
the frame under an unevenly distributed load can only be pre-
vented by the friction at the joints, the stiffness of the mem-
bers, and by the queens being rigidly fixed to AB at F and G.
Let W l be the load at F transmitted through the queen
FD to D.
Let W^ (< W^ be the load at G transmitted through the
queen GE to E.
If the frame is rigid, the reactions R l at A and R t at B,
which will balance these weights, can easily be found by taking
moments about B and A, successively. Thus,
and
w\
2
W
W,
2
IV
where AB = I and FG = c.
Draw the triangle of forces SHS^ for the joint A, SH rep-
resenting R^ .
The triangle SS^X is the reciprocal of the joint at /?, and the
tension in FD should, therefore, be XS, SH= R, . But the
30 THEORY OF STRUCTURES
tension in FD is actually W l , so that there is an unbalanced
force,
JW _ iv I c
- w i R i = 2 ~7~'
acting along FD.
To take up this unbalanced force and render the frame rigid
the diagonal DG is introduced, and the stress for which it
should be designed is evidently
W -- W I c 9
(W, - *,) sec FDG = - ' ', ^,
s being the length of the diagonal and d the depth of the truss.
The complete stress diagram is as shown in Fig. 46.
Cor. i. The manner in which distortion is prevented by the
stiffness of AB may be shown as follows:
Let x be the force of resistance which AB, by its stiffness,
can exert at F or G against any load which tends to make it
deviate from the horizontal.
If Wis the load at F t the actual downward pull upon D is
W x ; this must necessarily produce an equal upward pull at
E, which must be balanced by the force of resistance x at G,
and
W
' 2'
Thus the beam AB will be acted upon by an upward pull
W
at F and an equal downward pull at G, forming a couple
W
of moment c, and showing that equilibrium is impossible.
The upward reaction R^ at A is
l(-^+f W J^\- j^
1 ~~ / V 2 ~ 2 ~ ~2 ' 1 I ~~2~l
COMPOSITE FRAMES.
-- downward reaction at B, and the moment at F (or G)
Wcl-c We..
Cor. 2. Let a weight Wbe supported at the joint D of any
quadrilateral frame ADEB. Draw the reciprocal SS^ of D,
w
sv
FIG. 47.
FIG. 48.
S^S^ representing W. Draw 55 3 parallel to EB and intersecting
the vertical 5 t 5, produced in ,S 3 . The weight which can be
borne at E consistent with equilibrium is represented by S^S 3 .
19. Composite Frames or Trusses (i.e., frames made up
of two or more simple frames). An example of this class has
already been given in the case of the king-post roof (Art. 13).
Bent Crane. Fig. 49 shows a convenient form of crane
when much head-room is required near the post. The crane
is merely a semi-girder, and may be tubular with plate-webs if
the loads are heavy, or its flanges may be braced together as
in the figure for loads of less than ten tons. The flanges may
be kept at the same distance apart throughout, or the distance
may be gradually diminished from the base towards the peak.
Let the numbers in Fig. 50 denote the stresses ' \ the cor-
responding members. Three forces, S lt C 9 , and W, act through
the point (i), so that S l and 7 2 may be obtained in terms of
W; three forces, 5,, 5 2 , T 3 , act through (2), so that S 2 and T s
may be obtained in terms of Sj and therefore of W\ four
forces, S 2 , T 2 , 5 3 , 7 4 , act through (3), and the values of 5 2 , C t
3 2
THEORY OF STRUCTURES.
being known, those of S 9 , C 4 may be determined. Proceed-
ing in this way, it is found that of the forces at each succeed-
ing joint only two are unknown, and the values of these are
consequently determinate.
FIG. 50.
The calculations may be checked by the method of moments.
and by the stress diagram (Fig. 50).
E.g., let W 10 tons.
Take moments about the point (7). Then
or T, = - = 26 tons = (68) in Fig. 50.
No other forces enter into the equation of moments, as the
portion of the crane above a plane intersecting (68) and passing
through (7) is kept in equilibrium by the weight of 10 tons
and the stresses T 7 , 5 6 , C 6 ; the moments of S 6 and C 6 about
(7) are evidently zero.
In the stress diagram (Fig. 50) PaQ is the reciprocal of the
point i, abQ of the point 2, PcbQ of 3, Qbcd of 4, and so on.
Other examples of composite roof and bridge frames will
now be given.
20. Roof-trusses. A roof consists of a covering and of
trusses (or frames) by which it is supported. The covering
is generally laid upon a number of common rafters which rest
ROOF TRUSSES.
33
upon horizontal beams (or purlins), the latter being carried
by trusses spaced at intervals varying with the type of con-
struction but averaging about 10 ft. The truss rafters are
called principal rafters, and the trusses themselves are often
designated as principals.
In roofs of small span the trusses and purlins are sometimes
dispensed with.
Types of Truss. A roof-truss may be constructed of tim-
ber, of iron or steel, or of these materials combined. Timber
is almost invariably employed for small spans, but in the
longer spans it has been largely superseded by iron, in con-
sequence of the combined lightness, strength, and durability
of the latter.
Attempts have been made to classify roofs according to the
mode of construction, but the variety of form is so great as to
render it impracticable to make any further distinction than
that which may be drawn between those in which the reac-
tions of the supports are vertical and those in which they are
inclined.
FIG. 51.
FIG. 52.
FIG. 53.
FIG. 57.
FIG. 58.
FIG.
59-
Fig. 51 is a simple form of truss for spans of less that 30 ft.
Fig. 52 is a superior framing for spans of from 30 to 40 ft.;
it may be still further strengthened by the introduction of
struts, Figs. 53 and 54, and with such modification has been
employed to span openings of 90 ft. It is safer, however, to
limit the use of the type shown by Fig. 53 to spans of less than
34 THEORY OF STRUCTURES.
60 ft. Figs. 55, 56, 57, 58, and 59 are forms of truss suitable
for spans of from 60 to 100 ft. and upwards.
Arched roofs, Figs. 58 and 59, admit of a great variety of
treatrrent. They have a pleasing appearance, and cover wide
spans without intermediate supports. The flatness of the
arch is limited by the requirement of a minimum thrust at the
abutments. The thrust may be resisted either by thickening
the abutments or by introducing a tie. If the only load upon
a roof-truss were its own weight, an arch in the form of an
inverted catenary, with a shallow rib, might be used. But the
action of the wind induces oblique and transverse stresses, so
that a considerable depth of rib is generally needed. If the
depth exceed 12 in., it is better to connect the two flanges by
braces than by a solid web. Roofs of wide span are occasion-
ally carried by ordinary lattice-girders,
Principals, Purlins, etc. The principal rafters in Figs. 51
to 57 are straight, abut against each other at the peak, and are
prevented by tie-rods from spreading at the heels. When
made of iron, tee (T), rail, and channel (both single i i and
double ][) bars, bulb-tee (T) and rolled (I) iron beams, are all
excellent forms.
Timber rafters are rectangular in section, and for the sake
of economy and appearance, are often made to taper uniformly
from heel to peak.
The heel is fitted into a suitable cast-iron skew-back, or is
fixed between wrought-iron angle-brackets (Figs. 60, 61, 62),
and rests either directly upon the wall or upon a wall-plate.
FIG. 60. FIG. 61. FIG. 62.
When the span exceeds 60 ft., allowance should be made
for alterations of length due to changes of temperature. This
ROOF TRUSSES.
35
may be effected by interposing a set of rollers between the
skew-back and wall-plate at one heel, or by fixing one heel to
the wall and allowing the opposite skew-back to slide freely
over a wall-plate.
The junction at the peak is made by means of a casting or
wrought-iron plates (Figs. 63, 64, 65).
FIG. 63.
FIG. 64.
FIG. 65.
Light iron and timber beams as well as angle-irons are em-,
ployed as purlins. They are fixed to the top or sides of the
rafters by brackets, or lie between them in cast-iron shoes
(Figs. 66 to 71), and are usually held in place by rows of tie-
FIG. 70. FIG. 71. FIG. 72.
rods, spaced at 6 or 8 ft. intervals between peak and heel,
running the whole length of the roof.
The sheathing boards and final metal or slate covering are
fastened upon the purlins. The nature of the covering regu-
lates the spacing of the purlins, and the size of the purlins is
governed by the distance between the main rafters, which may
THEORY OF STRUCTURES.
vary from 4 ft. to upwards of 25 ft. But when the interval
between the rafters is so great as to cause an undue deflection
of the purlins, the latter should be trussed. Each purlin may
fee trussed, or a light beam may be placed midway between the
main rafters so as to form a supplementary rafter, and trussed
-as in Fig. 72.
Struts are made of timber or iron. Timber struts are
rectangular in section. Wrought-iron struts may consist of
J_-irons, T-bars, or light columns, while cast-iron may be em-
ployed for work of a more ornamental character. The strut-
heads are attached to the rafters by means of cast caps,
wrought-iron straps, brackets, etc. (Figs. 73 to 76), and the
strut-feet are easily designed both for pin and screw Conner
tions (Figs. 77 to 80).
FIG. 73-
FIG. 74.
FIG. 75.
FIG. 76.
FIG. 77 .
FIG. 78.
FIG.
FIG. 80.
Ties may be of flat or round bars attached either by eyes
and pins or by screw ends, and occasionally by rivets. The
greatest care is necessary in properly proportioning the dimen-
sions of the eyes and pins to the stresses that come upon them.
To obtain greater security, each of the end panels of a roof
snay be provided with lateral braces, and wind-ties are often
made to run the whole length of the structure through the
feet of the main struts.
ROOF WEIGHTS. 37
Due allowance must be made in all cases for changes of
temperature.
21. Roof-weights. In calculating the stresses in the
different members of a roof-truss two kinds of load have to be
dealt with, the one permanent and the other accidental. The
permanent load consists of the covering, \he framing, and ac-
cumulations of snow.
Tables at the end of the chapter show the weights of various
coverings and framings.
The weight of freshly fallen snow may vary from 5 to 20
Ibs. per cubic foot. English and European engineers consider
an allowance of 6 Ibs. per square foot sufficient for snow,
but in cold climates, similar to that of North America, it is
probably unsafe to estimate this weight at less than 12 Ibs.
per square foot.
The accidental or live load upon a roof is the wind-pressure,
the maximum force of which has been estimated to vary from
40 to 50 Ibs. per square foot of surface perpendicular to the
direction of blow. Ordinary gales blow with a force of from 20
to 25 Ibs., which may sometimes rise to 34 or 35 Ibs., and even
to upwards of 50 Ibs. during storms of great severity. Press-
ures much greater than 50 Ibs. have been recorded, but they
are wholly untrustworthy. Up to the present time, indeed, all
wind-pressure data are most unreliable, and to this fact may be
attributed the frequent wide divergence of opinion as to the
necessary wind allowance in any particular case. The great
differences that exist in all recorded wind-pressures are pri-
marily due to the unphilosophic, unscientific, and unpractical
character of the anemometers which give no correct informa-
tion either as to pressure or velocity. The inertia of the mov-
ing parts, the transformation of velocities into pressures, and
the injudicious placing of the anemometer, which renders it
subject to local currents, all tend to vitiate the results.
It would be practically absurd to base calculations upon
the violence of a wind-gust, a tornado, or other similar phe-
nomena, as it is almost absolutely certain that a structure
would not lie within its range. In fact, it may be assumed
that a wind-pressure of 40 Ibs. per square foot upon a surface
38 THEORY OF STRUCTURES.
perpendicular to the direction of blow is an ample and perfectly
safe allowance, especially when it is remembered that a greater
pressure than this would cause the overthrow of nearly all the
existing towers, chimneys, etc.
22. Wind-pressure upon Inclined Surfaces. The press-
ure upon an inclined surface may be obtained from the follow-
ing formula, which was experimentally deduced by Hutton,
viz. :
p n =p sin
p being the intensity of the wind-pressure in pounds per square
foot upon a surface perpendicular to the direction of blow, and
p n being the normal intensity upon, a surface inclined at an
angle a to the direction of blow.
Let/ A ,/ T be the components of /, parallel and perpen-
dicular, respectively, to the direction of blow^
'. PH = p n sin a, and p, p n cos a.
Hence, if the inclined surface is a roof, and if the wind blows
horizontally, a is the roof's pitch.
Again, let v be the velocity of a fluid current in feet per
second, and be that due to a head of h feet.
Let w be the weight of the fluid in pounds per cubic foot.
Let p be the pressure of the current in pounds per square
foot upon a surface perpendicular to its direction.
If the fluid, after striking the surface, is free to escape at
right angles to its original direction,
p = 2/lW = W.
Hence for ordinary atmospheric air, since w = .08 lb., approx-
imately,
-,._f.!'v
' 32 ~UoJ
DISTRIBUTION OF LOADS. 39
When the wind impinges upon a surface oblique to its
Iv sin /?\ 2
direction, the intensity of the pressure is ^ j , v being
the absolute impinging velocity, and ft being the angle between
the direction of blow and the surface impinged upon. (See
chapter on Bridges.)
Tables prepared from formulae A and B are given at the
end of the chapter.
23. Distribution of Loads. Engineers have been accus-
tomed to assume that the accidental load is uniformly dis-
tributed over the whole of the roof, and that it varies from 30
to 35 Ibs. per square foot of covered surface for short spans,
and from 35 to 40 Ibs. for spans of more than 60 ft. But the
wind may blow on one side only, and although its direction is
usually horizontal, it may occasionally be inclined at a con-
siderable angle, and be even normal to a roof of high pitch.
It is therefore evident that the horizontal component (/ A ) of
the normal pressure (p n ) should not be neglected, and it may
cause a complete reversal of stress in members of the truss,
especially if it is of the arched or braced type.
If P H is the total normal wind-pressure on the side of a
roof of pitch a, its horizontal component P n sin a will tend to
push the roof horizontally over its supports. This tendency
must be resisted by the reactions at the supports.
In roofs of small span, the foot of each rafter is usually fixed
to its support, and it may be assumed that each support exerts
the same reaction, which should therefore be equal to - !
In roofs of large span the foot of one rafter is fixed, while that
of the other rests upon rollers. The latter is not suited to with-
stand a horizontal force, and the whole of the horizontal com-
ponent of the wind-pressure must be borne at the fixed end,
where the reaction should be assumed to be equal to P n sin a.
In designing a roof-truss it is assumed that the wind blows
on one side only, and that the total load is concentrated at the
joints (or points of support) of the principal rafters.
E.g., let the rafters AB, AC of a truss be each supported at
40 THEORY OF STRUCTURES.
two intermediate points (or joints), D, E and F, G, respectively,
and let the wind blow on the side AB.
FIG. 81.
= FG = l^ EA = GA =/,; and
let / 1 + / 3 + / 8 = /; ' BC = 2! cos a, a being the angle
ABC.
Let W be the permanent (or dead) load per square foot of
roof-surface.
Let p n be the normal wind-pressure per square foot of roof-
surface.
Let d be the horizontal distance in feet from centre to
centre of trusses.
The total normal live load concentrated
2
at E = p n d ; at A =
j The total vertical dead load concentrated at D and F =
-- ; at E and G = wd- ; at A = wdl,.
Let R^ , R^ be the resultant vertical reactions at B and T,
respectively (i.e., the total vertical 'reactions less the dead
weights \wd-) concentrated at these points).
DISTRIBUTION OF LOADS. 4 1
Take moments about C.
-. R^.1 cos a = sum of moments of live loads about C-\- sum
of moments of dead loads about C,
= moment of resultant wind-pressure about C
-f- moment of resultant dead load about (7,
= PJd ~ + / cos 2a -f wd(l, + 2/, + 2/ 8 )/ cos a,
where + / cos 2a is the perpendicular from C upon the line
of action of the resultant wind-pressure which bisects AB
normally.
(N.B. The moment of the horizontal reaction at B or C
about C is evidently nil.)
R^ may be found by taking moments about B.
To determine the stresses in the various members of a roof-
truss two methods may be pursued :
(x) A single stress diagram may be drawn to represent the
combined effect of the live and dead loads. This will be found
to be the quickest and most useful method.
(y) The normal wind-pressure (/) may be resolved into its
vertical (p v ) and horizontal (/ A ) components ; p v may then be
combined with the dead load W, and a stress diagram drawn
for the vertical loads only. A second diagram may be drawn
for the horizontal loads. The resultant stresses will be the
algebraic sum of the corresponding stresses in the two dia-
grams.
A third method will be referred to in a subsequent article.
24. Ex. I. Method (x) applied to the roof-truss ABC,
Fig. 82.
The dead load = wld concentrated at A.
The live loads p n acting at each of the points A and
B, normally to AB.
The vertical reaction at B
wld p n ld /i cos 2d\
-- -- -- --- /
2 ' COS <*\ 2 /
THEORY OF STRUCTURES.
Let rollers be placed underneath C.
Tne total horizontal reaction = p n ld sin a, and is wholly
borne at B.
FIG. 82.
FIG. 83.
At B there z.rz five forces in equilibrium, of which three are
known, and the reciprocal of B may be thus described :
Draw SjS., to represent the normal wind-pressure \p n \
at B ^ 5 2 5 8 to represent R l ; 5 8 S 4 to represent the horizontal
reaction (p n ld sin a) ; S 4 S 6 parallel to BD ; 5 3 5 5 parallel to
AB.
The closed figure 5,5,5,5^5, is the reciprocal required, and
the stresses in BD, AB, at B, are represented by 5 4 5 3 , S } S 5 ,
respectively, being a tension and a thrust.
At D there are three forces in equilibrium, of which the
tension in DB has been found. Drawing 5 4 5 6 horizontally
and S 5 S 6 parallel to AD, the triangle 5 4 5 8 5 5 is evidently the
reciprocal of D, the stresses in DA, DE being represented by
5 5 5 6 , 5 6 5 4 , respectively, and being both tensions.
ROOF-TRUSSES. 43
Again, the triangle S t S 6 S, is the reciprocal of E, the stresses
in EC, EA being represented by 5 4 5 7 , 5 6 5 7 , respectively, and
being both tensions.
At A there are six forces in equilibrium, of which two, viz.,
the normal pressure, \p n J, and the dead weight, (wld\ are
given, while the stresses in AB, AD, AE have been found.
Draw 5 9 5i to represent p n , and S 6 S 9 to represent wld.
Five of the forces at A are therefore represented by the
following lines, taken in order : S 8 S d , S st S 1 , 5j5 5 , S b S 6 , S C S 7 .
Hence the closing line S,S S must necessarily represent in
direction and magnitude the force in AC at A, and it is a
thrust.
Also, S t S 6 S^ must be the reciprocal of C, and therefore S t S fs
represents the reaction at C.
The resultant reaction at B is represented in direction and
magnitude by 5 Q 5 4 .
The line 5 8 5 9 must pass through the point S 4 , as S 3 S 4 , the
horizontal reaction, is merely the horizontal projection of 5 9 5 2 ,
the total wind-pressure.
The dotted lines show the altered stresses if rollers are
under B, the end C being fixed. The stress in each member is
diminished, and as the truss should be designed to meet the
most unfavorable case, the stresses should be calculated on the
assumption that the rollers are on the leeward side.
This may be considered an invariable rule for roof-trusses.
Ex. 2. Method (x) applied to the roof-truss ABC, Fig. 84.
wld
The vertical dead load - at each of the points
F, A, G.
The live load, acting normally to AB, = at each of
4
the points B and A, and = -& at F.
The vertical reaction R^ at B
= l w ld + Pnld (COS 2<X + l).
2 cos a
44
THEORY OF STRUCTURES.
The horizontal reaction at B = p n ld sin a, rollers being
under C as before.
FIG. 85.
Describe the stress diagram in precisely the same manner
as in Ex. i.
Taking S,S 2 to represent the normal wind-pressure at B,
S^S 3 " " " vertical reaction R^ at B,
5 3 5 4 " " " horizontal reaction at B,
/. 5 I 5 a 5 3 5 4 5 6 5 1 is the reciprocal of B y
c c c c c c "7^
OiOsOp.OgO, F,
c c e c c n
^5 w - > 4^9 w5 65 J ^1
? C C " >4
> n o 12 o 7 A}
c <* /c
c e c c c
O 9 |J 4 W - > 13 V - > 109
C C C
O 4 O 14 13
" ,
" C.
S,S t is the horizontal projection of 5,5, + S,S, + 5,5,., , i.e.,
of the total normal wind-pressure, and therefore the vertical
through 5 la must pass through 5, .
ROOF TRUSSES.
45
The dotted lines show the altered stresses if rollers are
under B.
The resultant reaction at B is represented in direction and
magnitude by S z S t .
Ex. 3. Method (x) applied to the truss represented by
Fig. 86.
FIG. 86.
Data. Pitch = 30 ; AD = BD AE CE = 23 ft. ;
trusses 13 ft., centre to centre ; dead weight = 8 Ibs. per square
foot of roof-surface ; wind-pressure on one side of roof (say AB]
normal to roof-surface = 28 Ibs. per square foot; DF-=-DH
= EG = EK\ DF and EG are vertical ; rollers under one end,
say C\ span = 79 ft. ; AF=BH 21 ft, nearly ; FH = 3$ ft.,
nearly.
Total live load = 4459 Ibs. ^ = - 13.28] at each of the
points F, H,
and = 3822 lbs.( = . 13 . 28] at each of the
points A, B.
Total dead load = 1274 lbs.( = ~ . 13 . 8J at each of the
points F, H, K, G,
and = 2184 Ibs. ( = 21 . 13 . 8) at the point A.
THEORY OF STRUCTURES.
ROOF TRUSSES. 47
Resultant vertical reaction at B
= (4 X 1274 + 2184) + |TT^ = 13201.8 Ibs.
Horizontal reaction at B 16562 sin 30 = 8281 Ibs.
Let i inch represent 16,000 Ibs., and on this scale draw
S^S^ = 3822 Ibs., the normal wind-pressure at B ;
5 2 5 3 = 13201.8 Ibs., the vertical reaction at B ;
5 3 5 4 = 8281 Ibs., the horizontal reaction at B \
S 4 S 6 parallel to BD, and v & parallel to BA.
The figure S^S.S.S.S, is the reciprocal of JB.
The stress diagram can now be easily completed, the recip
rocals of the points H, F, D, A, G, K, , and C being
5 16 5 14 5 13 5 17 5 16 , S 17 S 18 S 19 S 16 S 17 , and S 19 S J8 S 4 S 17 , respectively.
S t , as before, is in the vertical line S 19 S 16 produced.
On the assumed scale,
S 4 S 5 = the tension at BD ; S,S b = thrust in BH;
5 M 5 W = " " " AD-, 5 6 5 8 = " " HF\
= " " " BE', S 9 S n = " " AF;
S 6 S = " DH-,
These are the maximum stresses to which the members of
one half of the truss can be subjected, and for which they
should be designed. It is also usual, except in special cases, to
make the two halves symmetrical.
5 2 5 4 is the resultant reaction at B.
If the end C is fixed and rollers placed under B, the reduced
stresses may be shown by dotted lines as in Exs. i and 2.
4 8
THEORY OF STRUCTURES.
Exs. 4 and 5. Method (x) applied to the trusses repre-
sented by Figs. 88 and 90.
It is assumed, as before, that there is a normal wind-press-
ure upon AB, and that rollers are under C.
Figs. 89 and 91 are the maximum stress diagrams corre-
sponding to Figs. 88 and 90, respectively, and are drawn in pre-
cisely the same manner as described in the preceding examples.
Remark on Fig. 88. The stresses at the joints F and D
are indeterminate, and it is assumed that the stress in FL
Si
FIG. 89.
is equal to that in FH. The reciprocal of F thus becomes
^io^ii^6^8^9^n^i3^io> Si*Si( = ^8^9) being the stress in FD. This
truss is an example of a frame with redundant bars, in which
the stresses can only be determined when the relative yield of
the bars is known.
ROOF TRUSSES.
49
Remark on Fig. 90. The stress-diagram, Fig. 91, for each of
the joints in the horizontal BC (Fig. 90) is closed by the return
of one side upon another. Thus at D the stress diagram is
, the closing line 5 4 5 9 (the tension in DE) returning
FIG. 91.
upon 5 5 vS 4 (the tension in DB). The total stress in AE is
evidently represented by 5 12 5 13 , the reciprocal of A being
c c c c c c
14 15 10 12 1314'
Ex. 6. A truss with curved upper and lower chords, the
portions, however, between consecutive joints being assumed
straight.
Under a uniformly distributed load the truss (Fig. 92) is
evidently incomplete, and the stress diagrams at the joints in
the lower chord will not close, so that equilibrium is impossible.
The frame is made complete and the stresses determinate by
introducing ties as in Fig. 93, the corresponding stress diagram
for one half the truss being shown by Fig. 94.
Next, let there be a wind-pressure on the side AB of the
truss. In order to prevent a reversal of stress in the diagonal
ties on the side AC (Fig. 93), additional ties DE, FG, called
counter -braces, are introduced as in Fig. 95. Fig. 96 gives the.
THEORY OF STRUCTURES,
stress diagram due to wind-pressure only, it being assumed
that the end C rests upon rollers and that B is fixed.
FIG. 92.
FIG. 93.
FIG. 94.
Note. 21 wind-press, at B =
23 = " "M =
34= "M =
45 =
56 =
67 " " A =
" O =
11 O =
FIG. 95.
wind-press, upon BM,
" BM,
" MO,
" " MO,
" OA,
" " OA.
\K vertical reaction, at B,
HK =. horizontal reaction at B.
Ex. 7. A single example will serve to illustrate method (y).
Take the truss represented by Fig. 97.
Fig. 98 is the stress diagram due to the vertical load upon
the roof, viz., the dead weight -{- vertical component of wind-
pressure. pq is the vertical reaction at B and is
ff
2AH)d.
ROOF- TR USSES.
qm is the weight at F and is
= weigh t at H =
Fig. 99 is the stress-diagram due to horizontal component
of wind-pressure, rollers being placed under B and the end C
being fixed.
p'o' = downward reaction at B =
p n ld sin 2 a
4 cos OL
o'q' =. horizontal force of wind at B = ^ BF\
q'm'
= m'n'= horizontal force of wind at ForH= -&ff.
2
Total resultant stresses in the members BF, FH, HA, DF,
DH, DB, JDA, DE are represented by qr q'r' , ms m's' y
nt n't', sr s'r', st st', pr p'r', tv /V, pv p V,
respectively.
Note. The stress diagrams for trusses with both of the lower
ends of the principal rafters fixed, are drawn in precisely the
same manner as described in the preceding examples.
THEORY OF STRUCTURES.
Thus, in Fig. 100, S^S^S^ is the reciprocal of A,
representing the portion of the horizontal wind-pressure borne
FIG. 100.
at A. Again, HS 6 S,S,H is the reciprocal of B, HS^ represent-
ing the portion of the horizontal wind-pressure borne at C.
HS Z = HS t + S 4 S 3 = total horizontal wind-pressure, 5 2 5 3 repre-
senting the vertical reaction at B, and HS^ that at C.
25. Bridge-trusses. A bridge-truss proper consists of
an upper chord (or flange), a lower chord (or flange], and an in-
termediate portion, called the web, connecting the two chords.
Its depth is made as small as possible consistent with economy,
strength, and stiffness. Its purpose is to carry a distributed
load, which, as in the case of roof-trusses, is assumed to be
concentrated at the joints, or panel-points, of the upper and
lower chord. Trussed beams are also employed for the same
object, and examples of simple frames of this class have already
been given.
The following are bridge-trusses of a more complex char-
acter.
Ex. I. The beam BC (Fig. 101) is supported at three points
by the vertical struts DF, AK, EG, which are tied at the feet
by the rods DB, DK, AB,AC, and EK, EC. Let W,, W,, W,
be the loads concentrated at the joints F, K, G, respectively.
Draw the line of loads S,S, , S,S t being W, , S 2 S 3 = W^ , and
C C _ TX7
3 4 VV i .
Describe the funicular polygon with any pole O, and draw
OH parallel to the closing line MN of this polygon. Then
BRIDGE TRUSSES.
53
} is the reaction at B and HS^ the reaction at 7 (Art. 3).
is the reciprocal of B, S^S & being the thrust along
FB, and ^/^ the tension along BD.
5 is the reciprocal of F t 5,5 2 being W 19 the
weight at F t 5 2 S 6 the thrust along KF, S e S 6 the
thrust along DF.
H is the reciprocal of D, 5 6 S 7 being the tension
along DK, and S,H the tension along Z^^.
is the reciprocal of A, S,S e being the thrust along
KA, and S^H the tension along AE.
So, S 2 S 3 S 9 S 8 S 7 S 6 S 2 , S 3 S 4 S 10 S 9 S 3 , S 9 S ]0 HS,S 9 , and S 4 //5 IO are
the reciprocals of K, G, E, and C, respectively, the closing line
5 10 5 4 being necessarily horizontal and representing the stress
in GC.
FIG. 101. FIG. 102.
This truss inverted is often used for bridge purposes in dis-
tricts where timber is plentiful, as it may be constructed
entirely of wood. The stresses in the several members of the
inverted truss are of course reversed in kind but unchanged in
magnitude, and are given by the same stress diagram.
Note. The reactions ffS lt HS^ may be obtained at once by
the method of moments. Thus, by taking moments about C,
the reaction R, at B is
and by taking moments about B, the reaction R^ at C is
54
THEORY OP STRUCTURES.
Ex. 2. In the truss represented in the accompanying figure,
the length of the beam AB is so great that the single triangu-
lar truss ACB with a single central strut CO is an insufficient
support. The two halves are therefore strengthened by the
simple triangular trusses AGO with a central strut GF and
BPO with a central strut PN.
Again, each quarter-length, viz., AF, FO, ON, NB, is simi-
larly trussed. The subdivisions may, if necessary, be carried
still farther. This truss in four, 'eight, sixteen, . . . divisions or
D F.SH L N Q B
panels is known as the Fink truss, and has been widely em-
ployed in America, the number of panels usually being eight
or sixteen.
The members shown by the dotted lines may be introduced
for stiffness, and the platform may be either at the top or
bottom. The weight directly borne by a strut is usually de-
termined from the loads upon the two adjacent panels by
assuming the corresponding portions of the beam to be inde-
pendent beams supported at the ends. Thus if there be a
weight PFat the point 5 in the panel FH, the portion of W
borne by the strut GF at F is
Sff
W
FH'
and the portion borne by the strut KH at H is
FS
W
FH
Let W lt W,, W,, W., W b , W 6 , W, be the weights upon
the struts (or posts) DE, FG, HK, OC, LM, NP, QR, respect-
ively.
Let P lt P tt P t9 P 4t P bf P tt P, be the compressions to which
these posts are severally subjected.
FINK TRUSS. 55
Let a, /?, y be the inclinations to the vertical of AE, AG,
AC, respectively.
Let T lt T^ T 3 , . . . be the tensions in the ties, as in Fig. 103.
The tensions in the ties meeting at the foot of a post are
evidently equal.
Each triangular truss may be considered separately.
From the truss AEF, 2T, cos a = P 1 W, ;
from the truss A GO, 2 T, cos /?=/>,= W,+( T,+ T,) cos a ;
from the truss FKO, 2 T 3 cos a = P 3 = W 3 \
from the truss ACB,
from the truss OMN, 2 T, cos a = P, = W^',
from the truss OPB, 2T 6 cos/3=P 6 = W 6 +(T b +T,) cos a ;
from the truss NRB, 2 T, cos a = P,= W, .
Hence
W
l Jw -1- W *+ W * I ^
4 " t "
GCC y,
= -- sec
and the values of P, , P 2 , P,, . . . can be at once found.
56 THEORY OF STRUCTURES.
Again, the thrust along AF= 7", sin a-\-T t sin /? +" jT 4 sin 7 ;
at ,F = T, sin )3 + T. sin 7 ;
along FO= T, sin /? + T t sin y +7; sin a -,
" at 6> = r 4 sin p ;
etc., etc.
If the truss carries a uniformly distributed load W,
W
, = - sec a,
; = T; = sec /?, T; = sec
o 4
If the above diagram is inverted, it will represent another
type of truss in which the obliques are struts and the verticals
ties.
Note. The stresses in the several members of each of the
trusses due to the weight it is designed to carry, may of course
be easily determined graphically in the manner already de-
scribed in previous articles.
Ex. 3. Fig. 104 represents a beam trussed by a number
of independent triangular trusses, the vertical posts being
FIG.
104.
equidistant. The weight concentrated at the head of each
post may be found by the method described in Ex. 2, which
in fact is generally applicable to all bridge and roof trusses.
Let 7*,, 7* a be the tensions in AE, BE, respectively.
Let W l be the weight at D.
WARREN TRUSS.
57
Let of l , # 2 be the inclinations of AE, BE, respectively, to
the vertical.
sin
1 sin (or, + or,) '
T =
D F
FIG. 105.
Similarly, the stress in any other tie may be obtained.
The compression in the top chord is the algebraic sum of
the horizontal components of all the stresses in the ties which
meet at one end.
The verticals are always struts and the obliques ties.
This truss has been used for bridges of considerable span,
but the ties may prove inconveniently long.
Ex. 4. The figure SANT represents an ordinary triangular
truss of the Warren type, supported at the ends 5 and T.
Draw the line of loads 16, 12 A c E .G L N
being the weight at B and 23, 34, AAAAAA
45, 56 the weights at D, F, K, M y \
respectively.
With any pole O describe the
funicular polygon and draw OP par-
allel to its closing line QR.
.'. [Pis the reaction at 5, and 6P
that at T.
The reciprocal of S is the triangle
Pi S l ; 15, being the tension in SB,
and 5,/^the compression in AS.
The reciprocal of A is the triangle Q
PS } S^ ; 5,5 2 being the tension in AB,
and S^Pthc compression in CA.
The reciprocal of B is the figure
5,i25 8 5 a 5, ; 25 3 being the tension in
BD, 5 3 5 2 the compression in CB, and
12 the weight at B. FIG. 107.
The reciprocal of C is the figure PS, 2 S 3 S 4 P', 5 3 5 4 being the
tension in CD, and Sf the compression in EC.
The reciprocal of D is the figure 5 3 235 5 5 4 5 3 ; 35 5 being the
tension in DF, 5 5 5 4 the compression in ED, and 23 the weight
at D.
/ \
\ /
V
S Q
FIG. 106.
THEORY OF STRUCTURES.
The reciprocal of E is the figure PS t S b S 6 P- S 6 S 6 being the
tension in EF, and S 6 P the compression in GE.
The reciprocal of F is the figure S^S, t S<.S. ; 48, being the
tension in FK, 5,5 6 the tension in FG, and 34 the weight at F.
And so on, the closing line PS n for the reciprocal of T being
necessarily parallel to NT.
The arrow-heads show the character of the stresses in the
several members of the truss.
Note. The reactions may also be at once determined by
the method of moments.
Thus
and
iP = |(I2) + 4(23) + 1(34) + 1(45) +
6P = 4(12) + f(2 3 ) + f (34) + (45) + f (56).
(12) (23) (34), (45) (56) (CV) (78)
FIG. 108.
Ex. 5. In the truss represented by the accompanying figure,
the joints in the upper as well as
those in the lower chord are loaded,
the weights being transmitted to the
former by means of vertical sus-
penders.
Fig. 109 is evidently the corre-
sponding stress diagram.
Note. In the trusses repre-
sented by Figs. 1 06 and 109, the floor
is carried upon the lower chords. If
the trusses are inverted, the floor
*
may be carried on the upper chords.
The stresses in the several members
are evidently the same in magnitude
and are only reversed in kind.
FlG - I0 9- Ex. 6. The Howe truss repre-
sented by Fig. no is very widely used and maybe constructed
of timber, of iron, or of timber and iron combined.
A C E G L N R
ciXAA*
/V/6
Qlp
B D 'F H K M Q
HOWE TRUSS.
59
Let there be a uniformly distributed load upon the truss
consisting of a weight W at each of the joints B, D, ... in the
lower chord.
The reaction at each support = 3^ W.
Fig. in is the stress diagram, and the several members of
the truss are indicated on the lines representing the stresses to
BS
i
%
W
W
w.
AC
~CE
W
w
MQ
W
QT
DF
#/ 8
FH
KM
FIG. in.
which they are subjected. The directions of these stresses at
the joints, and hence also their character, are easily determined
by following in order the sides of the reciprocals. The verti-
cals are evidently all ties and the diagonals all struts.
If the load is unevenly distributed, the stresses in different
members may be reversed. For example,
BD
T
DF
FIG. ii2.
-s 1 -
FIG. 113.
Let the truss carry a single weight P at any point D.
The reciprocal of D is S l S. t S^S t S^S l (Fig. 112), 5,5 2 represent-
6c
THEORY OF STRUCTURES.
ing P t and the arrow-heads showing the directions of the forces
now acting at D. Thus the force in DE at I), represented by
,S' 3 S 4 , acts from D towards E, and is, therefore, a tension.
Hence, in order that DE may not be subjected to a tensile
force, counterbraces CF, EH are introduced so that the por-
tion of P borne on the support at T may be transmitted
through the system CFEH to H and from H to T through the
rcgiilar system HGKLMNQRT. The reciprocal of D is now
5 1 5 2 5 8 5 6 (Fig. 1 1 3), and the reciprocal of Cthe figure HS,S,S,Sfl,
the arrow-heads showing the directions of the forces at C. It
will be at once observed that FC must be a strut.
In order to make provision for a varying load, as when a
train passes over a bridge, counterbraces are introduced in the
panels on both sides of the centre, and although they may not
be necessary in every panel, they will give increased stiffness to
the truss.
Note. Generally speaking, a panel is that portion of the
bridge-truss between two consecutive verticals, and the ends
of the verticals are called panel-points.
Ex. 7. Fig. 114 represents a Pratt truss, and is merely an
inverted Howe truss. The diagonals become ties and the
FIG. 114.
verticals struts. Counterbraces are introduced to resist th^
action of a varying load, precisely as described in Ex. 6.
Ex. 8. The bowstring truss in its simplest form is repre-
M Q
FIG. 115.
sented by Fig. 115. Assuming that the portions of the upper
chord between consecutive joints are straight, the stress dia-
BOWSTRING TRUSS.
61
gram for a uniformly distributed load and for one half the
truss is Fig. 116.
The panels, however, are incomplete frames, and if the truss
8. S.
1
J
S 8
FIG. 116.
FIG. 117.
has to carry an unequally distributed load, ties similar to that
shown by the dotted line MNmust be introduced in the several
panels in order to prevent distortion.
For example, let there be a single load P at the joint N,
and let there be no brace NM. The stress in the first vertical
is evidently nil. The reciprocal of ^V is S 1 S^S 3 S^S 5 S 1 , Fig. 117,
S 2 S 3 representing P. The reciprocal of L is HS^S^S^H, and the
arrow-heads show the directions of the forces at H.
Thus the force in OL, which is represented by S 4 S 6 , acts from
O towards L, and is, therefore, a compression. But, under a
uniformly distributed load, the diagonals are all ties, and NM
is introduced to take up that portion of P which would be
otherwise transmitted through LO in the form of a compression.
In this case the reciprocal of L is HS^S^H, since the stress in LO
due to P is assumed to be nil. Also the reciprocal of N is
S l S^S^S^S^S^S l . The stress in NM, represented by S 8 S 7 , acts
from N to M and is a tension.
Hence the diagonals NM are also ties, and the portion of
the weight P borne at L is carried to Q through the system
NMOQ.
Ex. 9. Fig. 118 is a bowstring truss with isosceles bracing.
Under an arbitrary load Fig. 119 is the stress diagram, the
loads at a, b, c, d, e, f, g being 12, 23, 34, 45, 56, 67, 78, respect-
ively. As in the Warren girder, the diagonals may, under the
action of a varying load, be subjected to both tensile and com-
62
THEORY OF STRUCTURES.
pressi-ve stresses. They must, therefore, be designed to bear
such reversal of stress.
a .b c d e f g
FIG. 118.
FIG.
It is assumed, as before, that the portions of the upper
chord between consecutive joints are straight.
Note. The design of bridge-trusses will be further con-
sidered in a subsequent chapter.
26. Method of Sections. It often happens that the
stresses in the members of a frame may be easily obtained by
the method of sections. This method depends upon the
following principle :
If a frame is divided by a plane section into two parts, and
if each part is considered separately, the stresses in the bars
(or members) intersected by the secant plane must balance the
external forces upon the part in question.
Hence the algebraic sums of the horizontal components,
2(X), of the vertical components, 2(Y\ and of the moments
of the forces with respect to any point, 2(M), are severally
zero ; i.e., analytically,
= o, S(Y) = o, and 2(M) = o.
These equations are solvable, and the stresses therefore
determinate, if the secant plane does not cut more than three
members.
EXAMPLES. 63
Ex. I. ABC is a roof-truss of 60 ft. span and 30 pitch.
The strut DF = GH = 5 ft.; the angle FDA =90. Also
AF=F = AG = GC.
The vertical reaction at .# = 5 tons. The weight concen-
trated at D 4f tons.
Let the angle ABF '= a.
AB = 30 sec 30 = 20 VI ; cot a = ? = 2 1/J,
.*. sm a = . _ ; cos a =
.
4/13
If the portion of the truss on the right of a secant plane
MNbe removed, the forces C, T^ , 7^ in the members AD,AF,
FG must balance the external forces 5 tons and 4^ tons in order
that the equilibrium of the remainder of the truss may be pre-
served.
Hence, revolving horizontally and vertically,
7;+ T, cos (a + 30) - ^sin6o =o;
T, sin (a + 30) C cos 60 + 5 - 4^ = 0.
Taking moments about F t
C.$- $BF cos (30 - ) + 4kDF sin 30 = o.
But
cos (+ 3 o) = sin (0300)=:, C os (30 )=
64 THEORY OF STRUCTURES.
BF = BD sec a 5 VT$, and DF = 5 tt.
T..
Hence 7= 15^ tons, 7^ = 9.89 tons, and 7", = 6.35 tons.
Ex. 2. The figure represents a portion of a bridge-truss cut
, off by a plane MN and supported at
409,400 Ibs
FiG.
the abutment at A.
The vertical reaction at A
= 409,400 Ibs.
T F The weight at B = 49,500 Ibs.
" C = 38,700 Ibs.
= BC= 24 ft. ; BD=24 ft.; CE = 29% ft.
The forces C', D' , T in the members met by MN must
balance the external forces at A, B, C.
Revolving horizontally and vertically,
T+D' cos a C cos/5 = o;
D' sin a + C sin ft 409400 + 495 +38/00 = o ;
a and /3 being the inclinations to the horizon of EF, DE,
respectively.
Taking moments about E y
T x 291 + 409400 x 48 49500 x 24 = o.
PIERS.
But
.-. sin a
Hence
= = and tan ft = = -.
24 9 24 9
II 9 2
, __ , cos a ^, sin p = -, cos p
202 4/202 1/85
T = 629,427^ Ibs. ;
981^00
a
^=
V8S
27. Piers. To determine the stresses in the members of
the braced piers (Fig. 122) supporting a deck bridge.
40 tons
FIG. 122.
FIG, 123,
Data. Height of pier = 50 ft. ; of truss = 30 ft. Width
of pier at top 17 ft. ; at bottom = 33! ft.
OO THEORY OF STRUCTURES.
The bridge when most heavily loaded throws a weight of
IPO tons on each of the points A and B.
Weight of half-pier = 30 tons.
The increased weight at each of the points C, D and E, F,
jfrom the portions AD and CF of the pier = 5 tons.
Resultant horizontal wind-pressure on train = 40 tons at
iS/j- feet above base.
Resultant horizontal wind-pressure on truss = 20 tons at
65 feet above base.
Resultant horizontal wind-pressure on pier = 2^ tons at
each of the points C and E.
With the wind-pressure acting as in the figure, the diagonals
CB, ED, and GF are required. When the wind blows on the
other side, the diagonals D to A, F to C, and H to E are
brought into play. The moment of the couple tending to
overturn the pier
= 40 X 8;J + 20 X 65 + 4 X 25 = 4900 ton-feet.
22.2.
The moment of stability = (200 -f 30) X ~- = 3871! ft.-tons.
Thus the difference, = 4900 3871! = IO28| ft.-tons, must
be provided for in the anchorage. The pull on a vertical
anchorage-tie at G == -jf 3r 5 oT tons -
33s
Again, if H be the horizontal force upon the pier at A due
to wind-pressure,
H X 50 = 40 x 8;J + 20 x 65 = 4800 ;
H = 96 tons.
The stress diagram can now be easily drawn.
The reciprocals of the points A, B, C, D, E, F are 4321,
2561, 11-10-4169, 65789, 13-12-11-98-14, and 87-15-16-14, respec-
tively. In the stress diagram 43 = 96 tons, 32 = 25 = 100 tons,
57 7-15=4-1011-12 6 tons, and 10-1 1 = 12-13 = 2 i
tons. The stress in EG is of an opposite kind to the stresses
in AC, CE.
WEIGHT OF ROOF-COVERINGS.
6 7
. In computing the stresses in the leeward posts of a
braced pier, it is usual in American practice to assume that the
maximum load is upon the bridge and that the wind exerts a
pressure of 30 Ibs. per sq. ft. upon the surfaces of the train and
structure, or a pressure of 50 Ibs. per sq. ft. upon the surface
of the structure alone. The negative stresses in the windward
posts of the pier are determined when the minimum load is on
the bridge, the wind-pressure remaining the same.
TABLE OF WEIGHTS OF ROOF-COVERINGS.
Description of Covering.
Weight of
Covering
in Ibs. per
sq. ft. of
Covered
Area.
Dead Weight of Roof in Ibs. per sq.
ft. of Covered Area.
Boarding ($~inch)
2c to "\
Boarding and sheet-iron
6 5
JC
8 to -1.25
Corrugated iron and laths
5e
Felt asphalted . .
^ tO 4.
Felt and gravel
8 to 10
Galvanized iron
I to 3
Laths and plaster ...
Pantiles
9 to 10
6 to 10
<; to 8
Sheet-zinc .
I 25 to 2
Sheet-iron (corrugated). ..
3 A
8 without boards and 11 with boards
K (i
3X
for spans up to 75 ft.
12 without boards and 15 with boards
Sheet-iron (16 W.G.) and laths.
Shingles (i6-inch) . .
5
2
for spans from 75 to 150 ft.
10 on laths for spans up to 75 ft
" (lone')
14 on laths for spans from 75 to
Sheathing (i-inch pine). ..?...
a
150 ft.
" (chestnut and maple)
(ash, hickory, oak),.
Slates (ordinary)
4
5
e to Q
13 without boards or on laths and 16
Slates (large) .
Q to 1 1
on i^-in. boards for spans up to
.75ft-
Slates and iron laths
IO
on i-in. boards for spans from
75 to 150 ft.
Thatch
6 S
Tiles
7 to 2O
Tiles and mortar
oe fO ^O
Timbering of tiled and slate
roofs (additional)
5c Jo 6 5
68
THEORY OF STRUCTURES.
WEIGHTS OF VARIOUS ROOF-FRAMINGS.
Description of Roof.
Location.
Cover-
ing.
Span.
Width
of
Bays.
Weight in Ibs.
per sq. ft. of
Covered Area.
Pitch.
Fram-
ing.
Cover-
ing.
Pent
ft. in
15
37 o
40 o
50 o
53 3
54 o
55 o
72 o
62 o
76 o
79 o
80 8
90 2
84 o
100
130 o
50 o
154
211
97 o
153
41 o
81 6
I2O O
72 o
240 o
45 o
ft. in.
5 o
12
IO O
II
14 o
6 6
20
12
25 o
13 c
ii 8
20 o
9 o
14 o
26 o
II O
26 o
24 o
13 o
26 o
16 o
24 o
29 4
14 6
3-5
4.6
5-5
3-0
2.085
9-5
ii. 6
7.0
3-013
2.6
3. 86
4.72
5.00
5.66
7.72
5.42
12. I
30
30
30
30
26 34'
j Liverpool \
1 Docks f
Felt
it
J Timber ")
^y\ rafters and 1
^^^^ | struts, iron f
ties
<
<
<
Common Truss . . . .
j Liverpool )
1 Docks 1
ii
ii
Zinc
Zinc
Zinc
Slates
13.6
3-5
7-0
6.4
9.6
4-9
II. O
12. O
15-0
10.7
16.8
ii. 8
ii. 3
24-5
11-5
> < i<
Bowstring . . .
Manchester
Lime Street
Birmingham
Strasburg
Paris
Dublin
Derby
Sydenham
St. Pancras
Cremorne
1,1
K
Arched
ii
WIND PRESSURES.
6 9
TABLE OF THE VALUES OF Pn, Pv , Ph, IN LBS. PER SQ. FT, OF
SURFACE, WHEN P - 40, AS DETERMINED BY
THE FORMULA P n = P . sin a'- 8 4 cosa-i.
Pitch of Roof.
Pn
Pv
Ph
5
5-0
4.9
4
10
9-7
9.6
i-7
20
18.1
17.0
6.2
30
26.4
22.8
13-2
40
33-3
25-5
21.4
50
38.1
24-5
29.2
60
40.0
20.0
34-0
70
41.0
I4.O
38.5
80
40.4
7.0
39-8
90
40.0
0.0
40.0
TABLE PREPARED FROM THE FORMULA
-(=)
Velocities in
feet per second.
Velocities in
miles per hour.
Pressure in
Ibs. per sq. ft.
10
6.8
25
20
13-6
1. 00
40
27.2
4.00
60
40.8
g.OO
70
47.6
12.25
80
54-4
16.00
90
61.2
20.25
100
68.0
25.00
no
74.8
30-25
1 20
81.6
36.00
130
88.4
42.25
150
102.0
56.25
JO THEORY OF STRUCTURES.
EXAMPLES.
1. Show that .the locus of the poles of the funicular polygons of
which the first and last sides pass through two fixed points on the clos-
ing line, is a straight line parallel to the closing line.
2. The first and last sides of a funicular polygon of a system of forces
intersect the closing line in two fixed points. Show that for any position
of the pole each side of the polygon will pass through a fixed point on
the closing line.
3. Four bars of equal weight and length, freely articulated at the
extremities, form a square ABCD, The system rests in a vertical plane,
the joint A being fixed, and the form of the square is preserved by
means of a horizontal string connecting the joints B and D. If W be
the weight of'each bar, show (a) that the stress at C is horizontal and
= , (b) that the stress on EC at B is WjL^L and makes with the ver-
2 2
tical an angle tan - J -J, (<r) that the stress on AB at B is W^ 1 ^ and
makes with the vertical an angle tan" 1 !, (d) that the stress upon
AB at A is f W, (e) that the tension of the string is 2 W .
4. Five bars of equal length and weight, freely articulated at the
extremities, form a regular pentagon ABCDE. The system rests in a
vertical plane, the bar CD being fixed in a horizontal position, and the
form of the pentagon being preserved by means of a string connecting
the joints B and E. If the weight of each bar be W, show that the
W
tension of the string is (tan 54 + 3 tan 18), and find the magni-
tudes and directions of the stresses at the joints.
5. Six bars of equal length and weight ( = IV}, freely articulated at
the extremities, form a regular hexagon ABCDEF.
First, if the system hang in a vertical plane, the bar AB being fixed
in a horizontal position, and the form of the hexagon being preserved
by means of a string connecting the middle points of AB and DE, show
W
that (a) the tension of the string is 3 W, (b) the stress at C is and
horizontal, (c} the stress at D is W 4/il and makes with the vertical
12
an angle cot - 1 2 1/3.
Second, if the system rest in a vertical plane, the bar DE being fixed
in a horizontal position, and the form of the hexagon being preserved
EXAMPLES. 71
by means of a string connecting the joints C and F, show that (a)
the tension of the string is W --, (b) the stress at C is W A/ 2 1 . and
V3 12
makes with C# an angle sin -'/f/JL, (V) the stress at B is F 4/-Z.
124 12
and makes with CD an angle sin ~ 1 -i/
Third, if the system hang in a vertical plane, the joint ^ being fixed,
and the form of the hexagon being preserved by means of strings con-
necting A with the joints E, D, and C, show that (a) the tension of
each of the strings AE and AC is W V^ , (6) the tension of the string
AD is 2 W; and determine the magnitudes and directions of the stresses
at the joints, assuming that the strings are connected with pins distinct
from the bars.
6. Show that the stresses at C and F in the first case of Ex. 5 remain
horizontal when the bars AF, FE, BC, CD are replaced by any others
which are all equally inclined to the horizon.
7. If the pole of a funicular polygon describe a straight line, show
that the corresponding sides of successive funicular polygons with re-
spect to successive positions of the pole will intersect in a straight line
which is parallel to the locus of the pole.
8. A system of heavy bars, freely articulated, is suspended from two
fixed points; determine the magnitudes and directions of the stresses at
the joints. If the bars are all of equal weight and length, show that the
tangents of the angles which successive bars make with the horizontal
are in arithmetic progression.
9. If an .even number of bars of equal length and weight rest in equi-
librium in the form of an arch, and if a\, a 2 , . . . a n be the respective
angles of inclination to the horizon of the ist, 2d, . . . ;zth bars count-
ing from the top, show that
27Z + I
tanar-i.j = tan a H .
2;z I
10. Three bars, freely articulated, form an equilateral triangle ABC.
The system rests in a vertical plane upon supports at B and C in the
same horizontal line, and a weight W is suspended from A. Determine
the stress in BC, neglecting the weight of the bars.
W
Ans. -.
2 V3
11. Three bars, freely articulated, form a triangle ABC, and the sys-
tem is kept in equilibrium by three forces acting on the joints. Deter-
mine the stress in each bar.
What relation holds between the stresses when the lines of action of
72 THEORY OF STRUCTURES.
the forces meet (a) in the centrpid, (b) in the orthocentre of the
triangle ?
12. A triangular truss of white pine consists of two equal rafters AB,
AC, and a tie-beam BC\ the span of the truss is 30 ft. and its rise is 7$
ft. ; the uniformly distributed load upon each rafter is 8400 Ibs. Deter-
mine the stresses in the several members.
Ans. Stress in BC ' 8400 Ibs., in AB = 4200 4/5 Ibs.
33. ABCD is a quadrilateral truss, AB and CD being horizontal and
15 and 30 ft. in length, respectively. The length of AC is 10 ft., and its
inclination to the vertical is 60. A weight WL is placed at C, and Wi
at D. What must be the relation between W\ and Wi so that the truss
may not be deformed ? For any other relation between W\ and W* ,
explain how you would modify the truss to prevent deformation, and
find the stresses in all the members.
Ans. Wi ssrW^J-Lti.
2
14. A Warren girder 80 ft. long is formed oifive equilateral triangles.
Weights of 2, 3, 4, 5, tons are concentrated, respectively, at the ist, 2d,
3d, and 4th apex along the upper chord. Determine the stresses in all
the members of the girder.
Ans. Tension Chord : Stress in ist bay = 2 4/3 ; 2d = 5$- 4/3 ;
3 d = 7 Vl ; 4th = 6i V~Z ; 5th = 2f V~3>
Compression Chord : Stress in ist bay = 41/3 ;
2d = 6f 4 / 3 ; 3d = 7-^3 ; 4th = 5^-4/3.
Diagonals : Stress in ist and 2d. =44/3;
3d and 4th = 2$ 4/3 ; 5th and 6th = f 4 / 3 :
7th and 8th =24/3; 9th and loth = 5i 4/3-
15. In a quadrilateral truss ABCD, AD is horizontal, AB and BC are
inclined at angles of 60 and 30 respectively to the horizontal, and CD
is inclined at 45 to the horizontal. What weight must be concentrated
at Cto maintain the equilibrium of the frame under a weight JV at J
If a weight W is placed at Cas well as at D, what member must be
introduced to prevent distortion? What will be the stress in that
member ?
Ans. First:
Second: Introduce braced/? and let BDA = a.
Then stress in ^=Jg 2 -/" 3) .
2 sin (60 + (x)
EXAMPLES. 73
1 6. The boom AB of the accompanying truss is supported at five
intermediate points dividing the length
into six segments each 10 ft. long. The
depth of the truss = 10 ft. Draw stress ^ ^\//^ h
diagrams for the following cases : FlG - I2 4-
(a) A weight of 100 Ibs. at each intermediate point of support.
(&) Weights of 100, 200, 300, 400, 500 Ibs. in order at these points.
Ans. (a) Stress in a = 375 ; b = 325 ; c = 375 ; // = 450 ;
m = 125 1/?3 ; n = 50 1/5 ; o - 504/5 ;
p = 25 4/73 Ibs.
() Stress in = 875 ; b 825; ^ = 925; k 1350; d= 1325,
*? = 1125; 7 = 1375; m = $oy$; w = 1004/5;
= 141! 1/?3 ; ^ = 8|/?3 ; r = 200^5 ;
s = 250 1/5 ; / = 458^ 4/?3 Ibs.
17. The rafters AB, AC of a factory roof are 18 and 24 ft. in length
respectively. The tie BC is horizontal and 30 ft. long. The middle
points of the rafters are supported by struts DE, DF from the middle
point D of the tie BC\ the point D is supported by the tie-rod AD.
The truss carries a load of 500 Ibs. at each of the points E, A, and F.
Find the stresses in all the members. Secondly, find the stresses in the
members when the rafter AB is subjected to a normal pressure of 300
Ibs. per lineal ft., rollers being at C.
Ans. Stress in BE = 1 1 12^ ; EA = 800 ; CF= ioi6f ; FA = 600 ;
BD = 667i ; CD = 813* ; DE = 312* ; DF = 416!;
AD = 502 Ibs.
Stresses due to 300 Ibs. in BE 1012^ ; EA 1800 ;
DE = 2812^ ; BD = 3847^ ; AD = 2250 ;
DC = 2160 ; v4C = 2700 ; Z>.F = o.
1 8. If it be assumed in the first part of the last, question that the
whole of the weight is concentrated at the points E and F, draw the
stress diagram.
19. A triangular truss consists of two equal rafters AB, AC and a tie-
beam BC, all of white pine ; the centre D of the tie-beam is supported
from A by a wrought-iron rod AD; the uniformly distributed load upon
each rafter is 8400 Ibs., and upon the tie-beam is 36000 Ibs. ; determine
(a) the stresses in the different members, BC being 40 ft. and AD 20 ft.
What (b) will be the effect upon the several members if the centre of the
tie-beam be supported upon a wall, and if for the rod a post be substi-
tuted against which the heads of the rafters can rest? Assume that the
pressure between the rafter and post acts at right angles to the rafter.
Ans. (a) Stresses in BD = 13200; AD = 18000; AB = 13200 |/2 Ibs.
(0) " " = 4200; " = 8400; " = 6300 |/2~lbs.
74 THEORY OF STRUCTURES.
20. A triangular truss of white pine consists of a rafter AC, a vertical
post AB, and a horizontal tie-beam BC\ the load upon the rafter is 300
Ibs. per lineal foot ; AC = 30 ft., AB = 6 ft. Find the resultant pressure
at C.
Ans. 4409 Ibs.
Find the stresses in the several members when the centre D of the
rafter is also supported by a strut from B.
Ans. Stress in BC = 4500 |/6; CD = 22500; DB 11250;
DA 1 1 250 ; AB = 2250 Ibs.
21. The rafters AB, AC of a. roof-truss are 20 ft. long, and are sup-
ported at the centres by the struts DE, DF; the centre D of the tie-
beam BC is supported by a tie-rod AD, 10 ft. long; the uniformly dis-
tributed load upon AB is 8000 Ibs., and upon AC is 2400 Ibs. Determine
the stresses in all the members.
What will be the effect upon the several members if AB be subjected
to a horizontal pressure of 156 Ibs. per lineal foot?
Ans. (a) Stress in BD = 4600 1/3 ; BE = 9200 ; EA = 5200 ;
ED = 4000 ; AD 2600 ; DF = 1 200 ;
AF = 5200 ; CF = 6400 ; CD = 3200 1/3.
() Tens, in BE = 520 4/3 ; AD = 260 1/3 ; compres. in
ED 520 1/3"; AC = 520 j/J; DC= 780.
No stresses in BD, AE.
22. Determine the stresses in all the members of the truss in the
preceding question, assuming the tie-beam to be also loaded with a
weight of 600 Ibs. per lineal foot.
Ans. Stress in AB increased by 6000 4/3 Ibs. ; in BC by 9000 Ibs.;
in AD by 6000 4/3 Ibs.
23. A horizontal beam is trussed and supported by a vertical strut at
its middle point. If a loaded wheel roll across the beam, show that the
stress in each member increases proportionately with the distance of the
wheel from the end.
W Wx
Ans. Stress in tie-beam (hor.) = rx cot 0; on tie = -, ;
/ /sin
W
% on strut = fix.
24. A frame is composed of a horizontal top-beam 40 ft. long, two
vertical struts 3 ft. long, and three tie-rods of which the middle one is
horizontal and 15 ft. long. Find the stresses produced in the several
members when a single load of 6000 Ibs. is concentrated at the head of
each strut.
Ans. Stress in horizontal members = 50000 Ibs.
" " sloping " = 51420 "
" " struts = 12000 "
EXAMPLES, 75
25. If a wheel loaded with 12000 Ibs. travel over the top-beam in the
last question, what members must be introduced to prevent distortion ?
What are the, maximum stresses to which these members will be sub-
jected ?
Ans. 19122 Ibs.
26. A beam of 30 ft. span is supported by an inverted queen-truss,
the queens being each 3 ft. long and the bottom horizontal member 10
ft. long. Find the stresses in the several members due to a weight W
at the head of a queen, introducing the diagonal required to prevent
distortion. Also find the stresses due to a weight W at centre of beam.
20 ^o
Ans. i. Stress \nAB = W\ AE=2.^2lV\ EF=~W;
BE=-W\ BF = i.i6W\ BC = W\
= \.\6W.
W
2. Stress \*AB =
-W.
27. A roof-truss of 20 ft. span and 8 ft. rise is composed of two
rafters and a horizontal tie-rod between the feet. The load upon the
truss = 500 Ibs. per foot of span. Find the pull on the tie. What would
the pull be if the rod were raised 4 ft.?
Ans. 3125 Ibs. ; 6250 Ibs.
28, The rafters AB, AC of a roof are unequal in length and are in-
clined at angles a, ft to the vertical ; the uniformly distributed load upon
AB = W\ , upon AC ' = Wi . Find the tension on the tie-beam.
+ Wi sin a sin ft
Ans.
sin (a + ft}'
29. In the last question, if the span = 10 ft., a = 60 and ft 45, find
the tension on the tie, the rafters being spaced 2^ ft. centre to centre,
and the roof-load being 20 Ibs. per square foot.
Ans. 198 Ibs.
30. The equal rafters AB, AC for a roof of 10 ft. span and 2\ ft. rise
are spaced 2\ ft. centre to centre ; the weight of the roof-covering, etc.
= 20 Ibs. per square foot. Find the vertical pressure and outward thrust
at the foot of a rafter.
Ans. Total vertical pressure = 125 4/5 Ibs. = horizontal thrust.
31. The lengths of the tie-beam and two rafters of a roof-truss are in
the ratios of 5 : 4 : 3. Find the stresses in the several members when
the load upon each rafter is uniformly distributed and equal to 100 Ibs.
Ans. Stress in tie = 48 Ibs. ; in one rafter = 60 Ibs.; in other = 80 Ibs.
/ THEORY OF STRUCTURES.
32. In a triangular truss the rafters each slope at 30 ; the load upon
the apex = 100 Ibs. Find the thrust of the roof and the stress in each
rafter.
Ans. 100 Ibs.; 86.6 Ibs.
33. A roof-truss is composed of two equal rafters and a tie-beam, and
the span = 4 times the rise; the load at the apex = 4000 Ibs. Find the
stresses in the several members.
Secondly, if a man of 150 Ibs. stands at the middle of a rafter, by
how much will the stress in the tie-beam be increased ?
Ans. i. Stress in tie = 4000 Ibs. ; in each rafter = 2000 1/5 Ibs.
2. 75 Ibs.
34. A king-post truss for a roof of 30 ft. span and 7^ ft. rise is com-
posed of two equal rafters AB, AC, the horizontal tie-beam BC, the
vertical tie AD, and the struts DE, DF from the middle point D of the
tie-beam to the middle points of the rafters ; the roof-load = 20 Ibs. per
square foot of roof-surface, and the rafters are spaced 10 ft. centre to
centre. Find the stresses in the several members.
Second, find the altered stresses when a man of 150 Ibs. weight stands
on the ridge.
Third, find the altered stresses when the tie-beam supports a celling
weighing 12 Ibs. per square foot.
Ans. i. Stress in BE = 56250 Ibs. ; BD = 2250 4/5 Ibs.;
AE 46875 Ibs. ; DE = 9375 Ibs.;
AD= 7500 4/5" Ibs.
2. Stresses in BD, BE, AE increased by 150 Ibs., 75 1/5
Ibs., and 75 \' '5 Ibs., respectively ; other stresses un-
changed.
3. Stresses in AD, tie-beam, and rafters increased by 1800,
1800. and 900 1/5 Ibs., respectively ; other stresses
unchanged.
35. The platform of a bridge for a clear span of 60 ft. is carried by
two queen-trusses 15 ft. deep; the upper horizontal member of the truss
is 20 ft. long; the load upon the bridge = 50 Ibs. per square foot of plat-
form, which is 12 ft. wide. Find the stresses in the several members.
Ans. Stress in vertical ^= 6000 Ibs.; in each sloping member
= loooo Ibs. ; in each horizontal member = 8000 Ibs.
36. If a single load of 6000 Ibs. pass over the bridge in the last ques-
tion, and if its effect is equally divided between the trusses, find (a) the
greatest stress in the members of the truss, and also (lij in the members
which must be introduced to prevent distortion. Also find (c) the
stresses when one half the bridge carries an additional load of 50 Ibs. per
square foot of platform.
EXAMPLES. 77
Ans. (a) In sloping end strut = 3333* Ibs.; horizontal tie =
2666f Ibs.; horizontal strut = 1333* Ibs.
(<:) In sloping end strut = 6250 Ibs.; horizontal tie =
5000 Ibs.; horizontal strut = 3000 Ibs.
(b) In case (a) = i666f Ibs.; in case (c) = 2500 Ibs.
37. A roof-truss consists of two equal rafters AB, AC inclined at 60
to the vertical, of a horizontal tie-beam BC of length /, of a collar-beam
DE of length , and of queen-posts DF, EG at each end of the
collar-beam ; the truss is loaded with a weight of 2600 Ibs. at the vertex,
a weight of 4000 Ibs. at one collar-beam joint, a weight of 1200 Ibs. at
the other, and a weight of 1500 Ibs. at the foot of each queen; the
diagonal DG is inserted to provide for the unequal distribution of load.
Find the stresses in all members.
Ans. Stress in BD= 11733*; BF = 5866f 4/3 ; DF = 1 500 ;
DA = 2600 ; DE = 3633* 4/3 ; DG = i866f ;
GC = 4933i 1/3 ; GE = 2433* ; CE = 9 866f ;
AE 2600 Ibs.
38. The rafters AB, AC are supported at the centres by the struts
DE, DF; the centre of the tie-beam is supported by the tie AD\
BC = 30 ft., AD = 7i ft. ; the load upon AB is 4000 Ibs., that upon AC
1600 Ibs. Find the stresses in all the members. By an accident the
strut DE was torn away; how were the stresses in the other members
affected ?
Ans. Case i : Stress in BE 2400 4/5"; BD = 4800 ;
DE = icoo 4/5"; AE = 1400 4/5 ;
AF = 1400 4/5"; DF = 400 V5 ;
FC = 1800 i/J"; DC 3600 Ibs.
Case 2: Stress in BA = 1400 4/5"; BD = 2800 ;
AD = 400 ; AF = 1400 4/5 ;
FC = 1800 4/5"; DF = 400 4/5";
DC = 3600.
39. The platform of a bridge for a clear span of 60 ft. is carried by
two trusses 15 ft. deep, of the type shown by the p n
accompanying diagram ; the load upon the
bridge is 50 Ibs. per square foot of platform,
which is 12 ft. wide. Find the stresses in the
several members.
Ans. Stress in BE = 13500; ^ = 67504/5"; EG = 4000
ED = 1 3500 ; CD = 2250 \/~^\ GA = 4500
AD = 9000 Ibs.
78 THEORY OF STRUCTURES.
40. If a single weight of 2000 Ibs. pass over a truss similar to that
shown in the preceding question, find the stresses in the several members
when the load is (i) at E, (2) at D.
Ans.Case i: Stress in BG = 1500 1/5 ; BE = 3000 ;
EG = 2000 ; ED 3000 ;
GD = looo 4/5 ; AG = 500 1/5 ;
AH 500^/5"; DH=o; FH-o\
DF 1000 ; FC = 1000 ;
77=5004/5" Ibs.
Case 2 : Stress in BA and CA = 1000 1/5 ;
BD and DC = 2000 ; AD = 2000 Ibs.,
and in other members = o.
41. A white-pine triangular truss consists of two rafters AB, -AC, of
unequal length, and a tie-beam BC. A vertical wrought-iron rod from
A, 10 ft. long, supports the tie-beam at a point D, dividing its length
into the segments BD = ioit. and CD = 20 ft. The load upon each
rafter is 300 Ibs. per lineal ft. ; the load upon the tie-beam is 18,000 Ibs.,
uniformly distributed. Determine the stresses in the several members.
Ans. In AB = 9650 V 2" Ibs. ; AC= 4825 4/5 Ibs. ; BD = CD = 9650 Ibs.
42. The post of a jib-crane is 10 ft. ; the weight lifted = W\ the jib
is inclined at 30, and the tie at 60, to the vertical. Find (a) the stresses
in the jib and tie, and also the B. M. at the foot of the post.
How (b} will these stresses be modified if the chain has four falls, and
if it passes to the chain-barrel in a direction bisecting the angle between
the jib and tie ?
Ans. (a) Stress in tie = W\ in jib = JFy J. B. M. = 5 4/3" ft. tons.
43. An ordinary jib-crane is required to lift a weight of 10 tons at a
horizontal distance of 9 ft. from the axis of the post. The hanging part
of the chain is in four falls; the jib is 15 ft. long, and the top of the
post is 16^ ft. above ground. Find the stresses in the jib and tie when
the chain passes (i) along the jib, (2) along the tie.
The post turns round a vertical axis. Find the direction and magni-
tude of the pressure at the toe, which is 3 ft. below ground.
Ans. (i) Stress in tie = 3_1_5 tons; in jib = iif tons.
ii
(2)
= [3 ^5 __ 2^) tons ; in jib = 9 T a T tons.
. \ ii /
Pressure on toe = 10 \/io tons, and is inclined to vertical at an angle
tan ~ l i.
EXAMPLES. 79
44. In the crane represented by the figure AB = AC = A
35ft.; 2?C=2oft.; D=2oi\..\ the weight lifted = 25 tons;
y4C slopes at 45 ; the chain hangs in four falls and passes
from A to D. Find the stresses in all the members and
the upward pull at D.
Ans. Stress in C=26; AC = 47.6 ; AB = 28.4 ; CD = 32.8 tons.
Vertical pull at D = 31.3 tons.
45. The figure represents the framing of an hydraulic crane. AB.BD
= DF = FG=^ HK = 5 ft.; KG = BC = 2*- ft. Find the
H f\~7^\ ^L stresses in the members of the crane when the weight
KG FOB" A (i ton) lifted is (a) at A\(&)a.\.B\ (c) at D. Also (d) find
FIG. 127. the stresses when there is an additional weight of ton
at each of the points B, D, F, and G.
Ans (a) Stress in tons in AB = BD 2 ; DF = FG = ;
9009
Iy
Stress in tons in ^^ = o = ^C; ^C= i ; CE = |/ 2 ;
=FG = ; GK =
13 II 2
M3
Stress in AB = o =
*-<
= 7^317;
80 THEORY OF STRUCTURES.
(d) Stress in tons in AB = o = AC; EC = - = EF ;
46. The inclined bars of the trape-
zoidal truss represented by the figure
make angles of 45 with the vertical ; a
load of 10 tons is applied at the top
joint of the left rafter in a direction of
^ 45 with the vertical. Assuming the
W "P reaction at the right to be vertical, find
FlG> I28> the stresses in all the pieces of the frame.
Ans. Vert, reaction at D - yT; stress in DE= |/2~;
j 3
,. 20 /- 10 ,
3 3
AC = I0 ; CE = - 1/2" tonr.
47. The post of a derrick-crane is 30 ft. high ; the horizontal traces of
the two back-stays are at right angles to each other, and are 15 ft. and
25 ft. in length. Show that the angle between the shorter trace and the
plane of the jib and tie, when the stress in the post is a maximum, is
30 58'.
Also find the greatest stresses in the different members of the crane
when the jib, which is 50 ft. long and is hinged at the foot of the post, is
inclined at 45 to the vertical, the weight lifted being 4000 Ibs.
Ans. Stress in jib = 6666f Ibs.; in tie =4768. 4 Ibs.; max.
thrust along post = 10991.5 Ibs. ; max. stress on long
back-stay = 7362.7 Ibs.; on short back-stay = 10539 Ibs.
48. A queen-truss for a roof consists of two horizontal members, the
lower 48 ft. long, the upper 16 ft. long; two inclined members AB, DC,
and two queens BE, CF, each 8 ft. long; the points E, ^divide AD into
three equal segments ; the load upon the members AB, BC, CD is
120 Ibs. per lineal foot. Find (a) the stresses in the several members.
How (b) will these stresses be modified if struts are introduced from the
EXAMPLES. Si
feet of the queens to the middle points G, H of the inclined members ?
In this latter case also, determine (c) the stresses due to a wind-pressure
of 120 Ibs. per lineal ft. normal to AB, assuming that the horizontal re-
action is equally divided between the two supports at A and D.
Ans.(a) Stress in Ibs. in AE 4066.56 = EF-DF = BC\
AB = 4546. $6 = CD; BE = 2033. 28 = CF.
() Stress in Ibs. in AE 5139.84 = DF\
BC = 4066.56 = AF; AG = 5746.56 - DH\
BG = 4546.56 = CH\ EG = 1200 = FH\
BE^ 536.64 = CF.
(c) Additional stress in AG = 1040.4/5"; BG 680 4/5";
GE- 600 4/5"; ^=2320 ; BE = 600 ; BC= 400 4/5";
BE= 400 4/5"; CF = 400 ; CB = 400 4/5"; EF = 1 120 ;
(In case (<:) the brace BF is introduced to prevent distortion.)
49. A pair of shear-legs, each 25 ft. long, with the point of suspension
20 ft. vertically above the ground surface, is supported by a tie 100 ft.
long ; distance between feet of legs = 10 4/5 ft. Find the thrusts along
the legs and the tension in the tie when a weight of 2 tons is being
lifted.
Ans. Tension in tie = 1.137 tons; compn. in each leg = 1.87 tons.
50. In the crane ABC, the vertical post AB = 15', the jib AC '= 23',
and the angle BAC '= 30. Find (a) the stresses in the jib and tie, and
also the bending moment at the foot of the post when the crane lifts a
weight, of 4 tons.
The throw is increased by adding two horizontal members CE, BD
and an inclined member DE, the figure BE being a parallelogram and
the diagonal CD coincident in direction with CA. Find (b) the stresses
in the several members of the crane as thus modified, the weight lifted
being the same.
In the latter case show (c) how the stresses in the members are affected
when the chain, which is in four falls, passes from E to B and then down
the post.
Ans. (a) Tension in tie = 3^ tons; thrust in jib = 6 T 2 7 tons;
(b) Stress in CE = 9.34 ; in ED = 10.16; in CB = 13.49;
in CD = 6.15; in DA = 10.7 ; in BD = 7 tons.
(c) Stress in CE = 8.9; in ED = 10.7; in CB = 12.9;
in CD = 5.8 ; in DA 10.7 ; in BD = 7.4 tons.
51. The horizontal traces of the two back-stays of a derrick-crane
are x and y feet in length, and the angle between them is fi. Show that
cos (ft 6) x
the stress in the post is a maximum when - - =0 beine the
cos 6 y
angle between the trace x and the plane of the jib and tie.
82
THEORY OF STRUCTURES.
52. The two back-stays of a derrick-crane are each 38' long, and the
angle between their horizontal traces 2 tan -I T 5 ^; height of the crane-
post = 32'; the length of the jib = 40' ; the throw of the crane = 20' ;
the weight lifted = 4 tons. Determine the stresses in the several mem-
.bers and the upward pull at the foot of each back-stay when the plane of
the jib and post (a) bisects the angle between the horizontal traces of
xhe back-stays, (b) passes through a back-stay.
Ans. In jib = 5 ; in tie = 2.52 tons ; in back-stay in (a) = 2.56,
in (b) 4.7 tons.
53. Find the stresses in the members of the crane represented
E 35' B by the figure; also find balance-weight
atC
Ans. Stress in BE = 25 ; DE = 26.9;
DB = 21.08; DA = 26.08;
BA = .24; BC = 1 8. 12 tons.
Counterweight at C= 15.14
tons.
54. Draw the stress diagram for the truss represented by the figure,
the load at each of the points and C being 500 Ibs.
FIG. 130.
Also, if the rafter AB is subjected to a nominal wind-pressure of 100
Ibs. per lineal ft., introduce the additional member required to prevent
deformation, and state in Ibs. the stress it should be designed to bear.
Draw the stress diagram of the modified truss, assuming that the foot A
is fixed, and that there are rollers at D.
(AB = AE =15'; BC = 10' ; angle BAD = 45 ; angle EAD 30.)
55. The post AB of a jib-crane is
20 ft. ; the jib AC is inclined at 30 and
the tie BC at 45 to the vertical ; the
weight lifted is 5 tons. Find the
stresses in the jib and tie when the
chain passes (a) along the jib, (b} along
the tie, (c) horizontally from C to the
post.
The chain has two falls.
56. In a mansard roof of 12 ft. rise, the upper triangular portion (of
4 ft. rise) has its rafters inclined at 60 to the vertical. The rafters of the
FIG. 131.
EXAMPLES. 83
lower portion are inclined at 30 to the vertical. If there is a load of
1000 Ibs. at the ridge, find the load at each intermediate joint necessary
for equilibrium, and the thrust of the roof.
A load of 2000 Ibs. is concentrated at each of the intermediate joints
and a brace is inserted between these points. Find the stress in the
orace.
Ans. 1000 Ibs. ; thrust = 500 4/3" Ibs.; 333!- |/J Ibs.
57. The horizontal boom CD is divided into eight segments, each
8 ft. long, by seven intermediate supports ;
the depth of the truss at each end = 16 ft.;
a weight of i ton is concentrated at C and
at D, and a weight of 2 tons at each of the
points of division. Determine the stresses
in the several members.
58. The figure is a skeleton diagram of a roof-truss of 72 ft. span and
12 ft. deep ; G, K, L, O, H are respectively the middle points of AE, EL,
EF, LF, FB ; AE = EL = LF= FB = 20 ft.;
the trusses are 12 ft. centre to centre ; the dead
.. weight of the roof = 12 Ibs. per sq. ft.; the
^ s normal wind-pressure upon AE may be taken
FIG. 133. _ 30 j bg> per S q f t . t h e enc j ^ j s fj xe( j an( j
B is on rollers. Draw a stress diagram. Show by dotted lines how the
stress diagram is modified with rollers under A, B being fixed.
59. The platform of a bridge of 84 ft. span B c D E
and 9 ft. deep is carried by a pair of trusses of A^^Tn P^P 55 ^
the type shown in the figure. If the load borne ^
by each truss is 300 Ibs. per lineal ft., find the FIG. 134.
stresses in all the members.
Ans. Stress in AB = 6000 ; AC = 1200 4/73 ; AD = 3600 1/17 ;
BC 4800 ; CD = 14400 ; DE = 28800.
Stress in horiz. chord = 288000 ; in each vertical = 3600
Ibs.
60. The figure represents the shore portion of one of the trusses for
a cantilever highway bridge. The depth of
t t B
truss over pier = 51 ft.; the length of each
panel = 17 ft.; the load at A (from weight of
is.sooibs. c ><^j/p centre span) 16800 Ibs.; the width of road-
way = 15 ft.; the load per sq. ft. of roadway
FIG. 135. _ go Ibs. Find the stresses in all the mem-
bers, assuming the reaction at the pier F to be vertical.
Ans. ti ti 28000 ; / 3 = 36500 ; / 4 = 45000 ; / 5 = 53500 ;
/ 6 = 55200 ; ti = 48400 ; / 8 = 41600 = / ; Ci = 5600 1/34";
<: 2 = 7300 V 34 ; vi = 10200 ; v t = 15300; z/ 8 = 20400 ;
8 4
THEORY OF STRUCTURES.
V* = 25500 ; 7/5 = 45900 J s/e = 20400 ; V-, 1 5300 ;
1/6 = 10200 ; <T 3 = 9000 4/34 ; <r 4 = 10700 1/34 ;
<r 5 = 12400 4/34 ; <r 6 = 77500 ; c-i 69000 ; <r 8 = 60500 ;
<r 9 = 52000 ; di = 1700 4/34 ; </ a = 1700 4/6? ;
rt?3 = 1700 4/106 ; </ 4 = 22100; d* = 1700 4/97";
dt = 3400 4/13 ; </ 7 = 8500 Ibs.
61. The inner flange of a bent crane forms a quadrant of a circle of
20 ft. radius, and is divided into four equal bays. The outer flange
forms the segment of a circle of 23 ft. radius. The two flanges are 5 ft.
apart at the foot, and are struck from centres in the same horizontal
line. The bracing consists of a series of isosceles triangles, of which the
bases are the equal bays of the inner flange. The crane is required to
lift a weight of 10 tons. Determine the stresses in all the members.
62. A braced semi-arch is 10 ft. deep at the wall and projects 40 ft.
The upper flange is horizontal, is divided into four equal bays, and carries
a uniformly distributed loatl of 40 tons. The lower flange forms the
segment of a circle of 104 ft. radius. The bracing consists of a series of
isosceles triangles of which the bases are the equal bays of the upper
flange. Determine the stresses in all the members.
63. The domed roof of a gas-holder for a clear span of 80 ft. is strength-
ened by secondary and primary trussing as in the figure. The points B
and C are connected by the tie BPC passing
beneath the central strut AP t which is 15 ft.
long, and is also common to all the primary
trusses ; the rise of A above the horizontal is 5
ft.; the secondary truss ABEF consists of the
equal bays AH, HG, GB, the ties BE, EF, FA, of which BE is horizon-
tal, and the struts GE, FH, which are each 2 ft. 6 in. long and are par-
allel to the radius to the centre of GH; the secondary truss ACLK is
similar to ABEF\ when the holder is empty the weight supported by
the truss is 36000 Ibs., which may be assumed to be concentrated at G,
H, A, M, N, in the proportions 8000, 4000, 1000, 4000, and 8000 Ibs., re-
spectively. Determine the stresses in the different members of the truss.
64. The figure is the skeleton diagram of a cantilever for a viaduct in
B
91 tons 42 tons 42 tons 42 tone 21 tool
FIG. 137.
EXAMPLES.
India. Determine graphically the stresses in the various members un-
der the loading indicated.
65. In the accompanying roof-truss AB = AC 30 ft., and the struts
ire all normal to the rafters. Find the
stresses in all the members, the load at
each of the joints in the rafters being 2
tons (angle ABC = 30 and angle DBC B ^y^LJ^J/^A^C
= 10). How will the stresses be mod-
ified if there is a force of 2 tons acting
at each of the points of support between
A and B at right angles to the rafter, and a force of i ton at A, assum-
ing that the end B is fixed and that C rests upon rollers?
66. The figure represents a portion of a Warren girder cut off by the
plane MN and supported upon the abutment at
A. The reaction at A =20 tons ; the load con-
/\ /\ /\ U c centrated at each of the points .# = 4 tons. Find
A/ \/ V vf the stresses in each of the members met by MN.
B B Bj
N Ans. Stress in tension chord = ^ tons ;
FIG. 139. 3
in compression chord = 32 1/3 tons ; compression in diagonal = 1/3
tons.
67. The figure represents a portion of a roof-truss cut off by a plane
MN and supported at A. The strut DC is
vertical ; AD = 23 ft., and the distance of D
from AC= 7i ft.; the angle between AC and
the horizontal = cos-'; the vertical reac-
tion at A 7 tons ; the horizontal reaction
at A = 2| tons ; at each of the points B and
C a weight of 4 tons is concentrated. Find
the stresses in the members met by MN.
(AD and 7\ make equal angles with the FIG. 140.
rafter.)
Ans. Ci = 13.2 tons ; T* = 2.1 tons ; 7\ = 10.8 tons.
68. The feet of the equal roof-rafters AB, A Care tied by rods BD t
CD which meet under the vertex and are joined to it by a rod AD. If
IVi , Wi are the uniformly distributed loads in pounds upon AB, AC,
respectively, and if 5 is the span of the roof in feet, find the weight of
metal (wrought-iron) in the ties.
A n s - 5_ ? 5 co t fi, f being inch-stress in pounds, and
6 /
ft the angle ABD.
(a) If AB = AC = 20 ft., AD 5 ft., the angle BAD = 60, find
86 THEORY OF STRUCTURES.
the stresses in the several members when a weight of 3500 Ibs. is con-
centrated at the vertex.
Ans. 7000 Ibs.; 6309.8 Ibs.; 3500 Ibs.
(b) The roof in (a) is loaded with 10 Ibs. per square foot on one side
and 33 Ibs. per square foot on the other; the trusses being 13 ft. centre
to centre. Determine (a) the stresses in the several members. Examine
(jj] the effect of a horizontal pressure of 14 Ibs. per square foot on the
most heavily loaded side, assuming that the reaction is equally divided
between the two supports.
Ans. (a) mSolbs.; 10077.65 Ibs.; 5590 Ibs.
69. In the truss represented in the accompanying figure, the load
on AB = IVi , on AC = W* ; the angle ABD ft ;
AD = BD = AE = CE. Find the total weight
of metal (wrought-iron) in the tie-rods.
5 Wi + Wi
FlG . 14t . Ans. - - S cot /?; 5 being the span
and /the inch- stress.
(a) If the stress in BD or EC is equal to the stress in DE, show that
ft = 60 ; a being the angle ABC.
(b) The trusses are 12 ft. centre to centre ; the span is 40 ft.; the hori-
zontal tie is 1 6 ft. long ; the rafters are inclined at 60 to the vertical ; the
dead weight of the roof, including snow, is estimated at lolbs. per sq. ft.
of roof-surface. Determine the stress in each member when a wind
blows on one side with a force of 30 Ibs. per sq. ft. normal to the roof-
surface, assuming that the horizontal reaction is equally divided between
the supports.
Ans. Stress in AB = 8956.8 Ibs. ; BD = 10015.2 Ibs. = EC\
AD = 2503.8 Ibs. = AE\ DE = 8196 Ibs.;
AC =11356.8 Ibs.
70. In the truss represented by the accompanying figure, the load
upon AB = W\ , upon AC = W* ; the angle ABD=
ft\ the span BC 5; the ties AD, BD, AE, CE are
equal ; /''and G are the middle points of the rafters.
Ftod the amount of metal in the tie-rods (wrought-
iron).
6 / sin ft cos ft
(a) The struts DF and EG are each 5 ft.; the angle ABC 30 ; the
dead weight of the roof, including snow, is 9 Ibs. per square foot of roof-
surface, and the trusses are 12 ft. centre to centre. Determine the
stresses in the several members when a wind blows with a force of 30
EXAMPLES. /
Ibs. per square foot of roof-surface normal to the side AB. The span
= 60 ft., and the end C rests upon rollers.
Secondly, determine the stresses produced in the members of the
truss in the preceding question when a single weight of 3000 Ibs. is sus-
pended from G.
Ans.(\) Stresses in BD ; DA ; DE; EA; EC;
31238.55; 19852.35; 12633.6; 8113.5; 24379.43;
BF; FA; FD; CG; GA ; GE.
29620.44; 28685.16; 7855.2; 22420.44; 21485.16; 1620 ibs.
(2) Stresses in BD; DA; DE; EA; EC;
375 t/39; 125 4/39; looo Vj; 875 \/&\ 11251/3^;
BF; FA; FD ; CG; GA; GE. _
2625; 2625; o; 7875; 6375; 1 500 4/3 Ibs.
(b) The rafters AB, AC are of unequal length and make angles of
60 and 45, respectively, with the vertical; the strut DF = -]\ ft.; the
tie DE is horizontal ; the dead load upon each rafter = 100 Ibs. per
lineal foot; the wind-pressure*normal to AB = 300 Ibs. per lineal foot;
rollers are placed at C. Find the stresses in all the members. The
rafter AB = 45 ft.
Show by dotted lines how the stress diagram will be modified:
(1) If the rollers aie placed at B.
(2) If the strut DF is omitted.
(3) If a single weight of 500 Ibs. is concentrated at D.
(V) If it is assumed that the horizontal reaction is equally divided be-
tween B and C, show that the stress in DE due to a horizontal wind-
pressure upon AB is nil ; the angle ABC being 30.
(d) In a given roof, the rafters are of pitch-pine, the tie-rods of
wrought-iron ; the span is 60 ft.; the trusses are 12 ft. centre to centre;
DF ' = 5 ft. =EG; the angle ABC = 30; the dead weight of the roof, in-
cluding snow, is 9 Ibs. per sq. ft. of roof-surface ; rollers are placed at C;
a single weight of 3000 Ibs. is suspended from F, and the roof is also
designed to resist a normal wind-pressure of 26.4 Ibs. per sq. ft. of roof-
surface on one side AB. Determine the stresses in the several
members.
71. In the truss represented in the accompanying figure, the struts
DF, DH, EG, EK are equal, and the ties BD, AD,
EA, EC are also equal ; the load upon AB is W\ ,
and upon AC is W* . Find the weight of metal
(wrought-iron) in the ties. FlG
Ans -5- - 4^i + 3W + ^ 2 )cos 2 /?
18 / cos ft sin ft
(a) AD = AE = BD = EC = 23 ft.; the angle ABC = 30 ; the span
= 79 ft.; the trusses are 13 ft. centre to centre; the heel B is free to
THEORY OF STRUCTURES.
slide on a smooth wall-plate ; the dead weight of the roof, including
snow, is 8 Ibs. per square foot of roof-surface. Determine the stress to
which each member is subjected when the wind blows horizontally with
a force of 40 Ibs. per square foot of vertical surface (i) upon the side
AB, (2) upon the side AC.
Ans. See Ex. 3, Art. 24.
(b) The rafters AB, AC are inclined at 60 to the vertical and are
each 40 ft. in length. The foot Crests
on rollers, and the foot B is fixed. The
strut DF is vertical, is 10 ft. long, and
is equal to the strut DE in length.
Also AF = HF=io ft. The dead load
carried by the rafters is 120 Ibs. per
lineal foot. Provision has also to be
made for a normal wind-pressure upon
AB of 300 Ibs. per lineal foot. Draw
the stress diagram, and show how it
will be'modified if the strut DF is re-
FlG - I44 ' moved.
Ans. Vertical reaction at B = 10528 Ibs. both before and after
DF is removed.
Horizontal reaction at B~= 6000 Ibs.
The dotted lines show the modified stresses for one half
of the truss.
72. The load upon a roof-truss of the accompanying type is 1000 Ibs. at
each joint; the span 100 ft.; the rise = 25 ft. Find the stresses in
A
C
G K N P
FIG. 145.
the different members. How will the stresses be affected by an addi-
tional load of 250 Ibs. at each of the joints between the foot and ridge
on one side?
Ans. Stress in BD = 5500 4/5~; DF = 5000 1/5";
' FH '= 4500 4/5 ; HL 4000-4/5'; LN '= 3500 4/5";
NA = 3000 1/5 ; DE o ; FG = 500 ; HK = 1000;
LM, = 1 500 ; NO = 2000 ; AP = 5000 ;
BE = i looo = EG- GK 10000 ; KM ~ 9000 ;
MO = 8000 ; OP = 7000 ; DG = 500 4/5";
FK = 1000 4/2 ; MM = 500 4/FJ ; LO 1000 4/5";
NP = 500 4/29 Ibs.
EXAMPLES. 89
73. The dead load upon a roof-truss of accompanying type consists
of looo Ibs. at F, 1000 Ibs. at K, and 500 Ibs.
a ^ G', the wind-pressure is a normal force of
3 lb s - P er sc l uare f ot of roof-surface upon
^ ; the span = 9 ft>; the rise = 2S ft-; the
trusses are 25 ft. centre to centre. Find the
FlG - 14<5t stresses in the several members when rollers
are (a) at C, (ff) at B.
Ans. (a) Reaction (vertical) at C= 12291$ Ibs.; vertical reaction
at ^ = 23958^ Ibs.; horizontal reaction at B =
18750 Ibs.
Tension in BD 48625 ; DL 34475 ; LE = 21675 '
EC= 22125; Z?//=786i; y2 = 15888! ;
A^ = 250 Ibs.
Compression in BF=^666^ 4/106 ; 7<y/=2788f 4/106 ;
/f,4 = 1977$ 4/106 ; ^A" = 2325 4/106 ;
ATS = 2408^ 4/106 ; C = 2458! 4/106 ;
= 1572! 4/106 ; Z/f= 1505! 4/181 ;
= 83^ 4/T8i ; ^6^ = 50 4/7^6 Ibs.
(<5) Only alteration in stresses is that each stress in the
different sections of the horizontal tie is diminished
by 18750 Ibs.; all the remaining stresses are un-
changed.
74. In the accompanying roof-truss, angle ABC = 30 ; the span = 90$-
ft.; DF = EG = io| ft.; each rafter is divided into
four equal segments by the points of support ;
the trusses are 20 ft. centre to centre ; the weight
of a bay of the roof = 24416 Ibs. Determine the
F IG - 147- stress in each member.
Also determine the stresses due to a wind-pressure of 30 Ibs. per square
foot of roof-surface acting normally to AB t when rollers are under (a) C,
75. The figure represents a bowstring truss of 80 ft. span, cut off by
the plane MN and supported at O. The upper
flange OCDE is an arc of a circle of 85 ft. radius ; M
OA = AB = etc. = 10 ft. ; the rise of the truss
= 10 ft. ; a load of 15 tons is concentrated at each
of the points A and B ; the reaction at O 45 "
tons. Find the stresses in the members cut by |N
the plane MN. FlG ' I48 '
THEORY OF STRUCTURES.
76. The figure is a portion of a bridge-truss cut off by the plane MN
and supported upon the abutment at A; AC
=CE= 14/4 ft.; the depth BC = DE = i;i ft. ;
in the third panel the compression in the upper
=64,600 ibs. chord is 64,600 Ibs. ; the tension in the lower
chord is 53,800 Ibs. Find the reaction at A, the
...,80011)8. equal weights supported at C and E, and the
diagonal stress T.
N Ans.- Reaction = 19,474 Ibs.; weight at C
FIG. 149.
and at E = 9737 Ibs. ; T= 17.977
77. The top beam of a roof for a clear span of 96 ft. consists of six
bars AB, BC, CD, DE, EF, FG, equal in length and so placed that
A, B, C, D, E, F, G are on circle of 80 ft. radius ; the lower boom also
consists of six equal rods AH, HK, KL, LM, MN, NG, the points H, K,
L, M, and N being on a circle of 148 ft. radius; B is connected with
//", C with K, D with L, E with M, and F with N\ the opposite corners
of the bays are connected by cross-braces ; the end A is fixed to its sup-
port, G being allowed to slide freely over a smooth bed-plate. Determine
graphically the stresses in the various members when there is a normal
wind-pressure per lineal foot of 460 Ibs. upon AB, 340 Ibs. upon BC, and
60 Ibs. upon CD.
78. A bowstring roof-truss, with vertical and diagonal bracing, of
50 ft. rise, and five panels, is to be designed to resist a wind blowing
horizontally with a pressure of 40 Ibs. per square foot. The depth of the
truss at the centre is 10 ft. Determine, graphically, the stresses in the
several members of the truss, assuming that the roof rests on rollers
at the windward support.
79. Determine the chord, vertical and diagonal stresses in a Howe
truss of 80 ft. span, 8 ft. depth, and ten panels, due to a load of 40 tons
(a) concentrated at the centre ; (ff) concentrated at the third panel point ;
(c) uniformly distributed ; (d) distributed so that 5 tons is at first panel
point, 10 tons at second, and 25 tons at third.
Ans. Panel stresses in tension chord :
ISt
2d
3d
4th
5 th
6th
7 th
8th
gth
loth
a
20
40
60
80
100
100
80
60
40
20
b
28
S6
84
72
60
60
48
36
24
12
c
18
32
42
48
50
50
48
42
32
18
d
30
55
70
60
50
50
40
30
20
10
EXAMPLES.
Panel stresses in compression chord :
a
20
40
60
80
80
60
40
20
b
28
56
84
72
48
36
24
12
c
18
32
42
48
48
42
32
18
d
30
55
70
60
40
30
20
10
Stresses in verticals :
a
2O
20
20
20
40
20
20
20
20
b
28
28
28
12
12
12
12
12
12
c
18
14
10
6
4
6
10
14
18
d
30
25
15
10
10
10
10
IO
10
Diagonal stresses :
a
20
4/~2
tons
in
each
diago
nal.
b
284/2
284/2
284/2
124/2
124/2
124/2
124/2
124/2
124/2
124/2
c
184/2
144/2
104/2
64/2
24/2
2 4/2
64/2
104/2
144/2
l8\/2
d
304/2
254/2
154/2
104/2
104/2
104/2
IO|/2
104/2
104/2
104/2
4/3;
4/3
80. A Warren girder of 60 ft. span, composed of six equilateral
triangles, carries upon its lower chord a weight of 2 tons at the first and
second joints, 15 tons at the centre joint, and Ji tons at the fourth and
fifth joints. Find the stresses in all the members.
Ans. Stresses in tension chord : ist bay ==
3d =
5th =
Stresses in compr. chord : ist bay =
3 d =
5th =_V ift
Diag. stresses ist and 2d bays = - 1 / 4/3 ' 3d and 4th = *$- 4/3";
5th and 6th = V- i/3"; ;thand 8th = - 1 / 4/3";
9th and loth = ^ 4/J; nth and I2th = ^- 4/3";
81. Determine the stresses in the members of a Fink truss of 240 ft.
span and sixteen panels; depth of truss = 30 ft.; uniformly distributed
load = W.
- 4/3"; 2d =
-/- 4/3"; 4th =
- Vl>\ 6th = || ^J.
-v/J"; 2d = % 3 - 4/3";
4/3 4th = S Vy,
92
THEORY OF STRUCTURES
Ans. A i LMNOPQRS' Stress in BA, BM, DM, DO, FO, FQ,
HQ, H$, same and =
C D E F G H
FIG. 150.
6 4
in
w w _
in EA, ES same and = -=- |/c ; in AK = y 17 ; i
5 4
W^ W
DN, FP, HR, same and = r ; in O/, (7(9 same and = -5- ;
10 o
W W
mEO = ; in KS = ; in AM, MO, OQ, QS same and
64
82. Determine the stresses in the members of a Bollman truss 100 ft.
long and 12^ ft. deep, under a uniformly distributed load of 200 tons, to-
gether with a single load of 10 tons concentrated at 25 ft. from one end.
Ans. Stress in AB = ^ ^2 ; BL = if* 4; ^Z? = y 4/5 ;
,DZ = - 2 / 4/37 ; AF = H 1 I/To; FL = y 1/26;
^^ ^ 3^ ^jy =///:; in ^C = 25 = /re =
HK = etc.; DE = 50 tons; compression along
^4 = 193! tons.
Note. Questions 53, 54, 57-59, 61, 66, 67, 70, 71, 73, and 74 can be
easily solved graphically.
83. Determine the stresses in the several members when the throw
of the crane in Question 55 is increased by the introduction of the new
members, shown by the dotted lines.
CHAPTER II.
SHEARING FORCES AND BENDING MOMENTS.
Note. In this chapter it is assumed that all forces act in one and the same
plane, and that the deformations are so small as to make no sensible alteration
either in the forces or in their relative positions.
I. Equilibrium of Beams. A beam is a bar of somewhat
considerable scantling, supported at two points and acted upon
by forces perpendicular or oblique to the direction of its length.
CASE I. AB is a beam resting upon two supports in the
same horizontal plane. The reactions Ro
R l and R^ at the points of support are
vertical, and the resultant P of the
remaining external forces must also
act vertically in an opposite direction
at some point C.
IA
FIG. 152.
According to the principle of the lever,
P AC
= P -r-f* , and
CASE II. AB is a beam supported or fixed at one end.
Such a support tends to prevent any deviation from the
straight in that portion of the beam,
and the less the deviation the more
'B perfect is the fixture.
The ends may be fixed by means of
two props (Fig. 153), or by allowing it
, B to rest upon one prop and preventing
upward motion by a ledge (Fig. 154), or
by building it into a wall (Fig. 155).
In any case it may be assumed that
the effect of the fixture, whether perfect
or imperfect, is to develop two unequal
forces, Q and R, acting in opposite di-
FIG. 155.
rections at points M and N.
These two forces are equivalent
93
94
THEORY OF STRUCTURES.
to a left-handed couple (Q, Q), the moment of which is
Q.MN, and to a single force R Q at N. Hence R Q
must = P.
CASE III. ^4/? is an inclined beam supported at A and
resting upon a smooth vertical surface
at B.
The vertical weight P, acting at the
point C, is the resultant load upon AB.
Let the direction of P meet the hori-
zontal line of reaction at B in the
point D.
The beam is kept in equilibrium by
the weight P, the reaction R^ at A, and the reaction 7v? 2 at B.
Now the two forces R^ and P meet at /), so that the force R^
must also pass through D.
Hence
cos
and
^ = P tan
. The same principles hold if the beam in Cases I and
II is inclined, and also whatever may be the directions of the
forces P and R^ in Case III.
CASE IV. hi general, let the beam AB be in equilibrium
under the action of any number of forces P lt P 2 , P 9 , . . . ,
Q\ * Qi > Qs > > f which the magnitudes and points of appli-
? &
f
M
?'
r
1
!
1^
FIG.
157-
v v
Q 3 Q 2
cation are given, and which act at right angles to the length of
the beam. Suppose the beam to be divided into two segments
by an imaginary plane MN. Since the whole beam is in equi-
librium, each of the segments must also be in equilibrium.
Consider the segment AMN.
EQUILIBRIUM OF BEAMS. 95
It is kept in equilibrium by the forces P lt P 9 , P 9 , . . . and
by the reaction of the segment BMN upon the segment AMN
at the plane MN; call this reaction^. The forces P, , P, ,
P 3 , . . . are equivalent to a single resultant R 1 acting at a point
distant r l from MN. Also, without affecting the equilibrium,
two forces, each equal and parallel to R t , but opposite to one
another in direction, may be applied to the segment AMN at
the plane MN, and the three equal forces are then equivalent
to a single force R, at MN, and a couple (R l , R,) of which
the moment is R l r l .
i * ! r R '
A. I !11 '
FIG. 158.
Thus the external forces upon AMN are reducible to a
single force R 1 at MN, and a couple (R lt RJ. These must
be balanced by E^ , and therefore l is equivalent to a single
force R t at MN and a couple ( R^ , R^.
In the same manner the external forces upon the segment
BMN are reducible to a single force R^ at MN, and a couple
(/?,, ^2) of which the moment is R^. These again must
be balanced by 3 , the reaction of the segment ^4 J/V upon
the segment BMN.
Now E l and 2 evidently neutralize each other, so that the
force R^ and the couple (R 1 , R } ) must neutralize the force
^ 2 and the couple (R 9 , R^). Hence the forced, and the
couple (R t , R^ are respectively equal but opposite in effect
to the force ^ 2 and the couple (R t , R 9 ) ; i.e.,
R l == RZ and R 1 r l = R^ ; /. r l = r. t .
The force R, tends to make the segment AMN slide over
the segment BMN at the plane MN, and is called the Shearing
THEORY OF STRUCTURES.
Force with respect to that plane. It is equal to the algebraic
sum of the forces on the left of MN,
So ^ 2 = 0, Q, Q 3 + . . . = 2(Q) is the algebraic sum
of the forces on the right of MN, and is the force which tends
to make the segment BMN slide over the segment AMN at
the plane MN. R^ is therefore the Shearing Force with respect
to MN, and is equal to R l in magnitude, but acts in an opposite
direction.
Again, let/, , / 2 , / 3 , . . . , ^ , q z , q z , . . . , be respectively the
distances of the points of application of P l , P t , P 3 , . . . , Q l , <2 2
<2, , . . . from MN.
Then R 1 r 1 , = the algebraic sum of the moments about
MN of all the forces on the left of MN,
is the moment of the couple (R lt R^).
This couple tends to bend the beam at the plane MN, and
its moment is called the Bending Moment with respect to MN
of all the forces on the left of MN.
So R 9 r 9 , = the algebraic sum of the moments about MN
of all the forces on the right of
is the Bending Moment, with respect to MN, of all the forces on
the rig/it of MN, and is equal but opposite in effect to R l r 1 .
It is seen that the Shearing Force and Bending Moment
change sign on passing from one side of MN to the other, so
that to define them absolutely it is necessary to specify the seg-
ment under consideration.
Remark. The reaction E l has been shown to be equivalent
to the force R l and the couple ( R l , R^. The Moment
of this couple may be called the Elastic Moment, the Moment
of Resistance, or the Moment of Inflexibility, and is equal in
magnitude, but opposite in effect, to the corresponding Bend-
ing Moment due to the external forces.
SHEARING FORCES AND BENDING MOMENTS.
97
FIG. 159.
2. Examples of Shearing Forces and Bending Mo-
ments. In each of the following
examples the beam is horizontal and
of length /. .
Ex. i. The beam OA, Fig. 159,
is fixed at A and carries a weight
Pat O.
The Shearing Force (S) at every
point of the beam is evidently con-
stant and equal to P.
Upon the verticals through A and take AB and OC each
equal or proportional to P; join BC. The vertical distance
between any point of the beam and the line BC represents the
shearing force at that point.
Again, the Bending Moment (M) at any point of the beam,
distant x from O is Px ; it is nil at <9, and PI at A.
Upon the vertical through A take AD equal or propor-
tional to PI; join DO. The vertical distance between any
point of the beam and the line DO represents the bending
moment at that point.
Ex. 2. The beam OA, Fig. 160, is fixed at A, and carries
a uniformly distributed load, of in-
tensity w per unit of length.
The resultant force on the right
of a vertical plane MN distant x from
O is wx and acts half-way between
) O and MN.
The Shearing Force (S) at MN is
therefore wx ; it is nil at (9, and wl
at A. Upon the vertical through A
take AB equal or proportional to wl\ join BO. The vertical
distance between any point of the beam and the line BO rep-
resents the shearing force at that point.
FIG. 160.
Again, the Bending Moment (M) at MN is ivx =
wx
it
is nil at
O, and at A.
Upon the vertical through A take
AC equal or proportional to .
9 8
THEORY OF STRUCTURES.
The bending moment at any point of the beam is repre-
sented by the vertical distance between that point and a pa-
rabola CO having its vertex at O and its axis vertical.
Ex. 3. The beam OA, Fig. 161, is fixed at A and carries
wx
FIG. 161.
a single weight P at O, together with a uniformly distributed
load of intensity w per unit of length.
The Shearing Force (S) at a plane MN distant x from O
is evidently P ' + wx \ it is P at (9, and P-\- wl at A.
Upon the verticals through O and A take OC equal or pro-
portional to P, and AB equal or proportional to wl -\- P\
join BC. The vertical distance between any point of the
beam and the line BC represents the shearing force at that
point.
Again, the Bending Moment (M) at MN is evidently
it is nil at O, and ---- \- PI at A.
Upon the vertical through A take AD equal or propor-
tional to -- 1- PL The bending moment at any point of
the beam is represented by the vertical distance between that
point and a parabola DOE having its axis EF vertical and its
SHEARING FORCES AND BENDING MOMENTS. 99
p
vertex at a point E, where OF = and EF is equal or pro-
portional to .
2w
Note. The ordinates of the line BC in Ex. 3, are equal to
he algebraic sum of the corresponding ordinates of the straight
lines BC and BO in Exs. I and 2. Also, the ordinates of the
curve DO in Ex. 3, are equal to the algebraic sum of the cor-
responding ordinates of the line DO in Ex. I, and the curve
CO in Ex. 2. Hence the same conclusions as in Ex. 3 are
arrived at by treating the weight P and the load wl inde-
pendently, and then superposing the respective results.
Ex. 4. The beam OA, Fig. 162, rests upon two supports
at O and A, and carries a weight P
at a point B, dividing the beam
into the two segments OB, J3A, of
which the lengths are a and b re-
spectively.
The reactions R l , R z at O and
A are vertical, and according to the
principle of the lever, 'F IG .
R, P~ t and R, = P*.
The Shearing Force (S) at every point between O and B is
constant and equal to R l = P-,. On passing B the shearing
force (S) changes sign, and its value at every point between B
-ind A is constant and equal to R^ P = P-, R^ . Upon
the verticals through O, B, and A take OC\ BE, each equal or
proportional to -y, and BF, AD, each equal or proportional
to -y ; join CE and DF. The shearing force at any point of
the beam is represented by the vertical distance between that
point and the broken line CEFD.
100 THEORY OF STRUCTURES.
Again, the Bending Moment (M) at any point between O
and B distant x from O is R l x=^P-.x- J it is nil at O, and
P^j at B.
The Bending Moment (M) at any point between B and A
distant x from O is R^x /%*; a) = P-.(l x) ; it is
ab
P-j at B, and nil at A.
Upon the vertical through B take BG equal or proportional
ab
to P~r\ join 06- and y^^. The bending moment at any
point of the beam is represented by the vertical distance be-
tween that point and the line OGA.
p
Cor. If Pbe at the centre of the beam, S = , and M
PI
at the centre = .
4
Ex. 5. The beam OA, Fig. 163, rests upon two supports
at O and A, and carries a uni-
formly distributed load of inten-
sity w per unit of length.
The reactions at O and A are
each equal to .
The resultant force between O
and a plane MN distant x from O
FlG - 1<5 3- is wx y and acts half-way between
O and MN. The Shearing Force (S) at MN is theiefoie
wl wl
wx ; it is at O, nil at the middle point B, and
2 2
- at A. Upon the verticals through O and A take OC
and ADj each equal or proportional to -- ; join CD. The
shearing force at any point of the beam is represented by the
vertical distance between that point and the line CD.
SHEARING FORCES AND BENDING MOMENTS.
Again, the Bending Moment (M) at MN is
101
wl
wl
wx"
.,
it is nil at O and at A ; it is a maximum and equal to - at
b
the middle point B. Upon the vertical through B take BE
equal or proportional to - . The bending moment at any
o
point of the beam is represented by the vertical distance be-
tween that point and a parabola OEA having its vertex at E
and its axis vertical.
Cor. i. The shearing force is a minimum and zero at the
Wl
centre, a maximum and at the ends, and increases uni-
formly with the distance from the centre.
Cor. 2. The bending moment is a minimum and zero at
w/ 2
the ends, a maximum and -=- at the centre, and diminishes
o
as the distance from the centre increases.
Ex. 6. The beam OA, Fig. 164, rests upon two supports at
O and A, and carries a weight P at
a point B, together with a uniformly
distributed load of intensity w per
unit of length.
Let the lengths of the segments
OB, BA be a and b, respectively.
The reactions R^ at O, and R^
at A, are vertical, and according to
the principle of the lever,
b wl
and
FIG. 164.
IO2 . THEORY OF STRUCTURES.
The Shearing Force (S) at any vertical plane between O and
B distant x from O is
b wl
R l vox = P-j -\ wx ;
/ 2
Pb wl Pb , w/
it is -T- -| -at (9, and j- -| w# at /?.
/ 2 / 2
The Shearing Force (S) at any plane between ^ and ^4
distant ;r from 6> is
R^-P-wx^P-f + ^-P-wx^^-P-j-wx-,
wl Pa Pa wl
it is j wa at B, and = at A .
21 I 2
Upon the verticals through O, B, and A take OC equal or
proportional to -, | , BD equal or proportional to
I 2
Pb wl wl Pa
j--\- wa, BE equal or proportional to -. wa,
wl Pa
and AF equal or proportional to -=- ; join CD and
2 /
EF. The shearing force at any point of the beam is rep-
resented by the vertical distance between that point and the
broken line CDEF.
wl Pa
If > -7- + wa, BE is positive, and therefore E is ver-
2 /
tically above B.
Again, the Bending Moment (M) at any point between
O and B is
it is nil at 0, and
SHEARING FORCES AND BENDING MOMEN7^S. 103
The bending moment (M) at any point between B and A
distant x from O is
/ b wl\ wx* I _a , wl\ wx*
(P-I + -J)*- -*\* -*) = (- p -i + -?)*T+ pa >
/ b wl\ wa*
it is \P-j + In -- at , and nil at A. ,
Upon the vertical through j take BG equal or propor-
(^ wl\ we?
P-j-\ -- Jtf -- . The bending moment at
any point of the beam between O and B is represented by the
vertical distance between that point and a parabola OGH
having its axis HT vertical and its vertex at a point //, where
and
HT is equal or proportional to \P-. -| -- ) .
2,1.U S / 2 '
The bending moment at any point between B and A is
represented by the vertical distance between that point and a
parabola AGK having its axis KV vertical and its vertex at a
point K, where
and
a wlY
KV is equal or proportional to 1 P-. -| j -|~
Cor, If the weight Pis at the centre,
P PI wl*
S = , and M at the centre = 1- -5-.
2 4 o
Note. The ordinates of the lines CD and EF in Ex. 6 are
equal to the algebraic sum of the corresponding ordinates of the
104
THEORY OF STRUCTURES.
lines CE, FD in Ex. 4, and the line CD in Ex. 5. Also, the
ordinates of the curves OG y AG are equal to the algebraic
sum of the corresponding ordinates of the lines OG, AG in Ex.
4, and the curve OEA in Ex. 5. Hence the same conclusions
as in Ex, 6 are arrived at by treating the weight P and the
load wl independently, and then superposing the respective
results.
Ex. 7. In fine, a beam, however loaded, may be similarly
treated, remembering that if the load changes abruptly at dif-
ferent points, the portions of the beam between the points of
discontinuity are to be dealt with separately. For example,
the beam OA, Fig. 165, rests upon two supports at and A,
and carries three weights P, , P 9 , P 3 at points C, D, E, of which
the distances from O are p^ / 3 ,-/ 3 , respectively. A point B
divides OA into segments OB = a and BA = #, which are
FIG 165.
uniformly loaded with weights of intensities w v and w^ per
unit of length, respectively. The reactions R } and R^ at O and
A are vertical, and according to the principle of the lever,
RJ = />,(/ - A) + Ptf - A} + />,(/ - A)
a
and
+ X4 + 4
SHEARING FORCES AND BENDING MOMENTS. 10$
To represent graphically the Shearing Force at different
points of the beam :
Upon the verticals through <9, C, B, D,E, A, take OF, CG,
CH, BK, DL, DM, EN, EV, and A T, respectively equal or
proportional to
R,, R, wj,, R - wj, - />,, R, - w,a -r- P lt
R, - w,a P, ?e> 2 (/ 2 - a\
R, - w,a, - P l - w 2 (A - a) - P
R, - w,a P l - wjp t - a) - P 9 ,
R v - ^v l a - P l - u>,(p, a) - P, - P 9 ,
and R, - w,a - P, - w,b - P, - P, = R,.
Join FG, HK, KL, MN, and VT. The shearing force at
any point of the beam is represented by the vertical distance
between that point and the broken line FGHKLMNVT.
To represent graphically the Bending Moment (M) at dif-
ferent points of the beam :
Mat = 0', Mat C=Rj> l - ;
Mat D = R. t p, - ?ey?( 2 -
106
THEORY OF STRUCTURES.
- ^ ( A_^T _ PI(A _ A) _ /> i(A _ A) .
and
MatA=0.
Upon the verticals through C, B, D, and take Ci, B2,
Z>3, and 4, respectively equal or proportional to the bending
moments at these points.
The bending moment at any point of the beam is repre-
sented by the vertical distance between that point and the
parabolic arcs Oi, 12, 23, 34, and 4.A. The axes of these pa-
rabolas are vertical, and the positions of the vertices may be
easily found from the several equations.
Ex. 8. A beam OA, Fig. 166, of which the weight may be
neglected, is 15 ft. long, is fixed
at (9, and carries a weight of 80
Ibs. at A. Determine the bend-
ing moment at a point distant 10
ft. from the free end. Also illus-
trate the shearing force and bend-
ing moment at different points of
the beam graphically. The re-
quired bending moment is 80 X
10 800 Ib.-ft.
The shearing force is the same at every point of the beam,
and equal to 80 Ibs. Choose a vertical scale of measurement
so that half an inch represents 160 Ibs.
Upon OA describe a rectangle OABC, in which OC = AB
= y. The ordinate from every point of BC to AO is J", or
80 Ibs., and is therefore the shearing force at the foot of such
ordinate.
Again, the bending moment at O is 80 X 15 = 1200 Ib.-ft.
Choose a vertical scale of measurement so that I inch repre-
sents 1 200 Ib.-ft. Upon the vertical through O take OD =
FIG. 166.
SHEARING FORCES AND BENDING MOMENTS.
107
I inch ; join DA. The ordinate from any point of DA to OA
is the bending moment at its foot. For example, at ii ft.
from O the ordinate is J", or 300 lb.-ft., and this is equal to
80 X 3i', i.e., the bending moment.
Ex. 9. A beam OA, Fig. 167, of which the weight may be
neglected, rests upon two supports
at and A, 30 ft. apart, and carries
a uniformly distributed load of 200
Ibs. per lineal foot, together with a
single weight of 600 Ibs. at a point
B dividing the beam into segments
OB, BA, of which the lengths are 10
and 20 ft. respectively. Determine
the shearing force and bending mo-
ment at the points C and D, distant
5 ft. from the nearest end. Also,
illustrate graphically the shearing
force and bending moment at differ-
ent points of the beam.
Let R l , R^ be the reactions at O
and A, respectively. Then
R, . 30 600 . 20 + 200 . 30 . 1 5 = 102000 ;
.-. R l = 3400 Ibs., and R^ 200 . 30 + 600 R l = 3200 Ibs.
The Shearing Force at C = 3400 200 . 5 2400 Ibs.
" >=3400-200. 25-600= 2200 Ibs.
The Bending Moment at C = 3400 . 5 200 . 5 . -
= 14,500 lb.-ft.
The Bending Moment at D = 3400. 25 200 . 25 . 600. 1 5
= 13,500 lb.-ft.
Next, considering the segment OB, the shearing force at O
is 3400 Ibs., and at B 1400 Ibs.
Considering the segment BA, the shearing force at A is
3200 Ibs., and at B 800 Ibs.
Choose a vertical scale of measurement so that i inch repre-
sents 3000 Ibs. Upon the verticals through O, B, A take OE
- -fa" , and AH = i y y ; join EF and
FIG. 167.
IOS THEORY OF STRUCTURES.
GH. The ordinate from any point of the broken line EFGH to
OA is the Shearing Force at its foot. For example, the ordinate
at D is ||", or 2200 Ibs.
Again, the bending moment at B is 3400. 10 200. 10. 5
= 24,000 Ib.-ft. Choose a vertical scale of measurement so that
I inch represents 24,000 Ib.-ft. Upon the vertical through B
take BK = i inch. Draw the parabolas OK, AK, with their
vertices at points determined as in Example (6). The ordi-
nate from any point of the curves OK, AK is the bending
moment at its foot.
For example, at a point 14 ft. from O the curve ordinate is
i-jJy", or 25,600 Ib.-ft., and this is the Bending Moment at the
same point, being also the greatest for the segment BA. The
vertex of AK is, therefore, vertically above the point of
which the horizontal distance from O is 14 ft.
3. Relation between Shearing Force and Bending
Moment. Let a beam AB be arbitrarily loaded with weights
w l , w 2 , w z , . . . concentrated at the points I, 2, 3, ...
A 1, 2, 3 r r l r r-+l n__ B
FIG. 168.
Let , , # 2 , # 3 , . . . be the lengths of the segments Ai, 12,
23, . . . , respectively.
Let M A , MB be the moments at A and B. These moments
are of course nil if the beam merely rests upon supports at its
ends.
The reaction R at A is given by the equation
Rl = /,(/-*,) + wJJ -a 1 -a,)+...+M B + M A)
I being the length of the beam.
The shearing force S, between A and i = R ;
3 O 12'
S n " -i " n = R 2(w)\
2(w) denoting the sum of the first ( i) weights.
SHEARING FORCE AND BENDING MOMENT. 1 09
The bending moment
M A -atA=M A ;
= M
, , A A
, " 2 = ^<X + a,) - v*, + M A = M,
M
Hence the difference between the bending moments at the be-
ginning and end of any interval is equal to the product of the
shearing force (S) for that interval by the length (a) of the in-
terval.
Let AM denote the difference between any two consecutive
bending moments ; then
This result has been deduced without any assumption as to
the number of the loads. They may therefore be infinite in
number and in the limit form a continuous load.
Thus, if 5 be the shearing force at a distance x from A,
dM
or, the shearing force at any point is equal to the rate of increase
of the bending moment per unit of length.
The above results may also be expressed as follows : The
shearing force at any point is measured by the tangent of the slope
at the corresponding point of the b ending-moment polygon or
curve.
The shearing force is positive, zero, or negative according as
R = 2(w) ;
IIO THEORY OF STRUCTURES.
2(w) being the sum of the weights up to the point under con-
sideration. In the case of a continuous load, of intensity GO,
-,(w) = I x wdx.
Thus the bending moment M at the same point is a maxi-
mum (or a minimum in certain special cases) when the shear-
ing force changes sign, i.e., when
Again, with an arbitrarily distributed load
and with a continuous load
Thus the difference between the ordinates of the bending-
moment diagram at any point and A is proportional to the
area of the shearing-force diagram between the same points.
From this result an important deduction may at once be
made.
The bending moment M x at any point between r and r + I
distant x from r is
M r + x(R w l w 2 . . . w r )
= M r + xS r+l .
Now M r+l = M r -f a r+l S r + iy and therefore S r+l is zero if
M r+l = M r , and also M x = M r = M r+1 .
Thus, the bending moment is the same at every point between
r and r -f- i, and the case is one of simple bending without shear,
as, e.g., with a carriage-axle.
EFFECT OF A ROLLING LOAD.
Ill
4. To Discuss the Effect of a Rolling Load. CASE I.
Let a single weight W travel from left to right over a girder
OA of length /, resting upon two
supports at O and A.
The reaction R l at (9, when W is
at B distant x from O, is W -, ,
FIG. 169.
and is the Shearing Force for all
points between O and B\ it is nil or
W according as the weight is at A
or 0. Upon the vertical through O
take OD equal or proportional to W; join DA. The shearing
force at any point of the beam between O and the weight, as
the latter travels from A towards (9, is represented by the ver-
tical distance between that point and the line AD.
Also, the shearing force at any point between B and A is
R^ W = Wjt and is equal or proportional to the vertical
distance between that point and the line OE where AE is equal
to OD.
Again, the Bending Moment at B, when W is at B, is
it is nil at O and at A ;
Wl
it is a maximum and = at the
4
middle point D. The bending mo-
ment at any point of the beam when
the weight is at that point is repre-
sented by the vertical distance be-
tween the point and the parabola OEA, having its axis vertical
Wl
and its vertex at E, where DE is equal or proportional to .
Note. The shearing and bending actions are symmetrical
on both sides of the centre, and it is therefore sufficient to deal
with one half of the girder only.
Cor. i. The shearing force and bending moment at any
point are maxima at the instant the weight passes that point.
For example, the shearing force at B for the segment OB,
FIG. 170.
112
THEORY OF STRUCTURES.
when the weight is at B, is equal or proportional to BC (Fig.
169), which is evidently greater than GH, representing the
shearing force at B, when the weight is at any other point G.
Again, the bending moment at B (Fig. 170), when W is at
B, is W -j x. If W is at any other point G distant a from
O, the bending moment at B is
or
- t
according as a < or > x, and in either case is greatest when
a x, i.e., when the weight is at B.
Cor. 2. In addition to the rolling load, let the girder carry
a permanent weight W at the centre.
Consider one half of the girder only, and, for convenience,
trace the shearing-force and bending-moment diagrams for W
below OA.
The compound diagram for maximum shearing forces is
W
DTLFD (Fig. 171), where KT is equal or proportional to ,
W
and KL OF is equal or proportional to .
The maximum shearing force at a point distant x from the
centre is represented by XY '== -r-(- -f- x\ -| .
/ \2 / 2
F Y
FIG. 171.
Again, the compound diagram for maximum bending mo-
ments is OEFO (Fig. 172), where DF is equal or proportional
W'l
to , and OF is a straight line.
EFFECT OF A ROLLING LOAD. 113
The maximum bending moment at a point distant x from
the centre is represented by
X Y=m-*\+*L\>
/ \4 / 2 \2
Or. 3. Theoretically, the total volume of material required
in the web of the girder in Cor. 2 is equal or proportional to
2 X ^^DTLF 3 Wl . i W'l
yj being the web unit stress.
So, if </be the effective depth of the girder, and f the unit
stress in one of the flanges, the total volume of metal in that
flange is equal or proportional to
2 X area OEFO 2 WF W'T I Wl* I W'P
fd ~
CASE II. Let a train weighing w per unit of length travel
over the girder from right to left, and let the total length o
the train be not less than that of p
the girder.
The reaction at A, Fig. 173,
when the front of the train is at
*7Jt)3u
B distant x from O, is -^ , and
is the shearing force for all points
between A and B. Upon the ver- FlG - J 73.
ticals through A and O take AD and OE each equal or pro-
portional to - . Thus between A and B the shearing force
at any point is represented by the vertical distance between
that point and a parabola having its axis vertical and its vertex
at O.
After the end of the train has passed O, the shearing force
at any point of the uncovered portion of the girder is evidently
represented by the vertical distance between that point and the
parabola AFE, having its axis vertical and its vertex at A.
in, as the train moves from O towards B, the reaction
114 f&EORY OF STRUCTURES.
at A, and consequently the bending moment at B, continually
increase. On passing B, the reaction at A still increases, and
the bending moment at B when the train covers a length a
of the girder is
we? w , wx
This expression is evidently a maximum when a(2l a) is a
maximum, i.e., when a ^ I. Hence the bending moment, and
therefore the flange stresses, at any point are greatest when the
moving load covers the whole girder.
Cor. I. The shearing force at any point B is a maximum
when the train covers the longest segment OB.
This is evidently the case until the train arrives at B, for
the reaction at A, and therefore the shearing force at B, will
continually increase up to this point. When the train passes
B and covers a length a(>x) of the girder, the shearing force
. wo*
at B is -j -- w(a x).
But this is < r , the shearing force at B when OB is
. c? x* f a + x
covered, if . < a x, i.e., if - -. < I, which is evi-
2/ 2/
dently the case.
Cor. 2. In designing the flanges of a girder, the rolling
load is supposed to cover the whole girder, and may be treated
as a uniformly distributed load.
Cor. 3. In addition to the roll-
ing load, let the girder carry a uni-
formly distributed load of w' per
unit of length.
r G As before, consider one half of
the girder only. Trace the shear-
ing-force diagram for the perma-
FlG - I74 ' nent load below OA. The com-
pound diagram is DHGK, where GH and AK are equal or
wl w'l
proportional to -~- and , respectively.
EFFECT OF A ROLLING LOAD. 115
The maximum shearing force at a point distant x from the
centre is represented by XY and is equal to
wfl_
27V 2
Again, the maximum flange-stresses are obtained by assum-
ing the total load upon the girder to be w -f- w' per unit of
length.
Ex. The two main girders of a single-track bridge are 80
ft. in the clear and 10 ft. deep. The dead load upon the
bridge is 2500 Ibs. per lineal foot. If the bridge is traversed by
a uniformly distributed live load of 3000 Ibs. per lineal foot,
determine the maximum bending moment and shearing force
at a point of the girder distant 10 ft. from one end.
The bending moment at any point is a maximum when the
train covers the whole of the bridge, in which case the total
distributed load is 5500 Ibs. per lineal foot, of which each girder
carries one half.
Thus the reaction at each support = . 80. ~ = 1 10,000
Ibs., and the bending moment at the given point = nooooX IO
10 X 2750 X 5 = 962,500 Ib.-ft.
The shearing force at the given point due to the dead load
= noooo 10x2750 82,500 Ibs.
The shearing force due to the live load is a maximum when
the live load covers the 70 ft. segment, and its value is then
1 500x70"
lbs -
Hence the total maximum shearing force
= 82,500 + 45,937* = 128,437^
THEORY OF STRUCTURES.
5. Moments of Forces with respect to a given Point Q.
First, consider a single force /*,.
Describe the force and fu-
. ....
nicular polygons, i.e., the line
S,S 9 and the lines AB, BC.
Through the point Q draw a
line parallel to S,S 6 , cutting the
lines AB and CB produced in x
and y.
Drop the perpendiculars BM
and ON upon yx and 5,5 6 produced. Then
.'. P,BM = xy . ON.
But .#lf is equal to the length of the perpendicular from Q
to the line of action of P lf and the product xy . ON is, there-
fore, equal to the moment of P l with respect to Q. Hence, if
a scale is so chosen that ON = unity, this moment becomes
equal to xy ; i.e., it is the intercept cut off by the two sides of
the funicular polygon on a line
drawn through the given point
parallel to the given force.
Next, let there be two forces,
P P
*ii *
Describe the force and fu-
nicular polygons vS^vSe and
ABCD.
Let the first and last sides
(AB and DC] be produced to
meet in G, and let a line through
the given point Q parallel to the line 5 a S 6 intersect these lines
in x and y.
Draw GM perpendicular to xy, and ON perpendicular to
S.S.. Then
xy _ S,S, _ resultant of P, and P a
~GM = ON = "ON
MOMENTS OF FORCES.
and hence
(the resultant of P, and P 2 ) X GM = xy . ON.
But GMis equal to the length of the perpendicular from Q
upon the resultant of P l and P^, which is parallel to S^S 6 and
must necessarily pass through G. Hence, if a scale is so chosen
that ON = unity, xy is equal to the moment of the forces with
respect to Q ; i.e., it is the intercept cut off by the first and last
sides of the funicular polygon on a line drawn through the given
point parallel to the resultant force.
A third force P 3 may be compounded with P l and P 9 , and
the proof may be extended to three, four, or any number of
forces.
The result is precisely trfe same if the forces are parallel.
The force polygon of the n parallel forces P l , P 9 , . . . P H
/
FIG. 177.
becomes the straight line 5 6 5,5 2 . . . S n . Let the first and last
sides of the funicular polygon meet in G. Drop the perpen-
diculars GM, ON upon xy and S 6 S n , xy, as before, being the
intercept cut off on a line through the given point Q parallel
to S C S H . Then
xy.ON = GM. S 6 S n . Hence, etc.
Thus the moment of any number of forces in one and the
same plane with respect to a given point may be represented
by the intercept cut off by the first and last sides of the funicu-
Il8 THEORY OF STRUCTURES.
lar polygon on a line drawn through the given point parallel to
the resultant of the given forces.
6. Bending Moments. Stationary Loads. Let a hori-
zontal beam AB, supported at A and B, carry a number of
weights P lt />, P. . . at the points N 19 N,, N 3 , . . .
I NX No N 3 L N 4 N 5 M
FIG. 178.
The force polygon is a vertical line 1234 . . . n, where
12 =/>, 23 = />, etc.
Take any pole (9 and describe the funicular polygon
A t A % A^ . . .
Let the first and last sides of this polygon be produced to
meet in G and to cut the verticals through A and B in the
points C and D.
Join Z>.
Let the vertical through G cut AB in Z, and CD in .";
is the line of action of the resultant.
Draw OH parallel to CD.
From the similar triangles OiH and GCK,
\H GK
~OH ~ CK '
BENDING MOMENTS.
From the similar triangles OnH and GDK,
nH_ GK
OH ~ DK '
iH DK BL R
R } , /? 2 being the reactions at A and B, respectively.
But iH+ nH = in = P, + P, + . . . = R, + R, .
Hence iH = R, and nH = R,.
Thus the line drawn through the pole parallel to the closing-
line CD divides the line of loads into two segments, of which the
one is equal to the reaction at A and the other to that at B.
Let it now be required to find the bending moment at any
point M of the beam, i.e., the moment of all the forces on one
side of J/with respect to M.
In the figure these forces are R t , P l , /> a , P s , P 4 , P 6 , and the
corresponding force polygon is Hi 23456. The first and last
sides of the funicular polygon of the forces are CD parallel to
OH, and A^A^ parallel to O6. If the vertical through M meet
these sides in x and y, then, as shown in Art. 5, the moment of
the forces R^ , P l , P 9 , P z , P t , P 6 with respect to M, i.e., the
bending moment at M, = ON .xy, ON being the perpendicular
from O upon iH produced.
Hence, if a scale is chosen so that the polar distance ON is
unity, the bending moment at any point of the beam is the inter-
cept on the vertical through that point cut off by the closing line
CD and the opposite bounding line of the funicular polygon.
7. Moving 1 Loads. Beams are often subjected to the
action of moving loads, as, e.g., in the case of the main girders
of a railway bridge, and it becomes a matter of importance to
determine the bending moments for different positions of the
loads. It may be assumed that the loads are concentrated on
wheels which travel across the bridge at invariable distances
apart.
At any given moment, let the figure represent a beam 1 1
120
THEORY OF STRUCTURES.
under the loads P lt P 9 , P 3 . . . Describe the corresponding
funicular polygon CC'C" . . . D, the closing line being CD.
Let the loads now travel from right to left. The result will
be precisely the same if the loads remain stationary and if the
supports 1 1 are made to travel from left to right.
Thus, if the loads successively move through the distances
C'
FIG. 179.
12, 23, 34, ... to the left, the result will be the same if the
loads are kept stationary and if the supports are successively
moved to the right into the positions 22, 33, 44, . . . The new
funicular polygons are evidently C' C" ...>', C"C" , . . D",
C'"C"" . . . D'", ... the new closing lines being CD', C"D",
C"D'", . . .
The bending moment at any point M is measured by xy for
the first distribution, xy' for the second, x"y" for the third,
etc., the position of M for the successive distributions being de-
fined by MM' = 12, M'M" = 23, M"M'" = 34, . ...
Similarly, if the loads move from left to right, the result
will be the same if the loads are kept stationary and if the sup-
ports are made to move from right to left.
It is evident that the envelope for the closing line CD for
all distributions of the loads is a certain curve, called the enve-
lope of moments. The intercept on the vertical through any
point of the beam cut off by this curve and the opposite bound-
MAXIMUM SHEAR AND BENDING MOMENT.
121
ary of the funicular polygon is the greatest possible bending
moment at that point to which the girder can be subjected.
EXAMPLE. Loads of 12 and 9 tons are concentrated upon a
horizontal beam of 12 ft. span at distances of 3 and 9 ft. from
the right-hand support. Find (a) the B. M. at the middle point
12'
3'
T
J9 tor
!12 tons-''
FIG. 180.
of the beam, and also (ft) the max, B. M. produced at the same
point when the loads travel over the beam at the fixed dis-
tances of 6 ft. apart.
Scales for lengths, -J in. = I ft. ; for forces, -^ in. I ton.
Take polar distance = in. 10 tons.
Case a. B.M. = xy X 10 = 3.15 X lOtons = 31 J ton-ft.
Case b. B. ^l.x'y' X 10 = 3.6 X 10 tons = 36 ton-ft.
8. Analytical Method of Determining the Maximum
Shear and Bending Moment at any Point of an Arbitrarily
Loaded Girder AB. At any given moment let the load con-
sist of a number of weights ;,,;,... w n , concentrated at
points distant a l , a^ , . . . a n , respectively, from B.
The corresponding reaction R^ at a is given by
RJ, it> l a l
wa
/being the length of the girder.
Let W n = w } + w 2 + + W M , the sum of the n weights.
" W r = ^v l + w 2 + . . . + w r , the sum of the first r w'ts.
The shear at a point P between the rth and the (r -\- i)th
weights is
5, = R, - w, - w, . . . - w r = R, W r .
122 THEORY OF STRUCTURES.
Let all the weights now move towards A through a distance
x, and let p of the weights move off the girder, q of the weights
be transferred from one side of P to the other, and s new
weights, viz., w n+1 , w n+ ^ , . . . w n+s , advance upon the girder,
their distances from B being a n+l , a n+y , . . . a n+s , respectively.
Let L = w^ -f- w z + . . . + w p , the total weight leaving the
girder.
Let T = w r+l -f- w r+t + . . . + w r+q , the total weight trans-
ferred from one side of P to the other.
Let R P l = w l a l -f- w a ^ 2 +
" Rl = w r a r Wr 2 ar
q
Thus R p , R q , R s are the reactions at A due, respectively, to
the weight which leaves the girder, the weight which is trans-
ferred, and the new weight which advances upon the girder.
The reaction R^ at A with the new distribution of the loads
is given by
RJ = w p+l (a p ^ + x) + Wp+ 2 (ap+ 2 + x) + . . . + w r (a r + x)
+ iVr+^a^ + x) + . . . + w n (a n + x) + zv n+1 a n+1 + .. .
+ ow^, - Rj - R P l +x(W n -L) + RJ,
and hence
Also, the corresponding j^^r at P is
w p ^ + . . . + w r +
Hence the shear at P with the first distribution of weights
is greater or less than the shear at the same point with the
second distribution according as
>
1 <
or ^ rt -
or T - L > R, -
MAXIMUM SHEAR AND BENDING MOMENT. 12$
Note. When no weights leave or advance upon the girder,
R s , R P , and L are severally nil, and hence
.S > 5"
according as - ^ - ;
i.e., according as the weight transferred divided by the distance
through which it is transferred is greater or less than the total
weight on the girder divided by the span.
Again, let z be the distance of P from B, and let
T) 7 I I I
J\. r l- '= W l & l | W^a^ f . . . j Wf/Zf. .
The bending moment at P with the first distribution of
weights is
J/j = R^(l z) ^(X z) w 2 (tf 3 z) ... w r (a r z)
The bending moment at the same point with the second dis-
tribution is
, = R % (1 z) w p+l (a^ + x z)~ w p+2 (a p + 2 + x z) . . .
w r (a r + x z) . . . w r+q (a r+q + x z)
= RJJL -z)- (R r l - R P l + R q l) -(x-z}(W r -L+ T).
Hence the bending moment at P with the first distribution
of weights is greater or less than the bending moment at the
same point with the second distribution according as
or
,(l -z)- (R r -X f + R q )l
-( x - z )(W r -L+T),
124 THEORY Of STRUCTURES.
or
*W r - (R f - R,y+ (x - *X W, - L + T) > (R, - *,)(/- *)
or
(B)
. If no weights leave or advance upon the girder R s ,
and L are severally nil, and
according as
If also the point P coincide with the rth weight, and the
distance of transfer, x l = a r a r+l , then
R q l tv r+l a r+I , T w r+l , and z = a r .
Hence J/i ^ J/ 2 , according as
or
i.e., according as the sum of the first r weights divided by the
length of the corresponding segment is greater or less than the
total weight upon the girder divided by the span,
If the weights are concentrated at the panel points of a
truss, the last relation may be expressed in the form
first (r) weights > total weight
r panels < total number of panels'
EXAMPLE. A series of loads of 3000, 23,600, 20,100, 21,700,
MAXIMUM SHEAR AND BENDING MOMENT. 125
22,900, 18,550, 18,000, 18,000, and 18,000 Ibs. travel, in order,
over a truss of 240 ft. span and ten panels.
Let ^AA . . . B be the truss, A A A being the
panel points. Let the loads travel from B towards A, and
compare the shear in the panel AA when the weight of 3000
Ibs. has reached A with the shear in the same panel when the
weights have advanced another 24 ft.
R ' = lo * I855 = l855 lbs " R > = ' 7 = To
W n 91300 lbs., L o, T = 3000 lbs.
Hence S^S^ according as (see A)
3000 - 0^1855 + -(91300 - 0)> 10985,
and
Let the weights again advance 24 ft.
R s = -- . 18000 = 1800 Ibs., R p = 9 X -j --,
10 / IO
W n 109,300 Ibs., L o, T ~ 23,600 Ibs.
Hence S t ^J S 2 , according as (see A)
23600 o^ 1800 o + -^(109300 o), or 23600^ 12730,
and
.-. s,>s,.
Hence the shear in the panel A A ' ls a maximum when the
weight of 3000 Ibs. is at A
Again, let the 3000 Ibs be at A> an d compare the bending
moment at A w ^h t ^ ie Den cling moment at the same point
when the weights have advanced first 24 ft. and then 48 ft.
towards A.
First, z 120 ft., L=o, T= 22,900 Ibs., AV 18000X24,
126
THEORY OF STRUCTURES.
R P = o, R q l = 22900 X 9 6 > x 24 ft., W r = 68,400 Ibs.,
W n 145,850 Ibs.
Hence M^M^ according as (see B)
120(0 22900 + JSoo o) + 22900 X 96 18000 X 24
+ 24(68400 - o + 22900) $ -(240 - 120X145850 - o),
240
or 1425600^1750200,
and
Second, z 120 ft., L = 3000 Ibs., T 18550) R s = o, R p l
= 3000 X 216, R q l= 18550X96, # = 24 ft., ^ = 91,300 Ibs.,
W n = 163,850 Ibs.
Hence M t ^ J/ Q , according as (see B)
120(3000 18550 + 3000. -} +240(18550. -^ -- o)
V 2407 \ 240 '
+ 24(91300 3000+ 18550)^ -(240- 120X163850 3000),
240
or 2155200^1930200,
and
Hence the bending moment at / 6 is a maximum when the
weight of 3000 Ibs. is at/ 1? i.e., when all the panel points are
loaded.
9. Hinged Girders. Any point of a girder at which the
bending moment is nil is termed a point of contrary flexure,
and on passing such a point the bending moment must neces-
sarily change sign.
Consider a horizontal girder resting upon supports at A, B,
C, D, and hinged at the points E and F in the side spans.
In order that there may be no distortion by the turning of
the hinges, the latter must not be subject to any bending
action ; i.e., they must be points of contrary flexure.
HINGED GIRDERS.
127
Let AE = a, EB = b, BC = c, CF = e, DF = d.
Let W^ W^ W z , W^ W, be the loads upon AE, EB, BC,
DF, FC, respectively, and let x l , x^ , x z , x^ , x b be the several
distances of the corresponding centres of gravity from the
points E, B, C, F, C.
Mt
FIG. 181.
The two portions AE and DF are evidently in precisely the
same condition as two independent girders of the same lengths,
carrying the same loads and supported at the ends. EF may
also be treated as an independent girder supported at B and C,
carrying the weights J/F 2 , W z , W & , and loaded at the cantilever
ends E and F with weights equal to the reactions at E and F
for the portions AE, DF assumed to be independent girders.
Let R^R^R^ R, be the reactions at A, B, C, A respec-
tively. Then
and
Hence, since R^ and R^ are always positive, there can be no
upward pull either at A or D, and no anchorage will be needed
at these points.
Next, taking EF as an independent girder,
the load at E = IV, - R, =
F= W,-R,=
128 THEORY OF STRUCTURES
Take moments about C and B. Then
- ( W, - *,X* + - Wfo + ) + ^ - >. +
and
-(IV,- Rtf - W^ + W 3 (c - x) - R,c
two equations giving R^ and R a , since R l and R 4 have been
already determined.
The pier moments P l at B and P t at C are
and
P =
,
'r values depending splely upon the loads on the spans contain-
ing the hinges.
The bending moment at any point in BC distant x from B
R,)(b + x] - W&, + x)-M
M being the bending moment due to the load upon the
length x.
The shearing-force and bending-moment diagrams for the
whole girder can now be easily drawn.
For any given loads upon the side spans, let AEH and DFL
be the bending-moment curves for the portions AB, CD ; BH
and CL representing the pier moments at B and C, respec-
tively. The bending moments for the least and greatest loads
upon BC will be represented by two curves HKL, HK'L, and
the distances TT' t VV through which the points of contrary
flexure must move, indicate those portions of the girder which
are to be designed to resist bending actions of opposite signs.
HINGED GIRDERS.
I2 9
Again, let the two hinges be in the intermediate span.
Let AB = a, BE = b, EF = c, FC = e, CD = d.
Let W,, W,, W,, W,, W, be the loads upon AB, BE, EF,
CD, CF, respectively, and let x l , # , x a , x^, x & be the several
distances of the corresponding centres of gravity from the
points B, B, F, C, C.
//
f
FIG. 182.
EF evidently may be treated as an independent girder sup-
ported at the two ends and carrying a load W 3 .
AE and DF may be treated as independent girders carry-
ing the loads W l , W^ and W^ , W^, respectively, and also loaded
at the cantilever ends E and F with weights equal to the reac-
tions at E and F due to the load W a upon girder EF, which is
assumed to be independent. Thus
the load at E = W,- ;
The pier moments P l at B and P 9 at C are
and
their values depending solely upon the loads on the span contain-
ing the hinges.
13 THEORY OF STRUCTURES.
Let R,, R,, R a , R t be the reactions at A, B, C, D, respec
tively, and take moments about the points B, A, D, C. Then
R -
R,d~ W,i - (, + d)- W.(x t + d) - W t (d - x t ) = O
- R t d - W,i - $ e - W tXi + W t x. = o
R n and ^ 8 are always positive;
R^ is positive or negative according as W^x, ^ P l ; and
t> t( tt tt " W r > P
A VV 1 X *<L 1 2
Thus there will be a downward pressure or an upward pull
at each end according as the moment of the load upon the ad-
joining span is greater or less than the corresponding pier mo-
ment. The ends must therefore be anchored down or they
will rise off their supports.
The shearing-force and bending-moment diagrams for the
whole girder can now be easily drawn.
Let HEFL be the bending moment curve for any given
load upon the span BC, BH and CL being the pier moments
at B and C, respectively.
The bending-moment curves for the least and greatest
loads on the side spans may be represented by curves A 777,
* A T'H and DVL, DV L, and the distances TT, W through
which the points of contrary flexure move indicate those por-
tions of the girder which are to be designed to resist bending
actions of opposite signs.
Reverse strains may, however, be entirely avoided by
making the length of EF sufficiently great as compared with
the lengths of the side spans.
The preceding examples serve to illustrate the mechanical
principles governing the stresses in cantilever bridges.
EXAMPLES. 13 l
EXAMPLES.
1. A beam 20 ft. long and weighing 20 Ibs. per lineal foot is placed
upon a support dividing it into segments of 16 and 4 ft., and is kept
horizontal by a downward force P at the middle point of the smaller
segment. Find the value of P and the reaction at the support.
Show that the required force P will be doubled if a single weight of
150 Ibs. is suspended from the end of the longer segment. Draw shear-
ing-force and bending moment diagrams in both cases.
Ans. 1200 Ibs. ; 1600 Ibs.
2. A man and eight boys carry a stick of timber, the man at the end
and the eight boys at a common point. Find the position of this point,
if the man is to carry twice as much as each boy.
Ans. Distance between supports = f length of beam.
3. A timber beam is supported at the end and at one other point ; the
reaction at the latter is double that at the end. Find its position.
Ans. Distance between supports = f length of beam.
4. Two beams ABC, BCD are bolted at B and C so as to act as one
beam supported at A and D\ AB 12 ft., C = 4 ft., CD = 16 ft, ; each
of the bolts will bear a bending moment of 100 Ib.-ft. Find the greatest
weight which can be concentrated on the portion BC. Draw diagrams
of maximum shearing force and bending moment when a wheel of the
same weight rolls over the beam.
Ans. I4 T 7 Ibs.
5. In the preceding question find the greatest uniformly distributed
load which the beam will bear.
Draw the shearing-force and bending-moment diagrams.
Ans. 25f Ibs.
6. A uniform beam 20 1/3 ft. in length rests with one end on the
ground and the other against a smooth vertical wall ; the beam is inclined
at 60 to the vertical and has a joint in the middle which can bear a
bending moment of 30,000 Ib.-ft. Find the greatest load which may be
uniformly distributed over the beam. Also find how far the foot of the
beam should be moved towards the wall in order that an additional
2000 Ibs. may be concentrated at the joint.
Draw curves of shearing force and bending moment in each case.
Ans. 8000 Ibs. ; distance = 10 ft.
I3 2 THEORY OF STRUCTURES.
7. A man of weight W ascends a ladder of length / which rests
against a smooth wall and the ground and is inclined to the vertical at
an angle a. The ladder has n rounds. Find the bending moment at
the rth round from the foot when the man is on the^th round from the
foot. (Neglect weight of ladder.)
AnSm M/ * sin "'
8. A regular prism of weight W and length a is laid upon a beam of
length 2/(>tf). If the prism is so stiff as to bear at its ends only, show
that the bending action on the beam is less than if the bearing were con-
tinuous from end to end of the prism.
Ans. ist. Max. B.M. = W(- -
2d.
= w( l -- a \
\2 &)
9. A railway girder, 50 ft. in the clear and 6 ft. deep, carries a uni-
formly distributed load of 50 tons. Find the maximum shearing stress
at 20 ft. from one end, when a train weighing \\ tons per lineal foot crosses
the girder.
Also find the minimum theoretic thickness of the web at a support
4 tons being the safe shearing inch-stress of the metal.
Ans. i6Jtons; .195 in.
10. A beam is supported at one end and at a second point dividing its
length into the segments / and n. Find the two reactions. Also find
the ratio of * to n which will make the maximum positive moment equal
to the maximum negative moment.
Ans, (w 2 - 2 ), (ni + ri)* ; m : n : : i + 4/3 : \/T>.
27/2 2in
11. One of the supports of a horizontal uniformly loaded beam is at
the end. Find the position of the other support so that the straining of
the beam may be a minimum.
length
Ans. Distance from end support = =b-.
4/2
12 A rolled joist 17 ft. long is supported at one end and at a point
13 ft. distant from that end. Two wagon-wheels 5 ft. apart and each
carrying a load of 1300 Ibs. pass over the joist. Find the maximum
positive and negative moments due to these weights, and also the corre-
sponding reactions.
Ans. Max. positive B. M. = 5512^ Ib.-ft. ;
reactions = 1550 and 1050 Ibs.
Max. negative B. M. = 5200 Ib.-ft. ;
reactions = 1700 Ibs. and 400 Ibs.
or = 2900 Ibs. and 300 Ibs.
EXAMPLES. 133
Denoting the distance from a support by x, the max. positive B. M.
diagram for each half of the 13-ft. span is given by M x = 100(21 2.x)x.
13. A uniformly loaded beam rests upon two supports. Place the
supports so that the straining of the beam may be a minimum.
Ans. Distance of each support from centre = /( i - 1.
14. Two bars AC, CB in the same horizontal line are jointed at C and
supported upon two props, the one at A, the other at some point in CB
distant x from C. The joint C will safely bear n Ib.-ft. ; the bars are
each /ft. in length and w Ibs. in weight. Find the limits within which
x must lie.
ivl 2
Ans. I .
yvl T 2
15. A uniform load PQ moves along a horizontal beam resting upon
supports at its ends A and B. Prove that the bending moment at a
given point O is a maximum when PQ occupies such a position that
OP : OQ : : OA : OB.
Draw curves of maximum shearing force and bending moment for all
points of the beam.
16. A beam is supported at the ends and loaded with two weights
m W and n W at points distant a, b, respectively, from the consecutive
supports. Show that the bending action is greatest at m W or n W
m > b
according as - ^ .
n "^ a
17. A wheel supporting 10 tons rolls over a beam of 20 ft. span. Place
the wheel in such a position as to give the maximum bending moment,
and find its value.
Ans. At the centre ; 50 ton-ft.
1 8. Two wheels a ft. apart support, the one m W tons, the other
n futons, m being > n, and roll over a beam of / ft. span. Show that
the bending moment is an absolute maximum at the centre or at a point
whose distance from the nearest support is according as
2 2(m + n)
^ a\ i + |/ ), and find its value in each
> \ ' m + nj
case.
mWl
m Wl . m + n rir ( na ) 2
Ans. ton-ft. ; -- W \ I I ton-ft.
4 4/ ( m + n j
19. Find the max. B. M. on a horizontal beam of length / supported
at the two ends and carrying a load which varies in intensity from w at
one end to w + px at the other.
134 THEORY OF STRUCTURES.
20. Four wheels each carrying 5 tons travel over a girder of 24 ft.
clear span at equal distances 4 ft. apart. Determine, graphically, the
max. B. M. at 8 ft. from a support, and also the absolute max. B. M. on
the girder.
Ans. 5p ton-ft. ; 80 ton-ft.
21. Two wheels each supporting 7 tons roll over a beam of 7$ ft.
span. Find the maximum bending moment for the whole span, and also
the curve of the maximum bending moment at each point when the
wheels are 4 ft. apart.
Ans. Abs. max. B. M. = -^V- ton-ft. at wheel at 2f ft. from one
end. Denoting the distance from support by x, the max.
B. M. curve for the first 3^ ft. is given by
M x = 1fi(ll -2X)X,
and for the remaining 4 ft. by
22. Two wheels supporting, the one 1 1 tons, the other 7 tons, travel
over a beam of 12^ ft. span. Find the maximum bending moment for the
whole span, and also the curves of the max. shearing force (both positive
and negative) and maximum bending moment at each point when the
wheels are 6 ft. apart.
Ans. Abs. max. B. M. = 37.2 ton-ft.
The max. positive shearing force at each point is given by
the equations
183-18^- 7(121-5)
s *= and s *= -
The max. negative shearing force at each point is given by
the equations
ix 45i- 1 8* ~ 42 + i%x iix
"' "' ~'
The max. B. M. curve is given by the equations
183-18*
M x = ---* and M x =
N.B. In the above cases x is measured from the support to the
nearest load.
23. In the preceding question show that the maximum negative shear
at 4^ ft. from a support, when the 7-ton wheel only is on the beam, is the
same as the maximum negative shear at the same point when both of
EXAMPLES. 135
the wheels are on the beam, and find its value. Also show that the
maximum negative shear at 9f ft. from a support is the same when only
the 1 1- ton wheel is on the beam as when the two wheels are on the
beam, and find its value.
Ans. ff- tons ; ffl- tons.
24. Solve question 22 when the beam carries an additional load of
1250 Ibs. (= % ton) per lineal foot.
Ans. Abs. max. B. M. is at 5.284 ft. f = ft.) from support.
Max. positive shearing- force diagram is given by S x =
18.54625 2.065.*- from x=o\,ox=6%it., and 6^=14.90625
1.505-*- from x = 6$ to x 12^ ft. The max. negative
shearing-force diagram is given by S x = .56^: from x = o
to x = 4/^ ft-; = 3.64 1.44^1: from x = 4^- to x = 6% ft. ;
= 7.54625 2.065.*- from x 6i to x = 6% ft. ; = 3.90625
1.505.* from x = 6%\.ox = 9f ft.; = 9.18625 2.065.*: from
x = 9f to x = I2 ft. Max. B. M. curve is given by M x
(18.54625 1.7525.*-)^, and M x = (14.90625 1.1925.*-).*-.
25. Three wheels, each loaded with a weight Wand spaced 5 ft. apart,
roll over a beam of 12 ft. span. Place the wheels in such a position as to
give the maximum bending moment, and find its value.
Ans. Middle weight at centre of beam ; 4 W.
26. Place (a) the wheels in the preceding question so that B.M. at
any point between the two hindmost wheels may be constant, and find
its value. Also (8) determine all the positions of the wheels which will
give the same bending moment at 6 and 12 ft. from one end, and find its
value.
das. (a) ist wheel at i ft. from support; B. M. = 7 W.
(b) When distance between end wheel and sup-
port is ^ 2 ft. and ^ 5 ft.; B. M. = 7 W.
27. Four wheels each loaded with a weight Wand spaced 5 ft. apart
roll over a beam of i8ft. span. Place the wheels in such a position as to
give the maximum bending moment, and find its value.
Ans. One wheel off the beam and middle wheel of remaining
three at the centre ; max. B. M. = 8 W. If all wheels
are on beam, max. B. M. = 8 W.
28. All the wheels in the preceding question being on the beam, the
B. M. at the centre for a certain range of travel is constant and equal to
that for a particular distribution of the wheels when only three are on
the beam. Find the range, the B. M., and the position of the three
wheels.
136 THEORY OF STRUCTURES.
Ans. While the end wheel travels 3 ft. from the support ; 8 W\
first wheel 5 ft. from the support.
29. A span of / ft. is crossed by two cantilevers fixed at the ends and
hinged at the centre. Draw diagrams of shearing force and bending
moment (i) for a single weight W at the hinge, (2) for a uniformly dis-
tributed load of intensity w.
Ans. Taking hinge for origin, the shearing-force and bending-
moment diagrams are given by
12)
30. A beam for a span of 100 ft. is fixed at the ends. Hinges are in-
troduced at points 30 ft. from each end. Draw curves of shearing force
and bending moment (i) when a weight of 5 tons is concentrated on
each hinge ; (2) when a uniformly distributed load of -J- ton per lineal foot
covers (a) the centre length, (fr) the two side lengths, (c} the whole span.
Ans. Take a hinge as origin ; the diagrams are given by
(i) For each side span S x = 5, M x = $x\
for centre span S x o, M x = o.
(2) (a) For each side span S x = , M x = x ;
'' . 5 x 5 x*
for centre span S x = ~ , M x = -x > .
X 1C X*^
(b) For each side span S x ^-, M x = x + -, ;
o A. I O
for centre span S x = o, M x o.
C Jf 2 C 3r^
(c} For each side span S x = + 5- , M x x + ;
2 o 4 IO
5 x 5 jr 2
for centre span S x = ^- , M x = x 7 ,
28 2 16
31. If the load on each of the wheels in question 27 is 5 tons, and if
the beam also carries a uniformly distributed load of 20 tons, and two
loads of 2 and 3 tons concentrated at points distant 5 and 9 ft., respec-
tively, from one end, find the maximum shearing force (both positive
and negative) and the maximum bending moment for the whole span ;
also find the loci for the maximum shearing force and bending moment
at each point.
EXAMPLES. 137
Ans. Denoting the distance from support by x, the max.
positive shearing-force diagram is given by S x =
$/- g-x from x = o to x = 3 ; S x = -*-/- %x
from x = 3 to x = 8 ; S* = &- %x from x = 8 to
;r = 13 ; S* = 5 ~ from x = 13 to x = 18 ft. The
max. negative shear ing- force diagram is given by
Sx T Vr from x = o to x = 5 ; S* f f f ^r from
;r = 5 to x 10 ; S r = V I-*" from x = 10 to .r = 1 5 ;
50 2O;r
5^ = -T- TT from .r = 15 to-r= 18. Max. positive
shear = - 3 /- tons ; max. negative shear = $/- tons ; max.
bending -moment curve is given by M x = ^/-x ^-x*
from x = o to x 3 ; M x = ^%x f f x" 1 from x = 3 to
x = 5 ; M x = %-^x Zf-x* 1 5 from x 5 to x = 8 ;
^T/r = s f y-.r 4| jf 2 + 12 from x = 8 to x 9 ; abs. max.
B. M. = 142 ton. -ft.
32. A rolled joist weighing 150 Ibs. per lineal foot and 20 ft. long
carries a uniformly distributed load of 6000 Ibs., and two wheels 5 ft.
apart, the one bearing 5000 Ibs. and the other 3000 Ibs., roll over the
joist. Find the maximum shears at the supports, at the centre, and at 5
ft. from each end.
Ans. 10,250 Ibs. ; 9750 Ibs. ; 3250 Ibs. ; 6750 Ibs.; 6250 Ibs.
33. A beam /ft. long and weighing w Ibs. per lineal foot has a load
of m W Ibs. at a ft. from the left end and a load of n W Ibs. at b ft.
from the right end. Find the shearing forces and bending moments at
the weights and at the middle of the beam, a and b being each < .
How will the result be affected if b > ?
34. A rolled joist weighing 450 Ibs. per lineal foot and 20 ft. long
carries the four wheels of a locomotive at 3, 8, 13, and 18 ft. from one
end. Find the maximum bending moment and the maximum shears,
both positive and negative, the load on each wheel being 10,000 Ibs.
Ans. Max. B. M. = 102,000 Ib.-ft. ; max. shears 19,000 Ibs.
and 21,000 Ibs.
35. Solve the preceding question when a live load of 2| tons per
lineal foot is substituted for the four concentrated weights on the
wheels.
36. The loads on the wheels of a locomotive and tender passing over
a beam of 60 ft. span are 14,180, 14,180, 21,260, 21,260, 21,260, 21,260,
16,900, 16,900, 16,900, 16,900 Ibs., counting in order from the front, the
IS THEORY OF STRUCTURES.
intervals being 5, 5!, 5, 5, 5, Sf, 5, 4, 5 ft. Place the wheels in such a
position as to give the maximum bending moment, and find its value.
Also find the maximum bending moments for spans of 30, 20, and 16
feet.
Ans. For 6o-ft. span, max. B. M. is at 5th wheel and = 1,559,925.4
Ib.-ft. when ist wheel is 7.95 ft. from
support.
For 3o-ft. span, max. B. M. at 5th wheel when 2d wheel is
.596 ft. from support and = 436,761.4
Ib.-ft.
For 2o-ft. span, max. B. M. at centre when 3d wheel is 2
ft. from support and = 212,600 Ib.-ft. =
max. B.M. at same point when 4th wheel
is 5 ft. from support.
For i6-ft. span, max. B. M. is at 5th wheel and = 132,875
Ib.-ft. when 4th wheel is 5 ft. from sup-
port.
37. If the 6o-ft. beam in the preceding question also carries a uni-
formly distributed load of 60,000 Ibs., find the curves of maximum
shearing force and bending moment at each point.
38. If a beam is supported at the ends and arbitrarily loaded, show
that the ordinate at the point of maximum moment divides the area of
the curve of loads into two parts which are equal to the supporting
forces. If a and b are the distances of the centres of gravity of the parts
from the ends of the beam, and if W is the total weight on the beam,
show that the maximum bending moment is W -s- I -f -r-
39. A span of / ft. is crossed by a beam in two half-lengths, sup-
ported at the centre by a pier whose width may be neglected. The suc-
cessive weightson the wheels of a locomotive and tender passing overthe
beam are 14,000, 22,000, 22,000, 22,000, 22,000, 14,000, 14,000, 14,000, 14,000
Ibs., the intervals being 7^, 4$, 4^, 4i, loj, 5, 5, 5 ft. Place the wheels in
such a position as to throw the greatest possible weight upon the centre
pier, and find the magnitude of this weight for spans of (i) 50 ft.; (2) 25
ft.; (3) 20 ft.; (4) 18 ft.
40. Loads of 3!, 6, 6, 6, and 6 tons follow each other in order over a
ten-panel truss at distances of 8, 5f, 4^, and 4| ft. apart. Apply the
results of Art. 8 to determine the position of the loads which will give
the maximum diagonal and flange stresses in the third and fourth panels.
41. A truss of 240 ft. span and ten panels, has loads of 12^, 10, 12, u,
9, 9, 9, 9, and 9 tons concentrated at the panel points. Find, by scale
measurement, the bending moments at the four panel points which are
the most heavily loaded, and determine by Art. 8 whether these are the
EXAMPLES. 139
greatest bending moments to which the truss is subjected as the weights
travel over the truss at the panel distances apart.
42. Loads of 7i, 12,12,12,12. tons are concentrated upon a horizontal
beam of 25 ft. span at distances of 18, 108, 164, 216, and 272 in.,
respectively, from the left support. Find, graphically, the bending
moment at the centre of the span. If the loads travel over the truss at
the given distances apart, find the maximum B. M. at the same section.
43. A beam ABCD is supported at four points A, B, C, and D, and
the intermediate span BC is hinged at the two points E and F. The
load upon the beam consists of 15 tons uniformly distributed over AB,
10 tons uniformly distributed over BE, 5 tons uniformly distributed over
FC, 30 tons uniformly distributed over CD, and a single weight of 5 tons
at the middle point of EF. AB = 15 ft.; BE =5 ft.; EF = 15 ft.,
FC = 10 ft. ; CD = 25 ft. Draw curves of B. M. and S. F., and find
points of inflexion.
44. Four wheels loaded with 4, 4, 8, and 8 tons are placed upon a
girder of 24 ft. span at distances of 3 in., 6 ft., 8f ft., and 9 ft. from the
left support. Find by scale measurement the bending moment at the
centre of the girder. If the wheels travel over the girder at the given
distances apart, find the maximum B. M. to which the girder is sub-
jected.
45. Three wheels loaded with 8. 9, and 10 tons and spaced 5 ft. apart,
are placed upon a beam of 15 ft. span, the 8-ton wheel being 3 ft. from
the left abutment. Determine graphically the B. M. at 6 ft. from the
left abutment. Also find the greatest B. M. at the same point when the
weights travel over the beam, and the abs. max. bending moment to
which the beam is subjected.
Am. 47f ton-ft.; 53^ ton-ft.; abs. max. B. M. = 56||^ ton-ft. at
2d wheel when ist is 2--- ft. from support.
CHAPTER III.
DEFINITIONS AND GENERAL PRINCIPLES.
I. Definitions. The science relating to the strength of
materials is partly theoretical, partly practical. Its primary
object is to investigate the forces developed within a body, and
to determine the most economical dimensions and form, con-
sistent with stability, of that body. Certain hypotheses have
to be made, but they are of such a nature as always to be in
accord with the results of direct observation.
The materials' in ordinary use for structural purposes may
be termed, generally, solid bodies, i.e., bodies which offer an ap-
preciable resistance to a change of form.
A body acted upon by external forces is said to be strained
or deformed, and the straining or deformation induces stress
amongst the particles of the body.
The state of strain is simple when the stress acts in one
direction only, and the strain itself is measured by the ratio of
the deformation to the original length.
The state of strain is compound when two (or more) stresses
act simultaneously in different directions.
A strained body tends to assume its natural state when the
straining forces are removed : this tendency is called its elas-
ticity. A thorough knowledge of the laws of elasticity, i.e., of
the laws which connect the external forces with the internal
stresses, is absolutely necessary for the proper comprehension
of the strength of materials. This property of elasticity is not
possessed to the same degree by all bodies. It may be almost
absolute, or almost zero, but in the majority of cases it has a
mean value. Hence it naturally follows that solid bodies may
be classified between two extreme, though ideal, states, viz.,
140
ELEMENTARY PRINCIPLES OF ELASTICITY. 14!
a perfectly elastic state and a perfectly soft state. Perfectly
elastic bodies which have been strained resume their original
forms exactly when the straining forces are removed. Per-
fectly soft bodies are wholly devoid of elasticity and offer no
resistance to a change of form.
Bodies capable of undergoing an indefinitely large deforma-
tion under stress are said to be plastic.
2. Stresses and Strains. Every body may be sub-
jected to five distinct kinds of stresses, viz. :
(a) A longitudinal pull, or tension.
(b) A longitudinal thrust, or compression.
(c) A shear, or tangential stress, which may be defined as a
stress tending to make one surface slide over another with
which it is in contact.
(d) A transverse stress.
(e) A twist or torsion.
Under any one of these stresses a body may suffer either an
elastic deformation, of a temporary character, or a plastic de-
formation, of a permanent character.
3. Resistance of Bars to Tension and Compression.
Let a straight bar of homogeneous material and length L be
stretched or compressed longitudinally by a force P uniformly
distributed over the constant cross-section A of the bar ; let
the line of action of P coincide with the axis of the bar, and
let / be the consequent extension or compression, i.e., the de-
formation.
If the transverse dimensions of the bar are small as com-
pared with the length, experiment shows that, within certain
limits, the force Pis directly proportional to the deformation /
and to the area A, and inversely proportional to the length L,
these quantities being connected by the relation
,
where E is a constant dependent upon the material of the bar
and is called the coefficient or modulus of elasticity. It is evi-
dently the force which will double the length of a perfectly
i p\
elastic bar of unit section. Denoting the unit stress LrJ by/,
\Ar
142 THEORY OF STRUCTURES.
and the strain per unit of length ( J by A,, the above equation
may be written
or the unit stress = E times the unit strain.
Thus the equation is the analytical expression of Hooke's
law, that for a body in a state of simple strain the strain is pro-
portional to the stress.
The longitudinal strain is accompanied by an alteration in
the transverse dimensions, the lateral unit strain being ,
m
where m is a coefficient which usually varies from 3 to 4 for
solid bodies and is approximately 4 for the metals of construc-
tion. In the case of india-rubber, if the deformation is small,
m is about 2.
Generally the deformation may be calculated per unit of
original length without sensible error, but for india-rubber it is
more accurate to make the calculation per unit of stretched
length {=
I lateral strain
The ratio = : : -.- : is called Poisson's ratio.
m longitudinal strain
If the transverse dimensions of a bar under compression
are small as compared with the length L, a slight disturbing
force will cause the bar to bend sideways, and the bar will be
subjected to a bending action in addition to the compression.
If the bar is to be capable of resisting a direct thrust only,
the ratio of L to its least transverse dimension should not
exceed a certain limit depending upon the nature of the ma-
terial. For example, experiment indicates that this limit
should be about 5 for cast-iron, 10 for wrought-iron, 7 for
steel, and 20 for dry timber.
If the temperature of the bar is raised t, the consequent
strain is at, a being the coefficient of linear dilatation ; and a
stress Eaf will be developed if a change of length is pre-
vented.
SPECIFIC WEIGHT. COEFFICIENT OF ELASTICITY. 143
4. Specific Weight ; Coefficient of Elasticity ; Limit
of Elasticity ; Breaking Stress. Before the strength of a
body can be fully known, certain physical constants, whose
values depend upon the material, must be determined.
(a) Specific Weight. The specific weight is the weight of a
unit of volume. The specific weights of most of the materials
of construction have been carefully found and tabulated. If
the specific weight of any new material is required, a conven-
ient approximate method is to prepare from it a number of
regular solids of determinate volume and weigh them in an
ordinary pair of scales. The ratio of the total weight of these
solids to their total volume is the specific weight. It must be
remembered that the weight may vary considerably with time,
etc. ; thus a sample of greenheart weighed 69.75 Ibs. per cubic
foot when first cut out of the log, and only 57 Ibs. per cubic
foot at the end of six months. When the strength of a timber
is being determined, it is important to note the amount of
water present in the test-piece, since this appears to have a
great influence upon the results.
The straining of a structure is generally largely due to its
own weight.
The total load upon a structure includes all the external
forces applied to it, and in practice is designated dead {perma-
nent} or live (rolling), according as the forces are gradually ap-
plied and steady, or suddenly applied and accompanied with
vibrations. For example, the weight of a bridge is a dead
load, while a train passing over it is a live load ; the weight
of a roof, together with the weight of any snow which may
have accumulated upon it, is a dead load ; wind causes at times
excessive vibrations in the members of a structure, and al-
though often treated as a dead load, should in reality be con-
sidered a live load.
The dead loads of many structures '(as masonry walls, etc.)
are so great that extra or accidental loads may be safely disre-
garded. In cold climates, great masses of snow and the pene-
trating effect of the frost necessitate very deep foundations,
which proportionately increase the dead weight.
(b) Coefficient of Elasticity. Generally speaking, a knowl-
edge of the external forces acting upon a structure, discloses
144
THEORY OF STRUCTURES.
the manner of their distribution amongst its various members,
but the deformation of these members can only be estimated
by means of the coefficient of elasticity, which expresses the
relation between a stress and the corresponding strain.
In practice it is usually sufficient to assume that a material
is elastic, homogeneous, and isotropic, and its deformation
under stress may be found, if the coefficients of elasticity, of
form, and of volume are known.
In a homogeneous solid there may be twenty-one distinct
coefficients of elasticity, which are usually classified under the
following heads :
(1) Direct, expressing the relation between longitudinal
strains and normal stresses in the same direction.
(2) Transverse, expressing the relation between tangential
stresses and strains in the same direction.
(3) Lateral, expressing the relation between longitudinal
strains and normal stresses at right angles to the strains; i.e., a
lateral resistance to deformation.
(4) Oblique, expressing other relations of stress and strain.
If a body is isotropic, i.e., equally elastic in all directions,
the twenty-one coefficients reduce to two, viz., the coefficients
of direct elasticity and of lateral elasticity. Such bodies, how-
ever, are almost wholly ideal. In a perfectly elastic body E
would be the same both for tension and compression. In the
ordinary materials of construction
it is slightly less for compression
than for tension ; but if the stresses
do not exceed a certain limit ( (e\
page 145), the difference is so slight
that it may be disregarded.
The equation f E\ may be
represented graphically by the
straight line MON, the ordinate at
any point representing the unit
stress required to produce the unit
strain respresented by the corresponding abscissa.
The angle MOY = tan ~*E = tan -'^V
Coefficients of elasticity must be determined by experiment.
FlG - l8 3-
LIMIT OF ELASTICITY. H5
The coefficients of direct elasticity for the different metals
and timbers are sometimes obtained by subjecting bars of the
material to forces of extension or compression, or by observing
the deflections of beams loaded transversely. The coefficients
for blocks of stone and masonry might also be found by trans-
verse loading ; they are of little, if any, practical use, as, on
account of the inherent stiffness of masonry structures, their
deformations, or settlings, are due rather to defective work-
manship than to the natural play of elastic forces.
The torsional coefficient of elasticity, i.e., the coefficient of
elastic resistance to torsion, has been shown by experiment to
vary from two fifths to three eighths of the coefficient of direct
elasticity.
(e) Limit of Elasticity. When the forces which strain a
body fall below a certain limit, the body, on the removal of the
forces, will resume its original form and dimensions without
sensible change (disregarding any effects due to the develop-
ment of heat) and may be treated as perfectly elastic. But if
the forces exceed this limit, the body will receive a permanent
deformation, or, as it is termed, a set.
Such a limit is called a limit of elasticity, and is the greatest
stress that can be applied to a body without producing in it an
appreciable and permanent deformation.
This is an unsatisfactory definition, as a body passes from
the elastic to the non-elastic state by such imperceptible
degrees that it is impossible to fix any exact line of demarca-
tion 'between the two states. Fairbairn defines the limit mo're
correctly, as the stress below which the deformation is approxi-
mately proportional to the load which produces it, and beyond
which the deformation increases much more rapidly than the
load. In fact, both the elastic and ultimate strengths of a ma-
terial depend upon the nature of the stresses to which they are
subjected and upon \k& frequency of their application. For ex-
ample, in experimenting upon bars of iron having an ultimate
tenacity of 46,794 Ibs. per sq. in. and a ductility of 20 %>
Wohler found that with repeated stresses of equal intensity,
but alternately tensile and compressive, a bar failed aft(br 56,430
repetitions when the intensity was 33,000 Ibs. per sq. in. ; a
146 THEORY OF STRUCTURES.
-second bar failed only after 19,187,000 repetitions when the
intensity was 18,700 Ibs. per sq. in. ; while a third bar remained
intact after more than 132,000,000 repetitions when the. inten-
sity was 16,690 Ibs. per sq. in. These experiments therefore
indicated that the limit of elasticity for the iron in question,
under repeated stresses of equal intensity, but alternately
'tensile and compressive, lay between 16,000 and 17,000 Ibs. per
sq. in., which is much less than the limit under a steadily ap-
plied stress. Similar results have been shown to follow when
the stresses fluctuate from a maximum stress to a minimum
stress of the same kind.
Generally speaking, then, the limit of elasticity of a ma-
terial subjected to repeated stresses, is a certain maximum
stress below which the condition of the body remains unim-
paired,
Bauschinger's experiments indicate that the application to
a body of any stress, however small, produces a plastic or
permanent deformation. This, perhaps, is sometimes due to a
want of uniformity in the material, or to the bar being not
quite straight initially. In any case, the deformations under
loads which are less than the elastic limit, are so slight as to be
of no practical account and may be safely disregarded.
The main object, then, of the theory of the strength of
materials, is to determine whether the stresses developed in any
particular member of a structure exceed the limit of elasticity.
As soon as they do so, that member is permanently deformed,
its strength is impaired, it becomes predisposed to rupture, and
the safety of the whole structure is threatened. Still, it must
be borne in mind that it is not absolutely true that a material
is always weakened by being subjected to forces superior to
this limit. In the manufacture of iron bars, for instance, each
of the processes through which the metal passes changes its
elasticity and increases its strength. Such a material is to be
treated as being in a new state and as possessing new properties.
The strength of a material is governed by its tenacity and
rigidity, and the essential requirement of practice is a tough
material with a high elastic limit.
This is especially necessary for bridges and all structures
BREAKING STRESS. 147
liable to constantly repeated loads, for it is found that these
repetitions lower the elastic limit and diminish the strength.
In the majority of cases, experience has fixed a practical
limit for the stresses, much below the limit of elasticity. This
insures greater safety and provides against unforeseen and
accidental loads, which may exceed the practical limit, but
which do no harm unless they pass beyond the clastic limit.
Certain operations have the effect of raising the limit of
elasticity: a wrought-iron bar steadily strained almost to the
point of its ultimate strength and then released from strain and
allowed to rest, experiences an elevation both of tenacity and
of the elastic limit.
If the bar is stretched until it breaks, the tensile strength
of the broken pieces is greater than that of the bar. A similar
result follows in the various processes employed in the manu-
facture of iron and steel bars and wires : the wire has a greater
ultimate strength than the bar from which it was drawn.
Again, iron and steel bars, subjected to long- continued com-
pression or extension, have their resistance increased, mainly
because time is allowed for the molecules of the metal to as-
sume such positions as will enable them to offer the maximum
resistance ; the increase is not attended by any ap-
preciable change of density. /
Under an increasing stress a brittle material will /
be fractured without any great deformation, while a /
tough material will become plastic and undergo a J/
large deformation. ^P
(d) Breaking Stress. When the load upon a /
material increases indefinitely, the material may /
merely suffer an increasing deformation, but generally /
a limit is reached at which fracture suddenly takes /
place. /
Cast-iron is perhaps the most doubtful of all /
materials, and the greatest care should be observed/
in its employment. It possesses little tenacity or
elasticity, is very hard and brittle, and may fail sud- FlG< l84 '
denly under a shock or an extreme variation of temperature.
Unequal cooling may predispose the metal to rupture, and its
148
THEORY OF STRUCTURES.
strength may be still further diminished by the presence of
air-holes.
Cast-iron and similar materials receive a sensible set even
under a small load, and the set increases with the load. Thus
at no point will the stress-strain curve be absolutely straight,
and the point of fracture will be reached without any great
change in the slope of the curve and without the development
of much plasticity.
Wrought-iron and steel are far more uniform in their be-
havior, and obey with tolerable regularity certain theoretical
laws. They are tenacious, ductile, have great compressive
strength, and are most reliable for structural purposes. Their
strength and elasticity may be considerably reduced by high
temperatures or severe cold.
When a bar of such material is tested, the stress-strain
jpurve (/= Ety, as has already been pointed out, is almost
absolutely straight within the elastic limit, e.g., from O to A in
tension and from O to B in com-
pression. As the load increases
beyond the elastic limit, the in-
creasing deformation becomes
plastic and permanent, and the
stress-strain diagram takes an ap-
preciable curvature between the
limits A and B and the points D
and E corresponding to the maxi-
mum loads. In tension, as soon
as the point D is reached, the bar
rapidly elongates and is no longer
able to sustain the maximum load,
its sectional area rapidly dimin-
ishes, and fracture ultimately takes
place under a load much less than
the maximum load. The point of
fracture is represented in the figure
by the point F the ordinate of F being the actual ultimate
final load on the bar
intensity of stress = section
FIG. 185.
BREAKING STRESS. 149
The exact form of the stress-strain curve between D and F
is unknown, as no definite relation has been found to exist be-
tween the stress and strain during the elongation from D to F.
Ordinarily, the breaking tensile stress has been defined to be
the maximum load applied divided by the initial sectional area
of the bar ; but this, although convenient, is manifestly in-
correct.
It is important to note that, as the deformation gradually
increases under the increasing load, the molecules of the ma-
terial require greater or less time to adjust themselves to the
new condition.
During the tensile test of a ductile material there is, at
some point beyond the elastic limit, an
abrupt break GH in the continuity of the
stress-strain curve, the curve again becom-
ing continuous from H to D.
The point G has been called the Yield
Point or the Breaking-down Point, and the
deformation from H onward is almost
wholly plastic or permanent.
In compression there is no local stretch
as in tension, and there is consequently
no considerable change in the curvature of FIG. iS6.
the compression stress-strain curve up to the point of fracture.
Timber is usually tested by being subjected to the action
of tensile, compressive, or transverse loads. Other character-
istics, however, must be known before a full conception of the
strength of the wood can be obtained. Thus the specific
weight must be found ; the amount of water present, the loss
in drying, and the corresponding shrinkage should be deter-
mined ; the structural differences of the several specimens, the
rate of growth, etc., should be observed.
The chief object of experiments upon masonry and brick-
work is to discover their resistance to compression, i.e., their
crushing strength. In fact, their stiffness is so great that they
may be compressed up to the point of fracture without sensible
change of form, and it is therefore very difficult, if not impos-
sible, to observe the limit of elasticity.
1 50 THEORY OF STRUCTURES.
The cement or mortar uniting the stones and bricks is most
irregular in quality. In every important work it should be an
invariable rule to prepare specimens for testing. The crushing
strength of cement and of mortar is much greater than the ten-
sile strength, the latter being often exceedingly small. Hence it
is advisable to avoid tensile stresses within a mass of masonry,
as they tend to open the joints and separate the stones from
one another. Attempts are frequently made to strengthen
masonry and brickwork walls by inserting in the joints tarred
and sanded strips of hoop-iron. Their utility is doubtful, for,
unless well protected from the atmosphere, they oxidize, to the
detriment of the surrounding material, and, besides this, they
prevent an equable distribution of pressure. They are, how-
ever, far preferable to bond-timbers.
The working load (or stress, or strength) is the maximum
stress which a material can safely bear in ordinary practice,
and depends both upon the character (see Art. 5, .below) of the
stress and upon the ultimate strength of the material, the ratio
of the ultimate or breaking stress to the working stress being
usually called a factor of safety. For example, the factor is
about
3 for long-span iron bridges, or bridges having great weight
as compared with the live load (a moving train).
4 for ordinary iron bridges.
5 for ordinary metal shafting.
8, 10, and even more for long struts and members subjected
to repeated stresses of varying magnitude.
10 is also generally taken to be the factor of safety for
timber.
Under a steady, or a merely statical load, even as great as
J- of the breaking stress, a member of a structure may prob-
ably not be unsafe.
5. Wohler's Law. It is now generally admitted that
variable forces, constantly repeated loads, and continued vibra-
tions diminish the strength of a material, whether they pro-
duce stresses approximating to the elastic limit, or exceedingly
small stresses occurring with great rapidity. Indeed many
engineers design structures in such a manner, that the several
WOHLER'S LAW. !$!
members are strained in one way only, so convinced are they
of the evil effect of alternating tensile and compressive stresses.
Although the fact of a variable ultimate strength had thus been
tacitly acknowledged and often allowed for, Wohler was the
first to give formal expression to it, and, as a result of obser-
vation and experiment, enunciated the following law:
" That if a stress t, due to a static load, cause the fracture
of a bar, the bar may also be fractured by a series of often-re-
peated stresses, each of which is less than t\ and that, as the
differences of stress increase, the cohesion of the material is
affected in such a manner that the minimum stress required to
produce fracture is diminished."
This law is manifestly incomplete. In Wohler's experi-
ments the applications of the load followed each other with
great rapidity, yet a certain length of time was required for the
resulting stresses to attain their full intensity ; the influence due
to the rapidity of application, to the rate of increase of the
stress, -and to the duration of individual strains still remains a
subject for investigation.
The experiments, however, show that the rate of increase of
repetitions of stress required to produce fracture, is much more
rapid than the rate of decrease of the stresses themselves, and
depends both upon the maximum stress and upon the differ-
ence or fluctuation of stress.
The effect of repeated stresses of equal intensity, but alter-
nately tensile and compressive, has been already pointed out
in Art. 4.
Bars of the same material repeatedly bent in one direction,
bore 31,132 Ibs. per square inch when the load was wholly re-
moved between each bending, and 45,734 Ibs. per square inch
when the stress fluctuated between 45,733 Ibs. and 24,941 Ibs.
The table on page 152 gives the results of similar experi-
ments on steel.
The axle-steel was found to bear 22,830 Ibs. per square inch,
when subjected to repeated shears of equal intensity but oppo-
site in kind, and 29,440 Ibs. per square inch, when the shears
were of the same kind. It would therefore appear that the
shearing strengths of the metal in the two cases are about
152
THEORY OF STRUCTURES.
of the strengths of the same metal under alternate bending and
under bending in one direction, respectively.
Character of Fluctuation.
Maximum Resistance to Repeated
Stresses in Ibs. per square inch.
Axle-steel.
Spring-steel (un-
hardened).
Alternating stresses of equal intensity ....
Complete relief from stress between each
29,000, 29,000
49,890, O
83,110, 36,380
52,000, o
93,500, 62,240
Partial relief from stress between each
From torsion experiments with various qualities of steel,
the important result was deduced, that the maximum resistance
of the steel to alternate twisting was |- of the maximum resist-
ance of the same steel to alternate bending.
Wohler proposed 2 as a factor of safety, and considered
that the maximum permissible working stresses should be in
the ratios of 1:2:3, according as members are subjected to
alternate tensions and compressions (alternate bending), to
tensions alternating with entire relief, or to a steady load.
The weakening of metal by repeated stresses has been
called fatigue, and is much more injurious to iron and steel
under tension than under compression. Egleston's investiga-
tions have shown that a fatigued metal may sometimes be
restored by rest or by annealing.
From the law, however, as it stands formulae may be de-
duced which, it is claimed, are more in accordance with the
results of experiment, give smaller errors, and insure greater
safety than the false assumption of a constant ultimate
strength.
The formulae necessarily depend upon certain experimental
results, but in applying them to any particular case, it must be
remembered that only such results should be employed, as
have been obtained for material of the same kind and under
the same conditions as the material under consideration. The
effects due to faulty material, rust, etc., are altogther indeter-
minate, so that no formula can be perfectly universal in its
application. Hence the necessity for factors of safety, with
values depending upon the class of structure, still exists.
LAUNHARDT'S FORMULA. I 53
A brief description of the principal of these formulae will
now be given, and in the discussion
/, the statical breaking strength, is the resistance to fracture
under a static or under a very gradually applied load.
?/, the primitive strength, is the resistance to fracture under
a given number of repeated stresses, the stress in each repeti-
tion remaining unchanged in kind, i.e., being due either to a
tension, a compression, or a shear.
s, the vibration strength, is the resistance to fracture under
alternating stresses of equal intensities, but different in kind,
due to a vibratory motion about the unstrained state of equi-
librium.
b is the admissible stress per unit of sectional area
/MS the effective sectional area and is
numericallv absolute maximum load
6. Launhardt's Formula. A bar of unit sectional area is
subjected to stresses (B) which are either wholly tensile, wholly
compressive, or wholly shearing, and which vary from a maxi-
mum #, ( = max. B) to a minimum # 2 (= min. If).
Let <7, <? Q d the maximum difference .of stress.
T a,. min. B
Let - 2 = -= = 0.
a l max. B
If a. 2 = o, #, = d u.
I f d o, a l a^t.
By Wohkr's law,
*,cc <*=/</, (i)
/being an unknown coefficient of which the value remains to
be determined.
If d o, a l = t and / = oo.
If d u, #, = d and f =. i.
154 THEORY OF STRUCTURES.
Launhardt's assumption, viz.,f= , satisfies these ex-
/ a 1
treme conditions, and also gives intermediate values of a l which
closely agree with the results of the most reliable experiments.
Hence (i) becomes
t H t-u
and
/ t u a 2 ] f * u\ , ^
.-. a .= u I -\ --- }=u(i-\ -- 0. . . (2)
\ u aj \ n m l
This is Launhardt's formula, and is an analytical expression
of Wohler's Law.
Wohler in his bending experiments upon Phcenix axle-iron
found that u = 2195* per cent. 2 * and t = 4020* per cent. 1 ;
/ u _ ^
~~u 6'
The same iron under tension gave u = 2195* per cent. 3 and
/ = 3290* per cent. 2 ;
t U !
U 2
Choosing the most unfavorable case, and, in order to insure
greater safety, taking u = 2100* per cent. 2 , equation (2) becomes
a, = 2100(1+-) ....... (3)
If 3 is the factor of safety,
6 = 700(1 + |). ....... (4)
* k per cent.' 2 is an abbreviation for kilogrammes per square centimetre.
One kilogramme per square centimetre is equivalent to 14.2232 Ibs. per sq. in.
LAUNHARDT'S FORMULA. 155
In his bending experiments upon Krupp cast-steel (untem-
pered) it was found that u = 3510* per cent. 2 and t = 7340* per
cent 2 .;
t u __ 7
~~u 6*
But steel varies considerably in strength, and great care
must be exercised in its use, especially in bridge construction.
For this reason take u 3300* per cent. 2 and t = 6ooo k per cent. 2 ;
/ u _ 9
~H~ = T?
and (2) becomes
, = 33oo(i + 0) (5)
If 3 is the factor of safety,
^= 1100(1+^-0) (6)
EXAMPLE I. The stresses upon a bar of Phoenix axle-iron,
normal to its cross-section, vary from a maximum tension of
50000* to a minimum tension of 20000*. Determine the admis-
sible stress per cent. 2 and the necessary sectional area
By (4),
f I 20000 \
b = 700(1 + 2^^-J = 8 40* Per cent 1 ,
and
50000 50000
.'. F = T = -Q - 59-5 2 sq. centimetres.
u 040
Let/ be the dead load and q the total load, per lineal unit
of length, upon the flanges of roof and bridge trusses.
THEORY OF STRUCTURES.
.-. = , and equations (4) and (6) respectively become
*= 700(1+-) ....... (7)
Ex. 2. Determine the limiting stress per cent. 2 for the
flanges of a wrought-iron lattice girder when the ratio of the
dead load to the greatest total load is =-.
By (7),
b = 700(1 +~) = 800*.
^ s*>*
7. Weyrauch's Formula. Let a bar of a unit sectional
area be subjected to stresses which are alternately different in
kind, and which vary from an absolute numerical maximum a'
(= max. B) of the one kind to a maximum a" (= max. B') of
the other kind.
L e t a' -\- a" = d = the maximum numerical difference of
stress.
a" max. B'
Let
If a" =0, a' = d = u.
T f f / ^ __
By Wohler's Law,
a 1 a </ =//, (9)
^" being an unknown coefficient of which the value remains to
be determined.
If a' = u, f I.
Tf /' nr C /= i
WEYRAUCH'S FORMULA. 157
Weyrauch's assumption, viz., f = - -, , satisfies
226 S ~~~ d
these extreme conditions, the most reliable results of the few
experiments yet recorded, and also Wohler's deduction that a'
diminishes as d increases and vice versa.
Hence (9) becomes
and
u-s
This is Weyrauch's formula, and it may be always applied
to those cases in which a member is subjected to stresses alter-
nating between tension* and compression, or due to shearing
actions in opposite directions.
In the Phcenix iron experiments already referred to it was
found that s = 1170* per cent. 2 ;
u s 7
u ~ 15"
Taking u = 2100* as before, and making - = , (10)
becomes
(II)
If 3 is the factor of safety,
* = 700(1 --) ....... (12)
Weyrauch considers 3 to be the proper factor of safety for
bridges and similar structures. It is also a suitable factor for
the parts of machines subjected to determinate straining
actions. A larger factor will be required when other con-
tingencies have to be provided against.
158 THEORY OF STRUCTURES.
In the steel experiments, Wohler found that s = 2050* per
cent. 3 ;
u-s = _5_
U 12
Taking u = 3300* and s 1800*,
- J _5
'11'
and (10) becomes
^ = 3300(1-^00 ....... (13)
If 3 is the factor of safety,
b= 1100(1 -- T 5 T 0') .......
If a very soft steel is employed in the construction of a
bridge, it may be advisable to diminish still further the ad-
missible stress per unit of sectional area. For example, it may
be assumed that t 5200*, u = 3000*, and s = 1500*, so that
(2) and (10) respectively become
a, = 3000(1 +|0) ...... (15)
and
a' = 3000(1 - 00. ..... (16)
EXAMPLE. The stresses in a wrought-iron bar normal
to its cross-section, vary between a tension of 40000* and a
compression of 30000*. Find the sectional area (disregarding
buckling).
By (12)
b = 700(1 - J. X t*W = 437-5" per cent. 9 .
40000
/. F= - - = 91.42 sq. centimetres.
437-5
Shearing Stresses. For shearing stresses in opposite direc-
tions Wohler found, in the case of Krupp cast-steel (untem-
WEYRAUCH'S FORMULA. 1 59
pered), that u = 2780* per cent. 2 and s = 1610* per cent. 2 , or
about - of the corresponding values for stresses which are
alternately tensile and compressive, and it may be generally
assumed, that the value of b for shearing stresses, is of its
value for stresses which are alternately tensile and compressive,
and which have the same ratio 0'.
8. Unwin has proposed to include all cases of fluctuating
stress in the formula
a being the actual strength, d the fluctuation of stress, / the
statical breaking strength, and n a coefficient whose value
remains to be determined.
When d = o, the load is steady and a' = t.
When d = a', the load alternates with entire relief and
a' 2t(Vi -\-tf- n).
When d = 20! ', the stresses are alternately tensile and com-
pressive and of equal intensity. The stress fluctuates from
a! to d ', and a' = .
2n
In these extreme cases, if n is made equal to 1.42 for
wrought-iron and to 1.66 for steel, results are obtained almost
identical with those given in Arts. 6 and 7. The formula may
therefore be assumed to be approximately correct for inter-
mediate cases.
The mean value of n for iron and steel seems to be f, so
that the formula may be written
EXAMPLE. One of the diagonals of a bowstring truss has
a sectional area of 3 square inches, and is subjected to stresses
l6o THEORY OF STRUCTURES.
which fluctuate between a tension of 14 tons and a compression
of 6 tons. Find the statical strength of the iron.
14 (6) 20
d = fluctuation of stress = - * -- - =
t = 10.17 tons per sq. in.
9. Remarks upon the Values of f, u, s, and 6. As yet
the value of u in compression has not been satisfactorily deter-
mined, and for the present its value may be assumed to be the
same both in tension and compression.
If, as Wohler states, "repeated stresses" are detrimental to
the strength of a material, then the values of u and s diminish
as the repetitions increase in number, and are minima in struc-
tures designed for a practically unlimited life.
Only a very few of Wohler's experiments give the values of
t, u, s, and a, so that Launhardt's and Weyrauch's assumptions
for the value of /must be regarded as tentative only, and re-
quire to be verified by further experiments. The close agree-
ment of Wohler's results from tests upon untempered cast-steel
(Krupp), with those given by Launhardt's formula, may be seen
from the following:
For /= iioo centners'* per sq. zoll, Wohler found that
u = 500 centners per sq. zoll. Thus (2) becomes
, =
and
.'. <2, a SOOa, 6ootf 2 o.
Hence for a 9 = o, 250, 400, 600, iioo,
Launhardt's formula gives
^ 500, 710, 800, 900, iioo;
* A centner = 110.23 pounds. A square zoll = 1.0603 square inches.
REMARKS UPON THE VALUES OF t, U, S, AND b. 1 6 1
while Wohler's experiments gave
a, = 500, 700, 800, 900, 1 100.
Again, with Phoenix iron, for t = 500 centners per sq. zoll,
u was found to be 300 centners per sq. zoll, and
or
a? 300^ 250^ o.
If a^ = 240, tfj 436.8, which almost exactly agrees with
the result given by the tension experiments,
In general, the admissible stress per square unit of sectional
area may be expressed in the form
b = v(i m<j>), ...... (17}
v and m being certain coefficients which depend upon the
nature of the material and also upon the manner of the loading.
Consider three cases, the material in each case being wrought-
iron :
(a) Let the stresses vary between a maximum tension and
an equal maximum compression ; then
0=1,
and
.. b = 700(1 i) 350* per cent. 2 .
(b) Let the material be subjected to stresses which are
either tensile or compressive, and let it always return to the
original unstrained condition ; then
min. B o, or max. B' = o, and .*. = o.
.-. b = 700(1 o) = 700* per cent. 2 .
(c) Let the material be continually subjected to the same
dead load ; then
min. B max. B
1 62 THEORY OF STRUCTURES.
and
..-. b =. 700(1 -|- J) = 1050* per cent.' = 14,934 Ibs. per sq. in.,
-which is one third of the ultimate breaking strength, viz.,
1050' per cent. 2 .
Thus in these three cases the admissible stresses are in the
-.ratios of 1:2:3, ratios which have been already adopted in ma-
chine construction as the result of experience.
Wohler, from his experiments upon untempered cast-steel
(Krupp), concluded that for alternations between an unloaded
condition and either a tension or a compression, b = iioo, and
for alternations between equal compressive and tensile stresses,
b = 580.
In America it has often been the practice to take
max. B + max. B' a' + a"
_~* " "
700 700
for stresses alternately tensile and compressive, it being as-
sumed that if the stresses are tensile only, their admissible
values may vary from o* to 700* per cent. 8 .
a ?ooF a' 700
Smce0 =r,.'.* ==>' and '* = = >- ( l8 )
Comparing this with (12),
for 0' = o, i, J, j, I,
(18) gives^ = 700, 560, 467, 400, 350,
and (12) gives b 700, 612, 525, 437, 350.
10. Flow of Solids. When a ductile body is strained
beyond the elastic limit, it approaches a purely plastic con-
dition in which a sufficiently great force will deform the body
indefinitely. Under such a force, the elasticity disappears and
the material is said to be in a fluid state, behaving precisely
like a fluid. For example, it flows through orifices and shows
a contracted section. The stress developed in the material is
called the fluid pressure or coefficient of fluidity.
The general principle of the flow of solids, deduced by
Tresca, may be enunciated as follows:
FLO W OF SOLIDS. 163
A pressure upon a solid body creates a tendency to the relative
motion of the particles, in the direction of least resistance.
This gives an explanation of the various effects produced
in materials by the operations of wire-drawing, punching, shear-
ing, rolling, etc., and in the manufacture of lead pipes. Prob-
ably it also explains the anomalous behavior of solids under
certain extreme conditions.
Rails which have been in use for some time are found to
have acquired an elongated lip at the edge. This is doubtless
due to the flow of the metal under the great pressures to which
the rails are continually subjected. Other examples of the flow
of solids are to be observed in the contraction of stretched bars
and in the swelling of blocks under compression. The period
of fluidity is greater for the more ductile materials, and may-
disappear altogether for certain vitreous and brittle substances.
In punching a piece of wrought-iron or steel, the metal is
at first compressed and flows inwards, while the shearing only
commences when the opposite surface begins to open. A case
brought under the notice of the author may be mentioned in
illustration of this. The thickness of a cold-punched nut was
1.75 inches, the nut-hole was .3125 inch in diameter, and the
length of the piece punched out was only .75 inch. Thus the
flow must have taken place through a depth of I inch, and the
shearing through a depth of .75 inch. Hence the surface
really shorn was TT x .3125 X 75 = -73^ sq. in. in area, and a
measure of the shearing action is the product of this surface
area and the fluid pressure. The nature of the flow may be
observed by splitting a cold punched nut in half and treating
the fractured surfaces with acid, after having planed them and
given them a bright polish. The metal bordering the core will
be found curved downwards, the curvature increasing from the
bottom to the top, and well-defined curves will mark the sepa-
rating planes of the plates which were originally used in piling
and rolling the iron.
Jn experimenting upon lead, Tresca placed a number of
plates, one above the other, in a strong cylinder, Fig. 188, page
165, with a hole in the bottom. Upon applying pressure the
lead was always found to flow when the coefficient of fluidity
164 THEORY OF STRUCTURES.
was about 2844 Ibs. per sq. in., the difference of stress being
double this amount. The separating planes assumed curved
forms analogous to the corresponding surfaces of flow when
water is substituted in the cylinder for the lead.
The flow of ductile metals, e.g., copper, lead, wrought-iron,
and soft steel, commences as soon as the elastic limit is ex-
ceeded, and in order that the flow may be continuous the dis-
torting stress must constantly increase. On the other hand,
in the case of truly plastic bodies, flow commences and con-
tinues under the same constant stress. It evidently depends
upon the hardness of the material, and has been called the co-
efficient of hardness. The longer the stress acts the greater is
the deformation, which gradually increases indefinitely or at a
diminishing rate.
Experiment shows that there is very little alteration in the
density of a ductile body during its plastic deformation, and
Tresca's analytical investigations are based on the assumption
that the body is deformed without sensible change of volume.
Consider a prismatic bar undergoing plastic deformation.
Let L be the length and A the section of the bar at com-
mencement of deformation.
Let L -f- x be the length and a the section of the bar at a
subsequent period.
Let/ be the intensity of the fluid pressure.
Since the volume remains unchanged,
LA = (L x}a, ...... . (i)
the positive or negative sign being taken according as the ba*-
is in tension or compression.
Let P l be initial force on bar.
Let P be force on bar when its length is L x. Then
P a L
= =
Hence P(L x) = P,L = a constant, .... (3)
FLOW OF SOLIDS.
I6 5
FIG. 187.
and the force diminishes as the bar stretches and increases ^s
the bar contracts under pressure.
If equation (3) be referred to rect-
angular axes, the ordinates repre-
senting different values of P and
the abscissae the corresponding
values of x, the stress-strain dia-
grams, tt in tension and cc in com-
pression, are hyperbolic curves,
having as asymptotes the axis of
x, XOX, and a line parallel to the
axis of y at a distance from it
equal to the length L of the bar.
Next consider a metallic mass
(e.g., lead) resting upon the end
CD of a cylinder of radius R, and filling up a space of depth
D. A hole of radius r is made at the centre
of the face CD, through which the mass
flows under the pressure of fluidity exerted
by a piston. When the mass has been com-
pressed to the thickness DO x, let y be
the corresponding length KE of the "jet."
First, assume that the specific weight of
the mass remains constant.
If dx be the diminution in the thickness
DO corresponding to an increase dy in the
length of the jet, then
TtR'dx -f- rtfdy O. . . (i)
Integrating eq. I, and remembering that
y o when x D,
R\D -x]- ry = o
(2)
Second, assume that the cylindrical portion EFGH is gradu-
ally transformed into NMPLKQN, of which the part PMNQ
is cylindrical, while the diameter of the part PLKQ gradually
1 66 THEORY OF STRUCTURES.
increases from the face of the cylinder to KL ( = EF\ at the
end of the jet. Then
r*)dx= amount of metal which flows into the
central cylinder
= znrdrx, (3)
dr being the depth to which the metal penetrates.
Third, assume that the diminution of the diameter of the
cylindrical portion PMNQ is directly proportional to the said
diameter.
Then, if 2 be the radius of the cylinder PQNM,
dr dz
7 = 7
By eqs. (3) and (4),
Integrating,
C 2 - r 2 ) log, x = 2r 2 log, g + c,
c being constant of integration.
When x = D, z r,
/. (^ 2 - r 2 ) log, -^ = 2r 3 log, |,
or
WORK. 167
By eqs. (2) and (5),
which is the equation to the profile PL or QK.
Note. If R- 3r 2 , eq. (6) represents a straight line.
" R* = 2r\ 4< " " parabola.
II. Work. Work must be done to overcome a resistance.
Thus bodies, or systems of bodies, which have their parts suit-
ably arranged to overcome resistances are capable of doing-
work and are said to possess energy. This energy is termed
kinetic or potential according as it is due to motion or to posi-
tion. A pile-driver falling from a height upon the head of a
pile drives the pile into the soil, doing work in virtue of its
motion. Examples of potential energy, or energy at rest, are
afforded by a bent spring, which does work when allowed to
resume its natural form ; a raised weight, which can do work by
falling to a lower level ; gunpowder and dynamite, which do
work by exploding ; a Leydenjar charged with electricity, which
does work by being discharged ; coal, storage batteries, a head
of water, etc. It is also evident that this potential energy
must be converted into kinetic energy before work can be
done. A familiar example of this transformation may be seen
in the action of a common* pendulum. At the end of the
swing it is at rest for a moment and all its energy is potential.
When, under the action of gravity, it has reached the lowest
point, it can do no more work in virtue of its position. It has
acquired, however, a certain velocity, and in virtue of this
velocity it does work which enables it to rise on the other side
of the swing. At intermediate points its energy is partly
kinetic and partly potential.
A measure of energy, or of the capacity for doing work, is
the work done.
The energy is exactly equivalent to the actual work done
in the following cases:
(a) If the effort exerted and the resistance have a common
point of application.
1 68 THEORY OF STRUCTURES.
(b) If the points of application are different but are rigidly
connected.
(c) If the energy is transmitted from member to member,
provided the members do not change form under stress, and
that no energy is absorbed by frictional resistance or restraint
at the connections.
Generally speaking, work is of two kinds, viz., internal work,
or work done against the mutual forces exerted between the
molecules of a body or system of bodies, and external work, or
work done by or against the external forces to which the body
or bodies are subjected. In cases (a), (b), (c) above, the inter-
nal work is necessarily nil.
As a matter of fact, every body yields to some extent under
stress, and work must be done to produce the deformation.
Frictional resistances tend to oppose the relative motions of
members and must also absorb energy. If, however, the work
of deformation and the work absorbed by frictional resistance
are included in the term work done, the relation still holds that
Energy = work done.
A measure of work done is the product of the resistance by
the distance through which it is overcome. When a man
raises a weight of one pound one foot against the action of
gravity he does a certain amount of work. To raise it two feet
he must do twice as much work, arid ten times as much to raise
it ten feet. The amount of work must therefore be propor-
tional to the number of feet through which the weight is
raised. Again, to raise two pounds one foot requires twice a.*
much work as to raise one pound through the same distance ;
while five times as much work would be required to raise five
pounds, and ten times as much to raise ten pounds. Thus the
amount of work must also be proportional to the weight raised.
Hence a measure of the work done is the product of the
number of pounds by the number of feet through which they
are raised, the resulting number being designated foot-pounds.
Any other units, e.g., a pound and an inch, a ton and an
inch, a kilogramme and a metre, etc., may be chosen, and the
work done represented in inch-pounds, inch-tons, kilogram-
OBLIQUE RESISTANCE.
169
metres, etc. This standard of measurement is applicable to all
classes of machinery, since every machine might be worked by
means of a pulley driven by a falling weight.
12. Oblique Resistance. Let a body move against a
resistance R inclined at an angle to the direction of motion
(Fig. 189). No work is done against the
normal component R sin 0, as there is
no movement of the point of applica-
tion at right angles to the direction of
motion. This component is, there-
fore. merely a pressure. The work
done against the tangential component
R . cos 6 between two consecutive
points M and N of the path of the body is R cos 6 . MN.
Hence the total work done between any two points A and B of
the path
= 2(R cos . MN) = S R cos 6ds,
FIG. i
s being the length of AB.
If AB is a straight line (Fig. 190), and if R is constant in
direction and magnitude,
the total work = R cos B .AB R.AC,
AC being the projection of the displacement upon the line of
action of the resistance. Let the path be the arc of a circle
Rsintf
FIG. 190. FIG. 191.
(Fig. 191) subtending an angle a at the centre. If R and 6 re-
main constant, the work done from A to B
= R cos arc AB = R cos . OA . a = R . OM cos 6 . a
I7O THEORY OF STRUCTURES.
p being the perpendicular from upon the direction of R, and
M = Rp being the moment of resistance to rotation.
If there are more resistances than one, they may be treated
separately and their several effects superposed. In such case,
Mwi\\ be the total moment of resistance and will be equal to
the algebraic sum of the separate moments.
The normal component R sin 8 produces a pressure.
13. Graphical Method. Let a body describe a path AB
B (Fig. I92)against a variable resistance of
such a character that its magnitude in
the direction of motion may be repre-
sented at any point M by an ordinate
'D MNto the curve CD. Let the path
AB be subdivided into a number of
parts, each part MP being so small
that the resistance from M to P may
be considered uniform. The mean
value of this resistance - , and the work done in
MN-\-PQ
overcoming it = . MP = the area MNQP in the
limit. Hence the total work done from A to B = the area
bounded by the curves AB, CD and the ordinates AC, BD.
14. Kinetic Energy. The velocity v acquired by a body
of weight w and mass m in falling freely from rest through the
vertical distance h is
v = \/2gh ;
w v v*
/. wh = = m .
g2 2
Thus an amount of work wh is done, and the body possesses
v*
the kinetic energy m .
Again, let v' be the velocity of the body after falling
through a further distance x, measured vertically. Then
w(/i -j- x)
KINETIC ENERGY.
and
Thus the work done in falling through the vertical distance x
is wx, and is equal to the corresponding change of kinetic
energy.
15. EXAMPLE I. Let it be required to determine the work
done in stretching or compressing a bar of length L and sec-
tional area A by an amount /.
Suppose that the force applied to the bar gradually in-
creases from o until it attains the value P\ its mean value is
P P
, and the work done is therefore /.
2' 2
But P = EA- ; E being the coefficient of elasticity.
E /" i IP\
/. the work done = A-? = 55-1-3"]
2 L E\Ai
AL_
2
/a
This formula is only true for small values of the ratio j.
In the case of a compressive force it is assumed that the bar
does not bend.
P
A suddenly applied force, , will do as much work as a
steady force which increases uniformly from o to P, and hence
it follows that a bar requires twice the strength to resist with
safety the sudden application of a given load than is necessary
when the same load is gradually applied.
If /is the proof stress or elastic limit per unit of sectional
area, 7^ is the corresponding proof strain, and the work done in
L
producing the latter is called the resilience of the bar. Accord-
f* AL f*
ing to the above, its value is ~ ; ~- is called the Modulus
of Resilience.
172 THEORY OF STRUCTURES.
Ex. 2. A wrought-iron tie-rod, 30 ft. in length and 4 sq.
in. in sectional area, is subjected to a longitudinal pull of
40,000 Ibs. Determine the unit stress, the strain, and the elon-
gation, the coefficient of elasticity being 30,000,000 Ibs.
40000
The unit stress is = 10,000 Ibs. per sq. in.
4
Also, from the elastic law, 10000 30000000 X strain.
.-. the strain = ~
and the elongation = ~ ^ ft.
Ex. 3. A steel rod is 15 ft. long and 2\ sq. in. in sec-
tional area. The proof strain of the steel is ^, and its coeffi-
cient of elasticity is 36,000,000 Ibs. Find the greatest weight
that can be safely allowed to fall upon the end of the rod from
a height of 27 ft?.
The proof stress = Ex proof strain = 36,000 Ibs. persq. in.
The compression of the rod under the proof-stress is
Jl _JL ft
1000 200 **
The resilience of the rod
_f*AL _ (36000)' 2J- X 15 X 12
~ E 2 36000000 2
= 8100 inch-lbs. = 675 ft.-lbs.
Again, let P^be the required weight in pounds.
The total distance through which it falls 27 ft. -f- com-
pression = (27 -f- TO) ^ eet > anc * the corresponding work is
W(2J + 4) ft.-lbs. This must of course be exactly equivalent
to the resilience of the rod, and
*
W(2 7 + ) = 675,
and W 24.9 Ibs.
KIXETIC ENERGY. 173
Tht resilience of the rod may also be at once found from
the fact that it is the product of one half of the total stress by
the compression, i.e., -J- . 2|- . 36000 X ^ = 675 ft.-lbs.
Ex. 4. Let w 1 , w 2 , w 3 , . . . w n be the weights of a system
of particles rigidly connected together and at distances ^ , jr a ,
JT S , . . . x n , respectively, from a given axis. Let the system
revolve around the axis with a uniform angular velocity A.
The kinetic energies of the several particles are
w xA* w x*A* w x*A*
and therefore the total kinetic energy of the system
r,
A* (
= - w.^i f
m^ , m^ , . . . m n being the masses of the particles.
The sum between the brackets is called the moment of in-
ertia of the system of particles about the axis and is usually
denoted by /.
A* I
.'. the total kinetic energy - .
Again, it appears from the definition that every moment of
inertia is the product of a mass and the square of a length.
This length is called the radius of gyration and is usually
designated by the symbol k.
If M be the total mass of the system, and W the total
weight,
W
M) 2 W (Ak?
and the total kinetic energy = M - - = - - , the re-
174 THEORY OF STRUCTURES.
suit being the same as if the particles were collected in a ring
of radius k, sometimes called the equivalent ring or fly-wheel.
Let I g be the moment of inertia of the system with respect
to a parallel axis through the centre of gravity, and let h be the
distance between the two axes. Then
IK = >,(/< - *,)' + mjji - O 2 + . . . + m n (h - x n f
Since the new axis passes through the centre of gravity,
2mx = Mh.
Also, 2(m) = M and 2(m^) = 7;
.'. I=I g +Mh\
So, if /' is the moment of inertia about another parallel axis
at the distance h' from the centre of gravity,
I' =I e +Mh'\
.-. I- Mh* = I' Mh'\
Hence, if the positions of two parallel axes relatively to the
centre of gravity are known, and if the moment of inertia about
one is given, the moment of inertia about the other can be
obtained by means of the last formula.
Note. Nothing has been said as to the number of the par-
ticles. They may be infinite in number and infinitely near
each other, forming in fact a solid body. The summation
2(mx*) is then best effected by integration.
16. Values of w.
1. For a rectangular plate of depth d with re-
spect to an axis through the centre , 2
perpendicular to the side d. .......... k* = .
2. For a circular plate of radius r with re- a
spect to a diameter .................. k* = -.
4
VALUES OF k\ 175
3. For an annulus of external radius r l and
internal radius r^ with respect to a
diameter ..........................
. If r 1 r^ = t, and the breadth
/ of the annulus is small as compared
with the radius r l , then
approx.,
4 ^
and the area
approx.
4. For the plates in (2) and (3) with respect
to an axis through the centre perpen-
dicular to the plates, the numerators
remain the same but the denominator
is in each case 2.
5. For a sphere of radius r with respect to a
diameter k* = r*.
6. For a solid cylinder of radius r with re- 3
spect to its axis k* = .
2
7. For an elliptic plate of which the major
and minor axes are 2b and 2d respec-
tively :
With respect to the major axis k* = .
4
ia
With respect to the minor axis J? = .
4
8. For a triangular plate of height h with re-
spect to an axis coinciding with the , a
base . .' k* = -.
6
THEORY OF STRUCTURES.
17. Momentum Impulse. A moving body of weight w
and mass m acted upon in the direction of motion for a time /
by a force F will acquire a velocity v which is directly propor-
tional to F and to /, and inversely proportional to w. Hence
Ft
v = n
n being some coefficient.
If F = w, the velocity generated in one second
'. g = n,
and
Ft Ft
''" = ^=-m>
or
mv = Ft.
This is the analytical statement of Newton's Second Law
of Motion, which has been expressed by Clerk Maxwell in the
following form : " The change of momentum (i.e., the product of
the mass and velocity) is numerically equal to the impulse (i.e.,
the product of the force and the time during which it acts)
which produces it, and is in the same direction"
Again, let / be the perpendicular from a fixed axis O upon
the direction of motion of the body, and let r be the radius OP
to the body. Then
mvp = Ftp = Fpt = Mt,
where M = Fp ; or the change of the moment of momentum, i.e.,
of the angular momentum, is equal to the moment of impulse.
The above results are also true for two or more bodies or
systems of bodies severally acted upon by extraneous forces,
and the equations may be written
In words, the total change of momentum in any assigned direction
is equal to the algebraic sum of the impulses in the same direction,
MOMENTUM IMPULSE. 1/7
and the total change of angular momentum is equal to the alge-
braic sum of the moments of the impulses.
Hence it follows that if two or more bodies or systems of
bodies mutually attract or repel each other, and if there are no
extraneous forces, the total momentum in ^ w
*^- fc *
any assigned direction is constant (the ~"7^^ \
principle of the conservation of linear / * s ***^p
momentum), and the angular momentum j /''
about a given axis is constant (the prin- / : r /'
ciple of the conservation of angular / /'
momentum). ,//' f 2 / a
Suppose that the velocity of the body 3^-- -^ y
of weight w and mass ;// changes from X^ /
v l to v z in the time / under the action of 2 ^X v/ /
a couple of moment M, and let/,,/, be FlG - ^s-
the corresponding values of/, and r lf r a those of r, Fig. 193.
or if w lt w^ are the components of v l , v^ in directions perpen-
dicular to r lt 7* a , respectively,
m(w 1 r l wj-^ Mt.
For example, a weight W of water passing through a turbine
of external radius r^ and internal radius r^ has its angular mo-
W W
mentum changed from w l r l to ^ a r a> w i-> w * being
o o
the tangential components of the velocity with which the
water enters and leaves the wheel. The water, therefore, exerts
W
upon the wheel a couple of moment ( l w l r l w 2 ^ 2 ), and if
o
the wheel rotates with an angular velocity A, the work done
upon the wheel by the water
w w
= ~A(wj-i wyj = (ze/.a, w,u,),
<b O
Ji l and u^ being the circumferential velocities corresponding to
r l and r a , respectively.
THEORY OF STRUCTURES.
18. Useful Work Waste Work. Let a body of mass m
.and weight w pass over the distance s under the action of a
force F acting in the direction of motion for a time /, and let
the velocity of the body change from v l to v 9 . Assume / to
.be so small that, for the interval in question, the velocity may
i) I ij
>be regarded as constant and of the average value - L - ;
But Ft = (mv t mVt).
m ,
2^ *
or
Fs = (,' - ,).
Thus Fs, the work done, is equal to the change of kinetic
energy in the given interval.
If the body is a material particle of a connected system, a
similar relation holds for every other particle of the system, and
the total work done = %(2mv* 2mv?).
A part of this work may be expended in doing what is
called effective work, i.e., in overcoming (i) an external resist-
ance, or in doing useful work, and (2) frictional resistance, or in
doing wasted work.
Denoting the total effective work by T e and the total motive
work by T m , the last equation may be written
and the difference between the total motive work and the total
effective work is equal to the total change of kinetic energy.
In the case of a machine working at a normal speed the
velocities of the different parts are periodic, being the same at
the beginning and end of any period or number of periods.
For any such interval, therefore, v t = v y , and /. T m T e , so
GENERAL CASE. 1 79
that there is an equality between the motive work and the
effective work.
19. General Case. Let x l , y lt z^ be the co-ordinates of
the C. of G. of a moving body of mass M with respect to three
rectangular axes at any given instant.
Let x t , y t , 2 be the co-ordinates of the same point after a
unit of time.
Let x, , y l , z, be the co-ordinates of any particle of mass m
at the given instant.
Let x^,y^, 2 be the co-ordinates of the same particle after
a unit of time.
, - y,} = 2m(y* - y,
or
M u = 2mu, Mv 2mv, Mw =
u, v, w being the component velocities of the C. of G. at the
given instant with respect to the three axes, and u, v, w the
component velocities of the particle m at the same instant.
From these last equations,
Mu = ^muu, Mv 2mvv, Mw =
.'. M(u -}-v'-\-w) = 2m(uu -\- vv
which may be written in the form
., 2 . 2
M(u + <v -\- w) + 2m\(u - uj + (v - v? + (w -
= 2m(u* + v* + w 2 ),
or
ISO THEORY OF STRUCTURES,
U being the resultant velocity of the C. of G. ; v, that of the
particle ; and F, that of the particle relatively to the C. of G.
The last equation may be written
MIT ^mV* _2mv>
O I 1 />
Thus the energy of the total mass collected at the centre
of gravity, together with the energy relatively to the centre of
gravity, is equal to the total energy of motion.
If the body revolves around an axis through its C. of G.
with an angular velocity A, the second term of the last equa-
tion becomes
r being the distance of the particle m from the axis, and / the
moment of inertia of the body with respect to the axis.
20. EXAMPLE I. The charge of powder for a 27-ton breech-
loader with a Q-ton carriage is 300 Ibs. ; the weight of the pro-
jectile is 500 Ibs., its diam. is 10 in., and its radius of gyration
3.535 in. ; the muzzle velocity is 2020 ft. per sec. ; the velocity
of recoil, i6J ft. per sec. ; the gun is rifled so that the projectile
makes one turn in 40 calibres.
Total energy of explosion = energy of shot -|- energy of recoil :
Energy of shot = energy of translation + energy of rotation
_ 5Q (2020)' 500 j. /V.-j-f 2020 y /3-535\ a
32.2'' 2 t "32.2'2*\ T 8 2- "4Q.|jy \ 12 I
= 31680124.2 +97758.6
= 31777882.8 ft.-lbs. ;
En.ro of recoil * . . 330652, ft,.b,
CENTRIFUGAL FORCE. l8l
Hence, if C be the energy of I Ib. of powder,
C . 300 = 31777882.8 + 330652.1
= 32108534.9 ft.-lbs.,
and hence
C 107028.45 ft.-lbs = 47.7 ft.-tons.
Ex. 2. Let Wbe the weight of a fly-wheel in Ibs,, and let
its max. and min. angular velocities be A lt A 9 , respectively.
The motion being one of rotation only, the energy stored up
when the velocity rises from A^ to A lt or given out when it
falls from A l to A^ is
I W W
- (A: - A:) = -k\A? - A:) = w - *.),
v l , v^ being the linear velocities corresponding to A lt A 2 , and
k being taken equal to the mean radius of the wheel.
It is usual to specify that the variation of velocity is not
to exceed a certain fractional part of the mean velocity.
Let V be the mean velocity, and the fraction. Then
V
Vi v* = '> also v l + v t
W F a
Hence the work stored or given out = .
* g P
21. Centrifugal Force. A body constrained to move in
a plane curve exerts upon the body which constrains it, a force
1 82 THEORY OF STRUCTURES,
called centrifugal force, which is equal and opposite to the de-
viating (or centripetal} force exerted by
the constraining body upon the revolving
body.
Let a particle of mass m move from a
\ point P to a consecutive point Q (Fig. 194)
\ of its path during an interval of time /
\ o under the action of a normal deviating
FIG. i 94 . force.
Let the normals at P and Q meet in O ; PQ may be con-
sidered as the indefinitely small arc of a circle with its centre
at O.
If there were no constraining force, the body would move
along the tangent at P to a point T such that PT = vt, v
being 1 the linear velocity at P.
Under the deviating force the body is pulled towards O
through a distance PN = %ff, f being the normal accelera-
tion, and QN being drawn perpendicular to OP.
Also, in the limit, PQ = PT = QN = vt.
But
being the radius OP; and hence
A being the angular velocity.
Hence the deviating force of the mass m
= mf = m -= = mA*r y
and is equal and opposite to the centrifugal force.
Again, if a solid body of mass M revolve with an angular
velocity A about an axis passing through its C. of G., the total
CENTRIFUGAL FORCE.
183
centrifugal force will be nil, provided the axis of rotation is an
axis of symmetry, or is one of the principal axes of inertia at
the C. of G.
If the axis of rotation is parallel to one of these axes, but
at a distance R from the C. of G.,
the centrifu-
, ,
gal force
W
g
r being the distance of a particle of mass m from the axis, and
f'Fthe weight of the body. Thus the centrifugal force is the
same as if the whole mass were concentrated at the C. of G.
If the axis of rotation is inclined at an angle to the prin-
cipal axis, the body will be con-
stantly subjected to the action of
a couple of moment 2E tan 6, E
being the actual energy of the body.
EXAMPLE. A ring of radius r
rotates with angular velocity A about
its centre O. Let p be the weight of
the ring per unit of length of periph-
ery. Consider any half-ring AFB.
The centrifugal force of any element
FlG - I9S>
The component of this force parallel to AB, is balanced by
an equal and opposite force at C" , the angle C" OB being
the angle CO A. Thus the total centrifugal force parallel to
A OB is nil.
The component of the force at C, perpendicular to AB,
g
sn
COD= P -A"r cos C'CE
_fCC C^_ A'r
Ar cc ,-p--DD.
184 THEORY OF STRUCTURES.
Hence, the total centrifugal force perpendicular to AB
= ?2(DD f ) = 2^4 V.
g g
If T is the force developed in the material at each of the
points A and B,
g
since the direction of T is evidently perpendicular to AB.
~~g ' F '
v being the circumferential velocity.
Let f be the intensity of stress at A and B, and w the
specific weight of the material.
Assuming that T is distributed uniformly over the sectional
areas at A and B,
Thus, the stress is independent of the radius .for a given
value of v, and the result is applicable to every point of a flex-
ible element, whatever may be the form of the surfaces over
which it is stretched.
22. Impact. When a body strikes a structure, or member
of a structure, the energy of the blow is expended in
(1) overcoming the resistance to motion of the body struck ;
(2) deforming the body struck ;
(3) the kinetic energy of either or of both of the bodies
after impact, if the motion is sensible ;
(4) deforming the striking body ;
(5) producing vibrations.
IMPACT. 185
Generally speaking, the energy represented by (5) is very
small and may be disregarded. Also, if the striking body is
very hard, the energy (4), absorbed in its deformation, is inap-
preciable and may be neglected.
First, let a body of weight P fall through a vertical dis-
tance h and strike a second body, the point of application
moving in the direction of the blow through a distance x
against a mean resistance R. Then
P(k -f- x) = work done = R'x.
Let Fbe the velocity of the striking body at the moment
of impact. Then
P V
energy of blow = - - = R'x = P(1i -\- x).
The actual resistance is directly proportional to the dis-
tance through which the point of application moves, so long as
tJie limit of elasticity is not exceeded. Its initial value is nil,
r>
and if R is its max. value, the mean value is R' = .
P V Rx
.'. - - = - - = P(h + x).
g 2 2 '
If h = o, R = 2P, or the sudden application of a load P
from rest, produces a pressure equal to twice the load, pro-
vided the limit of elasticity is not exceeded.
EXAMPLE. A i-oz. bullet moving with a velocity of 800 ft.
per sec. strikes a target and is stopped dead in the space of
,ft. inch (g = 32), Then
1 86 THEORY OF STRUCTURES.
.-. R ', the mean resistance overcome by the bullet, = 5000 Ibs.
The time in which the bullet is brought to rest
momentum y 1 -^ . % . 800 I
force 5000 ~ 3200 SeC *
Next, let a body of weight W^ moving in a given direction
with a velocity z\ strike a body of weight W^ moving in the
same direction with a velocity v^ . After impact let the bodies
continue to move in the same direction with a common ve-
locity v.
W, W,
P, + v+ = momentum before impact
= momentum after impact
or
, , ^
Energy before impact = + .
after " =1 ^ -) .
g
Energy lost by impact
%
IMPACT. 187
If either of the bodies is subjected to any constraint, energy
must be expended to overcome such constraint, and the loss of
energy by impact will be less.
EXAMPLE i. Let a weight of W^ tons fall h ft. upon the
head of a pile weighing W^ tons and drive it a ft. into the
ground against a mean resistance of R tons, the head of the
pile being crushed for an appreciable length x ft.
Let v be the velocity of the weight when it strikes the pile ;
n p n mean force of the blow;
" y " " distance through which pile moves during ac-
tion of blow ;
" / " " duration of the blow in seconds ;
" V " " common velocity of the pile and weight during
action of blow ;
" z " " distance through which pile moves after the
blow.
Px -\- Ry = work done in crushing the pile -f- work
done in overcoming ground-resistance
in time t = energy dissipated by blow
Also, considering the change of momentum first of weight
and then of pile,
Again,
W I W I 7 *
Rz = work done after blow = 1- - . (*\
g 2
Finally, y + 2 = a, . (4)
and
(5)
1 88 THEORY OF STRUCTURES.
Thus, if W lt W if h, a, and x are known, eqs. (i) to (5) will
give P, /, R, y, V, and z.
Ex. 2. Let a hammer weighing W l Ibs. moving with a ve-
locity of v ft. per sec., strike a nail weighing W^ Ibs. and drive
it x ft. into a piece of timber, of weight W 3J against a mean
resistance of R Ibs.
First, assume the timber to be fixed in position.
Let F, be the common velocity acquired by the hammer
and nail.
\ ~f" ^a) L ener gy expended in overcoming R
(i)
W W 4- W
But -v change of momentum = - - ? F 1 . . (2)
o o
3
x Wv
and the time of the penetration = JT>~ = -jr- sec. ... (4)
Second, let the timber be free to move, and let F 2 be the
common velocity acquired by the hammer, nail, anld timber.
F 2
(W\~\~ W^) = energy expended in overcoming R plus
the energy expended in producing the
velocity F 2
. ... (5)
W w _i_ w W 4- W -4- W
But ^^^TJ^F^-^^J --^ ^F 2 . . . (6)
g g g V ;
ON THE EXTENSION OF A PRISMATIC BAR. 189
Hence, substituting these values of F, and F 3 in eq. (5),
J?v . f>7\
= Kx\ . . . (7)
also, the time of the penetration
V_ . .
*
and the distance through which the timber moves
W ? W * jLft
U
23. On the Extension of a Prismatic Bar. The ele-
mentary law of extension is sometimes enunciated as follows:
A prismatic bar of length L and sectional area A is
stretched, and its length is L -\- x when the force of extension
isP- if dP is the increment of force corresponding to an in-
crement dx of length,
dP=EA d *.
L + x
Hence, the force producing an extension / is equal to
j = EA log.(i +~)=^, s
suppose.
But
/ A / i/A a i//\ 3 /
a PP rox '
/ A / i/A a i//\ 3 /
+ IJ = L ~ 2 (ll + jlzJ - = Z '
I QO THEORY OF STRUCTURES.
Corollary. From the last equation, ^ = , and -y-
is consequently a measure of the longitudinal stiffness of a bar,
so that for the same material, the stiffness varies directly as the
sectional area and inversely as the length, while for different
materials it also varies directly as the coefficient of elasticity.
Work of Extension. The force producing the increment dx
has for its least value P(=A^j, for its greatest value
P-\-dP, and for its mean value P-\ -- , so that the work done
is fP-\ -- \dx = Pdx, approximately.
Hence the work done in stretching the bar until its length
is L + /is equal to
24. On the Oscillatory Motion of a Weight at the End
o of a Vertical Elastic Rod. An elastic rod of natu-
* ral length L(OA) and sectional area A is suspended
from O, and carries a weight P at its lower end, which
elongates the rod until its length is OB = L -f- /.
Assume that the mass of the rod as compared
with P is sufficiently small to be disregarded, then
P=EA l - L .
A
If the weight is made to descend to a point C, and
| is then left free to return to its state of equilibrium, it
!Q must necessarily describe a series of vertical oscilla-
tions about B as centre.
FIG. 196. Take g as the or j gin) and at any time t i et t | ie
weight be at M distant x from B ; also let BC c.
Two cases may be considered.
First, suppose the end of the rod to be gradiially forced
down to C and then suddenly released.
OSCILLA TOR Y MO TION. i
According to the principle of the conservation of energy,
( -7- ) = the work done between C and M
2 \dtl
g
_EA
or
Pi (dxV Pi
and hence
z>, the velocity of the weight at M, = \ (? x*\\
V /
Now v is zero when x = c, so that the weight will rise
above B to a point C, where BC l ,==.; = .#7.
Again, from the last equation,
. / __^_
V 7 ~(V-^)*
and integrating between the limits o and x,
and the oscillations are therefore isochronous.
When .* = c,
x
and the time of a complete oscillation is
g
IQ 2 THEORY OF STRUCTURES.
Next, suppose the oscillatory motion to be caused by a
weight P falling without friction from a point D, and being
suddenly checked and held by a catch at the lower end of the
rod.
Take the same origin and data as before, and let AD = h.
The elastic resistance of the rod at the time t is
L
and the equation of motion of the weight is
Pcfx
~ g
or = -
Integrating,
!dx\? g
! j = jx* -f- c l , c l being a constant of integration.
dx P~
But -7- is zero when x = c, and c. =^c*.
at I
Hence
This is precisely the same equation as was obtained in the
first case, and between the limits o and x
rg . ,*
yf : sm 7-
OSCILLATORY MOTION. 1 93
so that the motion is isochronous, and the time of a complete
oscillation is
x4
Cor. i. When x = I,
and hence
f (<>-/<) =
or
Cor. 2. If h = o, i.e., if the weight is merely placed upon
the rod at the end A, c = /, and the amplitude of the
oscillation is twice the statical elongation due to P.
Cor. 3. The rod may be safely stretched until its length is
L + /, while a further elongation c might prove most injurious
to its elasticity, which shows the detrimental effect of vibratory
motion. If a small downward force Q is applied to Pwhen it
has reached the end of its vibration, it will produce a corre-
sponding descent, and the weight P will then ascend an equal
distance above its neutral position. At the end of the interval
corresponding to P's natural period of vibration, apply the
force again, and P will descend still further. This process
may be continued indefinitely, until at last rupture takes place,
however small Pand Q may be. If Q is applied at irregular
intervals, the amplitude of the oscillations will still be increased,
but the increase will be followed by a decrease, and so on con-
tinually. In practice the problem becomes much more com-
plex on account of local conditions, but experience shows that
a fluctuation of stress is always more injurious to a structure
than the stress due to the maximum load, and that the injury
194 THEORY OF STRUCTURES.
is aggravated as the periods of fluctuation and of vibration of
the structure become more nearly synchronous.
An example of a fluctuating load is a procession marching
in time across a suspension-bridge, which may strain it far
more severely than a much greater dead load, and may set up
a synchronous vibration which may prove absolutely dangerous.
In fact, a bridge has been known to fail from this cause.
Cor. 4. The coefficient of elasticity of the rod may be ap-
proximately found by means of the formula
/
y
g
T being the time of a complete oscillation. For suppose that
the rod emits a musical note of n vibrations per second, then
g
is the time of travel from C to C l ;
n-
and hence
Cor. 5. Suppose that the weight is perfectly free to slide
along the rod. When it returns to A, it will leave the end of
the rod and rise with a certain initial velocity. This velocity
is evidently V2gK t and the weight accordingly ascends to D,
then falls again, repeats the former operation, and so on. The
equations of motion are in this case only true for values of x
between x + c and x = /.
25. On the Oscillatory Motion of a Weight at the End
of a Vertical Elastic Rod of Appreciable Mass. Suppose
the mass of the rod to be taken into account, and assume :
(a) That all the particles of the rod move in directions par-
allel to the axis of the rod,
OSCILLATORY MOTION. 1 95
(b) That all the particles, which at any instant are in a plane
perpendicular to the axis, remain in that plane at all times.
As before, the rod OA of natural length L and sectional
area A is fixed at O and carries a weight P l at A.
Take O as the origin, and let OX be the axis of the rod.
Let ,-)- d, and x, x -j- dx, be respectively the actual
and natural distances from O of the two consecutive
sections MM, M'M'. o
Let p be the natural density of the rod, and p
the density of the section MM, distant from O.
The forces which act upon the rod are :
(a) The upward and constant force P at 0. M >
(b) The weight P l at A.
(c) The weight of the rod.
(d) A force X per unit of mass through the slice
bounded by the planes MM, M' M', distant % and
B> -\- d>, respectively, from O.
Suppose the rod, after equilibrium has been es-
tablished, to be cut at the plane M'M'. In order to
maintain the equilibrium of the portion OM ' M' it
will be necessary to apply to the surface of this plane a certain
force P, and the equation of equilibrium becomes
= o.
But if the thickness d$ of the slice MM' is indefinitely
diminished, P is evidently the elastic reaction, and its value is
Hence
= o.
IQ THEORY OF STRUCTURES.
Differentiating with respect to x,
~Kvt pdS ft^dx,
~ + p,gA = o,
or -
Also, p n AXdx is the resistance to acceleration arising from
the inertia of the slice, and is therefore equal to
so that
df
Hence
To solve this equation. In the state of equilibrium,
'*-')
is the tension in the section of which the distance from
OSCILLATORY MOTION. 197
is x, and counterbalances the weight P l and the weight
p Q A(l x)g of the portion AMN of the rod.
.\EA ~ -
or
Integrating,
+ *- ---- c>
There is no constant of integration, as x and % vanish
together.
This value of ^ is a particular solution of (i), and is inde-
pendent of /.
Put 5 = *
z being a new function of x and t. Then
_ _ p. z
*'' ~ g+ *' and
Hence, from eq. (i),
d*z Ed*z d*z E
75 = ~ T~^ = v \ 3~T > where v? = ,
d? p.dx* l dx* p
The integral of this equation is of the form
IQ8 THEORY OF STRUCTURES.
= A / ) being the velocity of propagation of the vibrations.
The full solution of (i) is therefore of the form
p,
26. Inertia Balancing. Newton's First Law of Motion,
called also the Law of Inertia, states that" a body will continue
in a state of rest or of uniform motion in a straight line unless
it is made to change that state by external forces."
This property of resisting a change of state is termed
inertia, and in dynamics is always employed to measure the
quantity of matter contained in a body, i.e., its mass, to
which the inertia must be necessarily proportional. Thus, to
induce motion in a body, energy must be expended, and must
again be absorbed before it can be brought to rest. The inertia
of the reciprocating parts of a machine may therefore heavily
strain the framework, which should be bolted to a firm foun-
dation, or must be sufficiently massive to counteract by its
weight the otherwise unbalanced forces.
EXAMPLE I. Consider the case of a direct-acting horizontal
steam-engine, Fig. 198. At any
given instant let the crank OP
and the connecting-rod CP make
angles and 0, respectively, with
the line of stroke AB.
Let v be the velocity of the
FIG. 198. crank-pin centre P, and let u be
the corresponding piston velocity, which must evidently be the
same as that of the end C of the connecting-rod.
Let OP produced meet the vertical through C in /.
At the moment under consideration, the points C and Pare
turning about / as an instantaneous centre.
1C sin
IP ~ COS
INpR TIA BA LA NCING.
Let W be the weight of the reciprocating parts, i.e., the
piston-head, piston-rod, cross-head (or motion-block), and a por-
tion of the connecting-rod.
Assume (i) that the motion of the crank-pin centre is uni-
form ;
(2) that the obliquity of the connecting-rod may
be disregarded without sensible error, and
/. o.
Draw PN perpendicular to AB, and let ON=.x\ ON is
equal to the distance of the piston from the centre of the
stroke, corresponding to the position OP of the crank.
The kinetic energy of the reciprocating parts
W 21* W v* sin 8 W V* i x^
g 2 "" g 2 " g 2 \ J " r*
r being the radius OP.
.'. the change of kinetic energy, or work done, corresponding
to the values x l , x^ of x,
W^<x;-x?\
~ r 2 r 2 )'
g 2
Let R be the mean pressure which, acting during the same
interval, would dp the same work. Then
W v* x? - x?
7 T~^~ =*(*'-*>'
and
Hence, in the limit, when the interval is indefinitely small,
#, = x^ = x, and the pressure corresponding to x becomes
wv
R = -r-ar.
'
200 THEORY OF STRUCTURES.
This is the pressure due to inertia, and may be written in the
form
R- C
- 6 7>
/ W v*\
C [= 1 being the centrifugal force of W assumed con-
centrated at the crank-pin centre. R is a maximum and equal
to C when x = r, i.e., at the points A, B, and its value at
intermediate points may be represented by the vertical ordi-
nates to AB from the straight line EOF drawn so that
AE = BF = C. In low-speed engines, C may be so small that
the effect of inertia may be disregarded, but in quick-running
engines, C may become very large and the inertia of the recip-
rocating parts may give rise to excessive strains.
Another force acting upon the crank-shaft is the centrifu-
gal force of the crank, crank-pin, and of that portion of the con-
necting-rod which may be supposed to rotate with the crank-
pin.
Let w be the weight of the mass concentrated at the crank-
pin centre which will produce the same centrifugal force as
these rotating pieces (i.e., wr sum of products of the weights
of the several pieces into the distances of their centres of gravity
from O).
TU -r i r e W V*
The centrifugal force of w = .
g *
Thus the total maximum pressure on the crank-shaft
A being the uniform angular velocity of the crank-pin.
This pressure may be counteracted by placing a suitable
balance-weight (or weights) in such a position as to develop in
the opposite direction a centrifugal force of equal magnitude.
INER TIA BA LA NCING. 20 1
Let W l be such a weight and R its distance from O. Then
or
from which, if R is given, W l may be obtained.
During the first half of the stroke an amount of energy
represented by the triangle AEO is absorbed in accelerating the
reciprocating parts, and the same amount, represented by the
triangle EOF, is given out during the second half of the stroke
when the reciprocating parts are being retarded.
During the up-stroke of a vertical engine the weights of the
reciprocating parts act in a direction opposite to the motion of
the piston, while during the down-stroke they act in the same
direction.
lr\AE produced (Fig. 199) take EE' to represent the weight
of the reciprocating parts on the same scale
as AE represents the pressure due to inertia. E
Draw E'O'F' parallel to EOF. A I
During the up-stroke the ordinates of
E'O' represent the pressures required to ac-
celerate the reciprocating parts, the pressures while they are
retarded being represented by the ordinates of O' F' .
The case is exactly reversed in the down-stroke.
N.B. The formula R = C may be easily deduced as
follows :
du dd v*
& = ^ sin 6; the acceleration = -77 = v cos ft -y- -;# ;
at at r
W du
.*. -- -j- = accelerating force = force due to inertia
g at
W v* *
= -- TX = C~.
g r* r
202 THEORY OF STRUCTURES.
Ex. 2. Consider a double-cylinder engine with two cranks
at right angles, and let d be the distance between the centre
lines of the cylinders (Fig. 2OO).
of Cylr.
Centre line
C.sin d
FIG. 200.
The pressures due to inertia transmitted to the crank-pins
when one of the cranks makes an angle 6 with the line of
stroke are
P,= C cos and P, = C sin V.
These are equivalent to a single alternating force
P = C(cos 6 sin 0)
acting half-way between the lines of stroke, together with a
couple of moment
M = P- = C-(cos sin 0).
2 2^
The force and couple are twice reversed in each revolution,
and their maximum values are
Cd
P,na X . = CV2 and M max , = Y2.
In order to avoid the evils that might result from the action
of the force and couple at high speeds, suitable weights are
introduced in such positions that the centrifugal forces due to
INER TIA BA LA NCING. 203
their rotation tend to balance both the force and the couple.
For example, the weights may be placed
upon the fly-wheels, or again, upon the driv-
ing-wheels of a locomotive.
Let a balance-weight Q be placed nearly
diametrically opposite to the centre of each
crank-pin (Fig. 201), and let R be the distance
from the axis to the centre of gravity of Q.
Let e be the horizontal distance between FlG> 20I>
the balance-weights.
The centrifugal force /^due to the rotation of Q
g (velocity of Q}" Q I R*_ , _ QR a
R ~ g ~Rr* V ~ gr* V >
and this force F is equivalent to a single force F acting half-way
between the weights and to a couple of moment
^
F . Let be the angle between the radius
to a balance-weight, and the common bisector
of the angle between the two cranks (Fig. 202).
Since there are two weights Q, there will
FIG. 202. be two couples each of moment F , and two
forces each equal to F acting half-way between the weights,
the angle between the axes of the couples being 180 20, and
that between the forces being 20. The moment of the result-
ant couple is Fe sin 0, and its axis bisects the angle between
the axes of the separate couples; the resultant force parallel
to the line of stroke 2.F cos 0.
Q and may now be chosen so that
2/ r cos = maximum alternating force = C ^2,
and
Fe sin = maximum alternating couple = 1/2.
d
.'. tan - ,
204 THEORY OF STRUCTURES.
and
or
_ A / ? a + *r a
g~?" ~ g r
and
W
Ex. 3. Again, the pressure C at a dead point may be
balanced by a weight Q diametrically opposite.
If R is the radius of the weight-circle, then
Wv* QR a
and
R
e -\- d
The weight Q may be replaced by a weight Q on the
near and a weight Q on the far wheel. Thus, since the
cranks are at right angles, there will be two weights 90 apart
on each wheel, viz., Q - in line with the crank and Q .
These two weights, again, may be replaced by a single weight
B whose centrifugal force is the resultant of the centrifugal
forces of the two weights. Thus
er-c* -c/^
^
B </_Y_ (Q e + d z/V (Q e-d z/V
R)"\fr 2e RJ " V 2e Rr
v f being the linear velocity at the circumference of the weight-
circle.
2e
CURVES OF PISTON VELOCITY.
205
or
B=V
If a is the angle between the radius to the greater weight
us,
Qe-dv'
e-\-d
Q and the crank radius,
tan a -
'
'*
R e ~
Note. In outside-cylinder engines e d is approximately
nil, and B = Q = W~.
/V
27. Curves of Piston Velocity. Consider the engine in
Ex. i.
a
FIG. 203.
Let CP produced intersect the vertical through in T, and
in OPtakeOT' = OT.
The piston velocity u and the velocity V of the crank-pin
centre are connected by the relation
sin(fl4-0)__<9:r
cos ~~OP~~~OP'
. . . (I)
If the velocity v is assumed constant, and if it is represented
by OP, then on the same scale OT' will represent the piston
velocity u. Drawing similar lines to represent the value of u
2O6
THEORY OF STRUCTURES.
for every position of the crank, the locus of T' will be found to
consist of two closed curves OGS, OUT, called the polar curves
of piston velocity. They pass through the point O and through
the ends 5 and T of the vertical diameter. On the side towards
the cylinder they lie outside the circles having OS and OT as
diameters, while on the side away from the cylinder they lie
inside the circles. If the connecting-rod is so long that its
obliquity may be disregarded,
= o and
u =
0,
and the curves coincide with the circles.
A rectangular diagram of velocity may be drawn as follows :
C N
FIG. 204.
C M
Upon the vertical through C, Fig. 204, take CL = OT\ the
locus of L is the curve required for one stroke. A similar
curve may be drawn for the return stroke either below MN or
upon the prolongation NR (= MN) of MN.
If the obliquity of the connecting-rod is neglected, the
curves evidently coincide with the semicircles upon MN and
NR, MN (= NR) defining the extreme positions of C. The
obliquity, however, causes the actual curve to fall above the
semicircle during the first half of the stroke, and below during
the second half.
Again, let the connecting-rod (/) = n cranks (r). Then
sin B I
z^i n,
sin r
and by eq. I,
u = v (sin -\- cos 8 tan 0) = v ( sin 8 -| -. ). (2)
\ #V sin 2 0/
CURVE OF CRANK-EFFORTCURVES OF ENERGY. 2O?
If the obliquity is very small,
and
sin
tan = sm = , approximately,
sin #cos
sin20\
).
28. Curve of Crank-effort. The crank-effort F for any
position OP of the crank is the component along the tangent
at P of the thrust along the connecting-rod.
This thrust =
cos <p'
COS
If the pressure P upon the piston is constant, and if it is rep-
resented by OP, then, on the same scale, OT' , Fig. 203, will rep-
resent the crank-effort. Thus, the curves of piston velocity
already drawn may also be taken
to represent curves of crank-effort.
If the pressure P is variable, as is
usually the case, let OP, the crank
radius, represent the initial value
of P. After expansion has begun,
take OP' in OP, for any position OP
of the crank, to represent the cor-
responding pressure which may be
directly obtained from the indicator-diagram. Draw P'T'
parallel to PT, and take OT" = OT . Then OT" will repre-
sent the required crank-effort, and the linear and polar diagrams
may be drawn as already described.
29. Curves of Energy Fluctuation of Energy. In the
curve of crank-effort as usually drawn, the crank-effort for any
position OP oi the crank is the ordinate S'H, the abscissa DH
being equal to the arc AP, i.e., to the distance traversed by the
FIG. 205.
208
THEORY OF STRUCTURES.
point of application of the crank-effort. Thus, DSE andEVG
being the curves,
DE = EG = semi-circumference of crank-circle ?rr.
If the obliquity is neglected, the curves of crank-effort are
the two curves of sines shown by the dotted lines.
H
The area DS'H also evidently represents the work done as
the crank moves from OA to OP, and the total work done is
represented by the area DSE in the forward and by EVG in
the return stroke.
Let F 9 be the mean crank-effort. Then
F X 2nr 2PX 2r,
assuming Pto be constant.
2 p
Draw the horizontal line 1234567 at the distance from
7t
DEG, and intersecting the verticals through D, E, and G in i,
4, and 7, and the curves in 2, 3, 5, and 6. The engine may be
supposed to work against a constant resistance R equal and
opposite to the mean crank-effort F .
From D to 2, R > crank-effort, and the speed must there-
fore continually diminish.
From 2 to 3, R < crank-effort, and the speed must contin-
ually increase.
Thus 2 is a point of min. velocity, and therefore also of
min. kinetic energy.
From 3 to E, R > crank-effort, and the speed must contin-
ually diminish.
CURVES OF ENERGY FLUCTUATION OF ENERGY. 2OQ
Thus 3 is a point of max. velocity, and therefore also of
max. kinetic energy.
Similarly, in the return stroke, 5 and 6 are points of min.
and max. velocity, respectively.
The change or fluctuation of kinetic energy from 2 to 3
area 283, bounded by the curve and by 23.
The fluctuation from 3 to 5 = area 3^5, bounded by 35 and
by the curve.
F Fr
Again, since -, the ordinates of the curves may be
taken to represent the moments of crank-effort, and the abscissae
are then the corresponding values of 0.
The work done between A and any other position P of
the crank-pin
= Pr(i cos + n \'n* si
If there are two or more cranks, the ordinates of the crank-
effort curve will be equal to the algebraic sums of the several
crank-efforts. For example, if the two cranks are at right
angles, and if F lt F 2 are the crank-efforts when one of the
cranks (F^ makes an angle 6 with the line of stroke,
sn
and
.-. F t -{-F 9 = P(sin + cos 0) = combined crank-effort,
P being supposed constant.
Note. In the case of the polar curves of crank-effort, if a
circle is described with O as centre and a radius = mean crank-
2 p
effort = - , it will intersect the curves in four points, which
are necessarily points of max. and min. velocity.
210
THEORY OF STRUCTURES.
U3
ffl
IN
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o'
c a
3-0
o o
o o -<t-
riTrM
OO ro (N O> O
-< a
000300000300 ro-i-b
C CCCCCCCCC
888
o_ q q_ q_
Com-
pressio
8 8
VO
is
vo~ vo" (xo rn cT -4- 10 to <> -r"--<?
*O vovovOioincKoooooo vo o
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rt
~ :::::::: .- :
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TABLES.
211
THE STRENGTHS, ELASTICITIES, AND WEIGHTS OF
VARIOUS ALLOYS, ETC.
Material.
Max. Load on Original Area
in Ibs. per sq. in.
Young's
Modulus,
(in Ibs.).
Coef-
ficient of
Rigidity,
G
(in Ibs.).
Weight
in Ibs.
per cu. ft.
Tension.
Com-
pression.
Shear-
ing.
A iuminum
28,800
17,600
52,000
30,000
60,000
36,000
1,850
3,100
57,000
4,980
7,5oo
4,000
10,380
58,000
7,100
9,600.000
9,100,000
15,000,000
14,000,000
15,000.000
15,000,000
17,000,000
9,900,000
710,000
996,000
14,000,000
5,690,000
13,500,000
25,000
3,600,000
3.400,000
5,600,000
5,700,000
5,700,000
5,250,000
2,140,000
160 to 166
487 to 524.4
$
555
529
712
456 to 468
424 to 449
1 ' (hammered)
Brass wire
Copper-plate, hammered
annealed
Gun-metal
Lead . .
Phosphor-bronze
Tin
Zinc
Leather
THE STRENGTHS, ELASTICITIES, AND WEIGHTS OF TIMBERS.
This table contains the results of the most recent and
most reliable experiments, but, generally speaking, only small
specimens of the material have been tested. It is found that
the strength, elasticity, and weight of a timber are affected by
the soil, age, seasoning, per cent of moisture, position in the
log, etc., and hence it is not surprising that specimens even
when cut out of the same log show results which often differ
very widely from the mean. Additional experiments on large
timbers are needed, and in each case should be accompanied
by a complete history of the specimen from the time of felling.
Description of
Timber.
Tensile
Strength
in tons
per sq. in.
Com-
pressive
Strength
in tons
per sq. in.
along
Fibres.
Shearing
Strength
in tons
per sq. in.
along
Fibres.
Young's
Modulus,
E
(in tons).
Coef-
ficient
of
Rigid-
*%
Coefficient
of Bending
Strength
in tons
per sq. in.
Weight
in Ibs.
per
cu. ft.
Acacia
Alder
4.5 to 6.3
3- J
Apple
88
Ash, Canadian..
Ash, English . ..
Beech
2-45
5.35107.58
4 9 to 9 8
2-5
3 4 6 7
.2 tO .312
620
723
607
47
43 to 53
Birch
6 60
Box "
4.6
807
Blue gum...
2. 7
3.078
509
c 86
Cedar
Chestnut
Ebony
Elm, Canadian..
2.23104.9
4-3
2.56
8.48
2.9
.308
217
509
IIOO
35 to 47
35 to 41
47
212
THEORY OF STRUCTURES.
THE STRENGTHS, ELASTICITIES, AND WEIGHTS OF TIMBERS
(Continued.}
Description of
Timber.
Tensile
Strength
in tons
per sq. in.
Com-
pressive
Strength
in tons
per sq. in.
along
Fibres.
Shearing
Strength
in tons
per sq. in.
along
Fibres.
Young's
Mod-lus,
E
(in tons).
Coef-
ficient
of
Rigid-
Coefficient
of Bending
Strength
in tons
per sq. in.
Weight
in Ibs.
per
cu. ft.
Elm, Eng
5-89
4.6
*
34 to 3;
Green heart ...
2.7 to 4.1
4.4610 6.5
759
58 to 72
Hawthorn
4.68
Hazel
8.48
Hornbeam
9.1
2.6
47*
Iron bark
7.12
4-54
8.15
Ironwood
4-3 1
5-21
Jarrah
3-2
4- J 3
Lancewood
3.6 to 6.7
42 to 63
Larch
3.92 to 4.55
1.42 to 2.45
32 to 38
Lignum vitae. . ..
5-26
4.46
446
7.18
41 to 83
Locust
4-5to6. 7
i-33
535
Mahogany, Span..
i. 7 to 7. 3
3-3
3-3
560 to 1339
53
'' Hond
1.3 to 3.6
712 to 879
35
Maple
4.7 to 7 . 7
2 23
N J
49
Mora
4-i
4-4
830
II
57 to 68
*0ak. Am
577
N-
2 712
Oak, Am red. ..
4-46
1.89 to 2.6
.324 to .446
.*
61
" white
8.8
2.84
335 to .431
tt
4 55
61
" Eng
5-4
4-4
664
*
49 to 58
Pine, Dantzic. ..
3-5
1025
3
36
" Memel
4-5
ij
34
" pitch
4.6
3-5
|
41 to 58
* " red
670
<*_
2 -93
34
" red
1.7 to 6.67
2. 4 to 3
962
5,
34
* " yellow
779
7*
3-255
41 yellow. . ..
2.2 tO 6.87
2.4 to 3.6
.227
900
4-Si
32
* " white
484
2.146
" white . .
1.3 to 5.1
2.24
. 119 to . 164
3-03
3
Plane
5 4
604
40
Poplar
2 -94
i 8
34
23 to 26
* Spvuce
594
2.18
2.99 to 5.97
438 to 737
1.63 to 2.86
n
. 113 to . 167
700
29 to 32
Sycamore
Teak ..
5-8
4.7 to 6.7
3-16
5-35
464
IOOO
36 to 43
41 to 52
Walnut
3-5
2-7
38 to 57
Willow
4.6 to 6.25
629
24 to 35
*The results for these timbers are deduced from experiments carried out by Bauschinger
Lanza, and others, on comparatively large specimens.
THE BREAKING WEIGHTS AND COEFFICIENTS OF BENDING
STRENGTH IN TOMS (OF 2240 LBS.) OF VARIOUS RECT-
ANGULAR BEAMS, THE WEIGHTS HAVING
BEEN UNIFORMLY DISTRIBUTED.
Material.
Clear
Span
between
Supports
in inches.
Breadth
in
inches.
Depth
in
inches.
Mean
Breaking
Weight of
each Joist
or Beam.
Coef-
ficient of
Bending
Strength.
g
5.66
.48
142
3*
II
7.89.
.08
" * '* 2 beams
126
14
15
60.97
.83
126
46 6
.6
142
3i
II
8.29
.08
g
5.7
49
Pine (Baltic), 2 beams
Baltic redwood deal (Wyberg), 2 joists ...
Spruce deals (St. John), 3 pairs with bridg-
ing oieces. . .
126
142
142
'Si
3
3
*3*
9
9
58-43
5-75
6.81
.24
52
2.98
TABLES.
213
THE BREAKING WEIGHTS AND COEFFICIENTS OF BENDING
STRENGTH IN TONS (OF 2240 LBS.) OF VARIOUS RECT-
ANGULAR BEAMS LOADED AT THE CENTRE.
Material.
Clear
Span
between
Supports
in inches.
Breadth
in inches.
Depth
in inches.
Breaking
Weight
in tons.
Coef-
ficient of
Bending
Strength.
Remarks.
Yellow tine ......
120
14
15
38.15
2.34
V
I2q
14
15
34
2.09
>i <
AC
c
7
c .q
1.62
I
4.:
5
7
5.7
I .57
Old timber
.( <
4?
5
5
3. 1
1.6;
i
AC
c
5
a.O^
1.64
*i <
1C
a|
3i
.025
2.O4
(3ld timber
.1 i
AC
2i
oi
I -07^
2. i7
Pitch pine . .
120
J 4
IE
CQ.2=;
3.64
<
I2Q
14.
1C
60.2^
3.7
,,
4;
5
7
7.8
2. 14
,
AS.
c
7
Q.7C
2.68
4D
4C
e
7
jo.Ct;
2. 02
46
5
7
ii
3.O3
(,
4e
2*
U
i 6
q c2
|,
AC.
2*
^i
i . 35
2.Q7
AS.
C
7
7
I . QI
AS.
c
7
8.;
2 71
,,
AS.
2i
u
I 125
2.48
.< (f
AS.
2^
ol
I 2
o 6 i
American elm . .
AS.
7
MQ
j. i
Old timber
AS.
e
7
IS 6
i 2Q
K tt
AC
ol
1
25c
. e 8.1
<
t <
AS.
2i
7!
2 6
57 }
X
Greenheart
AS.
C
7 56
AS.
2A
7
1 1 .4.5
6.^1
,
AS.
i 8^
952C
,
AS.
24
^4
400
8 Si
.4
R j iJ pine . . .
45
139
147
i
9
6
3i
8
12
3-55
24-5
7 5
7.82
8.87
I OI
147
6
12
8.45
2 15
A^.5. The results contained in the last two tables are mainly deduced from experi-
ments carried out under the supervision of W. Le Mesurier, M.Inst.C.E., Dock Yard, Liver-
pool.
214 THEORY OF STRUCTURES.
THE WEIGHTS AND CRUSHING WEIGHTS OF ROCKS, ETC.
Material.
Weight per cu. ft.
in Ibs.
Crushing- Weight in
Ibs. per sq. in.
Asphalt
6
S 300
184
17 2OO
" Greenstone* ...
181
1 6 SOO
" Welsh
172
800 to 1,400
Brick common ...... .
550 to 800
2 2^O
" Sydney N S
2 2OO
" yellow-faced (Eng ) ...... .
ioo to 135
I 44O
" Staffordshire blue
7 2OO
fire
I,7OO
" pressed (best) .
jen
IO 2OO
112
Cement Portland . ....
86 to 04
i 700 to 6 ooo
" Roman ........
IOO
Clav
I IQ
IIQ
460 to 775
" in cement ... . . . > . . .
I "\1
Earth
77 to 12^
112
19,600
Freestone
3 ooo to 3 500
Glass flint
IQ2
27 5OO
ic,7
31,000
" common green . .
1*8
31 ooo
172
16-3
10 800
' " red
16*
' Cornish , . .
1 66
14 ooo
' Sorrel
167
12 8OO
Irish
10 450
U S (Quincy)
1 5 ooo
' Argyll
10 900
06 to 175
19,600
Limestone . . .
154 to 162
7 500 to 9 ooo
Lime quick. ...
c a
86 to 119
1 20 to 240
" (average)
1 06
Masonry common brick . . . )
500 to 800
Il6 to 14.4
760
" rubble )
_4_. of cut stone
Marble statuary ... ..
I7O
3 200
168 to 170
8,000 to 9 700
Oolite Portland stone . . . . .
T C T
4 ioo
" Bath stone
I2 1 ?
177
" river . .
117
" pit
IOO
QS,
Sandstone red (Eng )
T O-J
57OO
ICQ
3 ioo
jc6 to 157
5 , 700 to 6 ooo
" Scotch
TC q to 155
5 300 to 7 800
U. S
Sqoo
Shingle ....
83
Slate Anglesea
1 7Q }
IO OOO
IC7 I
to
" Welsh
180 1
24 OOO
Trao .
I7O
TABLES.
21$
ACTORS OF SAFETY.
GOOD ORDINARY WORK.
Timber 4 to 5 for dead load, 8 to 10 for live load.
Metals 3 " 6 " " "
Masonry 4 " " " 8 " " "
EXPANSIONS OF SOLIDS.
Materials.
Linear Expansion per Unit of Length.
Expansion
in Bulk.
From 32 F. to
212 F.
From 32 F. to
572 F.
From 32 F.
to 212 F.
Brass
.001868 = ^ s
.00182 ^ T
.001075 = *>h
.001718 = ifa
.00352 = ^VT
.00861 = TT V r
.001466 = ifo
.00181 = T i,
.00144 = ^ T
.0002848= sk
.000746 = ^Vtf
.000884 = TT-VT
.001909 = 3 fo
.001079 = ^ 7
.00124 = B^T
.002173 = -flv
.001235 = gl 7
.001182 = g-^
.002941 = ^
.003108 = si 3
.001883 = ^
.001468 = ^ T
.0065
.0054
.0033
.0055
.0027
.0057
.0036
.0066
.0036
.0058
Bronze .
Fir
Glass . .
Gold
I ron wire
Lead
Oak . ....
Silver
Steel, unhardened
" hardened . . .
Tin . .
\Vrought-iron (bar) .
" (tor smith-work)..
Zinc, cast
" hammered
2l6 THEORY OF STRUCTURES.
EXAMPLES.
1. How many square inches are there in the cross-section of an iron
rail weighing 30 Ibs. per lineal yard? How many in a yellow-pin.e beam
of the same lineal weight? Ans. 3 sq. in.; 45 sq. in.
2. A vertical wrought-iron bar 60 ft. long and I in. in diameter is
fixed at the upper end and carries a weight of 2000 Ibs. at the lower end.
Find the factors of safety for both ends, the ultimate strength of the
iron being 50,000 Ibs. per sq. in. Ans. i9 T 9 f ; i8 T 3 7 2 T .
3. A vertical rod fixed at both ends is weighted with a load w at an
intermediate point. How is the load distributed in the tension of the
upper and compression of the lower portion of the rod?
Ans. Inversely as the lengths.
4. 'Find the length of a steel bar of sp. gr. 7.8 which, when suspended
vertically, would break by its own weight, the ultimate strength of the
metal being 60,000 Ibs. per sq. in. Ans. 17,723 ft.
5. The iron composing the links of a chain is % in. in diameter; the
chain is broken under a pull of 10,000 Ibs. What is the corresponding
tenacity per sq. in. ? Ans. 57,272^ Ibs.
6. A vertical iron suspension-rod 90 ft. long carries a load of 20,000
Ibs. at its lower end ; the rod is made up of three equal lengths square
in section. Find the sectional area of each length, the ultimate tenacity
of the iron being 50,000 Ibs. per sq. in., and 5 a factor of safety.
7. If the rod in the previous question is of a conical form, what
should be the area of the upper end ? Also find the intensities of the
tension at 30 and 60 ft. from the lower end.
Ans. 2.0407 sq. in.; 9999.612 Ibs., 9999.605 Ibs. per sq. in.
8. The dead load of a bridge is 5 tons and the live load 10 tons per
panel, the corresponding factors of safety being 3 and 6. If the two loads
are taken together, making 9 tons per panel, what factor of safety would
you use ? Ans. 5.
9. The end of a beam 10 in. broad rests on a wall of masonry. If it be
loaded with 10 tons, what length of bearing surface is necessary, the safe
crushing stress for stone being 150 Ibs. per sq. in. ? Ans. 13^- in.
10. Find diameter of bearing surface at the base of a column loaded
with 20 tons, the same stress being allowed as in the preceding question.
Ans. \ '380. 1 2.
EXAMPLES. 217
1 1 . In the chain of a suspension-bridge five flat links dovetail with four
alternately, and a cylindrical pin passes through the eyes. The pull on
the chain is 200 tons. Find the area of the pin, the bearing strength of
the metal being 6 tons per sq. in. Ans. if sq. in.
12. An iron bar of uniform section and 10 ft. in length stretches
.12 in. under a unit stress of 25,000 Ibs. Find E.
Ans. 25,000,000 Ibs.
13. A ship at the end of a 6oo-ft. cable and one at the end of a 5oo-ft.
cable stretch the cables 3 in. and 2$ in., respectively. What are the cor-
responding strains? Ans. y^ff.
14. A rectangular timber tie is 12 in. deep and 40 ft. long. If E =
1,200,000 Ibs., find the proper thickness of the tie so that its elongation
under a pull of 270,000 Ibs. may not exceed 1.2 in. Ans. J\ in.
15. A wrought-iron bar 60 ft. long is stretched 5 in. by a pull of
5000 Ibs. Find its diameter, E being 25,000,000 Ibs. Ans. .59 in.
1 6. A wrought-iron rod 984 ft. long alternately exerts a thrust and a
pull of 52,910 Ibs. ; its cross-section is 9.3 sq. in. Find the loss of stroke,
E being 29,000,000 Ibs. Ans. 4.632 in.
17. A wrought-iron bar 2 sq. in. in sectional area has its ends fixed
between two immovable blocks when the temperature is at 32 F. If
E = 29,000,000 Ibs., what pressure will be exerted upon the blocks when
the temperature is 100 F. ? Ans. 27388! Ibs.
1 8. What should be the diameter of the stays of a boiler in which the
pressure is 30 Ibs. per sq. in., allowing one stay to each i sq. ft. of
surface and a stress of 3500 Ibs. per sq. in. of section of iron ?
Ans. 1 1 in.
19. A force of 10 Ibs. stretches a spiral spring 2 in. Find the work
done in stretching it successively i in., 2 in., 3 in., up to 6 in.
Ans. f, -V-, -V-, - 8 /-> -!*, ip- in. -Ibs.
20. A roof tie-rod 142 ft. in length and 4 sq. in. in sectional area is
subjected to a stress of 80,000 Ibs. If E = 30,000,000 Ibs., find the
elongation of the rod and the corresponding work.
Ans. 1. 136 in.; 3786! ft.-lbs.
21. An iron wire i in. in diameter and 250 ft. in length is subjected
to a tension of 600 Ibs., the consequent strain being ^^. Find E, and
show by a diagram the amount of work done in stretching the wire
within the limits of elasticity. Ans. 14,661,818^ Ibs.
22. A timber pillar 30 ft. in length has to support a beam at a point
30 ft. from the ground. If the greatest safe strain of the timber is 3-^,
what thickness of wedge should be driven between the head of the
pillar and the beam? Ans. T ^ ft.
2l8 THEORY OF STRUCTURES.
23. An hydraulic hoist-rod 50 ft. in length and i in. in diameter is
attached to a plunger 4 in. in diameter, upon which the pressure is
800 Ibs. per sq. in. Determine the altered length of the rod, E being
30,000,000 Ibs. Ans. .0213 ft.
24. A short cast-iron post is to sustain a thrust of icoo Ibs., the ul-
timate crushing strength of the iron being 80,000 Ibs. per sq. in. and 10
a factor of safety. Find the dimensions of the post, which is rectangular
in section with the sides in the ratio of 2 to i. Ans. 4 in.; 2 in.
25. The length of a cast-iron pillar is diminished from 20 ft. to 19.97
ft. under a given load. Find the strain and the compressive unit stress,
E being 17,000,000 Ibs. Ans. .0015 ; 25,500 Ibs. per sq. in.
26. A rectangular timber strut 24 sq. in. in sectional area and 6 ft. in
length is subjected to a compression of 14,400 Ibs. Determine the
diminution of the length, E being 1,200,000 Ibs. Ans. .003 ft.
27. Find the height from which a weight of 200 Ibs. may be dropped
so that the maximum admissible stress produced in a bar of i sq. in.
section and 5 ft. long may not exceed 20,000 Ibs. per sq. in., the co-
efficient of elasticity being 27,000,000 Ibs.
Ans. -fa ft., or, more accurately, -/-fa ft.
28. Find the H. P. required to raise a weight of 10 tons up a grade
of i in 12 at a speed of 6 miles per hour against a resistance of 9 Ibs. per
ton. Ans. 31.3.
29. A square steel bar 10 ft. long has one end fixed ; a sudden pull of
40,000 Ibs. is exerted at the other end. Find the sectional area of the
bar consistent with the condition that the strain is not to exceed yl^.
E = 30,000,000 Ibs. Find the resilience of the bar.
Ans. 2 sq. in. ; 533^ ft.-lbs.
30. How much work is done in subjecting a cube of 125 cu. in. of
iron to a tensile stress of 3000 Ibs. per sq. in. ? Ans. 1 1^ ft.-lbs.
31. A signal-wire 2000 ft. in length and -J- in. in diameter is subjected
to a steady stress of 300 Ibs. The lever is suddenly pulled back, and the
corresponding end of the wire moves through a distance of 4 in. De-
termine the instantaneous increase of stress. Ans. 51 Iff Ibs.
32. If the total back-weight is 350 Ibs., what is the range of the sig-
nal end of the wire ? Ans. T ^f^ ft.
33. A steel rod of length L and sectional area A has its upper end
fixed and hangs vertically. The rod is tested by means of a ring weigh-
ing 60 Ibs. which slides along the rod and is checked by a collar screwed
to the lower end. A scale is marked upon the rod with the zero at the
fixed end. If the strain in the steel is not to exceed T ^- Gt what is the
reading from which the weight is to be dropped ? What should be the
reading of the collar? E = 35,000,000 Ibs.
Ans. Distance from point of suspension = (fJ \\A)L ; y
EXAMPLES. 219
34. A load of looolbs. falls i in. before commencing to stretch a sus-
pending rod by which it is carried. If the sectional area of the rod is
2 sq. in., length 100 in., and E = 30,000,000 Ibs., find the stress pro-
duced. Ans. 17,828 Ibs. per sq. in.
35. If the rod carries a load of 5000 Ibs., and an additional load of
2000 Ibs. is suddenly applied, what is the stress produced ?
Ans. 4500 Ibs. per sq. in.
36. Steam at a pressure of 50 Ibs. per sq. in. is suddenly admitted
upon a piston 32 in. in diameter. The steel piston-rod is 48 in. in
length and 2 in. in diameter, E being 35,000,000 Ibs. Find the work
done upon the rod. Ans. 117.69 ft. -Ibs.
37. What should be the pressure of admission to strain the rod to a
proof of .001 ? Ans. 68ff Ibs. per sq. in.
38. A boulder-grappler is raised and lowered by a wire rope i in. in
diameter hanging in double sheaves. On one occasion a length of 150
ft. of rope was in operation, the distance from the winch to the upper
block being 30 ft. The grappler laid hold of a boulder weighing 20,000
Ibs. What was the extension of the rope, E being 15,000,000 Ibs. ?
Ans. T jo ft.
39. The boulder suddenly slipped and fell a distance of 6 in. before
it was again held. Find the maximum stress upon the rope.
Ans. 50,452-^ Ibs. per sq. in.
40. What weight of boulder may be lifted if the proof-stress in the
rope is not to exceed 25,000 Ibs. per sq. in. of gross sectional area?
Ans. 78.571! Ibs.
41. The steady thrust or pull upon a prismatic bar is suddenly re-
versed. Show that its effect is trebled.
42. A weight W is suspended by a spring, which it stretches. The
weight is further depressed i ft., when it is suddenly released and allowed
to oscillate. Find its velocity at a distance x from the position of
equilibrium. / ~
Ans. y 10(1 - icur 2 )-^.
43. If a spring deflects .001 ft. under a load of i lb., what will be the
period of oscillation of a weight of 14 Ibs. upon the spring?
44. Show that the change of a unit of volume of a solid body under
a longitudinal stress is Af i 1, which becomes if m = 4, as in metals,
and nil when m = 2, as in india-rubber (page 142).
45. A steel bar stretches YgVffth of its original length under a stress of
20,oco Ibs. per sq. in. Find the change of volume and the work done
per cubic inch. Ans. ^V^th ; f ft.-lb. per cu. in.
220 THEORY OF STRUCTURES.
46. During the plastic deformation of a prismatic bar, show that the
change in sectional area is proportional to the deformation calculated on
the altered length of the bar.
47. A prismatic bar of volume V changes in length from L to L x
.inder the " fluid pressure"/. Find the corresponding work.
Ans. / Flog.CZ .r).
48. Show that the total work done in raising a number of weights
through to a given level is the product of the sum of the weights and the
vertical displacement of their centre of gravity.
49. An engine has to raise 4000 Ibs. icooft. in 5 minutes. What is its
H. P. ? How long will the engine take to raise 10,000 Ibs. 100 ft. ?
Ans. 24/3- H. P. ; i^ min.
50. How many men will do the same work as the engine in the pre-
ceding question, assuming that a man can do 900,000 ft. -Ibs. of work in
a day of 9 hours ? Ans. 480 men.
51. Determine the H. P. which will be required to drag a heavy rock
weighing 10 tons at the rate of 10 miles an hour on a level road, the
coefficient of friction being 0.8. What will be the speed up a gradient
of i in 50, the same power being exerted ?
Ans. 477|f ; 9f| miles per hour.
52. Two horses draw a load of 4000 Ibs. up an incline of i in 25 and
1000 ft. long. Determine the work done. Ans. 160,000 ft.-lbs.
53. At what speed do the horses walk if each horse' does 16,000 ft.-
lbs. of work per minute ? Ans. 2 T 3 T miles per hour.
54. A wrought-iron rod 25 ft. in length and i sq. in. in sectional
area is subjected to a steady stress of 5000 Ibs. What amount of live
load will instantaneously elongate the rod by i in., E being 30,000,000
Ibs. ? Ans. 6250 Ibs.
55. Determine the shortest length of a metal bar a sq. in. in sec-
tional area that will safely resist the shock of a weight of W Ibs. falling
n. distance of h ft. Apply the result to the case of a steel bar i sq. in. in
sectional area, the weight being 50 Ibs., the distance 16 ft., the proof-
strain Tj-^-g-, and E = 35,000,000 Ibs.
lEWah
Ans. ^^ ---- ,/ being the safe unit stress ; ^H^ ft.
,
56. A shock of N ft.-lbs. is safely borne by a bar / ft. in length and a
sq. in. in sectional area. Determine the increased shock which the bar
will bear when the sectional area of the last wth of its length is increased
to ra. f i i \
Ans. N(i --- 1 -- .
\ m rm]
57. The bar in Example 12 is i sq. in. in section. Determine the work
stored up in the rod in foot-pounds and compare it with the work which
EXAMPLES. 221
would be stored up if for half its length the rod has its section increased
to 4 in. Ans. 125 ft.-lbs.; f of 125 ft.-lbs.
58. If 25,000 Ibs. per sq. in. is the proof-stress, find the modulus of
resilience for the i-in. rod. Ans. 25 in in.-lb. units.
59. A steel rod 100 ft. in length has to bear a weight of 4000 Ibs. If
E 35,000,000 Ibs., and if the safe strain is .0005, determine the sectional
area of the rod (i) when the weight of the rod is neglected ; (2) when
the weight of the rod is taken into account. Also in the former case,
determine the work done in stretching the rod -fa in., -$ in., y 3 ^ in., . . .
T 6 in., successively.
Ans. & sq. in. ; -fflfo sq. in. ; 33$, 133^, 300, ... 1200 in. -Ibs.
60 A line of rails is 10 miles in length when the temperature is at.
32 F. Determine the length when the temperature is at 100 F.. and
the work stored up in the rails, E being 30,000,000 Ibs.
Ans. 10.008 miles ; 10.24 H. P.
61. A wrought- iron bar 25 ft. in length and i sq. in. in sectional area
stretches .0001745 ft. for each increase of i F. in the temperature. If
E = 29,000,000 Ibs., determine the work done by an increase of 20 F.
How may this property of extension under heat be utilized in straight-
ening walls that have fallen out of plumb ? Ans. 7.064 ft.-lbs.
62. Find the work done in raising a Venetian blind, w being the
weight of a slat, a the distance between consecutive slats, and n the
number of slats. n(n + i)
Ans. wa ---- .
2
63. How many |-in. rivets must be used to join two wrought-iron
plates, each 36 in. wide and in. thick, so that the rivets may be as
strong as the riveted plates, the tensile and shearing strength of
wrought-iron being in the ratio of 10 to 9? Ans. 17 rivets (16.3).
64. A horizontal string, without weight, of length 20, and sectional
area S, has its two ends fixed in the same horizontal plane. A weight
W suspended from its centre draws the string slightly out of the hori-
zontal. Show that, approximately,
t being the intensity of the tension, d the depression, and E the coef-
ficient of elasticity.
65. A heavy wire of length 2a, sectional area S, and weight IV has its
ends fixed in a horizontal plane and is allowed to deflect under its own
weight. Find the deflection d and the tenacity / (assumed uniform
throughout).
222 THEORY OF STRUCTURES.
66. A length 270 ft. of wire i sq. in. in section and of sp. gr. 7.8 is
subjected to the above conditions. Find the tenacity of the wire and the
deflection, the coefficient of elasticity, E, being 25.300,000 Ibs.
67. A brick wall 2 ft. thick, 12 ft. high, and weighing 112 Ibs. per
cu. ft. is supported upon solid pitch-pine columns 9 in. in diameter,
10 ft. in length, and spaced 12 ft. centre to centre. Find the compress-
ive unit stress in the columns (i) at the head; (2) at the base. The tim-
ber weighs 50 Ibs. per cu. ft. Ans. 507.03 Ibs. ; 510.5 Ibs.
68. If the crushing stress of pitch-pine is 5300 Ibs. per sq. in. and the
factor of safety 10, find the height to which the wall may be built.
Ans. 12.46 ft.
69. Determine the diameter of the wrought-iron columns which
might be substituted for the timber columns in question 67, allowing a
working stress in the metal of 7500 Ibs. per sq. in. Ans. 2.36 in.
70. Find the greatest length of an iron suspension-rod which will carry
its own weight, the stress being limited to 4 tons per sq. in. What will
be the extension under this load, E being 12,500 tons?
Ans. 2700 ft. ; .864 ft.
71. A horizontal cast-iron bar i ft. long exactly fits between two verti-
cal plates of iron. How much should its temperature be raised so that
it might remain supported between the plates by the friction, the coef-
ficient of friction being -\ ? Ans, ^ F.
72 The fly-wheel of a 40 H. P. engine, making 50 revolutions per
minute, is 20 ft. in diameter and weighs 12,000 Ibs. What is its kinetic
energy?
If the wheel gives out work equivalent to that done in raising 5000
Ibs. through a height of 4 ft., how much velocity does it lose?
The axle of the fly-wheel is 12 in. in diameter. What proportion of
the H. P. is required to turn the wheel, the coefficient of friction being
.08?
If the fly-wheel is disconnected from the engine when it is making
50 revolutions per minute, how many revolutions will it make before it
comes to rest ?
Ans. 511,260.4 ft. -Ibs. ; 1.04 ft. per sec. ; |ths ; 169.4.
73. The velocity of flow of water in service-pipe 48 ft. long is 64 ft.
per sec. If the stop-valve is closed in -^ of a sec., find the increase of
pressure near the valve. Ans. 375 Ibs. per sq. in.
74. Work equivalent to 50 ft.-lbs. is done upon a bar of constant
sectional area, and produces in it a uniform tensile stress of 10,000 Ibs.
per 'sq. in. Find the cubic content of the bar, E being 30,000,000.
Ans. 360 cu. in.
75. A fly-wheel weighs 20 tons and its radius of gyration is 5 ft. How
EXAMPLES. . 223
much work is given out while the speed falls from 60 to 50 revolutions
per minute ? Ans. 94TT 3 6T ft.-tons,
76. The resilience of an iron bar i sq. in. in section and 20 ft. long is
30,000 ft.-lbs. What would be the resilience if for 19 ft. of its length it
was composed of iron 2 sq. in. in section, the remaining foot being the
same size as before? Ans. 8625 ft.-lbs.
77. A particle under the action of a number of forces moves with a
uniform velocity in a straight line. What condition must the forces
fulfil ? Ans. Equilibrium.
78. Determine the constant effort exerted by a horse which does
1,650,000 ft.-lbs. of work in one hour when walking at the rate of 2$
miles per hour. Ans. 125 Ibs.
79. A train is drawn by a locomotive of 160 H. P. at the rate of 60
miles an hour against a resistance of 20 Ibs. per ton. What is the gross
weight of the train ? Ans. 50 tons.
80. A train of 292! tons is drawn up an incline of I in 75, 5^ miles
long, against a resistance of 10 Ibs. per ton, in ten minutes. Find the
H. P. of the engine. The speed on the level, the engine exerting 769.42
H. P., is 43.4 miles per hour. What is the resistance in pounds per ton ?
Ans. 1027 H. P. ; 22.7 Ibs. per ton.
81. The dead load upon a short hollow cast-iron pillar with a sec-
tional area of 20 sq. in. is 50 tons (of 2000 Ibs.). If the strain in the
metal is not to exceed .0015, find the greatest live load to which the
pillar might be subjected, E being 17,000,000 Ibs. Ans. 255,000 Ibs.
82. A steel suspension-rod 30 ft. in length and \ sq. in. in sectional
area carries 3500 Ibs. of the roadway and 3000 Ibs. of the live load. De-
termine the gross load and also the extension of the rod, E being
35,000.000 Ibs. Ans. yf^ ft.
83. A steel rod 10 ft. in length and \ sq. in. in sectional area is
strained to the proof by a tension of 25,000 Ibs. Find the resilience of
the rod, E being 35,000,000 Ibs. Ans. 178* ft.-lbs,
84. What form does the useful work done by a hammer take when a
nail is driven into any material ? What becomes of the rest of the energy
of the mass of the hammer after striking the blow?
85. A hammer weighing 2 Ibs. strikes a steel plate with a velocity of
10 ft. per sec., and is brought to rest in .0001 sec. What is the average
force on the steel ? Ans. 6250 Ibs.
86. A hammer weighing lolbs. strikes a blow of 10 ft.-lbs. and drives
a nail 5 in. into a piece of timber. Find the velocity of the hammer at
the moment of contact, and" the mean resistance to entry. Also find the
steady pressure that will produce the same effect as the hammer.
Ans. 8 ft. per sec. ; 240 Ibs. ; 480 Ibs.
224 THEORY OF STRUCTURES.
87. When a nail is driven into wood, why do the blows seem to have
little if any effect unless the wood is backed up by a piece of metal or
stone ?
88. In Question 86, taking the weight of the nail to be 4 oz. and the
weight of the piece of timber to be 100 Ibs., find the depth and time of
the penetration (a) when the timber is fixed ; (b) when the timber is free
to move.
Also in case (b) find the distance through which the timber moves.
Ans. (a) ff in. ; -fa sec.
(b) .44245 in.; .0009448 sec. ; .04113 in.
\ 89. Show that the greater part of the energy of impact is expended
in local damage at high velocities, and in straining the impinging bodies
as a whole at low velocities.
90. A pile-driver of 300 Ibs. falls 20 ft., and is stopped in -fa sec. What
is the average force exerted on the pile ? Ans. 3344 Ibs.
91. A weight falls 16 ft. and does 2560 ft.-lbs. of work upon a pile
which it drives 4 in. against a uniform resistance. Find the weight of
the ram, and the resistance. Ans. 160 Ibs. ; 7680 Ibs.
92. A pitch-pine pile 14 in. square is 20 ft. above ground, and is
being driven by a falling weight of 112 Ibs. It E = 1,500,000 Ibs., find
the fall so that the inch-stress at the head of the pile may be less than
800 Ibs.
Supposing that the pile sinks 2 in. into the ground, by how much
would it be safe to increase the fall?
Ans. 7.456 ft. ; 116.5 ft-
93. A weight of W\ tons falls h ft., and by n successive instantaneous
blows drives an inelastic pile weighing W* tons a ft. into the ground.
Assuming the pile and weight to be inelastic, find (a) the mean effective
resistance of the ground.
If the ground-resistance increases directly as the depth of penetration,
find (b) how far the pile will sink nnder the rth blow. If the head of the
pile is crushed for a length of x ft., x being very small as compared with
CL
the depth of penetration, find (i) the mean thrust, during the blow,
n
between the weight and hammer ; (2) the time of penetrating the ground ;
(3) the time during which the blow acts.
W^ nh a
l^/ r 1 -|. w^ a > n
(0 ~j*7~T~iT7 T-- ( 2 ) TTr
94. An inelastic pile weighing 788 Ibs. is driven 3^ feet into tile
ground by 120 blows from a weight of 112 Ibs. falling 30 ft. Find the
EXAMPLES. 22$
steady load upon the pile which will produce the same effect, assuming
the ground-resistance to be (a) uniform ; (b) proportional to the depth of
penetration. If the resistance is uniform, how long (c) does each move-
ment of the pile last ? How many blows (d) are required to drive the
pile the first half of the depth, viz., if ft., the ground-resistance being
7168 Ibs. ? How far (e) does the pile sink under the last blow?
Ans. (a) 14,336 Ibs. ; (b) 28,672 Ibs. ; (c) .0107 sec. ; (d} 30 ;
(e) .016 in.
95. A steamer of 8000 tons displacement sailing due east at 16 knots
an hour collides with a steamer of 5000 tons displacement sailing at 10
knots an hour. Find the energy of collision if the latter at the moment
of collision is going (i) due west ; (2) north-west ; (3) north-east.
96. A hammer weighing 2 Ibs. strikes a nail with a velocity of 15 ft.
per sec., driving it in in. What is the mean pressure overcome by the.
nail ? Ans. 673 Ibs.
97. A beam will safely carry i ton with a deflection of i in. From
what height may a weight of 100 Ibs. drop without injuring it, neglecting
the effect of inertia? Ans. 11.2 in.
98. A rifle-bullet 45 in. in diameter weighs i oz. ; the charge of pow-
der weighs 85 grains; the muzzle-velocity is 1350 ft. per sec. ; the weight.
of the rifle is 9 Ibs. Neglecting the twist determine the energy of i Ib. of
powder. If the bullet loses \ of its velocity in its passage through the
air, find the average force of the blow on the target into which the bullet:
sinks i in.
If there is a twist of i in 20 in., find the charge to give the same
muzzle- velocity, the length of the barrel being 33 in.
99. A leather belt runs at 2400 ft. per minute. Find how much its
tension is increased by centrifugal action, the weight of leather being
taken at 60 Ibs. per cubic foot. Ans. 2of Ibs.
100. Find the centrifugal force arising from a cylindrical crank- pin
6 in. long and 3^ in. in diameter, the axis of the pin being 12 in. from
the axis of the engine-shaft, which makes 200 revolutions per minute.
How would you balance such a pin ? Ans. 55.02 Ibs.
101. The pull on one of the tension-bars of a lattice girder fluctuates
from 12.8 tons to 4 tons. If 24 tons is the statical breaking strength of
the metal, 15 tons the primitive strength, determine the sectional area,
of the bar, 3 being a factor of safety. Ans. 2.15 sq. in. (Launhardt) ;
1.87 sq. in. (Unwin).
102. The stress in a diagonal of a steel bowstring girder fluctuates
from a tension of 15.15 tons to a compression of 7.65 tons. If the
primitive strength of the metal is 24 tons and the vibration strength 12:
226 THEORY OF STRUCTURES.
ions, find the proper sectional area of the diagonal, 3 being a factor of
safety. Ans. 2.53 sq. in. (Weyrauch) ;
1.7 sq. in. (Unvvin), 40 tons per sq. in. being statical
strength.
103. A wrought-iron screw-shaft is driven by a pair of cranks set at
^ight angles. Neglecting the obliquity of the connecting-rods, and
assuming that the pull on the crank-pin is constant, compare the coef-
rficients of strength (a' and /) to be used in calculating the diameter of
a. he shaft. How is the result affected by the stopping of the engine ?
Ans. a' .82*; a' /.
104. Taking/ = EA. as the ordinary analytical expression of Hooke's
Law, find the value of the modulus of elasticity when calculated (i) from
the actual stress and the elongation per unit of initial length ; (2) from
the actual stress and the elongation per unit of stretched length.
Ans. (i) E+f; (2) E+f(i + A) 2 = +/(! + 2A), if A is small.
105. In a fly-wheel weighing 12,000 Ibs. and making 50 revolutions
per minute, the centre of gravity is one seventeenth of an inch out of the
centre. Find the centrifugal force. Ans. 50.4 Ibs.
106. In the preceding question, if the axis of rotation is inclined to
the plane of the wheel at an angle cot-'.ooi, find the centrifugal
couple, the radius of gyration being 10 ft. Ans. 1028.9 ft.-lbs.
107. A cylinder and a ball each of radius R start from rest and roll
down an inclined plane without slipping. If V is the velocity of trans-
lation after descending through a vertical distance N. show that
F 2 = $(2g-/i) in the case of the cylinder,
and
J7 2 = %(2gh) in the case of the ball.
108. A wheel having an initial velocity of 10 ft. per sec. ascends an
incline of i in 100. How far will the wheel run along the incline, neg-
lecting friction ? Ans. 232.9 ft.
109. A wrought-iron fly-wheel 10 ft. in diameter makes 63 revolutions
per minute. Find the intensity of stress on a transverse section of the
rim, disregarding the influence of the arms. If the wheel, which weighs
W Ibs., gives out work equivalent to that done in raising W through a
height of 5i ft. in i sec., what velocity will it lose ? If the axle of the wheel
is 10 in. in diameter and if .08 is the coefficient of friction, show that it
W
will take H. P. to turn the wheel.
2500
Ans. 16,335 Ibs. ; 3.6 ft. per sec.
1 10. If the earth be assumed to be spherical, how much heat would
be developed if its axial rotation were suddenly stopped, a unit of heat
corresponding to 778 ft.-lbs. ?
EXAMPLES: 227
Weight of mass of earth = io 21 x 6.029 tons; diameter of earth
= 8000 miles.
in. A body weighing 50 Ibs. is projected along a rough horizontal
plane, the velocity of projection being 100 ft. per sec. What amount of
\vork will have been expended when the body comes to rest ?
If the coefficient of friction is i, how much work is done against fric
i ion in 4 sees., and in what time will the body come to rest ?
Ans. 7763.9 ft. -Ibs. ; 2298! ft.-lbs ; 24115- sees.
112. A chain / ft. in length and a sq. in. in sectional area has one
end securely anchored, and suddenly checks a weight of W Ibs. attached
to the other end, and moving with a velocity of V ft. per sec. away from
the anchorage. Find the greatest pull upon the chain.
f aEW
Ans. Pull = V\
o
113. Apply this result to the case of a wagon weighing 4 tons and
worked from a stationary engine by a rope 3 sq. in. in sectional area.
The wagon is running down an incline at the rate of 4 miles an hour,
and, after 600 ft. of rope have been paid out, is suddenly checked by the
stoppage or reversal of the engine (E = 15,000,000 Ibs.).
Ans. 26,884 Ibs.
1 14. A chain / ft. in length and a sq. in. in sectional area has one end
attached to a weight of W Ibs. at rest, and at the other end is a weight
of n W Ibs. moving with a velocity of V ft. per second and away from
the first. Find the greatest pull on the chain.
Ans. Pull = Vi/ aEWn .
Y lg(n + i)
115. A dead weight of io tons is to act as a drag upon a ship to
which it is attached by a wire rope 150 ft. in length and having an effec-
tive sectional area of 8 sq. in. If the velocity of the floating ship is
20 ft. per second, and if its inertia is equivalent to a mass of 390 tons,
find the greatest pull on the chain (E = 15,000,000 Ibs.).
Ans. 208 tons.
116. (a) A train weighing 160 tons (of 2240 Ibs.) travels at 30 miles
an hour against a resistance of io Ibs. per ton. What H. P. is exerted ?
(b) With the same H. P. what will be the speed up a gradient of i
in loo?
(c) If the steam is shut off, how far will the train run before stopping
(i) on the incline; (2) on the level?
(d) If the draw-bar suddenly breaks, in what distance would the
carriages (100 tons in weight) be stopped if the brakes are applied im-
mediately the fracture occur?, the weight of the brake-van being 20 tons
and the coefficient of friction .2?
228 THEORY OF STRUCTURES.
(e) If the engine (weight = 60 tons) continued to exert the same
power after the fracture, what would be its ultimate speed ?
(/) What resistance would be required to stop the whole train after
steam is shut off, in 1000 yards on the level ?
Ans. (a) 128; (b) 9^ miles per hour; (c) (i) 199.2 ft.,
(2) 6776 ft. ; (d) 680.3 ft. on the level, 52.9 ft.
on the incline ; (e) 80 miles an hour on the
level, 24.6 miles on the incline ; (/) 22.58 Ibs.
per ton.
117. A 4-in. X 3-in. diameter crank-pin is to be balanced by two
weights on the same side of the crank ; the length of the crank is 12 in. ;
the engine makes 100 revolutions per minute; the distance of the C. of
G. of each weight from the axis of the shaft is 6 in. Find the weights.
118. A shaft is worked with cranks at 120. Assuming the pressure
on the crank-pin to be horizontal and constant in amount, compare the
coefficients of actual and ultimate strength to be used in calculating the
diameter of the shaft. Ans. a' = .507^.
119. In a horizontal marine engine with two* cranks at right angles
distant 8 ft. from one another, weight of reciprocating parts attached to
each crank is 10 tons, revolutions 75 per minute, stroke 4 ft. Find the
alternating force and couple due to inertia.
Ans. 54.2 tons; 216.8 ft.-tons.
120. An inside-cylinder locomotive is running at 50 miles an hour;
the driving-wheels are 6 ft. in diameter; the distance between the centre-
lines of the cylinders is 30 in., the stroke 24 in., the weight of one piston
and rod 300 Ibs., and the horizontal distance between the balance-
weights 4! ft. ; the diameter of the weight-circle is 4i ft. Find the
alternating force and couple, and also the magnitude and position of
suitable balance- weights.
Ans. 7871 Ibs; 9839 ft.-lbs. ; 106.5 Ibs. 5 2 7f*
121. The pressure equivalent to the weight of the reciprocating parts
of an engine is 3 Ibs. per sq. in. ; the stroke is 36 in. ; the number of
revolutions per minute is 45 ; the back-pressure is 2 Ibs. per sq. in.; the
absolute initial steam- pressure is 60 Ibs. per sq. in.; the rate of expansion
is 3. Find the pressure necessary to start the piston, and also the effec-
tive pressure at each -J- of the stroke.
122. An engine with a 24-in. cylinder and a connecting-rod = six cranks
= 6 ft., makes 60 revolutions per minute. Show that the pressure re-
quired to start and stop the engine at the dead-points = ^ of the weight
of reciprocating parts.
123. Find the ratio of thrust at cross-head to tangential effort on
crank-pin when the crank is 45 from the line of stroke, the connecting-
rod being = four cranks.
EXAMPLES, 229
124. Draw the linear diagram of crank-effort in the case of single
crank, the connecting-rod being = four cranks. Assume the resistance
uniform and a constant pressure of 9000 Ibs. on the piston, the stroke
oeing 4 ft. and the number of revolutions per minute 55. Also find the
ductuation of energy in ft.-lbs. for one revolution.
125. An engine with a connecting-rod = six cranks = 6 ft. receives
steam at 70 Ibs. pressure per sq. in., and cuts off at one-quarter stroke.
Find the crank-effort when the piston has travelled one third of its for-
ward stroke. Diameter of piston = 2 ft. Also find the position of the
piston where its velocity is a maximum.
126. Data: Stroke = 3 ft. ; number of revolutions per minute = 60;
cut-off at one-half stroke; jnitial pressure = 56 Ibs. per sq. in. absolute;
diameter of piston = 10 in. ; weight of reciprocating parts = 550 Ibs. ;
back-pressure = i\ Ibs. per sq. in. absolute. Find the effective pressure
at each fourth of the stroke, taking account of the inertia of the piston.
Also find the pressure equivalent to inertia at commencement of
stroke.
127. A pair of 250 H. P. engines, with cranks at 90, and working
against a uniform resistance and under a uniform steam-pressure, are
running at 60 revolutions per minute. Assuming an indefinitely long
connecting-rod, find the maximum and minimum moments of crank-
effort, the fluctuation of energy, and the coefficient of energy.
128. An inside-cylinder locomotive runs at 25 miles per hour; its
drivers are 60 in. in diameter; the stroke is 24 in. ; the distance between
tne centre-lines of the cylinders = 30 in. ; weight of reciprocating
parts = 500 Ibs. ; horizontal distance between balance-weights = 59 in. ;
diameter of weight-circle = 42 in. Find the alternating force, alternat-
ing couple, and the magnitude and position of suitable balance-weights.
Ans. 226.8 Ibs.; 4113.8 ft.-lbs. ; = 26.
129. Draw a diagram of crank-effort for a single crank, the connect-
ing-rod being equal to four cranks, the stroke 4 ft., and the number of
Devolutions per minute 55. Assume a uniform resistance and a constant
ressure of 9000 Ibs. on the piston.
130. A vertical prismatic bar of weight Fi , sectional area A, and
length L has its upper end fixed, and carries a weight W* at the lower
end. Find the amount and work of the elongation.
Ans. Ext. = J^(-Y + W*\ ; work = *g( + ^ ^ +
131. A right cone of weight W and height h rests upon its base of
radius r. Find the amount and work of the compression.
. _ Wh i W*
Ans. Comp. = ; work = ~ =-..
2itEr* 8 Tthr*
132. A tower of height h, in the form of a solid of revolution about a
vertical axis, carries a given stir charge. If the specific weight cf the
230 THEORY OF STRUCTURES.
material of the tower is w, and the radius of the base a, determine the
curve of the generating line so that the stress at every point of the tower
may be/. If the surcharge is zero and the height of the tower becomes
infinite, show that its volume remains finite.
_-wx f
Ans. y = ae *f ; vol. of tower of infinite height = fta*.
w
133. Determine the generating curve when the tower in the last
question is hollow, the hollow part being in the form of a right cylinder
upon a circular base of given radius R.
Ans. y* - K* = (a? - K')e ~ T.
134. A heavy vertical bar of length / and specific weight w is fixed
at its upper end and carries a given weight Wat the lower end. Deter-
mine the form of the bar so that the horizontal sections may be pro-
portionate to the stress/ to Which they are subjected. (Note. Such a
bar is a bar of uniform strength.)
W, J (l ~ x]
Ans. Sectional area at distance x from origin = e
135. Find the upper and lower sectional areas of a steel shaft of uni-
form strength, 200 ft. in length, which will safely sustain its own weight
and 100 tons, 7 tons per sq. in. being the working stress.
Ans. 14.3 sq. in. ; 17.8 sq. in.
136. A vertical elastic rod of natural length L and of which the mass
may be neglected, is fixed at its upper end and carries a weight W\ at
the lower end. A weight W 2 falls from a height k upon W\ . Find the
velocity and extension of the rod at any time /,
")-()'
x being measured from mean position of ( W\ + W a ).
137. Determine the functions /''and/ in Art. 24 when P\ is zero, and
also when the rod is perfectly free ; i.e., when Po = o and Pi = o.
138. An elastic trapezoidal lamina ABCD, of natural length / and
thickness unity, has its upper edge AB (20) fixed and hangs verti-
cally. If a weight Wis suspended from the lower edge CD (2$), show
that, neglecting the weight of the lamina, the consequent elongation
= --TT ^7~llgT- If an additional weight is placed upon W and
then suddenly removed, show that the oscillation set up is isochronous
and that the time of a complete oscillation = n\ -=r 7- f .
Examine the case when a = b.
Ans. Ext. = ; time of oscillation = 7r,4/____
2<* Y
EXAMPLES. 231
139. If the specific weight of the lamina in the preceding question is
w, find how much it will stretch under its own weight, and also the
work of extension. Determine the result when a b.
i wbT b wra + d wl*
AnS ' 2~E ^7 10g ~a + iff ^ ; 2^'
ivl z ( 4 b* 2, a } "dFal*
= ^r^Jr \ - + f (a ~ d> - b !og < ? \ '' -W-
140. An elastic lamina in the form of an isosceles triangle ABC has
its base AB (= 20) fixed and hangs vertically. If its weight is W, find
its elongation. Take coefficient of elasticity = , thickness of lamina
= unity, and L the distance of C from AB. WL
Ans. --
141. A metal rod sq. in. in area and 5 ft. long hangs vertically with
its upper end fixed and carries a weight of 18 Ibs. at the lower end. On
striking the rod it emitted a musical note of 264 vibrations per second
(middle C of piano-forte). Find the coefficient of elasticity, the weight
of the rod being neglected. Am. 30,979,160 Ibs.
142. Diameter of a pipe is 18 in. ; at one point it is curved to an arc
of 6 ft. radius. Water flows round the curve with a velocity of 6 ft. per
second. Determine the centrifugal force per foot of length of elbow-
measured along the axis. Ans. 124.3 Ibs.
143. A disk of weight Fand area A sq. ft. makes n revolutions per
second about an axis through its centre, inclined at an angle to the
normal to the plane of the disk. Find the centrifugal couple.
WAn*
Ans. -- tan 6 ft. -Ibs.
5.12
144. In a circular pipe of internal radius r and thickness /, a column
of water of length /, flowing with a velocity due to the head h, is sud-
denly checked. Show that
E being the coefficient of elasticity of the material of the pipe, E\ the
coefficient of compressibility of the water, and A the extension of the
pipe circumference corresponding to E.
145. A heavy ball attached by a string to a fixed point O revolves in
a horizontal circle with a given uniform angular velocity GO. Find the
vertical depth of the centre of the ball below the point of attachment.
If a uniform rod be substituted for the ball and string, find its
position.
Also find the position when the ball is attached to the fixed point by
232 THEORY OF STRUCTURES.
a uniform rod ; r being the ratio of the weight of the rod to the weight
of the ball.
n
i +
GO"* ' i c3 2 ' Gfl 2 n
i +
3
146. The deflection of a truss of / ft. span is / x .001 under a station-
ary load W. What will be the increased pressure due to centrifugal force
when W 7 crosses the bridge at the rate of 60 miles an hour? ..-, ,
242 W
Ans. - - r-.
125 /
147. A fly-wheel 20 ft. in diameter revolves at 30 revolutions per
minute. Assuming weight of iron 450 Ibs. per cu. ft., find the intensity
of the stress on the transverse section of the rim, assuming it unaffected
by the arms. Ans. 96 Ibs. per sq. in.
148. Assuming 15,000 Ibs. per sq. in. as the tensile strength of cast-
iron, and taking 5 as a factor of safety, find the maximum working speed
and the bursting speed for a cast-iron fly-wheel of 20 ft. mean diameter
and weighing 24,000 Ibs., the section of the rim being 160 sq. in.
149. A 6o-in. driving-wheel weighs 3^ tons, and its C. of G. is i in.
out of centre. Find the greatest and the least pressure on the rails.
150. A wheel of weight W, radius of gyration k, and making n
revolutions per second on an axle of radius/?, comes to rest after having
made N revolutions. Find the coefficient of friction.
Ans. sin <f> = , and coeff. of fric. = tan 0.
Ng
151. A train starts from a station at^f and runs on a level to a station
at B, I ft. away. If the speed is not to exceed v ft. per sec., show that
the time between the two stations is
l_ Wv P + B
"r
v g 2 (P-R)(B + R)'
W being the gross weight of the train, P the mean uniform pull exerted
by the engine, R the road resistance, and B the retarding effect of the
brakes.
Also, if the speed is not limited, show that the least time in which
the train can run between the specified points is
^ W_ P + B
' ~i (P R)(B+R)
and that the maximum speed attained is
w ?, s --*-v~*
EXAMPLES. 233
152. A locomotive capable of exerting a uniform pull of 2 tons, with
a 24-in. stroke, 2o-in. cylinder, and 6o-in. driving-wheels, hauls a train
between two stations 3 miles apart. The gross weight of the train and
locomotive = 200 tons; the road resistance = 12 Ibs. per ton (of 2000
Ibs.) ; the brakes, when applied, press with two thirds of the weight on
the wheels of the engine and brake-van, viz., 90 tons, the coefficient of
friction being .18. Find (a) the least time between the stations; (b) the
distance in which the train is brought to rest; (c) the maximum speed
attained ; (d) the pressure of the steam ; (e) the weight upon the driving-
wheels.
Ans. 0) 513.8 sec. ; (6) 990 ft.; (c) 42 miles per hour; (d) 25
Ibs. per sq. in. ; (<?) i \\ tons.
153. If the speed in the last question is limited to 30 miles an hour,
find (a) the time between the stations; ()the distance in which the train
is brought to rest ; (c) the distance traversed at 30 miles an hour.
Ans. (a) 543i sec. ; (b) 504^ ft. ; (c) 7773^ ft.
154. If the steam-pressure in the above locomotive is increased to 50
Ibs. per sq. in., find (a) the weight of the heaviest train which can be
hauled between the stations in 10 minutes, the road-resistance being 20
Ibs. per ton (of 2000 Ibs.) and the braking power being sufficient to bring
the train to rest in a distance of 720 ft.
Also find (b) the braking power; (c) the weight thrown upon the
drivers, the coefficient of friction being % ; (d) the maximum speed
attained.
Ans. (a) 310^ tuns; (b) 15. 6 tons; (c) 24 tons; (d) 36 miles per hour.
155. The weight upon the driving-wheels (D in. in diameter) of a
locomotive is W tons; the adhesion = one fifth ; the cylinders have a
diameter of d in. and a stroke of / in. Find the steam-pressure re-
quired to skid the wheels. IV D
Ans. 400- Ibs. per sq. in.
156. Two trains, each with a brake-power of 190 Ibs. per ton (of 2000
Ibs.), run between Montreal and Toronto, a distance of 333 miles, against
an average resistance of 10 Ibs. per ton. One train runs through, and
the other stops at N intermediate stations. Show that the saving of
QN
fuel in the former is per cent ; the speed is not to exceed 30 miles
per hour.
157. If the end of a railway wagon exposes a surface of 6 x 4 ft. to
the wind, what is the greatest gradient up which a 20 Ib. to the sq. ft.
gale will drive it? Take the weight at 10 tons, the friction 10 Ibs. per
ton. Ans. I in 59.
1 58. A locomotive and tender weigh 70 tons, of which 26 tons are car-
ried by the driving-wheels. Taking the adhesion at , friction 10 Ibs. per
ton, what maximum gradient can the engine ascend? Ans. I in 16.
234 THEORY OF STRUCTURES.
1 59. Given a locomotive with two 18" x 26" cylinders, the connecting-
rod = 6 ft., the boiler-pressure = 140 Ibs., and driving-wheels of 7' o"
force at periphery
diameter, calculate the adhesion-friction, i.e., the ratio -- : - .
weight on drivers
160. A railway wagon weighing 20 tons, with two pairs of wheels
8' o" centre to centre, and with its centre of inertia 7' o" above top of
rails, has its wheels skidded while running. Take jn = 0.15. Required
the total retarding force and pressure of each wheel.
Ans. 7.375 ; 12.625, an d 3 tons on rail.
161, Find (a) the least time in which a locomotive exerting a uniform
pull of P tons can haul a train weighing W tons between two stations
/ ft. apart on an incline of I in m, the brake-power being B tons and the
road-resistance R tons.
Also find (V) the time between stations when the speed is limited to
v ft. per sec.
p + l Wv P + B
W
where A R + .
m
162. A locomotive exerting a uniform pull of 4 tons hauls a train of
200 tons up an incline of i in 200, between two stations 2 miles apart,
the greatest allowable speed being 30 miles an hour. If the road-resist-
ance is 10 Ibs. per ton (of 2000 Ibs.), and if the brakes are capable of ex-
erting a pressure of 100 tons, the adhesion being one fifth, find (a) the
time between the stations; (b) the distance in which the train is brought
to rest; (c) the distance traversed at 30 miles.
Also, if the speed is not limited to 30 miles, find (d) the least time in
which the distance can be accomplished ; (e) the maximum speed attained ;
(/) the distance in which the train is brought to rest.
Ans. (a) 5i min. ; (ff) 275 ft.; (c) 7260 ft.; (d) 4.87 min.;
(e) 53.8 miles per hour; (/) 880 ft.
163. With the same brake-power, adhesion, and road-resistance, find
the weight of the heaviest train which the locomotive in the preceding
question, exerting the uniform pull of 4 tons, can haul between the two
stations in 6 minutes. Ans. 360 tons.
164. If the locomotive has 6o-in. drivers and 24-in. x 2o-in. diameter
cylinders, find the weight required upon the drivers when the steam-
pressure is 50 Ibs. per sq. in. Ans. 20 tons.
CHAPTER IV.
STRESSES, STRAINS, EARTHWORK AND RETAINING-
WALLS.
1. Internal Stresses. The application of external forces
to a material body will strain or deform it, and the particles
of the body will be in a state of mutual stress.
In the following calculations it is assumed :
(a) That the stresses under consideration are parallel to one
and the same plane, viz., the plane of the paper.
(b) That the stresses normal to this plane are constant in
direction and magnitude.
(c) That the thickness of the plane is unity.
Def. The angle between the direction of a given stress and
the normal to the plane on which it acts is called the obliquity
of the stress.
2. Simple Strain. The solid ABCD (Fig. 207) of uniform
transverse section A is acted upon in the direction of its length
by a force P uniformly distributed over its end,
p
producing an intensity of stress - =/. At any |P
m
k/
\o/
-j/
other transverse section mn the intensity must be
the same in order that equilibrium may be main-
tained.
Draw an oblique plane m'n' , inclined at an
angle 6 to the axis. The total stress on m'n' = P ?
and necessarily acts in the direction of the axis. m
p
The intensity of the stress on m'n' = =
m'n'
p p
a = -T sin 6 = p sin 6. The normal com- FlG - 2 7-
mn cosec A
ponent of the intensity on m'n' = p sin a 6 = p n '.
235
c
236 THEORY OF STRUCTURES.
The tangential component or shear on m'ri
p sin 8 cos 8 = //.
^0, if m"n" is an oblique plane perpendicular to m'n' , the nor-
iiial component of the intensity on m"n" = p cos 2 8 = p n ".
The tangential component or shear on m"n"
= p cos 8 sin 8 = p" .
... A ' + p" = p and p{ = p;' = p sin 8 cos = ^5ll?.
The shear is evidently a maximum when 28 = 90 or
9 = 45.
3. Compound Strain. (a) First consider an indefinitely
small rectangular element OACB (Fig. 208) of a strained body,
p p p kept in equilibrium by stresses
\ \ v \ \ acting as in the figure.
^ ^ ^ L / is the intensity of stress on
\^ the faces OB, AC, and a its ob-
x^g liquity.
\ \ \ x \ B q is the intensity of stress on
\ \ \ \ the f aces OA, BC, and /? its ob-
FIG. 208. liquity.
OB .p cos a, the total normal stress on OB, is balanced by
AC .p cos a, the total normal stress on AC.
OB .p sin a, the total shear on OB, is equal in magnitude
but opposite in direction to AC .p sin a, the total shear on AC.
These two forces, therefore, form a couple of moment;
JB ./sin ex . OA.
Similarly, the total normal stresses on the faces OA,
BC balance and the total shears form a couple of moment
OA . q sin /3 . OB.
In order that equilibrium may be maintained the two
couples must balance.
.-. OB .p sin a . OA = OA . q sin /? . OB,
or
/ sin a q sin ft /, suppose.
COMPOUND STRAIN.
237
Hence, at any point of a strained body, the intensities of the
shears on any two planes at right angles to each other are equal.
(b] Next consider an indefinitely small triangular element
OAB (Fig. 209) of the strained body, bounded ' c
by a plane AB and two planes OA, OB at
right angles to each other.
Let / be the intensity of stress on OB,
a its obliquity.
Let q be the intensity of stress on OA,
/3 its obliquity. /\
Let / be the intensity of shear on each of B D
the planes OA, OB. Then FlG - 2 9 '
/ p sin a = q sin ft.
p n , the normal component of p, = p cos a.
q n , ' " " " q, = q cos ft.
Produce OA and take OC = p n . OB + t . OA = the total
force on OB in the direction of OA.
Produce OB, and take OD q n . OA + t . OB the total
force on OA in the direction of OB.
Complete the rectangle CD.
OE represents in direction and magnitude the resultant of
the two forces OC, OD, and must therefore be equal in magni-
tude and opposite in direction to the total stress on AB.
Let/ r be the intensity of stress on AB. Then
(p r . AB}* = OE* = OC* +
.fOB\* JOAy
.OA?
+ (q n .OA+t. 03? ;
OA . OB
or A- = A' < v -^j + q: \j- B ) + ** -jff-i Pn + ft) T ;
Let x be the angle between y4.# and OA. Then,
/V 2 = / w a sin 2 y + ? M 2 cos a y + 2t sin r cos 7 (p n + q n ) + / a .
This gives the intensity of stress on any plane AB inclined
at an angle y to OA, and in the limit AB is a plane through O.
238 THEORY OF STRUCTURES.
EXAMPLE. Consider an indefinitely small triangular ele-
ment abc (Fig. 210) of a horizontal beam bounded by a plane
- a b t.ab*8.db be inclined at to the vertical, the
/'^ horizontal plane ab, and the ver-
'
ac.
tf The element abc is kept in equ
FIG. 210. librium by the stress/ . ac upon ac,
the shear s.ab (= t.ab) along ad, the shear t .ac along. ac, and
the stress developed in the plane be. The weight of the element
is neglected as being indefinitely small as compared with the
forces to which it is subjected. Let the stress upon be be
decomposed into two components, the one X '. be normal and
the other Y. be tangential to be.
Resolving perpendicular and parallel to fc,
X .be = p .ac cos t . ab cos t . ac sin 6
and
Y. be = p . ac sin 6 t . ab sin B + t . ac cos 6,
or
X = PCQS* 6 t sin 20 . ....... (i)
and
The value of B for which X is a maximum is given by
- = o = / sin 26 2t cos 20, or tan 26 = -- . (3)
Substituting the value of B in eq. I, we have
max. value of X = - + A / + f. . . (4)
The value of B for which Y is a maximum is given by
dY P
o =/ cos 2/9 2t sin 20, or tan 2B = . (5)
COMPOUND STRAIN. 239
Substituting the value of 6 in eq. (2), we have
max. value of Y = \/- + t* , (6)
Eq. (4) gives the maximum intensity of stress of the same
kind as p. The maximum intensity of the opposite kind of
stress
o. x/ 4
Eq. (6) gives the maximum intensity of shear.
The position of the planes of principal stress (see following
2t
article) is given by tan 20 .
Let 0j , 2 be the values of for which X and Kare respec-
tively maxima. Then
tan 20, tan 20,= -'J*?**
and
*,-*, = 45-
Hence, at any point, the angle between the plane upon
which the normal intensity of stress is a maximum and the
plane upon which the tangential intensity of stress is a maxi-
mum, is equal to 45.
Again, t is zero when 0, = 90 or o, and p is zero when
*, = 45.
Thus, the curve of greatest normal intensity cuts the neutral
axis at an angle of 45, one skin surface at 90 and the opposite
at o, while the curve of greatest tangential intensity cuts
the skin surfaces at 45, and touches the neutral axis.
Fig. 211 serves to illustrate the curves of greatest normal
intensity. There are evidently two sets of these curves, re-
ferring respectively to direct thrust and direct tension.
240 THEORY OF STRUCTURES.
Fig. 212 illustrates the curves of greatest tangential in-
tensity.
V-
FIG. 211. FIG. 212.
4. Principal Stresses. Suppose that there is no shear
on AB, Fig. 209, and that the stress is wholly normal.
In such a case OE must be perpendicular to AB.
OC OC Pn-OB + t.OA
q n + t tan y
It
qnpn i tan* x
^E Two values of y satisfy this equation, viz.,
/\ y and y + 90.
Hence, at any point of a strained body, there
are two planes at right angles to each other, on
which the stress is wholly normal.
Such planes are called planes of principal
stress, and the stresses themselves principal
stresses.
5. Ellipse of Stress. At any point of the
strained body, consider a small triangular ele-
ment OAB (Fig. 213), OA and OB being the planes of principal
stress.
Let /, be the principal stress normal to OB.
" " " " " " " OA.
CONSTANT COMPONENTS OF p r .
241
Complete the construction as before, and let $ be the angle
between OE and OC. Then
CE OD p,OA
-
A...
(8)
p r sin w ., p r cos m
= i , sin v = ; and
cos =
A
(9)
Take 6^7? to represent / r in direction and magnitude.
Let X, Y be the co-ordinates of R with respect to O. Then
= A cos , = / r sn
and eq. (9) becomes
the equation to an ellipse with its centre at O, and its axesr
(equal to 2/>, and 2/> 2 ) lying in the planes of principal stress.
This ellipse is called the ellipse of stress, and the stress on any
plane AB at O is the semi-diameter of the ellipse drawn in a di-
rection making an angle ip with the axis OC, fy being given by
tan i/j = cot y. (Eq. (8).) . . . . (n)
6. Constant Components of p r . Take the planes of
principal stress as planes of reference (Fig. 214).
F
A
FIG.
242 THEORY OF STRUCTURES.
Draw ON perpendicular to AB, and take ON -^
Let the obliquity of OR = = RON = 90 - </> - y.
Join NR. Then
NR = OR" + ON 2 - 2OR . ON cos
= Pr + (^~} - A(A + A) sin (0 + 7
But A 8 = A 2 sin2 r +A 2 cos ' x> and
sin (^ + y) = sin ^'cos y + cos ^ sin y = ^ cos a y + -- sin" y.
(See eqs. (8).)
.-. NR* = A a sin 8 y + A' cos 2 r + ( 2
-(A+A)(Asin 8 y+Acos 8
(A+AV . . fA-AV
= V T ) -AA- I 2 |-
(12)
Hence, /^^ intensity of stress OR at any point O of the plane
A OB is the resultant of two constant intensities
and
the former being perpendicular to the plane.
THE ANGLE ONR. 243
7. The Angle ONK = 2y.
sin ONR OR p r
sin ~ NR ~~ p l A'
2
But
sin = cos (& + y) = cos ^ cos y sin ^ sin y
= ^~ sin r cos = sih 2 *
sn
2A 2
.-. sin OA 7 ^? = sin 2y, or (7^7? 2y. . . (13)
Let NR (Fig. 214) produced in both directions meet OA in
F and O in (7.
The angle OFN = 180 - ONR - NOF
= 180 - 2y - (90 - Y) = 90 - y = FON.
.'. NF = NO ; so, NG = NO =
.'. A 7 " is the middle point of F.
Also
RF=FN- NR = ON- NR =
and
RG = RN+NG = RN+ ON =
N. B. The shear at O
A -A
cos (2x - 90) = (A -A) sin Y cos X -
244 THEORY OF STRUCTURES.
8. Maximum Shear. ON has no component along AB.
Hence, the shear on AB is NR cos (angle between NR and
AB), and is evidently a maximum when the angle is nil. Its
value is then NR, or Pl ~ P \
9. Application to Shafting. At any point in a plane sec-
tion of a strained solid, let r be the intensity of stress, and 6
its obliquity.
At the same point in a second plane let s be the intensity
of stress, and 0' its obliquity.
By Art. 6, r and s are the resultants of two constant
stresses
and
+ /-/, + A) cos*'.
Subtracting one equation from the other,
First. Consider the case of combined torsion and bending,
as when a length of shafting bears a heavy pulley at some point
between the bearings.
Let p be the intensity of stress (compression or tension)
due to the bending moment M b .
APPLICATION TO SHAFTING.
245
Let q be the intensity of shear due to the twisting mo-
ment M t .
p and q act in planes at right angles to each other.
.-. r cos 6 = p y r sin 6 = q s, and 6' = 90.
/. r a =/ a + * and s = q.
Hence, by eq. (16),
and by eq. (15),
and
The max. shear =
__ A -
-V?
+2*;.
also
(17)
(19)
(20)
(21)
= (Chap. VI.) and f = r (Chap. IX.)
for a shaft of radius r.
t '\; . . . (22)
246
and
THEORY OF STRUCTURES.
(23)
&==(-
Perhaps the most important example of the application of
the above principle is the case of a
shaft acted upon by a crank (Fig. 215).
A force P applied to the centre C of
the crank-pin is resisted by an equal
and opposite force at the bearing JB,
forming a couple of moment P. CB = M.
This couple may be resolved into a
FIG. 215. bending couple of moment M b = P. AB
= P. BC cos d = J/cos tf, and a twisting couple of moment
M t = P . AC = P . BC sin d = M sin d ; tf being the angle
ABC.
', . . (24)
2M
and the max. shear = ,
nr
(25)
If the working tensile or compressive stress (/>,) and the
working shear stress ( ? ) are given, the corresponding
values of r may be obtained from eqs. (22) and (23) or eqs.
(24) and (25) ; the greater value being adopted for the radius
of the shaft.
Second. Consider the case of combined torsion and tension
or compression.
Let the tensile or compressive force be P.
P
p, the intensity of the tension or compression, = - t ;
<7, " " " shear = -5-4 .
PRINCIPAL AND CONJUGATE STRESSES. 247
and
10. Conjugate Stresses. Consider the equilibrium of an
indefinitely small parallelepiped
abed (Fig. 216) of a strained body,
the faces ab, cd being parallel to
the plane XOX, and the faces ad,
be to the plane YOY.
Let the stresses on ab, cd act
parallel to the plane YOY. The
total stresses on ab and cd are
equal in amount, act at the centres of the faces, are parallel to
YOY, and therefore neutralize one another.
Hence the total stresses on ad and be must also neutralize
one another. But they are equal in amount, and act at the
middle points of ad, be; they must therefore be parallel to
XOX.
Hence, if two planes traverse a point in a strained body,
and if the stress on one of the planes is parallel to the other
plane, then the stress on the latter is parallel to the first
plane.
Such planes are called planes of conjugate stress, and the
stresses themselves are called conjugate stresses.
Principal stresses are of course conjugate stresses as well.
Conjugate stresses have equal obliquities, each obliquity
being the complement of the same angle.
n. Relations between Principal and Conjugate
Stresses (Fig. 217). Take any line ON =
248 THEORY OF STRUCTURES.
P P
With N as centre and a radius = , describe a semi-
circle.
Let be the common obliquity of a pair of conjugate
stresses.
FIG. 217.
Draw ORS, making an angle with ON, and cutting the
semicircle in the points R and 5.
Join NR, NS.
OR and OS are evidently a pair of conjugate stresses.
Draw NV perpendicular to RS and bisecting it in V.
Draw the tangent OT\ join NT.
Let OR = r, OS = s. Then
rs = OR. OS = OT* =
and
=(A+A)c os 0. (29)
The maximum value of the obliquity, i.e., of 0, is the angle
TON.
Call this angle 0. Then
NT A -A / x
' ' ' ' ' (30)
PRINCIPAL AND CONJUGATE STRESSES.
249
Let OR, OR' be a pair of con-
jugate stresses (Fig. 218).
Let OG, OH be the axes of R ;
greatest and least principal stress,
respectively.
Draw ON normal to OR.
Let the angle GOR = #, RON
= 6, HON = GOR' = 7, as before. Then
FIG. 218.
and by eqs. (8),
~ cot y = tan f/> =cot
cot
,*_ /i ~A _ cot x cot (y + ^) sin 6
~ ~ ~ : sin 2
or
Hence,
angle GON 90 y = ^
and
angle HOR = y + & = {0 +
(32)
. . . (33)
250 THEORY OF STRUCTURES.
12. Ratio of r to s.
r_OR_ OV-RV _ OV-)/NR* -NV*
s ~" OS * OV+RV~ OV+VNR*-NV n -
ONcos 6 - VNR* ON* sin 3
ON cos e + VNR* - ON* sin 2 e
But
NR A - A NT
= sin TON = sin 0.
r cos 8 1/sin 3 <p sin 2
s cos 6* -{- I/sin 2 sin 3 6
cos <9 1/cos 2 6* cos 2
, . , . (34)
cos (9 + Vcos 2 6^ - cos 2
Let 2 = s i n a Then
cos ^
r i q: cos a a a
- = 7- = tan 3 - or =cot 2 -. (35)
j I cos a 2 2
If 8 = o, or = 90 0.
.-. j = tan 2 (45 - f ) or = cot 2 (45 - f ) (36)
If 6 = 0, = 90.
' (37)
RELATION BETWEEN STRESS AND STRAIN. 2$l
13. Relation between Stress and Strain. Let a solid
body be strained uniformly, i.e., in such a manner that lines of
particles which are parallel in the free state remain parallel in
the strained state, their lengths being altered in a given ratio,
which is practically very small. Lines of particles which are
oblique to each other in the free state are generally inclined at
different angles in the strained state, and their lengths are
altered in different ratios.
Let the straining of the body convert a rectangular portion
ABCD (Fig. 219) into the rectangle AB'C'D', where AB' =
(i + a)AB and AD' = (i + ft)AD.
Now a and ft are very small, so that their joint effect may
be considered to be equal to the sum of their separate effects.
Hence:
First. Let a simple longitudinal strain in a direction paral-
lel to AB convert the rectangle ABCD
into the rectangle AB'ED, where BB' D i _!_
= a.AB.
A line OF will move into the posi-
tion OF', where FF' = a . DF, and
OF' -OF O
the strain along OF = -=-=
Mr*
OF
8 being the angle OFD.
Also, the "distortion or deviation from rect angularity"
FF'sinB .
= angle FOF' = -- 7j/T~ QF -- ~ a COS sm
Second. Let a simple longitudinal strain in a direction
parallel to AD convert the rectangle ABCD into the rectangle
ABKD', where DD' ft. AD.
The line OF will move into the portions O'F n ', where
00' = ft.AO and F"F=DD' = ft. AD.
/-\f Tfli _
.'. the strain along OF
Ur
2$2 THEORY OF STRUCTURES.
Draw O'M parallel to OF. Then
O'F"- OF= O'F"- 0'M=F ff Msm = (F"F-FM) sin
= (DD r OO') sin = ft(AD - AQ) sin
= ft. O0 sin 0.
. sin
/. the strain along OF = - -- - = ft sin 2 0.
The distortion = the angle F"O'M
F"Mcos 6 ft. OP cos
OF OF
= ft sin cos 0.
Hence, when the strains are simultaneous, the line OF will
take the position O'F'" between O'F" and OF', and
the total strain along OF = a cos 2 + ft sin 2 ;
the total distortion = (a ft) sin 6 cos 0.
Again, draw a line OG perpendicular to OF.
The angle OGA = 90 0, and hence, from the above,
tlae total strain along OG = sin 2 # -f- ft cos 8 0,
and the corresponding distortion = (a ft) sin cos 0.
Denote the strain along <2F by *, , that along OG by ^ 2 , and
h of the equal distortions by /. Then
Again, if 0/% <9 are the sides of a rectangle enclosed in
the rectangle ABCD, the straining will convert the rectangle
into an oblique figure with its opposite sides parallel. The
lengths of adjacent sides are altered by the amounts e l and e^ ,
and the angle by 2t. The above results may also be consid-
ered to hold true if the straining, instead of being uniform,
varies continuously from point to point.
RELATION BETWEEN STRESS AND STRAIN. 253
Consider a unit cube ABCD subject to stresses of intensity
p l and / upon the parallel faces
AD, BC and AB, DC. By Art. 3,
Chap. Ill,
_A _ _A_
" E mE'
A , A
and the strain perpendicular to the
face ABCD = - - -
^?_
^'
If the stresses are of equal intensity but of opposite kind,
i.e., if the one is a tension and the other a compression,
p l = A =/, suppose.
.*. a /? = 7-, ( i + ), and the third strain is nil.
g\ ' mJ
Thus the volume of the strained solid
= (i 4- a)(i a )(i) = i 2 = i, approximately,
so that the volume is not sensibly changed.
Also, if OGHF is an enclosed square, O being the middle
point of AD, 6 = 45, and
e l = e^ = = o = strain along OF or OG,
and the distortion ~ change in angle O
a
254 THEORY OF STRUCTURES.
This result may be at once deduced from the figure. For
FOG OD i + ft i - a 790 - 2A
tan = -== = - = ; = tan I - ,
2 FD i + a l + a \ 2 f
or
i a _ i tan t _i t
i + a ~ l + tan / ~~ i +t*
since / is very small. Hence
t = a.
As already shown in Art. 3, shearing cannot take place
along one plane only, and at any point of a strained solid the
shears along planes at right angles are of equal intensity. The
effect of such stresses is merely to produce a distortion of
figure, and generally without sensible change of volume.
Thus, shears of intensity s along the parallel faces of the
unit square ABCD will merely distort the square into a rhom-
bus ABC'D' (Fig. 221). Denoting the change of angle by 2?,
and assuming that the "stress is propor-
tional to the strain,"
5 G.2t,
where G is a coefficient called the modulus
of transverse elasticity, or the coefficient of
A -< g ~~B rigidity, and depends upon a change of
FIG. 221. form.
Consider a section along the diagonal BD.
The stresses on the faces AB, AD, and on CB, CD, resolved
parallel and perpendicular to BD, are evidently equivalent to
nil and a normal force s 1/2, respectively. Thus, there is no
sliding tendency along BD, but the two portions ABD and
s \i *)
CBD exert upon each other a pull, or tension, of intensity -^-
_
V2~''
Similarly it maybe shown that there is no tendency to slide
along AC, but that the two portions ABC and ADC exert
RANKINGS EARTHWORK THEORY. 2$$
upon each other a pressure of intensity s. The straining due
to the shearing stresses is, therefore, identical with that pro-
duced by a thrust and tension of equal intensity upon planes
at 45. Hence, as proved above,
and
_E m
2 I m 2t
Now m rarely exceeds 4, and hence G is generally < \E.
Again, the coefficient of elasticity of volume, or cubic elasticity
(Art. 23), is
mE 2 (m
= -\-
$(m 2) 3 \m
and hence
6K+2G
m =
14. Rankine's Earthwork Theory. A mass of earth-
work tends to take a definite slope.
Rankine assumes, (i) that the stresses exerted in different
directions through a particle of a granular mass are subject
to the general principles enunciated in the preceding articles ;
(2) that the cohesion of the particles is gradually destroyed,
and that the stability of the mass ultimately depends on friction
only.
In the limit, therefore, the face of the mass is inclined to
the horizon at an angle equal to the angle of friction, or, as it
is sometimes called, the angle of repose.
Adopting for the present Rankine's assumptions, the equi-
librium of the mass requires that the direction of the mutual
pressure between the two parts into which the mass is divided
by a plane shall make an angle with the normal to the plane
less than the angle of friction.
256 THEORY OF STRUCTURES.
Denote the angle of friction by 0.
The maximum obliquity must be ^ 0.
By eq. (30),
or sin
/,ON
Thus, if a pressure of intensity /, acts through a mass of
earthwork, eq. (38) gives the least intensity of pressure/, acting
in a direction perpendicular to that of /, consistent with equi-
librium;
The limiting ratios of a pair of conjugate stresses in a mass
of earthwork may also easily be determined.
By eq. (34),
cos Vcos 2 cos 2
the ratio = f .... (39)
cos 6+ I/cos 2 0- cos 2
Hence the ratio cannot exceed
cos -f ^cos 2 6 cos 2
cos 6 Vcos 2 cos 2
nor can it be less than
cos & VCOS* B cos 2
cos + I/cos 2 cos 2
If 6 = o, the ratio becomes
I =f sin
sin
(40)
For example, let the ground-surface be horizontal.
The pair of conjugate stresses become a vertical stress
and a horizontal stress /, .
. A < i + sin0
RANKINGS EARTHWORK THEORY. 2$?
or
A > i - sin
A =i+sin0'*
as in eq. (38).
Pressure against a Vertical Plane. Let ACB (Fig. 222),
the ground-surface of a mass of earthwork, be inclined to the
horizon at an angle 6.
Consider a particle at a vertical depth CD = x below C.
Let s be the vertical intensity of pressure on the particle
at D.
Let r be the conjugate intensity of pressure on the particle
at D.
This conjugate pressure acts in the direction ED parallel ta
the ground-surface, and its obliquity is B.
Take DE so that
_r_ cos 1/cos' 8 cos*
DC '~ s " cos e _|_ |/ cos * e cos 2 0*
Then ze/ . ."/? represents in direction and
magnitude the intensity of pressure on the
vertical plane DC at the point D, w being the weight of a unit
of volume of the earthwork.
Join CE.
The intensity of pressure at any other point m is evi-
dently w . mn, mn being drawn parallel to DE.
Hence, the total pressure on the plane DC = weight of
prism DCE
w.DC.DE w.DC*r
cos 6 = cos
2 2 S
_ wx* cos 6 I/cos 2 6 cos 3
~2~ cos 6 + Vcos 2 - cos 2 ' ' * (43 '
Again, s is the pressure due to the weight of the vertical
column CD.
.-. s = wx cos 0, p ...... (44)
THEORY OF STRUCTURES.
and
cos 6 t/cos' 6 cos 2
r =. wx cos u . (45)
cos 6 + Vcos* V - cos 2
"By means of this last equation the total pressure on CD
imay be easily deduced as follows :
The pressure on an element dx at a depth x
^cos I/cos' cos' 7
= 77MT = wx cos - dx.
cos + l/cos' J cos 2
.'. total pressure = jrdx = etc.
The total resultant pressure is parallel in direction to the
ground-surface, and its point of application is evidently at
two thirds of the total depth CD.
15. Earth Foundations. CASE I. Let the weight of the
superstructure be uniformly distributed over the base, and let
p Q be the intensity of the pressure produced by it.
If p h is the maximum horizontal intensity of pressure cor-
responding to/ ,
/_ < I +sin
p h i sin '
In the natural ground, let p v be the maximum vertical in-
tensity of pressure corresponding to the horizontal intensity
ph Then
p^ < i + sin
p v -i sin '
Hence
+sin0\ 2
pi = VI sin
If ^ is the depth of the foundation, and w the weight of a
cubic foot of the earth,
p v = wx ;
A
(46)
wx = \i sin
EA R TH FO UNDA TIONS. 259
Let h + x be the height of the superstructure, and let a
cubic foot of it weigh w r . Then
Hence, a minimum value of x is given by
x) _ /i_+j|
" -~~ V7-si
= p, suppose;
wx ' \i sin 0/ ~ /&*'
';!#
*=^_^7F 2 (47)
CASE II. Let the superstructure produce on the base a
uniformly varying pressure of maximum intensity / : and mini-
mum intensity/.,.
By Case I,
W x =i- sn
( .
In the natural ground the minimum horizontal intensity of
pressure is
I sin
p h wx ; - - .
I + sin
When the foundation-trench is excavated, this pressure
tends to raise the bottom and push in the sides. The weight
of the superstructure should therefore be at least equal to the
weight of the material excavated in order to develop a hori-
zontal pressure of an intensity equal
A < * sin
' / 2 = I + sin '
Combining this with the last equation,
Combining (48) and (49),
(Rankine's Civil Engineering, Arts. 237, 239.)
200
THEORY OF STRUCTURES.
16. Retaining-walls. Consider a portion ABMN of a
wall (Fig. 223).
Let Wbe its weight, and let the di-
rection of W cut MN in C.
Let Pbe the resultant of the forces
externally applied to ABNM and tend-
ing to overthrow it. Let D be its
point of application, and let its di-
rection meet that of W in E.
Let F be the centre of pressure (or
resistance) at the bed MN.
FIG. 223. Let O be the middle point of MN.
Let MN t, OF qt, OC rt, q and r being each less
than unity.
Let x' and y', respectively, be the horizontal and vertical
co-ordinates of D with respect to F.
Let the inclination to the horizon of MN = a, of P's
direction = ft.
Conditions of Equilibrium. (a) The moment of P with
respect to F ^ the moment of W with respect to F, or
P(y' cos ft x' sin ft) < W(qt =F rt) cos a ;
(51)
the upper or lower sign being taken according as C falls on the
left or right of O.
In ordinary practice q varies from J to f .
EXAMPLE. A masonry wall (Fig. 224)
of rectangular section, x ft. high, 4 ft. wide,
weighing 125 Ibs. per cubic foot, is built
upon a horizontal base and retains water
(weighing 62^ Ibs. per cubic foot) on one
side level with the top of the wall.
FIG. 224.
= 125 X
= 4 ft.
RE TA IN ING- WALLS. 261
or
*? < 192? (52)
If 0=J, * 2 <48 and x < 6.928 ft.
If = f, *' < 72 and * < 8.485 ft.
() The maximum intensity of pressure at the bed MN must
not exceed the safe working resistance of the material to
crushing. The load upon the bed is rarely if ever uniformly
distributed. It is practically sufficient to assume that the in-
tensity of the pressure diminishes at a uniform rate from the
most compressed edge inwards.
Let / be the maximum intensity of pressure, and R the
total pressure on the bed.
Three cases^may be considered.
CASE I. Let the intensity of the pressure diminish uniformly
from / at M to o at N (Fig. 225).
Take MG perpendicular to MN and =f', join GN.
The pressure upon the bed Gp^
is represented by the triangle ^^
MGN. "\^
^-
MFC
The brdinate through the FlG - 22 s-
centre of gravity of the triangle, parallel to GM, cuts MN in
the centre of pressure F.
*- - - = g*.
CASE II. Let the maximum intensity /> MG in Case I.
Take MH =. f, and the triangle
MHK = R (Fig - 226 )-
The pressure on the bed is
^ now represented by the triangle
MHK.
R = MH. MK = . MK.
M F O K N
FlG - 22fi - The ordinate through the
262 THEORY OF STRUCTURES.
centre of gravity of the triangle MHK parallel to HM cuts MN
in the centre of pressure F.
.-. q t = OF = OM- MF=-- .
But
t 2 R
/. qt = --- - ; and hence
T O C' T
9 = ~ ~ -ft and is evidently > g. . Y (53)
CASE III. Let the maximum intensity/ < MG in Case I.
Take ML =f, and the trapezoid MLSN = R (Fig. 227).
The pressure on the bed is now represented by the trape-
zoid MLSN.
G .-. R = XML + NS)MN
r^.
and
MFC N 2R
FIG. 7 . ^V5 = - /.
The ordinate through the centre of gravity of the trapezoid
parallel to LM cuts MN in the centre of pressure F.
Draw ST parallel to NM.
The moment of MLSN with respect to O
= moment of MTSN with respect to O
+ moment of ZSiTwith respect to (9,
or
TS*
\(ML + NS)MN. OF =
Hence, q = ( t), and is evidently <%.... (54)
RE TAINING- WALLS. 263
Now W must be a function of x, the vertical depth of N
below B ; P also may be a function of x.
Hence if /is given, and the corresponding value of q from
(53) or (54) substituted in (52), x may be found.
When (53) is employed, the value of x found must make
?>>
When (54) is employed, the value of x found must make
r<>
EXAMPLE. The rectangular wall in (), the safe crushing
strength of the material being 10,000 Ibs. per square foot (=/).
R= W= 500*.
By (53),
~ 2 I2O
Substituting in (52),
Hence,
x < 9-03 ft.
Again, q > -- 2j > .4248, and is a fortiori > g.
If (54) is employed,
1/80 \
* = 5b V-
Hence, by (52),
By trial ;tr is found to lie between 12 and 13 ; each of these
values makes q > -J-, which is contrary to (54).
The first is therefore the correct substitution.
(c\ The angle between the directions of the resultant pres-
sure and a normal to the bed must be less than the angle of
friction.
264
THEORY OF STRUCTURES.
Let be the angle of friction, R the mutual normal pressure.
Resolving along the bed and perpendicular to it,
and
Pcos a 4- ft Wsm a < R tan
Psin a + J3+ Wcosa = R',
Pcos a-\- ft IV sin a
Psin
which reduces to
Wcosa
< tan 0,
P(cos ft-\-a cos sin /?-)- sin 0)> W^sin cos a-j-cos sin <*),
or
P cos /? -f or + < JFsin a + 0,
or
or
P(cos ft cos a -(- - sin ft sin or + 0) <
a -f- 0,
tan
Pcos
Psin/* +
. (55)
17. Rankine's Theory of Earthwork applied to Retain-
c ing-walls. Fig. 228 represents a
vertical section of a wall retaining
earthwork. AB is a vertical plane
cutting the ground-surface AC in the
point A.
Consider the equilibrium of the
whole mass of masonry and earth-
work in front of AB.
Let the depth AB x.
The total pressure on AB is, by
(43),
_ wx* cos 6 1/cos 2 6 cos 2
P = cos - .
2 cos + I/cos 3 cos 2
LINE OF RUPTURE. 265
Its point of application is D, and BD = .
Let W be the weight of the whole mass under considera-
tion, and let its direction cut the base of the wall in the point G.
Let F be the centre of pressure in the wall-base.
Taking moments of jPand W 7 about F,
p - cos 6 j[jf <f j) sin(0+a) | = W(qt ^ rt) cos a. (56)
The other conditions of equilibrium may be discussed as in
Art. 1 6.
18. Line of Rupture. Another expression for the press-
ure on AB may be obtained as follows :
If the whole mass in front of AB (Fig. 229) were suddenly
removed, some of the earthwork behind AB would fall away.
Suppose that the volume ABC would slip
along the plane CB.
The stability of ABC is maintained by the
reaction P on AB, the weight J^of ABC, and
the frictional resistance along BC.
Let the direction of P make an angle ft
with the horizon.
Let the angle CBA = i.
Let R be the mutual pressure on the plane BC.
Resolving along and perpendicular to BC,
- P cos (90 - i - /?) + Wcos i = R tan ;
and
P sin (90 -i- ft)+ W sin i = R.
.'. - Psin (ft+t) + W cos i = tan \Pcos (fl+ J)+ Wsin i j,
and
cos i sin t tan cos(/+0)
= ^ sin (ft + i} + cos (ft + i) tan ~ Y sin (0 + i + 0)'
266 THEORY OF STRUCTURES.
Also
BA . BC . . wx* cos sin i
W w sin i
2 2 cos
_ w^r 2 cos sin z cos (t + 0)
*'
2 cos (0+) sh(/J + * + 0)
The only variable upon which P depends is the angle i.
Differentiating the right-hand side of eq. 57 with respect to
i and putting the result equal to zero, a value of i is found in
terms of /?, and 0, which will make P a maximum.
The line inclined at this angle to the vertical is called the
line of rupture.
If the ground-surface is horizontal, 9 = o.
If the face retaining the earth is vertical, and if it is also
assumed that the friction between the face and the earthwork is
nil, P is horizontal and ft = o. Hence (57) becomes
P = tan i cot (i + 0) (58)
This is a maximum when 21 = 90 0, and then
/i -
WX* f 0\ / 0\ WX 1 !
= tan ( 45 - -) cot ( 45 + 2 -J= (
\ J *
or
__ wx* i sin
2 I + sin '
the same result as that obtained by Rankine's theory.
PRACTICAL RULES.
267
The following is an easy geometrical proof of eq. (59):
On any line KL (Fig. 230) de-
scribe a semicircle.
Draw KM inclined at the angle
to KL, and KN inclined at the
angle i to KM. .
Join NL, cutting KM in T.
Let O be the middle point of
the arc KM.
Join OL, cutting KM in Y.
Draw NV parallel to KM. Then
NT KN NT VY
FIG. 230.
VY
The ratio -== is evidently a maximum when N coincides
with 0, and hence tan i cot (i + 0) is a maximum when
coincides with KO.
Now the arc OK = the arc 6W, and hence the angle OKM
= the angle OLK.
Hence, if OKM= i, OLK must also = i.
But OKL + OLK= 90 = *'+ + * = 2*'+ 0.
19. Practical Rules. When the surface of the earthwork
is horizontal and the face of the wall against which it abuts ver-
tical, the pressure on the wall according to Rankine's theory is
sin
2 i + sin '
and the direction of P is horizontal.
This result is also identical with that obtained in Art. 18, on,
the assumption of Coulomb's wedge of maximum pressure
(Poncelet's Theory).
268 THEORY OF STRUCTURES.
Experience has conclusively proved that the theoretical value
of P given above is very much greater than its real value, so
that the thickness of a wall designed in accordance with theory
would be in excess of what is required in practice. In the
deduction of the formula, indeed, the altogether inadmissible
assumption is made that there is no friction between the earth-
work and the face of the wall. This is equivalent to the sup-
position that the face is perfectly smooth and that therefore
the pressure acts normally to it. Boussinesque, Levy, and St.
Venant have demonstrated that the hypothesis of a normal
pressure only holds true,
either, first, if the ground surface is horizontal and the wall-
face inclined at an angle of 45 to the vertical,
or, second, if the wall-face is vertical and the ground-surface
inclined at an angle to the horizon.
When the surface of the ground is horizontal and the face
of the wall vertical, and when = 45, the above formula gives
the correct magnitude of P. Its direction, however, is not hori-
zontal, but makes an angle with the vertical equal to the angle
of friction between the earth and the wall. The wall-face is gen-
erally sufficiently rough to hold fast a layer of earth, and in all
probability Boussinesque's assumption that the friction between
the wall and the earth is equal to that inherent in the earth is
a near approximation to the truth. The direction of P will
thus be considerably modified, leading to a smaller moment of
stability and a corresponding diminution in the necessary thick-
ness of the wall.
In practice the thrust P may always be made small by
carrying up the backing in well-punned horizontal layers.
In order to neutralize the very great thrust often induced
by alternate freezing and thawing and the consequent swelling,
a most effective expedient is to give a batter of about I in I to
the rear line of the wall extending below the line to which frost
penetrates.
The greatest difficulty in formulating a table of earth-thrusts
arises from the fact that there is an infinite variety of earth-
work. As an example of this, Airy states that he has found
PRACTICAL RULES. 269
the cohesive power of clay to vary from 168 to 800 pounds per
square foot, the corresponding coefficients of friction varying
from 1.15 to .36, and that even this wide range is less than
might be found in practice.
A correct theory for the design of retain ing-walls is as yet
wanting. According to Baker, experience has shown that with
good backing and a good foundation the stability of a wall
will be insured by making its thickness one-fourth the height,
and giving it a front batter of I or 2 in. per foot, and that under
no conditions of ordinary surcharge or heavy backing need its
thickness exceed one-half the height. Baker's usual practice in
ground of average character is to make the thickness one-third
the height from the top of the footings, and if any material is
taken out to form a face panel, three-fourths of it is put back
in the form of a pilaster.
General Fanshawe's rule for brick walls of rectangular
section retaining ordinary material is to make the thickness
24$ of the height for a batter of in 5 ;
or"
2/7
tt tt
O o it tt t( tt tt
~ II tt tl
in 6 ;
in 8 ;
in 10 ;
in 12 ;
in 24;
32 " " " " " for a vertical wall.
The thickness at the footing adopted by Vauban for walls
with a front batter of I in 5 or I in 6 and plumb at the rear, is
approximately given by the empirical formula
thickness = . 19/^ + 4 ft.,
H being the height of the wall above the footing. Counter-
forts were introduced at intervals of 15 feet for walls above 35
feet in height, and at intervals of 12 feet for walls of less height.
TT
The counterfort projects from the wall a distance of (-3 ft.
approximately, and the approximate width of the counterfort is
TT TT
-f- 3 ft., diminishing to + 2 ft.
2/0 THEORY OF STRUCTURES.
Brunei curved the face of the wall and made its thickness
one-fifth or one-sixth the height. Counterforts 2 ft. 6 in. in
thickness were introduced at intervals of 10 ft.
The vast importance of the foundation will be better appre-
ciated by bearing in mind that the great majority of failures
have been due to defective foundations. If water can percolate
to the foundation, a softening action begins and a consequent
settlement takes place, which is most rapid in the region sub-
jected to the greatest pressure, viz., the toe. In order to coun-
teract this tendency to settle, the toe may be supported by rak-
ing piles, the rake being given to diminish the bending action of
the thrust on the piles. It is also advisable to distribute the
weight as uniformly as possible over the base, a condition which
is not compatible with large front batters and deep offsets, as
they tend to concentrate weight on isolated points. In the
case of dock-walls, too, a large front batter will keep a ship
farther away from the coping and will necessitate thicker
fenders, as well as cranes with wider throws. As an objection
to offsets Bernays urges that, in settling, the backing is liable
to hang upon them, forming large holes underneath. He there-
fore favors the substitution of a batter for the offsets. On the
other hand, if water stands on both sides of the walls, the
hydrostatic pressure on the offsets will greatly increase its
stability.
Dock-walls are liable to far greater variations of thrust than
ordinary retaining-walls. The water in a dock with an im-
permeable bottom may stand at a much higher level than the
water at the back of the wall, and its pressure may thus even
more than neutralize the thrust due to the backing. With a
porous bottom the stability of a wall may be greatly dimin-
ished by an upward pressure on the base. The experience of
dock-wall failures has led to the conclusion that a large moment
of stability is not of so much importance as " weight with a
good grip on the ground." Many authorities, both practical
and theoretical, have urged the great advantages in economy
and strength attending the employment of counterforts. The
use of Portland cement, or cement concrete, will guard against
the breaking away of the counterforts from the main body of
RESERVOIR WALLS.
271
the wall, as has often happened in the case of the older walls.
But a uniform distribution of pressure as well as of weight is
important, and it therefore seems more desirable to introduce
the extra weight of the counterforts into the main wall. Be-
sides, the building of the counterforts entails of itself an in-
creased expense.
20. Reservoir Walls. Let /"be the maximum safe press-
ure per square foot of horizontal base, at inner face of a full
reservoir, at outer face when empty.
Let w be the weight of a cubic foot of the masonry.
Assume that the wall is to be of uniform strength, i.e., that
the section of the wall is of such form that
in passing from any horizontal section to the
consecutive one below, the ratio of the in-
crement of the weight to the increment of
the surface is constant and equal to/.
Let AB, Fig. 231, be the top of the wall.
Take any point O as origin, and the vertical
through O as the axis of x.
Let OA t,, O=r / 2 , and let
T = /, 4- t n --
B
For the profile AP consider a layer of thickness dx at a
depth x. Then
wydx
(0
or
w y
,^/-lc
w
c being a constant of integration.
When x = o, y = t l ;
' = - lo & 'i + '
2/2 THEORY OF STRUCTURES.
and hence
* = *"
which is the equation to AP and is the logarithmic curve.
It may be similarly shown that the equation to BQ is
Equations (2) and (3) may also be written in the forms
V)
y = t,ef x ..... ,>, , . . (4)
and
(5)
Corresponding points on the profiles, e.g., P and <2, have a
common subtangent of the constant value , for
= ..... (6)
dy) w
Area PNOA = ydx = f . - = ( F, - ,,), (7)
where /W= F,.
Area QNOB = f'yd = -( Y, - t,), (8)
where QN = Y,.
.: Area QPAB = -( K,+ F, - , + ^) = (T' - 7"), . (9)
where PQ=Y,+ Y,= T'.
RESERVOIR WALLS.
Thus the area of the portion under consideration is equal
to the product of the subtangent and the difference of thick-
ness at top and bottom.
Lines of resistance with reservoir empty. Let g l be the
point in which the vertical through the C. of G. of the portion
OAPN intersects PN. Then
Ng l X area OAPN = f* X ydx y - ;
I/O
So if gi be the point in which the vertical through the
C. of G. of the portion OBQN intersects QN,
Let G be the point in which the vertical through the
C. of G. of the whole mass ABQP intersects PQ. Then
NG X area ABQP = Ng, X area A ONP- Ng, X area BONQ,
or
IV A. W l A. W *
274
THEORY OF STRUCTURES.
The horizontal distance between G and a vertical through
the middle point of AB
=NG-\(f ^jy-'x-w-wjy-^-w*-**
=. one half of the horizontal distance between the verticals
through the middle points of AB and CD.
The locus of G can therefore be easily plotted.
Lines of Resistance with Reservoir Full. Let R be the centre
of resistance in PQ (Fig. 232).
Draw the vertical QS, and consider
the equilibrium of the mass QSAPQ.
Let w' = weight of a cubic foot of
water.
w'x* x
= moment of water-pressure
against QS about R
= moment of weight of QBS
about R -f- moment of
weight of QPAB about R t
or
W X
moment of QBS about R+ ~(T' - T)w.GR.
The first term on the right-hand side of this equation is
generally very small and may be 'disregarded, the error being
on the safe side.
In such case
i w' x*
GR -6jT'-T'
Also the mean intensity of the vertical pressure
w X area APQB
A = or>
RESERVOIR WALLS.
and the maximum intensity of the vertical pressure
275
2R
= */
I -2q
or
= -Jv(l + 6?) = /(! + 6?)(l - y,).
General Case. Let the profile be of any form, and consider
any portion ABQP, Fig. 233.
Take the vertical through Q as T
the axis of x, and the horizontal
line coincident with top of wall as
the axis of y.
The horizontal distance (y) be-
tween the axis of x and the vertical
through the C. of G. of the portion
under consideration is given by the
equation
/ being the width, dx the thickness,
and y the horizontal distance from OQ of the C. of G. of any
layer MN at a depth x from the top.
When the reservoir is empty, the deviation of the centre of
resistance from the centre of base
When the reservoir is full, let q'T be the deviation of the
centre of resistance from the centre of the base, and disregard
the moment of the weight of the water between OQ and the
profile BQ. Then
276
THEORY OF STRUCTURES.
moment of water pr. moment of wt. of ABQP
weight of ABQP
w I tdx
t/o
Hence
w'x*
w I tdx
t/o
21. General Equations of Stress. Let x, y, z be the
co-ordinates with respect to three rectangular axes of any point
O in a strained body.
Consider the equilibrium of an
element of the body in the form
of an indefinitely small parallele-
piped with its edges OA (= dx\
OB( dy), OC( dz] parallel to
the axes of x, y, z. It is assumed
that the faces of the element are
sufficiently small to allow of the
distribution of stress over them
being regarded as uniform. The
FlG - 234 ' resultant force on each face will
therefore be a single force acting at its middle point.
Let X l , Y l , Z^ be the components parallel to the axes x, y, z
of the resultant force per unit area, on the
face EC.
" X^ , Y 9 , Z 9 be the corresponding components for the
taceAC.
" -Yg , Y 3 , Z 3 be the corresponding components for the
face AB.
These components are functions of x, y, z, and therefore
become
for the adjacent face AD ;
GENERAL EQUATIONS OF STRESS.
for the adjacent face BD ;
for the adjacent face DC.
Hence, the total stress parallel to the axis of x
= X.dydz - (x, + d -^dx]dydz + X,dzdx - (x, + ^
IdX, . dX,
= ~(-&+W
Similarly, the total stress parallel to the axis of y
and the total stress parallel to the axis of z
t dZ t dZ
Let p be the density of the mass at O, and let P x ,P y , P g be
the components parallel to the axes of x, y, z of the external
force, per unit mass, at O.
pdxdydzP x is the component parallel to the axis of x of
the external force on the element ;
pdxdydzPy is the component parallel to the axis of y of
the external force on the element ;
pdxdydzP z is the component parallel to the axis of z of
the external force on the element.
278 THEORY OF STRUCTURES.
The element is in equilibrium.
dx l dy dz
dY, , dY, , dY,
dZ \ ,dZ^ >dZ*
dx" dy" dz~
These are the general equations of stress.
Again, take moments about axes through the centre of the
element parallel to the axes of co-ordinates, and neglect terms
involving (dxjdydz, dx(dyfdz, dxdy(dz}*.
and X=
(2)
Adopting Lamp's notation, i.e., taking
N l , N t , N 3 as the normal intensities of stress at O on planes
perpendicular to the axes of x, y, z ;
7", as the tangential intensity of stress at O on a plane
perpendicular to the axis of x if due to a stress
parallel to the axis of j/, or on a plane perpen-
dicular to the axis of y if due to a stress parallel
to the axis of x ; and 7^ , T a similarly, equa-
tions (i) become
^ + ~^'~ Jr ~^ 1
rf'+tL+W
dx dy dz
(3)
GENERAL EQUATIONS OF STRESS.
279
Next consider the equilibrium of a tetrahedral element
having three of its faces parallel to
the co-ordinate planes. Let /, m, n
be the direction-cosines of the normal
to the fourth face.
Also, let X, V, Z be the compo-
nents parallel to the axes of x, y, z
of the intensity of stress R on the
fourth face.
X = IN, + mT, + n T t + \pP x ldx.
But the last term disappears in FlG - 2 35-
the limit when the tetrahedron is indefinitely small, and hence
(4)
These three equations define R in direction and magnitude
when the stresses on the three rectangular planes are known.
Let it be required to determine the planes upon which the
stress is wholly normal. We have
Y=mR, ZnR..
(5)
Substituting these values of X, Y, Z in eqs. (4) and eliminat
ing /, m, n, we obtain
- T?- T?- T?
1 T,T S ) = o ; (6)
R* -
a cubic equation giving three real values for R, and therefore
three sets of values for /, m, and n, showing that there are three
planes at O on each of which the intensity of stress is wholly
normal. These planes are at right angles to each other and
are called principal planes, the corresponding stresses being prin-
cipal stresses. They are the principal planes of the quadric,
= c.
(7)
280
THEORY OF STRUCTURES.
For, the equation to the tangent plane at the extremity of a
radius r whose direction-cosines are /, m, n is
Xrx + Yry + Zrz = c, ....'. (8)
and the equation of the parallel diametral plane is
Xx + Yy + Zz = o (9)
The direction-cosines of the perpendicular to this plane are
X
-
Y
-
Z
-
so that the resultant stress R must act in the direction of this
perpendicular.
Hence the intensities of stress on the planes perpendicular
to the axes of the quadric (7) are wholly normal.
Refer the quadfic to its principal planes as planes of refer-*
ence. All the 7"s vanish and its equation becomes
(10)
Also, the general equations (3) become
%-*
Again,
RELATION BETWEEN STRESS AND STRAIN. 28 1
Consider X 9 Y, Z as the co-ordinates of the extremity of
the straight line representing R in direction and magnitude.
Equation (12) is then the equation to an ellipsoid whose semi-
axes are N lt N 91 N t . As a plane at O turns round O as a fixed
centre, the extremity of a line representing the intensity of
stress R on the plane will trace out an ellipsoid. This ellipsoid
is called the ellipsoid of stress.
Note I. The coefficients in the cubic equation (6) are in-
variants. Thus, NI -f N 9 + N, is constant, or the sum of three
normal intensities of stress on three planes placed at right
angles at any point of a strained body is the same for all
positions of the three planes.
Note 2. The perpendicular/ from O on the tangent plane,
equation (8),
Note 3. Let the stress be the same for all positions of the
plane at 0. Then N, = N^ = N z , and the ellipsoid (12) be-
comes a sphere. The stress is therefore everywhere normal,
and the body must be a perfect fluid. Conversely, if the
stress is everywhere normal, the body must be a perfect fluid,
the ellipsoid becomes a sphere, and therefore N l = N 9 = N a .
22. Relation between Stress and Strain. In Art. 13 it
was shown that when the size and figure of a body are altered
in two dimensions, there is an ellipse of strain analogous to the
ellipse of stress. If the alteration takes place in three dimen-
sions, it may be similarly shown that every state of strain may
be represented by an ellipsoid of strain analogous to the ellip-
soid of stress. The axes of the ellipsoid are the principal axes
of strain, and every strain may be resolved into three simple
strains parallel to these axes.
282
THEORY OF STRUCTURES.
It is assumed that the strains remain very small, that the
stresses developed are proportional
to the corresponding strains, and
that their effects may be superposed.
Consider an element of the un-
strained body in the form of a rect-
x angular parallelepiped, having its
edges PQ (= A), PR (= k}, PS (= I)
parallel to the axes of co-ordinates.
When the body is strained, the
element becomes distorted, the new
edges being P'Q', P'R', P'S'.
Let x, y, z be the co-ordinates of P.
Let x -f- u, y -f- v, z -\- w be the co-ordinates of P'.
By Taylor's Theorem the co-ordinates with respect to P' of
FlG> 236 '
du
dx
dv ,dw
*-, h
dx dx
r> , i.du J , dv \ ,dw
R are k, k( i +-=-) k ;
a \ a i a
du .dv . . dw\
r , , .
S'are/-, /- ,
dz dz
ay i ay
dw\
.
dz j
Hence, strain parallel to axis of x =
<fc ( y _
P'Q'-PQ du 1
PQ ~ ~dx
P'R'-PR dv
z =
P'S' -PS dw
PS ~~dz'
I SO TROPIC BODIES.
Again, cos QP'R
283
du
dv \dv dw dw
In the limit, this reduces to
Similarly,
dy dx
cos fi'P'S' = - +
... (16)
cos *'/"<?' =
Volume of unstrained element = hkl;
Volume of distorted element =
Difference of volume
multiplied by the cosines
of small angles
, du . dv . dw\
+ _ + _. + _),
in the limit.
du . dv . dw
' Vol. of unstrained element dx dy dz*
= the volume or cubic strain.
23. Isotropic Bodies, i.e., bodies possessing the same elas-
tic properties in all directions.
A normal stress of intensity N l parallel to the axis of x
N
produces a simple longitudinal strain -, and two simple
E
N
lateral strains, each = -r, parallel to the axes of y and z,
mh
284
THEORY OF STRUCTURES.
E being the ordinary modulus of elasticity and , Poisson's
ratio (Art. 3, Chap. III).
Normal stresses N 9 , N t parallel to the axes of y and a may
be similarly treated.
Let the three normal stresses act simultaneously and super-
pose the results. Then
, , a du ~\
total strain parallel to axis of x - l -- ?=
E mE dx '
mE dz '
The form in which these equations are given is due to
Grashof.
Solving for N t , N^ , N 9 ,
,, m(m i)E du mE tdv dw\
1 ~ (m + i)(m 2) dx " (m + i)(m- 2)<dy ' ~dzr
m(m I )E dv mE idw .du\
~ ~~
.,
~~
m(m i }E dw
mE
(m + i)(; 2) ~" (w + i)(m 2)\dx
The last equations may be written
.dv
(du >dv\
I 7Z~ i "717 / *
.
dy
ISOTROPIC BODIES.
28 5
where A, =
mE
, is the coefficient of dilatation, and
A = mm ~ l
~ (m+i)(m-2J
Again, the straining changes the angle RPS by an amount
-| - , producing two tangential stresses, each equal to
dy dz
G\ -| - I, parallel to the axes of y and z.
\dv dz>
dw dv
Similarly,
T -G(
2 " \dz
du
s-
(21)
G is called the coefficient of rigidity or transverse elasticity,
and is designated n in Thomson and Tait's notation, and >w in
Lamp's notation.
Relation between A, A, and G. Equations (20) and (21) pre-
serve the same forms whatever rectangular axes may be chosen.
Keep the axis of z fixed and turn the axes of x and y
through an angle a.
Let Nf be the normal stress parallel to the new axis of x.
.*. N{ N^ cos 2 a -|- N^ sin 2 a + 2 7" 3 sin a cos a. (22)
Let x 1 ', y' and z/, v' be the new co-ordinates and displace-
ments.
, T . A du' . ^Idv' . dw'\ ,. ..du' .
' N > = A ^' + A (^ + w) = (A ~ ^ + K6 -
_ du . dv . dw du' . dv' . dw' .
For -- + -- + -:-, = = y-, + + - - , is an invariant.
dx dy dz dx dy dz
.286 THEORY OF STRUCTURES.
The values of N{ given by eqs. (22) and (23) must be
identical. Now,
x = x' cos a y' sin a, y = x' sin a -)- y' cos a ; }
z/' = & cos a -f- ^ sin a', t;' = & sin a -\- v cos a . )
du du . dv
.'. _ = cos a + sin a
^,r dx dx
du 2 . dv . a /^ . ^/z/\ .
= -j- cos 2 a + - sin a + !_--[-__ jsm a cos a
^r dy \dy dxl
du . dv T.
- cos 2 a -f- - sin 3 a + ? sin a cos a ;
tfJtr y 6-
and by eq. (23),
AV = (A -*)(^ cos 8 a+^ sin 3 a+5 sin a cos a] + A0. (25)
>^r ay G- '
Also by eqs. (20) and (22),
= (A\)- cos 2 +- sin +r sin a cos a+A0. (26)
\d>x dy A ~~ A /
Eqs. (25) and (26) must be identical.
^ _ A w.fi'
IT = 5(^+1) = " = n - |iK * (27)
Adding together equations (20),
/ , flfe , d
^"dy^ ~dz'
It may be easily shown that the normal stresses can each
be separated into a fluid pressure p and a distorting stress.
AP PLICA TIONS. 287
Hence, putting
AT AT AT__^_ m & (d 11 I dV . 6
9 - 3 P -- ^ m __ ^ [^ -r j- -r -
t> mE
the cubic elasticity -. =- = . r = K. (28)
au dv aw ^(m 2)
dx dy dz
24. Applications. i. Traction. One end of a cylindrical
bar of isotropic material is fixed and the bar is stretched in the
direction of its length. The axis of the bar is the only line
not moved laterally by contraction.
Take this line as the axis of x.
The displacements u, v, w of any point x, y, z may be ex-
pressed in the form
u = ax, v = Py t w = PZ. . . , (29)
By eqs. (20) and (29),
.- 1 (30)
By eqs. (21) and (29), all the tangential stresses vanish.
Hence, since N lt N 9 , N 3 are constant, and since the equa-
tions of internal equilibrium contain only differential coeffi-
cients of the stresses, the hypothesis, eq. (29), satisfies these
equations.
First. Let N^ = o = N 3 ; i.e., let no external force act upon
the curved surface.
.-. - ft A + \(ft + ) = o,
or
P _ A _ I
a ~ T+Ji ~ m
Thus, the coefficient of contraction is less than the coefficient
of expansion.
288 THEORY OF STRUCTURES.
Again, by eqs. (30) and (32),
Z = A-2* = A-* = . (33)
a. am
Second. If the bar, instead of being free to move laterally,
has its surface acted upon by a uniform pressure P, then
N, = N 3 = P.
By eqs. (3 1) and (32),
ft APXN,
a ~ \(N, + 2/>) - AN, '
- * . (35)
For example, let P be sufficient to prevent lateral contrac-
tion. Then ft = o and, by eqs. (31) and (35),
aA = N, = -y = (m - i)P.
2. Torsion. (a) Let a circular cylinder (hollow or solid) of
length / undergo torsion around its axis (the axis of x), and let
t be the angle through which one end is twisted relatively to
the other. A point in a transverse section distant x from the
latter will be twisted through an angle x .
The displacements u, v, w of the point x,y, z in this section
may be expressed in the form
u = o, v = zx-y w = -\-yx- .
By eqs. (20) and (21),
and
APPLICA TIONS. 289
The algebraic sum of the moments of 7!, , J", with respect
to the axis
r being the distance of the point (x, y, z) from the axis.
Hence, the moment M, = Pp (Chap. IX), of the couple
producing torsion
_ ^ * / i J c* __ /~* T /~*f}T
~lJ ' 1
dS being an element of the area at (x,y, z\ I the polar moment
of inertia, and the torsion per unit of length of the cylinder,
or the rate of twist.
The torsional rigidity of a solid cylinder
= = GI = -nR*
6 2
R being the radius of the cylinder.
(b] Torsion of a bar of elliptic section.
The displacements u, v, w may now be expressed in the
form
u = F(y, z), v = 6x2, w = Qxy.
du _ dv _ dw^
' dx dy dz'
7-0, T.= G+
Hence, by the general eqs. (3),
* ....... (36)
2QO THEORY OF STRUCTURES.
.Also, the surface stresses are zero ;
-and hence, by eqs. '(3$),
~dy = Q(zdz -\-ydy) (38)
This equation must hold true at the surface.
Let the equation to the elliptic section be
dz fy
= - .^v,,, (40)
and by eq. (38),
2 dll . dlt _ _ / 72 _ 2X
u dyz satisfies this last equation and also eq. (36), if
Again, the algebraic sum of the moments 71, , T 3 with re-
spect to the axis of x,
2G6
T(<y + * v ) ........ (43)
APPLICATIONS. 291
The total moment (M] of the couple producing torsion
and the torsional rigidity
(c] Torsion of a bar of rectangular section.
As in case (b), u must satisfy the equation
Also, the equations of condition corresponding to eq. (38) are
and
du
-j- 02 = o when y = #, . . . . (46)
+ By = o when z = c\ ... (47)
2b and 2c (b < c) being the sides of the rectangle. The total
moment of torsion, viz.,/(7' 2 j T.z)dS is then found to be
,. ,
-'
tan h 2n i
If ^ = r, i.e., if the section is a square, eq. (48) becomes
M= .843462/0, ...... (49)
/( = f ^ 4 ) being the moment of inertia with respect to the axes..
(See Chap. IX).
2Q2 THEORY OF STRUCTURES.
If is very small, eq. (48) becomes
M= i6FcGO(--.2i-) (50)
The torsional rig'dity of a rectangular section is sometimes ex-
pressed by the formula
M 5 V
T>*4*q^ < 5I >
For the further treatment of this subject, the student is re-
ferred to St. Venant's edition of Clebsch, and to Thomson and
Tait's Natural Philosophy.
3. Work done in the small strain of a body (Clapeyron's
Theorem) M ultiply eqs. (3) \yyudxdy dz, v dx dy dz, wdxdydz,
and find the triple integral of their sum throughout the whole
of the solid.
The terms involving the components P x t P y , P 2 may be dis-
regarded, as the deformations due to their action are generally
inappreciable.
Also,
///s-*** ;
=ff(N x 'u x - N x "u x "}dydz -fffNSj-dxdydz\
N x ', N x " being the values of N l at the two points in which the
line parallel to the axis of x cuts the surface of the body, and
u x i u x" the corresponding values of u.
Let dS, dS' be the elementary areas of the surface at these
points, and /', I" the cosines of the angles between the normals
to these elements and the axis of x.
The double integral on the right-hand side of the last equa-
tion then becomes
ff(N x 'l'u x 'dS - NJ'l"H x "dS) =
APPLfCA TfOlfS 293
Treating the other terms similarly,
O = 2{(N l l+ T,m + 7a +(T,l+ N,m
Hence, the work done = $2(Xu + Yv + Zw)dS
- ^W, + N,N, + NJt t - T? - 77 - T.*) }
I
M + Njf t + ffjr t - T: - T; - T,'
E being the ordinary modulus of elasticity.
294 THEORY OF STRUCTURES.
EXAMPLES.
1. At a point within a strained solid there are two conjugate stresses,
viz., a tension of 200 Ibs. and a thrust of 150 Ibs. per square inch, the
common obliquity being 30. Find (a) the principal stresses ; (b) the
maximum shear and the direction and magnitude of the corresponding
resultant stress; (c) the resultant stress upon a plane inclined at 30 to
the axis of greatest principal stress.
Ans. (a) A tension of 204.65 Ibs. and a thrust of 146.95 Ibs. per sq. in.
(b) 175.8 Ibs. per sq. in.; 173.2 Ibs. in a direction making an
angle of 40 13' with the axis of greatest principal stress.
(<r) 163.3 Ibs. per sq. in.
2. A wall with a plumb rear face is to be 30 ft. high and 4 ft. wide at
the top ; the earth slopes up from the inner edge at the angle of 20,
30 being the angle of repose. Assuming Rankine's theory, determine
the proper width of the base, the masonry weighing 144 Ibs. per cubic
foot, and the earth 1 10 Ibs.
3. A wall 6 ft. wide at the bottom, plumb at the rear, and with a
front batter of i in 12, retains water level with the top. Find (a) the
limiting position of the centre of pressure at the base so that the stress
may be nowhere negative.
How (b) high may the wall be built when subjected to this condition?
(a cubic foot of masonry = 125 Ibs.).
Ans. (a} 12 in. from middle point of base; (b) height = 8.9 ft.
4. A wall is built up in layers, the water face being plumb and the
rear stepped. If / be the thickness of the n\.\\ layer and y the depth of
water above its lower face, show that width of layer x thickness of layer
= t/4^4 2 + 6Ats + mty* 2A ; A being the sectional area of the wall
above the layer in question, z the horizontal distance between the water
face and the line of action of the resultant weight above the layer, t the
layer's thickness, and m the ratio of the specific weights of the water and
masonry.
5. At a point within a strained solid, the stresses on two planes at
right angles to each other are a thrust of 30 ^2 Ibs. and a tension of 60
Ibs. per square inch, the obliquities being 45 and 30 respectively.
Determine (a) the principal stresses; (b) the ellipse of stress; (c) the
intensity of stress upon a plane inclined at 60 to the major axis.
Ans. (a) A thrust of 61.76 Ibs. and a tension of 39.80 Ibs.
(c) A thrust of 66.5 Ibs.
EXAMPLES. 295
6. If the principles of the ellipse of stress are applicable within a
mass of earth, and if at any point of the mass the stress upon a plane is
double its conjugate stress, the angle between the two stresses being
20 28', show that the angle of repose of the earth is 28. i.
7. The total stress at a point O upon a plane AB is 60 Ibs. per square
inch, and its obliquity is 30 ; the normal component upon a plane CD
at the point O is 40 Ibs. per square inch ; CD is perpendicular to AB.
Find (a) the total stress upon CD, and also its obliquity ; (b) the princi-
pal stresses at O ; (c) the equal conjugate stresses at O.
Ans. (a) tan -'(I) ; 50 Ibs.
(b) 76.57 Ibs. and 15.39 Ibs.
(c) 34.23 Ibs, ; obliquity = 41 42'.
8. Assuming Rankine's theory, find the pressure on the vertical face
of a retaining-wall, 30 ft. high, which retains earth sloping up from the
top at the angle of repose, viz., 30.
(Weight of masonry = 128 Ibs. per cubic foot.; weight of earth = 120
Ibs. per cubic foot.) Ans. 46,764 Ibs.
9. At a point within a strained solid the stress on one plane is a ten-
sion of 50 Ibs. per square inch with an obliquity of 30, and upon a
second plane is a compression of 150 ibs. per square inch with an ob-
liquity of 45. Find (a) the principal stresses ; (ff) the angle between the
two planes ; (c} the plane upon which the resultant stress is a shear, and
the amount of the shear.
Ans. (a) PI = 1 53. 8 Ibs. (comp.) ; p* = 20 Ibs. (tens.)
(*) 21 55'-
(c) 86.88 ibs.; y = 19 50'.
10. At a point within a strained solid the stress on one plane is a ten-
sion of loo Ibs. per square inch with an obliquity of 30, and on a second
plane a compression of 50 Ibs. with an obliquity of 60. Find (a) the
angle between the planes ; (b) the plane upon which the stress is wholly
a shear; (c) the planes of principal stress.
Ans. (a) 11 38'.
(b) 64.6 Ibs. ; y = 3 26'.
(c) pi = 106.46 (tens,) ; p* = 39.26 (compr.).
11. In the preceding question find the conjugate stresses at the given
point having the common obliquity 45. Ans. Impossible.
12. At a point within a strained mass the principal stresses at a given
point are in the ratio of 3 to i. Find the ratio of the conjugate stresses
at the same point having the common obliquity 30. Also find the in-
clination of the axis of greatest principal stress to the horizontal.
Ans. Equal ; 60.
13. A wall 3 feet thick, of rectangular section and weighing 125 Ibs.
per cubic foot, is subjected to a horizontal thrust of 800 Ibs. per foot rum
296 THEORY OF STRUCTURES.
at its top. What should be the height of the wall in order that all the
joints above the base may be f nationally stable ? Coefficient of friction
= unity. Ans. 12 ft.
14. A wall 12 ft. high, 2 ft. wide at the top, and 3 ft. wide at the bot-
tom, is constructed of masonry weighing 120 Ibs. per cubic foot. The
overturning force on the rear face of the wall, which is plumb, is a hori-
zontal force P acting at 4 ft. from the base. Find P so that the devia-
tion of the centre of pressure in the base may not exceed \ ft. The
centre of pressure being fixed at 2 in. from the middle of the base, show
that | of the section may be removed without altering its stability, and
find the increase in the inclination of the resultant pressure on the base
to the vertical, consequent on the removal.
Ans. 360 Ibs.; tangents of angles are in ratio of 5 to 3.
15. A reservoir wall is 4 ft. wide at top, has a front batter of i in 12, a
rear batter of 2 in 12, and is constructed of masonry weighing 125 Ibs.
per cubic foot; the maximum compression is not to exceed 12,800 Ibs.
per square foot. Find the limiting height of the wall.
Ans. 24 ft., q being f
1 6. A dock-wall, plumb at the rear and having a face with a batter of
i in 24, is 20 ft. high and 9 ft. wide at the base. Counterforts are built
at intervals of 12 ft., projecting 3 ft. from the rear and 6 ft. wide.
Determine the thickness of an equally strong wall without counterforts,
with the same face-batter and also plumb in the rear.
Ans. 10.95 ft-
17. If the walls in the preceding question are founded in earth weigh-
ing 112 Ibs. per square foot and having an angle of repose of 32,
find the least depth of foundation in each case, the masonry weighing
125 Ibs. per cubic foot. Ans. 2.72 ft. ; 2.71 ft.
1 8. A vertical retajning-wall is strengthened by means of vertical
rectangular anchor-plates having their upper and lower edges 18 and 22
ft., respectively, below the surface. Find the holding power per foot of
width, the earth weighing 130 Ibs. per cubic foot and having an angle of
repose of 30. Ans. 27,733^ Ibs.
19. Determine the limiting depths of foundation for (a) a wall of
rectangular section 20 ft. high ; (b} for a wall of trapezoidal section hav-
ing plumb rear and front faces 4 and 20 ft. high respectively. Angle
of repose of earth =30; weight of earth = 112 Ibs. per cubic foot;
of masonry = 140 Ibs. Ans. (a) 3.22 ft. ; (b) 1.93 ft.
20. A wall 20 ft. high and 6 ft. thick retains earth on one side level
with the top, and on the other the earth rises up the wall at its natural
slope, viz., 45, to the height of 5 ft. Will the wall stand or fall ?
(Weight of masonry per cubic foot = 130 Ibs.; of earth = 120 Ibs.)
Find the locus of the centres of pressure of successive layers.
EXAMPLES. 297
Ans. Overturning moment = 4128 ft.-lbs ; moment of stability
= 93600? + 750 ( 6 ?) = 369! 2 i ft.-lbs if q = f.
The wall is stable.
21. The upper half of the section of a masonry wall is a rectangle
4 ft. wide, and the lower half a rectangle 6 ft. wide, one face being plumb.
Find the height of the wall so that the stress on the base may nowhere
exceed 10,000 Ibs. per square foot when the wall retains water (a) on the
plumb face, (b) on the stepped face.
(Masonry weighs 125 Ibs. per cubic foot.)
Ans. (a) 13.08 ft.; (b) 9.8 ft.
22. A masonry dam h ft. high is a right-angled triangle ABC in sec-
tion, and retains water on the vertical face AB. Show that the thickness
4/ u
t of the base BC is given by f = - - , qt being the deviation of the
centre of pressure in the base from the middle point.
4/ 2
Also show that the thickness will be given by f = - - if the
rock upon which the wall is built is seamy, and if it is assumed that the
communication between the water in the seams, and that in the reservoir
produces an upward pressure upon the base BC, varying uniformly from
that equivalent to the head at B to nil at C. If q = $, show that, in
order that the wall may slide, the coefficient of friction must be less than
67 per cent in the first and 81 per cent in the second case.
(Weight of a cubic foot of masonry = 2 x weight of cubic foot of
water.)
23. A wall 30 ft. high is of triangular section ABC* the face AB being
plumb, and water being retained on the side AC level with the top of the
wall ; the masonry weighs 125 Ibs. per cubic foot. Find the thickness of /
the base BC (a) when q = f ; (b) when stress in masonry is not to exceed
10,000 Ibs. per square foot ; (c) when q = f and the wall also retains earth
on the side AB level with the top, the angle of repose being 30.
Ans. (a) 17.69 ft.; (b) 13.19 ft. ; (c) 17 ft.
24. A wall 4 ft. wide at the top, with a front batter of I in 8, and a
rear batter of i in 12, is 30 ft. high. Will the wall be stable or unstable
(i) when it retains water level with the top ; (2) when it retains earth?
(Weight of masonry per cubic foot = 125 Ibs.; of earth = 112 Ibs. ;
angle of repose = 30 ; and q = f .)
Ans. (i) Moment of wt. = 128,863 ft.-lbs. ; overturning moment
= 281,250 ft.-lbs., and wall is therefore unstable.
(2) Moment of wt. = 148,251 Ibs.; overturning moment
= 168,000 Ibs., and wall is therefore unstable.
25. The faces of a reservoir wall 4 ft. wide at top and 40 ft. high have
the same batter, and water rises on one side to within 6 ft. of the top.
Find the batter, assuming (a) that tbs pressure on the horizontal base is
298 THEORY OF STRUCTURES.
to be nowhere negative ; (b) that the pressure varies uniformly and at
no point exceeds 10,000 Ibs. per square foot.
(Weight of masonry =125 Ibs. per cubic foot.)
Ans. (a) 35.8 ft.; (b) 30 ft.
26. The faces AB, AC of a wall are parabolas of equal parameters hav-
ing their vertices at B and C\ water rises on one side to the top of the
wall. Determine the thickness of the horizontal base BC, (a) for a wall
50 ft. high ; (b) for a wall 100 ft. high, so that the pressure on the base
may at no point exceed 10,000 Ibs. per square foot. Also (c) compare
the volume of such wall with the volume of an equally strong wall of the
same height, but with a section in the form of an isosceles triangle with
its vertex at A.
(Weight of masonry = 125 Ibs. per cubic foot.)
Ans. () 32.44 ft. ; (b) 119. 17 ft.
(c} in case (a) ratio = 7 : |/i 18 ;
" (b) " =4/156:21.
27. The water-face AC of a wall has a batter of i in 10 ; the width of
the wall AD at the top is 6 ft. ; the rear of the wall DEF has two slopes,
DE, having a batter of 2 in 10, and EF, a batter of 78 in TOO; the masonry
weighs 125 Ibs. per cubic foot, and the maximum compression must not
exceed 85 Ibs. per square inch. Find the safe heights of the two portions
AE and EC.
28. The section ABCD of a retain ing-wall for a reservoir has a verti-
cal face BC and a parabolic water-face AD, with the vertex at D. The
width of the base DC 4 x width of the top AB. If AB = 6 ft., find
the height of the wall, and trace the curves of resistance (a) when the
reservoir is full ; (b) when empty.
(Cubic foot of masonry = 2 x cubic foot of water.)
Ans. 32 ft. if q = \, and then max. compn. = 8000 Ibs. per sq. ft.
29. The figure represents the section of the upper portion of a masonry
dam which has to retain water level with the top of
the dam. The face AC is plumb for a depth of 73 ft.
The width of the section is constant and = 22-J ft. for
a depth AB = 40 ft.
Find the maximum stress in the masonry at the
\ horizontal bed BF. With the same maximum stress,
\ what should be the width of the horizontal bed CG,
\ FG being straight ?
^ ~ Q (Masonry weighs 130 Ibs. per cubic foot.)
FlG - a 37- Ans. 20,720 Ibs. per sq. ft.
30. A wall of an isosceles triangular section with a base 36 ft. wide
has to retain water level with its top. How high may such a wall be
built consistent with the condition that the stress in the masonry is
nowhere to exceed 10,546! Ibs. per square foot ?
(Weight of masonry per cubic foot = 125 Ibs.)
Ans. 54 ft., and q = &.
EXAMPLES. 299
31. When a cylindrical bar is twisted, show that it is subjected to
shears along transverse and radial longitudinal sections, or to tensions
and compressions on helices at 45 to the axis.
32. Find the work done in gradually and uniformly compressing a
body of volume V\ to the volume F a , p being the final intensity of
pressure and k the modulus of compression. Also show that the
intensity of stress is constant throughout the body.
An,. .
33. A bar is stretched under a force of intensity/. If the bar is pre-
vented from contracting, find the lateral stress ; also find the extension.
An,. --'; t' n "- m -^.
m I h. m(in i)
34. Taking the value of the coefficient of elasticity (E) and the co-
efficient of rigidity (G) to be 15,000 and 5750 tons for steel, 13,950 and
5450 tons for wrought-iron, and 9500 and 3750 tons for cast-iron, find the
coefficient of elasticity of volume (AT), and also the values of the direct
elasticity (A) and the lateral elasticity (/I), assuming the metals to be
isotropic.
' m K A A
Ans. Steel ............ 3f 1277?!
Wrought-iron... 3 f 4 10559!
Cast-iron ........ 3! 6785^- ^-G \G
35. A body is distorted without compression or expansion ; find
the work done.
Ans. -1 C\NS + AV + N 3 * + 2 (TV + TV + T
36. Find the work required to twist a hollow cylinder of external
radius 7?i , internal radius RZ , and length / through an angle a.
ncr*
Ans. n(RS Rf).
4/
Prove that torsion is equivalent to a shear at each point.
37. Show that a simple elongation is equivalent to a cubical dilation
and a pair of shearing or distorting stresses.
38. Find the resultant shearing stress at any point in the surface of
the transverse section of an elliptic cylinder. (Art. 24, Case b.)
Ans. 26 c a , p being the perpendicular from the centre
upon the tangent to the ellipse at the given point, and
2b, 2.c the major and minor axes.
39*. A cylinder undergoes torsion round its axis. Show that the
curves of no traction are concentric circles.
CHAPTER V.
FRICTION.
I. Sliding Friction. Friction is the resistance to motion
which is always developed when two substances, whether solid,
liquid, or gaseous, are pressed together and are compelled to
move the one over the other. If P is the mutual pressure, and
if F is the force which must act tangentially at the point of
contact to produce motion, the ratio of ^to Pis called the co-
efficient of friction and may be denoted by/. The value of/
does not depend upon the nature of any single substance, but
upon the nature and condition of the surfaces of contact of a
pair of substances. It is not the same, e.g., for iron upon iron
as for iron upon bronze or upon wood ; neither is it the same
when the surfaces are dry as when lubricated.
The laws of friction as enunciated by Coulomb are :
(i) That / is independent of the velocity of rubbing ; (2)
that /is independent of the extent of surface in contact; (3)
that /depends only on the nature of the surfaces in contact.
The friction between two surfaces at rest is greater than
when they are in motion, but a slight vibration is often suffi-
cient to change the friction of rest to that of motion.
Morin's elaborate friction experiments completely verified
these laws within certain limits of pressure (from f Ib. to 128
Ibs. per square inch) and velocity (the maximum velocity being
10 ft. per second), and under the conditions in which they
were made.
A few of his more important results are given in the follow-
ing table :
300
SLIDING FRICTION.
301
Material.
State of Surfaces.
Coefficient of Friction.
Wood on wood
dry. .
.25 tO .5
Metal on wood
dry. .
.2 " .6
> < >
wet
22 " .26
Metal on metal
dry
.15 " .2
wet . .
.3
Metal and wood c
.15
on each other 1
or each on itself '
occasionally lubricated as usual. . . .
constantly lubricated
.07 to .08
.OS
The apparatus employed in carrying out these experiments
consisted of a box which could be loaded at pleasure, and
which was made to slide along a horizontal bed by means of a
cord passing over a pulley and carrying a weight at the end.
The contact-surfaces of the bed and box were formed of
the materials to be experimented upon. The pull was meas-
ured and recorded by a spring dynamometer.
More recent experiments, however, have shown that
Coulomb's laws cannot be regarded as universally applicable,
but that / depends upon the velocity, the pressure, and the
temperature. At very low velocities Morin's results have
been verified (Fleeming Jenkin). At high velocities f rap-
idly diminishes as the velocity increases. Franke, having
carefully examined the results of various series of experi-
ments, especially those of Poire"e, Bochet, and Galton, has
suggested the formula
v being the velocity and/ , a, coefficients depending upon the
nature and condition of the rubbing surfaces.
For example,
f n = .29 and a = .04 for cast-iron on steel with dry sur-
faces.
/ .29 and a = .02 for wrought-iron on wrought-iron with
dry surfaces.
/ = .24 and a .0285 for wrought-iron on wrought-iron
with slightly damp surfaces.
Ball has shown that at very low pressures f increases as
3C2 THEORY OF STRUCTURES.
the pressure diminishes, while Rennie's experiments indicate
that at very high pressures/ rapidly increases with the press-
ure, and this is perhaps partly due to a depression, or to an
abrasion of the rubbing surfaces.
2. Inclined Plane. Let a body of weight P slide uni-
formly up an inclined plane under a force Q inclined at an
angle ft to the plane.
Let F be the friction resisting the mo-
tion, R the pressure on the plane, and a the
plane's inclination.
The two equations of equilibrium are
F = Q cos j3 P sin a
and
R = Q sin ft + P cos ex.
F Qcos/3 Psma
/. -=r = 7=5 : 77-: n = coefficient of friction = /.
R Q sin /) -f- P cos a J
Let the resultant of F and R make an angle with the
normal to the plane. Then
_F_ Q cos ft P sin a Q _ sin (a + 0)
~R~ - Qsin P + Pcosa' P ~ cos (ft - $)'
is called the angle of friction. It has also been called the
angle of repose, since a body will remain at rest on an inclined
plane so long as its inclination does not exceed the angle of
friction.
If a = o = ft, then ^ = tan = /.
The work done in traversing a distance x = Q cos ft . x. If
Q is variable, the work done / Q cos ft . dx
3. Wedge. The wedge, or key, is often employed to con-
nect members of a structure, and is generally driven into posi-
WEDGE.
303
tion by the blow of a hammer. It is also employed to force
out moisture from materials by induc-
ing a pressure thereon.
The figure represents a wedge de-
scending vertically under a continuous
pressure P, thus producing a lateral
motion in the horizontal member C,
\yhich must therefore exert a pressure
Q upon the vertical face AB.
The member H is fixed, and it is
assumed that the motion of the machine
is uniform, so that the wedge and 7 are
in a state of relative equilibrium.
Let R l , R^ be the reactions at the faces DE, DF, respec-
tively, their directions making an angle 0, equal to the angle
of friction, with the normals to the corresponding faces.
Let a be the angle between DE and the vertical, a' the
angle between DF and the vertical.
Consider the wedge, and neglect its weight, which is usually
inappreciable as compared with P.
Resolving vertically,
R l cos (90 a + 0) + RS cos (90 a'
= P=R 1 sin (a + 0) + R, sin (a f -\- 0).
Resolving horizontally,
or
R, sin (90 a + 0) R^ sin (90 a' + 0) = o,
R l cos (or + 0) = ^ 2 cos (a' + 0) (2)
Consider the member C, and neglect its weight.
Resolving horizontally,
R, cos (a + 0) = Q = R, cos (a f + 0). .
(3)
Assuming the wedge isosceles, as is usually the case, a = a',
and hence,
by eq. (2), R l = R tt and by eq. (i), 2R, sin (a + 0) = P. (4)
3O4 THEORY OF STRUCTURES.
Hence, by eqs. (3) and (4),
Q _ cot (OL + 0) __ external resistance overcome
P~ 2 effort exerted " ' 5 '
(N.B. This ratio of resistance to effort is termed the mechan-
ical advantage, or put chase, of a machine.)
Suppose the motion of the machine reversed, so that Q be-
comes the effort and P the resistance.
The reactions R, , R^ now fall below the normals, and the
equations of relative equilibrium are the same as the above,
with substituted for 0.
Thus, = cot (a 0) (6)
The two cases may be included in the expression
-p =*cot(0) (7)
For a given value of P, Q increases with a.
If there were no friction, would be zero, and eq. (7) would
become
Q cot a
Thus, the effect of friction may be allowed for, by assuming
the wedge frictionless, but with an angle increased by 20 in the
first case, and diminished by 20 in the second case.
Again, when P is the effort and Q the resistance, eq. (5)
shows that if a -\- > 90, the ratio -y is negative, which is
impossible, while if a -\- = 90, -~ is zero, and in order to
WEDGE. 305
overcome Q, however small it might be, P would require to be
infinitely great. Hence,
a + must be < 90,
Q
and below this limit -p diminishes as increases.
Similarly, it may be shown from eq. (7) that when Q is the
effort and P the resistance,
must be < a,
and that below this limit 5- increases with 0.
Efficiency. During the uniform motion of the machine, let
any point a descend vertically to the point b. The correspond-
ing horizontal displacement is evidently 2bc.
The motive work = P . ab ;
" useful work =Q.2bc.
Hence, the efficiency = ~ 7- = -7,- . 2 tan a
= tan a cot (a -f- 0), by eq. (5).
This is a maximum for a given value of when
and the max. efficiency = tan ^45 j cot ^45 + -J
For the reverse motion, the efficiency
P.ab
3 o6
THEORY OF STRUCTUKES.
This is a maximum when a =. 45 -{- . Thus the
imax. efficiency = cot ^45 -f- -j tan 145 -) = -
sin
-{- sin 0'
4. Screws. A screw is usually designed to produce a
linear motion or to overcome a resistance in the direction of its
length. It is set in motion by means of a couple acting in a
plane perpendicular to its axis. A reaction is produced be-
tween the screw and nut which must necessarily be equivalent
to the couple and resistance, the motion being steady.
Take the case of a square * -threaded screw. It may be
assumed that the reaction is concentrated along a helical line,
whose diameter, d, is a mean between the external and internal
diameters of the thread, and that its distribution along this
line is uniform. It will also be supposed that the axes of the
couple and screw are coincident, so that there will be no lateral
pressure on the nut.
Let M be the driving couple.
" Q " " axial resistance to be over-
come.
" r " " reaction at any point a of the
helical line, and let be
angle between its direction
and the normal at a ; is
the angle of friction.
" a " " angle between the tangent
at a and the horizontal ; a
is called the pitch-angle.
Since the reaction between the screw and nut must be
equivalent to M and Q, then
* Square-threaded screws work more accurately than those with a V-thread,
but the efficiency of the latter has been shown to be very little less than that of
the former (Poncelet). On the other hand, the V-thread is the stronger, much
less metal being removed in cutting it than is the case with a square thread.
Again, with a V-thread there is a tendency to burst the nut, which does not
obtain in a screw with a square thread.
FIG. 240.
SCRE W 'S. SO/
Q algebraic sum of vertical components of the reac-
tions at all points of the line of contact,
= 2\r cos (a + 0)] = cos (a + <t>)S(r), .... (i)
and M = algebraic sum of the moments with respect to the
axis of the horizontal components of the reactions at all points
of the line of contact,
. . . (2)
Let the couple consist of two equal and opposite forces, P,
acting at the ends of a lever of length/, so that M = Pp.
Hence, by eqs. (i) and (2),
and the mechanical advantage
Q ip
If = o, -p = cot of, and the effect of friction may
be allowed for, by assuming the screw frictionless, but with a
pitch-angle equal to a -\- 0.
Again, let the figure represent one complete turn of the
thread developed in the plane of the
paper. CD is the corresponding length
of the thread ; DE the circumference
it d; CE, parallel to the axis, the pitch 5"
h ; and CDE the pitch-angle a. FG. 241.
The motive work in one revolution = M . 2?r = Pp . 2n.
The useful work done in one revolution = Qh.
Hence, the efficiency = ~^ = 22- co t (a + 0) j^
= cot ( a + 0) = tan a cot ( a + 0)- (4)
3O8 THEORY OF STRUCTURES.
This is a maximum when a = 45 -, its value then being
In practice, however, a is generally much smaller, efficiency
being sacrificed to secure a large mechanical advantage, which,
according to eq. (3), increases as a diminishes.
If a -\- = 90, = o, so that to overcome Q, however
small it may be, would require an infinite effort P.
Suppose the pitch-angle sufficiently coarse to allow of the
screw being reversed. Q now becomes the effort and P the
resistance. The direction of r falls on the other side of the
normal, and the relation between P and Q is the same as
above, being substituted for 0.
Thus,
and therefore the mechanical advantage
P
If a = 0, = o, and to overcome P, however small it
may be, Q would require to be infinite.
.-. a > 0.
If a < 0, reversal of motion is impossible, and the screw
then possesses the property, so important in practice, of serv-
ing to fasten securely together different structural parts, or of
locking machines.
ENDLESS SCREWS.
309
Again, it may be necessary to take into account the friction
between the nut and its seat, as well as the friction at the end
of the screw. The corresponding moments of friction with
respect to the axis are (Art. 8)
f
7
3 4 -
and / (
/ being the coefficient of friction, d, , d^ the external and inter-
nal diameters of the seat, and d' the diameter of the end of
the screw.
5. Endless Screws (Fig. 242). A screw is often made
to work with a toothed wheel, as, for ex-
ample, in raising sluice-gates, when the screw
is also made sufficiently fine to prevent, by
friction alone, the gates from falling back
under their own weight. The theory is very
similar to the preceding. Let the screw drive. (
A tooth rises on the thread, and the wheel
turns against a tangential resistance Q, which
is approximately parallel to the axis of the
screw.
FIG. 242.
Let Fig. 243 represent one complete turn of the thread
developed in the plane of the paper, a
being the pitch-angle as before.
Consider a tooth. It is acted upon
by Q in a direction parallel to the
axis, and by the reaction R between
the thread and tooth, making an angle
(the angle of friction) with the normal
to the thread CD.
FIG. 243.
.-. Q = R cos (a + 0).
Again, the horizontal component of R, viz., R sin (a -|- 0),
has a moment R sin (a -f- 0) with respect to the axis of the
3IO THEORY OF STRUCTURES.
screw, and this must be equivalent to the moment of the driv-
ing-couple, viz., Pp (Art. 4).
= * sin (a
Thus the relation between P and Q is the same as in the pre
ceding article.
Similarly if the wheel acts as the driver,
6. Rolling Friction. The friction between a rolling body
and the surface over which it rolls is called rolling friction.
Prof. Osborne Reynolds has given the true explanation of the
resistance to rolling in the case of elastic bodies. The roller
produces a deformation of the surfaces in contact, so that the
distance rolled over is greater than the actual distance between
the terminal points. This he verified by experiment, and con-
cluded that the resistance to rolling was due to the sliding of
one surface over the other, and that it would naturally increase
or diminish with the deformation. In proof of this he found,
for example, that the resistance to an iron roller on india-
rubber is ten times as great as the resistance when the roller is
on an iron surface. Hence the harder and smoother the sur-
faces, the less is the rolling friction. The resistance is not
sensibly affected by the use of lubricants, as the advantage of
a smaller coefficient of friction is largely counteracted by the
increased tendency to slip. Other experiments are yet re-
quired to show how far the resistance is modified by the
speed.
Generally, as in the case of ordinary roadways, the resist-
ance is chiefly governed by the amount of the deformation of
the surface and by the extent to which its material is crushed.
Let a roller of weight W (Y\g. 244) be on the point of motion
under the action of a horizontal pull R.
ROLLING FRICTION. 31 1
The resultant reaction between the surfaces in contact
must pass through the point of intersection of R and IV.
Let it also cut the surface in the point B.
Let d be the horizontal distance between B and W.
11 p " vertical " " B " R.
Taking moments about B y
Rp = Wd,
or R<
R = the resistance = W-.
P
Coulomb and Morin inferred, j w
as the results of a series of ex- FIG. 2 44 .
periments, that d is independent of the load upon the roller as
well as of its diameter,* but is dependent upon the nature of
the surfaces in contact.
* Dupuit's experiments led him to the conclusion that d is proportional to
the square root of the diameter, but this requires further verification.
Let n be the coefficient of sliding friction.
The resistance of the roller to sliding is /j, W, and " rolling " will be insured
d
if R < fj. W, i.e., if - < tan 0, which is generally the case so long as the direc-
/
tion of R does not fall below the centre of the roller.
Assume that R is applied at the centre. The radius r may be substituted
for/, since (/is very small, and hence
R = W-.
r
An equation of the same form applies to a wheel rolling on a hard roadway
over obstacles of small height, and also when rolling on soft ground. In the
latter case, the resistance is proportional to the product of the weight upon the
wheel into the depth of the rut, and the depth for a small arc is inversely pro-
portional to the radius.
Experiments on the tractional resistance to vehicles on ordinary roads are
few in number and incomplete, so that it is impossible to draw therefrom any
general conclusion.
From the experiments carried out by Easton and Anderson, it would appear
that the value of d in inches varies from 1.6 to 2.6 for wagons on soft ground,
and that the resistance is not sensibly affected by the use of springs. Upon
a hard road, in fair condition, the resistance was found to be irom to \ of
that on the soft ground, the average value of d being \ inch, and was very
sensibly diminished by the use of springs.
312
THEORY OF STRUCTURES.
7. Journal-friction. Experiments indicate that f is not
the same for curved as for plane surfaces, and in the ordinary
cases of journals turning in well-
lubricated bearings the value of /is
probably governed by a combina-
tion of the laws of fluid friction and
of the sliding friction of solids.
The bearing part of the journal
is generally truly cylindrical and is
terminated by shoulders resting
against the ends of the step in
which the journal turns.
Consider a journal in a semicircular bearing with the cap
removed. When the cap is screwed on, the load upon the
journal will be increased by an amount approximately equal
to the tension of the bolts. Let -Pbe the load.
Assume that the line of action of the load is vertical and
that it intersects the axis of the shaft. This load is balanced
by the reaction at the surface of contact, but much uncertainty
exists as to the manner in which this reaction is distributed.
There are two extremes, the one corresponding to a normal
pressure of constant intensity at every point of contact, the
other to a normal pressure of an intensity varying from a
maximum at the lowest point A to a minimum at the edge of
the bearing B.
Let / be the length of the bearing, and consider a small
element AS at any point C, the radius OC (= r) making an
angle 6 with the vertical OA.
First. Let / be the constant normal intensity of pressure.
p =
e. i)
2pir.
Frictional resistance =
= fpl^(AS}=fplnrfP- .
The frictional resistance probably approximates to this
limit when the journal is new.
JO URN A L- FRICTION. 3 1 3
Second. Let / = p Q cos 0,
so that the intensity is now proportional to the depth CD and
varies from a maximum / at A to nil at B. This, perhaps,
represents more accurately the pressure at different points
when the journal is worn.
2/ /r, A cos a . dO = / /r-
t/o 2
and the frictional resistance = 2(fj>4Sl) 2fpjr = fP-.
Hence, the frictional resistance lies between fP- and fP .
2 7t
It may be represented by j*P t IJL being a coefficient of friction
to be determined in each case by experiment.
The total moment of frictional resistance must necessarily
be equal and opposite to the moment M of the couple twisting
the shaft ; i.e.,
M =
Thus, the total reaction at the surface of contact is equiva-
lent to a single force P tangential to a circle of radius /*r having
its centre at O and called the friction-circle.
The work absorbed by axle-friction per revolution
= M.27t =. 2
The work absorbed by axle-friction per minute
N being the number of revolutions and v the velocity per
minute.
314 THEORY OF STRUCTURES.
The work absorbed by frictional resistance produces an
equivalent amount of heat, which should be dissipated at once
in order to prevent the journal from becoming too hot. This
may be done by giving the journal sufficient bearing surface
(an area equal to the product of the diameter and the length
of the bearing), and by the employment of a suitable unguent.
Suppose that h units of heat per square inch of bearing
surface (Id) are dissipated per minute.
Let / inches be the length and d inches the diameter of
the journal.
hdl = heat-units dissipated = heat-units equivalent to
frictional resistance
J being Joule's equivalent, or 778 ft.-lbs.
12/ft PN \2jJl Pv
.*. - = r- and -- = .
fJLTt I ^ Id
p
Let 7-7 =p = pressure per square inch of bearing surface.
Id
\2jh
= a constant.
In Morin's experiments af varied from 2 to 4 in., P from
330 Ibs. to 2 tons, and v did not exceed 30 ft. per minute; so
that/z/ was < 5000, and the coefficient of friction for the given
limits was found to be the same as for sliding friction.
Much greater values of pv occur in modern practice.
Rankine gives p(v + 20) = 44800 as applicable to locomo-
tives.
Thurston gives pv = 60000 as applicable to marine engines
and to stationary steam-engines.
Frictional wear prevents the diminution of /below a certain
JOURNAL FRICTION. 31$
limit at which the pressure per unit of bearing surf ace exceeds
a value/ given by the formula.
where
-
In practice k = % for slow-moving journals (e.g., joint-pins),
and varies from I-J to 3 for journals in continuous motion. The
best practice makes the length of the journal equal to four
diameters (i.e., k = 4) for mill-shafting.
Again, 'if the journal is considered a beam supported at
the ends,
q being the maximum permissible stress per square inch, and
C a coefficient depending upon the method of support and
upon the manner of the loading.
k
/. # oc .
i
For a given value of P, d diminishes as q increases. Also,
it has been shown that the work absorbed by friction is
directly proportional to d.
Hence, for both reasons, d should be a minimum and the
shaft should be made of the strongest and most durable
material. In practice the pressure per square inch of bearing
surface may be taken at about 2 tons per square inch for cast-
iron, 3! tons per square inch for wrought-iron, and 6J tons per
square inch for cast-steel.
It would appear, however, from the recent experiments of
Tower and others, that the nature of the material might become
of minor importance, while that of a suitable lubricant would be
of paramount importance. They show that the friction of
properly lubricated journals follows the laws of fluid friction
much more closely than those of solid friction, and that the
3l6 THEORY OF STRUCTURES.
lubrication might be made so perfect as to prevent any ab-
solute contact between the journal and its bearing. The
journal would therefore float in the lubricant, so that there
would be no metallic friction. The loss of power due to fric-
tional resistance, as well as the consequent wear and tear, would
be very considerably diminished, while the load upon the
journal might be increased to almost any extent.
Tower's experiments also indicate that the friction dimin-
ishes as the temperature rises, a result which had already been
experimentally determined by Him. It was also inferred by
Hirn that, if the temperature were kept uniform, the friction
would be approximately proportional to Vv, and Thurston
has enunciated the law that, with a cool bearing, the friction is
approximately proportional to Vv for all speeds exceeding
100 ft. per minute.
With a speed of 150 ft. per minute and with pressures vary-
ing from 100 to 750 Ibs. per square inch, Thurston found ex-
perimentally that /"varied inversely as the square root of the
intensity of the pressure. The same law, but without any
limitations as to speed or pressure, had been previously stated
by Hirn.
8. Pivots. Pivots are usually cylindrical, with the circular
edge of the base removed and sometimes with the whole of
the base rounded. Conical pivots are employed in special
machines in which, e.g., it is important to keep the axis of the
shaft in an invariable position. Spherical pivots are often
used for shafts subject to sudden shocks or to a lateral move-
ment.
(a) Cylindrical Pivots. If the shafts are to run slowly, the
intensity of pressure (/) on the step should not be so great as
to squeeze out the lubricant. Reuleaux gives the following
rules :
The maximum value of / in Ibs. per square inch should be
700 for wrought-iron on gun-metal, 470 for cast-iron on gun-
metal, and 1400 for wrought-iron on lignum-vitae.
For rapidly-moving shafts,
d=c
PIVOTS.
317
n being the number of revolutions per minute, c a coefficient
to be determined by experiment (=.0045),
and Pthe load upon the pivot.
Suppose the surface of the step to be
divided into rings, and let one of these
rings be bounded by the radii x, x -J- dx.
In one revolution the work absorbed
by the friction of this ring
= ji . 2nx . dx . 2nx.
Hence the total work absorbed in one revolution
where
FIG. 246.
1? - d*
and d l , d^ are the external and internal diameters of the sur-
face in contact.
If the whole of the surface is in contact, d^ = o, and the
work absorbed = \^,nPd r
Again, the moment of friction for the ring
.dx.x =
and the total moment
12
If d^ o, the moment = d l .
Thus, in both cases, the work absorbed by friction = 27t
times the moment of friction.
THEORY OF STRUCTURES.
Let D be the mean diameter of the surface in contact
Let 2y be the width of the surface in contact = d^ d^ .
Then
work absorbed = fj.nP \D -\- J.
Sometimes shafts have to run at high speeds and to bear
heavy pressures, as, e.g., in screw-propellers and turbines. In
order that there may be as little vibration as possible, p must
be as small as practicable, and this is to some extent insured
by using a collar-journal.
Let N be the number of collars, and let d^ , d^ be the exter-
nal and internal diameters of a collar.
Then work absorbed by friction per revolution per collar
L 2n X moment of friction.
o
According to Reuleaux, the mean diameter of a collar
n
~-
n being the number of revolutions per minute.
Also, the width of surface in contact = d l d^ = .48
and the maximum allowable pressure per square inch
(b) Wear. The wear at any point of the elementary ring
must necessarily be proportional to the friction ///>, and also to
the amount of rubbing surface which passes over the point in
a unit of time, i.e., the velocity Ax ; A being the angular ve-
locity of the shaft.
PIVOTS. 319
Hence, the wear at any point is proportional to
(c) Conical Pivots. As before, suppose
the surface of the step to be divided into a
number of elementary rings. Two cases will
be discussed :
First. Assume that the normal intensity
of pressure p at the surface of contact is
constant.
Let x,x-\-dx be the distances of D and
E, respectively, from the axis. IGZ
The total moment of friction
sin a
3 sin a
x^ , x^ being the radii of the top and bottom sections of the
step.
Also, P, the total load on the pivot,
= / pDE sin a . 2nx = 2np I xdx
U JC<t V X<i
= P(x? - *.)
2 U.P X* _ X 8
Hence total moment of friction = - -- - - -.
3 sin a x? x*
Second. Assume that the wear is of such a nature that every
point, e.g., D, descends vertically through the same distance.
Thus, the normal wear a sin a,
or up x a sn a,
or px oc sin a.
In the present case a is constant, and hence px = a con-
stant.
320 THEORY OF STRUCTURES.
Thus, total moment of friction
sm a
=-*>
r* 1
= I
t/.r a
/*,
I dx
IX * f
Also, P = I pDE sin a . inx
t/.r a
= 2npx
Hence total moment of friction = : - (x. + #,).
2 sm a v *
Schieles Pivots. The object aimed at in these pivots
is to give the step such a form that the wear
and the pressure are the same at all points.
Let be the angle made by the tangent at
-V--/D any point of the step with the axis.
Let y be the distance of the point from
the axis. Then
py a sin 0;
and hence if/ is constant,
y a sin 6 or y cosec = a const.
is the equation of the generating line of the step. This line is
known as the tractrix and also as the anti-friction curve. If
the tangent at D intersects the axis in T,
DT = y cosec 6 = a const.
The curve may be traced by passing from one point to an-
other and keeping the tangent DT of constant length.
The above equation may be written
ds
yr- = a const. = a,
' dy
BELTS AND ROPES.
321
or
ds a dy I fdy \
~d~x = ~~ ~y ~dx = V l + w '
which may be easily integrated, the result being the analytical
equation to the curve, viz.,
-f- Va* y + a const.
Schiele or anti-friction pivots are suitable for high speeds, but
have not been very generally adopted.
9. Belts and Ropes. Let the figure represent a pulley
movable about a journal at O, and let a belt (or rope), acted
upon by forces 7, , 7 2 at the ends, embrace a portion ABC
of the circumference subtending an angle a at the centre.
In order that there may be motion in the direction of the
arrow, 7, must exceed 7 2 by an amount sufficient to overcome
the frictional resistance along the arc of contact and the resist-
ance to bending due to the stiffness of the belt.
Consider first the frictional resist-
ance, and suppose the belt to be on the
point of slipping.
Any small element BB' ( ds) of
the belt js acted upon by a pull T tan-
gential to the pulley at B, a pull T dT
tangential to the pulley at B' , and by a
reaction equivalent to a normal torceRds
at the middle point of BB' , and a tan-
gential force, or frictional resistance,
j&ds.
Let the angle COB = 0, and the an-
gle BOB' = dB. ,
Resolving normally,
FIG. 249.
7/n
(T+ T-dT)sm Rds - o.
.322 THEORY OF STRUCTURES.
Resolving tangentially,
2 ^ '
-jj. being the coefficient of friction.
Now dti being very small, sin is approximately ,
dB
cos is approximately unity, and small quantities of the
second order may be disregarded.
Hence, eqs. (i) and (2) may be written
TdO Rds o, ...... (3)
= (4)
dT
, or
Integrating,
C being a constant of integration.
When 6 = 0, T = T 9 , and hence log.T; = C.
or = ^ ........ (6)
7 2
When = a, T = T l , and hence
^ being the number 2.71828, i.e., the base of the Naperian
system of logarithms.
BELTS AND ROPES. 323
If a is increased by /?, the new ratio of tensions will be
&& times the old ratio ; so that if OL increases in arithmetical
progression, the ratio of tensions will increase in geometrical
progression. This rapid increase in the ratio of the tensions,
corresponding to a comparatively small increase in the arc of
contact, is utilized in " brakes"
for the purpose of absorbing
surplus energy. For example :
A flexible brake consisting
of an iron or steel strap, or,
again, of a chain, or of a series
of iron bars faced with wood
and jointed together, embraces
about three-fourths of the cir-
cumference of an iron or wooden
drum. One end of the brake FlG - 25 '
is secured to a fixed point and the other to the end B of a
lever AOB turning about a fulcrum at O. A force applied at
A will cause the brake to clasp the drum and so produce fric-
tion which will gradually bring the drum to rest.
Let &9 be the angular velocity of the drum before the brake
is applied.
Let / be the moment of inertia of the drum with respect to
its axis.
The kinetic energy of the drum = .
When the brake is applied, the motion being in the direc-
tion of the arrow, let the greater and less tensions at its ends
be T lt T^ t respectively.
Let n be the number of revolutions in which the drum is
brought to rest. Then
i/fi? a = (7;- T^ndn, (8)
d being the diameter of the drum.
Also, if Pis the force applied at A, and if / and q are the
perpendicular distances of O from the directions of P and T 9 ,
respectively,
Pp=T# (9)
3 24 THEOR Y OF S 7^R UCTURES.
Again,
T* = TS*, ......... (10)
<x being the angle subtended at the centre by the arc of contact.
Hence, by eqs. (8), (9), (10), f
\ _ qltf
- i)7td ..... ' '
If the motion of the drum were in the opposite direction, q
would be the perpendicular distance of O from the direction of
T, , and then Pp = T,q.
Proceeding as before,
_
and therefore the number of turns in the second case, before
the drum comes to rest, is e^ a times the number in the first,
which is consequently the preferable arrangement.
The coefficient of friction JJL varies from .12 for greasy shop
belts on iron pulleys to .5 for new belts and hempen ropes on
wooden drums. In ordinary practice, an average value of >u
for dry belts on iron pulleys is .28, and for wire ropes .24; if
the belts are wet, >u is about .38.
Formulae (6) and (7) are also true for non-circular pulleys.
10. Effective Tension. The pull.available for the trans-
mission of power = T l 7!, = 5. Let HP be the horse-
power transmitted, v the speed of transmission in feet per sec-
ond, a the sectional area of the rope or belt, and s the stress
per square inch in the advancing portion of the belt.
Then, if T l and T t are in pounds,
= ., and
'
,
550 550'
The working tensile stress per square inch usually adopted
for leather belts varies from 285 Ibs. (Morin) to 35 5 Ibs. (Claudel),
EFFECT OF HIGH SPEED. 325
an average value being 300 Ibs. In wire ropes, 8500 Ibs. per
square inch may be considered an average working tension.
Hempen ropes for the transmission of power generally vary
from 4^ to 6J- in. in circumference.
II. Effect of High Speed. When the speed of trans-
mission is great, the effect of centrifugal force must be taken
into account.
wads if
The centrifugal force or the element ds = ---- , w being
the specific weight of the belt or rope, and r the radius of the
pulley.
Eq. (3) above now becomes
wads v*
Tdd Rds ---- = o,
g r
or
and hence, by eq. (4),
Integrating,
-r Wa
T
since T = T^ when 6 = o.
Also, T = T, when 6 = a, and therefore
g
or
_ V- i).
o
326 THEORY OF STRUCTURES.
The work transmitted per second
= (T, - 7> = T,v - *>(<- - i),
which is a maximum and equal to |7' 8 (^ a i) when
, and the two tensions are then in the ratio of
2 ^ _(_ i to 3.
The speed for which no work is transmitted, i.e., the limit-
ing speed, is given by
wa , i /*
\v v = o, or v = \f
12. Slip of Belts. A length / of the belt (or rope) becomes
/( i -j-^jon the advancing side and l[\ + -jj on the slack side,
T T
where p l = - and / = - , E being the coefficient of elasticity.
Thus, the advancing pulley draws on a greater length than is
given off to the driven pulley, and its speed must therefore
exceed that of the latter by an amount given by the equation
ifi] _
reduction of speed, or slip _ ~ El \ r E' _ p l
speed of driving pulley / p
The slip or creep of the belt measures the loss of work.
In ordinary practice the loss with leather belting does not ex-
ceed 2 per cent, while with wire ropes it is so small that it may
be disregarded.
PA N Y ' S D YNA MO ME TER.
327
13. Prony's Dynamometer. This dynamometer is one of
the commonest forms of friction-brake. The motor whose
power is to be measured turns a wheel E which revolves be-
tween the wood block B and a band of wood blocks A. To
/INP
FIG. 251.
the lower block is attached a lever of radius / carrying a
weight P at the free end. By means of the screws C, D the
blocks may be tightened around the circumference until the
unknown moment of frictional resistance FR is equal to the
known moment Pp.
The weight P, which rests upon the ground when the
screws are slack, is now just balanced.
The work absorbed by friction per minute = 2nRFn = 2nPpn v
n being the number of revolutions per minute.
14. Stiffness of Belts and Ropes. The belt on reaching
the pulley is bent to the curvature of the periphery, and is
straightened again when it leaves the pulley. Thus, an amount
of work, increasing with the stiffness of the belt, must be ex-
pended to overcome the resistance to bending. As the result
of experiment, this resistance has been expressed in the form
T-B, T being the tension of the belt, a its sectional area, R the
radius of the pulley, and b a coefficient to be determined.
According to Redtenbacher, b = 2.36 in. for hempen ropes.
" " " b = 1.67 " " " "
" " Reuleaux, b = 3.4 " " leather belts.
328
THEORY OF STRUCTURES.
Let the figure represent a sheave in a pulley-block turning
in the direction of the arrow about a
journal of radius r.
Let T t be the effort, T 9 the re-
sistance.
The resistance due to the stiff-
ness of the belt may be allowed for
*T*
qTs by adding -j- to the force 7", . The
ftR OK
FIG. 252. frictional resistance at the journal-
surface is P sin orfP, P being the resultant of T lt 7 1 , .
The motion being steady, taking moments about the centre,
or
If T; and T 9 are parallel, P = T> + T>, and the last equa
tion becomes
Let the pulley turn through a small angle 6.
The counter-efficiency of the sheave
motive work Tfl _ T^
useful work z= =
2fr
, a
~^~
R - fr~~ b R - fr'
In the case of an endless belt connecting a pair of pulleys
of radius R 19 R^ the resistance due to stiffness may be taken
equal to ^-\jf "f"j>"/* ^ Dem g tne mean tension [= ~~ L ~~^
The resistance due to journal-friction = frP\- +^~J-
The useful resistance = T, T^ = S.
WHEEL AND AXLE.
Hence, the counter-efficiency
329
In wire ropes the stress due to bending may be calculated
as follows:
Let x be the radius of a wire. The radius of its axis is
sensibly the same as the radius R of the pulley.
The outer layers of the wire will be stretched, and the inner
shortened, while the axis will remain unchanged in length.
Hence,
x change of length of outer or inner strands unit stress
R ~ length of axis E
x
and the unit stress due to bending = ~ .
15. Wheel and Axle. Let the figure represent a wheel
of radius / turning on an axle of radius r, under the action of
the two tangential forces P and Q, in-
clined to each other at an angle 6.
The resultant R of P and Q must
equilibrate the resultant reaction be-
tween the wheel and axle at the sur-
face of contact.
Let the directions of P and Q
meet in T.
If there were no friction, the re-
sultant reaction and the resultant R
would necessarily pass through O
and T.
Taking friction into account, the
direction of R will be inclined to TO.
Let its direction intersect the circumference of the axle in the
point A. The angle between TA and the normal AO at A,
the motion being steady, is equal to the angle of friction ; call
it 0.
FIG. 253.
330 THEORY OF STRUCTURES.
Taking moments about O,
Pp Qp Rr sin = o
Also,
. . (I)
R* = P* + Q + 2 PQ cos e (2)
Let/= sin =
=, fjL being the coefficient of friction.
Eq. (i) may now be written
Pp- Qp-fRr = 0. . ,
' If P and Q are parallel in direction,
8 = and R = P -4- Q.
Let the figure represent a wheel and axle.
(3)
254.
Let P be the effort and Q the weight lifted, the directions
of Pand Q being parallel.
Let IV be the weight of the " wheel and axle."
Let R 1 and R^ be the vertical reactions at the bearings.
Let/ be the radius of the wheel.
Let q " " " axle.
Let r " " " bearings.
Take moments about the axis. Then
Pp Qq RS sl ' n ~ RS sin = 0. . . (4)
TOOTHED GEARING. 331
But
...... (5)
Hence,
or
P(p-fr) = Q(q+fr)+fWr. ........ (6)
Efficiency. In turning through an angle 0,
motive work = PpQ,
useful work == QqQ,
effirier -
" e -~ = '
and the ratio -p is given by eq. (6).
16. Toothed Gearing. In toothed gearing the friction is
partly rolling and partly sliding, but the former will be disre-
garded, as it is small as compared with the latter.
Let the pitch-circles of a pair of teeth in contact at the
point B touch at the point A ; and consider the action before
reaching the line of centres a a , i.e., along the arc of
approach.
332 THEORY OF STRUCTURES.
The line AB is normal to the surfaces in contact at the
point B.
Let R be the resultant reaction at B. Its direction, the
motion being steady, makes an angle 0, equal to the angle of
friction, with AB.
Let B be the angle between O,O t and AB.
Let the motive force and force of resistance be respectively
equivalent to a force P tangential to the pitch-circle O l , arid
to a force Q tangential to the pitch-circle O, .
Let r l , r a be the radii of the two wheels.
The work absorbed by friction in turning through the small
arc ds
. (i)
Consider the wheel O v , and take moments about the centre,
Pr 1 = ^|r 1 sin(6/-0) + ^sm0f, ... (2)
where AB = x.
Similarly, from the wheel <9 3
0r a = tf K sin (0 - 0) - * sin 0}. ... (3)
Hence,
jtr
n sin (0 - 0) - sin 4>
= - -- - ...... (4)
sin (0 0) -f- - sin
f i
and therefore
'i i \
(r + rl* sin
'i
+ * sn
(5)
sin (0 0) -- sin
Hence, the work absorbed by friction in the arc ds
= Q-
sin (6 0) sin
7*0
TOOTHED GEARING. 333
In precisely the same manner it can be shown that, after
leaving the line of centres, i.e., in the arc of recess,
ft sin (0 -|- 0) sin
p - x
sin (0+0) + - sin
and the work absorbed by friction in the arc ds
= Q -- . .... (8)
sin (0+0) -- sin
^2
The ratio -p and the loss of work given by eqs. (4) and (6)
are respectively greater than the ratio ^~ and the loss of work
given by eqs. (7) and (8), and therefore it is advisable to make
the arc of approach as small as possible.
Again, by eq. (4), motion will be impossible if
sin (6 0) + - sin = o ;
**i
i.e., if cot = cot
r 1 sin
and this can only be true if the direction of R passes through (9 3 .
Simple approximate expressions for the lost work and
efficiency may be obtained as follows:
6 differs very little from 90, and x is small as compared
with r 2 and differs little from the corresponding arc s meas-
ured from A.
Hence the work absorbed by friction in the arc ds
= Q tan - *& =
334 .THEORY OF STRUCTURES.
and the work lost in arc of approach s 1
The useful work done in the same interval = Qs l .
The counter-efficiency (reciprocal of efficiency)
Similarly for the arc of recess s^ ,
the lost work = g/J- + - , - (i i)
*7*j T'j/ 2
and the counter-efficiency = i -}-/*( -J-- ). . (12)
\r, /y 2
If ^ = ,y a = pitch = / = - - 1 = - ? , n^ , 2 being the num-
^1 ?^ 3
her of teeth in the driver and the follower, respectively, the ex-
pressions for the lost work given by eqs. (9) and (11) are iden-
tical, and those for the counter-efficiency given by eqs. (10)
and (12) are also identical.
Thus, the whole work lost during the action of a pair of
teeth
....... (13)
and the counter-efficiency
<
This last equation shows that the efficiency increases with
the number of teeth.
EFFICIENCY OF MECHANISMS.
335
If the follower is an annular wheel, - - must be substi-
tuted for -+- in the above equations. Thus, with an an-
nular wheel the counter-efficiency is diminished and the
efficiency, therefore, increased.
It has been assumed that R and Q are constant, as their
variation from a constant value is probably small. It has also
been assumed that only one pair of teeth are in contact. The
theory, however, holds good when more than one pair are in
contact, an effort and resistance, corresponding to P and Q,
being supposed to act for each pair.
17. Bevel-wheels. Let I A, IB represent the develop-
ments of the axes of the pitch-
circles 77, , 77 3 of a pair of bevel-
wheels when the pitch-cones are
spread out fiat, O l , O z being the
corresponding centres.
The preceding formulae will ap-
ply to bevel-wheels, the radii being
6^7, <9 2 7, and the pitch being meas- ,!,''''
ured on the circumferences I A, IB.
18. Efficiency of Mechanisms.
Generally speaking, the ratio of
the effort P to the resistance Q in a.
mechanism may be expressed as a
function of the coefficient of fric-
tion yu. Thus,
t
If, now, the mechanism is moved so that the points of
application of P and Q traverse small distances Ax, Ay in the
directions of the forces,
.1. ^ .
the efficiency = j~- =
PAx
I Ay
336
THEORY OF STRUCTURES.
Ay
But the ratio -r- depends only upon the geometrical rela-
tions between the different parts of the mechanism, and will
therefore remain the same if it is assumed that /* is zero. In
such a case the efficiency would be perfect, or the motive work
(PAx) would be equal to the useful work (QAy\ and therefore
I =
I Ay
Hence, the efficiency
TABLE OF COEFFICIENTS OF AXLE-FRICTION.
>,
Q
Greasy and
Wet.
Ordinary
Lubrication.
Continuous
Lubrication.
Pure Carriage-
grease.
Lard and
Plumbago.
O
Bell-metal on bell-metal
Brass on brass .
Brass on cast-iron
Cast-iron on bell-metal
161
o6<;
.16
Cast-iron on brass ,
Cast-iron on cast-iron
14
Cast-iron on lignum-vitae.
xSe
Lignum-vitae on cast-iron
Lignum-vitae on lignum-vitae
Wrought-iron on bell-metal
Wrought-iron on cast-iron
.251
.189
.n6
075
J 7
.07
054
.09
.11
15
Wrought-iron on lignum-vitae. . .
.187
125
EXAMPLES. 337
EXAMPLES.
i. In a pair of four-sheaved blocks, it is found that it requires a force
P to raise a weight $P', and a force $P to raise a weight i $P'. Show
that the general relation between the force P and the weight W to be
raised is given by
Find the efficiency when raising the weights 5/" and
2. Find the mechanical advantage when an inch bolt is screwed up
by a i5-in. spanner, the effective diameter of the nut being if in., the
diameter at the base of the thread .84 in., and .15 being the coefficient of
friction.
3. A belt, embracing one-half the circumference of a pulley, transmits
10 H. P. ; the pulley makes 30 revolutions per minute and is 7 ft. in
diameter. Neglecting slip, find TV and Ti ; /a being .125.
4. A -in. rope passes over a 6-in. pulley, the diameter of the axis being;
% in. ; the load upon the axis == 2 x the rope tension. Find the efficiency
of the pulley, the coefficient of axle-friction being .08 and the coefficient
for stiffness .47.
Hence also deduce the efficiency of a pair of three-sheaved blocks.
5. If the pulleys are 50 ft. c. to c. and if the tight is three times the
slack tension, find the length of the belt, the coefficient of friction being
and the diameter of one of the pulleys 12 in.
6. Show that the work transmitted by a belt passing over a pulley
/~T
will be a maximum when it travels at the rate of A/ __ 2 ft. per sec., T*
yn
being the slack tension and m the mass of a unit of length of the belt.
The tight tension on a 2o-in. belt, embracing one-half the circum-
ference of the pulley, is 1200 Ibs. Find the maximum work the belt will
transmit, the thickness of the belt being .2 in. and its weight .0325 Ib.
per cubic inch. (Coefficient of friction = .28.)
7. In an endless belt passing over two pulleys, the least tension is
150 Ibs., the coefficient of friction .28, and the angle subtended by the
arc of contact 148. Find the greatest tension. The diameter of the
larger wheel is 78 in., of the smaller 10 in., of the bearings 3 in. Find
the efficiency. A tightening-pulley is made to press on the slack side
of the belt. Assuming that the working tension is to the coefficient of
elasticity in the ratio of i to 80, find the increment of the arc of contact
THEORY OF STRUCTURES.
on the belt-pulley, the tension of the slack side, and the force of the
tightening-pulley.
8. A belt weighing Ib. per lineal foot, connects two 42-in. pulleys, one
making 240 revolutions per minute. Find the limiting tension for which
work will be transmitted. Also find the tight and slack tensions and
the efficiency when the belt transmits 5 horse-power. Diameter of axle
2 in.; coefficient of friction = .28.
9. A circular saw makes 1000 revolutions per minute and is driven
"by a belt 3 in. wide and \ in. thick, itsweight per cubic inch being .0325
Ib. The belt passes over a lo-in. pulley, embracing one-half the cir-
cumference, and transmits 6 H. P. Find the light and slack tensions,
the coefficient of friction being .28.
10. A flexible band, embracing three-fourths of the circumference of
a brake-pulley keyed on a revolving shaft, has one extremity attached to
the end A of the lever AOB, and the other to \\\t fixed point O (between
A and J3) about which the lever oscillates. The pressure between the
band and pulley is effected by a force applied at right angles to the lever
at the end B. Show that the time in which the axle is brought to rest
is about i\ times as great when revolving in one direction as in the
opposite (/ = .2).
11. In a Prony-brake test of a Westinghouse engine, the blocks were
fixed to a 24-in. fly-wheel with a 6-in. face, and the balance-reading was
48 Ibs.; the distance from centre of shaft to centre of balance, measured
horizontally, was 30 in., and the number of revolutions per minute was
624. Find the H. P. Ans. 14.3.
12. An engine makes 150 revolutions per minute. If the diameter of
the brake-pulley is 45 in. and the pull on the brake is 50 Ibs., find the
B. H. P. Ans. 2.67.
13. A small water-motor is tested by a tail dynamometer. The pul-
ley is 18 in. in diameter; the weight is 60 Ibs.; the spring registers a
pull of 50 Ibs.; the number of revolutions per minute = 500. Find the
B. H. P. Ans. f
14. The power of an engine making n revolutions per minute is
tested by a Prony brake having its arm of length r connected with a
spring-balance which registers a force P. The arm is vertical and the
weight W of the brake is supported by a stiff spring fixed vertically
below the centre of the wheel. What error in B. H. P. would be intro-
duced by placing the spring x ft. away from the central position ?
Ans. Bl ^* , B being the B. H. P.
15. Find work absorbed by friction per revolution by a pivot 3 in.
long and carrying 6 tons, its upper face being 6 in. in diameter, coeffi-
cient of friction .04, and 2 a being 90.
EXAMPLES. 339
1 6. The diameter of a solid cylindrical cast-steel pivot is 2% in. Find
the diameter of an equally efficient conical pivot.
17. The pressure upon a 4-in. journal making 50 revolutions per min-
ute is 6 tons, the coefficient of friction being .05. Find the number of
units of heat generated per second; Joule's mechanical equivalent of
heat being 778 ft.-lbs.
1 8. A water-wheel of 20 ft. diameter and weighing 20,000 Ibs. makes
10 revolutions per minute; the gudgeons are 6 in. in diameter and the
coefficient of friction is .1. Find the loss of mechanical effect due to
friction. If the motive power is suddenly cut off, how many revolutions
will the wheel make before coming to rest ? Ans. f H. P. ; 10.9.
19. A fly-wheel weighing 8000 Ibs. and having a radius of gyration of
10 ft. is disconnected from the engine at the moment it is making 27
revolutions per minute ; it stops after making 17 revolutions. Find the
coefficient of friction, the axle being 12 in. in diameter. Ans. .2325.
20. A railway truck weighing 12 tons is carried on wheels 3 ft. in
diameter ; the journals are 4 in. in diameter, the coefficient of friction
jjV Find the resistance of the truck so far as it arises from the friction
of the journals. Ans. 37^ Ibs.
21. A tramcar wheel is 30 in. in diameter, the axle 2$- in.; the coeffi-
cient of axle-friction .08, of rolling friction .09. Find the resistance pet
ton. Ans. 28.37 Ibs.
22. A bearing 16 in. in diameter is acted upon by a horizontal force
of 50 tons and a vertical force of 10 tons ; the coefficient of friction is -fa.
Find the H. P. absorbed by friction per revolution. Ans. .906 H. P.
23. A steel pivot 3 in. in diameter and under a pressure of 5 tons
makes 60 revolutions per minute in a cast-iron step well lubricated with
oil. How much work is absorbed by friction, the coefficient of friction
being .08 ?
24. A pair of spur-wheels are 4 in. and 2 in. in diameter ; the flanks
of the teeth are radial ; the larger wheel has 16 teeth ; the arc of ap-
proach = arc of recess = | of the pitch. Show how to form the teeth,
and find their efficiency. (Coefficient of friction = .n.)
25. Find the work lost by the friction of a pair of teeth, the number
of teeth in the wheels being 32 and 16, and the diameter of the larger
wheel, which transmits 3 horse-power at 50 revolutions per minute, 3 ft.
26. The driver of a pair of wheels has 120 teeth, and each wheel has
an addendum equal to .28 times the pitch ; the arcs of approach and
recess are each equal to the pitch ; the tooth-flanks are radial. (Coeffi-
cient of friction = .106.) Find the efficiency.
CHAPTER VI.
ON THE TRANSVERSE STRENGTH OF BEAMS.
FIG. 257.
I. To determine the Elastic Moment. Let the plane of
-- |h the paper be a plane of symmetry
_ | c with respect to the beam PQRS.
If the beam is subjected to the
action of external forces in this
plane, PQRS is bent and as-
sumes a curved form P'Q'R'S'.
FIG. 258. The upper layer of fibres, Q'R', is
extended, the lower layer, P'S', is compressed, while of the
layers within the beam, those nearer P'S' are compressed and
those nearer Q'R' are extended. Hence, there must be a layer
M'N' between P'S' and Q'R' which is neither compressed nor
extended. It is called the neutral surface (or cylinder), and
its axis is perpendicular to the plane of flexure. In the present
treatise it is proposed to deal with flexure in one plane only,
and, in general, it will be found more convenient to refer to
M'N' as the neutral line (or axis), a term only used in refer-
ence to a transverse section.
If a force act upon the beam in the direction of its length,
the lower layer P'S', instead of being compressed, may be
stretched. In such a case there is no neutral surface within
the beam, but theoretically it still exists some-
where without the beam.
Let ABCD be an indefinitely small rect-
angular element of the unstrained beam, and
let its length be s. Let A'B'C'D', Fig. 260,
be the element after deformation by the external forces.
340
FIG. 259.
THE ELASTIC MOMENT. 34*
P'Q', the neutral line, being neither com- ,
pressed nor extended, is unchanged in length
and equal to PQ s.
Let the normals at P r and Q' to the neutral
line meet in the point O ; O is the centre of
curvature of P'Q'*
Also, as the flexure of the element is very
small, the normal planes through OP' and OQ'
may be assumed to be perpendicular to all the
layers which traverse the corresponding sec-
tions of the beam, so that they must coincide
with the planes A ' D' and B ' C' ', respectively.
The assumptions made in the above are :
(a) That the beam is symmetrical with
respect to a certain plane.
(b) That the material of the beam is homo-
geneous. FIG. 260.
(c) That sections which are plane before bending remain
plane after bending.
(d) That the ratio of longitudinal stress to the correspond-
ing strain is the ordinary (i.e., Young's) modulus of elasticity
notwithstanding the lateral connection of the elementary
layers.
(e) That these elementary layers expand and contract
freely under tensile and compressive forces.
Consider an elementary layer p'q', of length s', sectional
area a l , and distant y l from the neutral surface.
Let OP' = R = OQ'.
From the similar figures OP'Q and Op'q',
Op' p'q' R+y, s' y v s'-s
Also, if t l is the stress along the layer p'q' ,
34 2 THEORY OF STRUCTURES.
E being the coefficient of elasticity of the material of^ the
beam.
So, if / 2 , # 2 , 7 2 , / 3 , a 3 , jj/ 3 , . . . are respectively the stress,
sectional area, and distance from the neutral surface, of the
several layers of the element,
The total stress along the beam is the algebraic sum of all
these elementary stresses,
Again, the moment of /, about P' = t^ = ^7^ ;
and so on.
Thus, the Elastic Moment for the section A ' D' = the alge-
braic sum of the moments of all the elementary stresses in the
different layers about P' 9
THE ELASTIC MOMENT. 343
Now, 2 (ay*) is the moment of inertia of the section of the
beam through A 'D ', with respect to a straight line passing
through the neutral line and perpendicular to the plane of
flexure, i.e., the plane of the paper. It is usually denoted by
/ or AJ?, A being the sectional area, and k the radius of
gyration. Thus,
E E
the elastic moment = -7,- / = -^Ak*.
K. K.
But the elastic moment is equal and opposite to the bending
moment (M) due to the external forces, at the same section.
Hence
Note.\\. is necessary in the above to use the term alge-
braic, as the elementary stresses change in character, and
therefore in sign, on passing from one side of the neutral sur-
face to the other.
Cor. I. Bearing in mind assumption
(e), the figure represents on an exaggerated
scale the transverse section of the beam
at A'D ', the upper and lower breadths
of the beam, A' A" and H D" , being re-
spectively contracted and stretched, and
being also arcs of circles having a common
centre at O'.
Let R' be the radius of the arc P' P" ,
whose length remains unchanged.
Let mE be the lateral coefficient of
FlG - 2fii - elasticity, m being a numerical coefficient.
As before, for any layer at a distance y from P'P",
/. R' = mR.
344 THEORY OF STRUCTURES.
Thus, within the limits of elasticity, the curvature of the breadth
is that of the length, and does not sensibly affect the re-
sistance of the beam to bending. The influence, however,
upon the bending may become sensible if the breadth is very
large as compared with the depth, as, e.g., in the case of iron
or steel plates.
Cor. 2. If the resolved part of the external forces in the
direction of the length of the beam is nil,
E
the total longitudinal stress = -^(ay) = o, or ^(ay) = o,
showing that P f must be the centre of gravity of the section
through A' D' . Hence, when the external forces produce no
longitudinal stress in the beam, the neutral line is the locus of
the centres of gravity of all the sections perpendicular to the
length of the beam.
Cor. 3. If /, a, y be, respectively, the stress, sectional area,
and distance of a fibre from the neutral line, then
F E t
--ay = /, or -=,-y = - = intensity of stress =f y , suppose,
J\. rL a
EXAMPLE i. A timber beam, 6 in. square and 20 ft. long,
rests upon two supports, and is uniformly loaded with a weight
of 1000 Ibs. per lineal foot. Determine the stress at the centre
at a point distant 2 in. from the neutral line.
Also find the central curvature, E being 1,200,000 Ibs.
/T /^
/=- = 108, M = 1000 X io 1000 X 5 = 5000 ft.-lbs.
= 60,000 inch -Ibs., and y = 2 in.
INTERNAL STRESSES. 345
Hence from the above equations,
1200000 f y
x 108 = 60000 =
Thus R = 2160 in. = 180 ft., and f y nii-fr Ibs. per sq. in.
Ex. 2. A standpipe section, 33 ft. in length and weighing
5720 Ibs., is placed upon two supports in the same horizontal
plane, 30 ft. apart. The internal diameter of the pipe is 30
in., and its thickness j- inch. Determine the additional
uniformly distributed load which the pipe can carry between
the bearings, so that the stress in the metal may nowhere ex-
ceed 2 tons per square inch.
Let J^be the required load in pounds.
3
The weight of the pipe between the bearings = . 5720
= 5200 Ibs.
Thus, the total distributed weight between the bearings
= (^+5200 Ibs.)
Now M = f
and the stress in the metal is necessarily greatest at the central
section.
W+ 5200
M, at the centre, = .30. 12 inch-lbs. ;
f e = 2 X 2240 Ibs., and - = nr*t = . 1 5' .. -.
W-\- 52OO 22 , I
^ . 30. 12 = 2.2240.--. 15 - = 72000 X 22,
8 .72
and hence W 30,000 Ibs.
34-6 THEORY OF STRUCTURES.
Cor. 4. The beam is strained to the limit of safety when
either of the extreme layers A'B', D ' C is strained to the limit
of elasticity. In such a case, the least of the values of for
y
the extreme layers A'B', D' C is the greatest consistent with
the strength of the beam ; and if f c and c are the corresponding
intensity of stress, and distance from the neutral axis,
EXAMPLE. Compare the strengths of two similarly loaded
beams of the same material, of equal lengths and equal sectional
areas, the one being round and the other square.
Let r be the radius of the round beam ; f r9 the intensity of
the skin stress.
Let a be a side of the square beam ; f a9 the intensity of
the skin stress. Then
<2 2 ; /, for round bar, = , and for square bar = .
Also, since the beams are similarly loaded, the bending
moments at corresponding points are equal.
r 4 #12
2
so that
Thus, under the same load, the round beam is strained to
a greater extent than the square beam, and the latter is the
stronger in the ratio of \^SS to
BREAKING WEIGHTS. 347
Cor. 5. The neutral surface is neither stretched nor com-
pressed, so that it is not subjected to any longitudinal stress.
But it by no means follows that this surface is wholly free from
stress, and it will be subsequently seen that the effect of a
shearing force, when it exists, is to stretch and compress the
different particles in diagonal directions making angles of 45
with the surface.
bd* d
Cor. 6. For a rectangular beam// = , and c = .
,.*=// =/**=/..
c d 12 6
If the beam is fixed at one end and loaded at the other
with a weight W, the maximum bending moment = Wl.
If the beam is fixed at one end and loaded uniformly with
a weight wl = W, the maximum bending moment
wP Wl
If the beam rests upon two supports and carries a weight W
Wl
at the centre, the maximum bending moment ,
If the beam rests upon two supports and carries a uniformly
distributed load of wl = W, the maximum bending moment
Wl
Hence, in the first case, IV = =-\
" " second- W=2
" third W=^
" " fourth " W=^-
THEORY OF STRUCTURES.
In general, - W = g- q-j- ;
t/ being some coefficient depending upon the manner of the
loading.
Now, if the laws of elasticity held true up to the point of
rupture, these equations w^ould give the breaking weights (W 7 ),
corresponding to different ultimate unit stresses (/), but the
values thus derived differ widely from the results of experi-
ment. It is usual to determine the breaking weight (W) of a
rectangular beam from the formula W = C -T-, where C is a
constant which depends both upon the manner of the loading
and the nature of the material, and is called the coefficient of
rupture.
The modulus of rupture is the value of/ in the ordinary
bending-moment formula (M = /) when the load on the
beam is its breaking load.
The preceding equations, however, may be evidently em-
ployed to determine the breaking weights in the several cases
by making J -^-q = C. In this case /"is no longer the real stress,
but may be called the coefficient of bending strength.
The values of C for iron, steel, and timber beams, supported
at the two ends and loaded in the centre, are given. in the
Tables at the end of Chapter III.
The corresponding value of f is obtained from the equation
or
EXAMPLE. Determine the central breaking weight of a
EQUALIZATION OF STRESS. 349
red-pine beam, 10 in. deep, 6 in. wide, and resting upon two
supports 20 ft. apart.
The value of C for red pine is about 5700. Hence,
the breaking weight = W 5700 ll?_ = 14,250 Ibs.
2O X 12
2. Equalization of Stress. The stress at any point of a
beam under a transverse load is proportional to its distance
from the neutral plane so long as the elastic limit is not ex-
ceeded. At this limit materials which have no ductility give
way. In materials possessing ductility, the stress may go on
increasing for some distance beyond the elastic limit without
producing rupture, but the stress is no longer proportional to
the distance from the neutral plane, its variation being much
slower. This is due to the fact that the portion in compres-
sion acquires increased rigidity and so exerts a continually
increasing resistance (Chap. Ill) almost if not quite up to
the point of rupture, while in the stretched portion a flow of
metal occurs and an approximately constant resistance to the
stress is developed. Thus, there will be a more or less perfect
equalization of stress throughout the section, accompanied by
an increase of the elastic limit and of the apparent strength,
the increase depending both upon the form of section and the
ductility.
For example, if the tensile elastic limit is the same as the
compressive. the shaded portion of Fig. 262 gives a graphical
__
^'- J==
FIG. 262. FIG. 263. FIG. 264.
representation of the total stress in a beam of rectangular
section when the straining is within the elastic limit. Beyond
this limit, it may be represented as in Fig. 263, and will be
350
THEORY OF STRUCTURES.
intermediate between Fig. 262 and the shaded rectangle of
Fig. 264 which corresponds to a state of perfect equaliza-
tion.
3. Surface Loading.* It may be well to draw attention to
another important assumption upon which is based the mathe-
matical treatment of the problem of Beam Flexure.
It has been assumed that the external forces acting on a
beam can be so applied that they may be considered as dis-
tributed uniformly over the whole section. Thus when a beam
encastre" is loaded at the free end, Fig. 265, the load P is as-
FIG. 265.
sumed to be uniformly distributed over the section ab, i.e.,
each element in the section is supposed to experience the same
amount of strain due to the load, and the reaction of the wall
is also supposed to be uniformly distributed over each element
in the section cd.
It is clear that such suppositions must be far from the
truth.
In practice, the load P must be hung by some means from
the beam, say by a stirrup passing over the top. The whole
load is then concentrated at the line of contact of the stirrup
with the beam, and it is obviously untrue to say that every
* This article was kindly written by Professor Carus-Wilson and is an
abstract of a Paper presented by him to the Physical Society.
SURFACE LOADING. 35 1
element in the section ab is equally strained. But more
than this. It has been assumed that, taking the effect of
the load as distributed uniformly over the section ab, and
a certain deflection thereby produced, the effect of P on
each element of the section ab may be disregarded in com-
parison with the strains involved in the deflection which P
produces.
It will probably be difficult at first to grasp the fact that
certain measurable effects have been actually neglected, but
that this is so may be seen by supposing the beam in question
to be a pine beam, and the stirrup of iron. Experience proves
that with a very moderate load the beam will be indented at a.
But the theory shows that the longitudinal tension at a is
zero and increases to a maximum at d.
Thus, so far from the squeezing effect of the load being
distributed uniformly over the section ab, it is concentrated at
a, and hence it is impossible to neglect it.
Engineers have always recognized the existence of this
" surface-loading" effect in practice, and where possible, have
provided a good " bearing" in order to avoid such local
strains ; but this cannot always be done as, for instance,
in the case of rollers under bridge ends. The theory of flex-
ure is therefore manifestly incomplete if it cannot take into
account the actual manner in which the loads are and must be
applied.
FIG. 266.
It can be shown that the effect of placing a pressure of p
tons per inch run, say in the form of a loaded roller, on a beam
resting upon a flat surface, as in Fig. 266, to prevent it from
352
THEORY OF STRUCTURES.
bending, is to compress every element say along ab with an
intensity given approximately by the equation
where f is the pressure at a distance x from a, the point of con-
tact, and h = ab. This is the equation to a curve be which is
approximately an hyperbola.
When a beam is bent by the application of external forces, a
very close approximation to the true condition may be obtained
by superposing this surface-loading effect on that found for
bending.
Take the case of a beam supported at the ends and loaded
at the centre, and let it be required to find the condition along
ab, Fig. 267.
FIG. 267.
The effect of the bending is to produce compression above
and tension below the point c, and these effects may be repre-
sented by a right line de passing through c.
The surface-loading effect may be represented by an hyper-
bola giving the compression at any point along ab due to the
load. The hyperbola and straight line will intersect in two
SURFACE LOADING. 353
points h and/, which shows that at two points k ',/' along ab the
vertical squeeze produced by the load is of equal intensity to
the horizontal squeeze produced by the bending ; hence an
element at each of these points is subject to cubical compres-
sion only. From a to/' the beam is squeezed vertically, from
/' to h' it is squeezed horizontally, and from^' to b it is stretched
horizontally. The intensities are given at every point by the
difference between the ordinates of the line of bending de and
the curve of loading. It will appear that one effect of surface-
loading is to make the neutral axis rise up under the load and
pass through the point h f , for there is neither compression nor
tension at that point.
This can be verified by examining the condition of a bent
glass beam by polarized light. The neutral axis is pushed up
under the load and there is a black ring passing through the point
/'. If the span is diminished and the load kept constant, it is
clear that ae will become less, while the curve of loading remains
the same, until the line dee ceases to cut the curve ; every
element along ab will then be subjected to horizontal stretch,
and the stretch is greatest at a ; the result obtained by neglect-
ing the surface loading is that only elements from c to b are
stretched, the greatest stretch being at b. The position of the
" neutral points " is given by the equation
T6~m
where y is the distance from the top edge, h equals the depth
ab, m -^-y 4, and a = one-half of the span.
For all elements in ab to be stretched, the ratio of span to
depth, viz., -= , must be equal to or less than 4.25. In other
words, for any beam, and any load, if the span is less than 4^
times the depth, every element in the normal under the load
is stretched horizontally.
354
THEORY OF STRUCTURES.
4. Beam acted upon by a Bending Moment in a Plane
which is not a Principal Plane.
Let*XOX, YOYbe the principal axes of the plane section
^of the beam.
\y
FIG. 268.
Let the axis MOM of the bending moment M make an
angle ot with OX.
M may be resolved into two components, viz.,
M cos a X and M sin a =. Y.
These components may be dealt with separately and the
results superposed.
Thus, the total stress, /, at any point (x y y)
stress due to X + stress due to Y = ~- + =/,
1* -fy
I* , I y being the moments of inertia with respect to the axes
XOX, YOY, respectively.
If the point (xy) is on the neutral axis, then
or
being the angle between the neutral axis and XOX.
SPRINGS.
355
Also see Art. 6, Chap. VIII. In this article 6 is the angle
between the neutral axis and the axis of the couple, i.e.,
6 = ft - a.
5. Springs. (a) Flat Springs. If two forces, each equal
to P but acting in opposite directions in the same straight line,
are applied to the ends of a straight uniform strip of flat steel
spring, the spring will assume one of the forms shown below,
known as the elastic curve. This curve is also the form of the
linear arch best suited to withstand a fluid pressure, Chap.
XIII.
FIG. 274.
FIG. 275.
Consider a point B of the spring distant y from the line of
action of P. Then
Py bending moment at B = -=- ,
THEORY OF STRUCTURES.
R being the radius of curvature at B, and / the moment of
inertia of the section.
If E and /are both constant,
Ry = a constant
is the equation to the elastic curve.
(b) Spiral Springs (as, e.g., in a watch). Let the figure rep-
resent a spiral spring fixed at C and to an arbor at A, and
subjected at every point of its length
to a bending action only.
Consider the equilibrium of any
portion AB of the spring.
The forces at A are equivalent to a
couple of moment M, and to a force P
acting in some direction AD.
This couple and force must balance the elastic moment
at B.
.'. M-\- Py = El X change of curvature at B,
y being the distance of B from the line of action of P, or
R being the radius of curvature at B before winding, and R
that after winding.
Let ds be an elementary length of the spring at B.
Then, for the whole spring,
2(M + Py]ds = 12 - - - 12(40 -
or M2ds -f- P^yds El X total change of curvature between
A and C ;
.-. Ms -\- Psy = EI(B - ),
SP KINGS. 357
s being the length of the spring, y the distance of its C. of G.
from AD, 6 the angle through which the spring is wound up,
and 6 the " unwinding" due to the fixture at C. With a large
number of coils the distance between the C. of G. and A may
be assumed to be nil and then y = o.
Also, if the spring is so secured that there is no change of
direction relatively to the barrel,
= o, and Ms = Eld.
Let the winding-up be effected by a couple of moment
Qq = M, Q being a tangential force at the circumference of a
circle of radius q.
The distance through which Q moves (or deflection of Q)
= g# = s, since M = f,
/being the skin stress, and c the distance of the neutral axis
of the spring from the skin.
Thus, if b is the width of a spring of circular or rectangular
b
section, c = , and hence
the deflection = -j~s.
The work done = -Q x deflection.= -~q6 =
2 2 q 2
/' si y
2 E^ '2 EC* ' 2E a !
k* being the square of the radius of gyration, A the sectional
area of the spring, and Fits volume.
In case of spring of rectangular section - r = - .
c i rcu i ar 4 = -
* 4
358
THEORY OF STRUCTURES.
Again, the spiral spring in Fig. 277 is wholly subjected to
a bending action by means of a twisting couple of moment
M = Qq in a plane perpendicular to the axis of the spring.
Any torsion in the spring itself is now due to the coils not
being perfectly flat.
FIG. 277.
Let R = radius of a coil before the couple is applied.
" R = " " " " after " " u "
6 being the angle of twist ; or
El ~ El ~ R~~ ^ ~ v
N being the number of coils before the couple is applied, and
^y " after " " " u
The distance through which g acts, i.e., the " deflection,"
and the work done
fV V
_
8 E
for spring of rectangular section,
" u " circular
6. Beams of Uniform Strength. A beam having the
same maximum unit stress (/") at every section is said to be a
beam of uniform strength.
BEAMS OF UNIFORM STRENGTH.
359
At any section of a beam AB ( /) denote the bending
moment by M, the depth of the beam by y, and its breadth
by b. Then
being the distance of the skin from the neutral axis, and A
the area of the section.
Evidently c and k are each proportional to y, and A to by.
or
.-. fbf cc J/,
nfbf = M t
n being a coefficient whose value depends upon the form of
section.
Four cases will be considered.
CASE a. Assume that the breadth b is constant, and let
nfb = j. Then
or
y =
Thus AB may be either the lower edge of the beam, the
ordinates of the upper edge being the different values of y, or
it may be a line of symmetry with respect to the profile, in
which case the ordinates are the different values of -.
EXAMPLE I. A cantilever AB loaded at the free end with a
weight W^.
At a distance x from A t
y = pM p w,x.
Theoretically, therefore, the
beam, in elevation, is the area
ACD, the curve CAD being a
B
FIG. 278.
THEORY OF STRUCTURES.
parabola with its vertex at A and having a parameter
The max. depth = 2CB CD = Vp WJ.
The form of this beam is very similar to that adopted for
cranks and for the cast-iron beams of engines. In the latter,
the material is usually concentrated in the flanges, a rib being
reserved along the neutral axis for purposes of connection.
Again, geometrical conditions of transmission require the
teeth of wheels to be of approximately uniform strength.
A cantilever of approximately uniform strength may be ob-
tained by taking the tangents C, DF as the upper and lower
edges of the beam instead of the curves CA, DA. The depth
of the beam at A is then EF \CD = j- VpW~L Although,
theoretically, the depth at A is nil, practically the beam must
have sufficient sectional area at A to bear the shear due to W l ,
and the depth VpWJ will be found ample for this purpose.
Note. The dotted lines show the beams of uniform
strength, when the lower edge is the horizontal line AB.
Ex. 2. A cantilever AB carrying a uniformly distributed
load W,.
At a distance x from A,
or
FIG. 279.
The beam, in elevation, is there-
fore the area A CD, AC, AD being two straight lines, and the
maximum depth being
The sectional area at A is nil, as both the bending moment
and shear at that point are zero.
BEAMS OF UNIFORM STRENGTH.
361
Note. The dotted lines show the cantilever of uniform
strength when AB is the lower edge.
Ex. 3. A cantilever AB carrying a weight W l at the free
end A and also a uniformly distributed load W^,
FIG. 280.
At the distance x from A,
This equation may be written in the form
( ^/V
r + w l ) y
= i.
Theoretically, therefore, the beam, in elevation, is the area
ACD, the curve CAD being an hyperbola having its centre at
I W \
H ^where AH = -yn-l), and semi-axes equal to
U and
v IML
V w
The maximum depth CD = A. p\WJ-\- W~\ = 2BC.
362 THEORY OF STRUCTURES.
A cantilever of approximately uniform strength may be ob-
tained by taking the tangents CE, DF as the upper and lower
edges of the beam instead of the curves CA, DA. It may be
W
easily shown that the depth of this beam at A is
and this will give sufficient sectional area at A to bear the
shear due to W r
Note. The dotted lines show the cantilever of uniform
strength, when the lower edge is the line AB.
Ex. 4. A beam AB supported at A and B, and carrying a
load W^ at the middle point O.
\ At a distance x from 0,
G /^^ 1 JX E
Theoretically, therefore, the
beam, in elevation, is the area
ACBD, the curves CAD, CBD being two equal parabolas,
having their vertices at A and B, respectively, and having
parameters equal to \p W^ .
The maximum depth = CD ~ 2CO = % \ f ~pWJ .
A beam of approximately uniform strength may be ob-
tained by taking the tangents CE, CG as the upper edges
instead of the curves CA, CB, and the tangents DF t DH as
the lower edges instead of the curves DA, DB.
The depth of the beam at A and B is now EF = GH
2 '
and this depth will give a sectional area at the ends of the beam
sufficient to bear the shears at these point, viz., - .
Note. The dotted lines show the beam of uniform strength
when the line AB is the lower edge.
Ex. 5. A beam AB supported at A and B, and carrying a
uniformly distributed load W^
BEAMS OF UNIFORM STRENGTH.
363
At a distance x from the
middle point O,
This equation may be writ- D
^, c FIG. 282.
ten in the form
*4 _/_
1 pwj
4 8
Theoretically, therefore, the beam, in elevation, is an ellipse
ACBD, having its centre at O and axes
AB = / and CD
IpWJ
=v~
The maximum depth is of course the axis CD = 2 CO.
Practically, the beam must have a certain depth at A and
B in order to bear the shears due to the reactions at these
W
points, viz., -. If the horizontal tangents at C and at D are
substituted for the curves, the volume of the new beam is to
the volume of the elliptic beam in the ratio of 4 to n.
Note. The dotted line shows the beam of uniform strength
when its lower edge is the line AB.
Ex. 6. A beam AB supported at A and B, and carrying a
load W } at the middle point O and also a uniformly distributed
load W,.
At a distance x from (9,
This equation may be written in the form
. i WJ^
-;#*.
pi
= I
w-
THEORY OF STRUCTURES.
Theoretically, therefore, the
beam, in elevation, is the area
ACBD, the curves CAD and
CBD being the arcs of ellipses
having the centres at the points
K and L, respectively, where
( w/ W l\ *
The maximum depth CD = 20C = J>* ] '- + -~ \ .
(2 o )
A beam of approximately uniform strength may be obtained
by taking as the upper edge the tangents to the curves at C,
and as the lower edge the tangents to the curves at D.
It may be easily shown that the depth at the ends^4 and B
W A- W
is now CD ' ' ' and this depth will make allowance for
2 W \ -\- KK a
W _L w
the shear - a at these points.
Note. The dotted lines show the beam of uniform strength
when the lower edge is the line AB.
CASE b. Assume that the ratio of the breadth (b) to the
depth (y) is constant, i.e., that transverse sections are similar.
y oc b a ty M,
or the ordinates of the profile of the beam both in plan and
elevation are proportional to the cube roots of the ordinates
of the curve of bending moments.
For concentrated loads the bounding curves are evidently
cubical parabolas.
CASE c. Assume that the depth y is constant. Then
b a M t
so that the ordinates of the beam in plan are directly propor-
tional to the ordinates of the curve of bending moments.
CASE d. Assume that the sectional area yb is constant.
Then
y a 'M,
FLANGED GIRDERS, ETC, 365
and the ordinates in elevation are directly proportional to the
ordinates of the curve of bending moments.
In this beam, the distribution of the material is very de-
fective, as the breadth b ( = - J -) must be infinite when^ = o,
i.e., at the points at which the bending moment is nil.
Timber beams of uniform strength are uncommon, as there
is no economy in their use, the portions removed to bring the
beam to the necessary form being of no practical value.
6. Flanged Girders, etc. Beams subjected to forces, of
which the lines of action are at right angles to the direction of
their length, are usually termed Girders; a Semi-girder, or
Cantilever, is a girder with one end fixed and the other free.
It has been shown that the stress in the different layers of
a beam increases with the distance from the neutral surface, so
that the most effective distribution of the material is made by
withdrawing it from the neighborhood of the neutral surface
and concentrating it in those parts which are liable to be more
severely strained. This consideration has led to the introduction
of Flanged Girders, i.e., girders consisting of one or two flanges
(or tables), united to one or two webs, and designated Single-
webbed or Double-webbed ( Tubular] accordingly.
7
(
FIG. 284. FIG. 285.
T
j
IT T
i JL
FIG. 286. FIG. 287. FIG. 288. FIG. 289. FIG. 290.
The web may be open like lattice-work (Fig. 284), or closed
and continuous (Fig. 285).
The principal sections adopted for flanged girders are :
The Tee (Figs. 286 and 287), the I or Double-tee (Figs. 288
and 289), the Tubular or Box (Fig. 290).
Classification of Flanged Girders. Generally speaking,
flanged girders may be divided into two classes, viz.:
366 THEORY OF STRUCTURES.
I. Girders witJi Horizontal Flanges. In these the flanges
can only convey horizontal stresses, and the shearing force,
which is vertical, must be wholly transmitted to the flanges
through the medium of the web.
If the web is open, or lattice-work, the flange stresses are
transmitted through the lattices.
If the web is continuous, the distribution of stress, arising
from the transmission of the shearing force, is indeterminate,
and may lie in certain curves ; but the stress at every point is
resolvable into vertical and horizontal components. Thus, the
portion of the web adjoining the flanges bears a part of the
horizontal stresses, and aids the flanges to an extent depend-
ent upon its thickness.
With a thin web this aid is so trifling in amount that it
may be disregarded without serious error.
II. Girders with one or both Flanges Curved. In these the
shearing stress is borne in part by the flanges, so that the web
has less duty to perform and requires a proportionately less
sectional area.
Equilibrium of Flanged Girders. AB is a girder in equi-
o librium under the action of external
forces, and has its upper flange com-
pressed and its lower flange ex-
tended. Suppose the girder to be
FlG - 8 9 X - divided into two segments by an
imaginary vertical plane MN. Consider the segment AMN.
It is kept in equilibrium by the external forces on the left of
MN, by the compressive flange stress at N ( = C), by the
tensile flange stress at M ( T), and by the vertical and
horizontal web stresses along MN. The horizontal web
stresses may be neglected if the web is thin, while the vertical
web stresses pass through M and N, and consequently have no
moments about these points.
Let d be the effective depth of the girder, i.e., the distance
between the points of application of the resultant flange stresses
in the plane MN.
Take moments about Jfand //successively. Then
Cd = the algebraic sum of the moments about M of
FLANGED GIRDERS, ETC. 367
the external forces upon AMN =. the bending moment at
MN=M.
So, Td = M\ .'. Cd=M= Td, and C = T.
Hence, the flange stresses at any vertical section of a girder
are equal in magnitude but opposite in kind. The flange
stress, whether compressive or tensile, will be denoted by F.
EXAMPLE. A flanged girder, of which the effective depth
is 10 ft., rests upon two supports 80 ft. apart, and carries a uni-
formly distributed load of 2500 Ibs. per lineal foot. Determine
the flange stress at 10 ft. from the end, and find the area of
the flange at this point, so that the unit stress in the metal
may not exceed 10,000 Ibs. per square inch.
The vertical reaction at each support
80 X 2500
= -- - = 100,000 Ibs.
/. F. 10 = M looooo X 10 2500 X 10 X 5 = 875,000 ft.-lbs.
,:F = 87,500 Ibs.
87500
The required area = --- = 8.75 sq. in.
i oooo
Cor. i. Fd=M=^I = ^L
R y
Cor. 2. At any vertical section of a girder,
let #,,#, be the sectional areas of the lower and upper flanges,
respectively ;
fuft, be the unit stresses in the lower and upper flanges,
respectively. Then
and the sectional areas are inversely proportional to the unit
stresses.
This assumes that F is uniformly distributed over the
areas a lt a 9 , so that the effective depth is the vertical distance
between centres of gravity of these areas. Thus, the flange
stresses at the centres of gravity are taken to be equal to the
3^8 THEORY OF STRUCTURES.
maximum stresses, and the resistance offered by the web to
bending is disregarded. The error due to the former may
become of importance, and it may be found advisable to make
the effective depth a geometric mean between the depths from
outside to outside and from inside to inside of the flanges.
Thus, if these latter depths are h^ , /^ , the effective depth
= Vk& (Art. 7).
EXAMPLE I. At a given vertical section of a flanged girder
the sectional area of the top flange is 10 sq. in., and the cor-
responding unit stress is 8000 Ibs. per square inch. Find the
sectional area of the lower flange, so that the unit stress in it
may not exceed 10,000 Ibs. per square inch.
a t . 10000 ='F= 10 . 8000 ; /. a l = 8 sq. in. and F = 80,000 Ibs.
Ex. 2. A wrought-iron girder weighing w Ibs. per lineal
ft., of / ft. span and d ft. depth, has horizontal flanges and
a uniform cross-section. The weight of the web is equal to the
weight of the flanges. Show that if the coefficient of strength
is 9000 Ibs. per square inch, the limiting value of / is 5400^ ft.,
k being the ratio of depth to span.
wr
Maximum flange stress --,- ;
Area of each flange = -- 5-1 in. ;
9000 . Sa
and
4wT
Total sectional area = ^ in.,
total volume of girder in feet = ' 5--;'
9000. 8^. 144
Hence,
4wr .480
wl = total W e I ght = - 95 ^- w: ,
and
d
I = 5400^- = 5400^.
FLANGED GIRDERS, ETC. 369
Note. The compressive strength of cast-iron is almost six
times as great as the tensile strength, and therefore the area
of the tension flange of a girder of this material should be
about six times that of the compression flange. Considering,
however, the difficulty there is in obtaining sound castings,
and also the necessity to provide sufficient lateral strength, it
by no means follows, nor is it even probable, that the ratio of
ultimate strengths is the best for the working strengths. Some
authorities are of the opinion that girders should be designed
with a view to their elastic strength, and that therefore the
working unit stresses in the case of wrought-iron and steel
should be equal, if this will insure sufficient lateral stability,
and in the ratio of 2 to I or 3 to I for cast-iron, which will give
sufficient lateral stability and make allowance for defective
castings.
The formula W = Cr is often employed to determine the
strength of a cast- or wrought-iron girder which rests upon two x
supports / inches apart, d being its depth in inches, and a the
net sectional area of the bottom flange in square inches. C is
a constant to be determined by experiment. Its average value
for cast-iron is 24 or 26, according as the girder is cast on its
side or with its bottom flange upwards. An average value of
C for wr -ought-iron is 80.
Cor. 3. A girder with horizontal flanges, of length / and
depth d, rests upon two supports, and is uniformly loaded with
a weight w per unit of length.
The bending moment at a vertical plane distant x from the
centre is
. will \ I \i// \ wl
Also, M Fd afd, a being the sectional area of either
flange at the plane under consideration, and f the correspond-
ing unit stress.
wrt AX*\
= (i--).
3/O THEORY OF STRUCTURES,
Let A be the flange sectional area at the centre. Then
Hence
an expression from which the flange sectional area at any point
of the girder may be obtained when the area at the centre is
known.
Cor. 4. F represents indifferently the sum of the horizontal
elastic forces either above or below the neutral axis, and is
therefore proportional to A, the sectional area of the girder ;
d is the distance between the centres of resultant stress and is
proportional to D, the depth of the girder.
/. Mat AD = CAD,
a form frequently adopted for solid rectangular or round gird-
ers, but also applicable to other forms.
Remark. The effective length of a girder may be taken to
be the distance from centre to centre of bearings.
The effective depth depends in part upon the character of
the web, but in the calculation of flange stresses the following
approximate rules are sufficiently accurate for practical pur-
poses :
If the web is continuous and very thin, the effective depth
is the full depth of the girder.
If the web is continuous and too thick to be neglected, the
effective depth is the distance between the inner surfaces of
the flanges.
If the web is open or lattice-work, the effective depth is the
vertical distance between the points of attachment of the
lattices.
If the flanges are cellular, the effective depth is the distance
between the centres of the upper and lower cells.
EXAMPLES OF MOMENTS OF INERTIA. $Jl
7. Examples of Moments of Inertia. (a) Double-tee Sec-
tion. First, suppose the web to be so thin that
it may be disregarded without sensible error. ""T"^
K2'
Let the neutral axis pass through G, the cen- -k
tre of gravity of the section. J 1
Let a^ , a^ be the sectional areas of the lower A
and upper flanges, respectively, and assume that FlG ' 2Q2 '
each flange is concentrated at its centre line.
Let h l , h z be the distances of these centre lines from G.
Let //, + h, = d.
Approximately, / = aji* -\- aji*.
Also, (a l + a <i )/i l = a^d, and (a l + # 2 )/z 2 = a^.
' I = a \^aJ +a *
Again, if/,, / 2 are, respectively, the unit stresses in the
metal of the lower and upper flanges,
M = j-I f&d, and also = -/ = f^a^d.
If a, = a 9 = a, / = /,=/, suppose, and M = fad.
Second. Let the web be too thick to be neglected.
As before, let the neutral axis pass through , the centre
of gravity of the section.
Let a l , # 2 be the sectional areas of the lower
and upper flanges, respectively, and assume that
n each flange is concentrated at its centre line.
FIG. 2 9 *a. Let ^ ^ ^ be the sect j ona i areas o f t h e portions
of the web below and above G, respectively.
Let h^ , hi be the distances from G of the lower and upper
flange centre lines.
Let h l -f ^ 2 = d.
Approximately,
3/ 2 THEORY OF STRUCTURES.
Also, fa, + j^, = (a^ + U 8 , and this equation, together
with h l + hi = d, will give the values of ^ , &, ; hence the
value of / may be determined.
As before, / = J/=/V.
A'
Let #, = # 3 = A and # 3 = # 4 = . Then
Hence,
/being the unit stress in either flange.
Thus, the web aids the girder to an extent equivalent to the
increase which would be derived by adding one-sixth of the
web area to each flange. If the weight of the material remains
constant, M increases with*/. At the same time the thickness
of the web diminishes, its minimum value being limited by cer-
tain practical considerations (Art. 8). Hence it follows that
the distribution of material is most effective when it is concen-
trated as far as possible from the neutral axis (Art. 5).
N.B. It must be remembered that f l and / a are not the
maximum stresses. If /, , / a are the thicknesses of the lower
and upper flanges, respectively, then
and
maximum tension = ^, -
A + K
maximum compression =/ 2 - -, .
fin
EXAMPLES OF MOMENTS OF INERTIA. 3/3
Again, take moments about G. Then
or
which gives a relation between the flange and web areas if
/j , /, are known.
For example, take f t = 2^/j. Then
a formula which agrees very closely with modern practice in
cast-iron girders.
The principles of construction require a beam or girder
to be designed in such a manner as to be of uniform
strength, i.e., equally strained at every point. An exception,
however, is usually made in the case of timber beams or girders.
The fibres of this material are real fibres and offer the most
effective resistance in the direction of their length, so that if
they are cut, their remaining strength is due only to cohesion
with the surrounding material. Besides, there is no economy
to be gained by removing a lateral portion, as the waste is of
little, if any, practical value.
EXAMPLE. The lower and upper flanges of the section of a
girder are I in. and I J- in. thick, respectively, and are each 24 in.
wide ; the effective depth of the girder is 48 in., and the web
is in. thick. Determine the position of the neutral axis ; also
find the flange unit stresses when the bending moment at the
given section is 250 ft.-tons. Using the preceding notation,
a l = 24 sq. in., a z = 36 sq. in., and a 3 -f- # 4 = 24 sq. in.
The centre of gravity of the web is half-way between AB
and CD. Thus,
24/Z, + 24(//, - 24) = 36(48 - //,),
374 - THEORY OF STRUCTURES.
or
_// \AA >f
h i = and h^ --- , defining the position of G.
Again,
192 i 96 144 i 72
a = - . = sq. in. and a. = -- . - = sq. in.
727 7 2 7 4
Also,
M = 250 ft.-tons = 3000 inch-tons.
.-. /, = 2-g- 9 ^- tons per sq. in. and f t = i||-f tons per sq. in.
Third. It is often convenient to calculate the moment of
inertia of a built beam symmetrical with respect to the neutral
axis, as follows :
Let Fig. 293 represent the section of such a beam, com-
posed of equal flanges connected with the web by four equal
angle-irons.
Let the width AF oi the flange = a.
u the side BC(= DE) of an angle-iron = b.
" thickness GH(= KL) of an angle-iron =f.
EF _
, be the outside depth of the section.
, " " depth between flanges.
FIG. 293. h^ "
Let ^ 3 be the depth between the faces MN, M'N'.
" h, " " " " " " ATZ, A' 7 /: 7 .
EXAMPLES OF MOMENTS OF INERTIA.
375
In this value of /, the weakening effect due to the rivet-
holes in the tension flange has been disregarded.* If it is to be
taken into account, let/ be the diameter of the rivets.
The centre of gravity of the section is now moved towards
the compression flange from its original position through a
distance
and the moment of inertia of the net section with respect to
the axis through the new C. of G. is
A' being the net area of the section, and / having the value
given above.
Fourth. The value of 7 for a double-tee section may be
more accurately determined as follows :
Let the area of the top flange be A 1 , and
its depth h r
Let the area of the bottom flange be A^ ,
and its depth // 8 .
Let the area of the web flange be A^ , and
its depth /z 2 .
Let A,-\- Af{- A 3 = A, and /z,+ //+ h z =k. FIG. 294 .
Let G be the centre of gravity of the section.
" G l " " " " top flange. m
" web.
" bottom of flange.
Let y v be the distance of G from the upper edge of the
section.
Let y t be the distance of G from the lower edge of the
section.
Take moments about G s . Then
THEORY OF STRUCTURES.
or
GG = *,(*, + 2*. + *J + *.(*, + *.
2,A
So,
rr -A^+
^ ~
and
Hence,
h k h
2A
2A
+ kj - Afa + ^ 3 ) - A^ - h
2
So,
= GG, + ~= etc.
Again, /, with respect to G,
G,ff+A,. G,ff + A S . G,G',
A&* + A,&,* + A,A*
/, being equal to ' ' - ' ' ~ .
Hence,
7 = 7 ' + ^' +
\A^ + 2A, + A.) + A,(A, + /,,)'.
EXAMPLES OF MOMENTS OF INERTIA. ^77
A 1 A^ 1 +^+2A l A t A,(/l 1 +/l^ 1 +2^+^
+ A,A,'(A t + ky - 2A l A^i,(A, + A,)(A, + A,)
+ A,A,'(A, + A,)' + A,A'(A, + 2A, + A,)'
+ 2A 1 A,A,(A 1 + 2*. + h,)(h, + A,)
1+ 2A 1 A,A 3 (/i 1 + 2/1, + h^(h, + h, + h, + h,)
A.A^ + k$A + A,A,(/i, + k$A 1
+ (4& + AfAJfa + 2h, + h$
W* + WA
- A.A^/^ + 2// a + h$A
Hence, finally,
12
Cor. I. If 7/ t and h t are small compared with ^,, put
Then
y =
2A
h ,
= -- , nearly,
THEORY OF STRUCTURES.
and
A A' + A. (h' - k -^^] + AM
12
,,^A, , A,A,+A,A,
tttT" ~4A
4A
Note. If At is also very small, as in the case of an open
web, then
A A A
i/ 2 = h' j- and / = h' 2 j- 2 , approximately.
si. si.
Cor. 2. Let y a , y b be the distances of G from the upper and
lower edges, respectively ; let f a , f b be the corresponding
maximum working unit stresses.
From the preceding corollary, y b = y^ = - - - -,
2 2/i
or
A. + A. + A, __ y a +y b
2A. + A, 2y b
3 l yb * 2% 1 f b
Hence,
EXAMPLES OF MOMENTS OF INERTIA.
379
and
,,, 1 ,
~ lt 12 +
4A
12A
A + A: + 4 (A, +
+ A,
6 /. +/.
r&,
Fifth. T-section.
Let the area of the flange be A, , and its
depth h,.
Let the area of the web be A 9 , and its
depth h^.
LetAi + A, = A, and h, + h^ = h.
Let G be the centre of gravity of the sec-
tion, G^ of the flange, and G 9 of the web.
FIG.
295-
Let y l be the distance of G from foot of the web.
Then
. l -
2 2
3^0 THEORY OF STRUCTURES.
and
_ ^i + * I A A - A A = *i A A - AA
2 2(A l +A,) 2 ' 2A
Again, ,.
^ ^ ^ii/ AJi . ~ ~ h. AJi
G lG = ^ + k,- yt =^ and G.G =*;- = ..
Hence /, with respect to a horizontal line through G,
= A^+A,. G>G'+A/ + A..G.G; ''
which reduces to
.
12 4A
Cor. I. If h^ is very small as compared with ^ a , put
= AJf + A.(?- -) = (A, + ) k>, nearly,
then
or
and
1 =
or
12 ' 4A r
Cor. 2. Let y a be the distance of the compressed, or upper,
side from the neutral axis.
12 4A
TO DESIGN A GIRDER OF UNIFORM STRENGTH. 381
Let y b be the distance of the stretched, or lower, side from
the neutral axis.
Let/ a be the crushing unit stress, f b the tensile unit stress.
h' 2.A I A.
From the preceding, y a - '- ^- - ; but h' = y a -\-y b ;
m >, ^ = > =
2 A,+A t 2y b 2f b
Hence, /becomes
2 ~7 i 7"' 7
fa+fb fa
Note. Although the preceding approximate methods are
often useful, they can only be regarded as tentative and should
always be checked by an accurate determination of the moment
of inertia and of the position of the neutral axis.
8. To design a Girder of Uniform Strength, of an
I-section with equal Flange Areas, to carry a Given
Load.
Let y be the depth of the girder at a distance x from its
middle point.
Let A be the sectional area of each flange at a distance x
from its middle point.
Let A' be the sectional area of the web at a distance x from
its middle point.
Let M be the bending moment at a distance x from its mid-
dle point.
Let 5 be the shearing force at a distance x from its middle
point. Then *
/being the safe unit stress in tension or compression.
382 THEORY OF STRUCTURES.
Web. Assume that the web transmits the whole of the
shearing force. This is not strictly correct if the flange is
curved, as the flange then bears a portion of the shearing force.
The error, however, is on the safe side.
Theoretically, the web should contain no more material than
is absolutely necessary.
Let f s be the safe unit stress in shear. Then
A'- S
~
and the sectional area is, therefore, independent of the depth.
A.' S
The thickness of the web = = -7,
but this is often too small to be of any practical use.
Experience indicates that the minimum thickness of a plate
which has to stand ordinary wear and tear is about J or T 5 ^- in.,
while if subjected to saline influence its thickness should be
f or |- in. Thus, the weight of the web rapidly increases with
the depth, and the greatest economy will be realized for a cer-
tain definite ratio of the depth to the span.
The thickness of the web in a cast-iron girder usually
varies from I to 2 in.
In the case of riveted girders with plate webs of medium
size, all practical requirements are effectively met by specifying
that the shearing stress is not to exceed one-half of the flange
tensile stress, and that stiffeners are to be introduced at inter-
vals not exceeding twice the depth of the girder when the
thickness of the web is less than one-eightieth of the depth.
Again, it is a common practical rule to stiffen the web of a
plate girder at intervals approximately equal to the depth of
the girder, whenever the shearing stress in pounds per square
Crra \
I -\ ' j, H being the ratio of the
depth of the web to its thickness.
Flanges. First. Assume that the flanges have the same
sectional area from end to end of girder.
TO DESIGN A GIRDER OF UNIFORM STRENGTH. 383
If the effect of the web is neglected,
M
and the depth of the beam at any point is proportional to the
ordinate of the bending-moment curve at the same point.
For example, let the load be uniformly distributed and of
intensity w ; and let / be the span. Then
FIG. 296.
and the beam in elevation is the parabola ACJ3, having its
vertex at C and a central depth CO = TTT"> The depths thus
determined are a little greater than the depths more correctly
given by the equation
M
y =
Second. Assume that the depth y of the girder is constant.
Then
A 1 M
and, neglecting the effect of the web, the area of the flange at
any point is proportional to the ordinate of the curve of bend-
ing moments at the same point.
Let the load be uniformly distributed and of intensity w ;
also, let the flange be of the same uniform width b throughout.
A 1 2 3 5 6 7 B
FIG. 297,
The flange, in elevation, is then the parabola ACB, having
its vertex at C and its central thickness CO ^7-7. Such
Zfyb
THEORY OF STRUCTURES.
beams are usually of wrought-iron or steel, and are built up by
means of plates. It is impracticable to cut these plates in such
a manner as to make the curved boundary of the flange a true
parabola (or any other curve). Hence, the flange is generally
constructed as follows :
Draw the curve of bending moments to any given scale.
By altering the scale, the ordinates of the same curve will
represent the flange thicknesses. Divide the span into seg-
ments of suitable lengths.
From A to I and B to 7 the thickness of the flange is
\a 7/; from I to 2 and 7 to 6 the thickness is 2b = 6e ; from
2 to 3 and 6 to 5 the thickness is $c = ^d\ and from 3 to 5 the
thickness is CO.
The more correct value of A( = -p-1 is somewhat
V fy 67
less than that now determined, but the error is on the safe
side.
Again, at any section,
E 2f
- , and hence R oc y, the depth.
Thus the curvature diminishes as the depth increases, so
that a girder with horizontal flanges is superior in point of
stiffness to one of the parabolic form. The amount of metal
in the web of the latter is much less than in that of the
former. If great flexibility is required, as in certain dyna-
mometers, the parabolic form is of course the best.
9. Deflection of Girders. The principles of economy and
strength require a girder to be designed in such a manner that
every part of it is proportioned to the greatest stress to which
it may be subjected. When such a girder is acted upon by
external forces, it is uniformly strained throughout, and in
bending, the neutral axis must necessarily assume the form of
an arc of a circle, provided the limit of elasticity is not ex-
ceeded. It might be supposed that the curve of deflection is
dependent upon the character of the web, and this is doubtless
the case, but experiments indicate that so long as the flange
DEFLECTION OF GIRDERS. 385
unit stresses are unaltered in amount, the influence of the web
may be disregarded without sensible error.
Let f be the unit stress in the beam at a distance y from
the neutral axis ; let d be the depth of the beam. Then
/ M E
- = - T = -=. = a constant,
y 1 K
assuming that the neutral axis is an arc of a circle of radius R.
But y oc d, and
7 = Ak* a Ad\
Hence f a y a d\ and if the depth is constant, /is also con-
stant and the beam is of uniform strength.
If the area A is constant,
EXAMPLE I. A cantilever bent under the action of exter-
nal forces, so that its neutral axis AB assumes
the form of an arc of a circle having its centre
* O.
Draw the verticals OA, BF, and the horizon-
The vertical deviation of B from the hori-
zontal, viz., BF y is the maximum deflection. ,'/'
Denote it by D. 6
Let radius of circle = R.
Since the deflection is very small, BE is approximately
equal to AB ( = /), the length of the cantilever.
/. r = BE" = AE(2R - AE) = 2RD - D* = 2 RD,
as D 1 may be disregarded without much error.
Also, the deflection at any point distant x from A is evi-
dently ^. If /is the stress in the material at a distance y
from the neutral axis,
f__E L _^DE 2DEy
~y ~ R ~ ' S ~' f ~~ ~lf~ '
386
THEORY OF STRUCTURES.
Ex. 2. A girder resting upon two supports at A and B
is bent under the action of external
forces so that its neutral axis ACB
assumes the form of an arc of a circle
having its centre at O.
Draw the vertical OC, meeting the
horizontal AB in F.
CF is the maximum deflection ;
denote it by D.
Since D is very small, its square
may be disregarded and the horizontal AB may be supposed
equal to the length ACB ( = /) of the girder, without much
error. Then
FIG. 299.
- = AF* = FC(zR - FC) = 2RD - D 1 = 2RD.
Hence,
/ &
Also, since = -5,
y R
f ~
WEy
x
The deflection at a distance x from F = D .
Ex. 3. A timber beam of 20 ft. span, is 12 in. deep and 6
in. wide : what uniformly distributed load ( W) will deflect the
beam I in., E being 1,200,000 Ibs. ?
By Ex. 2,
(2 4 0) 2
.'. R 7200 in.
_E T _ 1200000
8 ' 12 "J^~' 7200
12
I20OOOO 6 . I2 9
720O 12
.-. W= 4800 Ibs.
DEFLECTION OF GIRDERS. 387
Ex. 4. Let s l , f l , d lt and j a , / 2 , 4 , respectively, be the
length, unit stress, and distance from the neutral axis of the
stretched and compressed outside fibres in Examples (i) and
(2).
Let d l -f- d^ = d the total depth of the girder.
Hence, from similar figures,
d, s, R-d,
- and 7- -
/ 2P "" Z?
* A'- ' A
Also,
. /i+/ IL '""* _ ^ + ^
Ex. 5. A truss of span 120 ft. and 15 ft. deep is strained
so that the flange tensile and compressive unit stresses are
10,000 and 8000 Ibs., respectively. Find the deflection, and
difference of length between the extreme fibres.
30000000 i 20
.'. s 1 J 3 = .864 in., and R = 25,000 ft.
Hence also D = = -864 in.
10. Camber. Owing to the play at the joints, a bridge-
truss, when first erected, will deflect to a much greater extent
388 THEORY OF STRUCTURES.
than is indicated by theory, and the material of the truss will
receive a permanent set, which, however, will not prove detri-
mental to the stability of the structure, unless it is increased
by subsequent loads.
If the chords were made straight, they would curve down-
wards, and, although it does not necessarily follow that the
strength of the truss would be sensibly impaired, the appear-
ance would not be pleasing.
In practice it is usual to specify that the truss is to have
such a camber, or upward convexity, that under ordinary loads
the grade line will be true and straight.
The camber may be given to the truss by lengthening the
upper or shortening the lower chord, and the difference of
length should be equally divided amongst all the panels.
The lengths of the web members in a cambered truss are
not the same as if the chords were horizontal, and must be
carefully calculated, otherwise the several parts will not fit
accurately together.
To find an approximate value for the camber, etc. :
Let d be the depth of the truss.
Let s l , ^ 2 be the lengths of the upper and lower chords,
respectively.
Let /! , / 2 be the unit stresses in the upper and lower
chords, respectively.
Let d l , d^ be the distances of the neutral axis from the
upper and lower chords, respectively.
Let R be the radius of curvature of the neutral axis.
Let / be the span of the truss.
d, *. //i d i l ~ ** / 3
- = j- = - and - = = - , approximately,
the chords being assumed to be circular arcs.
Hence, the excess in length, of the upper over the lower
chord,
STIFFNESS. 389
Let x l , x^ be the cambers of the upper and lower chords,
respectively. R-\- d l and R d^ are the radii of the upper
and lower chords, respectively.
By similar figures, the horizontal distance between the ends
r> 1
of the upper chord = -= V, and the horizontal distance be-
tween the ends of the lower chord = - 5~~V.
Hence,
R
and
( j
= x, . 2(R + O, approximately,
^) approximately.
i i * \ A I i
8^* R ' oR\ Ri
r iy
Hence, approximately, the camber = .
Note. The deflection of a well-designed and well-built
truss is often much less than, and should never exceed, I inch
per 100 ft. of span under the maximum load.
II. Stiffness. If D is the maximum deflection of a girder
W D
of span / under a load W, then -, or more usually , is a
measure of the stiffness of the girder.
In practice, the deflection of an iron or a steel girder, under
the working load, should lie between and ^ , i.e., it is
limited to i or 2 in. per 100 ft. of span, and rarely exceeds
, or 1.2 in. per 100 ft. of span.
1000
A timber beam should not deflect more than -^-, or i in.
360
per 30 ft. of span.
390 THEORY OF STRUCTURES.
Let M l be the bending moment at the most deflected
point. Then
Also,
f being a numerical coefficient (in Art. 9, Ex. \,p = i ; in Ex.
2,/ = i).
Thus
I
gives the bending moment M^ to which the girder of a speci-
fied stiffness y may be subjected.
Again, if the material is to bear a certain specified unit
stress y, the maximum bending moment M^ to which the
girder may be subjected is given by the equation
, = = -,
c qd
*
q being a numerical coefficient less than unity, depending upon
the form of the section.
Cczteris paribus, the ratio of depth to span may be fixed
by making the stiffness and strength of equal importance. Then
M M and therefore
*-* I ~ ] J r
~pl\l)~~ qd '
or
d~\l
DISTRIBUTION OF SHEARING STRESS.
391
In practice the proper stiffness of a girder is sometimes
secured by requiring the central depth to lie between and
, its value depending upon the material of which the girder
is composed, its sectional form, and the work to be done.
EXAMPLE. A cast-iron beam of rectangular section and of
20 ft. span carries a uniformly distributed load of 20 tons ; the
coefficient of working strength is 2 tons per sq. in. ; the stiff-
ness is .001 ; E is 8000 tons. Find the dimensions of the beam
viz., b the breadth and d the depth.
20.20 _f r _ bd* _bd*
jyj. !^ .12 j. ^^ 2 ., ^^ !
8 c 63
Also,
.-. bd* = 1800.
20 . 20 EIiD\ 8 . 8000 . bd*
--.12 = M= 7 (-.-) = .(.ooi);
Pl\l I 12 . 20. 12 V
.-. bd* = 27000.
8
Hence,
2700O
-f=
1800
= 15 in. and b = 8 in.
12. Distribution of Shearing Stress. Let Figs. 300 and
301 represent a slice of a beam bounded by two consecutive sec-
A
PX~ j > Q
-^ 4-2 *
0/~
*d i
B- B '
FIG. 300.
tions AB, A'B' , transverse to the horizontal neutral axis O O f .
Let the abscissae of these sections with respect to an origin
39 2 THEORY OF STRUCTURES.
in the neutral axis be x and x -f- dx, so that the thickness of
the slice is dx.
In the limit, since dx is indefinitely small, corresponding
linear dimensions in the two sections are the same.
Let / be the moment of inertia of the section AB (or A ' B'
in the limit) with respect to the neutral axis.
Let c be the distance of A (or A' in the limit) from the
neutral axis.
Let/j , f t be the unit stresses at A and A , respectively.
Consider the portion A CCA' of the slice, CC ' being
parallel to and at a distance Y from the neutral axis. Since
it is in equilibrium, the algebraic sum of the horizontal forces
acting upon it must be nil. These forces are:
The total horizontal force upon ACC,
11 A'C'C', and
shear along the surface CC'.
The horizontal force upon an element PQ of thickness dy
and at a olistance y from the neutral axis
2 being the width PQ. Thus the total horizontal force upon
"ACC
= (* yzdy = ^Ay,
c j Y c
A being the area of ACC, and y the distance of the centre of
gravity of this area from O0 r .
Similarly, "the total horizontal force upon A'C'C'
/ J J ~ ' ., J'
DISTRIBUTION OF SHEARING STRESS.
393
// f\ -
Hence, f -\Ay = difference of the horizontal forces
upon A CC and A'C'C',
=. horizontal shear along CC' ,
= qwdx ;
q being the intensity of this shear, and w the width of the
section at-CC.
Let M and M dM be the bending moments at the two
consecutive sections AB, A ' B '. Then
M=--I and M-dM=--I,
c c
and therefore
Hence,
dM
dM -
= (---)'
\C c I
J
= qwdx,
or
dM Ay S A -
qw = ~dx T = 7 A *>
since = shearing force at the section AB = 5.
dx
EXAMPLE I. Solid rectangular section of width b.
12
or
FIG. 302.
and the intensity of the shear at any point of AB may be rep-
resented by the horizontal distance of the point from the
parabola A VB, having its vertex at F, where OV ' - .
be
394 THEORY OF STRUCTURES.
The maximum intensity of shear is at O and its value is
<?^ = 3 -^-.
4 be
The value of the average intensity is
_ 5
~" b . 2c'
Ex. 2. A hollow rectangular section ; B and 2c being the
external and B' and 2c' the internal width and depth.
At the neutral axis,
g(B-B')^ S -\^- l
Thus, as in Ex. I, the intensity of shear is again greatest at
the neutral plane, i.e., when Y = o.
Ex. 3. Solid circular section of radius c.
*
Ay = y 2y Vc* -fdy = |(,' - Y*
w = 2 c -
and / =
and the intensity of the shear at any point of AB may be rep-
resented by the horizontal distance of the point from the pa-
rabola AVB, where OV= -^-.
3***
and
' Qmax. ' $av. ' '' 4 * 3'
DISTRIBUTION OF SHEARING STRESS. 395
Ex. 4. A double-flanged section, each of the flanges con-
sisting of five 8-in. X i-in. plates riveted to a 24-in. X i-in. web
by two 3-in. X 3-in. X i-in. angles.
To find the intensity of shear at the surface of contact
between the angles and the flange :
Ay = 20 X I3i = 265 ; w = 6J in. ; / = 8975!,
neglecting the effect of the rivet-holes in the tension flange.
Hence
2120
q = o .
466739
Let 5 = 49 tons. Then q .2226 ton per square inch.
Let the rivets have a pitch of 4 in., then
6 1
the total shear on each rivet = X 4 X .2226 = 2.8938 tons.
Let the coefficient of shearing strength be 4 tons per square
inch, and suppose that the surfaces of the angle-irons and of
the flange are close together ; then
2.8038
area of rivet = = - = .7234 sq. in.,
4
and its diameter = .96 in.
If the surfaces are not close together, so that the rivet may
be subjected to a bending action, then, by Ex. 3, the average
intensity of shear in a section =1.4=3 tons per sq. in., and
hence
area of rivet = - = .9646 sq. in. ;
its diameter is i.i in.
THEORY OF STRUCTURES.
13. Beam acted upon by Forces Oblique to its Direc-
tion, but lying in a Plane of Symmetry. In discussing the
equilibrium of such a beam the forces may be resolved into
components parallel and perpendicular to the beam, and their
respective effects superposed.
FIG. 304.
Let AB be the beam, P t , P 2 , P z , . . . the forces, and a l , <* a ,
a a the-lr respective inclinations to the neutral axis.
Divide"-the beam into any two segments by an imaginary
plane MN ^perpendicular to the beam, and consider the seg-
ment A MN.
It is kept in equilibrium by the external forces on the left
of MN and by the elastic reaction of the segment BMN upon
the segment AMN at the plane MN.
The resultant force along the beam is the algebraic sum of
the components in that direction, of P l , P t , P 3 , . . . ,
= P l cos a 1 + P 2 cos ,-{- = -SX/* cos a).
It may be assumed that this force acts along the neutral
axis, and is uniformly distributed over the section MN.
Thus, if A is the area of the section, - - ^ is the in-
tensity of stress due to this force.
Again, the components of P 1 , P % , P 3 , . . . , perpendicular to
the beam, are equivalent to a single force and a couple at MN.
The single force at MN is the Shearing Force, is per-
pendicular to the beam, and is the algebraic sum of P l sin a l9
P z sin 2 , . . . ,
= P l sin a l -\- P^ sin or a + . . . = 2"(P sin a).
BEAM ACTED UPON BY OBLIQUE FORCES. 397
This force develops a mean tangential unit stress of
2 - in MN, and deforms the beam, but so slightly as to
A
be of little account.
The moment of the couple is the algebraic sum of the
moments with respect to MN of P l sin a, , P 2 sin oc^ , . . . ,
= P l sin #,/! + PI sin a^p^ + . . . = 2(Pp sin a),
/,,/ being respectively the distances of the points of
application of P, , P 9 , . . . from MN.
Now, 2(Pp sin at) is the resultant moment of all the external
forces on the left of MN, for the resultant moment of the com-
ponents along the beam is evidently nil. Hence,
/ - *2(PJ> sin a),
is the unit stress in the material of the beam at a distance y
from the neutral axis due to the bending action at MN of the
external forces on the segment AMN.
Hence, also, the total um\. stress in the material in the plane
MN at a distance y from the neutral axis is
2(P cos a) 2(P cos a)
-~- 2 f y =~ -- 2
the signs depending upon the kind of stress.
It will be observed that this formula is composed of two
intensities, the one due to a direct pull or thrust, the other due
to a bending action. The latter is proportional to the distance
of the unit area under consideration from the neutral axis. It
is sometimes assumed that the same law of variation of stress
holds true over the real or imaginary joints of masonry and
brickwork structures, e.g., in piers, chimney-stacks, walls,
arches, etc. In such cases the loci of the centres of pressure
correspond to the neutral axis of a beam, and the maximum
THEORY OF STRUCTURES.
and minimum values of the intensity occur at the edges of the
joint.
EXAMPLE I. A horizontal beam of length /, depth d, and
sectional area A is supported at the ends, and carries a weight
W at its middle point. It is also subjected to the action of a
force H acting in the direction of its length.
First. Let the line of action of // coincide with the axis of
the beam.
The intensity of the stress in the skin at the centre
= 7".
But c oc d, and 7 = Ak* oc Ad\
I . , Ad
.-. - a Ad ,
c n
n being a coefficient depending upon the form of the section.
If the section is a circle, n = 8 ; if a rectangle, n = 6.
Hence,
HI , n W A
the skin stress = db ~2 I * 4~ ~ ~ZF 3)
si \ 4 " d/
Wl
since M = .
4
Wl
If the load W 7 is uniformly distributed, M = -5-.
o
Thus, a very small load on the beam may considerably in-
crease the intensity of stress, and this intensity will be still
further increased by the deflection of the beam under its load,
so that, in order to prevent excessive straining, it is often
necessary to introduce more supports than are actually required
to make the beam sufficiently stiff.
Second. If the line of action of H is at a distance h from
the neutral axis, an additional bending moment Hh will be in-
troduced.
BEAM ACTED UPON BY OBLIQUE FORCES.
399
Ex. 2. The inclined beam OA, carrying a uniformly dis-
tributed load of w per unit of length, is supported at A and rests
against a smooth vertical surface at O.
The resultant weight wl is vertical
and acts through the centre C of OA ;
the reaction R^ at O is horizontal.
Let the directions of wl and R^ meet
in B. For equilibrium, the reaction ^ 3
at A must also pass through B.
Let the vertical through C meet the
horizontal through A in D.
The triangle ABD is a triangle of forces for the three forces
which meet at B.
FlG< 3 5 '
R
- /
wl
AD AD
n T\ T*
BD 2. DC
Of,
being the angle OAD. Hence
wl
R, cot a.
Consider a section MN, perpendicular to the beam, at a dis-
tance x from O.
The only forces on the left of MN are R 1 and the weight
upon OM. This last is wx, and its resultant acts at the centre
x
of OM, i.e., at a distance - from MN.
The component of R t along the beam
wl cos 2 a
= R^ cos a = -- : - .
2 sin a
The component of R^ perpendicular to the beam
D wl
= R l sm a = cos a.
40O THEORY OF STRUCTURES.
The component of wx along the beam = wx sin a.
The component of wx perpendicular to the beam = wxcos a.
Hence,
wl cos a a
the total compression at NM = : -4- wx sin a = C- ;
2 sin or
the shearing force at MN = cos a wx cos a = S x
the bending moment at MN= x cos a wx cos a = M x \
and
These expressions may be interpreted graphically as already
described, C x , S x being represented by the ordinates of straight
lines, and M x , f y by the ordinates of parabolas.
f y , for example, consists of two parts which may be treated
independently. Draw OE and AF
perpendicular to OA,a.nd respectively
equal or proportional to
wl cos 3 a wl cos 2 a wl
-7 . and - : + r sin a.
2A sin a 2A sin a ' A
Join EF. The unit stress at any
FlG - 36- point of the beam due to direct com-
pression is represented by the ordinate (drawn parallel to OE
or AF) from that point to EF.
Upon the line GG' drawn through the middle point B per-
pendicular to OA, take .BG = BG ', equal or proportional to
y wl*
y-r- cos a. According as the stress due to the bending action
1 o
at any point of the beam is compressive or tensile, it is repre-
sented by the ordinate (drawn parallel to OE and AF) from
that point to the parabola OGA or OG'A ; G and ', respec-
tively, being the vertices, and GG' a common axis.
SIMILAR GIRDERS PRINCIPAL PROPERTIES. 4OI
By superposing these results, the parabolas EHF, EH'F are
obtained, the ordinates of which are respectively proportional
to the values of f y for the compressed and stretched parts of
the beam, i.e., for the parts above and below the neutral!
surface.
14. Similar Girders. Two girders are said to be similar
when the linear dimensions of the one bear the same constant
proportion to the corresponding linear dimensions of the other.
Thus, if ft, ft', d, $', A, X' are corresponding breadths, depths,
and lengths of two similar girders,
ft 6 X
~8' = <T' ~ A? = a constant " /* su Ppose.
15. To Deduce the Principal Properties of Similar
Girders. (a) The weight of a girder is proportional to the
product of an area and length, i.e., to the cube of a linear
dimension.
Hence, the weights of similar girders vary directly as the
cubes of their linear dimensions. Hence, too, the unit stresses
must vary directly as their linear dimensions.
(b) The Breaking Weight of a girder is calculated from a
formula of the form W '= Sy-j a being an area, da. depth, and
/ a length.
Now - is constant for similar girders, so that W is propor-
tional to a, i.e., to the square of a linear dimension.
Hence, the Breaking Weights of similar girders vary directly
as the squares of their linear dimensions.
EXAMPLE. A girder resting upon two supports 80 ft.
apart is 10 ft. deep and weighs 6 tons. Determine the length
and depth of a similar girder weighing 48 tons.
flengthV /depth y 48
\to~/ == \IS/ ~-~~ 6~ r=8 '
Hence, the length = 160 ft. and the depth = 20 ft. Also,
the unit stresses are in the ratio of 10 to 20, and the breaking
weights in the ratio of io 2 to 2O 2 .
4 2 THEORY OF STRUCTURES.
16. To Discuss the Relations between the Correspond-
ing Sectional Areas, Moments of Inertia, Weights, Bend-
ing Moments, etc., of two Girders which have the same
Sectional Form and are thus related :
The forces upon the one being P l , P 9 , P 9 , . . . with abscissae
x^ x t , x tt .* . . those upon the other are nP l , nP t , nP 3 , . . .
with abscissae px l , px t , px 9 .
The spans and corresponding lengths are in the constant
ratio /.
Corresponding sectional breadths are in the constant ratio q.
Corresponding sectional depths are in the constant ratio r.
Let A, A' be corresponding sectional areas ;
/, I' " " moments of inertia ;
<2, 0', " " weights ;
S, S f " " shearing forces ;
M, M' " " bending moments ;
/,/' " " flange unit stresses ;
j, s' " " web unit stresses ;
R, R' " " radii of curvature ;
A, 4 f " " deflections ;
W, W " " breaking weights.
/. () A a product of a breadth and depth ;
(/3) I a product of a breadth and the cube of a depth ;
.-./'= Iqr\
(y) Q a product of a length, breadth, and depth ;
... Q'= Qn= Qpqrp,
p being the ratio of the specific weights of the materials of the
girders.
If the materials are the same,
p = i and n = pqr.
SIMILAR GIRDERS. 43
($} S' = Sn = Spqrp, for, from (y), n = pqrp.
(e) M is the product of a force and a length
/. M' = Mnp Mp*qrp,
cM c'M'
(C) f=-f and /' = -;
f c'M' I i /
.. -^ = - = rnp , = p.
f c M I' ^qr* r r
(rf) s = - and *' = 7 ;
E f E' f
(6) = - and ^7- = , E and ' being the coef
JK. C \. C
ficients of elasticity of the respective girders ;
____ __ _ _.
" ~R ~f f c E~~~~ E p*p ~ = E / 2 p 9
(a length) 2
(0 ^ is proportional to j: - 7 - - - ;
v J radius of curvature
the product of an area and depth
(K) Wis proportional to- a length ~J
W_ _ qr.r _ q^_
w~- p '- p'
Hence, the values of A', /', Q f , . . . may be derived from
tho >e of Aj /, Q, . . . by means of certain constant multipliers.
404 THEORY OF STRUCTURES.
Cor. I. If the two girders are similar and of the same
material,
p = g = r = ju, E = J5', and p=i.
Hence,
from (y), Q = <2/A and the weights vary directly as the
cubes of the linear dimensions ;
" (e), M' = MJJ?, and the bending moments vary directly
as the fourth powers of the linear di-
mensions;
/' s/
" (C) and (;?), -j P = , and the flange unit stresses vary
/ s
directly as the web unit stresses;
" ( z )> "A = ^ 8 > an d t ^ ie deflections vary directly as the
squares of the linear dimensions ;
W r
" W ^M7 == ^ ' an< ^ ^ e breaking weights vary directly as
the squares of the linear dimensions.
Cor. 2. Let the girders be of the same material, of equal
length, of equal rectangular sectional areas, and equally loaded.
Let b, b lt and d, d,, respectively, be the breadths and
depths of the girders. Then
b l = qb and d l = rd.
Hence,
b^d^ = qrbd.
But b,d } = bd\ / qr i. Also, p i.
ALLOWANCE FOR THE WEIGHT OF A BEAM. 405
f i
Thus from C, y = - = q ;
R' A
0andi, -- = r' = ;
,- --- r ~~
W ~ ~ f '
b
If d l = b, then b l ^/, and r = -,.
_
Hence, -- - ?> and ^ == = =
17. To make Allowance for the Weight of a Beam. A
beam is sometimes of such length that its weight becomes of
importance as compared with the load it has to carry, and
must be taken into account in determining the dimensions of
the beam.
The necessary provision may be made by increasing the
width of the beam designed to carry the external load alone,
the width being a dimension of the first order in the expression
for the elastic moment.
Assume that the weight of the beam and the external load
are reduced to equivalent uniformly distributed loads.
Let W t be the external load.
" b e " " breadth of a beam designed to support this
load only. }
" B e " " weight of the beam.
" W " " total load, the weight of the beam being
taken into account.
" b " " corresponding breadth of the beam.
g weight " " "
Then W-B=W e ,
b B W W- B W e
and
b e ~ B~ w, " W.-B. ~ w e - B;
4O6 THEORY OF STRUCTURES.
wj>. wjs. w;
Hence, 6,-.-^-, B -- -^^ and W -- W^J;
EXAMPLE. Apply the preceding results to a cast-iron
girder of rectangular section resting upon two supports 30 ft.
apart. The girder is 12 in. deep and carries a uniformly dis-
tributed load of 30,000 Ibs.
Take 4 as a factor of safety ; b e is given by
I2OOOO bd*
where C 30,000 Ibs., d = 12 in., and / = 360 in. ;
/. b e 5 in.
Hence,
B e = -^7 -^ X 30 X 450 = 5625 Ibs. ;
144
W e B e 30000 5625 = 24,375 Ibs. ;
30000,5
24375
lbs> ;
W= ^ ^
EXAMPLES.
1. An iron bar is bent into the arc of a circle of 500 ft. diameter;
the coefficient of elasticity is 30,000,000 Ibs. Find the moment of resist-
ance of a section of the bar and the maximum intensity of stress in the
metal, (a) when the bar is round and i in. in diameter, (<) when the bar
is square having a side of r inch.
If the metal is not to be strained above 10,000 Ibs. per sq. in., find (c}
the diameter of the smallest circle into which the bar can be bent.
Ans.(a) ^Tt in. -Ibs.; 5000 Ibs. (b) 833^ in. -Ibs.; 5000 Ibs. (c) 250 ft.
2. A piece of timber 10 ft. long, 12 in. deep, 8 in. wide, and having a
working strength of 1000 Ibs. per sq. in., carries a load, including its
own weight, of w Ibs. per lineal foot. Find the value of w, (a) when the
timber acts as a cantilever ; (b) when it acts as a beam supported at the
ends. Find (c) stress in material 3 in. from neutral axis at fixed end of
cantilever and at middle of beam.
Am, (a) 320 Ibs.; (b) 1280 Ibs.; (c] 500 Ibs. per sq. in.
3. Is it safe for a man weighing 160 Ibs. to stand at the centre of a
spruce plank loft, long, 2 in. wide, and 2 in. thick, supported by vertical
ropes at the ends ? The safe working strength of the timber is 1200 Ibs.
per sq. in.
Ans. No ; the maximum safe weight at the centre is 53^ Ibs.
4. Compare the uniformly distributed loads which can be borne by
two beams of rectangular section, the several linear dimensions of the
one being n times the corresponding dimensions of the other. Also
compare the moments of resistance of corresponding sections.
Ans. n* ; n*.
5. A cast-iron beam of rectangular section, 12 in. deep, 6 in. wide, and
16 ft. long, carries, in addition to its own weight, a single load P\ the
coefficient of working strength is 2000 Ibs. per sq. in. Find the value of
P when it is placed (a) at the middle point; (b) at 2$- ft. from one end.
Ans. (a) 8475 Ibs. ; (b) 11,300 Ibs.
6. A round and a square beam of equal length and equally loaded are
to be of equal strength. Find the ratio of the diameter to the side of the
square. Ans. \/~$6 :
407
408 THEORY OF STRUCTURES.
7. Compare the relative strengths of two beams of the same length and
material (a) when the sections are similar and have areas in the ratio
of i to 4; (b} when one section is a circle and the other a square, aside of
the latter being equal to the diameter of the former.
Ans. (a) i to 8 ; (b) 56 to 33.
8. Compare the strength of a cylindrical beam with the strength of
the strongest (a) rectangular and (b) square beam that can be cut from it.
Ans. (a) 112 : 99 |/^~; (b) 33 : 14 fa
9. A boiler-plate tube 36 ft. long, 30 in. inside diameter, weighs 4200
IDS. and rests upon supports 33 ft. apart. Find the maximum intensity
of stress in the metal. What additional weight may be suspended from
the centre, assuming that the stress is nowhere to exceed 8000 Ibs. per
sq. in. ?
Ans. 741 i- Ibs. per sq. in.; 1 8,854 -|f| Ibs.
10. Compare the relative strengths of two rectangular beams of equal
length, the breadth (ft) and depth (d) of one being the depth (b) and
breadth (d) of the other. Ans. d* : P.
11. A yellow-pine beam 14 in. wide, 15 in. deep, and resting upon sup-
ports 129 in. apart, was just able to bear a weight of 34 tons at the cen-
tre. What weight will a beam of the same material, of 45 in. span and
5 in. square, bear ? Ans. \\\ tons.
12. A cast-iron rectangular girder rests upon supports 12 ft. apart
and carries a weight of 2000 Ibs. at the centre. If the breadth is one-
half the depth, find the sectional area of the girder so that the intensity
of stress may nowhere exceed 4000 Ibs. per sq. in.
Ans. 1 8 sq. in.; if weight of girder is to be taken into account,
the depth d is given by d s i.oi2$d* 216 = o.
13. Find the depth of a wrought-iron girder 6 in. wide which might be
substituted for the cast-iron girder in the preceding question, the coeffi-
cient of strength for the wrought-iron being 8000 Ibs. per sq. in.
Ans. 4.762 in.; if weight of girder is to be included, the depth d
is given by d* .54</ 2 108 = o.
14. An oak beam of circular section and 22 ft. long is strained to the
elastic limit (2 tons per sq. in.) by a uniformly distributed load of 2^
tons. Find the diameter of the beam. What load 2 ft. from one end
would strain the material to same limit? Ans. 7 in.; 3 T ^V tons.
1 5. A uniform beam of weight W\ crossing a given span can bear a uni-
formly distributed load Wt.. What load may be placed upon the same
beam if it crosses the span in n equal lengths supported at the joints by
piers whose widths may be disregarded ? Ans. n*( W^ + IVz) W\.
EXAMPLES. 409
16. A flat spiral spring .2 in. wide and .03 in. thick is subjected to a
bending moment of 10 in. -Ibs. Find its radius of curvature, E being
56,000,000 Ibs. Ans. 1.62 in.
17. Determine the diameter of a solid round wrought-iron beam rest-
ing upon supports 60 in. apart and about to give way under a load of 30
tons at 14 in. from one end. Take 5 as a factor of safety and 8960 Ibs.
per sq. in. as the safe working intensity of stress.
Ans. 5.47 in.; if weight of beam is taken into account, the diam-
eter (d) is given by 2019584 i375^ 2 12320^" =o.
1 8. A wrought-iron bar i-| in. wide and 20 ft. long is fixed at one end
and carries a load of 500 Ibs. at the free end. Find the depth of the bar,
so that the stress may nowhere exceed 10,000 Ibs. per sq. in.
Ans. 6.928 in.; if weight of bar is included, the depth d is given
by d* 4.8^ 48=0.
19. Compare the moments of resistance to bending of a rectangular
section and of the rhomboidal and isosceles sections which can be in-
scribed in the rectangle, the base of the triangle being the lower edge
of the rectangle. Ans. 6 : i : i or 6 : i : 2.
20. A stress of i Ib. per sq. in. produces a strain of 7^7777 in a beam
12 in. square and 20 ft. between supports. Find the radius of curvature
and the central deflection under a load of 2000 Ibs. at the middle point.
Ans. 2400 ft. ; \ in.
21. A piece of greenheart 139 in. between supports, 9 in. wide and 8
in. deep, was successively subjected to loads of 4, 8, and 16 tons at the
centre, the corresponding deflections being .32 in., .64 in. and 1.28 in.
Find E and the total work done in bending the beam.
What were the corresponding inch-stresses at f of the depth of the
beam? Ans. E = 5582682^ ; 1344 inch-tons ; ff Ib., |f Ib., Vr lb -
22. The effective length of the Conway tubular bridge is 412 ft. ; the
effective depths of a tube at the centre and quarter spans are 23.7 ft. and
22.25 ft- respectively ; the sectional areas of the top and bottom flanges
are respectively 645 sq. in. and 536 sq. in. at the centre and 566 sq. in.
and 461 sq. in. at the quarter spans ; the corresponding sectional areas
of the web are 257 sq. in. and 241 sq. in. Assume the total load upon a
tube to be equivalent to 3 tons per lineal foot, and that the continuity of
the web compensates for the weakening of the tension flange by the
rivet-holes. Find the flange stresses and the deflection at the centre and
quarter spans, E being 24,000,000 Ibs. What will be the increase in the
4IO THEORY OF STRUCTURES.
central flange stresses under a uniformly distributed live load of ton
per lineal foot ?
Ans. At centre = 181485 ; f t = 4.3799 tons per sq. in. ;
Deflection = 8.33 in.
At quarter span = 132774 ; f t = 1.414 " " "
f c =1.256 " " "
Deflection = 6.25 in.
The stresses and deflections are increased by the live load in
the ratio of 5 to 4.
23. A plate girder of 64 ft. span and 8 ft. deep carries a dead load of 2
tons per lineal foot. At any section the two flanges are of equal area, and
their joint area is equal to that of the web. Find the sectional area at
the centre of the girder, so that the intensity of stress in the metal may
not exceed 3 tons per sq. in. The deflection of the girder is f in. at
the centre. Find E and the radius of curvature.
Ans. 128 sq. in. ; 15,360 ft.; 25,804,800 Ibs.
24. Taking the coefficient of direct elasticity at 15,000 tons, the coeffi-
cient of lateral elasticity at 60,000 tons, and the limit of elasticity at 10
tons, determine the greatest deviation from the straight line of a wrought-
iron girder of breadth b and depth d. b*
Ans. 7.
24000^
25. Find the stress at the skin and also at a point 4 in. from the neu-
tral axis in a piece of 10" x8" oak, (a) with the 10" side vertical; (b} with
the 8" side vertical. The oak rests upon supports 3 ft. apart and carries a
load of 4900 Ibs. at its middle point. Also compare (c) the strength of the
beam with its strength when a diagonal is horizontal.
Ans. (a) 330! ; I32 T 3 7 Ibs. per sq. in.
_
(c) 4 : 1/41 or 5 : 4/41.
26. Find the uniformly distributed load which can be borne by a
rolled T-iron beam, 6" X4" x ", 10 ft. long, fixed at one end and free at
the other, the coefficient of strength being 10,000 Ibs. per sq. in.
Ans. 438 Ibs.
27. One of the tubes of the Britannia bridge has an effective length of
470 ft. , depth of 27^ ft. , and deflects 1 2 in. at the centre under a uniformly
distributed load of 1587 tons. Find E and the central flange stresses, the
sectional areas of the top flange, bottom flange, and web being 648 sq. in.,
585 sq. in., and 302 sq. in., respectively.
Ans. E == 22,910,496 Ibs.; f t = 5.37 tons per. sq. in.;
EXAMPLES. 411
28. Find the moment of resistance to bending, of a steel I-beam, each
flange consisting of a pair of 3-in. x 3 in. x ^ in. angle-irons, riveted to
a 12 in. x f in. web, the coefficient of strength being 5 tons per sq. in.
What load will the beam carry at 5 ft. from one end, its span being 20
ft. ? Find the central deflection, and also the deflection at the loaded
point, E being 15,000 tons.
Ans. 287!^ in. -tons ; 6ff- tons disregarding weight of beam, or
SyVA tons if weight of beam is taken into account ; deflection
at centre = f in., at loaded point = ^ in.
29. A shaft 5i in. deep x 5 in. wide x 98 in. long has one end abso-
lutely fixed, while at the other a wheel turns at the rate of 270 revolutions
per minute ; a weight of 200 Ibs. is concentrated in the rim, its C. of G.
being 2^ ft. from the axis of the shaft. Find the maximum stress in the
material of the shaft, and also find the maximum deviation of the shaft
from the straight, E being 27,000,000 Ibs.
30. The square of the radius of gyration of the equal-flanged section
of a wrought-iron girder of depth d is r V^ 2 ; the area of the section
= |d? 2 ; the span = 50 ft. In addition to its own weight it carries a uni-
formly distributed load of i V Ibs. per lineal foot ; the maximum intensity
of stress = 10,000 Ibs. per sq. in. Find the depth. Also determine the
stiffness, E being 25,000,000 Ibs. Ans. 3! in.; T fy.
31. The central section of a cast-iron girder is io in. deep; its web
area is five times the area of the top flange, and the moment of resistance
of the section is 360,000 in.-lbs.; the tensile and compressive intensities of
stress are 3000 and 7500 Ibs. per sq. in., respectively. Find the span and
load so that the girder may have a stiffness = .001, E being 17,000,000
Ibs.
Ans. <zi = 12^- sq. in.; a* = i|| sq. in.; a s + a t = 97||| sq.
in.; span = 136 ft.; uniformly distributed load = 1764^ Ibs.
32. A double-flanged cast-iron girder 14 in. deep and 20 ft. between sup-
ports carries a uniformly distributed load of 20 tons. Find suitable di-
mensions for the section, the tensile and compressive inch-stresses being
2 tons and 5 tons, respectively. Also find the stiffness of the beam, E
being 8000 tons.
Ans. Let thickness of web = i in.; a\ = 22f| sq. in.; a* = 4-^.-
sq. in.; stiffness = .001875.
33. The deflection of a uniformly loaded horizontal beam supported at
the ends is not to exceed i in. in 50 feet of span, and the stress in the ma-
terial is not to exceed 400 Ibs. per sq. in. Find the ratio of span to
depth, E being 1,200,000 Ibs. per sq. in., and the neutral axis being at
half the depth of the beam. Ans. 20.
34. Two equal weights are placed symmetrically at the points of tri-
section of a beam of uniform section supported at the ends. These weights
412 THEORY OF STRUCTURES.
are then removed and other two equal weights are placed at the quarter
spans. Find the ratio of the two sets of weights so that the maximum
intensity of stress may be the same in each case. Also show that the
stiffness of the beam is the same in each case. Ans. 3 to 4.
35. A cast-iron beam has a cruciform section with equal ribs 2 in.
thick and 4 in. long. If the intensity of longitudinal shear at the neutral
axis is i ton per sq. in., find the total shear which the section can
bear, and also find the moment of resistance, the least coefficient of
working tensile and compressive stress being i ton per sq. in.
Ans. 59.31 tons; 34.6 tons.
36. If a spiral spring is fastened to the barrel so that there is no
change of direction relatively to the barrel, show that the tendency to
unwind is directly proportional to the amount of winding up. (Con-
dition of perfect isochronism.)
37. Show that the modulus of rupture vi any material is 18 times the
load which will break a beam 12 in. long, i in. deep, and i in. wide when
applied at the centre.
38. Find the limiting length of a wrought-iron cylindrical beam 4 in.
in diameter, the modulus of rupture being 42,000 Ibs. What uniformly
distributed load will break a cylindrical beam of the same material 20 ft.
long and 4 in. in diameter ? Ans. 64.8 ft. ; 8800 Ibs.
39. A red-pine beam 18 ft. long has to support a weight of 10,000 Ibs.
at the centre. The section is rectangular and the depth is twice the
breadth. Find the transverse dimensions, the modulus of rupture being
3500 Ibs., and 10 being a factor of safety. (Neglect the weight of the
beam.) Ans. b = 9.84 in. ; d= 19.68 in.
40. A round oak cantilever 10 ft. long is just broken by a load of 600
Ibs. suspended from the free end. Find its diameter, the modulus of rup-
ture being 10,000 Ibs, (Neglect the weight of the beam.)
Ans. 4.185 in.
41. Determine the breaking weight at the centre of a cast-iron beam
of 6 ft. span and 4 in. square, the coefficient of rupture being 30,000 Ibs.
Ans. 26,666| Ibs.
42. The flooring of a corn warehouse is supported upon yellow-pine
joists 20 ft. in the clear, 8 in. wide, 10 in. deep, and spaced 3 ft. centre
to centre. Find the height to which corn weighing 48^ Ibs. per cu.
ft. may be heaped upon the floor, 10 being a factor of safety and 3000
Ibs. the coefficient of rupture. Ans. .68 ft.
43. A yellow-pine beam 14 in. wide, 15 in. deep, and resting upon sup-
ports 126 in. apart, broke down under a uniformly distributed load of
60.97 tons. Find the coefficient of rupture. Ans. 2731.456.
EXAMPLES. 413
44. Find the breaking weight at the centre of a Canadian ash beam
2-J- in. wide, 3! in. deep, and of 45 in. span, the coefficient of rupture being
7250. Ans. 4934uV
45. A timber beam 6 in. deep, 3 in. wide, 96 in. between supports, and
weighing 50 Ibs. per cu. ft., broke down under a weight of 10,000 Ibs.
at the centre. Find the coefficient of rupture. Ans. 891 1^.
46. A wrought-iron bar 2 in. wide, 4 in. deep, and 144 in. between sup-
ports, carries a uniformly distributed load W in addition to its own
weight. Find W, 4 being a factor of safety and 50,000 Ibs. the coeffi-
cient of rupture. Ans. 5235! Ibs.
47. Find the length of a beam of Canadian ash 6 in. square which
would break under its own weight when supported at the ends. The
coefficient of rupture = 7000 Ibs., and the weight of the timber = 30 Ibs.
per cu. ft. Ans. 230 ft.
48. The teeth of a cast-iron wheel are 3^ in. long, 2^ in. deep, and 7
in. wide. What is the breaking weight: of a tooth, the coefficient of
rupture being 5000 Ibs. ? Ans. 50,625 Ibs.
49. A wrought-iron bar 4 in. deep, in. wide, and rigidly fixed at one
end gave way at 32 in. from the load when loaded with 1568 Ibs. at the
free end. Find the coefficient of rupture. Ans. 4181^.
50. A cast-iron beam 12 in. wide rests upon supports 1 8 ft. apart, and
carries a 12-in. brick wall which is 12^ ft. in height and weighs 112 Ibs.
per cu. ft. Taking 63,000 as the modulus of rupture for a uniformly
distributed load and 5 as a factor of safety, find the depth of the beam,
(a) neglecting its weight ; (b) taking its weight into account.
Also (c) determine the depth of a cedar beam which might be sub-
stituted for the cast-iron beam, taking 11,200 Ibs. as the modulus of
rupture for the cedar. Ans. (a) 6 in. ; (fr) 6| in. ; (c) 14.23 in.
51. A cast-iron girder 27^ in. deep, rests upon supports 26 ft. apart.
Its bottom flange has an area of 48 sq. in. and is 3 in. thick. Find the
breaking weight at the centre, the ultimate tensile strength of the iron
being 15,000 Ibs. per sq. in. (Neglect the effect of the web.)
Ans. 253,846^ Ibs.
52. A beam of rectangular section, of breadth b and depth d, is acted
upon by a couple in a plane inclined at 45 to the axis of the section.
Compare the moment of resistance to bending with that about either
axis.
'
53. A 2-in. wrought-iron bar 10 ft. long is held at the ends and is
whirled about a parallel axis at the rate of 50 revolutions per minute.
If the distance between the axis of the bar and the axis of rotation is
10 ft., find the maximum stress to which the material is subjected.
Ans. 17148.5 Ibs. per sq. in.
4H . THEORY OF STRUCTURES.
54. A block of ice 3 in. wide and 4 in. deep has its ends resting upon
supports 30 in. apart and carries a uniformly distributed load of 4800
Ibs. An increase of pressure to the extent of 1125 Ibs. per sq. in. lowers
the freezing point i F. Assuming that the ordinary theory of flexure
holds good, find the temperature of the ice. Ans. 30 F.
55. Find the limiting length of a cantilever of uniform transverse
section,/ being the coefficient of strength, k the ratio of length to depth,
and u> the specific weight of the material.
Ans. ^ - > n being a coefficient depending upon the form of
the section.
56. If the beam in the preceding question is to be supported at its two
ends, what will its limiting length be ? Ans. , .
57. Find the limiting length,of a cedar cantilever of rectangular sec-
tion, k being 40, w = 36 Ibs. per cu. ft., and/ = 1800 Ibs. per sq. in.
Ans. 60 ft.
58. A steel cantilever 2 in. square has an elastic strength of 15 tons
per sq. in. What must its limiting length be so that there may be no
set f Ans. 23.4 ft.
59. Find the limiting length of a wrought-iron beam of circular sec-
tion, k being 64 and the elastic strength 8 tons per sq. in. What will this
length be if a beam of I-section, having equal flange areas and a web
area equal to the joint area of the flanges, is substituted for the circular
section ? Ans. 84 ft. ; 224 ft.
60. A rectangular cast-iron beam having its length, depth, and
breadth in the ratio of 60 to 4 to i, rests upon supports at the two ends.
Find the dimensions of the beam so that the intensity of stress under its
own weight may nowhere exceed 4500 Ibs. per sq. in.
Ans. I = 128 ft. ; d = 8-jV ft. ; b = 2 T ^ ft.
61. A beam supported at the ends can just bear its own weight W\.o-
W
gether with a single weight at the centre. What load may be placed
at the centre of a beam whose transverse section is similar but m* as
great, its length being n times as great ? If the beam could support only
its own weight, what would be the relation between m and n ?
(m* m^ n
Ans. W ; m = -.
\n 2 ) 2
62. The flanges of a rolled joist are each 4 in. wide by \ in. thick ;
the web is 8 in. deep by inch thick. Find the position of the neutral
axis, the maximum intensities of stress per square inch being 10,000
Ibs. in tension and 8000 Ibs. in compression. Ans. k\. = 3$ ; hi ~ 4$.
EXAMPLES. 415
63. A continuous lattice-girder is supported at four points, each of the
side spans being 140 ft. n in. in length, 22 ft. 3 in. in depth, and weigh-
ing .68 ton per lineal foot. On one occasion an excessive load lifted the
end of one of the side spans off the abutment. Find the consequent in-
tensity of stress in the bottom flange at the pier, where its sectional
area is 127 sq. in. Ans. 2.3893 tons per sq. in.
64. A railway girder is 101.2 ft. long, 22.25 ft- deep, and weighs 3764
Ibs. per lineal foot. Find the maximum shearing force and flange stresses
at 25 ft. from one end when a live load of 2500 Ibs. per lineal foot crosses
the girder.
65. A floor with superimposed load weighs 160 Ibs. per sq. ft. and is
carried by tubular girders 17 ft. c. to c. and 42 ft. between bearings.
Find the depth of the girders (neglecting effect of web), the safe inch-
stress in the metal being 9000 Ibs., and the sectional area of the tension
flange at the centre 32 sq. in. Ans. 24.99 in -
66. Design a timber cantilever of approximately uniform strength from
the following data: length = 12 ft.; square section; load at free end
= i ton ; coefficient of working strength = | ton per sq. in. What
must be the dimensions at the fixed and free ends so that the cantilever
might carry an additional uniformly distributed load of 2 tons ?
Ans. Side = 15.1 in. at fixed end and = 10 in. at free end;
side = 19.1 in. at fixed end and = (19.1 in.) at free end.
67. Show that the curved profile of a cantilever of uniform strength
designed to carry a load W at the free end, is theoretically a cubical
parabola. Also show that by taking the tangents to the profile at the
fixed end as the boundaries of the cantilever, a cantilever of approxi-
mately uniform strength is obtained having a depth at the free end
equal to two-thirds of the depth at the fixed 6nd.
68. Design a wheel-spoke 33 in. in length to be of approximately uni-
form strength, the intensity of stress being 4000 Ibs. per sq. in. ; the load
at the end of the spoke is a force of 1000 Ibs. applied tangentially to the
wheel's periphery, and the section of the spoke is to be (a) circular, (b)
elliptical, the ratio of the depth to the breadth being 2|.
Ans. (a) Depth at hub = 6.982 in., at periphery = 4.634 in.
(b) " = 9.435 " " = 9-35 in.
Breadth " " = 3.7/4 " " = 3-74 in.
69. A beam of 17 ft. span is loaded with 7, 7, 11, and n tons at
points 1,6, ii, and 15 ft. from one end. Determine the depths at these
points, the beam being of uniform breadth and of approximately uni-
form strength ; the coefficient of working strength = 2 tons per sq. in.;
the depth of the section of maximum resistance to bending = 16 in.
16065 277 x i6 s 1087 x i6 a
Ans ' * =
i 7 8 5
670 x i6 2
and d<? =
41 6 THEORY OF STRUCTURES.
70. Design a cantilever 10 ft. long, of approximately uniform strength,
to carry a load of 4000 Ibs. at the free end, the coefficient of strength
being 2000 Ibs. per sq. in., and the section (a) a rectangle of constant
breadth and 12 in. deep at the fixed end ; (b) a square.
How will the results be modified if it is to carry an additional uni-
formly distributed load of 4800 Ibs. ?
Ans. First, (a] b 10 in., d at free end = 6 in. ; (b) side
= ty 1440 at fixed end and = tyiSo at free end.
Second, (a) b = 16 in., */at free end = 6 in. ; (b) side = ty 2304
at fixed end and = ^288 at free end.
71. Design a cantilever 10 ft. long, of constant breadth, and of ap-
proximately uniform strength to carry a uniformly distributed load of 5000
Ibs. on the half of the length next the free end, the intensity of stress
being 2000 Ibs. per sq. in., and the section a rectangle 12 in. deep at the
fixed end. What must the dimensions be if loco Ibs. are concentrated
at 30 in. from fixed end ?
Ans. b = 9! in. ; d at centre = 6.928 in. ; at free end = o.
b= 10 in. ; depth =8.66 in. at 7^ ft. from free end, =6.928
in. at centre, and = o at free end.
72. A gallery 30 ft. long and 10 ft. wide is supported by four 9 in. by 5
in. cantilevers spaced so as to bear equal portions of the superincumbent
weight. What load per square foot will the gallery bear, the coefficient of
working strength being 700 Ibs. per sq. in. ? Find the depth of cast-iron
cantilevers 3 in. wide which may be substituted for the above, the coef-
ficient of working strength being 2000 Ibs. per sq. in. How should the
depth vary if the cantilevers are to be of uniform strength ?
Ans. 42 Ibs.; d* =,18.9; variation of depth for cast-iron canti-
lever is given by loood* = 189^, x being distance from free
end.
73. A span of 60 ft. is crossed by abeam hinged at the points of trisec-
tinn and fixed at the ends ; the beam has a constant breadth of 3 in. and
is to be of uniform strength ; the intensity of stress is 3 tons per sq. in.
Determine the dimensions of the beam when a load of ^ ton P er lineal
loot covers (a) the whole span ; (b) the centre span.
Ans. (a) Depth at support = 20 in., at centre = |/2a
(d) " " = VSo in., " " = |/2oT
74. In the following examples determine the position of the neutral
axis, the moment of resistance to bending, the resistance to shear, and
the ratio of the maximum to the average intensity of shear, the co-
efficients of strength being 4^ tons per sq. in. for tension and compression
and 3^ tons per sq. in. for shear.
(I) A rectangle 2 in. wide and 6 in. deep.
Ans. At centre ; 54 in. -tons ; 28 tons ; 3 to 2.
EXAMPLES.
417
(II) A circular section 4 in. in diameter.
Ans. At centre ; 28.2 in. -tons; 33 tons; 4 to 3.
(III) A regular hexagonal section with a diameter (a) vertical, (b)
horizontal, a being a side of the hexagon.
Ans. (a) At centre;
(b) At centre ;* 9 ;
16
1888
l 7 to 5.
: 1.258.
(IV) A triangular section 6 in. deep, with a base 6 in. wide, the sides
being equal. Ans. 4 in. from vertex ; 40.5 in. -tons ; 42^ tons ; 3 to 2.
(V) A double-tee section composed of a 3o-in. x f-in. web and four
angle-irons each 5 in. x 3^ in. x in.
Ans. At centre ; 1501.06 in. -tons ; 22. 36 tons: 4.916 to I.
(VI) A section having a semicircular top flange of 8 in. external
diameter and i in. thick, a web 14 in. deep and i in. thick, and a bottom
flange 8 in. wide and i inch thick.
(VII) A section having a semi-elliptic top flange 2 in. thick, the in-
ternal major and minor axes being 8 in. (horizontally) and 4 in. ^ver-
tically), respectively, a bottom flange 8 in. wide and 2 in. thick, and a
web 10 in. deep and 2 in. thick.
(VIII) A section having a semi-elliptic top flange 2 in. thick, the
external major and minor axes being 10 in. (horizontally) and 6 in.
(vertically), respectively, a trapezoidal web 8 in. deep having a width of 3
in. at the top and 6 in. at the bottom, and a bottom semicircular flange
of 10 in. external diameter and 2 in. thick.
(IX) The sections shown by Figs. 307, 308, and 309.
FIG. 309.
Also find the diameters of the rivets/? in Fig. i, neglecting the weak-
ening effect of the rivet-holes in the bottom flange. What is the ratio
of the maximum tensile and compressive stresses in each section ?
THEORY OF STRUCTURES.
(X) A trapezoidal section, the top side, bottom side, and depth h
(inches) being in the ratio of I to 2 to 4.
Ans. %h from top side ; T V 3 <^ 3 in.-tons.
(XI) A section in the form of a rhombus of depth ic and with a hori-
zontal diagonal of length 2b. Ans. ^bc* ; ^bc ; 9 to 8.
(XII) An angle-iron 2 in. x 2 in. x \ in.
Ans. Neutral axis divides depth into segments of J in. and f|
in. ; \\\l in.-tons ; ^or ton I ! 334 to 1369.
(XIII) A hollow circular section of external radius C and internal
radius C.
Ans .99 C* ~ C' 4 33 C 4 - C" 4 C 2 + CC' + C "
112 C '4 C' 2 + CC' + C' ' 3 C' 2 + C' 8
(XIV) A cruciform section made up of aflat steel bar 10 in. by i in.
and four steel angles, each 4 in. by 4 in. by in., all riveted together.
(Neglect weakening effect of rivet-holes.)
75. A girder of 21 ft. span has a section composed of two equal
flanges each consisting of two 3i-in. x 5~in. x i-in. angles riveted to a
39-in. x f-in. web ; the cover-plates on the flanges are each 12 in. x in.,
and the rivets in the covers alternate with those connecting the angles
and web ; the pitch of the rivets is 3^ in. Find the diameter and also find
the maximum flange stresses, (a) disregard ing the weakening effect of the
rivet-holes in the tension flange ; (b) taking this effect into account,
The load upon the girder is a uniformly distributed load of
20,800 Ibs. (including weight of girder) and a load of 50,000 Ibs. concen-
trated at each of the points distant 4^ ft. from the middle point of the
girder.
Ans. Diam. of rivets = .48 in. if tight, = .54 in. if subject to flexure.
(a) fi =/ 2 = 7762 Ibs. per sq. in.
(b) fi = 8248 Ibs. per sq. in.,/ 2 = 7847 Ibs. per sq. in.
76. A beam of triangular section 12 in. deep and with its base hori-
zontal can bear a total shear of 100 tons. If the safe maximum intensity
of shear is 4 tons per sq. in., find the width of the base. Ans. 6 in.
77. Assuming that the web and flanges of a rolled beam are rectangular
in section, determine the ratio of the maximum to the average intensity
of shear in a section from the following data : the total depth is times
the breadth of each flange, n times the thickness of each flange, and 2
times the thickness of the web. Show also that this ratio is V T B or y
according as the area of the web is equal to the joint area of the two
flanges or is equal to the area of each flange. How much of the shear--
ing force is borne by the web ? How much by the flange ?
3(# 2 + \in \2)(n + 6)
Ans. ratio = i- r i- : 70*1 85*.
EXAMPLES. 419
78. In a rolled beam with equal flanges, the area of the web is propor-
tional to the nth power of the depth. Find the most economical distribu-
tion of metal between the flanges and web, and the moment of resistance
to bending of the section thus designed. Also find the ratio of the aver-
age to the maximum intensity of shear.
Ans. Area of each flange : web area :: 2 i : 6 ;
B. M. = -
2 n
f being the coefficient of strength, S the total area of
section, and y the depth.
Max. intensity of shear : av. intensity :: (n + i)(4 + i) : 6n.
79. Find the moment of resistance to bending, the resistance to shear,
and the ratio of maximum to the average intensity of a shear in the
case of a section consisting of two equal flanges, each composed of a
pairof 5-in. x 3^-in. x f-in. angle-irons riveted to a 3ii-in x f-in. web,
the 5-in. sides of the angles being horizontal, and 4^ tons per sq. in.
being the coefficient of strength.
Ans. 1501.06 in. -tons ; 22. 36 tons; 4.916.
80. The floor-beam for a single-track bridge is 15 ft. between bearings,
and each of its flanges is composed of a pair of 2|-in. x 2f-in. x f-in.
angle-irons riveted to a 3o-in. x f-in. web. The uniformly distributed
load (including weight of beam) upon the beam is 4200 Ibs., and a weight
of 1600 Ibs. is concentrated at each of the rail-crossings, 2^ ft. from
the centre. Find (a) the maximum flange stress, (b) the ratio of the
maximum and average intensities of shear ; (c) the stiffness, E being
27,000,000 Ibs.
Ans. (a) 6523.4 Ibs ; (<) 2.037; (c) .00033.
/ = - ~ , neglecting effect of rivet-holes.
8 1. A beam 36 ft. between bearings is a hollow tube of rectangular sec-
tion and consists of a 24-in. x -in. top plate, a 24-in. x -in. bottom
plate, and two side plates each 35 in. x \ in. The plates are riveted
together at the angles of the interior rectangle by means of four 6-in.
x 4-in. x^-in. angle-irons, the 6-in. side being horizontal. Determine
(a) The intensity of shear at the surface between the angle-irons and
the upper and lower plates.
(b) The diameter of the rivets, the pitch being 4 in. and assuming an
effective width of 5^ in. in shear per rivet.
(c) The total shearing strength of the section, the safe intensity of
shear being 3^ tons per sq. in.
(d) The moment of resistance of the section, the coefficient of strength
being 4^ tons per sq. in.
(e) The uniformly distributed load which the beam will safely carry.
42O THEORY OF STRUCTURES.
Ans. (a) .11878 tons per sq. in.
(b) .97 in. if rivets are tight, 1.12 in. if liable to flexure.
(<:) i09 T y^ tons disregarding effect of riveting,
tons having regard to riveting.
(d) 408 5in.-tons disregard ing effect of riveting, 3838.9157
tons having regard to riveting.
(e) 75f| tons disregarding effect of riveting, 71.09 tons
having regard to riveting.
82. A cast-iron channel-beam having a web 12 in. wide and two sides
7 in. deep, the metal being everywhere i in. thick, crosses a span of 14
ft. If the tensile intensity of stress is i ton per sq. in., what uniformly
distributed load will the beam carry (a) with the web at the bottom ; (b)
with the web at the top? Find (c) the maximum compressive intensity
of stress to which the metal is subjected, and (d} compare the maximum
and average intensities of shear. Also, (e] what should be the area of
a rectangular section to bear the same total shear ?
Ans. 7 = i io ; (d) f f| tons ; (ff) fff tons.
83. A beam of rectangular section and of a length equal to 20 times the
depth is supported at the ends in a horizontal position, and is subjected
to a thrust //"whose line of action coincides with the axis of the beam.
Show that the maximum intensity of stress at the middle point will be
doubled 1 by concentrating at that point a weight Fequal to one thirtieth
of//.
84. The line of action of the thrust in a compression member is at a
distance from the axis equal to -th of the least transverse dimension.
Show that the maximum intensity of stress is doubled if the section is
rectangular and r = 6, or if the section is circular and r = 8.
85. A straight wrought-iron bar is capable of sustaining as a strut a
weight Wi,and as a beam a weight w* at the middle point, the deflection
being small as compared with the transverse dimensions. If the bar has
simultaneously to sustain a weight w as a strut and a weight w' as a
beam, the weight being placed at the middle of the span, show that the
beam will not break if
w\
IV + W < Wi.
IVl
86. A metal beam is subjected to the action of a bending moment
steadily applied beyond the elastic limit. Assuming that the metal acts
as if it were perfectly plastic, i.e., so that the stress throughout a trans-
verse section is uniform, compare the moment of resistance to bending
of a section of the beam with the moment on the assumption that the
metal continued to fulfil the ordinary laws of elasticity, (a) the section,
being a rectangle ; (b) the section being a circle.
EXAMPLES. 421
87. A lattice-girder of 100 ft. span carries 80 tons uniformly distributed;
the girder is 10 ft. deep and the safe working stress is 4 tons per sq. in.
If the width of the flange must be 20 in. to carry the load exclusive of
the weight of the girder, what must be the width of the flange when the
weight of the girder is taken into account ?
88. A plate-girder of double-tee section and of 3o ft. span is 8 ft. deep
and carries a uniformly distributed load of 80 tons. If the width of the
flange must be 12 in. to carry the load exclusive of the weight of the gir-
der, what must the width be when this weight is taken into account ?
89. If the plane of bending does not coincide with the plane of sym-
metry of a beam, show that the neutral axis is parallel to a line joining the
centres of two circles into which the beam would be bent by two com-
ponent couples whose axes are the principal axes of inertia of the section,
each couple being supposed to act alone.
90. The flanges of a girder are of equal sectional area, and their joint
area is equal to that of the web. What must be the sectional area to resist
a bending moment of 300 in. -tons, the effective depth being 10 in. and
the limiting inch-stress 4 tons? Ans. 22^ sq. in.
91. The effective length and depth of a cast-iron girder which failed
under a load of 18 tons at the centre were 57 in. and 5^ in., respectively ;
the top flange was 2.33 in. by .31 in., the bottom flange 6.67 in. by .66 in.,
and the web was .266 in. thick. Assuming that the ordinary theory of
flexure held good, what were the maximum intensities of stress in the
flanges at the point of rupture ?
Ans. ft = 12.36 tons per sq. in. ; f c = 44.9 tons per sq. in.
92. A railway bridge is supported upon two main girders each of span
51 ft. 4 in. ; at the centre the depth is 6 ft. 6 in., the gross sectional area
of the top flange 27 sq. in., and of the bottom flange 28 sq. in. Assum-
ing the efficiency of the tension flange is reduced one-fifth by the rivet-
holes, find the maximum flange intensities of stress under a uniformly
distributed load of 43 tons. Also find the uniformly distributed rolling
load which will increase these intensities by two tons.
Ans. .786 ton per sq. in. in compression ; .9475 ton per sq. in. in
tension ; 55/ T tons to increase compression ; S9j 2 ^ ^ ton s to
increase tension.
93. A lattice-girder of 80 ft. span and 8 ft. deep is designed to carry a
dead load of 5o tons and a live load of 120 tons uniformly distributed ;
at the centre the net sectional area of the bottom flange is 45 sq. in.,
and the gross sectional area of the top flange 56^ sq. in. Find the po-
sition of the neutral axis and the maximum flange intensities of stress.
If the live load travels at 60 miles an hour, what will be the increased
pressure due to centrifugal force?
4 22 THEORY OF STRUCTURES.
Ans. 3.546 ft. from top; 1120 Ibs. per sq. in.; 8920.35 Ibs. per
594000
sq. ID. ; -pj Ibs.
94. Determine the thickness of the metal in a cast-iron beam of 12 ft.
span and 8 in. deep which has to carry a uniformly distributed load of
4000 Ibs., the section being (a) a hollow square ; (b) a circular annulus.
The coefficient of working strength = 3000 Ibs. per sq. in. Also find the
limiting safe span of the beam under its own weight.
Ans. Neglecting weight of beam, (a) .281 in.; (b) 477 in. Taking
weight of beam into account, (a) .307 in. ; (b) .534 in. Limiting
span = 41.3 ft. in (a) and = 35.7 ft. in (b).
95. Determine suitable dimensions for a cast-iron beam 20 in. deep,
at a section subjected to a bending moment of 1200 in.-tons ; the coeffi-
cients of strength per square inch being 2 tons for tension and 8 tons for
compression. Take thickness of web = T 9 ^ in.
Ans. Sectional area of tension flange = 36 sq. in.; of compression
flange = 2j sq. in.
96. The thickness of the web of an equal-flanged I-beam is a certain
fraction of the depth. Show that the greatest economy of material is
realized when the area of the web is equal to the joint area of the
flanges, and that the moment of resistance to bending is ^/Ad,f being
the coefficient of strength, A the total sectional area, and d the depth.
97. In a double-flanged cast-iron beam the thickness of the web is a
certain fraction of the depth, and the maximum tensile and compressive
intensities of stress are in the ratio of 2 to 5. Show that the greatest
economy of material is realized when the areas of the bottom flange, web,
and top flange are in the ratio of 25 to 20 to 4, and that the moment of
resistance to bending is \fAd, where/ = -^ maximum tensile intensity
of stress.
98. Apply the results in the preceding question to determine the di-
mensions of a cast-iron beam at a section whose moment of resistance is
800 in.-tons and whose depth is 18 in., taking 2 tons per square inch as
the maximum tensile intensity of stress.
Ans. ai = ^fi- sq. in.; A' = - 4 ^ sq. in.; a^= |^ sq. in.
99. Determine suitable dimensions for a cast-iron girder of 20 ft. span
and 24 in. deep, carrying a load of 30,000 Ibs. at the centre, the
coefficients of working strength in tension and compression being respec-
tively 2000 and 5000 Ibs. per square inch.
Ans. ai = ijf 1 sq. in. ; A' = &3^ sq. in. ; a* = ty sq. in.
100. A cast-iron girder of 25 ft. span has a bottom flange of 36 sq. in.
sectional area. Find the most economic arrangement of material for the
web and top flange which will enable the beam to carry a load of 18,900
Ibs. at IG ft. from one end.
EXAMPLES. 423
Am. Depth = 2oJ in.; area of web = 28.8 sq. in.; area of top
flange 5.76 sq. in.
101. A double-flanged cast-iron girder has a sectional area of 93 sq.
in. ; the web is i in. thick and 21 in. deep ; the moment of resistance of
the section is 100,950 ft.-lbs. ; the coefficients of strength are 2100 Ibs.
per square inch in tension and 5250 Ibs. in compression. Find the
position of the neutral axis and the areas of the two flanges.
102. Determine the moment of resistance to bending of a section
of a beam in which the top flange is composed of two 34o-mm. x 12-mm.
plates and one 34o-mm. x lo-mm. plate, and- the bottom flange of one
340- mm. x lo-mm. plate and one 34o-mm. x 8-mm. plate, the flanges
being riveted to a i.4-m. x 7-mm. web plate by means of four
loo-mm. x loo-rnm. x 8-mm. angle-irons. The coefficient of strength
= 6 k. per mm. 2 .
103. Compare the moments of resistance to bending of the section in
the preceding question and of a section in which three 4oo-mm. x 15-mm.
plates are substituted for the top flange, and one 4oo-mm. x i5-mm.
plate is substituted for the bottom flange.
104. Floor-beams 4.4m. between bearings and spaced 2.548 m. c. to c.
have a section composed of two equal flanges, each consisting of two
85-mm. x 85-mm. x i2-mm. angle-irons riveted to a 49o-mm. x 7-mm.
web. A weight of 1 50 k. (due to longitudinals) and a weight of 1 50 k. (due
to rails, etc.), i.e., 300 k. in all, are concentrated at the rail-crossings, and
the ties have also to carry a uniformly distributed load of 400 k. due to
weight of floor-beam, 4000 k. due to weight of platform, and 4000 k. per
square metre of platform due to proof-load. Find the moment of resist-
ance to bending and the maximum flange intensities of stress.
Ans. /= .000438584615.
105. The section of a beam is in the form of an isosceles triangle
with its base horizontal. Show that the moment of resistance to
bending of the strongest trapezoidal beam that can be cut from it is
very nearly -^^fbd^, b being the width of the base and d the depth of
the triangle.
106. Taking/f,/ c as the tensile and compressive intensities of stress,
find the moment of resistance to bending of a section consisting of a
2o/-in. x 7^-in. top flange, an 8o/-in. x io/-in. bottom flange, and a
trapezoidal web 4/ in. thick at the top, 8/ in. thick at the bottom, and
I20/ in. deep. Also compare the maximum and average intensities of
shear.
107. Each of the flanges of a girder is a 35o-mm. by lo-mm. plate and
is riveted to a i.8-m. by 8-mm. web by means of two loo-mm. by
loo-mm. by 12-mm. angle-irons. Determine the moment of resistance
to bending, the coefficient of strength being 6k. per square millimetre,
424 THEORY OF STRUCTURES.
(a) disregarding the weakening effect of riveting ; (b) assuming that the
flange-plates are riveted to the angles by 2o-mm. rivets.
Ans. (a) 108661.04 km. '
108. The cross-tie for a single-track bridge is 4.1 m. between bearings,
the gauge of the rails being 1.51 m. ; each of the flanges is composed of a
148-mm. by 8-mm. plate riveted to a 55o-mm. by 8-mm. web by means of
two 7o-mm. by /o-mm. by 9-mm. angle-irons; a load of 296 k. (weight
of rails, etc.) is concentrated at each rail-crossing. What uniformly
distributed load will the tie safely bear, the metal's coefficient of strength
being 6 k. per square millimetre ? The load actually distributed over the
tie is 19782 k. Find the maximum intensity of stress.
Ans. 24162 k. ; 4.94 k. per sq. mm.
109. Design a longitudinal of .45 m. depth which is to be supported
at intervals of 3.3 m. and to carry at its middle point a weight of 7000 k.,
the coefficient of strength being 5 k. per square millimeter.
Ans. I = 259.875, and the /of a section with two equal flanges,
each composed of two 7o-mm. by 7o-mm. by 9-mm. angle-irons
riveted to a 45o-mm. by 8-mm. web is 259.102455.
no. Find the moment of resistance of a section composed of two
equal flanges, each consisting of two 6oo-mm. x 7-mm. plates riveted
to a i2OO-mm. x 8-mm. web plate by means of two loo-mm. x loo-mm.
x i2-mm. angle-irons; two 7o-mm. x 7o-mm. x 9-mm. angles are also
riveted to the lower faces of the flanges, the ends of the horizontal arms
being 24 mm. from the outside edges of the flanges ; the total depth of
the section = 3.228 m., and the interval between the two web plates,
which is open, is 2 m. ; coefficient of strength = 6 k. per mm.' 2 .
Ans. /= .093929232444 and moment = 349179.3018 km.
in. A longitudinal 2.548 m. between bearings consists of two equal
flanges, each composed of two 7o-mm. x 7o-mm. x 9-mm. angle-irons
riveted to a 35o-mm. x 7-mm. web plate. Find the flange intensity of
stress under a maximum load of 7000 k. at the centre.
Ans. I = .000139284508; stress = 5.6 k. per mm. 2 .
1 1 2. A cross-tie resting upon supports at the ends and 2.26 m. between
bearings is composed of two equal flanges, consisting of two 7o-mm.
x 7o-mm.X9-mm. angle-irons riveted at the top to a 45o-mm. x7-mm.
web plate and at the bottom to a 3oo-mm. x 7-mm. web plate, the
interval between the web plates, which is open, being 2.55 m. ; the tie is
designed to carry a uniformly distributed load of 676 k. per lineal metre
of its length, and also a load of 11644.8 k. at each of the points distant
.375 m. from the bearing. Find the position of the neutral axis and the
maximum flange stresses.
Ans. 1.516 m. from top flange; 7= .023194564198; maximum
B. M. = 4815.8161 km. ; maximum tensile stress = .37 k. per
mm. 2 ; maximum compressive stress = .314 k. per mm. 2 .
EXAMPLES. 425
113. Find the maximum concentrated load on a cross-tie for a single
track due to a six-wheel locomotive, the wheels being 2.3 m. centre to
centre, the ties being 3.2 m. centre to centre, and the weight on each
wheel being 7000 k. Ans. 10937.5 k.
114. The floor-beams for a double-track bridge are 8.3m. between
bearings and are spaced 2.58 m. centre to centre. The distance, centre
to centre, between track-rails is 1.5 m., and between inside rails is 2 m. ;
the tie has equal flanges, each consisting of two 7o-mm. x 7o-mm. x 9-mm.
angle-irons riveted to a 6oo-mm. x 7-mm. web; the maximum live
load upon the tie is that due to a weight of 7000 k. upon each of the six
wheels of two locomotives, the wheels being 2.4 m. centre to centre. If
the coefficient of working strength is 5! k. per square millimetre, what
uniformly distributed load will the tie carry?
115. Determine the safe value of the moment of inertia (/) of a
cross-tie for a double-track bridge ; the length of the tie between bearings
being 7. 624 m., its depth .6 m.,the gauge of the rails 1.5 m., the distance
between inside rails 2 m. The uniformly distributed load upon a tie
consists of 850 k. per square metre due to platform, etc., and of 1800 k.
due to weight of tie ; the ties are 3.584 m. centre to centre ; the live load
is that due to a weight of 7000 k. upon each of the centre wheels of a six-
wheel locomotive and a weight of 6000 k. upon each of the front and rear
wheels, the wheels being 2.4 m. centre to centre; the safe coefficient of
strength = 6 k. per square millimetre.
116. The upper chord of a Howe truss is 24 in. wide x 12 in. deep
and is made up of four 12-in. x 6-in. timbers; the lower chord is 24 in.
wide x 16 in. deep and is made up of four i6-in. x 6-in. timbers ; the
distance between the inner faces of the chords is 24 ft. Find the mo-
ment of resistance to bending, taking 800 Ibs. per square inch as the co-
efficient of tensile strength, and neglecting the effect of the web.
Ans. Neutral axis is 137^ in. from bottom face of lower chord ;
moment = 87441616 in. -Ibs.
117. The cross-ties of a single-track bridge consist of two equal
flanges, each composed of two 7o-mm. x 70 mm. x 9 mm. angle-irons
riveted to a 65o-mm. x 7-mm. web; the ties are 4.1 m. long, and each
carries 19,146 k. (viz., 384 k.for ties, 2762 k.for platform, and 16,000 k.for
proof load} uniformly distributed and 635 k. (due to longitudinals, rails,
etc} concentrated at each rail-crossing, i.e., at 755 mm. from the middle
point. Assuming that the cross-ties are merely supported at the ends,
find the maximum intensity of stress.
Ans. 5.7724 k. per mm. 2 ; = .0018423.
N.B. The fixture of the ends approximately doubles the strength.
426 THEORY OF STRUCTURES.
1 1 8. The longitudinals of the bridge in the last question consist of
two pairs of 7o-mm. x 7o-mm. x g-mm. angle-irons riveted to a 4m.
x 7 mm. web ; the cross-ties are 3.2 m. centre to centre. Determine the
maximum intensity of stress due to a load of 7000 k. concentrated on the
longitudinal half-way between the cross-ties, assuming that it is an inde-
pendent girder. What would the stress be if the ties were 3 m. centre to
centre ? /
Ans. = .00095458; 5.866k. per mm. 2 ; 5. 4994k. per mm. 2
119. The section for the Estressol bridge cross-ties is the same as that
for the Grande Baise (Ex. 117) bridge ties ; the load at each rail-crossing
is 335 k., and the uniformly distributed load is 18,062 k. Find the max-
imum intensity of stress in the flanges, assuming that the ties are
merely supported at the ends. Ans. 5.26 k. per mm. 2
120. In a rolled joist the sum of the two flange areas and the web
area is a constant quantity. Find the proportion between them which
will give a joist of maximum strength, the thickness of the web being
fixed by practical considerations. Ans. Flange area = f web area.
121. An aqueduct for a span of 26 ft. consists of a cast-iron channel-
beam 30 in. wide and 20 in. deep. Find the thickness of the metal so
that the water may safely rise to the top of the channel, the safe coeffi-
cient of strength being i ton per square inch. Find the safe limiting
span of the channel under its own weight.
122. A rolled beam with equal flanges and a web whose section is
equal to the joint section of the flanges has a span of 24 ft. and carries a
weight of 8 tons at the centre. If the stiffness is .001 and if the coefficient
of strength per square inch is 5 tons, find the depth of the beam and the
web and flange sectional areas.
123. A wrought-iron beam of I-section, 20 ft. between supports,
carries a uniformly distributed load of 4000 Ibs. and deflects .1 in.; the
effective depth =8 in.; ."=30,000,000 Ibs.; web area = joint area of
the equal flanges. Find the total sectional area. Also find the width of
a rectangular section 8 in. deep which might be substituted for the
above. Ans. /= 288 ; area = 27 sq. in.; width = 6f in.
124. A cast-iron beam of an inverted T-section has a uniform depth
of 20 in. and is 22 ft. between supports; the flange is 12 in. wide and 1.2
in. thick ; the web is i in. thick ; the load upon the beam is 4500 Ibs.
per lineal foot; = 17,000,000 Ibs. Find the deflection at the centre,
the moment of resistance to bending, the maximum tensile and com-
pressive intensities of stress, and the position of the neutral axis. Why
is the flange placed downwards ?
125. Find the sectional area of a wrought-iron beam of T-section
which may be substituted for the cast-iron beam in the preceding ques-
tion, the depth being the same and the coefficients of strength per
EXAMPLES. 427
square inch being 3 tons in compression and 5 tons in tension. Why
should the flange be uppermost ? What should the total sectional area
be if the flange and web are of equal area ?
126. A cast-iron girder 139 in. between supports and 10 in. deep had
a top flange 2^ in. x in., a bottom flange 10 in. x i J in., and a web f in.
thick. The girder failed under loads of 17^ tons placed at the two
points distant 3! ft. from each support. What were the central flange
stresses at the moment of rupture ? What was the central deflection
when the load at each point was 7! tons ? (E = 18,000,000 Ibs. ; weight
of girder = 3368 Ibs.; ton = 2240 Ibs.)
Ans. 182251.9 Ibs. = total flange stress; unit flange stresses
= 14,580, and 41,657 Ibs. persq. in.; deflection = .35".
127. A cylindrical beam of 2 in. diameter, 60 in. in length, and weigh-
ing \ Ib. per cubic inch, deflects T % in. under a weight of 3000 Ibs. at the
centre. Find E. Ans. E= 21,645,511 Ibs.
CHAPTER VII.
ON THE TRANSVERSE STRENGTH OF BEAMS. Continue*.
I. General Equations. The girder OA of length / carries
a load of which the intensity varies continuously and is / at a
point K distant x from O.
M+dM
K V "*
V
FIG. 310.
Consider the conditions of equilibrium of a slice of the
girder bounded by the vertical planes KL, K 'L ', of which the
abscissae are x, x + dx, respectively.
The load between these planes may, without sensible error,
be supposed to be uniformly distributed, and its resultant pdx
therefore acts along the centre line VV .
The forces acting upon the slice at the plane KL are equiv-
alent to an upward shearing force S, and a right-handed couple
of which the moment is M, while the forces acting upon the
slice at the plane K' L' are equivalent to a downward shearing
force 5 + dS t and a left-handed couple of which the moment
Since there is to be equilibrium,
S(S~\-dS)pdx= the algebraic sum of the vertical forces=o.
428
GENERAL EQUATIONS. 429
And, M-(M+JM) + S + (S+JS)^= the alge-
braic sum of the moments of the forces with respect to V or
The term - is disregarded, being indefinitely small as
compared with the remaining terms.
Equations (a) and (b) are the general equations applicable
to girders carrying loads of which the intensity is constant or
varies continuously. Their integration is easy, and introduces
two arbitrary constants which are to be determined in each
particular case.
Cor. i. From equations (a) and (3),
_dS_
* ~~~ ~ P '
dx* dx
Let p = wf(x), w being a constant, and f(x) some function
of x. Then
dM
and
M=c,
c l and , being the constants of integration, and o and x the
limits.
EXAMPLE. Let the girder rest upon two supports and
carry a uniformly distributed load of intensity w^ Then
and
M = c^ -[- cjc w l .
43 THEORY OF STRUCTURES.
But M is zero when x = o and also when x = /. Hence
Therefore,
and
a = o and c t = .
, l
M x -x\
2 2
-j W.X.
dx 2
. 2. The bending moment is a maximum at the point
/V 7l/f
defined by -7 = o = 5, i.e., at a point at which the shearing
force vanishes.
In the preceding example, the position of the maximum
bending moment is given by S o '= w^, or x = ,
. wj / w. r w.r
and its corresponding value is = ~^~-
22 24 o
W /
The shearing force is greatest and equal to when;r=O.
Cor. 3. Suppose that the load, instead of varying contin-
Nr Nr-fi uous ^y consists of a number of finite weights
at isolated points.
By reason of the discontinuity of the load-
FIG. 3 n. m g^ j-^e general equations can only be inte-
grated between consecutive points.
Let N rj N r+l , be any two such points, of abscissae x r ', x r+l ,
respectively.
Between these points equations (a) and (),become
dS dM
/. 5" = a constant = S, , suppose, between N r and N r+l .
GENERAL EQUATIONS. 431
Hence, -j- = 5 r , and ^f=S r ^+ , between N r and
_,_, , being a constant of integration.
Let M = .Af r when ;tr = ^r r . Then
c = M r SrX r , and J/ = S r (;r - # r
Also, if M M r+l when x = x r +^ ,
The terminal conditions will give additional equations, by
means of which the solution may be completed.
EXAMPLE. The girder OA, of length /, rests upon two
supports at 0, A, and carries weights P lt /*,, at points B, C,
R ?
Vl
FIG. 312.
dividing the girder into three segments, OB, BC, CA, of which
the lengths are r, s, t, respectively.
The reaction R, at O =
The reaction R, at A = P * r + P j( r + s \
Between O and B, S is constant = S r suppose, = R t .
there being no constant of integration, as M = o when x = o.
Also, when x = r, J/ *= .S r r.
43 2 THEORY OF STRUCTURES.
Between B and C, S is constant = 5 f suppose, = R t P v
c f being the constant of integration.
But M = S r r when x = r.
and M = S> + (S r - S,)r
Also, when x = r -{- s, M = S s s -\- S r r.
Between CandA, S is constant = S t suppose, =
and hence
c" being the constant of integration.
But M = SsS -\- S r r when x = r + s.
.-. c" = Sj + S r r - Sjr + s),
and
M= Ss + 5> + S r r - S t (r -f s).
Hence, at A t o = S t t + 5> + 5>.
6^r. 4. The equation - - = 5 indicates that the shearing
force at a vertical section of a girder is the increment of the
bending moment at that section per unit of length, and is an
important relation in calculating the number of rivets required
for flange and web connections.
2. On the Interpretation of the General Equations.
The bending moment M at any transverse section of a girder
El
may be obtained from the equation M= ,R being the
.A.
DEFLECTION OF BEAMS. 433
radius of curvature of the neutral axis at the section under
consideration. \
Let OA, in Figs. 313 and 314, represent a portion of the
neutral axis of a bent girder.
--^ Jy
FIG. 313. FIG. 314.
Take O as the origin, the horizontal line OX as the axis of
x, and the line O Y drawn vertically downwards as the axis of y.
Let x, y be the co-ordinates of any point P in the neutral
axis.
If R is the radius of curvature at P, then
dx* a d&
"
the sign being -f- or according as the girder is bent as in
Fig. 313 or as in Fig. 314, and being the angle between the
tangent at P and OX.
dy
Now, jk is the tangent of the angle which the tangent line
at P to the neutral axis makes with OX, and the angle is always
very small. Thus, -j- is also very small, and squares and
dy
higher powers of -7 may be disregarded without serious error.
Hence,
I ~ ,
= -jTT approximately,
and the bending-moment equation becomes
M = EI-jj.
434 THEORY OF STRUCTURES.
The integration of this equation introduces two arbitrary
constants, of which the values are to be determined from given
conditions. At the point or points of support, for example,
the neutral axis may be horizontal or may slope at a given
angle.
Let 6 be the slope at P. Since 6 is generally very small,
6 tan 6, approximately,
and hence
or
dB M
From this last equation
A" 0=
and the change of slope between any given limits is represented
by the corresponding area of the bending-moment curve.
dy r
Also, since = 0, y = J 6dx,
d>&
and the deflection is measured by the area of a curve repre-
senting the slope at each point.
Again, by Art. i,
Comparing eqs. (A) and (B), it will be observed that
M
y, Oj and -^-,, i.e., the deflection, slope, and bending moment,
are connected with one another in precisely the same manner
NEUTRAL AXIS OF A LOADED BEAM. 435
as M, S, and p, i.e., the bending moment, shearing force, and
load. Thus, the mutual relations between curves drawn to
represent the deflection, slope, and bending moment must be the
same, mutatis mutandis, as those between the curves of bending
moment, shearing force, and load.
For example, divide the effective bending-moment area into
a number of elementary areas by drawing vertical lines at con-
venient distances apart, and suppose these elementary areas to
represent weights. Two reciprocal figures connecting y, 0, and
M may now be drawn exactly as described in Chap. I, and it
at once follows that
(a) Any two sides of the funicular polygon, or, in the limit
(when the widths of the elementary areas are indefinitely
diminished), any two tangents to the funicular or deflection
curve, meet in a point which is vertically below the centre of
gravity of the corresponding effective moment area.
(b) The segments iH, nH into which the line of weights is
divided by drawing OH parallel to the closing line CD, give the
slopes (= ~2Mdx) at the supports.
N.B. In the case of a semi-girder, the last side of a
polygon is the closing line, and in gives the total change of
slope.
(c) If the polar distance is made equal to El, the intercept
between the closing line and the funicular or deflection curve
measures the deflection.
3. Examples of the Form assumed by the Neutral
Axis of a Loaded Beam.
EXAMPLE I. A semi-girder fixed at one end O so that the
neutral axis at that point is horizontal carries a weight W at
the other end A. At any point (x, y) of the neutral axis
x). . . (A)
Integrating,
l being a constant of integration. But FlG - 3*s-
the girder is fixed at O, so that the inclination of the neutral
THEORY OF STRUCTURES.
axis to the horizon at this point is zero, and thus, when x = o,
dy
-j- is o, and therefore c t = o.
Hence,
% : ^4-D ..... . . (B
Integrating,
2 being a constant of integration. But y = o when x = o,
and therefore a = o.
Hence,
dy
Equation (B) gives the value of , i.e., the slope, at any
point of which the abscissa is x.
Equation (C) defines the curve assumed by the neutral axis,
and gives the value of y, i.e., the deflection, corresponding to
any abscissa x.
Let <*! be the slope, and d l the deflection at A.
From (B),
w
and from (C),
i w r
d , ^~ T-
NEUTRAL AXIS OF A LOADED BEAM.
437
Ex. 2. A semi-girder fixed at one end O carries a uniformly
distributed load of intensity w.
At any point P (x, y) of the
neutral axis,
. . (A) Y
FIG. 316.
Integrating,
='* -<*+$+>..
l being a constant of integration.
dy
But -j- = o when x = o, and therefore ^ = O. Hence,
dx
(B)
Integrating,
a being a constant of integration.
But j = o when x = o, and therefore 2 = O. Hence,
2 2
(Q
Let a be the slope and d^ the deflection at A. Hence, from
(B),
and from (C),
i wr
43$ THEORY OF STRUCTURES.
Ex. 3. A semi-girder fixed at one end carries a uniformly
distributed load of intensity w, and also a single weight Wat the
free end. This is merely a combination of Examples I and 2,
and the resulting equations are :
x) + (l-xf (A)
Also, if A is the slope and D the deflection at the free end,
from (B),
i iwr wi\
tan A = -g^ 1- -^- ] = tan or, -j- tan <* a ;
and from (C),
. The slope (a) and deflection (</) of an arbitrarily
loaded semi-girder may be determined in the manner de-
scribed in Art. 2.
Let F be the area of the bending-moment curve. Its
centre of gravity is at the same horizontal distance ^ from the
vertical through A as the point T in which the tangent at A
intersects OX.
.'. --T= <* = angle A TX = .
Jj* J. 3v
In Ex. 3, e.g.,
F=m t + L^
2 32
and
NEUTRAL AXIS OF A LOADED BEAM. 439
Note. If the semi-girder in the three preceding examples
is only partially fixed at O, so that the neutral axis, instead of
being horizontal at the support, slopes at an angle 0, then
dy
when x = o, -y- = tan 6, and the constant of integration,
, , is also El tan 0. Thus, the left-hand side of eqs. (B) and
(C), respectively, become
\alc ~
EI ~ tan B and
Ex. 4. The girder OA rests upon two supports at <9,
A, and carries a weight W at the B
centre. r^- \V i ~ *
The neutral axis is evidently p J "
symmetrical with respect to the |y w
middle point C, and at any point FlG> 3I7 '
P (x, y) between O and C,
EI <fy = w_ x (A}
dx* ~' 2 ^ '
Integrating,
~ EI ~dlc"~'~^* i + c ^
^ being a constant of integration.
But the tangent to the neutral axis at C must be horizontal,
/ dy Wr
so that when x -, -y- o, and therefore c. = ^- .
2 dx 16
Hence,
~ EI ~d~x=~^ X * l6~ (B)
Integrating,
Ely = x* x -4- c n .
12 16
c 9 being a constant of integration.
44 THEORY OF STRUCTURES.
But y = O when x = o, and therefore a = o. Hence
(Q
Cor. Let a 1 be the slope at O, and d l the deflection at the
centre. Then,
i wr i wr
from (B), tan a 1 = -> -=rjr ; and from (C), ^ = -5 -py .
ID &i 48 /j/
a. B A EX * 5 ' The irder OA rests
-^Jj, | ^' x upon supports at <9, ^4, and carries
P~~~C" a uniformly distributed load of in-
tensity w.
At any point P (x, y) of the
FlG - 318. neutral axis,
Integrating,
P T dy __wl a wx*
~fcc''^~4~ X ~~6~ T~ C I>
\ being a constant of integration.
But -j- = o when x = -, and therefore c. .
dx 24
Hence,
dx " 4 6 24 '
Integrating,
wl , wx* wl*
Ely = x" x 4- c mt
12 24 24
:, being a constant of integration.
But 7 = when x = o, and therefore c^ = o.
Hence
7? = ^r 3 - ^- x. . (C)
12 24 24 V '
NEUTRAL AXIS OF A LOADED BEAM. 441
Let or a be the slope at O, and d^ the deflection at the centre.
Then,
from (B), tan a, = ^ ; and from (C), < = ^ ~.
Ex. 6. A girder rests upon two supports and carries a uni-
formly distributed load of intensity w, together with a single
weight W at the centre. This is merely a combination of
Examples 4 and 5, and the resulting equations are :
dx ' ~ 4 16 4 6
and
(B)
12 ID 12 24 24
Also, if A is the slope at the origin, and D the central
deflection, we have, from (B),
i fwr wr\
tan A = j(-rf- + J = tan ,+ tan a ;
and from (C),
*= +- -<+<
The slope and deflection of an arbitrarily loaded
girder resting upon two supports may be determined in the
manner described in Art. 2.
Let C be the lowest point of the deflection curve. The
tangents at C and O will intersect in a point T which is ver-
tically below the centre of gravity of the bending-moment
area corresponding to OC.
Denote this area by Fand the horizontal distance of centre
44 2 THEORY OF STRUCTURES.
of gravity from OY by x. Let OL be the angle between OT
and CT produced. Then
F d
d being the maximum deflection.
In Ex. 6, e.g., the girder being symmetrically loaded,
i wi i 2 wr i wr 2 i wr 5 /
F= sr- and Fx = ^ 4- ~- EId
242^382 1632 24 82
Ex. 7. Suppose that the end O of the girder in Ex. 5
R R ^ is fixed. The fixture introduces
\j(^\\ a left-handed couple at ; let its
K---.,_ a l ~~~^k * moment be M,.
P Let the reactions at O and A
y be R t , R^ , respectively.
At any point P (x, y] of the
neutral axis,
But M, i.e., 7-r4, is zero when x = /.
-^ ...... (2)
Integrating eq. (i),
There is no constant of integration, as -,- = o when x = o.
dx
Integrating eq. (3),
- Ely = R?*^ - M?- . (4)
1 6 24 J 2
There is no constant of integration, as x and y vanish to-
gether.
NEUTRAL AXIS OF A LOADED BEAM.
But y also vanishes when x = /, so that
443
Hence, by eqs. (2) and (5),
and so tf, = //. . (6)
Thus, the b ending-moment, slope, and deflection equations
are, respectively,
(*>
=<"* --
. I. The bending moment is nil at points given by
5 w , wl*
^f= o = -golx x* -- g-,
F
i.e., when x = or /. Take OF= .
4 4
Since -^-7 = o, .F is a point of inflexion.
If the girder is cut through at this point,
and a hinge introduced sufficiently
strong to transmit the shear (= %wl), FIG. 320.
the stability of the girder will not be impaired.
Hence, the girder may be considered as made up of two
independent portions, viz. :
(a) A cantilever OF-oi length , carrying a uniformly dis-
tributed load of intensity w, together with a weight wl at F.
444 THEORY OF STRUCTURES.
The maximum bending moment on OF is at O, and is
3 f / , wl I wr
= 8 W/ 4 + 78 =: l~'
(b) A girder FA of length , carrying a uniformly distrib-
uted load of intensity w.
The maximum, bending moment on FA is at the middle
This result may also be obtained from eq. (7) by putting
-- o. Whence
dx
O = %wl wx, or x = f/,
.and therefore
The shearing force and bending moment at different points
of the girder may be represented graphically as follows :
The shearing force at any point of which the abscissa is x is
S = f wl wx.
Take OB and A C, respectively equal or proportional to wl
and fw/; join BC. The line BC cuts OA in /?, where OP = f/.
The shearing force at any point is represented by the ordinate
between that point and the line BC.
The bending moment at the point (x, y) is
Take OG, DE, and OF, respectively equal or proportional
wr g I
to -5, --^w/ 2 , and -. The bending moment at any point is
o I2o 4
represented by the ordinate between that point and the parab-
ola passing through G, F, and A, having its vertex at E and
its axis vertical.
NEUTRAL AXIS OF A LOADED BEAM. 445
dy
Cor. 2. The deflection is a maximum when = o, i.e., when
or at the point given by x = -g05 1/33).
Substituting this value of x in eq. (9), the corresponding
value of jj/ may be obtained.
Ex. 8. If both ends of the girder in eq. (7) are fixed, the
wl
reaction at each support is evidently , and the equation of
moments becomes
T^jd^y _ wl wx*
Integrating,
dy wl w
El-f- = x* ** M.x (2)
dx \ 6
dy
No constant of integration is required, as -r- = o when x = o.
-j- is also zero when x = I (also when x = ).
dx V 21
and hence
^' = 1? (3)
Integrating eq. (2),
12 24 24
There is no constant of integration, as x and y vanish together.
The central deflection I i.e., when # = )=* ~^~~cj'
44-6 THEORY OF STRUCTURES.
If the load, instead of being uniformly distributed, is a
weight ^concentrated at the centre, then, for one half oi the
girder,
Integrating,
dv W
/-/- = x* - M,x ...... (6)
dx 4
dy
There is no constant of integration, as -j- = o when x = o.
dy I
-j- is also evidently zero when x = , and hence
W I Wl
= _/>_^-, or M$~-g-. V . (7)
Integrating eq. (6),
W x* W Wl
-Efy=x*-M l - = x*-~x\ ... (8)
12 '2 12 l6
There is no constant of integration, as x and y vanish together.
I Wl
The central deflection = --- p-=-.
192 El
4. Supports not in same Horizontal Plane. In the
preceding examples it has been assumed that the ends of the
girder are in the same horizontal plane. Suppose that one end,
e.g., A, falls below O by an amount jj>, , y l being small as com-
pared with /.
The abscissae of points in the neutral axis are not sensibly
changed, but the conditions of integration are altered. Con-
sider Ex. 4.
Between O and C,
W
NEUTRAL AXIS OF A LOADED BEAM. 447
Integrating,
c l being a constant of integration.
Integrating again,
W
- Ely = *> + c>* ...... (3)
There is no constant due to the last integration, as x and y
vanish together.
Between C and A,
W f A W
"*-
Integrating twice,
-EI d j- x = -~(l-xf + c,, ...... (5)
and
W
-Efy=(l-*r + cs + c.. ..... (6)
c^ , <:, being constants of integration.
The tangent at C is no longer horizontal, but makes a defi-
dy
nite angle 6 with the horizon, so that -j- is now tan 6 when
x . Also, the values of -=- and y at C, viz., tan and d, as
given by eqs. (2) and (3), must be identical with those given by
eqs. (5) and (6), while the value y, at A, as given by eq. (6)
when x = /, is equal to y l . Therefore
W W
? + ^ = - EI tan = - ^t' + c,,
r? + Ci = _ EId = p + C + c
90 '2 96 3 2
and
44^ THEORY OF STRUCTURES.
Hence,
v W v W W
'
fully defining both halves of the neutral axis.
dy
Again, in Ex. 6 it is no longer true that -^- = o when
x = , but the conditions of integration are y = o when x = o,
dy
and y = y l when x = /. These, together with -y- = o when
dx
x = o, are also the conditions in Ex. 7. Other cases may be
similarly treated.
5. To Discuss the Form assumed by the Neutral Axis
of a Girder OA which rests upon Supports at O and A,
and carries a Weight JP at a Point B, distant r from O.
Let OBA be the neutral axis of the deflected girder.
The reactions at O and A are
c . , ,
x / - r
P -j and P-j, respectively.
1 / /
Let BC, the deflection at C t
Let a be the slope of the neutral axis at B.
The portions OB, BA must be treated separately, as the
weight at B causes discontinuity in the equation of moments.
First, at any point (x, y) of OB,
Integrating,
c l being a constant of integration.
NEUTRAL AXIS OF A LOADED BEAM. 449
dy
But -r- = tan a when x = r, and therefore
/ rr*
-El tan a = P j + *, .
Hence,
Integrating,
Ti-T-'v- (3)
There is no constant of integration, as x and y vanish to-
gether.
Also, y = d when x = r.
- r tan a) = - P--. ... (4)
In the same manner, if A is taken as the origin, and AB
treated as above, equations similar to (i), (2), (3), and (4)
will be obtained, and may be at once written down by sub-
stituting in these equations n a for a, P T - for P^^-, lr
I I
for r, and r for / r.
Thus, the equation corresponding to (4) is
- El { d - (I - r) tan (* - a) } = - P '- iLU^ u ( 5)
o
Subtracting (5) from (4),
p
na==-r-r-2r); .... (6)
and from (4),
450 THEORY OF STRUCTURES.
Thus, eqs. (2) and (3) become
and
-Ely = ?'-*>-?(l-r)(2l-r)*,. . (9)
the latter being the equation to the portion OB of the neutral
axis, and the former giving its slope at any point.
Next, at any point (x,y) of BA,
z P j-x - P(x - r) (10)
Integrating,
c a being a constant of integration.
dy
But -j- = tan a when x = r.
dx
pl r ? + ' = ~ EI tan " = ~ f 7 (/ ~ r)(/ ~ 2r
and
Pl-r , , N
c 9 =---r(2t-r).
Hence,
- / = T-r 1 *' - ? ( ^ - r)i - S (/ - r)(2/ - r) -
Integrating,
- Ely = g 1 -^^' ~ g(* - ^ - /(' -
4 being a constant of integration.
But y d when x = r.
NEUTRAL AXIS OF A LOADED BEAM. 45 1
and 4 = o. Hence,
- Ely = > - (* - r)' - Pr (l-r)(2/-r)x, (12)
which is the equation to the portion BA of the neutral axis,
eq. (11) giving its slope at any point.
In the figure r < , and the maximum deflection of the
girder will evidently lie between B and A, at a point given by
dy
putting =o in eq. (12), which easily reduces to
and therefore
X = I
is the abscissa of the most deflected point. The corresponding
deflection is found by substituting this value of x in eq. (12).
If r > , the maximum deflection lies between O and B, at
dy
a point determined by putting -^- = o in eq. (8), which then
easily reduces to
from which
fr(2l - r)
"V" .<
452 THEORY OF STRUCTURES.
Substituting this value of x in eq. (9),
the maximum deflection = gj j I ) .
EXAMPLE. P = 15,000 Ibs., / = 100 ft., r = go ft.
The distance of most deflected point from O
/go X 1 10 ,
= y - = 57-44 ft.,
and the maximum deflection
_ 15000 10 too X iioy 500000
~ W X I5o V 3 / = ~~^T~ (33) *
6. To Discuss the Form of the Neutral Axis of a Girder
OA which rests upon Supports at O and A and carries
several Weights JP, , J* a , 1% , . . . , at points i, 2, 3, . . . , of
which the Distances from O are r r a , r, , . . . , respectively.
A-X
FIG. 322.
It may be assumed that the total effect of all the weights
is the sum of the effects of the separate weights, and thus each
may be treated independently, as in the preceding article.
Let a l , ar a , or 3 , . . . be the slopes at the points i, 2, 3, . . .
of the neutral axis.
Considering P t , the equation to Oi is
- Ely = '-V - '(/ -
NEUTRAL AXIS OF A GIRDER OA. 453
and to lA,
- Ely = ^V - (* - r,)' - T -j(l- r^(2l- r>.
Considering P, , the equation to O2 is
- Ely = ~V - ^/ -
and to 2A,
l
-6 T ~ - - / -
and so on for P z , P 4 , etc.
The total deflection F at any point (x, F) is the sum of the
deflections due to the several loads.
Take, e.g., a point between 3 and 4, and let d l , d^ , d^ , . . .
be the deflections of this point, due to P lt P 9 , P a , . . . , respec-
tively. Then
-Eld, =
-EU =
and so on. Hence,
-
(A)
454 THEORY OF STRUCTURES.
Again, the position of the most deflected point is found by
making -7- = o in the equation to that portion of the neutral
axis between two of the weights in which the said point lies.
The result is a quadratic equation, and the value of x derived
therefrom may be substituted in eq. (A), which then gives the
maximum deflection.
EXAMPLE. A girder of 100 ft. span supports two weights
of 20,000 Ibs. and 30,000 Ibs. at points distant respectively 20 ft.
and 60 ft. from one end.
The most deflected point must evidently lie between the
two weights, and the equation to the corresponding portion of
the neutral axis is
x* 2OOOO
EIY = ^(20000 X 80 + 30000 X 40) g (* 2 ) 3
- -^-(20000 X 20 X 80 X 1 80 + 30000 X 60 X 40 X 140)
14000 , i oooo.
= - x - (x 20) 26400000^.
Kis a maximum when
dY
- o 14000^ ioooo(^ 20) 26400000,
or
X* + lOO^r -7600 = O,
or
x = 50.497 ft.
Remark. Instead of assuming -jr = ^cEI-j-^ , it would be
77 T JA
more accurate to take -jz- = El cos 6 (Art. 2), and the
first integration would make the left-hand side of the slope
equation El sin instead of El tan 6.
MOMENT OF INERTIA VARIABLE.
455
7. Moment of Inertia variable. In the preceding exam-
ples the moment of inertia / has been assumed to be constant.
From the general equations,
-->. j 2 - ,
dx c
c being proportional to the depth of the girder at a transverse
section distant x from the origin.
Hence, for beams of uniform strength, the value of c in
terms of x may be substituted in the last equation, which may
then be integrated.
Again, let Fig. 323 represent a cantilever of length /, spe-
cific weight w, circular section,
and with a parabolic profile, the
vertex of the parabola being at A.
Let 2b be the depth of the
cantilever at the fixed end.
Let the cantilever also carry a
uniformly distributed load of in-
tensity/.
Consider a transverse section of radius z at a distance x
from the fixed end.
Let x, y be the co-ordinates of the neutral axis at the same
section. Then
But *' = -/_
FIG. 323.
or
Integrating,
4~ r
wnV
~6~7~
-7 + *T- - (0
45 6
THEORY OF STRUCTURES.
dy
There is no constant of integration, as - = o when x = o.
Integrating again,
px*
There is no constant of integration, as x and jj/ vanish to-
gether. Thus, equation (i) gives the slope at any point, and
equation (2) defines the neutral axis.
The slope at the free end (* = /) = -m +
The deflection
~&
8. Springs Fixed at One End and Loaded at the other
with a Weight W.
Data. Length = /; breadth = b, and depth = d at fixed
support ; V= volume of spring; f= maximum coefficient of
strength ; A = maximum deflection.
CASE a. Simple rectangular spring.
By Ex. i, Art. 39,
_
-"'
since
Wl _M _2f _ 1207
'
Also,
y 2//
61 '
Hence,
The work done = =
(3)
SPXJNGS.
457
CASE b. Spring of constant depth but triangular in plan.
Let b x be the breadth at a distance x from the fixed end.
Then
b I - x
and / at the same point
l-X
12
12
bd\
FIG. 325.
1207
EbcT'
Integrating twice,
and
Also,
dy _
dx~
12WI
(A\
6WI ,
V
y
A
E$3r* '
6wr /r
T/T7 A
Ebd* "Ed
bd*ffP f*bdl f*V
(5)
Hence,
/. V
The work done =
61 Ed 6E ' $'
r
WA f*V
(6)
N.B. The results I to 6 are the same if the springs are
compound ; i.e., if the rectangular spring is composed of n
simple rectangular springs laid one above the other, and if the
triangular spring is composed of n triangular springs laid one
above the other.
458 THEORY OF STRUCTURES.
CASE c. Spring of constant width but parabolic in elevation.
Let d x be the depth at a distance x from the fixed end.
Then
dl - /
and /at the same point =
FIG. 326.
bd: wti-*\i
12 ' 12 I / /'
d *y W (J \ l2Wl * f , x
,.^ r = _(/_*)=__.(/_,)
Integrating twice,
^ = T ^ ^ * (/ ~ * }l ~ 2/t(/
and hence
*KJL. 4// a
E bd*~ ->K<1 v)
Also,
The work done = =-2
2 ~6 '
9. Girder Encastre at the Ends. The girder BCDEFG
rests upon supports at the ends, is held in position by blocks
forced between the ends and the abutments, and carries a uni-
formly distributed load of intensity w.
GIRDER ENCASTRE AT THE ENDS.
459
It is required to determine the pressure that must be devel-
oped between the blocks and the girder so that the straight
portion between vertical sections at points O and A of the
W.I
FIG. 327.
neutral axis may be in the same condition as if the girder were
fixed at these sections.
Let / be the length of OA.
Let R be the reaction at the surface BC> and r its distance
from O.
Let /fbe the reaction between the block and the end CD,
and h its distance from O.
Let P be the weight of the segment on the left of the ver-
tical section O, and / its distance from O.
Then for the equilibrium of the segment on the left of the
section at O,
2 * 12
/? P
.\K-:*>*~ 2 ,
and
I wl\ wl*
(P \r-Pp--
\ 21 ' 12
the required pressure.
Again, take O as the origin, OA as the axis of x, and a
vertical through O as the axis of _y.
At any point (x, ^/) of the neutral axis,
_ E = x _ _ . ( s ee Ex. 8.)
dx* 2 2 12 ^
460 THEORY OF STRUCTURES.
10. On the Work done in bending a Beam. Let
A'B'C'D' be an originally rectangular ele-
\q' ment of a beam strained under the action of
jfg. external forces.
Let the surfaces AD', B'C' meet in G>;
O is the centre of curvature of the arc P Q of
the neutral axis.
Let OP = R = OQ f .
Let the length of the arc P Q' = dx.
Consider any elementary fibre/'/, of length
dx'y of sectional area a, and distant y from the
neutral axis.
Let t be the stress in p'q' .
The work done in stretching/'/
But --- and
-jr--4
dx 9 -dx y
The work done in stretching/'/ = -
and the work done in deforming the prism A'B'C'D'
iE
Hence, the total work between two sections of abscissae
-^a>
C'-'iEI , El r'"dx
= .T.*- I?"
I J/
But - \ therefore the work between the given limits
TRANSVERSE VIBRATIONS. 461
This expression is necessarily equal to the work of the ex-
ternal forces between the same limits, and is also the semi vis-
viva acquired by the beam in changing from its natural state
of equilibrium.
Cor. If the proof load P is concentrated at one point of a
p
beam, and if d is the proof deflection, the resilience = d.
If a proof load of intensity w is uniformly distributed over
the beam, and if y is the deflection at any point, the resili-
ence = / wydx, the integration extending throughout the
whole length of the beam.
The case of the single weight, however, is the most useful
in practice.
ii. On the Transverse Vibrations of a Beam resting
upon Two Supports in the same Horizontal Plane.
It is assumed
(a) That the beam is homogeneous and of uniform sectional
area.
(b) That the axis (neutral) remains unaltered in length.
(c) That the vibrations are small.
(d) That the particles of the beam vibrate in the vertical
planes in which they are primarily situated. In reality, these
particles have a slight angular motion about the horizontal axis
through the centre of gravity of the section, but for the sake
of simplicity the effect of this motion is disregarded.
Jif M44M
/I A
dx
A-X
Y
Ci Y C'
i S+dS
w dx
FIG. 329.
Let OA be the beam.
Take O as the origin, the neutral line OA as the axis of x,
and the vertical O Fas the axis of y.
Consider an element of the beam, bounded by the vertical
4^2 THEORY OF STRUCTURES.
planes BC, B ' C ', of which the abscissae are x and x -f- dx,
respectively.
Let w be the intensity of the load per unit of length ; hence
wdx is the load upon the given element, and acts vertically
through its centre.
Let 5 be the shearing force at B ; 5 -f- dS the shearing
force at B'.
Let M be the bending moment at B] M -\- dM the bend-
ing moment at B'.
iv d* v
Also, the resistance of the element to acceleration = -.
g at
Hence, at any time t,
w d y
dx -j-f + 5 (S -f- dS) wdx = o,
or
_
d? w dx
-<r = o (O
Again, taking moments about the middle point of BB f or
CO,
dx
T = '
or
dx
But M = EI--^, . Therefore
dM
--T- = S. (2)
^ '
vT j ds KT
EI-^-., and -r- = EI~.
dx* ' dx dx*
Hence, from (i),
CONTINUOUS GIRDERS. 463
This equation does not admit of a finite integration, but
may be integrated in the form of a partial differential equation.
12. Continuous Girders. When a girder overhangs its
bearings, or is supported at more than two points, it assumes a
wavy form and is said to be continuous. The convex portions
are in the same condition as a loaded girder resting upon a
single support, the upper layers of the girder being extended
and the lower compressed. The concave portions are in the
same condition as a loaded girder supported at two points, the
upper layers being compressed and the lower extended. At
certain points, called points of contrary flexure, or points of in-
flexion, the curvature changes sign and the flange stresses are
necessarily zero. Hence, apart from other practical 1 considera-
tions, the flanges might be wholly severed at these points with-
out endangering the stability of the girder.
13. The Theorem of Three Moments. It is required to
determine a relation between the bending moments at any three
consecutive points of support of a loaded continuous girder of
several spans.
Rr-1 Rr
r !
A i x
Rr+1
!v
1 -3J
V
FIG. 330.
Let O, X, V be the (r i)th, rth, and (r + l)th supports,
respectively.
Let OX=t r ,XV=t r + l .
CASE A. Let w r be the load per unit of length on OX,
w r + 1 the load per unit of length on XV.
Let R r .i, R r , ft r + i be the reactions at O,X, V, respectively.
Let M r ^, M r , M r+l be the bending moments at O, X, V,
respectively.
Let a be the angle which the tangent to the girder at X
makes with OV.
Consider the segment OX, and refer it to the rectangular
axes Ox, Oy.
4 6 4 THEORY OF STRUCTURES.
The equation of moments at any point (x, y) is
R r . lX -w r ~ + M r ., = M. . . (i)
lr
.;R r _ l t r -w r -- + M r _ l = M r (2)
Similarly, the segment XV gives
Combining (2) and (3),
r _ r1Ir r
= J/X4 + /, + ,) (4)
Integrating (i),
-'% = Xr-^-*>r% + Jfr.S + C, (5)
c being a constant of integration.
dy
When x l ry -j- = tan a.
.-. - El tan a = R r _, |- - v|r + ^.^ + c, . . (6)
Integrating (5),
- Ely = Rr-^ - v>r ~ + M r _t ~ + ex. . (7)
There is no constant of integration, as x and y vanish together.
Also, y = o when x l r .
or
c = R r 1 4- -\- v>r M r , -. (8)
r izr r 4 ^ r 1 / .. \/
THE THEOREM OF THREE MOMENTS. 4 6 5
Substituting this value of c in eq. (6),
- EI tan a = R r ., ^ - w r + Jf r . t j ..... ( 9 )
Similarly, the segment XV gives
Adding eqs. (9) and (10), transposing, and simplifying,
R r -Jr + Rr + fr + i
= WS + fV + ,/',+ . - \M r _J r - PV r+I . (ii)
Finally, combining eqs. (4) and (u),
+ w r+I r r+l ). (12)
If the girder is supported at points, there are w 2 equa-
tions connecting the corresponding bending moments, and two
additional equations result from the conditions of support at
the ends. For example, if the ends merely rest upon the sup-
dy
ports M l o and M n = o ; if an end is fixed, -=- = o at that
point.
The point of maximum bending moment, the points of
inflexion, and the point of maximum deflection in any span are
dM dv
found by making =- = o, M o, and -j- o, respectively.
Thus, for the span OX,
dM
o = R r ^ WyX ;
/. x = f , and maximum B.M. = -- -f- M r . t \
"Wr 2 W r
a quadratic giving x ;
466 THEORY OF STRUCTURES.
c,
a cubic from which x may be found by trial. The maximum
deflection is obtained by substituting the value of x in eq. (7) ;
c being given by eq. (8).
CASE B. Let the loads upon OX, XV, respectively, consist oi
a number of weights P l , P 9 , P 3 , . . . , distant p l , / 2 ,/ 3 , . . . from
O, and Q l , <2 3 , <2, , . . . , distant #,,,,&,... from F. Refer
the neutral axis OA X to the rectangular axes Ox, Oy.
It may be assumed that the total effect of all the weights
is the algebraic sum of the effects of the weights taken sepa-
rately.
Consider the effect of P t at A.
The equation of moments at any point (x, y) of the neutral
axis between O and A is
Integrating,
c l being a constant of integration.
Integrating again,
-Efy^Rr-^+Mr-^ + cs. . . (3)
There is no constant of integration, as* and y vanish together.
The equation of moments at any point (x, y) between A and
(4)
THE THEOREM OF THREE MOMENTS.
Integrating,
-El4jL = R r .-^(x-p l )'+M r -* + c t . ... (5)
Integrating again,
- Ely = R r .~ - (* - A) 3 + M r _ + v + ,, . (6)
dy
Now, at the point A, the values of -7- and jj/, as given by
m&
eqs. (5) and (6), are identical with those given by eqs. (2) and
(3) ; also, in equation (6), y = o when x = l r .
Hence,
and
o = R^ - (/ r - A) 3 + ^r-~ + cj r + c 9 ;
so that
3 = 0,
and
, = *.= -^-, + J(/,-/ 1 ) t -^-,j. (7)
Let a be the slope at X\ then, by eqs. (5) and (7),
- EI tan a = R_ - (/, - py(2l r +/,) + Jf r _. (8)
Similarly, the segment XV gives
-EI^(n- a } = Rj + M r+ ...... (9)
468 THEORY OF STRUCTURES.
Adding eqs. (8) and (9), and transposing,
*,_,/,' + JWV* = & - A)'(2/ r +A)
-f^.^-fJf^/^. (10)
Again, taking moments about Jf,
Rr j r _ />('r-A) + -^-i = ^ = ^WH-X + ^H-:> 00
whence
R r _,/; + ^H-x'V+x
and finally, by eqs. (10) and (12),
M r j r + 2M r (l r + /, +1 ) + M r+1 t r+I = - Pfa - A 2 )- (13)
The effect of each weight may be discussed in the same
manner, and hence the relation between M r _ iy M r , and M r+l
may be expressed in the form
r _ j r + 2 M r (l r + / r+1 ) + M r+l t r+l = - 2fa - p ,
Cor. I. The relation between M r _ T , M r , M r + l for a uni-
formly distributed load maybe easily deduced from eq. (14).
For example, let a uniformly distributed load of intensity w r
cover a length 2a(<l r ) of the span OX, and let z be the distance
of its centre from O. Then
V _ /} = f + ^,(4< -/) = ^; - * -
^r *J z- a r
which reduces to ~-f when z = a = .
THE THEOREM OF THREE MOMENTS. 469
Cor. 2. Considering the rth span and taking moments
about the rth support,
M being the moment of the load on the span, and the reaction,
or shear,
M M r M r _,
Hence, the shear at the (r i)th support for the rth span
= the reaction at the same support, supposing the span
an independent girder, i.e., cut at its supports,
+ the difference of the forces, or reactions, equivalent
to the moments at the supports.
Again, let M x be the moment of the load on the segment
x with respect to the point (x,y).
Hence, the total moment about (x,y)
'M , r \ , M r _,,, . , M.
= the moment at the same point supposing the span
an independent girder.
+ the reactions equivalent to the moments M r _^ , M r ,
multiplied respectively by the segments l r x and x.
In Fig. 331, OX being the rth span, let OB X be the curve
FIG.
of bending moments, supposing OX an independent girder, i.e.,
cut at O and X. On the same scale as this curve is drawn,
take the verticals OE and XF to represent M r _^ and M r , re-
47 THEORY OF STRUCTURES.
spectively, and join EF. The curve OBX corresponds to the
(M \
portion \-j-x M x } of the above equation, and the line EF
T
M r _J \ M r
to the remainder, i.e., j\l r xj -f- -j-x. The actual bend-
ing moment at any point of OX is represented by the algebraic
sum of the ordinates of the curve and line at the same point,
which will be the intercept between them, since they represent
bending moments of opposite kinds.
Let A be the effective moment area, or the algebraic sum of
the areas for the load and for the moments at O and JT, and
let x be the horizontal distance of its centre of gravity
from O.
Let A r be the area for the load, i.e., the area of the curve
OBX, and let z r be the horizontal distance of its centre of
gravity from O. Then
l ' , I,
Ax = A r z r + M r .= + -(M r - M r _W
= A r z r -f- \M r .j; -f- lMrl r \
This result will be referred to in a subsequent article.
14. Applications. EXAMPLE i. Swing-bridges of two
spans revolving about a single support at the pivot pier.
This is a case of a girder of two spans, OX(=. l^) y XV(= / 2 ),
resting upon supports at O and F, and continuous over a
pier at X.
The bending moments at O and Fare
X V both nil.
T~~ T~ 't Let M be the bending moment at X.
FIG. 332.
For a uniformly distributed load,
or = -
w t being the intensity of the load on OX, w, that on XV.
SWING-BRIDGES. 47 l
For an arbitrarily distributed load,
l^ = -A-B, or j|f=_I-
where A = 2-(/* - /) and B = 2(tJ - f).
f, * 2
Let R l , ^ 2 , -/? 3 be the reactions at O, X, V, respectively.
For a uniformly distributed load,
P - w & \ M - 3 w * / * > ~
*~ "-
2 "/- 8/^
i ^-
r -
2
For an arbitrarily distributed load,
K ^(A~/) , M _ P(l t -p) I
^ = ~
R = 2^_Zl} +~= ^ Q(/ * ~ q) - - A+B
If w l = O, or if P and hence ^4 o, then R l is negative.
So if w a = o, or if Q and hence B o, then ^? 3 is nega-
tive.
Hence, if either of the spans is unloaded, the reaction at
the abutment end of the unloaded span is negative and that
end is subjected to a hammering action. This evil may be
obviated :
(a) By loading the spans sufficiently to make R^ and R^
zero or positive.
This result is attained for R,
if 3Wl /,' + 4W/.Y, > a/4 1 , or
and for ^ 3
if 4 vW + 3 ^/, 3 > ^y, 8 , or if
47 2 THEORY OF STRUCTURES.
(b) By using a latching apparatus to keep the ends from
rising.
(c) By employing suitable machinery to exert an upward
pressure, at least equal to the corresponding negative reaction
upon each end, which is thus wholly prevented from leaving its
seat.
Cor. i. When the load is uniformly distributed, the dis-
tance x of the point of inflection in OX from O is given by
M = o = R,x -- , and therefore x - .
2 W 1
Similarly, the distance of the point of inflection in XV from
K= 2 A
U\
If / t = /, = /, then
M=-^ l+ w, R t = 7 -^^ * = =
And if w l = 2/ a = w, then
In the latter case - = / = - 3 , and thus a hinge may
w 1 w 2
be introduced in each span at a distance from the centre pier
equal to one fourth of the span, without impairing the stability
of the girder. Hence, also, the continuous girder of two equal
spans may be considered as consisting of two independent
girders, each of length |/, resting upon end supports, and of
two cantilevers each of length .
Ex. 2. Swing-bridges with two points of support at the
' t' f
h {> I*
FIG. 333.
pivot pier, as, e.g., when they are carried upon rollers running
in a circular path.
A P PLICA TIONS. 47 3
This is a case of a continuous girder of three spans.
Let I, , / 2 , / be the lengths of the spaces, w, , /, , ze/ 3 the
corresponding intensities of the loads, which are assumed to
be uniformly distributed.
Let R lt R^j R 3 , R t be the reactions at the supports ;
M l , M t , M a , M t the corresponding bending moments. Then
-$(w 1 t: + wj;)', . (i)
-}(>X + wA). . (2)
Let the ends of the girders rest upon the supports, and
assume, as is usually the case in practice, that the centre span
is unloaded, i.e., that w^ = o. Then
M l = o and M^ = o.
From (i) and (2),
A = -*/,', ". (3)
and
M,l, + 2 !&,(!, + /,)=-&,!: ..... (4)
Hence,
/./.V,
'
34"
and
_ ^/. 3 4 - W,V, + 4)
- 4(444 + 34' + 444 + 444)'
Taking moments about the second support,
6/,V > + 6/ 1 V J ' + 8/ 1 '4/ 1 ) + Wj 4V,_ _ (7)
4(444 + 34' + 444 + 444)
Taking moments about the third support,
6/.V, + 6/.V.' + 8/,V,Q + w,n (K .
()
4(444 + 34' + 444 + 444)
474 THEORY OF STRUCTURES.
Thus R t and jR t are both positive for all uniform distribu-
tions of load over the side spans, and no hammering action
can take place at the ends.
Again, if the span on the left is unloaded, i.e., if w 1 = o,
M 9 is positive and M 3 negative ; and if the span on the right is
unloaded, i.e., if w 3 = o, M, is negative and M 3 positive.
Thus, at the piers, the flanges of the girder will be sub-
jected to stresses which are alternately tensile and compressive,
and must be designed accordingly. The same result is also
true for arbitrarily distributed loads.
Ex. 3. The weights on the wheels of a locomotive passing
over a continuous girder of two spans, each of 50 ft., taken in
order, are as follows : 15,000 Ibs., 24,000 Ibs., 24,000 Ibs., 24,000
Ibs., 24,000 Ibs. The distances of the wheels, centre to centre,
taken in the same order, are 90 in., 56 in., 52 in., 56 in. Let
it be required to place the wheels in such a position as to give
the maximum bending moment at the centre pier.
The pier must evidently lie between the third and fourth
wheels.
Let x be the distance, in inches, of the weight of 15,000 Ibs.
from the nearest abutment. The remaining two weights on
the span are respectively x -f- 90 and x -(- 146 in. from the
same abutment.
The two weights on the other span are 142^ and 198 ;r
in 1 ., respectively, from the nearest abutment.
Hence, by Case B, Art. 13, if M is the bending moment at
the centre pier,
' - (x + 90)'
APPL1CA TIONS.
475
Making o for maximum value of M t and simplifying,
dx
and therefore
+ 27648* = 2518848,
x = 87.39 in. = 7.28 + ft.
Thus, the B. M. at the centre pier is a maximum when the
first wheel is 7.28 ft. from the nearest abutment.
The maximum B. M. in inch-pounds is obtained by substi-
tuting x = 87.39 in. in the above equation.
15. Maximum Bending Moments at the Points of Sup-
port of Continuous Girders of n equal Spans.
Let the figure represent a continuous girder of n spans, I,
2, 3, ... I being the n I intermediate supports.
o 123 r I r r-\-i I n
"A A A A A A A A A A A~
CASE I. Assume all the spans to be of the same length /,
and let w l , w t , . . . w n _ l , w n be the intensities of loads uni-
formly distributed over the 1st, 2d, ...( i)th and nth spans,
respectively.
By the Theorem of Three Moments,
-(
4 l
_ _/_'
4 2
_/_'
4
/ 2
? 8 + 4^ 4 + ^ 6 = -(w. + w 6 ) ;
(O
(2)
(3)
(4)
(5)
4/6 THEORY OF STRUCTURES.
r
m n _ 3 + 4m n _ 2 + /_, = - -(w n _ 2 + w n . r ) ; (n 2)
r
m n . 2 + 4w_ T = -<>-i + Wn)- (n i)
4
?# and ;# are both zero, as the girder is supposed to be rest-
ing upon the abutments at o and n.
From these (n i) equations, the bending moments m l ,
m^ , . . . m n _i may be found in terms of the distributed loads.
Eliminating m^ from 2 and 3,
r
mi _ x 5 ^ 3 _ 4 ^ 4 = _ - j ( Wf + ^3) - 4K + ^4)}- (*0
Eliminating w g from 4 and x^ ,
Eliminating MI from 5 and
m
.
Finally, by successively eliminating w 6 , m 6
-! -X
/ a
--U w + w 3)-4K +
the upper or lower sign being taken for the terms within the
brackets according as n is odd or even, and the coefficients
^-i , #*- 2 , tf- 3 , bein'g given by the law,
APPLICATIONS. 477
a, = 4a t a, = 209 ;
# a = 4*1 = 4 ;
, = I.
Commencing with equations n 3 and n 2, and proceed-
ing as before,
0_x*i *-i
/i
= - \ a n - 9 (u>i + /,) rf_ 3 (w a + ze; 3 ) + d_ 4 (w s + ;) . . .
1 5O_ 4 + w_ 3 ) =F 4(w-3 + >_,) (^- 2 + V_ f ) } , (^)
the upper or lower sign being taken for the terms within the
brackets according as n is odd or even.
Solving the two equations y and 2,
r
,(* f -xi)= - -|^-i^- 2
(<* n -i + a-3a n -*Wn-i =F
and
^-i(tfVi - i) =
/ a (
- 1 a n _ 2 w,
4
Hence, since w 19 w tt ...w n are positive integers, the value of
m H will be greatest when ;, , / a , w 4 , w e , o; 8 , . . . are greatest
and w a , w 5 , w n , . . . are least ; and the value of m n _ v will be
greatest when w w , w n ^ , w_ 3 , w M _ 4 , . . . are greatest, and w n _ 2 ,
w M - 4 , W M _, . . . are least. In other words, the bending moments
at the 1st and (n i)th intermediate supports have their maxi-
47 8 THEORY OF ^STRUCTURES.
mum values when the two spans adjacent to the support in
question, and then every alternate span, are loaded, and the re-
maining spans unloaded.
#?,,#?,,... m n -i ma y now be easily determined.
Thus, by eq. (i),
r
m * = 7^ + w ^ ~ 4w i
4
r ( 4
= - -7 J K + wj - -i - -a
4 l "
a + . . . I-
But a n _, = 4a n _ 2 a n _ y
and is greatest when w 9 , w 3 , w t , w 1f . . . are greatest and
w 4 , w, , w s , . . . are least.
Similarly, by eqs. (i) and (2),
CONTINUOUS GIRDERS. 479
and is greatest when w l , w a , w^ , w t , w 6 , . . . are greatest and
w a , w 6 , w 7 , w 9 , . . . are least.
Thus, the general principle may be enunciated, that " in a
horizontal continuous girder of n equal spans, with its ends
resting upon two abutments, the bending moment at an inter-
mediate support is greatest when the two spans adjacent to
such support, and the alternate spans counting in both direc-
tions, carry uniforrnly distributed loads, the remainder of the
spans being unloaded."
CASE II. The principle deduced in Case I also holds true
when the loads are distributed in any arbitrary manner.
Consider the effect of a weight w in the rth span concen-
trated at a point distant/ from the (r i)th support.
By the Theorem of Three Moments,
4^+^ = 0; (i)
m 1 + 4m, + m 3 = 0' t (2)
w* + 4^3 + ^4 = o ; . , , , . . (3)
m r . a + 4m r ., + m r = w -* /) = A, suppose ; (r i)
r + m r+l = -w
= w (I p)(2l /)=: B, suppose ; (r)
m r + 4m r+1 + m r+2 = o ; . . . . (r + i)
_i = O ; ....( 2)
= o (n i)
4^0 THEORY OF STRUCTURES.
By equations (i), (2), (3), ... (r - 2),
ill I
-
the upper or lower sign being taken according as r is even or
odd.
By equations (n i), (n 2), (n 3), . . . (r + i),
. m r+2 m r+l m r
= -f - - = -
a n-r-i
The coefficients a are given by the same law as for the co-
efficients a in Case I. Thus,
a r ... a n _ r
m r _ 2 = -- m r _ t and m r+l = - m r .
Substituting these values of m r _ 2 and m r+l in the (r i)th
and rth equations,
and
rtr-i + M r \A - ^A = B = m r .,
a n-r+2 '
where
<*r- 2
and c = ~
Hence, solving the last two equations,
Ac - B Bb-A
* r _ I ^-. _ and ^=_ 7 - .
The ratios - - and ^- are each less than unity, and
&r-i &n-r+i
hence b and c are each < 4 and > 3.
CONTINUOUS GIRDERS. 481
It may now easily be shown that Ac B and Bb A are
each positive. Hence, m r _^ and m r are both of the same sign*
The bending moment m q at any intermediate support on
the left of r i is given by
m q = -| -- m r -i if 9 an d r are the one even and the other odd,
or
= -- --m r -i if 9 an< 3 r are both even or both odd.
Thus the bending moment at the q\.\\ support is increased
in the former case and diminished in the latter.
If q is on the right of r,
m q = -\ *~ q+l m r if q and r are both even or both odd,
a n-r+i
or
m q -=. -- ^^~m r if q and r are the one even and the other odd,
and the bending moment on the ^th support is increased in the
former case and diminished in the latter.
Thus, the general principle may be enunciated, that, "in a
horizontal continuous girder of n equal spans, with its ends
resting upon two abutments, the bending moment at an inter-
mediate support is greatest when the two spans adjacent to
such support, and the alternate spans counting in both direc-
tions, are loaded, the remainder of the spans being unloaded."
CASE III. The same general principle still holds true when
the two end spans are of different lengths.
E.g., let the length of the first span be kl, k being a
numerical coefficient, and let 2(1 -\- k) = x.
Eq. (i) now becomes
m^x -^m^ Q.
482 THEORY OF STRUCTURES.
Proceeding as before,
.m^ _ m^ _ _m* _ _ m, _
}>'* ~ V' " ^3 : " b, ~~
the coefficients b lt b^, b^, . . . being given by the same law as
before, viz.,
=..*;
, = 4 3 - ^ = 4* i ;
4 = 4A ^= 1 5* -4J
5 = 4^ 3 b % = 56^ 15;
The two sets of coefficients (a) and (b) are identical when
x = 4; and when x > 4, all the coefficients b except the first
(b l i) are numerically increased.
Hence, the same general results will follow.
N.B. The equations giving m q are simple and easily ap-
plicable in practice. They may be written
a q B Ac .
m a = - -j if q is on the left of r,
a r _i DC I
and
m q = "Ir + i ~L - if ^ is on the right of r.
If there are several weights on the rth span,
') and B
EXAMPLE. The viaduct over the Osse consists of two end
spans, each of 94 ft., and five intermediate spans, each of 126 ft.
The platform is carried by two main girders which are con-
MAXIMUM BENDING MOMENTS. 483
tinuous from end to end. The total dead load upon the girders
may be taken at one ton (of 2000 Ibs.) per lineal foot.
Denote the supports, taken in order, by the letters a, b, c, d,
e,f,g, h, and let it be required to find the maximum bending
moment at d when the bridge is subjected to an additional
proof load of ij tons per lineal foot.
The spans ab, cd, de, fgol each girder carry ij tons per
lineal foot.
The spans be, ef, gh of each girder carry \ ton per lineal foot.
Denoting the bending moments at a, b, c, d, e, f t g, h, re-
spectively, by m l ,;//,... m s , the intermediate spans by /, the
end spans by /, and remembering that m l = o = m 6 , we have
a = -~
4
r
, + 4m 3 + m< = - -
r
, + 4m t + m,= - -(i J- + i J) ;
/ a
5 + 4^6 + ^ = - -(4 +
But k = T 9 ^- = f , very nearly.
/a 2^
.-. 7m, + 2m, = - - . ^ ; ..... (i)
4 5
-.- ..... (2)
THEORY OF STRUCTURES.
2m t + 7m,
From eqs. (i), (2), (3),
From eqs. (4), (5), (6),
l l L.
(V)
4 2 '
/^ *7
CA>
4 2'
7.
CO
4 2'
(6}
Hence, ; 4 , the maximum required,
16. General Theorem of Three Moments. The most
general form of theorem of three moments may be deduced as
follows :
Oi 2 3
FIG. 334.
Let O, X, V, the (r i)th, rth, and (r -\- i)th supports of a
continuous girder of several spans, be depressed the vertical
distances d, (= O,O\ d, (=O,X), and d, (= 6> 3 F), respectively,
below the proper level Ofl^O^ of the girder.
GENERAL THEOREM OF THREE MOMENTS. 48$
d^ , d^, d z are necessarily very small quantities.
Let OCXDVbe the deflection curve, and let the tangent at
X meet the verticals through O and V in E and F, and the
tangents at O and V in T t and 7", .
Let #, be the change of curvature from O to X (= OTf).
" ^ " " " " " " Fto X(= FT^V).
Let A lf A i be the effective moment areas for the spans OX,
XV, respectively.
Let .TJ be the distance (measured horizontally) of the centre
of gravity of A l from O.
Let x^ be the distance (measured horizontally) of the centre
of gravity of A^ from F.
Let 0,E =?>, Of = y % .
By Art. 2,
, _
r +I h /, + J~^/ 1 4 "4 +I /
But
7i - < _ < - J 8 Z. I _L A-^ . ^
7 7 > r 7* "t" 7 ~~ / IT
/ r / r+I / r / r+I / r /r+x
d^ d^d^ di i (A,x A^\
-JT / r+I -^A / r -/ ;
Again, by Art. 13, Cor. 2,
^
and
^4 r , y4 r+I being the areas of the bending-moment curves for
the spans OX, XV, respectively, on the assumption that they
are independent girders, or cut at O, X, and V, and z rj z r ^
486 THEORY OF STRUCTURES.
being the horizontal distances of the centres of gravity of these
areas from O and V. Hence,
M r _J r + 2M r (l r + / r+I ) + M r+l l r+l
KA* r 6 A * r+l -U
. = bA r j -- vSl r+l -j -- 1-
*r tr+i
Note. If O, X, or V is above O,O^ then d, , af a , or d^ is
negative.
Cor. The forms of the Theorem of Three Moments given in
Cases A and B, Art. 13, may be, immediately deduced from the
last equation.
CASE A.
A ---
^r-
' r+I
CASE B.
17. Advantages and Disadvantages of Continuous
Girders. The advantages claimed for continuous girders are
facility of erection, a saving in the flange material, and the re-
moval of a portion of the weight from the centre of a span to-
PROPERTIES OF CONTINUOUS GIRDERS. 487
wards the piers. Circumstances, however, may modify these
advantages, and even render them completely valueless. The
flange stresses are governed by the position of the points of in-
flexion, which, under a moving load, will fluctuate through a
distance dependent upon the number of intermediate supports
and upon the nature of the loading. In bridges in which the
ratio of the dead load to the live load is small the fluctuation is
considerable, so that for a sensible length of the main girders,
a passing train will subject local members to stresses which are
alternately positive and negative. This necessitates a local
increase of material, as. each member must be designed to bear
a much higher stress than if it were strained in one way only.
Again, the web of a continuous girder, even under a uni-
formly distributed dead load, is theoretically heavier than if
each span were independent, and its weight is still further in-
creased when it has to resist the complex stresses induced by
a moving load.
Hence, in such bridges the slight saving, if there be any,
cannot be said to counterbalance the extra labor of calculation
and workmanship.
In girders subjected to a dead load only, and in bridges in
which the ratio of the dead load to the live load is large, the
saving becomes more marked, and increases with the number
of intermediate supports, being theoretically a maximum when
the number is infinite. This maximum economy may be ap-
proximated to in practice by making the end spans about four-
fifths the intermediate spans.
In the calculations relating to the Theorem of Three Mo-
ments, it has been assumed that the quantity El is constant,
while in reality E, even for mild steel, may vary 10 or 15 per
cent from a mean value, and / may vary still more. It does
not appear, however, that this variation has any appreciable
effect if the depth of the girder or truss changes gradually, but
the effect may become very marked with a rapid change of
depth, as, e.g., in the case of swing-bridges of the triangular
The graphical method of treatment may still be employed
by substituting, for the actual curve of moments, a reduced
488 THEORY OF STRUCTURES.
curve, formed by changing the lengths of the ordinates in the
ratio of the value of El at a datum section to EL
It is often found economical to increase the depth of the
girder over the piers, which introduces a local stiffness and
moves the points of inflexion farther from the supports. A
point of inflexion may be made to travel a short distance by
raising or depressing one of the supports.
In order to insure the full advantage of continuity the ut-
most care and skill are required both in design and workman-
ship. Allowance has to be made for the excessive expansion
and contraction due to changes of temperature, and the piers
and abutments must be of the strongest and best description
so that there shall be no settlement. Indeed, the difficulties
and uncertainties to be dealt with in the construction of con-
tinuous girders are of such a serious if not insurmountable
character that American engineers have almost entirely dis-
carded their use except for draw-spans.
Much, in fact, is mere guesswork, and it is usual in prac-
tice to be guided by experience, which confines the points of
inflexion within certain safe limits.
Under these circumstances it may prove desirable to fix
the points of inflexion absolutely, and the advantages of doing
so are (a) that the calculation of the web stresses becomes
easy and definite^ instead of being complicated and even in-
determinate ; (b) that reversed stresses (for which pin-trusses
are less adapted than riveted trusses) are almost entirely
avoided ; (c) that the stresses are not sensibly affected by
slight inequalities in the levels of the supports ; (d) that the
straining due to a change of temperature takes place under
more favorable conditions.
The fixing may be thus effected :
(a) A hinge may be introduced at the selected point.
The benefit of doing so is very obvious when circumstances
require a wide centre span and two short side spans.
(b) If the web is open, i.e., lattice-work, the point of inflex-
ion in the upper flange may be fixed by cutting the flange at
the selected point and lowering one of the supports so as to
produce a slight opening between the severed parts. The
ADVANTAGES OF CONTINUOUS GIRDERS. 489
position of the point of inflexion in the lower flange is then
defined by the condition that the algebraic sum of the hori-
zontal components of the stresses in the diagonals intersected
by a line joining the two points of inflexion is zero.
It must be remembered, however, that this fixing of the
points of inflexion, or the cutting of the chords, destroys the
property of continuity, and, indeed, is the essential distinction
between a continuous girder and a cantilever.
Four methods may be followed in the erection of a contin-
uous girder, viz.:
1. It may be built on the ground and lifted into place.
2. It may be built on the ground and rolled endwise over
the piers. As the bridge is pushed forward, the forward end
acts as a cantilever for the whole length of a span, until the
next pier is reached. This method of erection is common in
France.
3. It may be built in position on a scaffold.
4. Each span may be erected separately, and continuity pro-
duced by securely jointing consecutive ends, having drawn to-
gether the upper flanges. A more effective distribution of the
material is often made by leaving a little space between the
flanges and forming a wedge-shaped joint.
49 THEORY OF STRUCTURES.
EXAMPLES.
1. Two angle-irons, each 2 in. x 2 in. x in., were placed upon sup-
ports 12 ft. 9 in. apart, the transverse outside distance between the bars
being 9^ in., and were prevented from turning inwards by a thin plate
upon the upper faces. The bars were tested under uniformly distributed
loads, and each was found to have deflected 2 T 5 ^ in. when the load over
the two was 1008 Ibs. Find E and the position of the neutral axis.
Ans. /= J/\ ; E= 17,226,139 Ibs.; neutral axis $ in. from*
upper face.
2. Both bars in the preceding question failed together when the
total load consisted of ioi cwts. (cwt. = 112 Ibs.) uniformly distributed,
and 3 cwts. at the centre. Find the maximum stress in the metal.
Ans. Compressive unit stress = 20,323 Ibs. ;
Tensile unit stress = 39,577 Ibs.
3. Show that the moments of resistance of an elliptic section and of
the strongest rectangular section that can be cut out of the same are in
the ratio of 99 ^3 to 112, and that the areas of the sections are in the
ratio of 33 to 14 |/2.
4. Show that the moments of resistance of an isosceles triangular
section and of the strongest rectangular section that can be cut out of
the same are in the ratio of 27 to 16, and that the areas of the two
sections are in the ratio of 9 to 4.
5. An angle-iron, 3 in. x 3 in. x ^ in., was placed upon supports
12 ft. 9 in. apart, and deflected i- in. under a load of 8 cwts. uniformly
distributed and 2 cwts. at the centre. Find E and the position of the
neutral axis.
Ans. E = 16,079,611 Ibs.; neutral axis iff in. from upper face.
6. The effective length and central depth of a cast-iron girder resting
upon two supports were respectively n ft. 7 in. and 10 in. ; the bottom
flange was 10 in. wide and i J in. thick ; the top flange was 2^ in. wide
and in. thick; the thickness of the web was f in. The girder was
tested by being loaded at points 3f ft. from each end, and failed when
the load at each point was 17^ tons. What were the total central
flange stresses at the moment of rupture?
What was the central deflection when the load at each point was 7|
tons? (E = 18,000,000 Ibs., and the weight of the girder = 3368 Ibs.)
Ans. 164,747.4 Ibs.; .368 in.
EXAMPLES. 491
7. A tubular girder rests upon supports 36 ft. apart. At 6 ft. from
one end the flanges are each 27 in. wide and 2f in. thick, the net area of
the tension flange being 60 in., while the web consists of two fa-in.
plates, 36 in. deep and 18 in. apart. Neglecting the effect of the angle-
irons uniting the web plates to the flanges, determine the moment of
resistance.
The girder has to carry a uniformly distributed dead load of 56 tons,
a uniformly distributed live load of 54 tons, and a local load at the
given section of 100 tons. What are the corresponding flange stresses
per square inch?
How many |--in. rivets are required at the given section to unite the
angle-irons to the flanges?
Ans. 238.13 x coeff. of strength ; 3.3186 tons ; 3.896 tons.
8. A yellow-pine beam, 14 in. wide and 15 in. deep, was placed upon-
supports 10 ft. 9 in. apart, and deflected f in. under a load of 20 tons at
the centre. Find E, neglecting the weight of the beam.
Ans. E = 1,272,112 Ibs.
9. What were the intensities of the normal and tangential stresses at
2 ft. from a support and 2\ in. from neutral plane, upon a plane inclined
at 30 to the axis of the beam in the preceding question?
Ans. 132.83 and 218.91 Ibs.
10. A beam is supported at the ends and bends under its own weight.
Show that the upward force at the centre which will exactly neutralize
the bending action is equal to f or \ of the weight of the beam (w) r
according as the ends are free or fixed.
Find the neutralizing forces at the quarter spans.
Ans. Ends free ^-gw at each or faw at one of the points of
division. *
Ends fixed ^w at each or %w at one of the points of
division.
11. A beam 8 in. wide and weighing 50 Ibs. per cubic foot rests upon
supports 30 ft. apart. Find its depth so that it may deflect f in. under
its own weight. (E i ,200,000 Ibs.) Ans. 9.185 in.
12. A rectangular girder of given length (/) and breadth (&) rests
upon two supports and carries a weight P at the centre. Find its depth
so that the elongation of the lowest fibres may be l Vff of the original
length.
Ans
13. A yellow-pine beam, 14 in. wide, 15 in. deep, and weighing 32 Ibs.
per cubic foot, was placed upon supports 10 ft. 6 in. apart. Under
uniformly distributed loads of 59,734 Ibs. and of 127,606 Ibs. the central
49 2 THEORY OF STRUCTURES.
deflections were respectively .18 in. and .29 in. Find the mean value
of E.
Also determine the additional weight at the centre which will increase
the first deflection by ^ of an inch. Ans. 2,552,980 Ibs.; 24,121 Ibs.
14. In the preceding question find for the load of 59,734 Ibs. the
maximum intensities of thrust, tension, and shear at a point half-way
between the neutral axis and the outside skin in a transverse section at
one of the points of trisection of the beam. Also find the inclinations
of the planes of principal stress at the point.
Ans. 1609.255, 169.562, 119.364 Ibs. ; = 3 48!'.
15. A pitch pine beam, 14 in. wide, 15 in. deep, and weighing 45 Ibs.
per cubic foot, is placed upon supports 10 ft. 9 in. apart, and carries a
load of 20 tons at the centre. Find the deflection and curvature, E
being 1,270,000 Ibs. What stiffness does this give ?
What amount of uniformly distributed load will produce the same
deflection? Ans. ^J T ; 32 tons.
16. In the preceding question find the maximum intensities of thrust,
tension, and shear at points (a) half-way between the neutral axis and
the outside skin, (b) at one third of the depth of the beam, in a trans-
verse section at one of the quarter spans. Also find the inclinations of
the planes of principal stress at these points.
Ans. (a) 95 I - 8 53> 292.969, 329.442 Ibs.; 6=9 34$'.
(b) 658.774, 171.108, 243.833 Ibs.; 6 = 15 5of.
17. A piece of greenheart, 142 in. between supports, 9 in. deep, and
5 in. wide, was tested by being loaded at two points, distant 23 in. from
the centre, with equal weights. Under weights at each point of 4480
Ibs., 11,200 Ibs., and 17,920 Ibs. the central deflections were .13 in., .37
in., .67 in., respectively. Find the mean coefficient of elasticity. The
beam broke under a load of 32,368 Ibs. at each point. Find the
coefficient of bending strength.
1 8. A sample cast-iron girder for the Waterloo Corn Warehouses,
Liverpool, 20 ft. 7-J- in. in length and 21 in. in depth (total) at the centre,
was placed upon supports 18 ft. i in. apart, and tested under a
uniformly distributed load. The top flange was 5 in. x i in., the
bottom flange was 18 in. x 2 in., and the web was ij in. thick. The
girder deflected .15 in., .2 in., .25 in., and .28 in. under loads (including
weight of girder) of 63,763 Ibs., 88,571 Ibs., 107,468 Ibs., and 119,746 Ibs.,
respectively, and broke during a sharp frost under a load of 390,282 Ibs.
Find the mean coefficient of elasticity and the central flange stresses at
the moment of rupture.
Ans. 7 = 3309.122; E= 17,427,327 Ibs.; 20,121 Ibs., 47,168 Ibs.
19. A steel rectangular girder, 2 in. wide, 4 in. deep, is placed upon
EXAMPLES. 493
supports 20 ft. apart. If E is 35,000,000 Ibs., find the weight which, if
placed at the centre, will cause the beam to deflect i in.
Ans. 1296^7 Ibs.
20. A timber joist weighing 48 Ibs per cubic foot, 2 in. wide x 12 in.
deep x 14 ft. long, deflected .825 in. under a load of 887 Ibs. at the
centre. Find E. Ans. 397,880 Ibs.
21. A beam of span / is uniformly loaded. Compare its strength and
stiffness (a) when merely resting upon supports at the ends ; (b) when
fixed at one end and resting upon a support at the other ; (c) when fixed
at both ends. In case (c) two hinges are introduced at points distant y
from the centre ; show that the strength of the beam is economized to
the best effect when y = , and that the stiffness is a maximum when
y = very nearly.
4
Ans. Cases (a) and (). mi :m*m:i
Cases (a) and (c). mi : m a :: 3 : 2
Case (c). Max. economy, mi : m*
Max. stiffness, mi : m*
: D* :: i : .416.
: Da :: 10 : 3.
2 : i
5:24/2.
4:3;
15 : 4 (approx.).
22. A beam AB of span /, carrying a uniformly distributed load of
intensity TV, rests upon a support at B and is imperfectly fixed at A, so
i iif/ a
that the neutral axis at A has a slope of g -j=j . The end B is lower
than A by an amount jzj . Find the reactions. How much must B
be lowered so that the whole of the weight may be borne at A ? Find
the work done in bending the beam.
21 ii 7 wl*
Ans. wl, ivl ; -~ -==.
32 32 48 1
23. A round wrought-iron bar / ft. long and d in. in diameter can
just carry its own weight. Find / in terms of d, (a) the allowable de-
flection being i in. per 100 ft. of span, E being 30,000,000 Ibs. ; (ff) the
allowable stress being 8960 Ibs. per square inch; (V) the stiffness given by
(a) and the strength given by (b) being of equal importance.
Ans. (d) /= |/2 50^? 2 ; (b) I = 4/224^; (c) /= \\d.
24. A square steel bar i ft. long and having a side of length d in. can
just carry its own weight; its stiffness is y^r and the maximum allow-
able working stress is 7 tons per square inch. Find / in terms of d, E
being 13,000 tons. /(in ft.) 13
Ans. -77: : \ ==
<T(in in.) 7
494 THEORY OF- STRUCTURES.
25. A uniformly loaded beam with both ends absolutely fixed is
hinged at the quarter-spans. Show that the slope is suddenly doubled
on passing a hinge.
26. A horizontal beam with both ends absolutely fixed is loaded with
a weight W at a point dividing the span into two segments a and b.
W ( ab \ 3
Show that the deflection at the point is ^. I j , and find the
work done in bending the beam. W* I ad
HS '
27. Determine the isosceles section of maximum strength which can
be cut out of a circular section of given diameter, and compare the
strengths of the two sections.
28. A 3-in. x 3-in. x -in. angle-iron, with both ends fixed and a clear
span of 20 ft. .carries a uniformly distributed load of 500 Ibs. which causes
it to deflect 2 in. Find E. What single load at the centre will produce
the same deflection ? Find the work done due to bending in each case.
Ans. E = 20,775,415 Ibs.; 250 Ibs.
29. A steel plate beam of uniform section and 30 ft. span has both
ends fixed and is freely hinged at the points of trisection. Determine
the neutral axis(#) for a uniformly distributed load of 60,000 Ibs.; (b) for
a single load of 10,000 Ibs. concentrated,^?/ 1 , 7^ ft. and, second, 15 ft.
from a support.
Ans. (a) Side span, y = -~=- ( 100 S x H ) >
/ N
centre span,^> = -f -2 f 100 2o;r 2 -f jr 8 j.
(b) First. Loaded span between support and weight,
Loaded span between weight and hinge,
l2 5 ar ~ 703125.
Unloaded side span horizontal ; centre span
straight between hinges.
Second. Side span.jK = - -f %x ' ) ;
El \ 6 /
x2\ 5000000
centre span,, = -5- -j +
EXAMPLES. 495
30. A uniformly loaded beam, with both ends absolutely fixed, is
hinged at a point dividing the span into segments a and b. Draw curves
of shearing force and bending moment, and compare the strength and
stiffness of the beam when the hinge is (a) at the middle point ; (b) at
a point of trisection ; (c) at a quarter-span. Also, determine the slope of
the segments of these points.
w $a* + 8at>* + 3^ 4 w 3# 4 4- %a*b + 5^ 4
Ans ' * 1= ~ ~ : * 2= - ^
_ 3^ */ _ wa $a* + ^a*b + 4
" 8 " a 3 + P ' 8 s + 8
J/' : J/" : M'" :: 14 : 14 : ii ; >' : >'' : >'":: 6.25 : 3.29 : 2.66.
Slopes in (#) = - ; in (b) = - for segment a,
c 6 E c 81
and = - for segment ;
in (c) = f for segment a, and
E c 176
/ / 9 2 r
= for segment b.
EC 891
31. A horizontal beam rests upon two supports and is loaded with a
weight W at a point dividing the span into segments a and b. Find
the deflection at this point and the work done in bending the beam.
W
Ans -
a*t>* IV* a*P I W \
; = x deflection .
EI(a + b) 6EI a + <H 2 /
32. A wrought-iron beam of rectangular section and 20 ft. span is
16 in. deep, 4 in. wide, and is loaded with a proof load at the centre. If
the proof strength is 7 tons per square inch, find the proof deflection and
the resilience, E being 12,000 tons. Ans. .029 ft. ; 650 ft.-lbs.
33. Design a wooden cantilever 12 ft. long, of circular section and
uniform strength, to carry a uniformly distributed load of 2 tons, the
coefficient of working strength being i ton per square inch. Also, find
the deflection of the free end.
Ans. Taking fixed end as origin and z being radius in inches at
distance x ft. from origin, then \\z* = 14(12 ;r) a .
Deflection at end = -^- in.
E
34. A girder fixed at both ends carries (2n -f- i) weights ^concen-
trated at points dividing the length of the girder into 2n + 2 equal
divisions. Find the total central deflection. n + i Wl*
Ans. =r r -.
192 El
THEORY OF STRUCTURES.
35. A girder 30 m. long has both ends fixed and carries a uniformly
distributed load of 5800 k. per lineal metre. Find the deflection and
the work of flexure. 567675000000
Ans. gj km.
36. A steel beam of circular section is to cross a span of 15 ft. and
to carry a load of 10 tons at 5 ft. from one end. Find its diameter, the
stiffness being such that the ratio of maximum deflection to span is
.00125. E 13,000 tons. Ans. 10.3 in.
37. Determine the dimensions of a beam of rectangular section
which might be substituted for the round beam in the preceding ques-
tion, the stiffness remaining the same and the coefficient of working
strength being 7^ tons per square inch. Ans. bd* = 320.
38. The flange of a girder consists of a pair of angle-irons and of a
plate which extends over the middle portion of the girder for a certain
required distance. Show that the greatest economy of material is
secured when the length of the plate is f of the span and the sectional
areas of the plate and angle-irons are as 4 to 5 (the girder being
uniformly loaded).
39. The flange of a uniformly loaded girder is to consist of two-
plates, each of which extends over the middle portion of the girder for
a certain required distance, and of a pair of angle-irons. Show that
the greatest economy of material is realized when the lengths of the
plates and angle-irons are in the ratio of 12 : 18 : 23, and when the areas
of the plates are in the ratio of 4 : 5.
What should be the relative lengths of the plates if they are of equal
sectional area ? AnSm I . ^. j( ^ + i).
40. An elastic beam rests upon supports at its ends, and a weight
placed at a point A produces a certain deflection (d) at a point B.
Show that if the weight is transferred to B the same deflection (d) is
produced at A.
41. A uniform beam is supported by four equidistant props, of which
two are terminal. Show that the two points of inflexion in the middle
segment are in the same horizontal plane as the props.
42. Find the slope and deflection at the free end of the following
cantilevers when bending under their own weight, / being the length,
2b the depth at the fixed end, iv the specific weight, and E the coefficient
of elasticity :
(a) Of constant thickness / and with profile in the form of atrapezoid
with the non-parallel sides equal and of depth 20. at the free end.
(b) Of circular section and with profile in the form of an isosceles
triangle.
EXAMPLES. 497
(<r) Of constant thickness and with profile in the form of a parabola
symmetrical with respect to the axis and having its vertex at the free
end.
wl* yvl (a* a P- at>* + 8a*6 - 2a* ) '
; rfv=*f \ T^a iog 'b + " ~6F~ : \
wl* i wl* 2wl*
'
43. Deduce the slope and deflection at the free end
(d) When the depth 2a in (a) of the preceding question is nil, i.e.,
when the profile is an isosceles triangle.
(e) Due to a uniformly distributed load of intensity p over the
cantilever (a). Hence, also, deduce the slope and deflection when the
depth 2a is nil.
(/) Due to a weight W at the free end of (a).
(g) Due to a uniformly distributed load of intensity p upon the
cantilever (<:).
Wl* Wl*
a (2? + $ a b - a*)(b - a)
)
I 90 '
3 pl*
~4~EtP'
^' g 3*-
b-a
2 EPt* 10
44. A cantilever of fength /, specific weight -w, and square in section,
a side of the section being 2b at the fixed and 2a at the free end, bends
under its own weight. Find the slope and deflection of the neutral axis
at the free end. Hence, also, deduce corresponding results when the
cantilever is a regular pyramid.
(b
45. If the section of the cantilever in the preceding question, instead
of being square, is a regular figure with any number of equal sides, show
that the neutral axis is a parabola with its vertex at the point of fixture.
46. The section of a cantilever of length / is an ellipse, the major axis
(vertical} being twice the minor axis. Find the deflection at the end
49^ THEORY OF STRUCTURES.
under a single weight W,f being the coefficient of working strength
and E the coefficient of elasticity. / 297 /V 5 \ J
Ans. --- m?7T
\7ooo h 3 W I
47. A cast-iron beam of an inverted T-section rests upon supports
22 ft. apart; the web is I in. thick and 20 in. deep; the flange is 1.2 in.
thick and 12 in. wide; the beam carries a uniformly distributed load of
99,000 Ibs. Find the maximum deflection, E being 17,920,000 Ibs.
Ans. .822. in. (/ = 1608.65).
48. Find the maximum deflection of a cast-iron cantilever 2 in. wide
x 3 in. deep x 120 in. long under its own weight, E being 17,920,000 Ibs.
Ans. H in.
49. A girder of uniform, strength, of length /, breadth b, and depth d,
rests upon two supports and carries a uniformly distributed load of w Ibs.
per unit of length, which produces an inch-stress of / Ibs. at every point
n 2 f\t b \\
of the material. Show that the central deflection is -- (~ / t
2 E \ 3y
when b is constant and ^variable. Find the deflection when d is con-
stant and b variable. fr
Ans. +.
4,Ed
50. A semi-girder of uniform strength, of length /, breadth b, and
depth d, carries a weight W at the free end which produces an inch-
stress of/ Ibs. at every point of the material. Prove that the maximum
deflection is - [ -) when b is constant and d variable, and that
3 E \6 W)
it is twice as great as it would be if the section were uniform throughout
and equal to that at the support.
What would be the maximum deflection if the semi-girder were
subjected to a uniformly distributed load of w Ibs. per unit of length ?
Ans. .
E J $w
51. The neutral axis of a symmetrically loaded girder, whose moment
of inertia is constant, assumes the form of an elliptic or circular arc.
Show that the bending moment at any point of the deflected girder is
inversely proportional to the cube of the vertical distance between the
point and the centre of the ellipse or circle.
52. A vertical row of water-tight sheet piling, 12 ft. high, is
supported by a series of uprights placed 6 ft. centre to centre and
securely fixed at the base. Find the greatest deviation of an upright
from the vertical when the water rises to the top of the piling. What
will the maximum deviation be when the water is 6 ft. from the top ?
wblf 3110400 wb wbc 218720
Ans. - = ~ : =r. (h - cY + =-,(* - cY = -~- .
3o7 El y>EI 24 El hi
EXAMPLES. 499
53. A vertical row of water-tight sheet piling, 30 ft. high, is supported
by a series of uprights placed 8 ft. centre to centre and securely fixed at
the base, while the upper ends are kept in the vertical by struts sloping
at 45. If the water rises to the top of the piling, find (a) the thrust on a
strut ; (d) the maximum intensity of stress in an upright ; (c) the amount
and position of the maximum deviation of an upright from the vertical.
iuh*
Ans. 45000 |/2 Ibs. ; max. B. M. = -- , and max. intensity of
stress = . 1 - - -7- -- - ; deflection is a max. when
A ( 10 / I5| / 5 j
h 30 7vh* 32
x = = =, and its amount = ^j ---- - .
V5 VS ^ 7SV5
54. The piling in the preceding example is strengthened by a second
series of struts sloping at 45 from the points of maximum deviation.
Find the normal reactions upon an upright and the bending moment at
its foot.
What will be the reactions and bending moment if the second row of
struts starts from the middle of the uprights?
Ans. .00754W/* 2 ; .lyjwk* ; .^2O27 i w/t 3 ; -^wk* ; If^rf* ; fH^// 9 .
55. A continuous girder of three spans, the outside spans being
equal, is uniformly loaded. What must be the ratio of the lengths of
the centre and a side span so that the neutral axis may be horizontal
over the intermediate supports? Ans. 4/T : y'T.
56. What should the ratio be if the centre span is hinged (a) at the
centre; (ff) at the points of trisection ? Ans. (a) 4/2" : i ; () 3 : 2 \/~ 2 .
57. Four weights, each of 6 tons, follow each other at fixed distances
of 5 ft. over a continuous girder of two spans, each equal to 50 ft. If the
second and third supports are i in. and i in., respectively, vertically
below the first support, find the maximum B. M. at the intermediate
support. / El \ ,
Ans. .9855 -- 1 ft. -tons.
\ 40000,1
58. A continuous girder of two equal 5o-ft. spans is fixed at one of
the end supports. The girder carries a uniformly distributed load of
loco Ibs. per lineal foot. Find the reactions and bending moments at
the points of support. How much must the intermediate support be
lowered so that it may bear none of the load ? How much should the
free end be then lowered to bring upon the supports the same loads as at
the first ?
Ans. Reactions = 23,214^, 57,142^, 19,642! Ibs. ;
Bending moments = 178,571^, 267,8574 ft. -Ibs;
500 THEORY OF STRUCTURES.
59. Four loads, each of 12 tons and spaced 5,4, and 5 ft. apart, travel
in order over a continuous girder of two spans, the one of 30 and the other
of 20 ft. Place the wheels so as to throw a maximum B. M. upon the
centre support, and find the corresponding reactions.
Draw a diagram of B. M., and find the maximum deflection of each
span.
60. The loads upon the wheels of a truck, locomotive, and tender,
counting in order from the front, are 7, 7, 10, 10, 10, 10, 8, 8, 8, 8 tons,
the intervals being 5, 5, 5, 5, 5, 9, 5, 4, 5 ft. The loads travel over a
continuous girder of two 5o-ft. spans AB, BC. Place the locomotive,
etc., (a) on the span AB so as to give a maximum B. M. at B\ (8) so as
to give an absolute maximum B. M. at B.
61. A continuous girder of two spans AB, BC has its two ends A
and C fixed to the abutments. The load upon AB is a weight P distant
/ from A, and that upon BC a weight Q distant g from C. The length
of AB = l\ , of BC /a . The bending moments at A, B, C are Mi , M* ,
Ms, respectively. The areas of the bending-moment curves for the
spans AB, BC assumed to be independent girders are^i, At, respect-
ively. Show that
Mill + M t (li + / a ) + M*h = - 2(Ai + A*),
and M*(li + /,) =
If /, = / 3 = /, show that Mi is a maximum if
2l(Pp - Qg) =
62. A continuous girder of two spans AB, BC rests upon supports at
A, B. A uniformly distributed load EF travels over the girder. G\ is
the centre of gravity of the portion BE upon AB, and G* that of the
portion./?/'' upon BC. If the bending moment at B is a maximum, show
that
AE.EB _ Ad
CF . FB ~ CG*'
63. An eight-wheel locomotive travels over a continuous girder of
two loo-ft. spans ; the truck-wheels are 6 ft. centre to centre, the load
upon each pair being 8000 Ibs. ; the driving-wheels are 8 ft. centre to
centre, the load upon each pair being 16,000 Ibs. ; the distance centre to
centre between the front drivers and the nearest truck-wheels is also
8 ft. Place the locomotive so as to throw a maximum B. M. upon the
centre support, and find the corresponding reactions.
64. If an end of a continuous girder of any number of spans is fixed,
show that the relation between the moment of fixture (Mi) and the
W/ 2
bending moment (M*) at the consecutive support, is 2Mi + M? = ,
4
or 2J/i + M* = ~^S[Pp(l p)(2l /)], according as the load upon
EXAMPLES. 5OI
the span (/) between the fixed end and the consecutive support is of
uniform intensity or consists of a number of weights Pi, P*, Pa, . . .
concentrated at points distant A, /, /s, . . . from the fixed end.
65. A continuous girder of two spans AB, BC, carrying a load of
uniform intensity, has one end A fixed, and the other end rests upon the
support at C. If the bending moments at A and B are equal, show that
the spans are in the ratio of 1/3 to f/2, and find the reactions at the
supports, Wi being the load upon AB, and W* that upon BC.
Ans. At A reaction =
B " =
C " =tJF..
66. A viaduct over the Garonne at Bordeaux consists of seven spans,
viz., two end spans, each of 57.375 m., and five intermediate spans, each
of 77.06 m. ; the mam girders are continuous from end to end, and are
each subjected to a dead load of 3050 k. per lineal metre. Determine the
absolute maximum bending moment at the third support from one end.
Also find the corresponding reactions, the points of inflexion, and the
maximum deflection in the first and second spans.
67. A continuous girder consists of two spans, each 50 ft. in length;
the effective depth of the girder is 8 ft. If one of the end bearings
settles to the extent of i in., find the maximum increase in the flange
and shearing stress caused thereby, and show by a diagram the change
in the distribution of the stresses throughout the girder. (Assume the
section of the girder to be uniform, and take E = 25,000,000 Ibs.)
Ans. Increase of maximum B. M. = s -Af[ ~ f i I,
\2i6rf /
" shearing force = ff/,
w being weight per unit of length, and /the moment of inertia.
68. A girder carrying a uniformly distributed load is continuous over
four supports, and consists of a centre span (/ 2 ) and two equal side
spans (/i). Find the ratio of l\ to / 2 , so that the neutral axis at the
intermediate supports may be horizontal. Also find the value of the
ratio when a hinge is introduced (a) at the middle point of the centre
span ; (fr) at the points of trisection of the centre span ; (c) at the middle
points of the half lengths of the centre span.
A I_i ^-! !_! A '-3
' /,- 3 ' / 2 2 ~2'/ 2 '- 9 ; ^~4'
69. In a certain Howe truss bridge of eight panels, the timber cross-
ties are directly supported by the lower chords, and are placed suffi-
ciently close to distribute the load in an approximately uniform manner
over the whole length of these chords, thus producing an additional
stress due to flexure. Assuming that the chords may be regarded as
girders supported at the ends and continuous over seven intermediate
5O2
THEORY OF STRUCTURES.
supports coincident with the panel points, and that these panel points
are in a truly horizontal line, determine (a) the bending moments and
reactions at the panel points ; (b) the maximum intermediate bending
moments; and (c) the points of inflection, corresponding to a load of w
per unit of length, / being the length of a panel.
Ans. (a) At i st support ; 2d support; 3d support;
B. M. = o ; - VW 2 ;
Reaction = R, = a// ; R* = ivl ; R z =
At 4th support ; 5th support.
B. M. = -^W 2 : -H^l\
Reaction = Ri= fffw/; Rs> f ffw/.
Maximum intermediate B. M. = - '-
in istspan ;
6208. 5
= W" / 4
(<:) Points of inflexion in the four spans are given by
* = = ^/; *i(/ + *)
W 388
x)
x)
-*0 2 = o ;
xf = o.
70. A. continuous girder of two equal spans \&fixefl $& one of the end
supports. The girder carries a uniformly distributed load of intensity
w. If the length of each span is /. find the reactions and moment of
fixture. How much must the intermediate support be lowered so that
it may bear none of the load ? How much should the free-end support
then be lowered to bring upon the supports the same loads as before ?
ii 16 13 w/ 2 c wl* 5 //*
wl tvl -
Ans. wl, wl, tvl\
28 14 28
14
24
71. Each of the main girders of the Torksey Bridge is continuous
and consists of two equal spans, each 130 ft. long. The girders are
double-webbed ; the thickness of each web plate is i in. at the centre
and f in. at the abutments and centre pier ; the total depth of the gir-
ders is 10 ft., and the depth from centre to centre of the flanges is 9 ft.
4 in.---FTtidX/?) the reactions at the supports, and also (b) the points of
inflexion, when 2oc\tpns of live load cover one span, the total dead load
upon each span being. 180 tons uniformly distributed. The top flange is
EXAMPLES.
503
cellular ; its gross sectional area at the centre of each span is 51 sq in ,
and the corresponding net sectional area of the bottom flange is 55 sq.
in. Determine (6-) the flange stresses and (d} the position of the neutral
axis. (7=372,500.) Also (e) determine the reactions when, first, B and,
second, Care lowered i in. (E '= 16,900 tons.)
Ans. (a) 155, 350, 55 tons.
() io6 r ^ and 79! ft. from end support.
(c) 6 7 and 7.3 tons per sq. in. in loaded span; 1.13 and
1.22 tons per sq. in. in unloaded span.
(d) 58.3 in. from centre line of top flange.
(e) First. R, =
i EI
Second. Ri = 1 55 ~ ^TT
lEI
= 350- sTTi
I 7
- S ? + ;T-
lEI
= 350+--^-;
4 *
i EI
'=55-o
Where
}./
18625
11232
72. Two tracks, 6 ft. apart, cross the Torksey Bridge, and are sup-
ported by single-webbed plate cross-girders 25 ft. long and 14 in. deep.
If the whole of the weight upon a pair of drivers, viz., 10 tons, be directly
transmitted to one of these cross-girders, draw the corresponding shear-
ing-force and bending-moment diagrams (i) if the ends of the cross-
girder are fixed to the bottom flanges of the main girders ; (2) if they
merely rest on the said flanges. Find the maximum deflection of the
cross-girder and the work done in bending it, in each case.
Ans. (i)
- at 13.208 ft. from one end.
EI
Total work of flexure = .-- ft -tons.
73. A swing-bridge consists of the tail end AB, and of a span ffC, of
length i ft., the pivot being at B. The ballast-box of weight ?Fextends
over a length AD (= 2c ft.), and the weight of the bridge from D to B
is w tons per lineal foot. If DB = x, if p is the cost per ton of the
bridge, and if q is the cost per ton of the ballast, show that the total cost
is a minimum when x + c = (- ), and that the corresponding
weight of the ballast is wx( - i J
504 THEORY OF STRUCTURES.
74 Compare, graphically, the shearing forces and bending moments
along the span BC of the bridge in the preceding question when the
bridge is closed, with their values when the bridge is open. What pro-
vision should be made to meet the change in the kind of stress ?
75. Each of the main girders of a railway bridge resting upon two
end supports and five intermediate supports is fixed at the centre sup-
port, is 3 ft. deep throughout, and is designed to carry a uniformly dis-
tributed dead load of t ton and a live load of | ton per lineal foot. The
end spans are each 51 ft. 8 in. and the intermediate spans each 50 ft. in
the clear. Find the reactions at the supports. The girders are single-
webbed and double-flanged ; the flanges are 12 in. wide and equal in
sectional area, the areas for the intermediate spans being 13 sq. in. and
17 sq. in. at the centre and piers respectively. Find the corresponding
moments of resistance and flange stresses, the web being in. thick.
Ans. Reaction at istand 7th supports = I5HMM at 2C ^ an ^ 5th
supports = 43fff til; at 3d and 5th supports = 35|ffl ;
at 4th support = 3%'?/4\ 8 s tons.
At piers = 693 and flange stresses are 3.59 tons per sq. in.
at 2d support, 2.45 at 3d, and 2.83 at 4th.
At centre == 549 and flange stresses in istspan = 3.2 tons
per sq. in., in 2d = 1.3, and in 3d = 1.78.
76. A continuous beam of four equal spans carries a uniformly dis-
tributed load of w intensity per unit of length. The second support is
depressed a certain distance d below the horizontal, and the reaction at
the 2d support is twice that at the ist. Show that the reactions at the
ist, 2d, 3d, 4th, and 5th supports are in the ratio of the numbers 15, 30,
36, 34, and 1 3 ; find d. With this same value of d find the reactions when
one end infixed.
i wP
Ri =
77. A continuous girder of two equal spans (/) is uniformly loaded.
Show that the ends will just touch their supports if the centre support
w/ 4
is raised
78. If ^i , di , d 3 , di are respectively the deflections of the ist, 2d, 3d,
and 4th panel points in question 69, show that the bending moment at
the middle panel point (J/ 4 ) is given by
EXAMPLES. 505
79. A girder supported at the ends is 30 ft. in the clear and carries
two stationary loads, viz., 7 tons concentrated at 6 ft. and 12 tons at 18
ft. from the left support. Find the position and amount of the maxi-
mum deflection, and also the work of flexure. The girder is built up of
plates and angle-irons and is 24 in. deep. If the moment of resistance
due to the web is neglected, and if the intensity of the longitudinal stress
is not to exceed 5 tons per sq. in., _ what should be the flange sectional
area corresponding to the maximum bending moment.
Ans. Max. deflection = f fr 8 %(x 6) 3 ^g&x, where
x = 1 5.34 ft.
67161.6 ,
Work = ----;,_ ft.-tons.
J^LJ.
Sect, area = 10.32 sq. in.
80. Determine the work of flexure and the necessary flange sectional
area at the centre if the girder in the preceding question is subjected
to a uniformly distributed load of 40 tons, instead of the isolated loads.
1540000,
Ans. Work = - - ft. -tons; sect, area = 15 sq. in,
81. (a) The bridge over the Garonne at Langon carries a double
track, is about 695 ft. in length, and consists of three spans, AB, BC, CD.
The two main girders are continuous and rest upon the abutments at
A and D and upon piers at B and C. The effective length of each of
the spans AB, CD is 208 ft. 6 in., and of the centre span J3C 24.3 ft. The
permanent load upon a main girder is 1277 Ibs. per lineal foot, and the
proof load is 2688 Ibs. per lineal foot. Find the reactions at the sup-
ports (i) when the proof load covers the span AB ; (2) when the proof
load covers the span BC '; (3) when the proof load cover the spans AB
and BC '; (4) when the proof load covers the whole girder.
Draw shearing-force and bending-moment diagrams for each case.
(b) At the piers the web is in. thick and 18 ft. in depth, and each
flange is made up of four plates in. thick and 3 ft. wide. Determine
the flange stresses for cases (i) and (3).
(c) The angle-irons connecting the flanges with the web at the pier
are riveted to the former with i^-in. rivets and to the latter with i-in.
rivets. How many of each kind are required in one line per lineal foot
on both sides of the pier at B, 8000 Ibs. per square inch being safe
shearing stress ?
(d) The effective height of the pier at B is 41 ft., its mean thickness
is 14 ft. 9 in., its width is 42 ft. 9 in., and it weighs 125 Ibs. per cubic
foot. If there is no surcharge on the bridge, and if the coefficient of
friction between the sliding surfaces at the top of the pier is taken at
.15, show that the overturning moment due to the dilatation of the
girders is about -fa of the amount of stability of the pier.
506 THEORY OF STRUCTURES.
(e) Find the points of inflexion and also the maximum deflections in
Case 3.
What practical advantage is derived from the calculation of the
deflection ?
Ans.(a) Case i. R\ = 353469.95; R? = 656955.7;
Rs = 280612.55 ; Ri = 109608.77 Ibs. ;
Mi= 12247115.3; M 3 = 4823424.5 ft.-lbs.
Case 2. R, = 68185 2 = R, Ibs. ;
A> 2 = 6791783 = ^? 3 Ibs.;
M*= 13439537.7 ft.-lbs. = Aft.
Case 3. RI = 312982.65 ; 7? 2 = 1024035 ;
Rs = 647691 ; R t = 69121.47 Ibs. ;
Mi= 1565031.2 ; Aft = 8226621.2 ft.-lbs.
Case 4. .#1=422591.42 =y? 4 lbs.;
7? 2 = 1304647. 55 = R 3 Ibs. ;
J/ 2 = 15455566 = J/a ft.-lbs.
(0) /= 2130816; in case i,/ 2 = 7448.9 Ibs. per sq. in.
/ = 2 933-6 " " " "
in case 3,/ 2 = 9400.3 " " " "
73 = 5003.5 " " " "
(Weakening effect of rivet-holes in tension flange is
neglected.)
(?) 9.1 per lineal foot; 11.5 pe^r lineal foot.
(d) Moment of stability = 23833291^11 ft.-lbs. ;
overturning moment = 1919408.8 ft.-lbs.;
ratio = 12.4.
(e) Points of inflexion: in AB, 157.8 ft. from A', in BC,
at a distance x from B given by x* 2581^+10426^
= o; in CD, at 54.1 ft. from D.
Max. deflections:
In AB, (165/5 jr 4 5 2l6 3-7* 8 + 227693091.6*),
where* is given by 66o|* 2 156491. 3* + 227693091. 6=0 ;
In BC, 7(1 65.2** 853829*" + 9977485. 9* 2 10327286968),
xS/
where * is given by * 2 3876* + 30196^ = o.
82. A beam AB of span / carrying a uniformly distributed load of
intensity w is fixed at A and merely supported at B. The end B is
w/ 4
lowered by an amount -r-=- r . Find the reactions. How much must B
ioA/
be lowered so that the whole of the weight may be borne at A?
i w/ 4
Ans. f|w/ at A, -&wl at. ^ ; - .
8 L1
EXAMPLES.
83. Solve the preceding question supposing the fixture at A to be
imperfect, the neutral axis making with the horizontal an angle whose
tangent is -7^7. Ans. %wl, \wl\ -~ -777.
40 L1 4 '*
84. A wrought-iron girder of I-section, 2 ft. deep, with flanges of equal
area and having their joint area equal to that of the web, viz., 48 sq. in.,
carries ^ ton per lineal foot, is 100 ft. long, consists of five equal spans,
and is continuous over six supports. Find the reactions when the third
support is lowered |- in. How much must this support be lowered so
that the reaction may be nil at (a) the ist support ; (b) the 3d ; (c) the
5th ? How much must the support be raised so that the reaction may
be nil at (d} the 2d ; (e) the 4th ; and (/) the 6th support ? E = 16,500
tons.
Ans. R^ = 2 If ; 7? 2 = 1 5f > ^ 8 = STF
j? 4 = Hff ; Rs = 9 T *V ; -#6 = 4 T 5 tons.
() if in.; mff in.; (c) 2ft in. ;
(d) ijft in. ; 00 2^ in. ; (/) 6* in.
85. If the three supports of any two equal consecutive spans of a
continuous girder of any number of spans are depressed below the
horizontal, show that the relation between the three bending moments
at the supports will be unaffected if the depression of the centre support
is a mean between the depressions of the other two supports.
86. A girder consists of two spans AB, BC, each of length /, and is
continuous over a centre pier B. A uniform load of length 2a (< /) and
of intensity W travels over AB. Find the reactions at the supports for
any given position of the load, and show that the bending moment at
aivl f a"* \ i
the centre pier is a maximum and equal to --.[-I -s] when the
3 1/3 V rj
centre of the load is at a distance I j from A.
87. A continuous girder rests upon three supports and consists of
two unequal spans AB (= /,), BC (= / a ). A uniform load of intensity iv
travels over AB, and at a given instant covers a length AD (= r) of the
span. If R lt R 3 are the reactions at A and C, respectively, show that
2 _ / 2 3 T r *
Draw a diagram showing the shearing force in front of the moving
load as it crosses the girder.
88. If the live load in the preceding question may cover both spans,
show that the shearing force at any point D is a maximum when AD
and BC SLVQ loaded and BD unloaded.
Illustrate this force graphically, taking into account the dead load
upon the girder.
508 THEORY OF STRUCTURES.
89. A continuous-girder bridge has a centre span of 300 ft. and two
side spans, each of 200 ft. The dead load upon each of the main girders
is 1250 Ibs. per lineal foot. In one of the side spans there is also an
additional load of 2500 Ibs. per lineal foot upon each girder. Find the
reactions and points of inflexion. How much must the third support
from the loaded end be lowered so that the pressure upon it may be just
zero ?
Ans. Let W = weight on loaded span = 750,000 Ibs.
R, = -ftfc W Ibs. ; Ri = |ff I W Ibs.;
R, = -/WV ^ Ibs. ; Rt = tffr W Ibs.
M, = \V/- W ft.-lbs.; M*= ViV- ^ ft. -Ibs.
Distance of point of inflexion in loaded span from nearest
end support = i62ff ft.
Distance of point of inflexion in unloaded end span from
nearest end support = I45ff ft.
Distance of point of inflexion in intermediate span from
end support in unloaded span is the value of x in the
equation x* f p,r + AJU^QJUL _ o
56350000 W
3d support must be lowered a distance = - ^ -- .
87 El
90. A continuous girder AC consists of two equal spans AB, BC, each
of length /, and carries a uniformly distributed load of intensity w\ upon
AB, and of intensity / 2 upon BC. Determine the bending moments at
the supports, the maximum intermediate bending moments, and the re-
actions (a) when both ends of the girder are fixed ; (b) when one end A
is fixed and the other free.
Ans. Denoting the reactions and bending moments at A, B, C
by Ri , Mi , y? 2 , Mi , R s , M 3 , respectively :
r / 2
(a) Mi = ( 50/1 + w a ) ; M 9 = -- (wi + /a);
48 24
r RS
M 3 = (wi 5^2); M max . in AB -- 1- Mi , in BC
48 2Wi
* r>
= 4-
2W 2 ID
R* = (o/i + 90/2).
ID
/ 2 / a
(b) Mi = -- 3(3^1 o/O; Mi = ^-(o/i + 20/2);
28 28
/? a /? 2
M 3 = o ; M max . mAB = + M lt in BC =
22/i 20/
I20/a).
EXAMPLES. 509
91. In the preceding question, if Wi = iv* = w, find the points of in-
flexion and the maximum deflection in each case and for each span.
Ans. (a) Points of inflexion for AB or BC are given by
6.r>-6;r/ + / 2 = o.
Max. deflection for AB or BC is given by
Ely- (2tx-x*-r),
in which the value of x is found from
(b) Points of inflexion in AB are given by
I4.r 2 i3.r/ + 2/ 2 = o, and in BC by x =
Max. deflection for AB is given by
- Ely = -==-
I DO
and
28.T 2 39/JT + I2/ 2 =0.
Max. deflection for BC is given by
and
28* 3 33.^ + 4/ 3 = o.
92. A continuous girder ^4C consists of two equal spans AB, BC of
15 m. each. Determine the bending moments at the supports, the maxi-
mum intermediate bending moments, and the reactions (a) when the
load upon each span is 3000 k. per metre ; (&) when the load per metre is
3000 k. upon AB and 1000 k. upon BC. Call M\ , M*, M* the bending
moments and R\ , Ri , R 3 the reactions at A, B, C, respectively, and con-
sider three cases, viz., when both ends of the girder are free, when both
ends are fixed, and when one end is free and the other fixed.
Ans. Case I :
(a) M! = o = M 9 ; M* = 84375 km. ; M max . in AB or BC
47460.9375 km.
fii = R 3 = I6875/& ; 7? 2 = 56250^.
(&) Mi = o = M 3 ; M* = 56250 km. ; M ma x. in AB
= 5 8 593-75 km - in BC 7031.25 km.
R l = 18750 k. ; y? 2 = 37500 k.; 7? 8 = 3750 k.
Case II:
(a) Mi = M-! = M = 56250 km.; M max . in AB or BC
= 28125 km.
r>
7?i = - = 7? 3 = 22500 k.
510 THEORY OF STRUCTURES.
(b) Mi 65625 km. ; M* = 37500 km. ; M 3 = 9375
km. M max . in AB = 33398.4375 km., in BC
= 6445.3125 km.;
/?! = 24375 k.; R* = 30000 k.; R 3 = 5625 k.
Case III :
(a) Mi = 48214!- km.; M 9 = 72321$ km.; M s = o.
M ma . v . in AB= 24537111 km., in BC = 52o88-j-f| km.;
R, = 20892^ k.; 7? 2 = 51428^ k.; R 3 = 482 if k.
(b} Mi = 64285! km.; Mi = 401 78$ km.; M, = o.
Mmax. in AB = 32573-^5- km., in BC = 11623-^
km. ;
A'i = 24107! k.; Ri = 3 107 1 } k., R 3 = 4821! k.
93. Show that a uniformly loaded and continuous girder of two equal
spans, with both ends fixed, is 2.08 times as stiff as if the ends were free
and merely rested on the supports.
94. A single weight travels over the span AB of a girder of two equal
spans, AB, BC, continuous over a centre pier B. Show that the reaction
at C is a maximum when the distance of the weight from A is -~ if the
ends A and C rest upon their supports, and when the distance is \AB if
the two ends are fixed. Find the corresponding bending moments at
the central pier.
PJ ">
Ans. --- -PL
95. A girder with both ends fixed carries two equal loads W at points
dividing the girder into segments a, b, c. Determine the reactions and
bending moments at the supports.
+ babe + 3#V + 2<r 3 + 6a^ + 6bc*
(a + b + c} 3
4-
sty yy -
If ll 7 "*
(a + b + cf
c + 2abc 4- be* + ab^
M W 2(ZC
(a + b + r) 2
2 4- 2cibc + <%*b + b"*c
(a + b +
96. A bridge a ft. in the clear is formed of two cantilevers which.
meet in the centre of the span and are connected by a bolt capable of
transmitting a vertical pressure from the one to the other. A weight
W is placed at a distance b from one of the abutments. Find the pres-
sure transmitted from one cantilever to the other, and draw the curve
of bending moments for the loaded cantilever.
Ans. Ri= W\ --, +
'
;.
2 -- );
a*)
EXAMPLES. 5 1 1
97. The weights 7, 7, 10, 10, 10, 10, 8, 8, 8, 8 tons, taken in order pass
over a continuous girder of two spans, each of 50 ft. and fixed at both
ends, the successive intervals being 5, 5, 5, 5, 5, 9, 5, 4, 5 ft. Place the
wheels so as to give the maximum bending moment at the centre sup-
port, and find its value.
Arts. First wheel 25.8399 ft. from nearest abutment ;
Max. B. M. = 306.62 ft.-tons.
98. The bridge over the Grande Baise consists of two equal spans of
19.8 m. ; each of the main girders is continuous and rests upon abut-
ments at the ends. Find the position of the points of inflexion, the
bending moment at the centre support, the maximum intermediate
bending moment, and the maximum flange stress (a) under the dead
load of 1700 k. per lineal metre; (b) under the same dead load together
with an additional proof load of 2000 k. per lineal metre on one span.
The depth of the girder = 3.228 m., and /= .093929232444.
Ans. (a) 14.85 m.from the abutments ; 83308.5 kilogrammetres
(km.); 46,861^ km.; 1.4315 k. per sq. mm.
(b) 16.18 m. from abutment on loaded side; 11.876 m.
from abutment on unloaded side; 132313.5 km.;
101991.65625 km.; 2.27356 k. per sq. mm.
99. The Estressol viaduct consists of four spans of 25 m. ; the main
girders are continuous and their ends rest upon abutments ; the dead
load upon each girder is 1700 k. per lineal metre. Determine the
position of points of inflexion in each span, the reactions and bending
moments at the supports when an additional load of 2000 k. per lineal
metre crosses (a) the ist span; (b) the ist and 2d spans; (<r) all the
spans.
Also, find the absolute maximum bending moments at the inter-
mediate supports.
Ans. Call Xi , x* x*, x^ the distances of points of inflexion in
ist, 2d, 3d, and 4th spans from the ist, 2d, 4th, and
5th supports, respectively; Ri , Ri, R 3 , R 4 , R 5 the
reactions; Mi (= o), M<t, M 3 , M*, Ms> (= o) the
bending moments.
(a) XT. = 20.72 m. ;
x* is given by I7oojr 2 2 424017^2 + 395089! = o;
x z by I7oo.r 3 2 47767^3 + 238839! =o ; jr 4 = 19.38 m.
l?i = 38348 T V k.; R* = 81 i6of k. ; y? 3 = 341071 k.;
R, = 499 I0 f k. ; ^ = 16473^ k.
M-i 197544^ km. ; M 3 = 5357if km. ;
km.
$12 THEORY OF STRUCTURES.
(b) x l = 19.556 m. ; 37ocxr 2 2 103000^3 + 50357^ = o;
I700.T3 2 41071^3 + 205357^ = o ; XL = 20. 168 m.
7?! = 361787 k. ; ^ 2 = 107821^ k.; fi s = 62964! k.;
/? 4 = 45892^ k. ; ^ 5 = i7H2f k.
M* = 258928^ km. ; Ms = 120535!- km.;
M t = 102678^ km.
(e) xi = 19.64 m. ; -r 2 and jr a are given by
i4^ 2 -375.r-f 1875 = 0.
Mt Mi = 247767% km. ; M a = 165178^ km.
Abs. max. B. M. at 2d support (= max. B. M. at 4th sup-
port) occurs when ist, 2d, and 4th spans are loaded,
and = 264508^1 km.
Abs. max. B. M. at 3d support occurs when 2d and 3d
spans are loaded and = 209821^ km.
100. In the preceding question find the absolute maximum flange
unit stress at the piers, / being .093929232444. Ans. 4.5 k. per sq. mm.
101. The Osse iron viaduct consists of seven spans, viz., two end
spans of 28.8 m. and five intermediate spans of 38 m. ; each main girder
is continuous and carries a dead load of 1450 k. per lineal metre. Find
the bending moments at the supports when a proof load of 2250 k. per
lineal metre for each girder covers all the spans; and also find the
absolute maximum bending moment at the fourth support. Is the
following section of sufficient strength ? two equal flanges, each com-
posed of a 6oo-mm. x 8-mm. plate riveted by means of two loo-mm.
x ico-mm. x i2-mm. angles to a 6oo-mm. x lo-mm. vertical web plate
and two 8o-mm. x 8o-mm. x ii-mm. angles riveted to each horizontal
plate with the ends of the horizontal arms 15 mm. from the edges of
the plates; the whole depth of the section being 4.016 m., and the dis-
tance between the web plates, which is open, being 2.8 m. If insuf-
ficient, how would you strengthen it?
Ans. My = 416,518 km.; Ms = 452,790 km.; M^ = 443,722 km.
Max. B. M. = 542,199 km. /= .14074440467.
.-. = .07009183,
cM*
and max. flange stress = = 7.73 k. per sq. mm.
This is much too large. The section may be strengthened
by adding two 6oo-mm. x 8-mm. plates to each flange.
/ is thus increased by .0783425536, and the flange unit
stress becomes 5 k. per sq. mm.
CHAPTER VIII.
PILLARS.
1. Classification. The manner in which a material fails
under pressure depends not merely upon its nature but also
upon its dimensions and form. A short pillar, e.g., a cubical
block, will bear a weight that will almost crush it into powder,
while a thin plank or a metal coin subjected to enormous com-
pression will be only condensed thereby. In designing struts
or posts for bridges and other structures, it must be borne in
mind that such members have to resist buckling and bending in
addition to a direct pressure, and that the tendency to buckle
or bend increases with the ratio of the length of a pillar to its
least transverse dimension.
Hodgkinson, guided by the results of his experiments,
divided all pillars with truly flat and firmly bedded ends into
three classes, viz. :
(A) Short Pillars, of which the ratio of the length to the
diameter is less than 4 or 5 ; these fail under a direct pressure.
(B) Medium Pillars, of which the ratio of the length to the
diameter exceeds 5, and is less than 30 if of cast-iron or tim-
ber, and less than 60 if of wrought-iron ; these fail partly by
crushing and partly by flexure.
(C) Long Pillars, of which the ratio of the length to the
diameter exceeds 30 if of cast-iron or timber, and 60 if of
wrought-iron ; these fail wholly by flexure.
2. Further Deductions from Hodgkinson's Experi-
ments. A pillar with both ends rough from the foundry so
that a load can be applied only at a few isolated points, and a
pillar with a rounded end so that the load can be applied only
513
514 THEORY OF STRUCTURES.
along the axis, are each one-tJiird of the strength of a pillar of
class B, and from one-third to two-thirds of the strength of a
pillar of class C, the pillars being of the same dimensions.
The strength of a pillar with one end flat and the other
round is an arithmetical mean between the strengths of two
pillars of the same dimensions, the one having both ends flat
and the other both ends round.
Disks at the ends of pillars only slightly increase their
strength, but facilitate the formation of connections.
An enlargement of the middle section of a pillar sometimes
increases its strength in a small degree, as in the case of solid
cast-iron pillars with rounded ends which are made stronger
by about one-seventh ; hollow cast-iron pillars are not affected.
The strength of a disk-ended pillar is increased by about one-
eighth or one-ninth when the middle diameter is lengthened by
50 per cent., but for slight enlargements the increase is imper-
ceptible.
The strength of hollow cast-iron pillars is not affected by a
slight variation in the thickness of the metal, as a thin shell is
much harder than a thick one. The excess above or deficiency
below the average thickness should not exceed 25 per cent.
3. Form. According to Hodgkinson, the relative strengths
of long cast-iron pillars of equal weight and length may be
tabulated as follows :
(a) Pillars with flat ends.
The strength of a solid round pillar being 100,
" " square " is 93 ;
" " triangular " is 1 10.
(#) Pillars with round ends, i.e., ends for hinging or pin
connections.
The strength of a hollow cylindrical pillar being 100,
" " an H-shaped . " is 74.6;
" " a +-shaped " is 44.2.
The strengths of a long solid round pillar with flat ends,
and a long hollow cylindrical pillar with round ends, are ap-
proximately in the ratio of 2.3 to I.
The stiff cst kind of wrought-iron strut is a built tube, the
THE FAILURE OF PILLARS. 515
section consisting of a cell or of cells, which may be circular,
rectangular, triangular, or of any convenient form.
In experimenting upon hollow tubes, Hodgkinson found
that, other conditions remaining the same, the circular was the
strongest, and was followed in order of strength by the square
in four compartments j+j ; the rectangle in two compartments,
FT! ; the rectangle, a ; and the square.
The addition of a diaphragm across the middle of the rect-
angle doubled its resistance to crippling.
4. Modes of Failure. The manner in which the crush-
ing of short pillars takes place depends upon the material, and
the failure may be due to splitting, bulging, or buckling.
(a) Splitting into fragments is characteristic of such crys-
talline, fibrous, or granular substances as glass, timber,
stone, brick, and cast-iron.
The 'compressive strength of these substances is
much greater than their tensile strength, and when they
fail they do so suddenly.
A hard vitreous material, e.g., glass or vitrified
brick, splits into a number of prisms (Fig. 335).
A fibrous material, e.g., timber, and granular materials, e.g.,
cast-iron and many kinds of stone and
brick, shear or slide along planes oblique
to the direction of the thrust, and form
one or more wedges or pyramids (Figs.
336, 337, 238)-
Sometimes a granular or a crystalline substance will sud-
denly give way and be reduced to powder.
(b) Bulging, i.e., a lateral spreading out, is characteristic of
blocks of fibrous materials, e.g., wrought-iron, copper, lead, and
timber, and fracture occurs in the form of longitudinal cracks.
All substances, however, even the most crystalline, will
bulge slightly before they fail, if they possess some degree of
toughness.
(c) Buckling is characteristic of fibrous materials, and the
resistance of a pillar to buckling is always less than its resist-
ance to direct crushing, and is independent of length.
Thin malleable plates usually fail by the bending, pucker-
THEORY OF STRUCTURES.
ing, wrinkling, or crumpling up of the fibres, and the same
phenomena may be observed in the case of timber and of long
bars.
Long plate tubes, when compressed longitudinally, first
bend and eventually fail by the buckling of a short length on
the concave side.
The ultimate resistance to buckling of a well-made and
well-shaped tube is about 27,000 Ibs. per square inch section of
metal, which may be increased to 33,000 or 36,000 Ibs. per
square inch by dividing the tube into two or more compart-
ments.
A rectangular wrought-iron or steel tube offers the greatest
resistance to buckling when the mass of the material is con-
centrated at the angles, while the sides consist of thin plates
or lattice-work sufficiently strong to prevent the bending of
the angles.
Timber offers about twice the resistance to crushing when
dry that it does when wet, as the presence of moisture dimin-
ishes the lateral adhesion of the fibres.
5. Uniform Stress. Let a short pillar be subjected to a
w | pressure of W Ibs. uniformly distributed over its
end and acting in the direction of its axis.
Let 5 be the transverse sectional area of the pil-
lar.
W
Let / = be the intensity of stress per unit
O
of area of any transverse section AB.
TIG. 339. J
Let A'B' be any other section of area S', in-
clined to the axis at an angle 0. The intensity of stress per
W W
unit of area of A'B' = = sin 6 = p sin 0, which may be
o o
resolved into a component/ sin 2 normal to A'B', and a com-
ponent/ sin 6 cos 6, i.e.,/ - , parallel to A'B'. The last
intensity is evidently a maximum when 6 = 45, so that the
plane along which the resistance to shearing is least, and there-
fore along which the fracture of a homogeneous material would
tend to take place, makes an angle of 45 with the axis.
UNIFORM L Y'VAR YING STRESS.
517
None of the materials of construction are truly homo-
geneous, and in the case of cast-iron the irregularity of the
texture and the hardness of the skin cause the angle between
the plane of shear and the direction of the thrust to vary
from 32 to 42. Brick chimneys sometimes fail by the shear-
ing of the mortar, the upper portion sliding over an oblique
plane.
Hodgkinson's experiments upon blocks of different mate-
rials led him to infer that the true crushing strength of a ma-
terial is obtained when the ratio of length to diameter is at
least i-J-; for a less ratio the resistance to compression is un-
duly increased by the friction at the surfaces between which
the block is crushed.
6. Uniformly Varying Stress. The load upon a pillar is
rarely, if ever, uniformly distrib-
uted, but it is practically sufficient
to assume that the pressure in
any transverse section varies uni-
formly.
Any variable external force ap-
plied normally to a plane surface
AA of area 5 may be graphically
represented by a cylinder AABB,
the end BB being the locus of the
extremities of ordinates erected
upon A A, each ordinate being pro-
portional to the intensity of press-
ure at the point on which it is
erected.
Let P be the total force upon *AA, and let the line of its
resultant intersect A A in C\ C is the centre of pressure of A A,
and the ordinate CC necessarily passes through the centre of
gravity of the cylinder.
Again, the resultant internal stress developed in AA is P,
and may of course be graphically represented by the same
cylinder AABB.
Assume that the pressure upon AA varies uniformly; the
surface BB is then a plane inclined at a certain ano-le to AA.
FIG. 340.
5 i8
THEORY OF STRUCTURES.
Take 0, the centre of figure of AA, as the origin, and AA
as the plane of x, y.
Let O y, the axis of y, be parallel to
that line EE of the plane BB which is
parallel to the plane AA.
Through EE draw a plane DD par-
allel to AA, and form the cylinder
A ADD.
The two cylinders A ABB and A ADD
are evidently equal in volume, and OF,
the average ordinate, represents the mean
pressure over AA ; let it be denoted
by A-
At any point R of the plane AA,
erect the ordinate RQP, intersecting the
planes DD, BB, in Q and P, respect-
Y/ ively.
FlG - 341- Let x, y be the co-ordinates of R.
The pressure at R
a being a constant depending upon the variation.
Note. The sign of x is negative for points on the left of O,.
and the pressure at a point corresponding to R\sp Q ax.
Let x n , y be the co-ordinates of the centre of pressure C.
Let AS be an elementary area at any point R.
Then pAS is the pressure upon AS, and 2(J>4S) is the
total pressure upon the surface AA, 2 being the symbol of
summation.
Hence,
and
But / = / + ax.
and
axy)AS\
UNIFORMLY VARYING STRESS.
519
Now is the centre of figure of AA, and therefore 2
and ~2(ydS} are each zero.
Also, 2(4 S) = S, 2(x*4S) is the moment of inertia (/) of
A A with respect to OY, and 2Z(xydS) is the product of inertia
(K) about the axis OZ.
.'. x p S = al x P
and
(l)
(2)
Cor. I. Ip any symmetrical
section y Q is zero, and X Q is the
deviation of the centre of pres-
sure C from the centre of fig-
ure a
Let x l be the distance from
O of the extreme points A of
the section.
The greatest stress in A A is/ + ax l = /\, suppose.
FlG " 342 '
A-!
or
^>
A
It is generally advisable, especially in masonry structures,
to limit x by the condition that the stress shall be nowhere
negative, i.e., a tension. Now the minimum stress is/ ax^ ,
so that to fulfil this condition,
p f > or = ax l , But p l = ax,
< or = 2/..
520 THEORY OF STRUCTURES.
Hence, by eq. (3),
A, i
and therefore
I
< or = I ; i.e., x, < or = -.
Cor. 2. The uniformly varying stress is equivalent to a
single force P along the axis, and a couple of moment
= a Vr + K\
Cor. 3. The line CO is said to be conjugate to OY.
If the angle COX = 0, then cotV = = ~.
7. Hodgkinson's Formulae for the Ultimate Strength of
Long and Medium Pillars. When a Jong pillar is subjected to
a crushing force it first yields sideways, and eventually breaks
in a manner apparently similar to the fracture of a beam under
a transverse load. This similarity, however, is modified by the
fact that an initial longitudinal compression is induced in the
pillar by the superimposed load.
Hodgkinson deduced, experimentally, that the strength of
long solid round iron and square timber pillars, with flat and
firmly bedded ends, is given by an expression of the form
W being the breaking weight in tons of 2240 Ibs. ;
d " " diameter or side of the pillar in inches ;
I " " length of the pillar in feet;
n and m being numerical indices ;
FORMULAE FOR ULTIMATE STRENGTH OF PILLARS. $21
A being a constant varying with the material and with the
sectional form of the pillar.
. For iron pillars ............... n = 3.6 and m = 1.7
" timber pillars ............ n = 4 and m = 2
" cast-iron ............................ A = 44.16
" wrought-iron ........................ A = 133.75
" dry Dantzic oak ..................... A 10.95
" dry red deal ......................... A 7.81
" dry French oak ...................... A 6*9
The strength of long hollow round cast-iron pillars was
found to be given by
- d
w -44.34 -r-
d being the external and d l the internal diameter, both in
inches.
Thus, the strength of a hollow cast-iron pillar is approxi-
mately equal to the difference between the strengths of two
solid cast-iron pillars whose diameters are equal to the external
and internal diameters of the hollow pillar.
The strength of medium pillars may be obtained by the
formula
w Wfs
-
W being the breaking weight in tons of 2240 Ibs. ;
W " " " " " " " " " as derived
from the formula for long pillars ;
/being the ultimate crushing strength in tons per square inch ;
S being the sectional area of the pillar in square inches.
Again, if the ends of a cast-iron pillar are rounded, the
above formulae may be still employed to determine its strength,
A being 14.9 for a solid and 13 for a hollow pillar.
522 THEORY OF STRUCTURES.
8. Gordon's Formula for the Ultimate Strength of a
Pillar. The method discussed in the preceding articles, being
practically very inconvenient, is not generally used,
and the present article will treat of Professor Gordon's
formula, which has a better theoretical basis and is
easier of application.
The effect of a weight W upon a pillar of length /
;2/ n and sectional area 5 may be divided into two parts :
(a) A direct thrust, which produces a uniform com-
W
pression of intensity -= = p l .
(b) A bending moment, which causes the pillar to
FIG. 343. yield in the direction of its least dimension (k).
Letjy be the greatest deviation of the pillar from
the vertical.
The bending moment M at the point of maximum stress
may be represented by Wy.
Let / 2 be the stress in the extreme layers due to this bend-
ing moment.
Now
c being the distance of the layer under consideration from
the neutral axis, ju. a constant depending upon the sectional
form, and b the dimension perpendicular to the plane of flexure.
Wv
and
Butj/cc. (Art. 9, Chap. VI.)!
wr wr / a
' and
VALUES OF a AND / GORDON'S FORMULA. $2$
a being some constant to be determined by experiment.
Hence, the total stress in the most strained fibre is
or
which is Gordon's formula.
Cor. If the weight- upon the pillar causes the stress in any
transverse section to vary uniformly, the direct thrust in the
W\ *2' S ] W
extreme layers is -~r\i -| * / instead of -~, (Cor. i, Art.
J 1 O
6,) ;F O being the greatest deviation of the line of resultant
thrust from the axis of the pillar.
Let k be the radius of gyration of the cross-section. Then
and the expression for the direct thrust may be written
W
Hence, Gordon's formula becomes
/
-
5 A
9. Values of a and /. The following table, giving the
values of the constants a and/* in Gordon's formula, has been
524
THEORY OF STRUCTURES.
prepared by taking an average of the best known results, and
is applicable to round and square pillars with square ends.
f in Ibs.
per sq. in.
a
For cast-iron solid rectangular pillars
80 ooo
-r\r*
" " " round "
80 ooo
ff
T JU
" " hollow rectangular "
80 ooo
TOU
*4
" " " round "
80 ooo
,ri*
For wrought-iron solid rectangular pillars . .
TTO
-JU.
" " " round "
36 ooo
^(TTJT
JL*
" " thick hollow round "
36 ooo
*^M
6? 2OO
^FffT
_JL-
" round " ',
67 2OO
^OtTTT
*JU.
67 2OO
TTTTTT
, 1,^.
For strong-steel solid rectangular pillars
1 14 ooo
^?oiy
d&
" " " round "
1 14 ooo
1400
JL-
" " hollow round "
114 ooo
^DTT
i .
5 ooo
T^ffTJ
_1I
" " round "
5 ooo
^^^
_1_
7 2OO
2s ff
_i
^0
If Gordon's formula is applied to pillars with pin ends, 4*2
takes the place of a ; and if to pillars with one pin end and one
square end, |^z takes the place of a.
10. Graphical Comparison of the Crushing Unit Strength
of Solid Round Cast-iron, Wrought-iron, and Mild-steel
Pillars.
The crushing unit stress is given by/ =
Take the different values of j as abscissae, and the corre-
sponding values of / as ordinates ; the resulting curves are
shown in Fig. 344.
Hence, the strength of a mild-steel pillar always exceeds
that of a wrought-iron pillar but is less than that of a cast-iron
pillar when j < 10.7 ; a wrought-iron pillar is stronger or weaker
than a cast-iron pillar according as r > or < 28.5.
APPLICATIONS OF GORDON'S FORMULA.
80,000 Ibs.
28,486.4 Ibs.
in
40 50 60
FIG. 344.
ii. Application of Gordon's Formula to Pillars of other
Sectional Forms.
In any section whatever, the least transverse dimension for
calculation (i.e., ti) is to be measured in the plane of greatest
flexure.
Thus, it may be taken as the least diameter of the rectangle
circumscribing tee (Fig. 345), channel (Fig. 346), and cruciform
(Fig. 347) sections, and as the perpendicular from the angle to the
opposite side of a triangle circumscribing angle(F\g. 348) sections.
FIG. 345 .
FIG. 34 6.
FIG. 347.
FIG. 348.
From a series of experiments upon wrought-iron pillars of
these sections, f was found to be 42,500 Ibs., and a, .
QOO
In cast-iron struts of a cruciform section /= 80,000 Ibs.
and a -.
400
526 THEORY OF STRUCTURES.
These results are only approximately true, and apply to
pillars fixed at both ends.
12. Rankine's Modification of Gordon's Formula. The
factor a in Gordon's formula is by no means constant, and not
only varies with the nature of the material, with the length of
the pillar, with the condition of its ends, etc., but also with the
sectional form of the pillar. The variation due to this latter
cause may be eliminated, and the formula rendered somewhat
more exact, by introducing the least radius of gyration instead
of the least transverse dimension.
If k is the least radius of gyration,
/ mbh* __ m
r* *
- / T
mass non n
m and n being constants which depend upon the sectional
form. Thus, Gordon's formula for pillars with square ends
may be written
.+:, j
in which , is independent of the sectional form, all variations
of the latter being included in k*. This modified form of
Gordon's formula was first suggested by Rankine.
4a l is substituted for a l if the pillar has two pin ends, and
Q
-a t or 2#, is substituted for a l if the pillar has one pin end and
one square end.
Rankine gives
for wrought iron, f =. 36000 Ibs., - = 36000;
for cast-iron, /= 80000 Ibs., - = 6400 ;
for dry timber, /= 7200 Ibs., - == 3000;
a,
KANKINE'S MODIFICATION OF GORDON'S FORMULA. $2?
In good American practice the safe working unit stress in
bridge compression members is determined by the formula
Safe working unit stress =
f being 8000 Ibs. for wrought-iron and 10,000 Ibs. for steel,
and being 40,000 for two square ends, 30,000 for one square
and one pin end, and 20,000 for two pin ends.
Another formula often employed is,
H\ . /'
/ Tf\
Working stress in Ibs. persq. in. X (4 + ) ==
H being the ratio of length to least breadth, where, in the case
of wrought-iron,
f = 38,500 Ibs. and = 5820 for two square ends;
f 38,500 " " - = 3000 " one square and one pin end.
/' = 37,800 " " - = 1900 " two pin ends.
r_r
^\\e factor of safety, viz., 4 -| , increases with H, and par-
tially provides for the corresponding decrease in the strength
to resist side blows.
EXAMPLES. According to Rankine the ultimate compres-
sive strength of wrought-iron struts, in pounds per square
inch, is
W 36000
36000 *
528 THEORY OF STRUCTURES.
If the section is a solid rectangle, k? = , and hence
36000
r
1 3000 h*
If the section is a solid circle, / 2 = -^, and hence
36000
A=
7-.-
1 2250 tf
If the section is a thin annulus, k* = -^ nearly, and hence
36000
1 +4500^
_If - is small, W=fS.
If is large, ^. ,.
Comparing the last result with eq. (5), Case 4, Art. 16,
which gives a theoretical value of a t , the actual value being
somewhat different.
13. Values of fc a for Different Sections.
(a) Solid rectangle : k* = -~ = , h being the least dimen-
O 1 2
sion.
tx\ if n i v 7 I ( bh * - b ' h '*\
(b) Hollow rectangle: k = -^ = 7^ bh b'h'r ' g
the greatest and least outside dimensions, and b', h' the great-
est and least inside dimensions, respectively.
VALUES OF &* FOR DIFFERENT SECTIONS. $2$
Let / be the thickness of the metal. Then
b' = b 2t and h' = h 2t,
and hence
I bh*-(b-2i)(h-2t}* V
~ 12 bh (b 2t)(h 2t)' 12 b + h>
approximately, when t is small compared with h, i.e., for a thin
hollow rectangle.
For a square cell, J? = -g-.
(c) Solid triangle : ft* = -~- = =, h being the height.
(d] Hollow triangle : 1? = -^- = - , . ^ , 7 . , , ^, ^ being the
base and height of the outside triangle, and ', h' the base and
height of the inside triangle, respectively. Also, T> = 77.
_ ^ P - V* i__tf IP + b f *\
- k ^i$F -&'*?- is\ y r
Hence, for a thin triangular cell, &* = .
r ;r2
(e) Solid cylinder : * = = ;, k being the diameter.
(/) Hollow cylinder: ff = ^ = ~^( A ' + h '^ h and h ' bein g
the external and internal diameters, respectively.
Hence, for a thin cylindrical cell, & = , approximately.
EXAMPLE. Gordon's formula for hollow cylindrical cast-
iron pillars is
w__ _L _ _ _ /_
~ " a>
.
h 3
500 / 3 4000
530 THEORY OF STRUCTURES.
The relation /, = -- - may be assumed to hold for
hollow square struts and also for struts of a cruciform section.
Ex. i. For a hollow square having its diagonal equal to
the internal diameter of the hollow cylinder, i.e., k ',
=' and ' =
Ex. 2. If the side of the square is equal to the external
diameter, i.e., h, then
k* = -z- , and /, =
6 ' ^ ~ 3 /' '
(g) Cruciform section, the arms being equal:
_
/ = h ; S zbk ^ .
12 ' 12 12
~ = ' nearly -
FIG. 349.
Hence, the formula for a cast-iron pillar of cruciform section
may be written
(/z) Angle-iron of unequal ribs, the greater being and the
less A :
VALUES OF &* FOR DIFFERENT SECTIONS.
53 1
ra
Hence, if b = h, i.e., if the ribs are equal, ft? .
(i) Channel-iron, the dimensions being as in Fig. 350 :
2bht\h + tY
T
12 ~ 4(2At + 6t)
2ht . 2bhf
( 2t
= h \ 71 + 4
Also, 5 = bt + 2ht.
=] IF
' nearly '
_ _ ___
1 2(2 At + bt) 4(2 At + bt}* ) '
Let the area of the two flanges A = 2ht, and let the
area of the web =. B bt. Then
_ -
12(A
(fc) \\-iron, breadth of flanges being b, length of web h, and
thickness of metal / :
73j 1 .3 73.
/ = 2~ + = 2--, nearly ; 5=
At.
*' /P ~ I22bt
2bt
I2A+B'
A being the area of the flanges, and .5 the area of the web.
(/) Circular segment, of radius r and length rO :
Hence, for a semicircle, since = ?r,
532 THEORY OF STRUCTURES.
(m) Barlow rail: tf = -- , nearly.
(n) Two Barlow rails, riveted base to base: a = .393;-*,
nearly.
14. American Iron Columns. In 1880 Mr. G. Bouscaren
read before the American Society of Civil Engineers a paper
containing the results of a series of experiments made for the
Cincinnati Southern Railroad upon Keystone, square, Phoenix,
and American Bridge Co.'s columns.
EYSTONE SQUARE PHOENIX ' AM. BRIDGE CO.
FIG. 351. FIG. 352. FIG. 353. FIG. 354.
These experiments show, as those of Hodgkinson and
others have also shown, that the strength of iron and steel
columns is not only dependent on the ratio of length to diam-
eter, and on the forfh of the cross-section, but also on the
proportions of parts, details of design and workmanship, and
on the quality of the material of which the columns are con-
structed.
Further, they seem to lead to the conclusions that Gordon's
formula is more correct as modified by Rankine, and that, in
the case of columns hinged at both ends, Rankine's formula,
with tfj assumed at double the value it has when the formula is
applied to columns with flat ends, is practically correct.
The subjoined table gives the values of the constants
#, and / as deduced from Bouscaren's experiments by Prof.
W. H. Burr.
In 1 88 1 Messrs. Clarke, Reeves & Co. presented to the
American Society of Civil Engineers a paper containing the
results of experiments upon twenty Phcenix columns, which
appeared to show that neither Gordon's nor Rankine's formula
expressed the true strength of a column of the Phcenix type.
In the discussion that followed the reading of this paper, how-
ever, it was demonstrated that, within the range of the experi-
ments, the strength of intermediate lengths and sections of
AMERICAN IRON COLUMNS.
533
/ in Ibs.
For keystone columns with flat ends swelled
" " " " " " straight (open or
closed)
" " " " " " open (swelled straight)
" pin ends swelled
For square columns with flat ends
" " " " pin ends
For Phoenix columns with flat ends
" " '* " round ends
" " " " pin ends
For American Bridge Co.'s columns with flat ends
" " " " " " round ends
" " 4< " " " pin ends
36,000
39>500
38,300
38,300
39,000
39.000
42,000
42,000
42,000
36,000
36,000
36,000
18300
1
18300
1
12000
1
35000
1
17000
1
50000
1
12500
1
22700
1
46000
1
11500
1
21500
Phoenix columns can be obtained either from Rankine's for-
mula by slightly changing the constants, or from very simple
new formulae.
Mr. W. G. Bouscaren showed that by making a. =
i ooooo
W
and/= 38000, the calculated values of -^- agree very nearly
o
with the actual experimental results.
Mr. D. J. Whittemore gave the following (only applicable
for lengths varying from 5 to 45 diameters) as expressing the
probable ultimate strength of these columns :
Wlbs. = (1200 /T)3
H being the ratio of length to diameter.
Mr. C. E. Emery stated that the ultimate strength in each
case is approximately represented by the formula
M/1K 355Q63 + 3095Q#
H+6.i 7 $ -'
H being the ratio of length to diameter.
534 THEORY OF STRUCTURES.
Taking the different values of H as abscissae, and of W as
ordinates, this is the equation of an hyperbola. It agrees very
accurately with the experimental results from 20 diameters
upwards; at 15 diameters the calculated values of W arc
greater than those given by the experiments ; for a less num-
ber of diameters the experimental results are the higher, but
the variations are slight, and are provided for in the factor of
safety.
The following very simple formulae, due to Prof. W. H.
Burr, give results agreeing closely with those obtained in the
experiments :
For values of j < 30, the ultimate strength in pounds per
square inch
= 64700 4600. IJL
V k '
For values of T between 30 and 140, the ultimate strength
in pounds per square inch
= 39640 46 j,
k being the radius of gyration.
15. Long Thin Pillar. Let ACB be the bent axis of a
thin pillar of length /, having two pin ends and carry-
ing a load W at B.
Let d be the greatest deviation of the axis from the
vertical. Then
E
Wd bending moment = ^,-7, . . (i)
being the curvature of the pillar and 7 the moment
K
v of inertia of the most strained transverse section.
FJG. 355 . This equation is only true on the assumptions that
(1) initially, the pillar is perfectly straight ;
(2) initially, the line of action of the load coincides with the
axis of the pillar ;
LONG THIN PILLAR. 535
(3) the material of the pillar is homogeneous.
These assumptions cannot be fulfilled in practice, and varia-
tions from theoretical accuracy may, perhaps, be provided for
by supposing that the line of action of the load is at a small
distance x from the axis of the pillar. The bending-moment
equation then becomes
f^ being the skin stress due to bending at a distance c from the
neutral axis.
Again, assuming that the bent axis is in the form of an arc
of a circle,
f ='/!, ... . . (4)
and consequently
Wx
d =-pI^W> ' (5)
where
' "'^ ' " ''' ''^" P =^f- (Q
If the line of action of the load W coincided with the axis
of the pillar, then x would be nil.
Hence, by eq. (5), so long as the load is less than P, d o,
and the failure of the pillar would be due to direct crush-
ing. If the load is equal to P, d would become indeterminate
I ) and the pillar would remain in a state of neutral equi-
librium at any inclination to the vertical.
It is impossible that W should exceed P, as d would then
be negative ; and therefore a load greater than P would cause
the pillar to bend over laterally until it broke.
536 THEORY OF STRUCTURES.
Thus, P ~ JT- must be the theoretical maximum compres-
sive strength of the pillar.
Again, let A be the area of the section under consideration ;
" p be the total intensity of the skin stress at the
section ;
" f be the intensity of the direct stress due to W
W_
~~ A ;
" /", be the intensity of the stress due to P
P
Then
the sign of f^ being positive for the compressed side of the
pillar and negative for the side in tension.
(8)
k being the radius of gyration.
Let h be the least transverse dimension of the section in
the plane of flexure. Then
c oc h and k also a h.
c _n
*' fi = ~h '
n being a coefficient depending upon the form of the section.
For a rectangle, n = 6 ; for a circle, n = 8 ; also,
Px
LONG THIN PILLAR. 537
Thus, however small x may be,/ continually increases as
the difference between f l and f diminishes. The pillar will
therefore fail for some value of / less than the theoretical
maximum. This is in accordance with experience, as it is
found that a small load causes a moderate flexure in a long
pillar, and that this flexure gradually increases with the load
until fracture takes place.
In no case should / exceed the elastic limit, as in such
case a set would be produced and the deviation x would be
increased.
If the tensile strength of the material of the pillar is small,
as in the case of cast-iron, failure may arise from the tearing of
the stretched layers.
Cor. i. The above also applies to the case of a pillar with
one end fixed and the other free, but the value of P is
2EI
then -^- .
Cor. 2. According to Euler (see following article), the
7T 2
more correct value of P is pEIjt, ^ being I, 2, J, or 4, accord-
ing as the pillar has two pin ends, one fixed end and one end
guided in the direction of the thrust, one fixed and one free
end, or two fixed ends.
El & fh^
P evidently a ^ a EAj$ a -A\j
Hence, (a) the strength of a long pillar is proportional to
the coefficient of elasticity ; (b) the strengths of similar pillars
are as the sectional areas.
Again, f t oc -^ oc d.
But Wd = --/ a / 2 a d.
Hence W is approximately constant, and the weight which
produces moderate flexure is approximately equal to the break-
ing weight.
EXAMPLE. Find the crushing load of a solid mild-steel
pillar 3 in. in diameter and 10 ft. long, with two pin ends.
THEORY OF STRUCTURES.
Also find the deviation (x) of the line of action of a load of
20,000 Ibs. from the axis of the pillar, so that the maximum
intensity of stress may not exceed 10,000 Ibs. per square inch.
By Gordon's formula and the table, page 524,
the crushing load = --r- Viiw = 85292.3 Ibs.
~T TTTFTV -3 y
Again, the theoretical maximum compressive strength P
8 X 28000000 n(tf
~5T 6l875lbs -
/, 61875 99
* P- w /,-/ 41875 67*
Hence
20000 /
10000 = -3-3-
or x = .65 in.
16. Long Columns of Uniform Section. (Ruler's Theory.)
CASE I. Columns with both ends hinged.
f The column OA of length / is bent
[k , under a thrust P and takes the curved
form OMA.
Take O as the origin, the vertical
through O as the axis of x, and the hori-
zontal through O as the axis of y.
Consider a section at any point M
(x,y). If there is equilibrium and if the
line of action of P coincides with the
TT~" axis of the column, the equation of mo-
FIG. 35 6. ments at M is
; :; : ; . : - EI & = M =**> ::.::.,'_,
or
d'y P
LONG COLUMNS OF UNIFORM SECTION. 539
dy
Multiplying each side of the equation by - and integrating,
(2)
b being a constant of integration.
dy adx
Integrating,
sm-'.l-J =
or
y = b sin (ax + c), . . . . (3)
c being a constant of integration.
When x = o, y is also o, and hence b = o or c = o.
If b = o, y is always o, and lateral flexure is impossible.
Take c = o. Then
jj/ = sin ax (4)
Also, when x OA = OMA, nearly, = /, y = o.
.*. o = b sin al,
or
and hence
P=n*El"t (5)
Now the /ft&rf value of P evidently corresponds to n = i r
and hence the minimum thrust which will bend the column
laterally is
540 THEORY OF STRUCTURES.
Cor. I. If the column is made to pass through N points
dividing the vertical OA into N + I equal divisions, then
y = o when x =
and therefore, by eq. (4),
al
or
al
and hence
i) 3 .
FIG. 357.
As before, the least value of P corresponds to n = I, and
is the least force which will bend the column laterally.
Hence, the strength of the column is increased in the ratio
of 4, 9, 16, etc., by causing it to pass through points which divide
its length into 2, 3, 4, etc., equal parts, respectively.
Cor. 2. The value of b may be approximately determined
as follows : \
Let ds = length of element at M.
Let = inclination to vertical of tangent at M.
Then
pressure upon ds P cos 6 = P-r-
and the
. ., S , ,
compression of ds = ds = -=r-;dx,
A being the sectional area of the column.
Hence, the total diminution of the length of the column
!'
. 7 />
: / rf*__ /.
LONG COLUMNS OF UNIFORM SECTION. 541
Again, the length of the column
= f'( l + ^^) dx = f (l + a ' 6 ' cos ' ax} ' dx '
= / ( l H -- cos2 ax\dx, approximately,
Hence, if L is the initial length of the column, i.e., the
length before compression,
and consequently
EIL-l\ I
b = 2 ~
CASE 2. Columns with one end fixed and the other constrained
to lie in the same vertical.
Assume that the lateral deviation is prevented ^
by means of a horizontal force H at the top of a
column. Then
TO ! \
M
-*).. . . (I)
MW
A particular solution of this is
I
,, - FIG. 358.
^ y d u
54 2 THEORY OF STRUCTURES.
and eq. (i) becomes
<T*
~ EI d? = Pu >
or
d*u
-&=-* *, *VV (2)
The solution of the last equation is
y / = u = b sin (ax + c), . . : , ' " v" . (3)
b and c being constants of integration.
TT
(4)
dy
But - = o when x = o,
and j = o when ^r = o and when x /.
rr
.* 0= , -\-abcosc\
TT
O = sn
Hence
<2/-|- ^ = o and al = tan c = tan at,
and therefore
= 4-493
= / V /'
which may be written in the form
L
/
(5)
LONG COLUMNS OF UNIFORM SECTION.
It is sufficiently approximate to write
El
543
P= 27T 5
CASE 3. Columns with one end fixed and the other free.
A rigid arm AB is connected with the
free end A of a column, and a vertical B
force P applied at B bends the column
laterally, until its axis assumes the curved
form OMA.
Let AB q, A C = p, and let / be the
length of the column, OC, nearly.
The inclination of AB to the horizon
is so small that the difference in length
between AB and its horizontal projection
may be disregarded. The moment equa-
tion at any point M (x, y) is
or
(i)
dy
Multiplying each side by 2 and integrating,
b being a constant of integration.
dy
But -T- = o when y = o, and hence b = o.
(2)
or
544 THEORY OF STRUCTURES.
Integrating,
c being a constant of integration, or
4) 4- a y
But y = o when x = o, and hence c = o.
.*. p = cos^. .<,;..... . . (4)
Also, y ~ p when ^- = /.
.*. -: cos a/. (5)
If q is very small or nil, the term ~rq~ may be disregarded,
and then
o cos al.
being a whole odd number.
The least value of P corresponds to n = i, and the minimum
pressure which will cause the column to bend laterally is
Cor. I. By eq. (5) the deviation of the top of the column
from the vertical is
i cos al
LONG COLUMNS OF UNIFORM SECTION. 54$
Cor. 2. Let the force applied at B be oblique and let its.
vertical and horizontal components be P and //, respectively.
The moment equation now becomes
A particular solution of this is
*) ..... (10)
Let y = y 1 -f- u.
Substituting in eq. (9),
or
The solution of this equation is
u = b sin (ax -f- c) y y' t
b and c being constants of integration.
TT
(12)
When x = o, / and ^ are each = o ; and when x = /, y = q+
TT
Hence, o =/ + ?+ /-(- ^ sin ^;
rr
O = ^ -\- al> cos c ;
546 THEORY OF STRUCTURES.
three equations giving b, c, and p, and therefore fully deter-
mining y.
CASE 4. Column with both ends fixed.
Let IJL be the end moment of fixture. Then
or
-*> + * = (*- JO, . . (l)
,
where b = -p.
FW. 360. Multiplying each side of the equation by 2^ and
integrating,
d being a constant of integration.
But -- = o when ^ o, and hence d
y*) ...... (2)
or
I Integrating,
or
-^ = cos (** + <:), (3)
c being a constant of integration.
LONG COLUMNS OF UNIFORM SECTION. 547
But jj/ = o when x o and when x I. Hence
i = cos c and I = cos (al-\- c).
and therefore c = o and al =
n being a whole number. Hence,
or
The least value of P corresponds to n = I, and the mini-
mum thrust which will cause the column to bend laterally is
17. Remarks. From the preceding it appears that the
maximum theoretical compressive strength of a column per
unit of area may be expressed in the form
k being the radius of gyration, and A a coefficient whose value
is i, 2, J, or 4, according as the column has two hinged ends,
one end fixed and the other guided in the direction of thrust, one
end fixed and the other free, or two fixed ends.
This formula is easy of application, but Hodgkinson's
experiments show that the value of P as derived therefrom is
too large. This may be partly due to the assumption that the
elasticity of the material is perfect.
The factors of safety to be used with this formula vary
from 4 to 8 for iron and steel and from 4 to 15 for timber.
The objection to the use of flat bars as compression mem-
bers has sometimes been overestimated.
Consider, e.g., the case of a flat bar hinged at both ends.
THEORY OF STRUCTURES.
Let the coefficient of elasticity of the material be 25,000,-
ooo Ibs.
Let the working stress per square inch be 8000 Ibs.
The bar will not bend laterally under pressure so long as
7T
the unit stress < Etft, and
jrV /
8000 < 25000000- 7 f , or -7 < 50.7.
Hence, the length of a flat bar in compression seems to be
comparatively limited. If, however, both ends are securely
fixed, the strength is quadrupled and the admissible length of
bar is doubled, while it may be still further increased by fixing
the bar at intermediate points as indicated in Corollary i, page
540. This shows the marked advantage to be gained by rivet-
ing together the diagonals of lattice-girders at the points where
they cross each other.
p
The value of f . . r (Art. 15) must not exceed the elastic
./i
limit. It is difficult to define with any degree of accuracy the
elastic limit of cast-iron and timber. It is claimed, indeed, that
the latter has no elastic limit, properly so called, but that a
permanent set is produced by every elastic change of form. It
may be assumed, however, that the elasticity of these materials
is practically unaffected so long as they are not loaded to more
than one half of the ultimate crushing load.
Hence, taking
E = 29,000,000 Ibs. and f = 20,000 Ibs. for wrought-iron r
" = 29,000,000 " " /= 33,600" " soft steel,
E 29,000,000 " " / = 56,000 " " hard steel,
=17,000,000 " " / = 40,000 " " cast-iron,
= 1,500,000 " " /= 3,600 " " dry timber,
the pillars will not bend laterally unless the ratio of -> or
a 2r
LONG COLUMNS OF UNIFORM SECTION. 549
(d being the shortest side of a rectangular section and r the
radius of a circular section) exceeds the values given in the
following table :
Material. Value of -4- Value of . Formula.
a 27"
Wrought-iron 34.5 29.9
Soft steel 26.6 24.3
Hard steel 20.3 17.9 } f = ' =
Cast-iron 18.7 16.2
Dry timber 18.5 16
Wrought-iron 48.8 42.3
Soft steel 37-7 34-4
Hard steel 28.8 25.3 V ^ _ _
Cast-iron 26.4 22.9
Dry timber 26. i 22. 7
Wrought-iron 17.2 14.9
Soft steel 13.3 12. i
Hard steel 10.1 8.9
Cast-iron 9.3 8.1
Dry timber 9.2 8
Wrought-iron 69 59.9
Soft steel 53.3 48.7
Hard steel 40.7 35.7 \ f ^.
Cast-iron 374 32.4
Dry timber 37 32
Baker has deduced by experiment the following formulae
for the strength of wrought-iron and steel pillars of from 10 to
30 diameters in length and with fixed ends, the tensile strength
of the metals ranging from 20 to 60 tons (2240 Ibs.) per square
inch :
Let t be the tensile strength of the iron or steel, and H the
ratio of length to diameter.
Then the ultimate compressive resistance, in pounds per
square inch,
for solid round pillars = (.4 .oo6Ii)(t + 18) ;
for thin tubes = (.44 .OO4//X/ + 18) ;
for tubes with stiffening ribs = (.44 .OO2//)(V 4- 18) ;
for girder sections = (.4 - .oO4//)'/-f- 18).
55O THEORY OF STRUCTURES.
18. Weyrauch's Theory of the Resistance to Buckling.
In order to make allowance for buckling, Weyrauch pro-
poses the two following methods :
METHOD I. Let F l be the necessary sectional area, and .
the admissible unit stress for a strut subjected to loads vary-
ing from a maximum compression B l to a minimum com-
pression By
Let/ 7 ' be the necessary sectional area, and b' the admissible
unit stress for a strut subjected to loads which vary between a
given maximum tension and a given maximum compression,
B' being the numerically absolute maximum load, and B" the
maximum load of the opposite kind.
According to Art. 7, Chap. Ill, if there is no tendency to
buckling,
* = !=T-^IH ..... c>
1 4+-.f) - ,. .
and
"
If there is a tendency to buckling, let / be the length of
the strut, F its required sectional area, and T the mean unit
.stress at the moment of buckling.
Then, according to the theory of long struts,
(3)
d being a coefficient depending upon the method adopted for
securing the ends, E the coefficient of elasticity, and / the
least moment of inertia of the section.
Also, let t be the statical compressive strength of the ma-
terial of the strut, and take t = j*T. Then
RESISTANCE TO BUCKLING WE YRA UCH'S THEORY. 55 1
where <r = ................ (5)
If the strut under a pressure B were not liable to buckling,
it would be subjected to a direct thrust only. The required
B
sectional area of the strut would then be , and the unit stress
B
for an area F would be -=; .
r
If the strut under the pressure B is liable to buckling, its
ry
required sectional area will be = , since T is the mean unit
stress at the moment of buckling. Let x be the unit stress at
the moment of buckling, for the area F.
Assuming that the unit stresses in the two cases are in the
same ratio as the required sectional areas, then
.... -
* '' F :: T '' /'
B t B
The force which, when uniformly distributed over the area
F, will produce this stress, is Fx juB.
Hence, allowance may be made for buckling by substitut-
ing for the compressive forces in equations (i) and (2), their
values multiplied by yu. Thus, equation (i) becomes
'= f =
and equation (2) becomes
F' = r = -. - P"' ^ B' 1S a com P ress i n >
552 THEORY OF STRUCTURES.
and
n/ D/
F ,, = - ~~\> if B " is a compression. (9)
If /* < I, equations (i) and (2) give larger sectional areas
than equations (7), (8), and (9), so that the latter are to be ap-
plied only when /* > I.
METHOD II. General formulae applicable to all values of /*
may be obtained by following the same line of reasoning as
that adopted in the proof of Gordon's formula. It is there
assumed that the total unit stress in the most strained fibre is
72\
A ( l +*!/ A Dem g tne stress due to direct compression, and
r
p^a-j^ that due to the bending action.
So, instead of employing equations (i) and (2) when ju < i,
and equations (7), (8), and (9) when ^ > i, formulae including
all cases may be obtained by substituting for the compressive
forces in equations (i) and (2) their values multiplied by I
Thus, equation (i) becomes
and equation (2) becomes
F= 7 g?7 {, if B' is a compression, (n)
V '(,_ m '
or
B'
com P ress i n -
Equations (7), (8), (9), respectively, give larger values of F
than the corresponding equations (10), (u), and (12).
RESISTANCE 7V BUCKLING IVEYRAUCH' S THEORY. 553
Note. For wrought-iron bars it may be assumed, as in Arts.
5, 6, Chap. Ill, tha't z/, = z/ == 700 k. per sq. cm., and m^ = m
=1.
The value of <r is given by formula (5), but is unreliable, and
varies in practice from 10,000 to 36,000 for struts with fixed
ends.
When the ends are fixed, 8 47r a , according to theory.
Hence,
o- = 4*' _.
Therefore, if E =. 2,000,000 k. per sq. cm., and t 3300 k.
per sq. cm., $ 23,926, or in round numbers 23,900 ; 24,000
is the value usually adopted by Weyrauch.
EXAMPLE. The load upon a wrought-iron column 360 cm.
long varies between a compression of 50,000 k. and a compres-
sion of 25,000 k. Calculate the sectional area of the column,
assuming it to be first solid and second hollow, allowance being
made for buckling.
First. By eq. (i),
50000 _ _400
700(1 +ixttW)~' 7
r being the radius of the section.
Also, /= .
4
r 50
Hence, by eq. (4),
360 X 360 ii
P - - x = 1.188.
24000 50
Thus }JL > i, and by eq. (7) the required sectional area is
/^ X 1.188 = AGO. x IsI 88 _ 5 7<9 sq> cm<
Second. F, = Af& = ^(r t f r 2 2 ),
r, being the external and r 2 the internal radius of the section.
554 THEORY OF STRUCTURES.
Let r l 9 cm. and r, = 7.92 cm. Then
*(r? ~ O - 5743 sq. cm.
Also,
' I r' + rf" 143-7264
Hence, by eq. (4),
_ 360X360 4 .
24000 143,7264
Thus, in the latter case, since ft < I, there is no tendency
to buckling.
If the area is determined by equation (10), its value becomes
1.15 X $JL = 65 sq. cm.
19. Flexure of Columns. In Art. 16 the moment equa-
tion has been expressed in the form
and this is sufficiently accurate if the deviation of the axis of the
strut from the vertical is so small that [-T-] may be neglected
without sensible error.
The more correct equation is
p being the radius of curvature.
Consider, e.g., the strut in Art. 16, Case I. Then
P i dO dO
-j^- T y = - = = -j- sin 6,
Er p ds dy
FLEXURE OF COLUMNS. 555
being the inclination of the tangent at M to the axis of *,
and ds an element of the bent strut at M.
... _ tfydy sin Odd.
Integrating,
, ..... (i)
being the value of at a strut end.
a
Let sin = p and sin - = p sin 0. Then
or
y = cos ( 2 )
^
Let Y be the maximum deviation of the axis of the strut
from the vertical, i.e., the value of y when = o or = o.
Then
2 sin ~
-- 2 - ......... (3)
a
Again,
i dB
a Vi // sin 2
Hence, if /be the length of the strut,
d<l> 2 ,
(4)
.F M (0) being an elliptic integral of the first kind.
Let P be the least thrust which will make the strut bend.
As shown in Art. 16,
556 THEORY OF STRUCTURES.
and, by eq. (4), the corresponding value of the modulus yu is
given by
F^}=- 2 (5)
*
Let the actual thrust on the strut be
P=n'P', .-..- . (6)
H* being a coefficient > unity.
The corresponding value of the modulus is given by
By reference to Legendre's Tables it is found that a large
n
increase in the value of //, i.e., of sin or # , is necessary in
order to produce even a small increase in the value of
and therefore of n{ A / f I. Hence, as soon as the thrust
(-V?)
P exceeds the least thrust which will bend the column, viz.,
P', rapidly increases.
The total maximum intensity of stress in the skin of the
strut at the most deflected point
P MB P PYz
z being the distance of the skin from the neutral axis, and /
p
being equal to r.
A
The last term of this equation includes the product fE,
n
which is very large, and also the factor sin , which increases
with # so that the ultimate strength of the material is rapidly
approached, and, in fact, rupture usually takes place before the
column has assumed the position of equilibrium defined by the
slope # at the ends.
FLEXURE OF COLUMNS. 557
If there were no limit to the flexure, the column would
take its position of equilibrium only after a number of oscilla-
tions about this position, and the maximum stress in the
material would be necessarily greater than that given by
eq. (8).
Again,
I (i 2;*' sin 2 0X0
dx = ds cos 6 = .
a V I p sin a
Let X be the vertical distance between the strut ends. Then
0) being an elliptic integral of the second kind.
Hence, the diminution in the length of the strut
20. Flexure of Columns (Findlay). In a paper on the
flexure of columns read before the Canadian Society of Civil
Engineers (Vol. IV, Part I), Findlay expresses the moment
equation in the form
p and being the values of p and when M = o.
558 THEORY OF STRUCTURES.
Hinged Ends. It is assumed that the line of action of the
thrust P is at a distance d from the axis of the strut. Then
'. - w
or
where a* = -~j, p = total stress at the distance z from the
neutral axis, and /= stress due to direct thrust f -jj, so that
.the stress due to bending = p f.
It is also assumed that the form of the axis of the column
before it is acted upon by the thrust P, is a curve of sines
defined by the equation
nx
.y. = Jcos T ,........ (4)
the'origin being half-way between the ends of the strut, and A
being the maximum initial deviation of the axis from the ver-
tical, i.e., the value of y^ when x = o.
d*y An* nx
.*. r-^ = j$- cos -y,
and hence, by eq. (3),
^2.- V 4-d\- A-
A solution of this equation is
nx
, cos ax . f COS T~
cos-
FLEXURE OF COLUMNS. $59
Now - - is always small for such values of f as would con-
stitute a safe working load, and therefore
al aT
cos - i --, approximately,
so that eq. (6) becomes
COS
or
-f </ = ^ cos 0*i + -yj + J cos -j-i + -^r, approx. (7)
Let Fbe the maximum value of y, i.e., the value of y when
= o. Then
Hence, by eq. (3), the total maximum intensity of stress
+4(rH. (9)
where b = - ( -I ^ ] and c = -
z
Eq. (9) is a quadratic from which /may be found in terms
of/. As a first approximation,/ may be substituted for f in
the last term of the portion within brackets, the error being in
the direction of safety.
Fixed Ends. Let M l be the moment of fixture.
Eq. (3) now becomes
560 THEORY OF STRUCTURES.
Assuming again that the initial form of the axis is a curve
of sines, the solution of the last equation is
nx
cos
MI I j , Mf} cos ax /
cos-
Initially,
v n = A cos
and -j- is equal to -7- when x = or = .
dx dx 22
Hence,
al
sin
al
COS I
or
Again, the value of y at the point x =. o is
Also, if/j, p 9 are the total maximum intensities of stress at
the end and at the most deflected point, then
' = etc.,
ZJtL \
and
M\.
two equations from which / may be found as before.
The following conclusions are drawn from the above inves-
tigation :
First. The actual strength of a column depends partly upon
FLEXURE OF COLUMNS. 56 1
known facts as to dimensions, material, etc., and partly upon
accidental circumstances.
Second. Experiments upon the crippling or destruction of
columns cannot be expected to give coherent results when
applied to the determination of the constants in such an equa-
tion as No. (9).
Third. It is a question whether p should be made the
elastic limit of the material and the working load a definite
fraction of the corresponding value of f derived from eq. (9),
or whether/ should be the allowable skin working stress, and
the working stress /^ be found by means of the same equation.
The former seems to be the more logical assumption.
Fourth. It would appear that the strength of hinged col-
umns is likely to be much more variable than the strength of.
columns with fixed ends, as it depends upon two variable
elements d and z/, while the end fixture eliminates d.
Note. The Tables on the following page give the numerical
values of elliptic integrals of the first and second kind, and are
useful in applying the results of Art. 18.
THEORY OF STRUCTURES.
FIRST ELLIPTIC INTEGRAL,
J>
5
i<r
15
20
o.ooo
0.087
0.175
0.262
0-349
0.000
0.087
0.175
0.262
0.349
o.ooo
0.087
0.175
0.262
0-349
o.ooo
0.087
0.175
0.262
0.350
0.000
0.087
0.175
0.262
0.350
o.ooo
0.087
0.175
0.263
0.351
o.ooo
0.087
0.175
0.263
0.352
o.ooo
0.087
0.175
0.263
0-353
o.ooo
0.087
0.175
0.264
0.354
0.000
0.087
0.175
0.264
0-355
0.000
0.087
0.175
0.265
0.356
25
30
35
40
45
0.436
0.524
0.611
0.698
0.785
0.436
0.524
0.611
0.699
0.786
0.437
0.525
0.612
0.700
0.789
0.438
0.526
0.614
0.703
0.792
0.439
0.527
0.617
0.707
0.798
0.440
0.529
0.620
0.712
0.804
0.441
0.532
0.624
0.718
0.814
0-443
0.536
0.630
0.727
0.826
0.445
0.539
0.636
0.736
0.839
0.448
0-544
0.644
0.748
0.858
0.451
o 549
0.653
0.763
0.88 1
So
65
70
0.873
0.960
.047
.134
.222
0.874
0.961
1.049
137
.224-
0.877
0.965
054
.143
.232
0.882
0.972
.062
.153
.244
0.889
0.981
.074
.168
.262
0.898
0-993
.090
.187
.285
0.911
.010
.112
.215
320
0.928
034
.142
254
370
0.947
I 060
1.178
1.302
1.431
0.974
1.099
1-233
1.377
1-534
on
.154
.317
.506
.735
75
80
85
90
.309
.396
.484
571
.312
.400
.487
575
.321
.410
.499
.588
.336
.427
.519
.610
357
452
.547
.643
.385
485
.585
.686
.426
534
.643
.752
.488
.608
731
1.854
1.566
1.705
1.848
1.993
1.703
1.885
2.077 '
2.275
2.028
2.436
3.131
CO
SECOND ELLIPTIC INTEGRAL.
*
It =
/* =.i
ft = .2
V- = -3
tt .4
f* - -5
It = .6
M = -7
M = -8
M =-9
M = I
5
10
O.OOO
0.087
0.175
0.262
0-349
0.000
0.087
0.175
0.262
0.349
0.000
0.087
0.174
0.262
0.349
O.OOO
0.087
0.174
0.262
0.348
o.ooo
0.087
0.174
0.261
0.348
o.ooo
0.087
0.174
0.261
0.347
O.OOO
0.087
0.174
0.261
0-347
O.OOO
0.087
0.174
0.26O
0.346
O.OOO
0.087
0.174
0.260
0.345
0.000
0.087
0.174
0.259
0-343
O.OOO
0.087
0.174
0.259
0.342
25
30
35
40
45
0.436
0.524
o.6n
0.698
0.785
0.436
0.523
0.610
0.698
0.785
0.436
0.523
0.609
0.696
0.782
0.435
0.521
0.607
0.693
0.779
0.434
0.520
0.605
0.690
0.773
0.433
0.518
0.602
0.685
0.767
0431
0.515
0.598
0.679
0-759
0.430
0.512
0-593
0.672
0.748
0.428
0.509
0.588
0.664
0.737
0.425
0.505
0.581
0.654
0.723
0.423
O.5OO
0-574
0.643
0./07
i?
60
65
70
0.873
0.960
.047
134
.222
0.872
o.959
i.04b
.132
.219
0.869
0-955
.041
.126
.212
0.864
0.948
1.032
.116
.200
0857
0.939
.021
.103
.184
0.848
0.928
.008
.086
.163
0.837
0.914
0.989
.063
.135
0.823
0.895
0.965
1.033
I.0<)9
0.808
0.875
0.940
I.OOI
1. 060
0.789
0.850
0.007
O.QdO
I.OOS
0.766
0.819
0.866
0.906
0.940
g:
.309
.396
.484
I-57I
.306
-393
.480
.566
.297
.383
.468
1-554
-283
.367
450
533
.264
344
.424
504
.240
.316
392
.467
.207
277
347
.417
I.l"63
1.227
1.289
I-35I
1.117
1.172
1.225
1.278
1-053
1.095
I-I35
I.I73
0.966
0.985
0.996
I.OOO
EXAMPLES. 563
EXAMPLES.
1. A Phoenix column in four segments, each weighing 17 Ibs. per
lineal yard, carries a load of 68,000 Ibs. What is the compressive unit
stress ? Ans. 10,000 Ibs. per sq. in.
2. The sectional area of a pillar is 144 sq. in., and the pillar carries
a load of 4000 Ibs. Find the normal and tangential intensities of stress
on a plane inclined at 20 to the axis. Ans. 3.25 Ibs.; 8.93 Ibs.
3. A short hollow square column has to support a load of 120,000 Ibs.,
the allowable stress being 15,000 Ibs. per square inch. Find the thick-
ness of the metal, an external side of the column being 6 in.
Ans. .36 in.
4. A solid cast-iron pillar 9 ft. in height and 4 in. in diameter sup-
ports a load of 55,000 Ibs. Find the normal and shearing intensity of
stress per square inch in a plane section inclined at 30 to the axis.
If the ends of the pillar are flat and firmly bedded, determine its
breaking weight, both by Hodgkinson's and by Gordon's formula.
Ans. 1093! Ibs. ; 1894.375 Ibs. ; 141^ tons by H.; 159 tons by G.
5. A cylindrical pillar 6 in. in diameter supports a load of 400 Ibs.,
of which the centre of gravity is $. in. from the axis. Determine the
greatest and least intensities of stress upon any transverse section of
the pillar. Ans. 25$$ Ibs. ; 2jjff Ibs.
6. Compare the breaking weights of round cast-iron, wrought-iron,
and mild-steel pillars with flat and firmly bedded ends, each being 9 ft.
in length and 6* in. in diameter.
Ans. 1,250,197 Ibs. ; 890,109 Ibs. ; 1,543,572 Ibs.
7. A hollow cast-iron pillar with an external diameter of 9 in. is to
be substituted for the solid pillar in the preceding question. Determine
the thickness of the metal. Ans. f in.
8. Determine the breaking weight of a solid round pillar with both
ends firmly secured, 10 ft. in length and 2 in. in diameter, (r) if of
cast-iron ; (2) if of wrought-iron ; (3) if of steel (mild).
Ans. 25142.8 Ibs.; 43516.48105.; 54^36 Ibs.
9. A hollow cast-iron pillar 12 ft. in height has to support a steady
load of 33,000 Ibs.; its internal diameter is 5^ in. Find the thickness
of the metal, the factor of safety being 6. Ans. .28 in.
10. A solid wrought-iron pillar is to be substituted for the pillar in
the preceding question. Find its diameter. Ans. 3^ in.
564 THEORY OF STRUCTURES.
11. What is the breaking weight of a hollow cast-iron pillar 9 ft. in
length and 6 in. square, the metal being i in. thick ?
Ans. 970873.6 Ibs.
12. Compare the breaking weight of a solid square pillar of wrought-
iron 20 ft. long and 6 in. square with that of a solid rectangular pillar
of the same material, the section being 9 in. by 4 in.
Ans. 845,217 Ibs.; 589,090 Ibs.
13. Compare the breaking weights, as derived from Hodgkinson's and
Gordon's formulae, of a solid round cast-iron pillar 20 ft. in length and
10 in. in diameter, (i) both ends being securely fixed ; (2) both ends
being imperfectly fixed.
Ans. (i) 951.4 tons by H.; 1150.05 tons by G.
(2) 280.6 tons by H.; 415 tons by G.
14. Determine by Hodgkinson's formula the diameter of a solid
wrought-iron pillar equal in length and strength to that in the preceding
question. Ans. 7.35 in.
15. A solid or hollow pillar of cast-iron, wrought-iron, or mild steel
is to be designed to carry a steady load of 30,000 Ibs. Determine the
necessary diameter in each case, 6 being a factor of safety. (The pillar
is to be 12 ft. high, and the metal of the hollow pillar is to be f in,
thick.) Ans. Solid: 3.42^.; 3.25 in.; 2.8 in.
Hollow: 4. 5' in.; 4.75 in.; 3.5 in.
16. Determine the load in the preceding question that will produce
a maximum stress of 9000 Ibs. per square inch in the solid steel pillar.
17. A pillar of diameter D supports a given load ; if N pillars, each
of diameter d, are substituted for this single pillar, show that d must lie
D , D
between - and -.
TV* N*
Is it more economical to employ few or many pillars of given height
to support a given load?
1 8. A solid round pillar of mild steel, 16 ft. high, supports a steady
load of 20,000 Ibs. If the factor of safety is 6, what is its diameter ?
Ans. 3 in.
19. Find the diameter of each of four pillars of the same material
which may be substituted for the single pillar in the preceding question.
Ans. 2.04 in.
20. What is the breaking weight of a cast-iron stanchion of a regular
cruciform section and 15 ft. in height, the arms being 24 in. by i in. ?
Ans. 2,811,215 Ibs.
21. Each of the pillars supporting the lowest floor of a refinery is
6$ ft. high, is of a regular cruciform section, and carries a load of
EXAMPLES. 565
240,000 Ibs. ; the total length of an arm is 26 in. Determine its thick-
ness, the factor of safety being 10. Ans. 2.558 in.
22. Find the breaking stress per square inch of a 4-in. x 4-in. solid
wrought-iron pillar for lengths of 5, 10, 15, and 20 ft., the two ends
being absolutely fixed.
Ans. 33,488 Ibs.; 27,692 Ibs.; 21,492 Ibs.; 16,363 Ibs.
23. Find the diameter of a wooden column 20 ft. long, to support a
load of 10,000 Ibs., 10 being a factor of safety and both ends of the
column being absolutely fixed. Ans. 8.55 in.
24. The external and internal diameters of a hollow cast-iron column
12 ft. in length are D and |Z? ( respectively; the load upon the column
is 25,000 Ibs. If the factor of safety is 4, find D, (a) when both ends of
the column are absolutely fixed ; () when both ends are hinged.
Ans. (a) 4.3 in.; (b) 5.4 in.
25. Determine the breaking weight of an oak pillar 9 ft. high, n in.
wide, and 5 in. thick. Ans. 138,160 Ibs.
26. What weight will be safely borne by a pillar of dry oak subject
to vibration, 10 ft. high and 6 in. square, 10 being a factor of safety?
Ans. 9969 Ibs.
27. The web members of a Warren girder are bars of rectangular
section and 10 ft. in length. One of the bars has to carry loads varying
between a steady maximum tension of 20.2 tons and a maximum tension
of 40.4 tons, and another to carry loads varying between a maximum
compression of 8.7 tons and a maximum tension of 14.4 tons. Find the
sectional area in each case, allowance being made for buckling in the
latter.
28. Determine the sectional area of a double-tee strut which is to
carry a load varying between a maximum tension of 80,000 Ibs. and a
maximum compression of 60,000 Ibs. Each flange consists of two 6-
in. x 6-in. x f-in. angle-irons riveted to a i2-in. x f in. web plate.
The length of the strut is to be (a) 6 ft.; (8) 12 ft.
29. A steel strut 10 ft. long consists of two tees back to back, each
4 in. X4in. x ^in. Taking/ = 60,000 Ibs., ai = 77 ^ VT (page 526), and 6
as a factor of safety, find the working load (a) when the strut has two
pin ends; (b) when it has two fixed ends. (E = 29,000,000 Ibs.)
Also, find the deviation of the axis of the load from the axis of the
strut so that the maximum stress in the metal may not. exceed 10,000
Ibs. per square inch.
Ans (a) 25, 585 Ibs. ; (ff) 52,229 Ibs.
Deviation = .55 in. in (a} and .158 in. in (&).
30. A solid wrought-iron strut 20 ft. high and 4 in. in diameter has
one end fixed and the other perfectly free. Find the deviation of the
566 THEORY OF STRUCTURES.
line cf action of a load of 10,000 Ibs. from the axis, so that the stress
may not exceed 10,000 Ibs. per square inch, E being 27,000,000 Ibs.
A?is. .88 in. if P = ^ - ; 1.8 in. if P -EI^.
31. A hollow cast-iron column with two pin ends is 24 ft. high and
has a mean diameter of 12 in. ; it carries a load of 80,000 Ibs. Find the
proper thickness of the metal, 10 being a factor of safety. If the
deviation of the line of action of the load from the axis is i in., find
the maximum stress per square inch in the metal, E being 17,000,000
Ibs. Ans. 1.28 in.; 2236 Ibs. per sq. in.
32. Find the crushing load of a solid wrought-iron pillar 3 in. in
diameter, 10 ft. high, and fixed at both ends. Calculate the deviation
which will produce a maximum stress in the metal of 9000 Ibs. per
square inch under loads of (a) 15,000 Ibs.; () 30,000 Ibs.; E being
29,000,000 Ibs. Ans. 148,775 Ibs. (a) 1.158 in. ; () .38 in.
33. Solve the preceding question on the assumption that the column
has two pin ends. Ans. 66,218 Ibs. ; (a) .985 in.; (b) .261 in.
34. A pier consists of N rows of posts equidistant from each other,
N being even; d is the distance from centre to centre of the outside
rows; W is the gross vertical load upon the pier; H is the greatest
horizontal thrust, and acts upon the pier at a height y above the base.
Assuming the principle of a uniformly varying stress, the portion of the
load borne by the n-th row of posts measured from the centre line is
1 . . Find the value of the coefficient a in terms of d, H,
N -2 N \
y, and N, and determine the best value for d.
35. Prove that the flexural rigidity of a straight beam, of sectional
area A, under a thrust P per unit of area, is EAk*l i ). and that the
beam will bend if its length, when unstrained, exceeds
Ak* being the moment of inertia of the section, and E the coefficient of
elasticity of the material.
36. Find the safe load on a rolled tee-iron strut 6 in. x 4 in. x % in.,
10 ft. long, fixed at one end, free at the other.
37. In Art. 19, show how equations (3) and (6) will be modified if the
line of action of P is distant a + ft from one end and a ft from the
other end of the column's axis. Also, if the coefficient of elasticity, E,
is variable and equal to in n- at a point distant z from the axis, r being
EXAMPLES. 567
the .maximum value of z, and m and n coefficients, show that
n k"
y + - must be substituted for y in eq. (3).
38. In one of Christie's experiments an angle-bar 2 in. x 2 in. x -fa in.,
with hinged ends, for which had the value 154, deflected .01 in. for an
increase in the load of 3000 Ibs. Show that - Q H 9 = .01 in.
o n
39. A long column with pin ends is bent laterally until the angular
deviation (0 ) at the ends is 4. Find the total maximum intensity of
stress, the section of the column being (a) a circle ; (^) a square.
= 29,000,000 Ibs., and the stress due to direct thrust = 1500 Ibs. per
square inch. Ans. (a) 30,615 Ibs.; () 26,715 Ibs.
40. With the same maximum stress as in the last question, find
the angular deviation at the ends so that the stress due to direct thrust
may be 10,000 Ibs. per square inch. Ans. (a) i 5'; (b) i 33'.
41. Show that the load required to produce an angular deviation of
14 at the two pin ends of a long column is only one per cent greater
than that which just produces flexure.
CHAPTER IX.
TORSION.
I. TORSION is the force with which a thread, wire, or pris-
matic bar tends to recover its original state after having been
twisted, and is produced when the external forces which act
upon the bar are reducible to two equal and opposite couples
(the ends of the bar being free), or to a single couple (one end
of the bar being fixed), in planes perpendicular to the axis of
the bar. The effect upon the bar is to make any transverse
section turn through an angle in its own plane, and to cause
originally straight fibres, as DE, to assume helicoidal forms, as
FG or DC. This induces longitudinal stresses in the fibres,
/ p p \
p p 'f
I
FIG. 360. FIG. 361.
and transverse sections become warped. It is found suf-
ficiently accurate, however, in the case of cylindrical and regu-
lar polygonal prisms, to assume that a transverse section which
is plane before twisting remains plane while being twisted.
In order that the bar may not be bent, its axis must coincide
with the axis of the twisting couple.
2. Coulomb's Laws. The angle turned through by one
transverse section relatively to another at a unit distance from
it, is called the Antrle of Torsion, and Coulomb deduced from
568
TOR SIGNAL STRENGTH OF SHAFTS. 569
experiments upon wires, that this angle is directly proportional
to the moment of the twisting couple, and inversely propor^
tional to the fourth power of the diameter.
Thus, if a force P, at the end of a lever of radius p, twists a
cylindrical bar of length L and radius R, and if 6 is the circu-
lar measure of the angle of torsion, then
oc Pp, and also oc -,
so that 6 = C -~^ C being a constant depending only upon the
nature of the material.
Let T be the total angle of torsion, in circular measure,
i.e., the angle turned through by one end of the bar relatively
to the other. Then
T _ r P P
3. Torsional Strength of Shafts (see Art. 23, Chap.
IV). Consider a portion of the shaft bounded
by the planes CE and MN 9 Fig. 361. It is kept
in equilibrium by the couple (P, P\ and by
the elastic resistance at the section MN. Hence,
this elastic resistance must be equivalent to a
couple equal and opposite to (P, P). FIG. 362.
Let Fig. 362 be the transverse section at MN 9 on an en-
enlarged scale, and let abb' a' be any elementary area (= 4A^
(P 9 P) of the surface bounded by the radii OA, OB, and by
the concentric arcs aa' , bb' .
Let x l be the distance of AA l from O.
It is assumed, and is approximately true, that the resist-
ance of any element abb' a' to torsion is directly proportional
to the angle of torsion (#), to its distance from the axis (#,),
and to its area (^M,), and also that it acts at right angles to
the radial line of the element, i.e., to OA or OB.
Thus, the resistance of abb' a' to torsion = G9x^AA\^ G be-
ing a constant to be determined by experiment.
The corresponding moment of resistance about the axis =
14,. Similarly, if x^ jr 8 , x^ 9 ... are the distances
5/0 THEORY OF STRUCTURES.
from the axis of any other elements, 4A Z , 4A a , 4A 4 , . . . t
respectively, the corresponding moments of resistance are
Gtix*AA^> G8x*dA z , . . . Hence, the total moment of resist-
ance of the section
= G6I,
I being the moment of inertia with respect to the axis.
But this moment of resistance (M) is equal and opposite to
the moment of the couple (P,P). Hence,
M=GOI=Pp.
The twisting moment will of course vary with a variable
resistance, and the last equation gives its mean value.
The shaft, however, must be designed (see Cor. 4) for the
maximum couple to whi<*h it may be subjected, and the moment
of this couple (= J/,) may be expressed in terms of the mean
by the equation
ju. being a coefficient to be determined in each case. In a series
of experiments with different engines, Milton found that //
varied from 1.3 to 2.1, but doubtless the variation is often be-
tween still wider limits.
Cor. i. Let /be the stress at the point farthest from the
axis. For a solid round shaft, of diameter D,
and = G9 .
3 2
Let T be the total torsion in degrees. Then
i nT ,
z
J^OR SIGNAL STRENGTH OF SHAFTS.
and hence
D
" L 1 80 2'
or
_
D~ fi.60
Taking the following mean values of G
Material. G f
Cast-iron .............. 6,300,000 5,6oo
Wrought-iron .......... 10,500,000 7,200
Steel .................. 12,000,000 1 1,200
2j = 9.8T for cast iron, = 12.7^ for wrought-iron, = 9.3^
for steel.
Thus, the twist is i each 9.8 diameters in length for cast-
iron, each 12.7 diameters in length for wrought-iron, and each
9.3 diameters in length for steel. This is often much too small,
and in practice the twist is usually limited to T ^- per lineal foot
of length. For a hollow round shaft, D being the external and
D l the internal diameter,
If the thickness (7") of the hollow shaft is small compared
with D,
& - Z> 3 4 = D' - (D - 2Ty = SD*T, approximately,
\
and
5/2 THEORY OF STRUCTURES.
The use of compressed steel admits of shafts being made
hollow. For a solid square shaft, H being the side of the square,
IT
and/, the stress at the end of a diagonal, = G6-=-
H 6
and
GI GH*
In these results it is assumed that G0[ or = =r~ I
\ D H /
is constant at different points of the cross-section, which, how-
ever, is only true for circular sections.
In non-circular sections the stress is more generally greatest
at points in the bounding surface which are nearest to the axis
and least at those points which are farthest from the axis.
St. Venant, who first called attention to this fact, gave the fol-
lowing, amongst others, as the results of his investigations.
Designating by unity the torsional rigidity \^= j-j of a shaft
with circular section, the torsional rigidity of a shaft of equal
/ 271 /~fr
sectional area is .8863, .8863 X \ / n * . l , .7255, or \ / -, ac-
cording as the section is a square, a rectangle with sides in
the ratio n to I, an equilateral triangle, or an ellipse whose
major and minor axes are 2a and 2b, respectively.
Cor. 2. The torsional stress per unit of area- at a distance x
from the axis is GOx.
Hence, if 9 i and x = i, G is the force that will twist a
TORSIONAL STRENGTH OF SHAFTS. 573
unit of area at a unit of distance from the axis through an angle
unity.
Cauchy found analytically that in an isotropic body G is
two-fifths of the coefficient of direct elasticity.
Experiments indicate that G is about three-eighths or one-
third of the coefficient of direct elasticity.
f -f (1 r>4
Cor. 3. For a solid cylinder, Pp = - , R being the radius,
and therefore R* oc -^T> If the shaft is to have a certain speci-
Gv
fied stiffness, i.e., if is fixed, R* oc -?r, and for a given twisting
moment R* oc . Now G is nearly the same for wrought-iron
G
and steel, so that there is little if any advantage to be gained
by the use of the latter.
After passing the elastic limit, the stress varies much more
slowly than as the distance from the axis, and there will be a
partial equalization of stress, the apparent torsional strength
being increased.
Cor. 4. In any transverse section of a solid cylindrical shaft,
the maximum unit stress
M l being the moment of the maximum twisting couple.
This relation is true so long as the stress does not exceed
the elastic limit, and agrees with the practical rule that the
diameter of a cylindrical shaft subjected to torsional forces is
proportional to the cube root of the twisting couple.
The rule is usually expressed in the form
M, = KD\ so that K = ^.
Wohler's experiments show that the value of f depends,
to some extent, upon its fluctuation under the variable twist-
574 THEORY OF STRUCTURES.
ing moment. Ordinarily it should not exceed 7200 Ibs. per square
inch for wrought-iron, in which case K i\ J^ X --f- = I4H-
(Note. If P l is the torsional breaking weight,
_
" D* ~~ I} 3
is the coefficient of torsional rupture?)
Cor. 5. Let* Wbe the work transmitted to a shaft of D in.
diameter, in foot-pounds per minute, N being the correspond
in- number of revolutions. Then
M
12 W = inch-pounds transmitted = 2nMN 2it TV
KD
M l
since M = mean twisting moment = -. Hence,
w
Let HP be the horse-power transmitted per minute. Then
W = 33000 HP. Also for wrought-iron K i| x - ? f.
Hence p -- -^- = D\ and if p = 1.43,
HP
N
-i formula agreeing with the best practice in the case of
wrought-iron shafts subjected to torsional forces only. Such
shafts should, therefore, carry no pulleys.
Cor. 6. The resilience of a cylindrical axle is the product of
one half of the greatest moment of torsion into the correspond-
ing angle of torsion.
Cor. 7. It often happens in practice that a shaft (or beam)
is subjected to a bending as well as to a torsional action.
DISTANCE BETWEEN THE BEARINGS OF SHAFTING. $?$
The combined bending and twisting moments are equiv-
alent (Art. 8, Chap. IV) to the moment
where M b = nM t , M b being the bending and M t the twisting
moment at the given section.
Hence, remembering that the maximum twisting moment
J/, is equal to pM tJ we have for a wrought-iron shaft,
If n = .36 -f- , this becomes
HP
a formula agreeing with the best practice in the case of trans-
mission with bending, as, e.g., in the crank-shafts of marine
engines.
It often happens that n has a still larger value, as, e.g., in
the case of head shafts properly supported against springing.
The usual formula is then
D
3 I HP
= 5V W'
corresponding to n = .72 -{- .
4. Distance between Bearings. The distance between
the bearings of a line of shafting is limited by the considera-
tion that the stiffness of the shaft must be such as will enable
it to resist excessive bending under its own weight and under
any other loads (e.g., pulleys, wheels, etc.) applied to it.
For this reason, the ratio of the maximum deviation of the
axis of the shaft from the straight to the corresponding dis-
tance between bearings should not exceed a certain fraction
whose value has been variously estimated by different writers.
Let / be the distance in feet between bearings, a the
diameter of the shaft in inches, w the weight of the ma-
576 THEORY OF STRUCTURES.
terial of the shaft per cubic foot, and let the applied load be
equivalent to a load per lineal unit of length m times that
of the shaft. Assume a stiffness of T ^Vir an ^ tna t the axis of
the shaft is truly in line at the bearings. The maximum de-
flection of the shaft is given by the formula (Art. 3, Ex. 8.
Chap. VII)
_ I (m + i)(weight of shaft)/ 3 . 1728
- ~
I nd* i 64 / 3 . 1728
-s-UH + l) -- tt>/-~r - ~ - .
384 v 4 144 ar*/ 4 ^
/ " IOO 2E
or
_ s / Ed*
~ V 5 ow ( m +
EXAMPLE. For wrought-iron, E = 3o,ocxD,ooo Ibs. and
w 480 Ibs.
If the applied load, instead of being uniformly distributed
is concentrated at the centre, the maximum deflection
i (m + j)( weight of shaft)/ 3 . 1728
~ 192 El ~~'
and hence
~, Ed'
[QOw(m -f- i
EXAMPLE. For wrought-iron /= 8.5
CYLINDRICAL SPIRAL SPRINGS.
577
5. Efficiency of Shafting. Let it require the whole of
the driving moment to overcome the friction in the case of a
shaft of diameter d and length L. The efficiency of a shaft of
the same diameter and length / = I -= .
But
,. = (Pp) = moment of friction = /* L
w being the specific weight of the material of the shaft, and
the coefficient of friction. Hence,
w
wl
and the efficiency = I 2yw--.
Hence,
S= 2nyn, approx.
Wy =
fnr* GQnr*
2 2
r being the radius of the spring.
6. Cylindrical Spiral Spring. Let the figure represent
a cylindrical spiral spring of length s, supporting
a weight W. Consider a section of the spring at
any point B.
At this point there is a shear W^and a torque
Wy, y being the distance of B from the axis of
the spring, i.e., the radius of the coil.
The effect of W may generally be neglected
as compared with the effect of the moment Wy,
and it may be therefore assumed that the spring
is under torsion at every point. Let there be n
coils. Then
FIG. 363.
578 THEORY OF STRUCTURES.
The elongation of the spring
Syf 2nfny*
__
: ~~
Wy a Wy'S fnr'S
The work done OS = -
A weight hung at the lower end tends to turn as well as
lengthen the spring, and this is due to a slight bending action.
According to Hartnell, f =. 60,000 Ibs. per square inch for
f-in. steel, f = 50,000 Ibs. per square inch for -J-in. steel, and
G varies from 13,000,000 Ibs. for J-in. steel to 11,000,000 Ibs.
for f-in. steel.
Also for wire less than f in. in diameter,
Wny*
and the deflection
EXAMPLE. A wrought-iron shaft in a rolling-mHl makes 50
revolutions per minute and transmits 120 H. P., which is sup-
plied from a waterfall by means of a turbine. Determine the
diameter of the shaft (i) if the maximum stress in the metal
is not to exceed 9000 Ibs. per square inch ; (2) if the angle of
torsion is not to exceed ^ per lineal foot.
As a matter of fact, the diameter of the shaft is 3 j in. at
the bearings and 4 in. in the intermediate lengths. What are
the corresponding maximum inch-stresses in the metal?
Let the twisting couple be represented by force P at the
end of an arm /. Then
P X 27tp x 95 = 120 X 33,ooo ft.-lbs.
120 X 33000 126000 , 126000
.% Pp = ll4-_ _ _ _ ft.-lbs. = X 12 in.-lbs.
27r X 95 19 J 9
126000 X 12
First, - -
and hence
EXAMPLE. 579
, 126000
Second, -- X 12 = Pp =
But 6 -- ~ X y- X ~ ; take G = io,5OO,cxx). Then
126000 X 12 10500000 22 I I I 22
~ T 9 -- = 2 x 7 X 785 x n X 77 x 7^-
Hence,
Z^ 4 =689.45, and D= 5.12 in.
Third, the maximum stresses in the real shaft at the bear-
ings and in the intermediate lengths are respectively given by
126000 stress 22
_ XI2 = __ X _
and
126000 stress 22
^- X 12, -^- Xy X(4).
From the former, the maximum stress = 7682 Ibs. persq. inch.
" latter, " " =6330 " " " "
580 THEORY OF STRUCTURES.
EXAMPLES.
1. A steel shaft 4 in. in diameter is subjected to a twisting couple
which produces a circumferential stress of 15,000 Ibs. What is the stress
(shear) at a point I in. from the centre of the shaft ?
Determine the twisting couple. Ans. 7500 Ibs.; 23,57 if Ibs.
2. A weight of 2| tons at the end of a i-ft. lever twists asunder a
steel shaft if in. in diameter. Find the breaking weight at the end of a
2-ft. lever, and also the modulus of rupture.
Ans. 1 1 tons ; 23, 510 Ibs.
3. A couple of A^ft.-tons twists asunder a shaft of diameter d. Find
the couple which will twist asunder a shaft of the same material and
diameter 2.d. Ans. &N.
4. Compare the couples required to twist two shafts of the same
material through the same angle, the one shaft being / ft. long and
d in. in diameter, the other 2/ ft. long and id in. in diameter.
Compare the couples, the diameter of the latter shaft being .
Ans. i to 8 ; 32 to I.
5. A shaft 1 5 ft. long and 4^ in. in diameter is twisted through an angle
of 2 under a couple of 2000 ft.-lbs. Find the couple which will twist
a shaft of the same material 20 ft. long and i\ in. in diameter through
an angle of 2^. Ans. 12,288 ft. Ibs.
6. A round cast-iron shaft 15 ft. in length is acted upon by a weight
of 2000 Ibs. applied at the circumference of a wheel on the shaft ; the
diameter of the wheel is 2 ft. Find the diameter of the shaft so that the
total angle of torsion may not exceed 2. Ans. 3.53 in.
7. A wrought-iron shaft is subjected to a twisting couple of 12,000 ft.-
lbs. ; the length of the shaft between the sections at which the power is
received and given off is 30 ft.; the total admissible twist is 4. Find
the diameter of the shaft, ju (page 570) being f , and m 10,000,000 Ibs.
Ans. 7.74. in.
8. A wrought-iron shaft 20 ft. long and 5 in. in diameter is twisted
through an angle of 2. Find ihe maximum stress in the material, m
being 10,500,000 ft.-lbs. Ans. 3819.2 ibs. per sq. in.
EXAMPLES. 58l
9. A crane chain exerts a pull of 6000 Ibs. tangentially to the drum
upon which it is wrapped. Find the diameter of a wrought-iron axle
which will transmit the resulting couple, the effective radius of the drum
being 7^ in. ; the safe working stress per square inch being 7200 Ibs.
Ans. 3.17 in.
10. Find the diameter and the total angle of torsion of a 12-ft.
wrought-iron shaft driven by a water-wheel of 20 H. P., making 25
revolutions per minute, m being 10,000,000 Ibs., and the working stress
7200 Ibs. per square inch. Ans. 5.6 in.; 2. 2.
n. A turbine makes 1 14 revolutions per minute, and transmits 92
H. P. through the medium of a shaft 8 ft. 6 in. in length. What must be
the diameter of the shaft so that the total angle of torsion may not ex-
2
ceed , m being 10,500,000 Ibs. ? Ans. 4.7 in.
Determine the side of a square .pine shaft that might be substituted
for the iron shaft.
12. A steel shaft 20 ft. in length and 3 in. in diameter makes 200
revolutions per minute and transmits 50 H. P. Through what angle is
the -shaft twisted ?
A wrought-iron shaft of the same length is to do the same work at
the same speed. Find its diameter so that the stress at the circumference
may not exceed f of that at the circumference of the steel shaft.
Ans. 2. 6; 3.556 in.
13. A vertical cast-iron axle in the Saltaire works makes 92 revolu-
tions per minute and transmits 300 H. P.; its diameter is 10 in. Find
the angle of torsion. Ans. .0144 per lineal foot.
14. In a spinning-mill a cast-iron shaft 8J in. in diameter makes 27
revolutions per minute; the angle of torsion is not to exceed per
lineal foot. Find the work transmitted. Ans. 62.19 H. P.
15. A square wooden shaft 8 ft. in length is acted upon by a force of
200 Ibs., applied at the circumference of an 8 ft. -wheel on the shaft.
Find the length of the side of the shaft, so that the total torsion may
not exceed 2 (in = 400000). What should be the diameter of a round
shaft of equal strength and of the same material ?
Ans. 4.96 in.; 5.09 in.
16. A shaft transmits a given H. P. at ^revolutions per minute with-
out bending. Find the weight of the shaft in pounds'per lineal foot.
/H.P.M
Ans. 2.- .
582 THEORY OF STRUCTURES.
17. The working stress in a steel shaft subjected to a twisting couple
of looo in.-tons is limited to 11,200 Ibs. per square inch. Find its diam-
eter; also find the diameter of the steel shaft which will transmit 5000
H. P. at 66 revolutions per minute,// being f. Ans. 10 in. ; 6.88 in.
1 8. A wrought-iron shaft is twisted by a couple of 10 ft. -tons. Find
its diameter (a) if the torsion is not to exceed i per lineal foot, (ft) if the
safe working stress is 7200 Ibs. per square inch, m = 10,000,000 Ibs.
Ans. (a) 3.7 in.; (&) 5.7 in.
19. A steel shaft 2 in. in diameter makes 100 revolutions per minute
and transmits 25 H. P. Find the maximum working stress and the tor-
sion per lineal foot, m being 10,000,000 Ibs. Also find the diameter of a
shaft of the same material which will transmit 100 H. P. with the same
maximum working stress. Ans. io,o22 T \ Ibs. ; .0478 ; 3.17 in.
20. The crank of a horizontal engine is 3 ft. 6 in. and the connecting-
rod 9 ft. long. At half-stroke the pressure in the connecting-rod is 500
Ibs. What is the corresponding twisting moment on the crank-shaft ?
Ans. 1716^ ft.-lbs.
21. If the horizontal pressure upon the piston end of the connecting
rod in the previous question is constant, find the maximum twisting mo-
ment on the crank-shaft.
/ . sin 0cos0 \
Ans. Pi sin + _ -- j , being given by
n* cos 2 + 4 (sin 2 cos 2 i) + 2 sin 4 0(i + sin 2 0) sin < = o
where n = */- = \ 8 -.
N.B. If sin 2 is neglected as compared with 2 ,
the maximum moment = P sin 0(1+
V n
being very nearly 72.
22. Show that a hollow shaft is both stiffer and stronger than a solid
shaft of the same weight and length.
23. Find the percentage of weight saved by using a hollow instead of
a solid shaft.
Ans. If of equal stiffness = .
m* + i
If of equal strength = 100 \ i /J m \ m * ~ !
1 (" + O a )
m being the ratio of the external to the internal diameter
of hollow shaft.
24. A hollow cast-iron shaft of 12 in. external diameter is twisted by
a couple of 27,000 ft.-lbs. Find the proper thickness of the metal so that
the stress may not exceed 5000 Ibs. per square inch. Ans. .619 in.
EXAMPLES.
25. The external diameter of a hollow shaft is/ times the internal.
Compare its torsional strength with that of a solid shaft of the same ma-
terial and weight.
Ans. -
/ J + i
26. If the solid shaft is 10 in. in diameter, and the internal diameter
of the hollow shaft is 5 inches, find the external diameter and compare
the torsional strengths. AnSt 5 j/g i n< . ^ to 3.
27. A hollow steel shaft has an external diameter d and an internal
d
diameter . Compare its torsional strength with that of (a) a solid
steel shaft of diameter d\ (b) a solid wrought-iron shaft of diameter d\
the safe working stresses of steel and iron being 5 tons and 3! tons re-
spectively. Ans (a) -fa ; () |-|.
28. What twisting moment can be transmitted by a hollow steel shaft
of 8 in. internal and 10 in. external diameter, the working stress being
$ tons per square inch ? Ans. 184^ in.-tons.
29. If/i is the safe torsional working stress of a shaft, and / a is the
safe working stress when the shaft acts as a beam, show that the tor-
sional resistance of the shaft is to its bending resistance in the ratio of
2/ a tO/,.
30. The wrought-iron screw shaft of a steamship is driven by a pair
of cranks set at right angles and 21.7 in. in length; the horizontal pull
upon each crank-pin is 176,400 IDS., and the effective length of the shaft
is 866 in. Find the diameter of the shaft so that (i) the circumferential
stress may not exceed 9000 Ibs. per square inch ; (2) the angle of torsion
r d
may not exceed per lineal foot ; m being 10,000,000 Ibs. The actual
diameter of the shaft is 14.9 in. What is the actual torsion ?
Ans. (i) 14.53 m 'J ( 2 ) H-89 in.; (3) total torsion = 5. 545.
31. The ultimate tensile strength of the iron being 60,000 Ibs. per
square inch, find the actual ultimate strength under unlimited repetitions
of stress. Ans. 54,899 Ibs. (Unwin's formula).
32. What is the torsion in the preceding question when one of the
cranks passes a dead point ?
33. A steel shaft 300 feet in length makes 200 revolutions per minute
and transmits 10 H. P. Determine its diameter so that the greatest
stress in the material may be the same as the stress at the circumference
of an iron shaft i in. in diameter and transmitting 500 ft. -Ibs.
Ans. .807 in. (= in.)
34. Determine the coefficient of torsional rupture for the shaft in
Question 33, 10 being the factor of safety.
35. A wrought-iron shaft in a rolling-mill is 220 feet in length, makes
95 revolutions per minute, and transmits 1 20 H. P. to the rolls; the main
body of the shaft is 4 in. in diameter, and it revolves in gudgeons 3f in.
584 THEORY OF STRUCTURES.
in diameter. Find the greatest shear stress in the shaft proper and in
the portion of the shaft at the gudgeons. Ans. 6330.2 Ibs.; 7508 Ibs.
36. Power is taken from a shaft by means of a pulley 24 inches in
diameter which is keyed on to the shaft at a point dividing the distance
between two consecutive supports into segments of 20 and 80 in. ; the
tangential force at the circumference of the pulley is 5500 Ibs. If the
shaft is of cast-iron, determine its diameter, taking into account the
bending action to which it is subjected. Ans. 4.7 in.
37. Show that the resilience of a twisted shaft is proportional to its
weight. / 2 Volume
Ans. Resilience = .
in 4
38. If a round bar of any material is subjected to a twisting couple,
show that its maximum resilience is two-thirds the maximum resilience
of the material.
39. Determine the diameter of a wrought-iron shaft for a screw
steamer, and the torsion per lineal foot; the indicated H. P. = 1000, the
number of revolutions per minute = 150, the length of the shaft from
thrust bearing to screw = 75 ft., and the safe working stress = 7200 Ibs.
per square inch. Ans. 6.67 in. ; io.5.
40. In a spinning-mill a cast-iron shaft 84 ft. long makes 50 revolu-
tions per minute and transmits 270 H. P. Find its diameter (i) if the
stress in the metal is not to exceed 5000 Ibs. per square inch ; (2) if the
, o
angle of torsion per lineal foot is not to exceed .
Also (3) in the first case find the total torsion.
Ans. (i) 7.02 in. ; (2) 10.23 in. ; (3) 28. 8.
41. A circular shaft is twisted beyond the limit of elasticity. If the
equalization of stress is perfect, show that for a given maximum stress the
twisting couple is greater than it would be if the elasticity were perfect,
in the ratio of 4 to 3.
42. Determine (a) the profile of a shaft of length / which at every
point is so proportioned as to be just able to bear the power it has to
transmit plus the power required to overcome the friction beyond the
point under consideration. Find (ft) the efficiency of such a shaft, and
(c) the efficiency of a shaft made up of a series of n divisions, each of
uniform diameter.
Ans. (a) The radius/ of any section distant x from the driving end
X
is y = re ?> L , r being the radius of the driving end and
L the length of a shaft of uniform diameter, such that
the whole driving moment is required to overcome its
own friction.
EXAMPLES. 585
43. A steel shaft carries a 5~ft. pulley midway between the supports
and makes 6 revolutions per minute, the tangential force on the pulley
being 500 Ibs. Taking the coefficient of working strength at 11,200 Ibs.
per square inch, find the diameter of the shaft and the proper distance
between the bearings.
44. A steel shaft 4 inches in diameter and weighing 490 Ibs. per cubic
foot makes 100 revolutions per minute. If the working stress in the
metal is 11,200 Ibs. per square inch, find the twisting couple and the dis-
tance to which the work can be transmitted ; the coefficient of friction
being .05, and the efficiency of the shaft f.
Ans. 140,800 in. -Ibs. ; 8228^ ft.
45. If the shaft is of steel, and if the loss due to friction is 20 per cent,
find the distance to which work may be transmitted, ju being .05.
Ans. 6582! ft.
46. A wrought-iron shaft 220 ft. between bearings and 4 in. in diam-
eter can safely transmit 120 H. P. at the rate of 95 revolutions per
minute. What is the efficiency of the shaft? (jit = fa.) Ans. .976.
47. The efficiency of a wrought-iron shaft is ; the working stress in
the metal is 7200 Ibs. per square inch ; the coefficient of friction is .125.
How far can the work be transmitted ? Ans. 4320 ft.
48. A spring is formed of steel wire ; the mean diameter of the coils
is i inch ; the working stress of the wire is 50,000 Ibs. per square inch ;
the elongation under a weight of 19/3- Ibs. is 2 inches; the coefficient of
transverse elasticity is 12,000,000 Ibs. Find the diameter of the wire and
the number of coils.
49. Find the weight of a helical spring which is to bear a safe load of 6
tons with a deflection of I inch, G being 12,000,000 Ibs., and f 60.000 Ibs.
50. Find the time of oscillation of a spring, the normal displacement
under a given load being A. AJ
Ans. TI\ ~.
5 1 . Find the deflection under the weight W of a conical helical spring
(a) of circular section ; (<) of rectangular section, the radii of the extreme
coils being ji and j a , and the radial distance from the axis to a point of
the spring at an angular distance </> from the commencement of the spiral
r 2 R <p
being given by the relation --- = - . (n = number of coils.)
yi yi 27zv/
.n(y, + yd(y?+yS)W f ny*lV
Ans. (a) - ~~Gr r ~ V Gr*~' ^ = ^ ~
b and h being the sides of the rectangular section.
52. The efficiency of an axle is \ ; the working stress in the shaft is
9000 Ibs. per square inch ; the coefficient of friction is .10. How far may
work be transmitted ?
585*
THEORY OF STRUCTURES.
B
Fig. A shows the distortion produced by twisting 1 a round %-in. iron bar.
Fig. B shows the distortion produced by twisting a square %-in. iron bar.
DISTORTION OF IRON BARS BY TWISTING. 585^
The above figures show the distortion produced by twisting a 1% x fc" iron bar.
CHAPTER X.
STRENGTH OF CYLINDRICAL AND SPHERICAL BOILERS.
i. Thin Hollow Cylinders ; Boilers ; Pipes.
Let r be the radius of the cylinder.
Let t be the thickness of the metal.
Let/ be the fluid pressure upon each unit
of surface.
Let/^ be the tensile or compressive unit
stress, according as p is an internal or exter-
FIG. 364. nal pressure.
Assume (i) that the metal is homogeneous and free from
initial strain ;
(2) that / is small as compared with r ;
(3) that the pressures are uniformly distributed
over the internal and external surfaces ;
(4) that the ends are kept perfectly flat and rigid ;
(5) that the stress in the metal is uniformly dis-
tributed over the thickness.
The last assumption is equivalent to supposing that it is
the mean circumferential stress which is governed by the
strength of the metal, while in reality it is the internal or maxi-
mum circumferential stress which is so governed.
The figure represents a cross-section of the cylinder of
thickness unity.
A section made by any diametral plane, as AB, must de-
velop a total resistance of 2tf, and this must be equal and
opposite to the resultant of the fluid pressure upon each half,
i.e., to 2pr. Hence,
2tf=2pr, or tf=pr (i)
586
THIN HOLLOW CYLINDERS; BOILERS; PIPES. 587
This formula may be employed to determine the bursting,
proof, or working pressure in a cylindrical or approximately
cylindrical boiler, provided that/", instead of being the tensile
or compressive unit stress, is some suitable coefficient which
has been determined by experiment. If r/ is the efficiency of
a riveted joint, the formula
may be employed to determine the working pressure in a cylin-
drical or approximately cylindrical boiler.
In ordinary practice the values of rf and /are given by the
following table :
Material.
Joint.
it
/in Ibs. per sq. in.
Wrought-iron
Single-riveted
- *!*>
8000 to oooo
Double-riveted
7
H
Treble-riveted
8 to 85
,,
Steel
Single-riveted
. ^^
I2OOO to I3OOO
14
Double-riveted
7
(,
Treble-riveted
8 to 85
tt
For cast-iron cylinders the working value of /may be taken
at about 2000 Ibs. per square inch.
The total pressure upon each of the flat ends of the cylinder
The longitudinal tension in a thin hollow cylinder
K^P P^ " ,
" 2nrt ~ 2t' ^ }
and is one half of the circumferential stress/.
Cor. i. Let the cylinder be subjected to an external pressure
p' as well as to an internal pressure p. Then
fl=pr-p'r>,
(3)
r' being the radius of the outside surface of the cylinder, /is
a tension or a pressure according as pr
588 THEORY OF STRUCTURES.
Generally, r r' is very small, and the relation (3) may be
written
ft = r(p-p').
2. Thick Hollow Cylinder. If / is large, the stress is no
longer uniformly distributed over the thickness. Suppose that
the assumptions (i) and (3) of Art. I still hold, also that the
cylinder ends are free, and that the annulus forming the section
of the cylinder is composed of an infinite number of concentric
rings. Under these conditions the straining of the cylinder
cannot affect its cylindrical form. Hence, right sections of the
cylinder in the unstrained state remain planes after the strain-
ing, so that the longitudinal strain at every point must be the
same. Two methods will be discussed.
FIRST METHOD. Let dx be the thickness of one of the
rings of radius x, and let dq be the intensity of the circum-
ferential stress.
pr p'r' = difference between the total pressures from
within and without = total circumferential stress = / *" dq.
If it be assumed that the thickness (= r' r) remains un-
changed under the pressure, then the circumferential extension
of each of the concentric rings must be equal to the same con-
stant quantity A, and therefore
dq = Edx - ,
*
E being the coefficient of elasticity. Hence,
. , E\ C r 'dx 1 r'
prp'r'=- I = log,-.
27t J r X 27t *' r
en /= E
2
elastic limit is not exceeded, and therefore
Let / be the tensile unit stress. Then /= E - if the
2nr
THICK HOLLOW CYLINDER. 589
or
r' pr-p'r' i(pr
,. -= x + _- + -( - f - , approx., (4)
if p is small as compared with/; and hence,
t r ' pr-p'r' ilr
= ~ ~~ h
r
In most cases which occur in practice/' is so small as com-
pared with/ that it may be disregarded.
Hence, making/' zero in equation (5),
Formulae (5) and (6) may be employed even if the elastic
limit is exceeded, if f is considered a coefficient of strength
to be determined by experience.
Cor. Rankine, in his Applied Mechanics, obtains by
another method,
//_P
'-~\ f-P'
if/' be neglected. Hence,
, *3
== i +y + 2 7^ ' a PP roximatel y'
if p is small as compared with /, and therefore
59O THEORY OF STRUCTURES.
an equation identical with (6).
SECOND METHOD. Consider a ring bounded by the radii
x, x + dx, at any point.
Let q be the normal (i.e., radial) intensity of stress.
Let /be the intensity of stress tangential to the ring.
" s " " " " " perpendicular to the plane
of the ring.
Let a-, /?, y be the corresponding strains.
Let E and mE be respectively the coefficients of direct and
lateral elasticity.
Then, since E, f, s are principal stresses (Chap. IV),
_ _
~
f <? + * f+g
_ _ _
E mE' ~E ' mE ' ~ E ~ mE ' v
But y is constant. Also, since the ends are free, the total
pressure on a transverse section is nil, and hence it might be
inferred that s is zero at every point. Adopting this value of s,
By eq. (i),
f-\-g= a constant = c ....... (2)
Again,
d(qx) fdx = xdq + qdx ...... (3)
By eqs. (2) and (3),
xdq + 2qdx = cdx.
.'. d(x*q) = cxdx.
Integrating,
\ / V *V = f + < / , . ...... (4)
c' being a constant of integration.
When x = r, the internal radius, q =/.
" j; = r ; , the external radius, q =p f .
SPHERICAL SHELLS. 59 1
Hence, by eq. (4),
_ = , = ,.,_;
and therefore
Hence, by eq. (4),
r*p r 1 8 / p P' r'r'
L' a .J a
r _ r i ; (5)
and, by eq. (2),
r *p _ r f *p' p p'
f= * -7r- + ^-r
J *** /*+' * I >%*
r' r' ' x' r r'
3. Spherical Shells. Let the data be the same as before.
The section made by any diametral plane must develop a total
resistance of 2nrtf. Then
2nrtf 7tr*p,
or
*f = pr - (0
Hence, a spherical shell is twice as strong as a cylindrical
shell of the same diameter and thickness of metal, so that the
strongest parts of egg-ended boilers are the ends.
Cor. I. Let the shell be subjected to an external pressure
p' as well as to an internal pressure p. Then
2n -tf= nrp* nr' *#'.
/"is a tension or a pressure according as r*p ^ r'*p'.
Generally, r' r is very small, and the relation (2) may be
written
ft = r -(.P-p'} (3)
592 THEORY OF STRUCTURES.
Cor. 2. For a thick hollow sphere, Rankine obtains
P == 2 Sr" + 2r" a PP roximatelv - (4)
4. Practical Remarks. A common rule requires that the
working pressure in fresh-water boilers should not exceed one-
sixth of the bursting pressure, and in the case of marine boilers
that it should not exceed one-seventh.
An English Board of Trade rule is that the tensile working
stress in the boiler-plate is not to exceed 6000 Ibs. per square
inch of gross section, and French law fixes this limit at 4250
Ibs. per square inch.
The thickness to be given to the wrought-iron plates of a
cylindrical boiler is, According to French law,
/ = .0036/2;- + .1 in.;
according to Prussian law,
/ (^3 _ i) r + i in. .oo$nr + .1 in., approximately,
r being the radius in inches, and n the excess of the internal
above the external pressure in atmospheres.
The thickness given to cast-iron cylindrical boiler-tubes is,
according to French law, five times the thickness of equivalent
wrought-iron tubes ; according to Prussian law,
t _ ^.om _ i)r + -J in. = .01 nr + i in., approximately.
Steam-boilers before being used should be subjected to a
hydrostatic test varying from \\ to 3 times the pressure at
which they are to be worked.
Fairbairn conducted an extensive series of experiments
upon the collapsing strength of riveted plate-iron flues, by
enclosing the flues in larger cylinders and subjecting them to
hydraulic pressure. From these experiments he deduced the
following formula for a ivrought-iron cylindrical flue or tube :
Collapsing pressure I = ^ =403 1 50 ^
in pounds per square inch of surface i "
PRACTICAL RULES FOR BOILERS AND FLUES. 593
t "being the thickness and r the radius in inches, and / the length
in feet.
This formula cannot be relied upon in extreme cases and
when the thickness of the tube is less than f in.
Note. In practice, t 2 may be generally used instead of t 2 - 19 .
The experiments also showed that the strength of an elliptical
tube is almost the same as that of a circular tube of which the
radius is the radius of curvature at the ends of the minor axis.
Hence, if a and b are the major and minor axes of the ellipse^
the above formula becomes
b / 2 ' 19
p = 403150 - a -j-.
By riveting angle- or T-irons around a tube, its length is
virtually diminished and its strength is therefore increased, as
it varies inversely as the length.
The thickness of tubes subjected to external pressure is,
according to French law, twice the thickness of tubes subjected
to interior pressure, but under otherwise similar conditions ;
according to Prussian law the thickness of heating pipes is
/ = .0067^ Vn -f- .05 in., if of sheet-iron,
and
t .Old Vn + .07 in., if of brass.
According to Reuleaux, the thickness (f) of a round flat
plate of radius r, subjected to a normal pressure, uniformly dis-
tributed and of intensity/, is given by the formula
7 t _ A7
according as the plate is merely supported around the rim or
is rigidly fixed around the rim, as, e.g., the end plates of a
cylindrical boiler ; /", as before, is the coefficient of strength.
The corresponding deflections of the plate are
-(-V
6\t>
and -, - M=.
594 THEORY OF STRUCTURES.
EXAMPLES.
1. What should be the thickness of the plates of a cylindrical boHer
6 ft. in diameter and worked to a pressure of 50 Ibs. per square inch, in
order that the working tensile stress may not exceed 1.67 tons per square
inch of gross section ? Ans. .42 in.
2. A cylindrical boiler with hemispherical ends is 4 ft. in diameter
and 22 ft. in length. Determine the thickness of the plates for a steam-
pressure of 4 atmospheres.
3. What is the collapsing pressure of a flue 10 ft. long, 36 in. in
diameter, and composed of i-in. plates? Also of a flue 30 ft. long, 48 in.
in diameter, and T \ in. thick? Ans. 490.84 Ibs.; 91.59 Ibs.
4. Determine the thickness of a 2-in. locomotive fire-tube to support
an external pressure of 5 atmospheres.
5. A copper steam-pipe is 4 in. in diameter and \ in. thick. Find the
working pressure, the safe coefficient of strength for copper being 1000
Ibs. per square inch. Ans. 125 Ibs. per square inch.
6. A /-ft. boiler of ^-in. plates was burst at a longitudinal double-
riveted joint by a pressure of 310 Ibs. per square inch. Find the coef-
ficient of ultimate strength. Ans. 29,760 Ibs.
7. A 50- in. cylindrical boiler of -^ in. plates is made of wrought-
iron whose safe coefficient of strength is 4000 Ibs. per square inch. Find
the working pressure. Ans. 50 Ibs. per square inch.
8. A lo-in. cast-iron water-pipe is subjected to a pressure of 250 Ibs.
per square inch. Find its thickness, the coefficient of working strength
being 2000 Ibs. per square inch. Ans. i in.
9. A steel spherical shell 36 in. in diameter and f in. thick is sub-
jected to an internal fluid pressure of 300 Ibs. per square inch. Find its
coefficient of strength. Ans. 7200 Ibs.
10. A thin, hollow, spherical, elastic envelope, whose internal
radius is R, was subjected to a fluid pressure which caused it to expand
gradually until its radius became R\ . Determine the work done.
11. The plates of a cylindrical boiler 5 ft. in diameter are \ in. thick.
Find to what pressure the boiler may be worked so that the tensile stress
in the plates may not exceed i| tons per square inch of gross section.
EXAMPLES, 595
12. Show that the assumption of a uniform distribution of stress in
the thickness of a cylindrical or spherical boiler is only admissible when
the thickness is very small.
13. A metal cylinder of internal radius r and external radius nr is
subjected to an internal pressure of p tons per square inch. Show that
the total work done in stretching the cylinder circumferentially is
-~~ -j - ft. -tons per square foot of surface, E being the metal's co-
efficient of elasticity.
14. The cast-iron cylinder of an hydraulic press has an external
diameter twice the internal, and is subjected to an internal pressure of
/tons per square inch. Find the pHncipal stresses at the outer and inner
circumferences. Also, if the pressure is 3 tons per square inch, and if the
internal diameter is 10 in., find the work done in stretching the cylinder
circumferentially, E being 8000 Ibs.
Ans. At inner circumference, q =p, a thrust, and/= ^p, a tension.
At outer circumference, q = o, and/ = f /, a tension.
Woik = 126 ft.-lbs. per square foot of surface.
15. The chamber of a 2/-ton breech-loader has an external diameter
of 40 in. and an internal diameter of 14 in. Under a powder pressure of
1 8 tons per square inch, find the principal stresses at the outer and inner
circumferences, and also the work done ; E being 13,000 Ibs.
Ans. At inner, q = 18 tons, compression ; at outer, q = o.
At inner,/ = 23^ tons, tension ; at outer,/ = 5^ tons,
tension.
Work = \\ ft.-tons per sq. ft. of surface.
16. What should be the thickness of a o-in. cylinder (a) which has
to withstand a pressure of 800 Ibs. per square inch, the maximum allow-
able tensile stress being 24,000 Ibs. per square inch ; (ff) which has to
withstand a pressure of 6000 Ibs. per square inch ; the maximum allow-
able tensile stress being 10,000 Ibs. per square inch?
Ans. (a) 1.86 in. ; (b) 4^ in.
17. Show that the radial (or) and hoop (fl) strains in thick hollow
cylinders and spheres are connected by the relation a - .
1 8. Prove that the relation in Ex. 1.7 is satisfied by the values ob-
tained for/ and q in the Second Method of Art. 2, Chap. X.
19. A thick hollow sphere of internal radius r and external radius
nr is subjected to an internal pressure p and an external pressure p .
Determine the principal stresses at a distance x from the centre.
_P'n*-p p_-Jf V . p*?_-_p _ p_-_ n*S_
~ - " 3 ' J ~ " -
THEORY OF STRUCTURES.
20. Assuming that the annulus forming the section of a cylindrical
boiler is composed of a number of infinitely thin rings, show that the
JL
pressure at the circumference of a ring of radius r is per unit of
surface, and that the circumferential stress is h , A and B de-
r mr m + I
noting arbitrary constants, and m being the coefficient of lateral con-
traction. Find the values of A and B, p* and pi being respectively the
internal and external pressures.
21. Show that in the case of a spherical boiler the pressure and cir-
cumferential stress are respectively and + . Find
' r i+m r i (0 l)f** ~ 3
A and B.
22. Solve Questions i, 2, 6, 7, 8, 9, and n on the supposition that / is
not small as compared with r.
23. Taking/ = 4000 Ibs. per square inch and E = 30,000,000 Ibs.,
Find the thickness and deflection of the end plates of the boiler in Ques-
tion 7.
CHAPTER XL
BRIDGES.
1. Classification. Bridges may be divided into four gen-
eral classes, viz. : (A) Bridges with horizontal girders ; (B) Can-
tilever bridges (Art. 15) ; (C) Suspension bridges (Chap. XII);
(D) Arched bridges (Chap. XIII). The present chapter treats
of bridges in Classes A and B only.
2. Comparative Advantages of Curved and Horizontal
Flanges in Girders for Bridges of Class A. The depth is
sometimes varied for the sake of appearance, and it is also
claimed that, an economy of material is effected by giving the
chord a slope, as, e.g., in the case of the Sault Bridge (Art. 19).
Such a truss is intermediate between a truss with horizontal
flanges and one of the parabolic form. The curved or para-
bolic form is not well suited to plate construction, and a dimi-
nution in depth lessens the resistance of the girder to distor-
tion. Again, if the bottom flange is curved, the bracing for
the lower part of the girder is restricted within narrow limits,
and the girder itself must be independent, so that in a bridge
of several spans any advantage which might be derivable from
continuity is necessarily lost. Generally speaking, the best and
most economical form of girder is that in which the depth is
uniform throughout, and in which the necessary thickness of
flange at any point is obtained by increasing the number of
plates.
3. Depth of Girder or Truss (Class A). The depth usu-
ally varies from one-fifteenth to one-seventh (and even more) of
the span. It is generally found advisable to give large girders
597
598
THEORY OF STRUCTURES.
an increased depth, and they should, therefore, be designed to
have a specified strength. If the span is more than twelve times
the depth, the deflection becomes a serious consideration, and
the girder should be designed to have a specified stiffness. The
depth should not be more than about i^ times the width of
the bridge, and is therefore limited to 24 ft. for a single and to
40 ft. for a double-track bridge.
4. Position of Platform. The platform may be supported
either at .ike-'top^r bottom flanges, or in some intermediate
positidfu In favor/ oiVthe last it is claimed that the main girders
may be braced together below the platform (Fig. 365), while
the upper portions s'tfrve as parapets or guards, and also that
,-the vibration communicated by a passing train is diminished.
J.The position, however, is not conducive to rigidity, and a large
amount of metal is required to form the connections.
pi A
FIG. 365.
FIG. 366.
The method of supporting the platform on the top flanges
(Fig. 366) renders the whole depth of the girder available for
bracing, and is best adapted to girders of shallow depth.
Heavy cross-girders may be entirely dispensed with in the case
of a single-track bridge, and the load most effectively distrib-
uted, by laying the rails directly upon the flanges and vertically
above the neutral line. Provision may be made for side spaces
by employing sufficiently long cross-girders, or by means of
short cantilevers fixed to the flanges, the advantage of the
POSITION OF PLATFORM.
599
former arrangement being that it increases the resistance to
lateral flexure, and gives the platform more elasticity.
Figs. 367, 368, 369 show the cross-girders attached to the
bottom flanges, and the desirability of this mode of support
increases with the depth of the main girders, of which the cen-
tres of gravity should be as low as possible. If the cross-girders
are suspended by hangers or bolts below the flanges (Fig. 369),
the depth, and therefore the resistance to flexure, is increased.
FIG. 368.
FIG. 369.
In order to stiffen the main girders, braces and verticals,
consisting of angle- or tee iron, are introduced and connected
with the cross-girders by gusset pieces, etc. ; also, for the same
purpose, the cross-girders may be prolonged on each side, and
the end joined to the top flanges by suitable bars.
When the depth of the main girders is more than about
5 ft., the top flanges should be braced together. But the
minimum clear headway over the rails is 16 ft., so that some
other method should be adopted for the support of the plat-
form when the depth of the main girders is more than 5 ft.
and less than 16 ft.
Assume that the depth of the platform below the flanges is
2 ft., and that the depth of the transverse bracing at the top is
i ft. ; the total limiting depths are 7 ft. and 19 ft., and if I to 8
is taken as a mean ratio of the depth to the span, the corre-
sponding limiting spans are 56 ft. and 152 ft.
600 THEORY OF STRUCTURES.
5. Comparative Advantages of Two, Three, and Four
Main Girders. A bridge is generally constructed with two
main girders, but if it is crossed by a double track a third is
occasionally added, and sometimes each track is carried by two
independent girders.
The employment of four independent girders possesses the
one great advantage of facilitating the maintenance of the
bridge, as one-half may be closed for repairs without interrupt-
ing the traffic. On the other hand, the rails at the approaches
must deviate from the main lines in order to enter the bridge,
so that the width of the bridge is much increased, and far
more material is required in its construction.
Few, if any, reasons can be urged in favor of the introduc-
tion of a third intermediate girder, since it presents all the
objectionable features of the last system without any corre-
sponding recommendation.
The two-girder system is to be preferred, as the rails, by
such an arrangement, may be continued over the bridge with-
out deviation at the approaches, and a large amount of ma-
terial is economized, even taking into consideration the in-
creased weight of long cross-girders.
6. Bridge Loads. In order to determine the stresses in
the different members of a bridge truss, or main girder, it is
necessary to ascertain the amount and character of the load to
which the bridge may be subjected. The load is partly dead,
partly live, and depends upon the type of truss, the span, the
number of tracks, and a variety of other conditions.
The dead load increases with the span, and embraces the
weight of the main girders (or trusses), cross-girders, platform,
rails, ballast, and accumulations of snow.
As to the live load see Art. 19.
7. Trellis or Lattice Girders. The ordinary trellis or
lattice girder consists of a pair of horizontal chords and two
series of diagonals inclined in opposite directions (Fig. 370).
The system of trellis is said to be single, double, or treble, ac-
cording to the number of diagonals met by the same vertical
section.
TRELLIS OR LATTICE GIRDERS.
60 1
Vertical stiffeners, united to the chords and diagonals, may
be introduced at regular intervals.
FIG. 370.
371, 372, 373, 374 show appropriate sections for the
top chord ; the bottom chord may be formed of fished and
riveted plates, or of links and pins.
-ii ir
FIG. 371. FIG. 372. FIG. 373. FIG. 374.
The verticals and diagonals may be of an L, T, I, H, LJ , or
other suitable section, but the diagonals, except in the case of
a single system of trellis, are usually flat bars, riveted together
at the points of intersection.
An objection to this class of girder is the number of the
joints.
The stresses in the diagonals are determined on the assump-
tion that the shearing force at any vertical section is equally
distributed between the diagonals met by that section, which
is equivalent to the substitution of a mean stress for the differ-
ent stresses in the several bars.
E.g., let w be the permanent load concentrated at each apex
in Fig. 370.
Let 9 be the inclination of the diagonals to the vertical.
The reaction at A fyw, and the shearing force at the
section MN = y^w ^.w = 2>2 W -
This shearing force must be transmitted through the diag-
onals.
Hence, the stress in ab due to the permanent load
sec
-w sec
602 THEORY OF STRUCTURES.
Again, let w' be the live load concentrated at an apex.
The greatest shear at mn due to the live load occurs when
every apex between a and 7 is loaded.
This shear = corresponding reaction at I = -f-J-w', and the
stress in ab due to the live load
= i X -f- f ' sec f f w' sec 8.
Hence, the total maximum stress in ab = (|w -f- J J ze/) sec 0.
The greatest stress of a kind opposite to that due to the
dead load is produced in ab when the live load w' is concen-
trated at every apex between I and b.
The shear to be transmitted is then 2\w due to the dead load,
and -j-fw' due to the live load, and the resultant stress in ab
sec 6 = (f w l%w') sec 6.
This stress may be negative, and must be provided for by
introducing a counter-brace or by proportioning the bar to
bear both the greatest tensile and the greatest compressive
stress to which it may be subjected.
The stress in any other bar may be obtained as above.
The chord stresses are greatest when the live load covers
the whole of the girder, and may be obtained by the method
of moments, or in the manner described in the succeeding
articles.
In the above it is assumed that the members of the girder
are riveted together. If they are connected by pins, each of
the diagonal systems may be treated as being independent.
Thus, the system I 2^^34567 transmits to the supports
the stresses due to loads at a, 3, and 5.
The shear due to the dead load, transmitted through ab,
3 w
= reaction at I load at a = w w .
2 2
W
Hence, the stress in ab due to the dead load = - sec 6.
The stress in ab due to the live load is greatest when w' is
concentrated at each of the points 3 and 5,
WARREN GIRDER.
603
The maximum shear due to live load transmitted through ab
= ^w f = fee/,
and the corresponding stress in ab = / sec 9.
Hence, the total maximum stress in ab
as compared with (fcw + |- f /) sec obtained on the first as-
sumption.
8. Warren Girder. The Warren girder consists of two
horizontal chords and a series of diagonal braces forming a
single triangulation, or zigzag. Fig. 375.
N-1
024 n N-2
FIG. 375.
The principles which regulate the construction of trellis
girders are equally applicable to those of the Warren type.
The cross-girders (floor-beams) are spaced so as to occur at
the apex of each triangle.
When the platform is supported at the top chords, the re-
sistance of the structure to lateral flexure may be increased by
horizontal bracing between the cross-girders and by diagonal
bracing between the main girders.
When the platform is supported on the bottom chords,
additional cross-girders may be suspended from the apices in
the upper chords, which also have the effect of adding to the
rigidity of the main girders.
Let w be the dead load concentrated at an apex or joint.
.' u live tt u " " *' " **
" / " " span of the girder.
" k " " depth ". "
" s " " length of each diagonal brace.
" 6 " " inclination of each diagonal brace to the ver-
tical.
" N -\- i be the number of joints.
604 THEORY OF STRUCTURES.
Then
Two cases will be considered.
CASE I. All t he joints loaded.
Chord Stresses. These stresses are greatest when the live
load covers the whole of the girder.
Let S n be the shearing force at a vertical section between
the joints n and n -f- i.
Let H n be the horizontal chord stress between the joints
n I and n -f- I-
The total load due to both dead and live loads
= (w-\-w')(N - i).
The reaction at each abutment due to this total load
The shearing forces in the different bays are
UU _!_ it)'
w A- w'
-~-
Tfl -J 7/
(N i), between o and i ;
3), ! .< 2 .
2 x - 5), 2 3 ;
and
The corresponding diagonal stresses are
5 sec 0, .V, sec 0, S M sec 0.
WARREN GIRDER. 605
The last stresses multiplied by sin 6 give the increments of
the chord stresses at each joint. Thus,
H l = tension in o 2 = 5 tan
+ 1 j i / / r
IV / IV f W l> 1\ I
ff t = compression in i 3 = ,S tan 0-|- .S, tan
w + w' IN i w -\- w' I N 3
_ W + tt/ / 2(7V ~ 2)
-fiT 3 = tension in 2 4 = //, + 5", tan 0-\- S 9 tan
-3)
_ _
2 k N
compression in 3 5 = //!, + ^a tan ^ + $* tan
4)
and H n = horizontal stress in chord, between the joints
w -\- w' / n(Nn)
n i and n + i ! -- T - X7 ' being a tension for a
2t K J.\
bay in the bottom chord, and a compression for a bay in the
top chord.
Note. The same results may be obtained by the method
of moments ; e.g., find the chord stress between the joints
n r and n -|- i.
Let a vertical plane divide the girder a little on the tight
of n.
The portion of the girder on the left of the secant plane is
kept in equilibrium by the reaction at the left abutment, the
horizontal stresses in the chords, and the stress in the diagonal
from ;/ to n -\- i.
Take moments about the joint n. Then
606 THEORY OF STRUCTURES.
n _ - _ _
w-\- w' N n
~ 2 N
/. H n = etc.
Diagonal Stresses due to Dead Load.
Let d n be the stress in the diagonal n, n -\- I, due to the
dead load.
The shearing forces in the different bays due to the dead
load are
w w
-(N i), between o and I ; -(N 3), between I and 2 ;
2 2
W W
-(^-5), " 2 " 4; -(N-j), 3 " 4;
and (N 2n i), between n and n -j- I.
The corresponding diagonal stresses are :
w w s
a compression -(N i) sec 6 = -(N i)-, = d^ inoi ;
2 A?
a tension (TV 3) sec 6 = (N $)-r = d l in 1 2
2 2 A?
a compression (A 7 " 5) sec -(N 5)7- ^ 2 in 23 ;
2 2 rt
and the stress in the nth diagonal between n and n + i is
WARREN GIRDER. 607
being a tension or a compression according as the brace slopes
down or up towards the centre. <i
Diagonal Stresses due to Live Load. The live load produces
the greatest stress in any diagonal (n, n -\- i), of the same kind
as that due to the dead load, when it covers the longer of the
segments into which the diagonal divides the girder. Repre-
sent this maximum stress by D n .
The live load produces the greatest stress in any diagonal
(n, n -|- i), of a kind opposite to that due to the dead load, when
it covers the shorter of the segments into which the diagonal
divides the girder. Represent this maximum stress by D n f .
The shearing force at any section due to the live load, as it
crosses the girder, is the reaction at the end of the unloaded
segment, and the corresponding diagonal stress is the product
of this shearing force by sec 6, or -r.
The values of the different diagonal stresses are :
> = compression in o i when all the joints are loaded
sw'N(N- i)
~~k^~ N~
D l = tension in I 2 when all the joints except one are loaded
SW '(N-
~ 7~ N ~
^ = compression in 2 3 when all the joints except I and 2 are
D^ = tension in 34 when all the joints except I, 2, and 3 are
In^H ^'(^-3)(^-4)
loaded = J -- - Jr -.
D n = stress in n, n -f- i when all the joints except i, 2, 3, . . .
sw f (N-n)(N-n-i)
and n are loaded = - - ~v~~
DI stress in o I before the load comes upon the girder = 9.
608 THEORY OF STRUCTURES.
s w'
J}/= compression in I 2 when the joint I is loaded = -7 i.
K IN
/Y = tension in 2 3 when the joints I and 2 are loaded = - 3.
>,'= compression in 34 when the joints I, 2, and 3 are loaded
sw'
= kN 6 '
JD n '= stress inn,n-{-i, when the joints I, 2, . . . and ware loaded
s w' n(n-\- i)
= 'k~N~~~2~*
The total maximum stress in the th diagonal of the same
kind as that due to the dead load = d n + D n .
The resultant stress in the nth diagonal when the load
covers the shorter segment = d n D n '.
This resultant stress is of the same kind as that due to the
dead load so long as d n > D n ', and need not be considered since
d n + D n is the maximum stress of that kind.
If D n ' > d M1 it is necessary to provide for a stress in the
given diagonal of a kind opposite to that due to d n -\- D H , and
equal in amount to D n ' d n .
This is effected by counterbracing or by proportioning the
bar to bear both the stresses d H -)- D H and D n ' d n .
CASE II. Only joints denoted by even numbers loaded.
2 ' 4 N~ 2 N
vvvvvv
1 3 N-1
FlG - 376-
2 ^ N-2 N
3 N-T
FIG. 377-
Chord Stresses. The stresses are greatest when the live load
covers the whole of the girder.
WARREN GIRDER.
The total load due to both dead and live loads
609
The reaction at each abutment due to this total load
To find //i, take moments about I. Then
w I w' /
To find H % , take moments about 2.
To find //g, take moments about 3.
u k = _?L( N _ 2 ) 3 __ _ (w
To find HI, take moments about 4.
w 4- w' /
-(7^- 2)4^ -
To find H n , take moments about n, an
Then
let ^^ be even.
and
_
~
(A 7 ' 2)^ ( 2}
4~ 4
|- w' / (#" ~ *)
6 10 THEORY OF STRUCTURES.
Next, let n be odd. Then
- (w + w')x\(n - 2) + ( - 4) + + 5 + 3 +
' L \ ^ N ~ 2 ) n _ (* ~ I )'
')~N\ 4 4
and
w J rW ' I ny 2)n (n i) s
N
N
Note. If is even,
ryy _!_ / I
HN_, the stress in the middle bay, = - g -rtf.
N .
If is odd,
w + w' I N* 4
HN_, the stress in the middle bay, = g T ^7
Diagonal Stresses due to the Dead Load. The shearing forces
w
in the different bays due to the dead load are (N ' 2) between
4
w "W
o and 2, -(N 6) between 2 and 4, -(TV 10) between 4 and
4 4
^, etc.
The corresponding diagonal stresses are
s w
d in o i = T -W 2) = d l in I 2 ;
4 v
i,m 23 =( N ~ 4) =4 in 34;
/ 4 in 4 5 = j ^- 10) n= d b in 56;
* 4
etc., etc., etc.
HOWE TRUSS.
611
Thus the stresses in the diagonals which meet at an unloaded
joint are equal in magnitude but opposite in kind.
Diagonal Stresses due to the Live Load. These are found
as in Case I, and
-A,
N
If is odd, there is a single stress at the foot of each of
these columns.
The maximum resultant stress due to both dead and live
loads is obtained as before.
E.g., the maximum resultant stress in 3 4 when the longer
segment is loaded
and the maximum resultant stress in 3 4 when the shorter seg-
ment is loaded
= 4- D; = A - A'-
Note. 6 is generally 60, in which case s 2 ^..
9. Howe Truss. Fig. 378 is a skeleton diagram of a
Howe truss.
FIG. 378.
The truss may be of timber, of iron, or of timber and iron
combined.
The chords of a timber truss usually consist of three or
more parallel members, placed a little distance apart so as to
allow iron suspenders with screwed ends to pass between them
(Figs. 379 and 380).
6l2
THEORY OF STRUCTURES.
Each member is made up of a number of lengths scarfed
or fished together (Figs. 381 and 382).
The main braces, shown by the full diagonal lines in Fig.
378, are composed of two or more members.
The counter-braces, which are introduced to withstand the
effect of a live load, and are shown by the dotted diagonal
lines in Fig. 378, are either single or are composed of two or
more members. They are set between the main braces, and
are bolted to the latter at the points of intersection.
The main braces and counters abut against solid hard-wood
or hollow cast-iron angle-blocks (Fig. 380). They are designed
to withstand compressive forces only, and are kept in place by
tightening up the nuts at the heads of the suspenders.
At
FIG. 380.
FIG. 381.
FIG. 382.
FIG. 383.
FIG. 384.
The angle-blocks extend over the whole width of the chords ;
if they are made of iron, they may be strengthened by ribs.
If the bottom chord is of iron, it may be constructed on the
same principles as those employed for other iron girders. It
often consists of a number of links, set on edge, and connected
by pins (Figs. 383 and 384). In such a case the lower angle-
blocks should have grooves to receive the bars, so as to prevent
lateral flexure.
If the truss is made entirely of iron, the top chord may be
formed of lengths of cast-iron provided with suitable flanges
by which they can be bolted together. Angle-blocks may also
be cast in the same piece with the chord.
To determine the stresses in the different members, the
HOWE TRUSS. 613
same data are assumed as for the Warren girder, except that
N is now the number of panels.
Chord Stresses. These stresses are greatest when the live
load covers the whole of the girder.
Let H n be the chord stress in the nth panel.
The total load due to both dead and live loads
= (w + w'}(N i).
The reaction at each abutment due to this total load
Let a plane MM' divide the truss as in Fig. 378. The por-
tion of the truss on the left of the secant plane is kept in
equilibrium by the load upon that portion, the reaction at the
left abutment, the chord stresses in the nth panels, and the
tension in the nth suspender.
First, let the load be on the top chord and take moments
about the foot of the nth suspender. Then
or
w -f- w' n(N n)
_ w + w' I n(N ri)
ff - ~~~~'
Next, let the load be on the bottom chord and take
moments about the head of the nth suspender. Then
w + w' n(N n)
H n k - L / -- -, as before.
Thus, H n is the same for corresponding panels, whether
the load is on the top or bottom chord.
Diagonal Stresses due to the Dead Load. Let V M f be the
6l4 THEORY OF STRUCTURES.
shearing force in the nth panel, or the tension on the nth sus-
pender due to the dead load.
First, let the load be on the top chord. Then
w (N i \
V H ' = (N i) nw = w\ - n\.
Next, let the load be on the bottom chord. Then
lit //V-U I \
V m ' = --(N _,)_(_ i) w = w(t- - n].
The corresponding diagonal stresses are
and
N+i
Diagonal Stresses due to the Live Load. Let V n " be the
shearing force in the nth panel, or tension on the nth suspen-
der, when the live load covers the longer segment.
First, let the load be on the top chord.
The greatest stress in the nth brace, of the same kind as that
produced by the dead load, occurs when all the panel points
on the right of MM' are loaded. With such load, V n ", the
shearing force on the left of MM ', = the reaction at o
/' N-n
--(7VT _ n i) TT ,
2^ N
and the corresponding diagonal stress, D n ,
___
- k 2 (1V I} N
Hence, the resultant tension on the nth suspender due to
both dead and live loads = V n = V n r + V n " .
N- \ \ w '
- - n + ~(N- n -
HOWE TRUSS. 615
and the resultant maximum compression on the nth brace due
to both dead and live loads
V N-n
The live load tends to produce the greatest stress in the
nth counter when it covers the shorter segment up to and in-
cluding the nth panel point. Even then there will be no stress
in the counter unless the effect of the live load exceeds that
of the dead load in the (n + i)th brace.
The shearing force on the right of MM' = the reaction at N
_ w' n(n -\- i)
Hence,
ZV tne corresponding diagonal stress, = - - 7T ,
w 2 M V
and the resultant stress in the counter = D n f d n+t
s ( w' n(n + i) JV-
Next, let the load be on the bottom chord. Then
and
D = SW -( N ri N - n + l
k 2 ^ N
Hence,
and
Also, the stress in the ^th counter is
N i
6i6
THEORY OF STRUCTURES.
.K common value of is 45, when sec 6 = = 1.414,
and tan 6 = -rr/, l -
NK
The end panels and posts, shown by the dotted lines in
Fig. 378, may be omitted when the platform is suspended from
the lower chords.
10. Single and Double Intersection Trusses. Fig.
385 represents the simplest form of single-intersection (or
FIG. 385.
Pratt) truss ; i.e., a truss in which a diagonal crosses one panel
only. It may be constructed entirely of iron or steel, or may
have the chords and verticals of wood. The verticals are in
compression and the diagonals in tension. The angle-blocks
are therefore placed above the top and below the bottom
chord. Counter-braces, shown by the dotted diagonals, are in-
troduced to withstand the effect of a live load.
If the truss is inverted it becomes one of the Howe type,
and the stresses in the several members of both trusses may
be found in precisely the same manner.
Fig. 386 represents a double-intersection (or Whipple)
XXXXXXX
/
7
FIG. 386.
truss, i.e., a truss in which a diagonal crosses two panels. It
may be constructed entirely of iron or steel. It is of the pin-
connected type, and the two diagonal systems may be treated
independently.
Let 0' be the inclination of AB to the vertical.
" * " " " Ai, CD, . .. to the vertical.
Chord Stresses. These stresses are greatest when the live
load covers the whole of the girder.
The reaction at A from the system A BCD . . . = 4(w j r w') ;
" " A " - 4*23... = (H-*'l
SINGLE AND DOUBLE INTERSECTION TRUSSES. 6l/
Wj w' being the dead and live loads concentrated at the panel
points C, 2, E, 4, . . .
The shearing forces in the different bays are :
4(w + w') in AC, from the system ABCD . . .
(w + w') in AC, " " " ^123...
3w w n <2,
|( w _|_ w ') in 2E, " " " ^4 I 2 3
2 ( w + /') in 4, " " "
K W + w ') in 46, " " "
i(w + z/ r ) in G6, " " c<
j(, _j. ,') in 67, " " " ^123...
The corresponding diagonal stresses are :
4(w + w'} sec 6^ in ^.5 ; $%(w + w') sec 6 in ^4 I ;
3(0; -f w/ ) sec ^ in ^^ J 2 i( w + w ') sec ^ in 2 3 J etc -
Hence, the /<?/ chord stresses are :
in AC = 4w + w') tan r + 3^(0; + */) tan 6^;
C a in C2=C,+ 3(w + w') tan
= 4(0; + w') tan 0' + 6J(w + w'} tan ;
C, in 2^ = (T 2 + 2\(w -\- w') tan
= 4(w + w r ) tan 6' + 9(0; + w') tan 6^ ; etc.
The bottom chord stresses are :
7; in j5i = 4(w + w') tan ^ ;
T 9 in iZ? = T; + 3i(w + w/ ) tan e
= 4(w + w') tan 7 + 3i(w + /) tan 0=C,.
So, T Z = C,, T 4 = C 3 , etc., etc.
Again, the stress in any diagonal 4 5 of the system A I 2 . . .
due to the dead load = \\w sec 0.
The live load produces the greatest stress in 4 5, of the same
6l8 THEORY OF STRUCTURES.
kind as that due to the dead load, when it is concentrated at
all panel points of the system A I 2 3 ... on the right of 4.
The reaction at A is then ^w' , and the corresponding
diagonal stress = iw' sec 0.
Hence the maximum resultant stress in 45 = (f-> -f- J /ze/)
sec ft
The live load tends to produce the greatest stress in any
counter 5 8 when it is concentrated at all the panel points of
the system A I 2 3 ... on the left of 8.
The reaction at the right abutment is then /, and the
corresponding stress in the counter = ze/ sec 0. Thus, the
resultant stress in the counter = (f w' %w) sec 0, \w sec 6 being
the stress in 6 7 due to the dead load.
Similarly, the stresses in any other diagonal and counter
may be found.
The Pratt truss composed entirely of iron and with some
of the details of the Whipple truss is sometimes called a
Murphy-Whipple truss. The Linville truss is a Whipple truss
made of wrought-iron, the verticals being tubular columns.
II. Post and Quadrangular Trusses. The peculiarity of
the Post truss (Fig. 387) is that the
struts are inclined at an angle of
about 22 30' to the vertical, with a
FlG 387- view to an economy of material.
The ties cross two panels at an angle of 45 with the vertical.
In the quadrangular truss
(Fig. 388) the bottom chord has
additional points of support half- FIG. 3 88.
way between the panel points.
The Bollman, Fink, and other bridge-trusses have been
referred to in a previous chapter.
12. Bowstring Girder or Truss. The bowstring girder
in its simplest form is represented by Fig. 389, and is an excel-
lent structure in point of strength and economy.
The top chord is curved, and either springs from shoes
(sockets) which are held together by a horizontal tie, or has its
ends riveted to those of the tie.
The strongest bow is one composed of iron or steel cylin-
BOWSTRING TRUSS.
619
drical tubes, but any suitable section may be adopted, and the
inverted trough offers special facilities for the attachment of
verticals and diagonals.
The tie is constructed on the same principles as those em-
ployed for other iron girders, but in its best form it consists of
flat bars set on edge and connected with the shoes by gibs and
cotters.
The platform is suspended from the bow by means of ver-
tical bars which are usually of an I section, and are set with
the greatest breadth transverse, so as to increase the resistance
to lateral flexure. In large bridges the webs of verticals and
diagonals may be lattice-work.
If the load upon the girder is uniformly distributed and
stationary, verticals only are required for its suspension, and
the neutral axis of the bow should be a parabola. An irregu-
larly distributed load, such as that due to a passing train, tends
to change the shape of the bow, and diagonals are introduced
to resist this tendency.
A circular arc is often used instead of a parabola.
To determine the stresses in the different members, assum-
ing that the axis ABC of the top chord is a parabola :
Let w be the dead load per lineal foot.
" w' " " live "
" / " " span of the girder.
" k " " greatest depth BD of the girder.
Chord Stresses. These stresses are greatest when the live
load covers the whole of the girder.
The total load due to both dead and live loads (w -j- w')l.
The reaction at each abutment due to this total load
/.
620
THEORY OF STRUCTURES.
Let H be the horizontal thrust at the crown.
T " " " tension in the tie.
Imagine the girder to be cut by a vertical plane a little on
the right of BD. The portion ABD is kept in equilibrium by
the reaction at A, the weight upon AD, and the forces H
and T.
Take moments about B and D. Then
Tk =
and
= Hk,
Let H' be the thrust along the chord at any point P.
Let x be the horizontal distance of P from B.
The portion PB is kept in equilibrium by the thrust H at
B, the thrust H' at P, and the weight (w + w')x between P
and B. Hence,
H* sec' i = H'* = H* + (w + ^v}*x\
i being the inclination of the tangent at P to the horizontal,
and
tw + w'\t r
the thrust at A = +
Diagonal Stresses due to Live Load. Assume that the load is
concentrated at the panel points, and let it move from A
towards C.
If the diagonals slope as in Fig. 390, they are all ties, and
the live load produces the greatest stress in any one of them,
as QS, when all the panel points from A up to and including
Q are loaded.
BOWSTRING TRUSS. 621
Let ;r, y be the horizontal and vertical co-ordinates, respec-
tively, of any point on the parabola with respect to B as
origin.
The equation of the parabola is
Let the tangent at the apex P meet DB produced in L,
and DC produced in E.
Draw the horizontal line PM.
From the properties of the parabola, LM = 2BM.
Let PM x and BM = y.
From the similar triangles LMP and LDE,
.-
~'~~
x+QE'
A,,,
. &L _ l ~ 2x
'' QE~ 1+2X'
Draw EF perpendicular to QS produced, and imagine the
girder to be cut by a vertical plane a little on the right of PQ.
The portion of the girder between PQ and C is kept in
equilibrium by the reaction R at C> the thrust in the bow at P,
the tension in the tie at Q, and the stress in the diagonal QS.
Denote the stress by D n , and let the panel OQ be the nth.
Let be the inclination of QS to the horizontal.
Take moments about E. Then
RX CE,
or
D n = R* cosec B ....... (3)
622 THEORY OF STRUCTURES.
Let N be the total nuntber of panels. Then
is a panel length, and w'-r= is a panel weight.
Also, x = njj -, and hence
l-2x _CE___ Nn
7+ 2x ~ QE ~~ ~~^~*
R, the reaction at C when the n panel points preceding T
are loaded,
n(n
Thus, equation (3) becomes
A-y/(+i)^^cosecft .... (4)
Again, by equation (i),
_
" 2 ^ 2
N*
and
ST' ST
Hence, finally,
/ H -\TI \ (* # I )(^ ~T~ I ) I
M AT 2 < v /\
- (5)
This formula evidently applies to all the diagonals between D
and C.
BOWSTRING TRUSS. 623
Similarly, it may be easily shown that the stress in any
diagonal between D and A is given by an expression of pre-
cisely the same form.
Hence, the value of D n in equation (5) is general for the
whole girder.
A load moving from C towards A requires diagonals in-
clined in an opposite direction to those shown in Fig. 390.
Stresses in the Verticals due to the Live Load. Let V H be the
stress in the ?zth vertical PQ due to the live load. This stress
is evidently a compression, and is a maximum when all the
panel points from A up to and including O are loaded.
Imagine the girder to be cut by a plane S' S" very near PO,
Fig. 390. The portion of the girder between S'S" and C is kept
in equilibrium by the reaction R' at C, the thrust in the bow
at P, the tension in the tie at O, and the compression V n in
the vertical.
Take moments about E. Then
V n QE = X'X CE, or V n = R'^^-- ,
and R', the reaction at C when the (n i) panel points from A
to and i
Hence,
up to and including O are loaded, = / -
Vn = l-
2
a general formula for all the verticals.
Let v n be the tension in the nth vertical due to the dead
load. The resultant stress in it when the live load covers AO
is v^r- V H , and if negative, this is the maximum compression
to which PQ is subjected.
\iv n V n is positive, the vertical PQ is never in compression.
The maximum tension in a vertical occurs when the live
load covers the whole of the girder and = w'--*j + the tension
due to the dead load.
Note. The same results are obtained when N is odd.
624 THEORY OF STRUCTURES.
13. Bowstring Girder with Isosceles Bracing.
Diagonal Stresses due to the Dead Load. Under a dead load
the bow is equilibrated and the tie is subjected to a uniform
tensile stress equal in amount to the horizontal thrust at the
crown. The braces merely serve to transmit the load to the
bow and are all ties.
Let T l , 7", be the tensile stresses in the two diagonals
meeting at any panel point Q. Let 0, , # 2 be the inclinations
of the diagonals to the horizontal.
Let W be the panel weight suspended from Q.
The stress in the tie on each side of Q is the same, and
therefore T l , 7^, and Ware necessarily in equilibrium.
Hence,
T* 1X7 I J T* 117
T t W - -TT. ; 77-7, and T z = W-
Sin \y j -J- c/ 2 ) Sin ^i/j p t/ 2 y
Diagonal Stresses due to the Live Load. Let TV be the num-
ber of half panels.
2l
The length of a panel = -ry ; the weight at a panel point
Let the load move from A towards C. All the braces in-
clined like OP are ties, and all those inclined like QP are struts.
The live load produces the greatest stress in OP when it
covers the girder between A and O.
Denote this stress by D n ; OG is the th half-panel.
As before,
RX CE (i)
BOWSTRING GIRDER WITH ISOSCELES BRACING. 62$
I w'nl(n-\-2]
The load upon AO = nw -j-=, and hence R = -j~r' - TT .
The ratio of CE to EF is denoted by the same expression
as in the preceding article. Thus,
_i-
_w'ln-\-2N n[. N
" == ~8~;&+i AT N-n- I ~'.' '
The live load produces the greatest stress in OM when it
covers the girder up to and including D.
Denote the stress by D n f ; DG is now the nth half panel.
Let R f be the reaction at C.
As before,
CE
cosec 6, ...... (3)
being tjae angle MOD.
The weight upon AD = (n i)ze/-r~
and hence
2 A"' '
It may be easily shown, as in the preceding article, that
CE N-n-, A .
TvS - i - y an d cosec = N - TT^TT - -
OE n+i 4nk(Nn)
7"") ^ "^ f \
n ~~~~~%k n N N-n ~
Hence, when the load moves from A towards C, eq. (2)
gives the diagonal stress when n is even, and eq. (4) gives the
stress when n is odd.
If the load moves from C towards A, the stresses are re-
versed in kind, so that the braces have to be designed to act
both as struts and ties.
626
THEORY OF STRUCTURES.
. By inverting Fig. 391, a bowstring girder is obtained
with the horizontal chord in compression and the bow in
tension.
14. Bowstring Suspension Bridge (Lenticular Truss}.
This bridge is a combination of the ordinary and inverted
bowstrings. The most important example is that erected at
Saltash, Cornwall, which has a clear span of 445 feet. The
bow is a wrought-iron tube of an elliptical section stiffened
at intervals by diaphragms, and the tie is a pair of chains.
A girder of this class may be made to resist the action of a
passing load either by the stiffness of the bow or by diagonal
bracing.
B Q
In Fig. 392, let BD = k, B'D = k'.
Let //be the horizontal thrust at B, and T the horizontal pull
at B' ', when the live load covers the whole of the girder. Then
fjei _] not' 7 2
rr 1
L i z./ *
8
First, iet k = k' . Then
TT y
f
1 6
which is one half of the corresponding stress in a bowstring
girder of span / and depth k.
One half of the total load is supported by the bow and one
half is transmitted through the verticals to the tie. Hence,
the stress in each vertical -(w' -\- w"\
w" being the portion of the dead weight per lineal foot borne
by the verticals, and A^the number of panels.
CANTILEVER TRUSSES. 627
The diagonals are strained only under a passing load.
Let PP' be a vertical through , the point of intersection
of any two diagonals in the same panel, and let the load move
from A towards O.
By drawing the tangent at P and proceeding as in. Art. 13,
the expression for the diagonal stress in QS becomes, as before,
Similarly, the stress in the vertical QQ is
/ w'n(n i)l -
Next, let k and k' be unequal.
Let Wbe the weight of the bow, W the weight of the tie.
Then, under these loads,
~%"k = H ^ H ' : ~8~' r W'~~k'* ' ' ^
The verticals are not strained unless the platform is attached
to them along the common chord ADO. In such a case, the
weight of the platform is to be included in W'
The tangents at P and P' evidently meet AO produced in
the same point O ', for EO' is independent of k or k'. Hence,
the stresses in the verticals and diagonals due to the passing
load may be obtained as before.
15. Cantilever Trusses. A cantilever is a structure sup-
ported at one end only, and a bridge of which such a structure
forms part may be called a cantilever bridge. Two cantilevers
BEUD.GE.OVER ST. LAWRENCE AT NIAGARA.
FIG. 393.
may project from the supports so as to meet, or a gap may be
left between them which may be bridged by an independent
628
THEORY OF STRUCTURES.
girder resting upon or hinged to the ends of the cantilevers.
The form of the cantilever is subject to considerable variation.
SUKKUR BRIDGE
FIG. 394.
\ / \ /^T*"^^*^_
FORTH BRIDGE.
FIG. 395-
Figs. 396 to 401 represent the simplest forms of a cantilever
frame. If the member AB has to support a uniformly dis-
FIG. 396.
A J, i 4, -IB
FIG. 398.
FIG. 397.
FIG. 400.
FIG. 401.
FIG. 403.
tributed load as well as a concentrated load at B, intermediate
stays may be introduced as shown by the full or by the dotted
CANTILEVER TRUSSES. 629
lines in Figs. 398 and 399. Should a live load travel over
AB, each stay must be designed to bear with safety the
maximum stress to which it may be subjected.
Figs. 400 and 401 show cantilever trusses with parallel
chords. If the truss is of the double-intersection type, Fig.
401, the stresses in the members terminating in B become in-
determinate. They may be made determinate by introducing
a short link BD, Fig. 402. Thus, if, in DB produced, BG be
taken to represent the resultant stress along the link, and if
the parallelogram HK be completed, BK will represent the
stress along BE, and BH that along BF.
This link device has been employed to equalize the pressure
on the turn-table TT of a swing-bridge (Fig. 403). An " equal-
izer" or " rocker-link" BD, Fig. 404, conveys the stresses trans-
mitted through the members of the truss terminating in D to
the centre posts BT.
Theoretically, therefore, the pressure over TT will be evenly
distributed, whatever the loading may be, if the direction of
BD bisects the angle TB T and if friction is neglected.
The joint between the central span and the cantilever re-
quires the most careful consideration and should fulfil the
following conditions:
(a) The two cantilevers should be free to expand and con-
tract under changes of temperature.
(b) The central span should have a longitudinal support
which will enable it to withstand the effect of the braking of a
train or the pressure of a wind blowing longitudinally.
(c) The wind-pressure on the central span should bear
equally on the two cantilevers.
(d) The connections at both ends should have sufficient
lateral rigidity to check undue lateral vibration. Conditions
(a) and (c) would be fulfilled by supporting the central span
like an ordinary bridge-truss upon a rocker bolted down at one
end and upon a rocker resting on expansion rollers at the
other. This, however, would not satisfy condition (b). It is
preferable to support the span by means of rollers or links at
both ends, and to secure it to one cantilever only on the
central line of the bridge with a large vertical pin, adapted to
630 THEORY OF STRUCTURES.
transmit all the lateral shearing force. A similar pin at the
other end, free to move in an elongated hole, or some equiva-
lent arrangement, as, e.g., a sleeve-joint bearing laterally and
with rollers in the seat, is a satisfactory method of transmitting
the shearing force at that end also. (If there is an end post, it
may be made to act like a hinge so as to allow for expansion,
etc.) The points of contrary flexure of the whole bridge under
wind-pressure are thus fixed, and all uncertainty as to wind-
stresses removed.
Where other spans have to be built adjacent to a large can-
tilever span, it should not be hastily assumed that it is neces-
sarily best to counterbalance the cantilever by a contiguous
cantilever in the opposite direction. If it is possible to obtain
good foundations arid if piers are not expensive, it might be
cheaper to build a number of short independent side spans and
to secure the cantilever to an independent anchorage. If this
is done, care must be taken to give the abutment sufficient sta-
bility to take up the unbalanced thrust along the lower boom
of the cantilever.
Suppose that the cantilever is anchored back by means of
a single back-stay.
Let W = weight necessary to resist the pull of the back-
stay ;
h = depth of end post of cantilever ;
z = horizontal distance between foot of post and
anchorage ;
M bending moment at abutment = Wz.
If it is now assumed that the sectional areas of the post
and back-stay are proportioned to the stresses they have to bear
(which is never the case in practice), the quantity of material in
these members must be proportional to
!?* Wh = W** =
which is a minimum when z
If a horizontal member is introduced between the feet of
CANTILEVER TRUSSES. 631
the back-stay and the post, the quantity of material becomes
proportional to
h h
which is a minimum when z h, i.e., when the back-stay slopes
at an angle of 45. By making the angle between the back-
stay and the horizontal a little less than 45, a certain amount
of material may be saved in the joints of the back-stays and
also in the anchors, which more than compensates for the in-
creased weight of the anchors themselves.
(Note. In these calculations it is also assumed that the top
chord is horizontal, and that the feet of the post and back stay
are in the same horizontal plane. This is rarely the case in
practice.)
According to the above the weight of material necessary
for the back-stay is directly proportional to the bending moment
a the abutment and inversely proportional to the depth of the
cantilever, other things being equal. A double cantilever has,
in general, no anchorage of any great importance.
If the span is very great, a cantilever bridge usually re-
quires less material than any other rigid structure of equal
strength, even though anchorage may have to be provided.
If two large spans are to be built, a. double cantilever, requir-
ing no anchorage, may effect a very considerable saving in
material, although a double pier, of sufficient width for stability
under all conditions of loading, will be necessary.
Again, where false-works are costly or impossible, the
property of the cantilever, that it can be made to support
itself during erection, gives it an immense advantage. If the
design of the cantilever is such that it can be built out rapidly
and cheaply, it will often be the most economical frame in the
end, even if the total quantity of material is not so small as
that required for some other type of bridge. In all engineering'
work, quantity of material is only one of the elements of cost,
and this should be carefully borne in mind when designing a
cantilever bridge, because a want of regard to the method of
632 THEORY OF STRUCTURES.
erection may easily add to its cost an amount much greater
than can be saved by economizing material.
In ordinary bridge-trusses the amount of the web metal is
greatest at the ends and least at the centre, while the amount
of the chord metal is least at the ends and greatest at the
centre. Thus, the assumption of a uniformly distributed dead
load for such bridges is, generally speaking, sufficiently ac-
curate for practical purposes. In the case of cantilever
bridges, however, the circumstances are entirely different. In
these the amount of the metal both in the web and in the
chords is greatest at the support and least* at the end. For
example, the weight of the cantilevers (exclusive of the weight
of platform, viz., -J ton per lineal foot) for the Indus Bridge,
per lineal foot, varies from 6J tons at the supports to I ton at
the outer ends. Hence, the hypothesis of a uniformly dis-
tributed dead load for such structures cannot hold good.
The weight of a cantilever for a given span may be approxi-
mately calculated in the following manner :
Determine the stresses in the several members, panel by
panel
(A) For a load consisting of
(1) a given unit weight, say 100 tons, at the outer end ;
(2) the corresponding dead weight.
(B) For a load consisting of
(1) the specified live load ;
(2) the corresponding panel dead weight.
Thus, the whole weight of a panel will be the sum of the
weights deduced in (A) and (B), and the total weight of the
cantilever will be the sum of the several panel weights.
This process evidently gives at the same time the weights
of cantilevers of one, two, three, etc., panel lengths, the loads
remaining the same.
The panel dead weights referred to in (A) and (B) must, in
the first place, be assumed. This can be done with a large de-
gree of accuracy, as the dead weight must necessarily gradiially
increase towards the support, and any error in a particular
panel may be easily rectified by subsequent calculations.
CAN TILE VER TR USSES.
633
Again, the preceding remarks indicate a method of finding
the most economical cantilever length in any given case.
Take, e.g., an opening spanned by two equal cantilevers and
an intermediate girder. Having selected the type of bridge to
be employed for the intermediate span, estimate, either from
existing bridges or otherwise, the weights of independent
bridges of the same type and of different spans. Sketch a
skeleton diagram of the cantilever, extending over one-half of
the whole span, and apply to it the processes referred to in (A)
and (B).
If L is the length of the cantilever and P that of a panel,
the following table, in which the intermediate span increases
by two panel lengths at a time, may be prepared :
c
1 1 1)
S-, j
P-a
||6
c
rt
u
rt2
li:ll
Jf2.|4
.11
0.23
|||
'"Id
llll
*o C
ss
||
ill?
JltJl
Is
o c
3"'
aT" 5
^ ,
a
&"
&""
f*
H C
L
2P
L-2P
6P
L - 6P
SP
L - SP
etc.
etc.
Weight in col. 3 = one-half vi the weight of the intermediate
girder
+ one-half of the live load it carries if uni-
formly distributed. (The proportion will
be greater than one-half for arbitrarily
distributed loads, and may be easily de-
termined in the usual manner.)
Col. 5 gives the weights obtained as in A.
weight on end of cantilever
Col. 6 = col. 5 X -.
Col. 7 gives the weights obtained as in B.
Col. 8 = col. 2 -f- col. 6 -f col. 7.
It is important to bear in mind that an increase in the weight
of the central span necessitates a corresponding increase in the
634
THEORY OF STRUCTURES.
weights of the cantilevers. Hence, in order that the weight of
the structure may be a minimum, the best material with the
highest practicable working unit stress should be employed for
the centre span.
The table must of course be modified to meet the require-
ments of different sites. Thus, if anchorage is needed, a column
may be added for the weights of the back-stays, etc.
16. Curve of Cantilever Boom. Consider a cantilever
with one horizontal boom OA, and let x, y be the co-ordinates
of any point P in the other boom, O being the origin of co-or-
-a J
\y
FIG. 405.
FIG. 406.
dinates and A the abutment end of the cantilever.
Let Wbe the portion of the weight of an independent span
supported at O.
Let w be the intensity of the load at the vertical section
through P.
Assume (i) that there are no diagonal strains, and, hence,
that the web consists of vertical members only ;
(2) that the stress H in the horizontal boom is
constant, and therefore the bending moment
at P = Hy ;
(3) that the whole load is transmitted through the
vertical members of the web.
Let k be such a factor that kTl is the weight of a member
of length /, subjected to a stress T.
(Note. If / is in feet and T in tons, then k for steel is about
.0003, allowance being made for loss of section or increase of
weight at connections.)
w consists of two parts, viz., a constant part p, due to the
weight of the platform, wind-bracing, etc., which is assumed to
CURVE OF CANTILEVER BOOM. 635
be uniformly distributed ; and a variable part, due to the
weight of the cantilever, which may be obtained as follows:
Weight of element dx of horizontal boom = kHdx.
" " web corresponding to dx = kwydx.
" " element of curved boom corresponding to dx
-).
Hence the variable intensity of weight
and
w =
Again, if M is the bending moment and 5 the shearing
force at the vertical section through P, then
d'M ttS d*y
kH
Integrating twice,
A and B being constants of integration.
dv
When x = o, y = o, and ffg- = W.
Thus, A=o and B = W.
636 THEORY OF STRUCTURES.
Hence,
is the equation to the curve of the boom, and represents an
ellipse with its major axis vertical, and with the lengths of the
(p + 2kH\*
two axes in a ratio equal to ^ -777 -- J .
The depth of the longest cantilever is determined by the
vertical tangent at the end of the minor axis, and corresponds
to the value of y given by making = o in the preceding
equation, which gives y = .
For a given value of H the curve of the boom is independ-
ent of the span. Again, for a given length of cantilever with
a boom of this elliptic form, a value of H may be found which
will make the total weight a minimum, and which will there-
fore give the most economical depth. Such an investigation,
however, can only be of interest to mathematicians, as the
hypotheses are far from being even approximately true in
practice, and the resulting depth would be obviously too great.
Assumption (i) on page 634 no longer holds when a live
load has to be considered. Diagonal bracings must then be
introduced, which become heavier as the depth increases, in
consequence of their increased length. The diagonal bracings
are also largely affected by the length of the panels. If the
panels are short, and if a great depth of cantilever, diminishing
rapidly away from the abutment, is used, the angles of the
diagonal bracing, near the abutment, will be unfavorable to
economy. This difficulty may be avoided by adopting a
double system of triangulation over the deeper part of the
cantilever only, or even a treble system for some distance in
a large span. The objections justly urged against multiple
systems of triangulation in trusses lose most of their force in
large cantilevers. In the first place, the method of erection
by building out insures that each diagonal shall take its proper
share of the dead load ; and in the second place, it should be
CURVE OF CANTILEVER BOOM. 637
remembered that only in large spans could a double system
have anything to recommend it, and then only near the abut-
ment where the stresses are greatest : in such cases the moving
load only produces a small portion of the entire stress in the
web. In practice, a compromise has to be made between dif-
ferent requirements, and the depth must be kept within such
limits as will admit of reasonable proportions in other respects,
while the diagonal ties or struts may be allowed to vary in in-
clination, to some extent, from one panel to another.
Again, in fixing the panel length, care must be taken that
there is no undue excess of platform weight, as this will pro-
duce ?, corresponding increase in the weight of the cantilever.
An excessive depth of cantilever generally causes an in-
crease in the cost of erection.
Both theory and practice, however, indicate that it will be
more advantageous to choose a greater depth for a cantilever
than for an ordinary girder bridge.
An ordinary proportion for a large girder bridge would be
one-ninth to one seventh of the span, and if for the girder were
substituted two cantilevers meeting in the middle of the span,
the depth might with advantage be considerably increased
beyond this proportion at the abutment, if it be reduced to nil
where the cantilevers meet. When a central span is introduced,
resting upon the ends of the two cantilevers, the concentrated
load on the end gives an additional reason for still further in-
creasing the depth at the abutment proportionally to the hngth
of the cantilever. The greatest economical depth has probably
been reached in the Indus bridge, in which the depth at the
abutment = .54 X length of cantilever. Probably the propor-
tion of one-third of the length of the cantilever would be
ample, except where the anchorage causes a considerable part
of the whole weight, but each case must be considered on its
own merits. The reduction of deflection obtained by increas-
ing the depth is also an appreciable consideration.
If a depth be chosen not widely different from that which
makes the quantity of material a minimum, the weight will be
only slightly increased, while it is possible that great structural
advantages may be gained in other directions. In recommend-
638 THEORY OF STRUCTURES.
ing a great depth for a cantilever at its abutment, it is assumed
that the depth will be continuously reduced from the abutment
outwards. If the load were continuously distributed, it is by
no means certain that a cantilever of uniform depth would re-
quire more material than one of varying depth, but it has
already been pointed out to what extent the weight of the
structure itself necessarily varies, and if the concentrated load
at the end were separately considered, the economical truss
would be a simple triangular frame of very great depth. From
economic considerations, it would be well to reduce the depth
of the cantilever at the outer end to nil, but in many cases it
is thought advisable to maintain a depth at this point equal to
that at the end of the central span, so that the latter may be
built out without false-works, under the same system of erection
as is pursued in the case of the cantilever. The post at the
ends of the central span and cantilever is sometimes hinged to
allow for expansion.
17. Deflection. A serious objection urged against can-
tilever bridges is the excessive and irregular deflection to which
they are sometimes subject. They usually deflect more than
ordinary truss-bridges, and the deflection is proportionately
increased under suddenly applied loads. In the endeavor to
recover its normal position, the cantilever springs back with
increased force and, owing to the small resistance offered by
the weight and stiffness at the outer end, there may result,
especially in light bridges, a kicking movement. It must, how-
ever, be borne in mind that the deflection, of which the impor-
tance in connection with iron bridges has always been recog-
nized, is not in itself necessarily an evil, except in so far as it is
an indication or a cause of over-strain.
18. The Statical Deflection, due to a quiescent load,
must be distinguished from what might be called the dynamical
deflection, i.e., the additional deflection due to a load in motion.
The former should not exceed the deflection corresponding to
the statical stresses for which the bridge is designed. The
amount of the dynamical deflection depends both upon the
nature of the lo^ds and upon the manner in which they are
applied, nor are there sufficient data to determine its value
LIVE LOAD. 639
even approximately. It certainly largely increases the statical
stresses and produces other ill effects of which little is known.
Hitherto, the question as to the deflection of framed struc-
tures has received but meagre attention, and formulae deduced
for solid girders have been employed with misleading results.
It would seem to be more scientific and correct to treat each
member separately and to consider its individual deformation.
19. Rollers. One end of a bridge usually rests upon nests
of turned wrought-iron or steel friction rollers running between
planed surfaces. The diameter of a roller should not be less
than 2 inches, and the pressure upon it in pounds per lineal
inch should not exceed 500 Vd if made of wrought-iron, or
600 tfa if made of steel, d being the diameter in inches.
20. Live Load. It is a common practice with many en-
gineers to specify the live load for a bridge as consisting 01 a
number of arbitrary concentrated weights which are more or
less equivalent to the loads thrown upon the locomotive and
car axles.
Figs. 407, 408, and 409 are examples of such practice.
90'
FIG. 407.
FIG. 408.
f
FIG. 409.
With such a live load, the determination of the position of
the locomotive and cars which will give a maximum shear and
a maximum bending moment at any section is much facilitated
by the principles enunciated in Art. 8, Chap. II.
640 THEORY OF STRUCTURES.
If the chords* are parallel, and if 5 is the maximum shear
transmitted through a diagonal inclined at an angle 6 to the
vertical, the maximum stress in that diagonal = vSsec 0, and
the corresponding stress transmitted to a chord through the
diagonal
= Ssec#sin# = 5 tan 6.
A modification is necessary when the chords are not paral-
lel. Consider, e.g., a truss with a horizontal bottom chord
and a top chord composed of inclined members. Retain the
same notation as in the article referred to, and let />, , D 9 be
the stresses corresponding to \hefirst and second distributions,
respectively, in a diagonal met by a vertical section between
the rth and (r -\- i)th weights. Also, let the member of the
upper chord cut by the same section be produced to meet the
horizontal chord produced in the point C.
FIG. 410.
Let AC = h, and let / be the perpendicular from C upon
the diagonal in question.
Taking moments about C,
D,p RJi w,
and
w r (h + / a r x) w r+l (h + / a r ^ x) . . .
- w r+q (A + I a r+q x).
It. is assumed, for simplicity, that no weights leave or ad-
vance upon the bridge.
/. A A ,
LIVE LOAD.
641
according as
a,) . . . w r (k + la r ) =
RJt w,(h + I a, x) w^(h + / a, x) . . .
- w r (h -\-l-a r x) tv r+l (h + I a r+I x) . . .
- w r+q (h + I - a r+q -f x\
or
=
where R q '(l -\- h) = algebraic sum of the moments, with respect
to C f of the weights transferred,
and
Hence,
=
according as
Take, e.g., the truss represented by the accompanying dia-
gram (Sault Ste. Marie Bridge), the live load being that shown
by Fig. 411, i.e., the loading from a Standard Consolidation
engine with four drivers and one leading wheel.
FIG. 411.
Span 239 ft.
Length of centre verticals 40 ft.; of end verticals 27 ft,
THEORY OF STRUCTURES.
Applying the principles referred to in the preceding it is
found that the distributions of live load, concentrated at the
panel points, which will give the maximum stresses in the
several members, may be tabulated as below :
End
Distribu-
Reac-
Load
Load
Load
Load
Load
Load
Load
Load
Load
tions.
tion
at A-
at/ 2 .
at/ 8 .
at/> 4 .
at A-
at/ 6 .
at/ 7 .
at/e-
at/,.
at A.
Case i
2
187990
162920
495<
38700
495oo
459 2 5
38700
43750
45925
36225
43750
36000
36225
36000
36000
36000
36000
36000
36000
3
124230
6400
47209
40200
43400
45800
37100
36000
36000
6400
47200
40200
43400
45800
37100
69410
6400
47200
40200
4^400
45800
37100
6
47400
6400
47200
40200
43400
45800
7
29100
6400
47200
40200
43400
" 8
15380
6400
47200
40200
Dead weight
121500
27000
27000
27000
27000
27000
27000
27000
27000
,27000
N.B. These numbers are convenient whole numbers within
about one-half of one per cent of the calculated results. The
panel length is also assumed to be 24 ft.
In Cases I arid 2 the third driver is at a panel point ; in the
remaining cases the second driver is at a panel point.
The dead weight includes the weight of the ironwork and
flooring. The panel loads may be easily calculated, either
analytically or graphically. For example, let A, B, C, D be
four consecutive panel points, and let the third driver be at B.
Panel load at A
,
= 7500 g + 12000
Panel load at B
/io8 2\
J = 11823, say 11,900 Ibs.
8o+236+288+232
> sa y 49*500 Ibs.
Panel load at C
87+21
288
= 38445, say 38,700 Ibs.
LIVE LOAD.
643
Or, graphically, upon the vertical through B (Fig. 412) take
BM to represent 7500 Ibs., and join AM. Let the vertical
through #, meet AM in , , and the horizontal through AM
in c t . Then a^ represents the portion of 7500 Ibs. borne at
B, and b l c l the portion borne at A.
Also, take BN to represent 12,000 Ibs. ; join ^4 N, CN. Let
the verticals through a^, a 9 , <z 4 meet AN, CN in b^b^,bi, and
the horizontal through ^V in a , 8 , 4 . Then #.,&,, # 3 ^ s > #A
FIG. 412.
represent the portions of each 12,000 Ibs. borne at B, while
AA> ^3 represent the portions borne at A, and 4 4 the portion
borne at C.
Finally, take BO to represent 10,625 Ibs., and join CO. Let
the verticals through a 6 , a 6 meet CO in < 6 , ^ 6 , and the horizon-
tal through O in 6 , c % . Then # B ^ 5 , ajb^ are the portions of each
10,625 Ibs. borne at B, while b^c^, b 6 c 6 are the portions borne at
C. Thus the total weight at B
= <* A + "A + "A
A
A
It is open to grave question whether the extremely nice
calculations required by the assumption of arbitrary weight
calculations are not unnecessary except for floor systems. The
constantly increasing locomotive and car weights and the
variety in type of locomotive would seem to render such cal-
culations, based as they are upon one particular distribution of
load, of no effect.
On the other hand, if it is assumed that the standard live
load consists of a uniform load of, say, 3000 Ibs. to 3600 Ibs.
per lineal foot, with a single weight of, say, 25,000 Ibs. to
35,000 Ibs. for each truss, at the head or at any other specified
644
THEORY OF STRUCTURES.
point, i.e., rolling on the uniform load, the calculations would
be much simplified and the resulting stresses would be at least
as approximately accurate.
Let E be the single concentrated load, T the panel train
load, and D the panel dead load.
Consider a truss of N panels with a single diagonal system,
Fig. 413, and let E be at the rth panel point.
N-t N
FIG. 413.
The shear immediately in front of due to
the shear at same point due to T
~ N 2
the shear at same point due to D
D N(N 2r + I)
-W~ ~^ ~*
Diagonal Stresses. The maximum diagonal stresses may
now be easily tabulated as follows :
TABLE T.
2
O
S-o
o
" ' J5
%
^
%
V
3
Q ^
o
o
6
o
T
1
>
flj
%
Z
5
2 . |
Q
o
Diagonal.
Multiplier (N -
Max. Vertical S
E transmitted
(1)
Multiplier
(N-r-i
M
!p
Total Maximut
Shear due to
transmitted.
Secant 6.
Max. Diagonal S
Live Load.
Multiplier
N(N - 2r
Vertical Shear d
Load transmit
D
N
Diag. Stress due t
j
fj t
A
N.N-T
N-I.N--L T
JV i^
N.Ni
N.N-i D
a 1
*w
2
2 N
1 JV
2
2 N
N-a.N-i T
~~l N
LIVE LOAD. 645
Col. I designates the several diagonals.
Col. 2 gives the multiplier N r for different values of r.
Gol. 3 gives the maximum vertical shears due to E trans-
mitted through the several diagonals. This shear for any
given diagonal is the product of the corresponding multiplier
in col. 2 and -r^r.
Cols. 4 and 5, 9 and 10 give similar quantities for the live
and dead loads.
Col. 6 gives the sums of the shears in cols. 3 and 5, i.e., it
gives the total maximum vertical shears due to live load.
Col. 8 gives the maximum diagonal stresses due to live
load. For any specified diagonal it is the product of the cor-
responding shear in col. 7 and the secant of the angle between
the vertical and the diagonal in question.
Col. 1 1 in like manner gives the maximum diagonal stress
due to dead load.
Col. 12 gives the total maximum diagonal stresses due to
both live and dead loads.
Another column might be added giving the sectional areas
of the diagonals.
In the above table the diagonal stresses due to the live and
dead loads are separately determined, as different coefficients
of strength are sometimes specified for the two kinds of load.
With a suitable compound coefficient of strength, cols. 6, 8,
and ii may be replaced by a column giving the sums of the
corresponding shears in cols. 3, 5, and 10. These sums, multi-
plied by secant #, give the maximum diagonal stresses.
Stresses in the Verticals. The maximum stress in any ver-
tical, say at the rth panel point, is evidently the vertical com-
ponent of the maximum diagonal stress in the rth panel, i.e., it
is the maximum vertical shear in the rth panel.
To be more accurate, this amount should be diminished by
the portion of the weight of the lower chord borne at the foot
of the vertical in question.
Chord Stresses. Take the load at each panel point
= ~
646
THEORY OF STRUCTURES.
TABLE II. (COMPRESSION CHORD.)
-
T3X!
p
/
N
C
1
I
+
o
sl+
ll* 1
rtC H
^
-
11 -
5*21
^1
PH
I-
a
*
fl
gs^s
a
rt
^6^
SsS
-1
2
>
H
C/5
H
Col. i designates the chord panel length.
Col. 3 gives the several vertical shears transmitted to the
chords through the diagonals. They are the product of
W\W ~^~ ^ ~^~ i
Col. 5 gives the chord stresses due to these shears, i.e., the
product of the shears in col. 3 and the corresponding values of
tan in col. 4.
Col. 6 gives the total chord stresses in the several panels.
In any given panel the total chord stress is equal to the chord
stress due to the shear in that panel plus the total chord stress
in the preceding panel.
Another column for the sectional areas of the several
lengths of chord may be added if required, each length being
designed as a strut, hinged or fixed at the ends, according to
the method of construction.
A precisely similar
table may be prepared
for the tension chord.
EXAMPLE i. An
7V^/-panelled deck-truss
of 108 ft. span and 18 ft. deep, with a single diagonal system.
Concentrated load E for each truss = 25,000 Ibs.
Train load T for each truss = 1600 Ibs. per lineal ft.
= 21,600 Ibs. per panel.
Bridge (dead) load D for each truss = 800 Ibs. per lineal ft.
= 10,800 Ibs. per panel.
sec 8 = , tan 6 = f .
c 4 a
LIVE LOAD. 647
TABLE OF MAXIMUM DIAGONAL STRESSES. (See TABLE I.)
uS
T
1
8
fx
2
.2 g-o
i
8?
II
i
(I
co
II
u
"t3
i
00 o
i
k
Q
k
8!
3?*
c
C
x^>
. * J 'd
2 *
* rt
1
&
00
1
v8 oo
OWJiJ
5^3
0>
00
00
OQ
Q
00
N
Y
1
r 1
C/)
S^
*
M
Q"
H
4
7
21875
21
56700
78575
li
982,8}
28
37800
4725
145468}
d<i
6
18
7 So
15
40500
5925
I*
74062^
20
27
000
33750
107812$
d^
5
is
62=;
10
27000
42625
12
16
200
20250
7353 I i
<'*
4
12
Soo
6
16200
28700
t*
35875
4
5400
6750
42625
1i
3
2
9375
6250
3
i
8100
2700
J7475
8950
:j
21843*
11187*
- 4
54o
16200
- 6750
20250
15093*
d,
'
3125
3125
3906*
20
27000
-33750
It will be observed that in the fifth panel there is a maxi-
mum positive shear of 17,475 Ibs. and a negative shear of 5400
IDS., the former due to the live and the latter to the dead load.
The resultant shear of 12,075 I DS - which is opposite in kind to
that due to the dead load, is provided for by means of the
counterbrace ab. No counterbraces are theoretically required
in the sixth and seventh panels, but they are often introduced
in order to stiffen the truss.
TABLE OF MAXIMUM STRESSES IN THE VERTICALS.
^ = 78575 + 378oo = 1 16,375 Ibs.
v^ 59250 + 27000 = 86,250 "
i\ 27000 -|- 16200 = 43,200 "
v^~ 16200+ 5400= 21,600 "
Chord Stresses. Load at each panel point
= -+ T+D= 35,525 Ibs.
TABLE OF MAXIMUM STRESSES IN COMPRESSION CHORD.
Member.
4(9 - ar).
Sf=<4
Tanfl.
Chord Stress due
to Shear.
Total Max.
Chord Stress.
fl
28
I24337i
1
93253i
93253i
<*
2O
888i2i
f
666ogf
159862^
c*
12
5328;i
*
399 6 5f
199828^
ft
4
17762^
f
I332i|
213150
648 THEORY OF STRUCTURES.
TABLE OF MAXIMUM STRESSES IN TENSION CHORD.
Member.
4(9 - vr).
-*
Tanfl.
Chord Stress due
to Shear.
Total Max.
Chord Stress.
ti
28
124337!
f
93253*
93253*
ti
20
88812!
f
666o 9 f
159862!
12
532871
f
399 6 5t
199828^
The above figures may be checked by the method of
moments.
Note. If the truss is inverted it becomes one of the Howe
type. The stresses are the same in magnitude, but reversed
in kind.
Ex. 2. An eight-panel through-bridge of the double-inter-
section type (Fig. 414), hav-
ing the same span, depth,
and loading as in Ex. i.
The systems 1234...
FlG - 4 ' 4 - and I a b c . . . are inde-
pendent. It is assumed that the load at the foot of the end
vertical (gh) is divided equally between the two systems.
TABLE OF MAXIMUM STRESSES IN END-POSTS AND
DIAGONALS.
o
o
m>a
10
.
8
^ OJ-O
nf^-d
1
L
il
v
m
_
N
S 3 g
.
S 3
fc
1
^
c^Q
si
a
II
71
II
u 1 -^
_
*^i-3
'a
I
\
o .
C/5
a
V
H
O 1
8 00
"5
O 1
Jloo
"5J.I
i
K^>
rt's'l
2.2 1
i
X
&l
X
51
H^^
in
|c^HJ
i
C" ^
H 5
!
7
6
5
4
21875
18750
15625
12500
21
6*
^}
56700
17550
9450
6750
78575
36300
25075
19250
I
982I8J
45375
45135
34650
28
"i
4i
37800
16875
"475
6075
47250
21093!-
20655
10935
i 454 68f
66465*
65790
45585
*4
3
9375
J-
1350
10725
1 r
J 935
1.
)j C
1215
20520
?
2
i
6250
1562*
1562*
J 3S
7600
1562*
15624
ii
13680
28124
1953*
- 34
~ll*
- 4725
10125
-'5525
- 8505
-18225
2 7945
5'75
The counterbrace cf is required to take up the resultant
shear of 6250 4725 = 1525 Ibs., which is opposite in kind to
that due to the dead load.
The first line in the table gives the maximum thrust along
the end post (/). It is made up of the stresses transmitted
LIVE LOAD.
649
through the two systems of diagonals when the 25,000 Ibs. is
at the first panel point.
TABLE OF MAXIMUM STRESSES IN VERTICALS.
The maximum stress in an end vertical evidently occurs
when the 25,000 Ibs. is concentrated at its foot.
z/j = 25000 + 10800 = 35800 Ibs. (tension);
^19250+ 6075 = 25325 " (compression);
v 3 = 10725 + 675 = 11400 "
v 4 = 6250 4725= 1525 " "
Chord Stresses. Load at each panel point
= ~+T+D = 35525.
TABLE OF MAXIMUM STRESSES IN COMPRESSION CHORD.
Total
Member
Multiplier.
25|5 = 444 of .
Tan 0.
Chord Stress
due to Shear.
Maximum
Chord
Stress.
( 28
I24337&
93253*
j
Cl
j +12*
55507^1
37745*
I
4163011
566171*
| T 9i50i$i
191501$!
c?
4*
19982^
.*
29974-^
221476^5
Cl
*
j
3330*|
224806^
Note. c l is made up of the thrusts transmitted through
TABLE OF MAXIMUM STRESSES IN TENSION CHORD.
Member.
Multi-
plier.
35525
Tang.
Chord Stress
due to Shear.
Total Maximum
Chord Stress.
8
/I = /S
28
124337*
i
93253*
93253^
/ 3
19}
i
''
37745 T 5 ir
i
56617!*
191501!*
Ex. 3. A through-bridge of the Warren type (Fig. 415)
having the same span and loading as in Exs. i and 2.
FIG. 415.
650 THEORY Of STRUCTURES.
TABLE OF MAXIMUM STRESSES IN DIAGONALS.
$
8
.
1
iii
1
0.
II
if.
t^
M
II
.a
13.
II
-"1
c.
11.
g eg
i
u
1
8 OO
"3
as
00
"3
00 OO
III
1
|II
d\ ~=-di
7
21875
21
56700
28
37800
116375
1.155
I344I4
dsdt
6
18750
15
40500
20
27000
86250
99619
df, = d*
5
15625
10
27OOO
12
16200
58825
67943
dT=d*
4
12500
6
16200
4
5400
34100
39386
dg=div
3
9375
3
8100
4
- 5400
12075
13947
dn=d^
2
6250
i
27OO
12
16200
d\$~=-d\i
I
3125
20
27000
The resultant stresses, d 9 = d l9 , are of an opposite kind to
the corresponding stresses due to the dead load. Thus, the
diagonals upon which they act must be designed so as to bear
both tensile and compressive stresses. The stresses d lt d s ,
d t , . . . are compressions, and d 9 , d^ , </ 6 , . . . tensions.
TABLE OF MAXIMUM STRESSES IN COMPRESSION CHORD.
Mem-
ber.
Multi-
plier.
3 -f^ = 4440L
Shear
transmitted.
Tanfl.
Chord Stress.
Total
Maximum
Chord
Stress.
c\
j 28
i 28
Through d* 124337!
d* 124337!
| 248675
577
143485 +
143486
Ci
( 20
/ 2O
d t 88812!
d, 88812!
| 177625
577
102489 +
245976
Cz
< 12
( 12
</ 6 53287!
^8 53287!
| 106575
577
61493 +
307469
Cl
\ 4
\ 4
^7 17762!
<** 17762!
| 35525
577
20497 +
32/967
TABLE OF MAXIMUM STRESSES IN TENSION CHORD.
Total
Mem-
ber.
Multi-
plier.
^f = 444of
Shear
transmitted.
TanQ.
Chord Stress.
Maximum
Chord
Stress.
/i
28
Through d\ 124337!
124337!
577
71742 +
7
/I
-
28
20
d* 124337!
d* 88812!
[213150
577
122987 +
19.
*
20
12
^ 4 88812!
d* 53287*
I 142100
577
81991 +
276722
f
12
4
d* 53287!
' di 17762^
| 71050
577
40995 +
3i77i8
WIND PRESSURE. 651
21. Wind-pressure. Numerous experiments to deter-
mine the pressure and velocity of the wind have been made
by means of feathers, cloud-shadows, anemometers of various
kinds, wind-gauges, pendulum, tube, and spring instruments.
The results, either through errors of observation, errors of con-
struction, or for other occult reasons, are almost wholly unre-
liable and give the engineer no accurate information upon
which to base his calculations as to the effect of wind upon a
structure. Theoretical investigations on the subject are equally
unsatisfactory. The formulae expressing the relations between
the speed of the anemometer, the velocity of the wind and its
pressure, are of a purely empirical character, and are only
applicable to a specific series of recorded observations.
Smeation inferred from Rouse's experiments that the aver-
age pressure in pounds per square foot = (velocity in miles per
hour)" -r- 200, or
... : ,,,.. :
200
According to Dines the formula should be
2000
The Wind-Pressure Commission (Eng.) recommended the
formula
F'
=
.as giving with tolerable accuracy the maximum pressure.
,,Stokes considers that the actual wind velocities should be
^ 2.4 4
ut = of the values recorded by anemometers, so
t a velocity of 64 miles per hour recorded as corresponding
t.o : & maximum pressure of 40.6 Ibs. per square foot (the aver-
age oi five observed pressures) would be reduced to 51.2 miles
per hour. The average pressure corresponding to 51.2 miles
652 THEORY OF STRUCTURES.
per hour would be 13.1 Ibs. per square foot according to
Smeaton's rule and only 9.18 Ibs. according to Dines.
Again, certain experiments at Greenwich indicated that
the pressure was increased by the stiffness of the copper wire
connecting the recording pencil with the pressure plate, and a
flexible brass chain was therefore substituted for the wire.
Thus modified, a pressure of 29 Ibs. per square foot was regis-
tered as corresponding to a velocity of 64 miles per hour,
whereas with the copper wire a pressure of 49^- Ibs. per square
foot had been registered with a velocity of only 53 miles per
hour.
These facts tend to show that the actual pressure is much
less than that given by a recording instrument, and that the
very high pressures, as, e.g., 80 Ibs. per square foot and even
more, must be due to gusts or squalls having a purely local
effect. This opinion seems to be confirmed by Sir B. Baker's
experiments at the Forth Bridge, which also indicate that the
pressure per square foot diminishes as the area acted upon
increases. No engineering structure could withstand a press-
ure of 80 Ibs. per square foot of surface, and a pressure of 28
Ibs. to 32 Ibs. would overturn carriages, drive trains from the
track, and stop all traffic.
It is, of course, well known that wind-forces sufficiently
powerful to uproot huge trees and to demolish the strongest
buildings are occasionally developed by whirlwinds, tornadoes,
and cyclones, but these must be classed as acta Dei and can
scarcely be considered by an engineer in his calculations.
Numerous observations as to the effect of wind upon struc-
tures in different localities must yet be made before any useful
and reliable rules can be enunciated. In the case of existing
bridges the elongation of the wind-braces during a storm can
easily be measured within -^-^ of an inch. Investigations
should be made as to the action of the wind upon surfaces of
different forms and upon sheltered surfaces, as, e.g., upon the
surfaces behind the windward face in bridge-trusses. Again,
it is quite possible, if not probable, that many of the recorded
upsets have been due to a combined lifting and side action,
requiring a much less flank-pressure than would be necessary
EMPIRICAL REGULATIONS. 653
if there were no upward force, and hence further light should
be obtained on this point.
Under any circumstances, the wind-stresses should be as
small as possible, compatible with safety, seeing how largely
they influence the sections of the several members, especially
in bridges of long span.
22. Empirical Regulations.
Wind-Pressure Commission Rules. For railway bridges and
viaducts assume a maximum pressure of 56 Ibs. per square foot
upon an area to be estimated as follows :
A. In c/ose-girder bridges or viaducts the area
= area of windward face of girder
-|- area of train surface above the top of the same gir-
der.
B. In 0/^-girder bridges or viaducts the area for the wind-
ward girder
= area of windward face, assumed close, between rails
and top of train
-|- calculated area of windward surface above the top
of the train
-j- calculated area of windward surface below the rails.
For the leeward girder or girders the area
= calculated area of surface of one girder above the
top of the train and below the level of the rails,
the pressure being 28, 42, or 56 Ibs. per square
foot, according as this area < f5, > \S and <fS,
or >{5, where 5 is the total area within the out-
line of the girder. The assumed factor of safety
is to be 4.
American Specifications. (a) The lateral bracing in the
plane of the roadway is to be designed so as to bear a pressure
of 30 Ibs. per square foot upon the vertical surface of one
truss and upon the surface of a train averaging 12 sq. ft. per
lineal foot, i.e., 360 Ibs. per lineal foot ; this latter is to be re-
garded as a live load. The lateral bracing in the plane of the
other chord is to be designed so as to bear a pressure of 50 Ibs.
per square foot upon twice the vertical surface of one truss.
(&) The portal, vertical, and horizontal bracing is to be
654
THEORY OF STRUCTURES.
proportioned for a pressure of 30 Ibs. per square foot upon
twice the vertical surface of one truss and upon the surface of
a train averaging 10 sq. ft. per lineal foot, i.e., 300 Ibs. per
lineal foot, the latter being treated as a live load.
(c) Live load in plane of roadway due to wind-pressure
= 300 Ibs. per lineal foot.
Fixed load in plane of roadway due to wind-pressure
= 150 Ibs. per lineal foot.
Fixed load in plane of other chord due to wind-pressure
= 150 Ibs. per lineal foot.
Lateral Bracing. Consider a truss-bridge with parallel
chords and panels of length/. Let A be the area of the ver-
tical surface of one truss.
According to (a), the lateral bracing in the plane of the
roadway is subjected to (i) a panel live load of 360^ Ibs. and
(2) a panel fixed load of $oA Ibs., while in the plane of the
other chord it is subjected to a panel fixed load of
50 X 2A 100^4 Ibs.
Thus, if the figure represent the bracing in the plane of the
roadway of a ten-panel truss, and if the wind blow upon the
30|A
30 A
80JA 36CJjP
3o|A
S6 P
FlG. 416.
36g P 360 P 360 P 36<j[ P
side AB, the maximum horizontal force for which any diagonal r
e.g- CD, is to be designed is
= 4$A Ibs. due to the horizontal force of 30^ Ibs. at
each panel point
+ 756^ Ibs. due to the horizontal force of 360^ Ibs. at
each panel point between C and B.
The dotted lines show the bracing required when the wind
blcws on the opposite side.
It is sometimes maintained that the wind-forces in the
CHORDS. 65$
plane of the upper chords of a through-bridge or the lower
chords of a deck-bridge are transmitted to the floor-bracing
through the posts. This can hardly be correct in the case of
long posts, as they do not possess sufficient stiffness. It has,
however, been pointed out by Mr. W. B. Dawson that, in
through-bridges, the cumulative effect of the wind-pressure at
the ends of the bridge might produce a serious bending action
in the end posts. This action would have to be resisted by
additional plating on the end posts below the portals, or by an
increase of their sectional area.
Under wind-pressure the floor^beams act as posts ; hence,
if the wind-bracing is attached to the top or compression flange
of a floor-beam, the flange's sectional area must be propor-
tionately increased. If the bracing is attached to the lower
or tension flange, the stresses in the latter will be diminished.
23. Chords. The wind-pressure transmitted through the
floor-bracing increases the stresses in the several members, or
panel lengths, of the leeward chord, the greatest increments
being due to a horizontal force of (360^ -f- $oA) Ibs. at each of
the panel points in AB. The corresponding chord stresses in
the ten-panel truss-bridge referred to above are :
,=0;
C 9 = 4^(360;) -f- 30^) tan 6 Ibs. ;
C 3 = C, + 3i(3<x>/ + 30^) tan 8 = 8(360^ + 30^) tan V Ibs. ;
C 4 C 3 + 2|(36o/ + $oA) tan B = \o%($6op + 30^) tan Ibs. ;
tan = \26o 0^ tan fl Ibs.
90 8 being the angle between a diagonal and a chord.
Again, the wind-pressure tends to capsize a train and throws
y
an additional pressure of P-=. Ibs. per lineal foot upon the lee-
ward rail, P being the pressure in pounds per lineal foot on the
train surface, y the vertical distance between the line of action
of P and the top of the rails, and G the gauge of the rails.
THEORY OF STRUCTURES.
Thus, the total pressure on leeward rail
(
w v\
2 +P G) lbS> Pei " Hneal f 0t >
and the total pressure on windward rail
(w y\
- P ~j Ibs. per lineal foot,
w being the weight of the train in pounds per lineal foot.
Hence, the total vertical pressure at a panel point of the
leeward truss
S being the distance between the trusses.
24. Stringers. Each length of stringer between consecu-
tive floor-beams may be regarded as an independent girder
resting upon supports at the ends, and should be designed to
bear with safety the absolute maximum bending moment to
which it may be subjected by the live load. If the beams are
not too far apart, the absolute maximum bending moment will
be at the centre when a driver is at that point. Again, in the
case of the Sault Ste. Marie Bridge, it may be easily shown
that the maximum bending moment is produced when the four
pairs of drivers are between the floor-beams.
Let/ distance of first driver from nearest point of support.
The reaction at this support
= 4tr(824 - 47) = 4^(206 -y).
The bending moment is evidently a maximum at the second
or third driver, and at the second driver
o.o.( 2 o6 y)($6 +7) 12000 X 56 ;
MAXIMUM ALLOWABLE STRESS.
at the third driver
= 104(206 7X108 +7) 12000(52 + 108).
In the first case it is an absolute maximum when y = 75" ;
" " second " " " " " " y 49" ;
its value in each case being 2,i88,i66| in.-lbs.
Hence, the bending moment is an absolute maximum and
equal to 2,i88,i66f in.-lbs., at two points distant 75 in. from
each point of support.
Also, if /! is the moment of inertia of the section of the
stringer at these points, , the distance of the neutral axis from
the outside skin, and/ t the coefficient of strength, then
-(2i88i66f) /i for the inner stringer,
and
-(2i88i66f) = /,- for the outer stringer.
3 c \
The continuity of the stringers adds considerably to their
strength.
25. Maximum Allowable Stress. Denoting by A and B,
respectively, the numerically greatest and least stresses to
which a member is to be subjected, the following rules will give
results which are in accordance with the best practice :
I. Members subjected to Tensile Stresses only.
For ivr ought-iron, maximum stress per square inch
= 10000 Ibs. 8000^1 + -jj Ibs. = ^3.81 + 1.9 -jj tons.
For steel, maximum stress per square inch
= 12000 Ibs. = 10000(^1 -f- -rj Ibs. = [,5.08 -f- 2.54 J tons.
658 THEORY OF STRUCTURES.
II. Members subjected both to Tensile and Comprcssivc Stresses.
For wr ought-iron, maximum stress per square inch
= 8000(1 - -jj Ibs. = (3 8l -- 1.9^) tons.
For steel, maximum stress per square inch
= 10000(1 - j) Ibs. = (5.08 - 2.54-) tons.
III. Members subjected to Compressive Stresses only.
Denote the ratio of the length (/) to the least radius of
gyration (k) by r.
The maximum stress per square inch = 2 Ibs.,
/being 8000 Ibs. for wrought-iron and 10,000 Ibs. for steel, and
being 40,000, 30,000, or 20,000, according as the member has
a
two square (fixed) ends, one square and one pin end, or two
pin ends.
Again, the maximum stress per square inch for steel struts
(/? \
1 + A! ft>s.;
A I
" " square ends (10000 4Or) I + -j Ibs.;
\ A'
" pin ends = (5 - ^)(i + ~) tons ;
" " square ends = ( 5 ^Jl 1 4~ ~/r] tons.
In the last two expressions r < 40. These expressions may
be also employed in the case of alternating stresses, but the
(T)
I -|
CAMBER. 659
26. Camber. Owing to the play at the joints, a girder or
truss will deflect to a much greater extent than is indicated by
theory, and the material will receive a permanent set, which,
however, will not prove detrimental to the stability of the
structure unless it is increased by subsequent loads. If the
chords were initially made straight, they would curve down-
wards ; and although it does not necessarily follow that the
strength of the truss would be sensibly impaired, the appear-
ance would not be pleasing.
In practice it is often specified that the girder or truss is to
have such a camber or upward convexity that under ordinary
loads the grade line will be true and straight ; or, again, that a
camber shall be given to the span by making the panel lengths
of the top chord greater than those of the bottom chord by
.125 in. for every 10 ft.
The lengths of the web members in a cambered truss are
not the same as if the chords were horizontal, and must be care-
fully calculated so as to insure that the several parts will fit
together.
To find an Approximate Value for the Camber, etc.
Let d be the depth of the truss.
Let s 1 , s y be the lengths of the upper and lower chords, re-
spectively.
Let /"j , yj be the unit stresses in upper and lower chords,
respectively.
Let d l , d^ be the distances of the neutral axis from the
upper and lower chords, respectively.
Let R be the radius of curvature of the neutral axis.
Let / be the span of the truss,
Then
4 -/ /I 4 /-*. /,
= - = and = - =
the chords being assumed to be circular arcs.
Hence, the excess in length of the upper over the lower
chord
660 THEORY OF STRUCTURES.
Let ;r, , x^ be the cambers of the upper and lower chords,
respectively ; R -f- d^ and R d^ are the radii of the upper
and lower chords, respectively.
By similar triangles,
the horizontal distance between ) R-\- d
the ends of the upper chord j " R
I ._ K-a
\-^ L
the horizontal distance between i R
the ends of the lower chord
Hence,
\1 "
and
j = x, . 2(R + d,\ approximately,
(- ^ -/) = x^ . 2(R d^), approximately.
,
and ** = SR I --R]-
27. Rivet-connection between Flanges and Web.
The web is generally riveted to angle-irons forming part of
the flanges.
The increment of the flange stress transmitted through
the web from point to point tends to make the angle-irons slide
over the flange surfaces.
Denote the increment by F, and let h be the effective depth
of the girder or truss.
Then, if 5 be the shearing force at any point,
Fh = the increment of the bending moment per unit
of length
ldM\
= l = = o in the case ot a close web,
\dx I
and Fh the increment of the bending moment
= (4M) = Sa in the case of an open web ;
a being the distance between the two consecutive apices or
panel points within which ^ lies.
EYE-BARS AND PINS.
66 1
Hence, if TV be the number of rivets per unit of length for
the close web, or the number between the two consecutive
apices for the open web,
N f s = F = Y for the close web,
and
= T- for the open web,
d being the diameter, of a rivet, and f s the safe coefficient of
shearing strength.
28. Eye-bars and Pins. Eye-bars connected with pins
have been commonly employed in the construction of suspen-
sion cables, the tension chords of ordinary trusses and canti-
levers, and the diagonals of web systems. The requisite sec-
tional area is obtained by placing a number of bars side by
side on the same pin, and, if necessary, by setting two or more
tiers of bars one above another.
FIG. 417.
rFh
rf=h
FIG. 418.
FIG. 419.
The figures represent groups of eye-bars as they often
occur in practice.
If two sets of 2n bars pull upon the pin in opposite direc-
tions, as in Figs, 418 and 419, the bending moment on the pin
will be nPp, P being the pull upon each bar, and / the distance
between the centre lines of two consecutive bars.
THEORY OF STRUCTURES.
Hence,
f being the stress in the material of the pin at a distance c from
the neutral axis, and / the moment of inertia.
In general, the bending action upon a pin connecting a
number of vertical, horizontal, and inclined bars may be de-
termined as follows :
Consider one-half of the pin only.
Let F, Fig. 420^ be the resultant stress in the vertical bars.
It is necessarily equal in magnitude but opposite in direc-
tion to the vertical component of the resultant of the stresses
h-*-
FIG. 420.
in the inclined bars. Let v be the distance between the lines
of action of these two resultants. The corresponding bending
action upon the pin is that due to a couple of which the mo-
ment is Vv.
Let h be the distance between the lines of action of the
equal resultants H of the horizontal stresses upon each side of
the pin. The corresponding bending action upon the pin is
that due to a couple of which the moment is Hh.
Hence, the maximum bending action is that due to a couple
of which the moment is the resultant of the two momenjts Vv
and Hh. viz.,
Eye-bars. In England it has been the practice to roll bars
having enlarged ends, and to forge the eyes under hydraulic
EYE-BARS AND PINS. 663
pressure with suitably shaped dies. In America both hammer-
forged and hydraulic-forged eye-bars are made, the latter being
called weldless eye-bars. Careful mathematical and experimen-
tal investigations have been carried out to determine the proper
dimensions of the link-head and pin, but owing to the very
complex character of the stresses developed in the metal around
the eye, an accurate mathematical solution is impossible.
Let d be the width and /
the thickness of the shank of
the eye-bar represented in Fig.
421. Let 5 be the width of
the metal at the sides of the
eye, and H the width at the
end. Let D be the diameter
of the pin.
The proportions of the head
are governed by the general condition that each and every part
should be at least as strong as the shank.
When the bar is subjected to a tensile stress the pin is
tightly embraced, and failure may arise from any one -of the
following causes :
(a) The pin may be shorn through.
Hence, if the pin is in double shear, its sectional area should
be at least one-half that of the shank.
It may happen that the pin is bent, but that fracture is pre-
vented by the closing up of the pieces between the pin-head
and nut ; the efficiency, however,of the connection is destroyed,
as the bars are no longer free to turn on the pin.
In practice, D for flat bars varies from f^to %d, but usually
lies between %d and ^d.
The diameter of the pin for the end of a round bar is gen-
erally made equal to \\ times the diameter of the bar.
The pin should be turned so as to fit the eye accurately,
but the best practice allows a difference of from -fa to ^4^ f
an inch in the diameters of the pin and eye.
(&) The link may tear across MN.
On account of the perforation of the head, the direct pull
on the shank is bent out of the straight and distributed aver
664 THEORY OF STRUCTURES.
the sections S. There is no reason for the assumption that
the distribution is uniform, and it is obviously probable that
the intensity of stress is greatest in the metal next the hole.
Hence, the sectional area of the metal across MN must be at
least equal to that of the shank, and in practice is always
greater.
5 usually varies from .55^ to .625^.
The sectional area through the sides of the eye in the head of
a round bar varies from i-J times to twice that of the bar.
(c) The pin may be torn through the head.
Theoretically, the sectional area of the metal across PQ
should be one-half that of the shank. The metal in front of
the pin, however, may be likened to a uniformly loaded girder
with both ends fixed, and is subjected to a bending as well as
to a shearing action. Hence, the minimum value of H has been
fixed at \d, and if H is made equal to d, both kinds of action
will be amply provided for.
(d) The bearing surface may be insufficient.
If such be the case, the intensity of the pressure upon the
bearing surface is excessive, the eye becomes oval, the metal is
upset, and a fracture takes place. Or again, as the hole elon-
gates, the metal in the sections 5 next the hole will be drawn
out, and a crack will commence, extending outwards until frac-
ture is produced.
In practice, adequate bearing surface may be obtained by
thickening the head so as to confine the maximum intensity of
the pressure within a given limit.
(e) The head may be torn through the shoulder at XY.
Hence, ^TFis made equal to d.
The radius of curvature R of the shoulder varies from I \d
to J.6d.
Note. The thickness of the shank should be , or d at
4 7
least.
The following table gives the eye-bar proportions common
in American practice :
STEEL EYE-BARS.
66 5
Value of 6".
of d.
of D.
Weldless
Hammered
Bars.
Bars.
.00
67
1-5
33
.00
75
1-5
33
.00
1. 00
1-5
50
.00
1.25
1.6
50
.00
1-33
1-7
.00
1.50
1.8 5
.67
.00
i-75
2.OO
67
.00
2.OO
2.25
75
Also, in weldless bars, //= S\ in hammered bars, H ' = d.
29. Steel Eye-bars. Hydraulic-forged steel eye-bars are
now being largely made. The steel has an ultimate tenacity of
from 60,000 to 68,000 Ibs. per square inch, an elastic limit of not
less than 50 per cent, and an elongation of from 17 to 20 per cent
in a length equal to ten times the least transverse dimension.
The Phcenix Bridge Co. and the Edge Moor Iron Co. give
the following tables of steel eye-bar proportions :
Phoenix Bridge Co.
Edge Moor Iron Co.
Mini- i Excess of
Width of
bar*/.
Diameter of
Pin-hole.
Diam-
eter of
Head.
Width
of
baroT.
Diameter
of
Pin-hole.
Diameter
of
Head.
mum
Thick-
ness
Sectional Area
of Head along
PP over Sec-
of Bar. 1 tion of Bar.
3
2&. 2 lf
7
2
Ii
*4
I
33#
3
3iV 3lf
8
2
2|
si
I
33*
4
3 T V
9
2i
2i
5+
I
33#
4
311, 4rV 4tt
10
2i
3i
6i i
33#
5
5
5
3 tf ' 4
4H 5 T V
sl-5, V*
ii
12
13
3
3
4
2i
4
4f
6*
8 f
9* f
33#
33#
33
6
4H- 4lt
13
4
3
ioi f
33^
6
5f, 5iV 5lf
14
5
41
i f
37^
6
6ft, 6tt. 6 Tf
15
5
5|
I2i
f
37^
7
5 A
15
6
5i
I3i
|
37
7
5lf, 6ft. 6H
16
6
6i
J4i
37^
8
6'if, 7ft, 7U
6f*
17
17
7
7
5*
7i
1 51
17
i
40^
40^
8
6ft, 6tf, 7 A
18
8
5f
17
i
40^
8
7li 8f
19
8
6f
18
i
4<$
8
ai,9t
20
9
9
7 T \- 7H
8|, 8|
2O
21
10
8|
22
10
8J,9f
23
10
10, 10^
24
In both the Phoenix and Edge Moor bars the thickness of the head is the
same as that of the body of the bar, or does not exceed it by more than -fa in.
O66 THEORY OF STRUCTURES.
30. Rivets. A rivet is an iron or steel shank, slightly
tapered at one end (the tail), and surmounted at the other by
a cup or pan-shaped head (Fig. 422). It is used to join steel or
iron plates, bars, etc. For this purpose the rivet is generally
heated to a cherry-red, the shank or spindle is passed through
D
FIG. 422. FIG. 423. FIG. 424. FIG. 425. FIG. 426.
the hole prepared for it, and the tail is made into a button, or
point. The hollow cup-tool gives to the point a nearly hemi-
spherical shape, and forms what is called a snap-rivet (Fig.
423). Snap-rivets, partly for the sake of appearance, are com-
monly used in girder-work, but they are not so tight as conical-
pointed rivets (staff-rivets), which are hammered into shape
until almost cold (Fig. 424).
When a smooth surface is required, the rivets are counter-sunk
(Fig. 425). The counter-sinking is drilled and may extend
through the plate, or a shoulder may be left at the inner edge.
Cold-riveting is adopted for the small rivets in boiler work
and also wherever heating is impracticable, but tightly-driven
turned bolts are sometimes substituted for the rivets. In all
such cases the material of the rivets or bolts should be of su-
perior quality.
Loose rivets are easily discovered by tapping, and, if very
loose, should be at once replaced. It must be borne in mind,
however, that expansions and contractions of a complicated
character invariably accompany hot-riveting, and it cannot be
supposed that the rivets will be perfectly tight. Indeed, it is
doubtful whether a rivet has any hold in a straight drilled hole,
except at the ends.
Riveting is accomplished either by hand or machine, the latter
being far the more effective. A machine will squeeze a rivet,
at almost any temperature, into a most irregular hole, but the
exigencies of practical conditions often prevent its use, except
for ordinary work, and its advantages can rarely be obtained
STRENGTH OF PUNCHED AND DRILLED PLATES. 667
where they would be most appreciated, as, e.g., in the riveting
up of connections.
31. Dimensions of Rivets. The diameter (d) of a rivet
in ordinary girder-work varies from f in. to I inch, and rarely
exceeds i in.
The thickness (/) of a plate in ordinary girder-work should
never be less than J in., and a thickness of f in., or even -^ in.,
is preferable.
Let T be the total thickness through which a rivet passes.
According to Fairbairn,
When / < % in., d should be about 2t.
When / > -J in., d should be about \\t.
According to Unwin,
When / varies from J in. to I in. and passes through
two thicknesses of plate, d lies between f t -f- jV and
When the rivets join several plates, d -- (- .
8 8
According to French practice,
Diameter of head if d.
Length of rivet from head = T -f- \\d.
According to Rankine,
Length of rivet from head = T -f- 2\d.
The rise of the head = \d.
The- diameter of the rivet-hole is made larger than that of
the shank by from ^ to -J in., so as to allow for the expansion
of the latter when hot.
There seems to be no objection to the use of long rivets,
provided they are properly heated and secured.
32. Strength of Punched and Drilled Plates. Experi-
ment shows that the tenacity of iron and steel plates is con-
siderably diminished by punching. This deterioration in
tenacity seems to be due to a molecular change in a narrow
annulus of the metal around the hole. The removal of the
annulus largely neutralizes the effect of the punching, and,
hence, the holes are sometimes punched -J in. less in diameter
than the rivets and are subsequently rimered or drilled out to
the full size. The original strength may also be almost
668 THEORY OF STRUCTURES.
entirely restored by annealing, and, generally, in steel work,
either this process is adopted or the annulus referred to above
is removed.
Punching does not sensibly affect the strength of Landore-
Siemens unannealed plates, and only slightly diminishes the
strength of thin steel plates, but causes a considerable loss of
tenacity in thick steel plates ; the loss, however, is less than
for iron plates.
The harder the material the greater is the loss of tenacity.
Iron seems to suffer more from punching when the holes
are near the edge than when removed to some distance from
it, while mild steel suffers less when the hole is one diameter
from the edge than when it is so far that there is no bulging at
the edge.
The injury caused by punching may be avoided by drilling
the holes. In important girder-work and whenever great
accuracy of workmanship is required, a uniform pitch may be
insured and the full strength of the metal retained by the use
of multiple drills. Drilling is a necessity for first-class work
when the diameter of the holes is less than the thickness of
the plate, and also when several plates are piled. It is impos-
sible to punch plates, bars, angles, etc., in spite of all ex-
pedients, in such a manner that the holes in any two exactly
correspond, and the irregularity becomes intensified in a pile,
the passage of the rivet often being completely blocked. A
drift, or rimer, is then driven through the hole by main force,
cracking and bending the plates in its passage, and separating
them one from another.
The holes may be punched for ordinary work, and in plates
sf which the thickness is less than the diameter of the rivets.
\Vhenever the metal is of an inferior quality, the holes should
be drilled.
33. Riveted Joints. In lap joints (Figs. 427 and 430) the
plates overlap and are riveted together by one or more rows
of rivets which are said to be in single shear, as each rivet has
to be sheared through one section only.
In fish (or btitf) joints (Figs. 428 and 429) the rivets are
in double shear, i.e., must be each sheared through two sections.
RIVETED JOINTS.
669
Thus they are not subjected to the one-sided pull to which
rivets in single shear are liable.
FIG. 427
In fish joints the ends of the plates meet, and the plates
are riveted to a single cover (Fig. 428), or to two covers (Fig.
429), by means of one or more rows of rivets on each side of the
joint.
A fish joint is properly termed a butt joint when the plates
are in compression. The plates should butt evenly against one
another, although they seldom do so in practice. Indeed, the
mere process of riveting draws the plates slightly apart, leav-
ing a gap which is often concealed by caulking. A much
better method is to fill up the space with some such hard sub-
stance as cast-zinc, but the best method, if the work will allow
of the increased cost, is to form a jump joint, i.e., to plane the
eyes of the plates carefully, and then bring them into close
contact, when a short cover with one or two rows of rivets will
suffice to hold them in position.
The riveting is said to be single, double, triple, etc., according
as the joint is secured by one, two, three, or more rows of rivets.
o o o
o o o
O O
o o o
o o o
o o o
^o
00
00
00
oo
ZIGZAG
FIG. 432.
CHAIN
FIG. 431.
Double, triple, etc., riveting may be chain (Fig. 431) or zig-
zag (Fig. 432). In the former case the rivets form straight
lines longitudinally and transversely, while in the latter the
rivets in each row divide the space between the rivets in adja-
cent rows. Experiments indicate that chain is somewhat
stronger than zigzag riveting.
670
THEORY OF STRUCTURES.
Figs. 433 to 435 show forms of joint usually adopted for
bridge-work. In boiler-work the rivets are necessarily very
close together, and if the strength of the solid plate be assumed
to be IOO, the strength of a single-riveted joint hardly exceeds
50, while double-riveting will only increase it to 60 or 70. Fair-
o o
o o o
o o o
o o
FIG. 433.
FIG. 434.
FIG.
435-
bairn proposed to make the joint and unpunched plate equally
strong by increasing the thickness of the punched portion of
the plate, but this is somewhat difficult in practice.
The stresses developed in a riveted joint are of a most com-
plex character and can hardly be subjected to exact mathe-
matical analysis. For example, the distribution of stress will
be necessarily irregular (a) if the pull upon the joint is one-
sided ; (b) when local action exists, or the plates stretch, or in-
ternal strains are in the metal before punching ; (c) if there is a
lack of symmetry in the arrangement of the rivets, so that one
rivet is more severely strained than another; (d) when the
workmanship is defective.
The joint may fail in any one of the following ways :
(1) The rivets may shear.
(2) The rivets may be forced into and crush the plate.
(3) The rivets may be torn out of the plate.
(4) The plate may tear in a direction transverse to that of
the stress.
The resistance to rupture should be the same in each of the
four cases, and always as great as possible.
The shearing and tensile strengths of plate-iron are very
nearly equal. Thus, iron with a tenacity of 20 tons per square
inch has a shearing strength of 18 to 20 tons per square inch.
Rivet-iron is usually somewhat stronger than plate-iron.
Again, the shearing strength of steel per square inch varies
THE ORE TIC A L DED UC TIONS. 67 1
from about 24 tons for steel, with a tenacity of about 30 tons,
to about 33 tons for steel, with a tenacity of about 50 tons ; an
average value for rivet-steel with a tenacity of 30 tons being
24 tons.
Hence, if 4 be a factor of safety, the working coefficient?
become
For wrought-iron \ * tons P er SC * uare inch in shear ' and
( 5 " " " ' ; tension.
_ . (6 tons per square inch in shear, and
F rSteel \n " " "tension
Allowance, however, must be made for irregularity in the dis-
tribution of stress and for defective workmanship, and in
riveting wrought-iron plates together it is a common practice
to make the aggregate section of the rivets at least equal to
and sometimes 20 per cent greater than the net section f the
plate through the rivet-holes.
Hence, the working coefficients are reduced to
4 or 4^ tons per square inch for wrought-iron,
and
5 or 5J " " " " " steel,
according to the character of the joint.
There is very little reliable information respecting the in-
dentation of plates by rivets and bolts, and it is most uncertain
to what extent the tenacity of the plates is affected by such
indentation. Further experiments are required to show the
effect of the crushing pressure upon the bearing area (i.e., the
diameter of the rivet multiplied by the thickness of the plate],
although a few indicate that the shearing strength of the rivet
diminishes after the intensity of the bearing pressure exceeds
a certain maximum limit.
34. Theoretical Deductions.
Let 5 be the total stress at a riveted joint ;
/i>/2> / a >/4> De the safe tensile, shearing, compressive,
and bearing unit stresses, respectively ;
/ be the thickness of a plate, and w its width ,
?2 THEORY OF STRUCTURES,
N\*t the total number of rivets on one side of a joint;
it be the total number of rivets in one row ;
p be the pitch of the rivets, i.e., the distance centre to
centre ;
d be the diameter of the rivets ;
x be the distance between the centre line of the nearest
row of rivets and the edge of the plate.
Value of x. It has been found that the minimum safe
value of x is d, and this in most cases gives a sufficient overlap
( 2x\ while x = \d is a maximum limit which amply pro-
vides for the bending and shearing to which the joint may be
subjected. Thus the overlap will vary from 2d to ^d.
x may be supposed to consist of a length x^ to resist the
shearing action, and a length x^ to resist the bending action.
It is impossible to determine theoretically the exact value of
JF,, as the straining at the joint is very complex, but the metal
in front of each rivet (the rivets at the ends of the joint ex-
cepted) may be likened to a uniformly loaded beam of length
d, depth x^ -- , and breadth /, with both ends fixed. Its
moment of resistance is therefore -?t Lr, j , / being the
maximum unit stress due to the bending. Also, if P is the
load upon the rivet, the mean of the bending moments at the
P
end and centre is ^d.
o
Hence, approximately,
* '--
It will be assumed that the shearing strength of the rivet
is equal to the strength of a beam to resist cross-breaking.
Single-riveted lap and single-cover joints (Figs. 427 and 428).
~~A = (p-d)tf^dtf^ ..... (I)
THEORETICAL DEDUCTIONS. 6/3
*xjj[i = -f % .
d , i / d'f, . .
-* = 2+ 4\/**-J7 ' ' (3)
As already pointed out, these joints are weakened by the
bending action developed, and possibly also by the concentra-
tion of the stress towards the inner faces of the plates.
Single-riveted double-cover joints (Fig. 429).
(4)
tf t = 2 /..
4
(5)
3 rfr 21 - 2 4
These joints are much stronger than joints with single
covers. Also, equation (4) shows that the bearing unit stress
in a double-cover joint is twice as great (theoretically] as in a
single-cover joint (eq. i), so that rivets of a larger diameter
may be employed in the latter than is possible in the former,
d
for corresponding values of -.
THEORY OF STRUCTURES.
Chain-riveted joints (Fig. 431).
5 = N -- / 9 when there is one cover only ; . . (8)
5 = N - / 2 when there are two covers. . . . (9)
This class of joint is employed for the flanges of bridge
girders, the plates being piled as in Figs. 436, 437, 438, and n
being usually 3, 4, or 5.
In Fig. 437 the plates are grouped so as to break joint, and
opinions differ as to whether this arrangement is superior to
the full butt shown in Fig. 438. The advantages of the latter
FIG. 436.
FIG. 437. FIG. 438.
are that the plates may be cut in uniform lengths, and the
flanges built up with a degree of accuracy which cannot be
otherwise attained, while the short and awkward pieces accom-
panying broken joints are dispensed with.
A good practical rule, and one saving much labor and ex-
pense, is to make the lengths of the plates, bars, etc., multiples
of the pitch, and to design the covers, connections, etc., so as
to interfere with the pitch as little as possible.
The distance between two consecutive joints of a group
(Fig. 437) is generally made equal to twice the pitch.
An excellent plan for lap and single-cover joints is to
arrange the rivets as shown in Figs. 431 to 435.
The strength of the plate at the joint is only weakened by
one rivet-hole, for the plate cannot tear a^t its weakest section,
COVERS. 675
i.e., along the central row of rivets (##), until the rivets be-
tween it and the edge are shorn in two.
Let there be m rows of rivets, i 1,22,
3 3, ... (Fig. 439)-
The total number of rivets is evi- (_
dently m\ O^O^O^O
Let /i i & > ? 3 , , , ... be the unit ten- O,XO I !
sile stresses in the plate along the lines I I,
2 2 > 3 3 respectively. Then
i
+
U
FIG. 439.
5 = (w d}tf^ = nfft , for the line i i ;
n <? t \s
( m - i)/;, 22;
4
3 3;
44;
Assume that /", = ^ a . Then ze; = (n? -\- \]d.
Hence, by substituting this value of w in the first of the
above relations, - -- . Since q z , ^ 4 , . . . are each less than
t 1 1 j <i
fi , the assumption is justifiable.
35. Covers. In tension joints the strength of the covers
must not be less than that of the plates to be united. Hence,
a single cover should be at least as thick as a single plate ; and
if there are two covers, each should be at least half as thick.
When two covers are used in a tension pile it often
happens that a joint occurs in the top or bottom plate, so that
the greater portion of the stress in that plate may have to be
676 THEORY OF STRUCTURES.
borne by the nearest cover. It is, therefore, considered advis-
able to make its thickness five-eighths that of the plate.
The number of the joints should be reduced to a minimum,
as the introduction of covers adds a large percentage to the
dead weight of the pile.
Covers might be wholly dispensed with in perfect-jump
joints, and a great economy of material effected, if the dif-
ficulty of forming such joints and the increased cost did not
render them impracticable. Hence, it may be said that covers
are required for all compression joints, and that they must be as
strong as the plates ; for, unless the plates butt closely, the
whole of the thrust will be transmitted through the covers.
In some of the best examples of bridge construction the ten-
sion and compression joints are identical.
36. Efficiency of Riveted Joints.* The efficiency of a
riveted joint is the ratio of the maximum stress which can be
transmitted to the plates through the joint to the strength of
the solid plates.
Denote this maximum efficiency by rf.
Let / be the pitch of the rivets ;
d " diameter of the rivets ;
t " thickness of the plates ;
f t " tenacity of the solid plate ;
mf t " " " " riveted plate ;
f s " shearing strength of the rivets ;
N " number of rivets in a pitch length ;
e " ratio of the strength of a rivet in double
shear to its strength in single shear.
Then
77, = efficiency as regards the plates - J
eN-d'f,
77, = efficiency as regards the rivets - -7 . (2)
* From an article by Professor Nicolson in the Engineer, Oct. 9, 1888.
EFFICIENCY OF RIVETED JOINTS.
The efficiency of the joint is, of course, the smaller of these
two values ; and the joint is one of maximum efficiency when
rj i = % = rf ; that is, when
m - -- =
Ptft
or
In this expression the quantities m f t , N, and e are con-
stants for any given joint, being of necessity known, or having
been fixed beforehand ; and the equation thus expresses one
condition governing the relations of the three variables /, d,
and / to each other. It is obvious, however, that, in order to
determine the values of any two of these variables in terms of
the third, another relation between them must be postulated.
In short, in designing a joint, the value of one of the three
ratios --, -, and must be fixed.
at t
P
CASE I. Suppose that the ratio -, has a certain value.
This is very frequently the quantity predetermined ; but it is
most usually done by fixing the value of 77, rj very obviously
P / d\
involving -- in fact y m\i -}.
Equation (3) may be written
4 t
or
If the ratio b denoted by k, then
678 THEORY OF STRUCTURES.
m(p d)
But since tj -,
P
dm rf' * '
Therefore, substituting in (5),
= eNk-^--: , (7)
m rf 4 mf t
and, ultimately,
4 mf t r)
^~eNn'J^'m~^ ( 8 )
The process of designing a joint of maximum, efficiency for
a boiler of given diameter and pressure of steam, when rj
or the ratio is fixed, is then as follows: Settle the number
di
of rivets per pitch (i.e., N)', the value to be allowed for e (de-
pending on the nature of the shearing stress on the rivets) ;
and the values of m,f t , and f f Then k is known from equa-
tion (8).
But / may be found from the relation,
pressure X diameter = rf X 2tf t ,
or
pressure X diameter
Hence, since k = is known, d may be found ; and since
d m
m i /* i
is known, p is also fixed.
CASE II. When -, the ratio of rivet pitch to plate thick-
ness, is given, equation (5) must be otherwise manipulated.
EFFICIENCY OF RIVETED JOINTS. 679
d
Multiplying it by , and substituting for d its value kt, we
have
4 P **ftP
Putting this in the form of a quadratic equation in k,
2
For brevity, substituting A for -^- , T for ^-', and R for -,
and solving the quadratic,
.. . . (12)
The method of designing the joint is, then, as follows :
A, T, and R being known, k may be found by substituting
their values in equation (12), the positive sign of the second
term being taken.
Now,
( d \ I kt \ ( k \
r) = m(i J = m\i I = m\i -=\ ;
and since both k and R are now known, the thickness of
plate (f) may be found, as in Case I, by equation (9). The
values of the diameter and pitch of rivets follow at once from
the known values of k and R.
This method of designing a joint appears to be the most
rational of the three. For the greatest pitch for which a joint
will remain steam-tight depends mainly on the relation of pitch
of rivets to thickness of plates ; although it is also affected by
the relative size of rivets and of rivet-heads.
CASE III. If y, or k, be predetermined, the value of rj
must first be obtained, in order that the plate thickness may
680 THEORY OF STRUCTURES.
be found by means of equation (9). Now, rj = m - may
be put into the form
md
m
and if this value is substituted for/ in equation (4),
md leNn f, \
= I k > ~J- I la.
m rj \ 4 mft^t
From this is finally deduced
eNnkf
The plate thickness may now be found by equation (9) ;
the diameter of rivet from d = kt, and the pitch from
md
p = - . In the above investigations no account has been
taken of the effect of the bearing pressure on the rivets or
plate.
If f c be the allowable bearing pressure per projected square
inch of rivet surface, the following relation must obtain :
. . . . . (14)
This may be written
fc ~ Nd
Then if f e be estimated by this equation, and if it should 1
greater than 43 tons per square inch in a lap joint, or 45 fr
tons in a butt joint, such joint will fail by the rivets shec
before the full strength of the plate is exerted, as Kent
experiments show that with these values of f c the rive
not attain their natural ultimate shearing strength (viz.,/ 5 ,,
fail at shearing stresses much below this.
EFFICIENCY OF RIVETED JOINTS. 68 1
Ar;ain, the maximum allowable ratio of (i.e., k] as the
preliminary datum for the design of a joint, may be fixed by
using the expression
<->
deduced from the obvious relation similar to (14)
eN-d*f s = Ndtf c .
4
win suggests the relation </=
~ Tin designing the joint by any of the methods given above,
any value obtained for k greater than that supplied by (16)
should be rejected.
1 Note. The following Tables of the Weights of Bridges
have been prepared from data supplied by the engineers of the
bridges in question.
r
v.
.
682
THEORY OF STRUCTURES.
V) ^
W -o
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TABLE OF ACTUAL WEIGHTS OF MODERN BRIDGES. 683
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684
THEORY OF STRUCTURES.
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TABLE OF ACTUAL WEIGHTS OF MODERN BRIDGES. 68$
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686
THEORY OF STRUCTURES.
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TABLE OF ACTUAL WEIGHTS OF MODERN BRIDGES. 68?
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688 THEORY OF STRUCTURES.
TABLE OF LOADS FOR HIGHWAY BRIDGES.
Spaa in Feet.
City and Suburban
Bridges liable to
Heavy Traffic.
Bridges in Manu-
facturing Districts.
Ballasted Road.
Bridges in Country
Districts.
Unballasted Roads.
joo and under
100 to 200
200 to 300
300 to 400
above 400
100 Ibs. per sq. ft.
80 " "
70 " "
60 " "
50 " "
90 Ibs. per sq. ft.
60 " "
50 " "
50 " "
50 " "
70 Ibs, per sq. ft.
60 " "
50 " "
45 " " "
45 " " "
f I
EXAMPLES. 089
EXAMPLES.
I. A bridge of N equal spans crosses a span of L ft.; the weights in
tons per lineal foot of the main girders of the platform, permanent way,
etc., and of the live load, are iu\ , w* , w s , respectively. Show that
LA
where A = w^pk + r) + iv*(pk + q) and B = pk + r,
k being the ratio of span to depth, and p, g, r numerical coefficients.
Hence also determine the limiting span of a girder.
If X is the cost of a pier and if Y is the cost per ton of the super-
structure, find the value of 7V which will make the total cost per lineal
foot a maximum, and prove that this is approximately the case when the
spans are so arranged that the cost of one span of the bridge structure
is equal to the cost of a pier.
Ans. A span < - - Cost is a minimum when N LB = Z4/fU_,
and the minimum cost of the span = X( i j = X, approx.
2. A car of weight W for a gauge of 4 ft. 8 in. is 33 ft. long, 6 ft. deep,
and its bottom is 2 ft. 6 in. above the rails. Find the additional weight
thrown upon the leeward rail when the wind blows upon a side of the
car with a pressure of 20 Ibs. per square foot. Also find the minimum
pressure that will blow the car over. Ans. 4625.84 Ibs. ; .428 W.
3. A lattice girder 200 ft. long and 20 ft. deep, with two systems of
right-angled triangles, carries a dead load of 800 Ibs. per lineal foot.
Determine the greatest stresses in the diagonals and chords of the fourth
bay from one end when a live load of 1200 Ibs. per lineal foot passes over
the girder.
Ans. If riveted : Diagonal stress = 37,2004/2 Ibs. ;
Chord stress = 450,000 Ibs.
If pin-connected : Diagonal stress = 44,800 4/2 and 29,600 4/2 Ibs.;
Chord stress = 460.000 Ibs.
690 THEORY OF STRUCTURES.
4. A lattice-girder 80 ft. long and 8 ft. deep carries a uniformly dis-
tributed load of 144,000 Ibs. Find the flange inch-stresses at the centre,
the sectional area of the top flange being 56^ sq. in. gross, and of the
bottom flange 45 sq. in. net.
What should be the camber of the girder, and what extra length
should be given to the top flange, so that the bottom flange of the loaded
girder may be truly horizontal ? (E = 29,000,000 Ibs.)
Ans. 3185.8 Ibs. ; 4000 Ibs.
ri = -29735 : ** = -2987 J S* ~ s* = Ttffifr.
5. A lattice-girder 80 ft. long and 10 ft. deep, with four systems of
right-angled triangles, carries a dead load of 1000 Ibs. per lineal foot.
Determine the greatest stresses in the diagonals met by a vertical plane
in the seventh bay from one end when a live load of 2500 Ibs. per lineal
foot passes over the girder. Design the flanges, which are to consist of
plates riveted together.
The lattice-bars are riveted to angle-irons. Find the number of f-iri,
rivets, required to connect the angle-irons with the flanges in the first
bay, 10,000 Ibs. per square inch being the safe shearing strength of the
rivets.
Ans. \iriveted: Diagonal stress = 10,664^1/2 Ibs.
If pin-connected ': " = 9062^4/2; 62501/2; 15,4681/2;
11,8751/2 Ibs.
22 rivets
6. The bracing of a lattice-girder consists of a single system of tri-
angles in which one of the sides is a strut and the other a tie inclined to
the horizontal at angles of a and ft respectively ; in order to give the
strut sufficient rigidity its section is made k times that indicated by
theory, the coefficient k being > unity. Show that the amount of ma-
terial in the struts and ties is a minimum when
tan a = k tan ft.
7. A lattice-girder of 40 ft. span, 5 ft. depth, and with horizontal
chords has a web composed of two systems of right-angled triangles and
is designed to support a dead and a live load, each of -J- ton per lineal
foot. Determine the maximum stresses in the members of the third
bay from one end met by a vertical plane.
Ans. \iriveted: Diagonal stress = V// 2 tons J
Chord stress = 27 tons.
If pin-connected : Diagonal stress = ff 1/2 and f ^2 tons ;
Chord stress = 26 tons.
8. A lattice-truss of 100 ft. span and 10 ft. depth has a web composed
of four systems of right-angled triangles. The maximum stress in the
EXAMPLES. 691
diagonal joining the sixth apex in the upper chord to the fourth apex in
the lower is 16 tons. Find the dead load, the live load being i ton per
lineal foot, assuming the truss to be (a) riveted, (b) pin-connected.
Ans. (a) .554 ton ; (b) 1.062 tons.
9. A lattice-girder of 40 ft. span has a web composed of two systems
of triangles (base = 10 ft.) and is designed to carry a live load of 1600
Ibs. per lineal foot and a dead load of 1200 Ibs. per lineal foot. Defin-
ing the stress-length of a member to be the product of its length into
the stress to which it is subjected find the depth of the truss so that its
total stress-length may be a minimum. Ans. 10.19 ft-
10. Determine the maximum stresses in the members of a lattice-
truss of 40 ft. span and 4 ft. depth, with two systems of triangles (base
= 8 ft.), (a) when riveted together; (b) when pin-connected. Dead load
= J ton per lineal foot, live load = $ ton per lineal foot.
Ans. Bays ist; 2d ; 3d; 4th; 5th.
(a) Diags. 6f 4/2~; 5.534/2"; 4.054/2"; 2.854/2"; if 4/2~tons ;
Tens.chd. 6| ; i8f; 2;f; 33!; 36$.
Comp. chd. Same.
( 7.5^2 p.rVS 4.7^2 j 2.34/1 v\ 2.34/2^
Tens.chd. 6; 18; 27; 33; 36 "
Comp. chd. Same.
ii. The platform of a single-track bridge is supported upon the top
chords of two Warren girders ; each girder is 100 ft. long, and its brac-
ing is formed of ten equilateral tfiangles (base 10 ft.); the dead weight
of the bridge is 900 Ibs. per lineal foot. ; the greatest total stress in the
seventh sloping member from one end when a train crosses the bridge
is 41,394.8 Ibs. Determine the weight of the live load per lineal foot.
Prepare a table showing the greatest stress in each bar and bay when a
single load of 1 5,000 Ibs. crosses the girder.
Ans. t\ t z >t 3 h ft
WWWXAAA/ 277H Ibs. per lin. ft.
FIG. 440.
Stresses in diagonals : </, = of, = 9 4/J; d a d* = 8 1/3 ;
d* = d = 7 |/5 ; <*, = // = 6 j/3 ;
d* =d= 5 1/3 tons.
Stresses in compression : c\ = 9 1/3 ; c<i = 16 4/3"; c =21 4/3";
<r 4 = 24 1/3 tons.
Stresses in torsion : /i = 4i I/J : /a = 1 2 |/J; / 3 = 1 7^ 1/3 ;
/ 4 = 21 4/3 ; / 5 = 22 4/3 tons.
692
THEORY OF STRUCTURES.
12. A Warren girder with its bracing formed of nine equilateral tri-
angles (base = 10 ft.) is 90 ft. long, and its dead weight is 500 Ibs. per
lineal foot. Determine the maximum stresses in each member when
a live load of 1350 Ibs. per lineal foot, preceded by a concentrated
load of 18,000 Ibs., crosses the girder, assuming that every joint is loaded.
The diagonals and verticals are riveted to angle-irons forming part of
the flanges.
How many f-in. rivets are required for the connection of the several
members meeting at the third apex in the upper chord? (23, 6, and 13.)
How many are required in the first bay of each chord to prevent longi-
tudinal slip? (15 in tension chord and 18 in compression chord.)
Ans.
FIG. 441
Tension chord stresses :
t 9 =/, =
V3
333125
V3
415125
V3
39462 5
164000 287000 -360000
Compression chord stresses: c\ = - ;=--; c* - ; c a =
V3 V3 V3
410000
V~3
178500 159500 141250
Stresses in sloping members : di= ;=- ; d*= 7=- ; a s = ^= ;
T 3 r 3 T 3
I2 375o 107000 91000
^3~ ^3 "4/3"
70250
47500
Ibs.
__^ d _ I075Q
^3" VJ ' 4/3"
The stresses ^/jo, d\\ , da, are
max. stresses of an opposite kind
to those due to dead load.
Verticals : Max. load on each vertical = 20, 500 Ibs.
13. If a force of 5000 Ibs. strike the bottom chord of the girder in the
preceding question at 20 ft. from one end and in a direction inclined at
30 to the horizontal, determine its effect upon the several members.
EXAMPLES. 693
14. A Warren girder for a single-track railway bridge consists of
eight equilateral triangles and has to cross a span of 96 ft. ; the platform
is on the bottom chord ; the loads per lineal foot for which the truss is to
be designed are 2250 Ibs. due to engine, 1 500 Ibs. due to train, and 450 Ibs.
due to bridge. Determine the maximum stresses (both tensile and com-
pressi ve) in the members met by vertical planes immediately on the right of
the second, third, and fourth apices in the compression chord. Also,
find how many |-in. rivets are required to connect the diagonals met by
these planes with the chords and to prevent any tendency to longitudinal
slip between the support and the first apex, and between the first and
second apices in the tension chord.
15. The accompanying figure represents the half-truss fora bridge of
80 ft. span. Show how to determine the stresses
in the several members. Depth at centre = 12 ft.;
at =12 ft.; at .4 = 6 ft.
FIG. 442.
16. A Warren girder composed of eight equilateral triangles has its
upper chord in tension and has every joint loaded with a weight of 2 tons,
the loads being transmitted to the joints in the lower chord by means of
vertical struts. The span = 80 ft. Find the stresses in all the members.
Ans. Bays in tension chord : ist = 5 4/3 ; 2d == 13 4/3 ;
3d = i8i 4/3 ; 4th = 21 4/3 tons.
Bays in compression chord : ist = 9i 1/3 ; 2d = 16 4/3 ;
3d = 20 4/3 ; 4th = 21^ 4/3 tons.
Stresses in verticals : In each vertical = 2 tons.
Stresses in diagonals : ist = 10 4/3 ; 2d = 8f 4/3
3d -.= 7*4/3; 4th =64/3;
5th = 4f 4/3 ; 6th = 3 fc 4/3 ;
7th = 24/3; 8th = 4/3 tons.
17. A Warren girder, with the platform on the lower boom, carries a
load of 20 tons at the centre. Find the stress in each member, and also
find the weight at each joint of lower boom which will give the same
stresses in the centre bays.
There are six bays in the lower chord.
Ans. Stress in each diagonal = 2 4/3 tons.
Tens, chord : stress in ist bay = y> 4/3 ; 2d = 10 4/3 ; 3d 5 / -^3
tons.
Cornp. chord : stress in ist bay = 2 F 4/3 ; 2d = 4 ? 4/3 ; 3d = 20 ^3
tons.
Weight at each joint = 5{f tons.
694 THEORY OF STRUCTURES.
1 8. The accompanying truss of 240 ft. span and 30 ft. deep is to be de-
signed for a panel engine load of 24,000 Ibs., a panel train load of 18,000
i Ibs., and a panel bridge load of 12,000
Ibs, Determine graphically the maximum
// stresses in the members met by the ver-
tical MN. Also, draw a stress diagram
for the whole truss when it is covered
FIG. 443. with a uniformly distributed live load of
180,000 Ibs.
19. Loads of 3f , 6, 6, 6, and 6 tons follow each other in order over a ten-
panel truss at distances of 8, 5f, 4^, and 4! ft. apart. Determine the posi-
tion of the loads which will give the maximum diagonal and chord
stresses in the third and fourth panels. Span = 120 ft.
20. Determine the moment of resistance of a floor-beam for the Sault
Ste. Marie Bridge from the following data : Floor-beams, 16' 6" long
and 23' lof" apart ; the dead weight of the flooring, stringers, etc.
= 800 Ibs. per lineal foot of floor-beam; the live load as given in Fig.-
407, Art. 20, page 639 ; the load is transmitted to the floor-beam by four
lines of stringers so spaced as to throw two-thirds of the load upon the
inner pair, which are 3 ft. C. to C.
21. In a truss-bridge the panels are 17 ft. and the floor-beams 13 ft.
in length. Loads of 8, 12, 12, 12, 12, 10, 10, 10, and 10 tons follow each
other in order over the bridge at the distances of 7^, 4^, 4^, 4!, 7^, 5^, 6,
and 5^ ft. apart. Determine the moment of resistance of the beam,
taking the load due to the platform, etc., to be 500 Ibs. per lineal foot.
22. If the bridge in the preceding question is of the riveted type with
a single diagonal system, and with verticals at the panel points, the num-
ber of the panels being ten, find how many i-in. rivets are required in
the third panel from one end to connect the web with the chords, assum-
ing the panel live load to be 30,000 Ibs. and the panel dead load to be
10,000 Ibs.
23. With the loading given by Fig. 409, Art. 20, page 639, design a
floor-beam for a single-track bridge with panels 22 ft. long, the weight of
the platform being 450 Ibs. per square yard, and of each longitudinal 200
Ibs. per lineal yard.
24. Prepare a table giving the stresses of the several members of a
double-intersection through-truss of 154 ft. span, 20 ft. depth, and with
eleven panels. The panel engine, live, and bridge loads are 56,000,
35,000, and 16,800 Ibs., respectively.
25. Prepare a table giving the stresses in the several members of a
double-intersection deck-truss of 342 ft. span, 33 ft. depth, and with
eighteen panels. (Double-track bridge.) The panel engine, train
(or live), and dead loads are 96,000, 54,000, and 36,200 Ibs., respectively.
26. Prepare a table giving the stresses in the several members of a
.EXAMPLES.
.695
through-truss for a double-track bridge of 342 ft. span, 40 ft. depth, and
with nineteen panels. The panel engine, live, and dead loads are 96,000,
53,000, and 43,200 Ibs., respectively (double-intersection).
Cl C2 C3 C4 C5 C6 C7 C8 C9
Ans.
FIG. 444.
Chord
Panel.
Mult.
7326
Mult.
5063
Total Shear
transmitted
to Panel.
Tan-
gent.
Panel Stress
from Shear.
Total
Panel
Stress.
t = t.
18
131868
i53
774639
906507
45
407929
407929
t^
- 7326
8oi
407571*
400245^
45
180110.475
588039
t^
72*
9
* 6
6ij-
9
53*
9
2
42*
9
t 6
- 7326
34*
1 74673*
167347*
9
150612$
tg
23*
9
'10 j -.
9
(
jeo
45
1
c \
i
1
<
80*
45
V
921696
1
72*
9
|
Cn
_
61*
9
c a
53*
9
C 4
42*
9
f|
34*
9
C 6
23*
9
C-r
_
15*
9
c
4*
9
'*
- 3*
9
1
Shear
Diag.
Mult.
553
Mult.
2790
due to
Live
Mult.
2274
Total
Shear.
Secant
Max.
Stress.
Load.
p
18
153
171
.0965
4t
J 7
63*
.0965
^2
16
56*
72*
345
\
i5
75795
70742
48*
42*
118575
2III IO
189317
53*
139851
121659
350961
310976
345
345
12
65689
60636
35*
30*
99045
85095
164734
42*
34*
96645
78453
261379
224184
345
345
301528
di
II
24*
23*
345
dg
10
20^
15*
345
dg
9
4*
345
^10
8
40424
I 2 *
34875
75299
- 3*
- 7959
345
90572
a'll
7
35371
8*
-14*
- 32973
345
35122
/1 = 139200 Ibs.
>2 = 350961 "
/3 = 310976 "
' 4 = 261379 Ibs.
Z' 5 = 224184 "
"6 = 177377 "
= 142972 Ibs.
= 98955 "
= 67340 "
27. Prepare a table showing the stresses in the several members'
(including counters) of a ten-panel double-track through railway bridge
of 184^ ft. span and 34 ft. depth, the live and dead loads being respect-
ively 2250 Ibs. and 1100 Ibs. per lineal foot. (Thamesville Bridge.)
28. Determine the minimum stress-length (stress in a member multi-
plied by its length) for a double-intersection Pratt truss of 154 ft. span and
with eleven panels. The panel loads for engine = 44,000 Ibs., for train =
696 THEORY OF STRUCTURES.
27,500 Ibs., for bridge = 13,200 Ibs. ; coefficient of working strength = 8000
Ibs. per square inch for both compression and tension.
29. A six-panel single-intersection Pratt truss is uniformly loaded.
Assuming the same coefficient of strength both for compression and
tension, show that the economy of material will be greatest when the
diagonals are inclined at 32 25' to the vertical.
30. A double-intersection truss for a single-track through-bridge of
204 ft. span is 29 ft. deep, 20 ft. wide, and has twelve panels. Find the
stresses produced in the members of the leeward truss by a panel wind-
pressure of 5000 Ibs. acting 8 ft. above base of rails (5-ft. gauge).
Ans. Sloping members : ist = 27500 sec oc ; 2d = 12708$ sec a ;
3d = 10208$ sec ft ; 4th = 7708$ sec ft\
5th = 5208$ sec a ; 6th = 2708$ sec ;
7th = 208$ sec ft.
Tension chord : ist panel = 27500 tan a = 2d ; 3d = 40208$ tan a;
4th = 40208$ tan a + 10208$ tan ft;
5th = 40208$ tan a + 17916! tan ft ;
6th = 40208$ tan a + 23125 tan ft.
Compression chord : ist = 40208$ tan a + 10208$ tan ft ;
2d = 40208$ tan a + 17916$ tan ft ;
3d = 40208$ tan n + 23125 tan ft ;
4th = 40208$ tan a + 25833$ tan ft ;
5th = 40208$ tan a + 26041! tan ft.
Verticals : ist = 5000 ; 2d = 7708$ ; 3d = 5208$;
4th = 2708$ ; 5th = 208$ Ibs.
tan a = $| ; tan ft = ff .
31. In the preceding question find the maximum stresses in the
members of the fourth panel met by a vertical plane ; engine panel load
= 85,000 Ibs., train panel load = 40,800 Ibs., bridge panel load 22,500
Ibs.
Ans. Stresses in tension chord = 456^430.45 Ibs. ; in compression
chord 645,311.77 Ibs. ; in sloping members = 206,242.5 Ibs.;
and 139,705.62 Ibs.
32. Each of the two Pratt single-intersection five-panel trusses for a
single-track bridge is 55 ft. centre to centre of end pins and 1 1 ft. 6 in.
deep. Timber floor-beams are laid upon the upper chords 2f ft. centre
to centre ; the width between the chords = 10 ft. Find the proper scant-
ling of the floor-beams for the loading given in Fig. 407, page 639.
Also determine the maximum chord and diagonal stresses in the centre
panel due to the same live load.
33. Prepare a table giving the stresses in the several members of a
double-intersection deck-truss of 342 ft. span, 40 ft. depth, and with
EXAMPLES. 697
nineteen panels. (Double-track bridge.) The panel engine, train (or
live), and dead loads are 96,000, 53,000, and 43,200 Ibs., respectively.
34. Prepare a table giving the stresses in the several members of a
deck-truss for a double-track bridge of 342 ft. span, 33 ft. depth, and
with eighteen panels. The panel engine, live, arid dead loads are 96,000,
54,000, and 36,000 Ibs., respectively.
35. The two trusses for a 16 ft. roadway are each 100 ft. in the clear,
17 ft. 3 in. deep, and of the, type repre-
sented in the figure ; under a live load of
1 1 20 Ibs. per lineal foot the greatest total
stress in AB is 35,400 Ibs. Determine the
permanent load. FIG. 445.
The diagonals and verticals are riveted to angle-irons forming
part of the flanges. How many f-in. rivets are required for the con-
nection of AB and BC at B ? Also, how many are required between
A and C to resist the tendency of the angle-irons to slip longitudinally ?
Working-shear stress = 10,000 Ibs. per square inch.
Ans. 708.9 Ibs. ; 8; 4; n.
36. The compression chord of a bowstring truss is a circular arc of
80 ft. span and 10 ft. rise ; the bracing is of the isosceles type, the bases
of the isosceles triangles dividing the tension chord into eight equal
lengths. Determine the maximum stresses in the members met by a
vertical plane 28 ft. from one end. The live and dead loads are each
i ton per lineal foot.
37. Design a parabolic bowstring truss of 80 ft. span and 10 ft. rise
for a dead load of ton and a live load of i ton per lineal foot. The
joints between the web and the tension chord are to divide the latter
into eight equal divisions.
38. The compression chord of a bowstring truss is a circular arc.
The depth of the truss is 14 ft. at the centre and 5 ft. at each end ; the
span = 100 ft. ; the load upon the truss = 840 Ibs. per lineal foot. Find
the stresses in all the members. Determine also the maximum stresses
in the members met by a vertical 25 ft. from one end when a live load
of looo Ibs. per lineal foot crosses the girder. What counter-braces are
required ?
39. A Pratt truss with sloping end posts has a length of 150 ft. centre
to centre, and a height of 30 ft. centre to centre, with panels 15 ft. long ;
the dead load is 3000 Ibs. per lineal foot, and the live load 1200 Ibs.
Determine the maximum stresses in the end posts, in the third post from
one end, in the middle of the bottom chord, and in the members of the
third panel met by a vertical plane.
40. Design a cross-tie for a double-track open-web bridge, the ties
698 THEORY OF STRUCTURES.
being 18 ft. 5 in. centre to centre, and the live load for the floor system
being 8000 Ibs. per lineal foot.
41. A bowstring roof-truss of 50 ft. span, 15 ft. rise, and five panels is
to be designed to resist a wind blowing horizontally with a pressure of
40 Ibs. per square foot* The depth of the truss at the centre is 10 ft.
Determine, graphically, the stresses in the several members of the truss,
assuming that the roof rests on rollers at the windward support.
42. A bowstring truss of 120 ft. span and 15 ft. rise is of the isosceles
braced type, the bases of the isosceles triangles dividing the tension
chord into twelve equal divisions ; the dead and live loads are ton and
i ton per lineal foot, respectively. Find the maximum stresses in the
members met by vertical planes immediately on the right of the second
and fourth joints in the tension chord.
43. The figure is a skeleton diagram of the Sault Ste. Marie Bridge
(C P. R.). Span = 239 ft. ; there are ten panels, each of 23.9 ft.,, say 24
493000499,000
275,000 275,000 416,000 478,000 484,000
FIG. 446.
ft. ; the length of the end verticals = 27 ft., of the centre verticals = 40
ft.; width on truss centres = 17^ ft. The bridge is designed to bear the
loading given by Fig. 407, page 639. Show that
(a) The stresses in every panel length of each chord are greatest when
third driver is at a panel point ; and find the value of the several
stresses.
: (b} The stresses in the verticals a and the diagonals b are greatest
when the third driver is at a panel point ; and find their values.
/ (f) The stresses in the remaining members of the truss are greatest
when the second driver is at a panel point ; and find their values.
(d) The maximum stresses in the verticals a! vary from a tension of
64,000 Ibs. to a compression of 11,000 Ibs.
(e) The stress in the counter-brace c is nil.
Am. The values of thestresses in the several members are marked
on the diagram. They are deduced from the distributions
given in the table on page 642, and are correct within a very
small percentage.
44. The figure represents a counterbalanced swing-bridge, 16 ft. deep
and wholly supported upon the turn-table at A
and B '; the dead weight is 650 Ibs. per lineal foot
of bridge ; the counterpoise is hung from C and D.
FIG. 447? Find its weight, assuming (a) that the whole of it
is transmitted to B ; (b) that a portion of it is transmitted to A through
EXAMPLES.
699
a member BE, sufficient to make the reactions at A and B equal.
determine the stresses in the several members of the truss.
Also,
Ans. Counterpoise in case (a) = 26,162$ Ibs. ;
in case (b) = 22,186$% Ibs.
Stress transmitted through BE in case ()= 24,012 Ibs.
45. The figure represents a counterbalanced swing-bridge ; the dead
load upon the bridge is 650 Ibs. per lineal
foot ; the counterpoise is suspended from
CD. Find its value, the joint at E being
so designed that the whole of the load
upon the bridge is always transmitted FIG. 44 b.
through the main posts EA, EB, and is evenly distributed between the
points of support at A and B.
Ans. 20,694.3 Ibs.
46. Find the stresses in the several members of the truss in the pre-
ceding question (a) when the bridge is open ; (b) when the bridge is
closed and is subjected to a live load of 3000 Ibs. per lineal foot.
Height of truss at E = 16 ft., at F = 8 ft.
47. Prepare a table giving the stresses in the several members of a
a cs c 3 C4 C6 . single-intersection through-truss of
y\ \ 2 N 3 J\ 4 JX 5 X 6 \ \ \ \ J 54 ft - span> 20 ft - de P th ' and with
/ fi\ff2\f3\ffAfo\j \J \j \J \J \ eleven panels. The panel engine,
IQ U& 3 14 tfl 1C
FIG. 449. live, and dead (or bridge) loads
27,500, 17,600, and 8470 Ibs., respectively.
Ans.
Diag.
Mult.
2500
Mult.
1600
Sum.
Mult.
770
Sum.
sec.
Total M%
Stress. Y
p
IO
25000
45
72000
97000
55
42350
139350
22
170007
d l
9
22500
36
57600
80100
44
3380
113980
22
139056
dn
g
200 oo
28
44800
64800
33
25410
90210
22
110057
<^3
7
17500
21
33600
51100
22
16940
68040
22
83009
6
15000
15
24000
39000
11
8470
47470
22
579 J 4
</6
5
12500
IO
16000
28500
28500
22
34770
*6
4
10000
6
9600
19600
II
-8470
"130
22
13579
Panel.
Mult.
3270
Mult.
2370
Sum.
tan.
Panel
Stress.
Total Panel
Stress.
'i = t*
'3
'4
'5
*6
10
I
I
I
I
32700
- 3270
- 3270
3270
- 3270
45
45
34
2 3
12
106650
106650
80580
545 10
28440
139350
103380
77310
51240
25170
f?
97545
72366
54"7
35868
17619
97545
169911
224028
259896
2775*5
700
THEORY OF STRUCTURES.
Panel.
Mult.
3270
Muit.
2370
Sum.
tan.
Total
Stress.
Total Max.
Stress.
c\
10
32700
- 3 2 7
45 I
45 j
10^650 I
106650 f
242730
*
169911
169911
c i
3270
34
80580
773*0
"
54"7
224028
^3
- 3*70
2 3
545 10
51240
**
35868
259896
^4
^270
12
28440
25170
"
17619
2775 X 5
c*
3270
i
2370
- 900
- 630
276885
v\ 35*97 Ibs. 5 v * 90,210 ; v s = 68,040 ; z> 4 = 47,470 ; ^ 5 = 28,500 Ibs.
48. Compare the relative amounts of iron required in the webs of a
single- and a double-intersection Pratt deck-truss of zoo ft. span and
having eight panels. Panel live load = L, panel dead load = D.
49. The figure represents a pier, square in plan, supporting the ends
of two deck -trusses, each 200 ft. long and 30 ft. deep. The height of the
pier is 50 ft. and is made up of three panels, the
upper and lower being each 17 ft. deep. Ten
square feet of bridge surface and ten square feet
of train surface per lineal foot are subjected to
a wind-pressure of 40 Ibs. per square foot. The
centre of pressure for the bridge is 68 ft., and
for the train 86 ft., above the pier's base. The
wind also produces a horizontal pressure of 4000
Fia. 450. Ibs. at each of the intermediate panel points on
the windward side of the pier. Width of pier = 17 ft. at top and 33! ft.
at bottom. The bridge load = 1600 Ibs. per lineal foot, live load = 3000
Ibs. per lineal foot. Determine
(a) The overturning moment (3180 ft.-tons).
(b) The horizontal force due to the wind at the top of the pier.
(61.6 tons.)
(c) The tension in the vertical anchorage ties at 5 and T. (Nil.}
(d) The vertical and horizontal reactions at T. (275 and 65.6 tons.)
Draw a diagram giving the wind-stresses in all the members, and in-
dicate which are in tension and which in compression.
Ascertain whether the wind-pressure of 40 Ibs. per square foot upon
a train of empty cars weighing 900 Ibs. per lineal foot will produce a
tension anywhere in the inclined posts. What will be the tension in the
anchorage ties ? (20.75 tons.)
Find the stresses in the traction bracing (i) when a loaded train trav-
elling at 30 miles an hour is braked just as the engine is over the pier
and brought to rest in a length of 300 ft. ; (2) when a loaded train with
the engine over the pier is started by a sudden admission into the cylin-
ders of steam at 100 Ibs. per square inch. Stroke of cylinder = 16 in.,
diameter of drivers = 5 ft.
EXAMPLES.
701
50. The figure represents one half of one of the piers of the Bouble
Viaduct. The spans are crossed by two lattice-gir-
ders, 14' 9" deep and having a deck platform. The
height of the pier is 183' 9" and is made up of eleven
panels of equal depth. Width of pier at top = 13'
ii", at bottom = 67' 7". With wind-pressure at
55.3 Ibs. per square foot, the total pressure on the
girder, train, and pier have been calculated to be 20,
16.2, and 20 tons, acting at points 196.2, 210.3, and
92.85 ft., respectively, above the base. The dead
weight upon each half pier is 222^ tons, of which 60
tons is weight of half span, 120 tons the weight of
the half pier, and 42^ tons the weight of the train.
Assuming that the wind-pressure on the pier is a
iiorizontal force of 2 tons at each panel point on the
windward side, and that the weight of the pier may
be considered as a weight of 6 tons at each panel
point, determine
(a) The overturning moment. FIG. 451.
(b) The total horizontal force at the top of the pier due to the wind.
(c) The tension in each of the vertical anchorage ties at 5 and T due
to the wind-pressure.
(d) The vertical and horizontal reactions at T.
Show that the greatest compressive stress occurs in the member RT,
and that it amounts to 422 tons.
Draw a stress diagram giving the stresses in all the members, indi-
cating which are in tension and which in compression. Width of pier
at A = 20 ft., at = 23$ ft., at C = 36^ ft.
What will be the effect of braking the train when running at 30 miles
an hour, so as to bring it to rest within a distance of 220 ft. ? Width
of pier in direction of bridge = 9! ft. at top and = 20 ft. at bottom.
Ans. (a) 9188 ft.-tons; (ti) 39.9 tons; (c} 24! tons, (d) Hori-
zontal reaction = 59.9 tons ; vertical reaction = 247 tons.
The accompanying figure represents a portion of a cantilever truss,
the horizontal distances of the points A, B t C
from the free end being A , / a , I* , respectively.
The boom ABC is inclined at an angle a, and the
boom XYZ at an angle ft, to the horizon. Find
the deflections at the end of the cantilever due to
X Y z (a) an increase kAB in the length of AB\ (2) an
FIG. 452. increase kiB Y in the length of B Y ; (3) a decrease
k*XY in the length of XY\ (4) a decrease kBX in the length of BX.
w
Ans. (i)
BX sin ABX
702
THEORY OF STRUCTURES.
(3)
BX
(4)*
BX cos a
sin
- / 9 (cot BXY - cot
In the preceding question, if k l = k* = k* = ki = k, and if A W is
parallel to ^Jf, and AX to ^ F, show that the angle between WX and
xY F after deformation
= 2>&(cot ABX + cot B YX).
Hence also, if the truss is of uniform depth d, show that the " deviation "
2k
of the boom per unit of length is constant and equal to
d
52. Six bars have to be arranged upon a steel pin ; each bar is I in.
wide and is subjected to a stress of 64,000 Ibs. Should the bars be ar-
64,000 Ibs.
FIG. 454. Method 2.
ranged according to method i or method 2 ? Why ? Determine the di-
ameter of the pin.
53. The accompanying sketch represents one of the pin connections
in a certain bridge which was recently overthrown. The two innermost
bars are web members inclined to the horizon at an angle whose cosine
"^^^
-[42,100 Ib8.
FIG. 455.
is .815. The thickness of the bars and the maximum stresses to which
they are severally subjected are shown on the diagram. Is the 3-in.
wrought-iron pin sufficiently strong?
CHAPTER XII.
SUSPENSION-BRIDGES.
I. Cables. The modern suspension-bridge consists of two
or more cables from which the platform is suspended by iron
or steel rods. The cables pass over lofty supports (piers), and
are secured to anchorages upon which they exert a direct pull.
Chain or link cables are the most common in England and
Europe, and consist of iron or steel links set on edge and
pinned together. Formerly the links were made by welding
the heads to a flat bar, but they are now invariably rolled in
one piece, and the proportional dimensions of the head, which
in the old bridges are very imperfect, have been much im-
proved.
Hoop-iron cables have been used in a few cases, but the
practice is now abandoned, on account of the difficulty attend-
ing the manufacture of endless hoop-iron.
Wire-rope cables are the most common in America, and
form the strongest ties in proportion to their weight. They
consist of a number of parallel wire ropes or strands, compactly
bound together in a cylindrical bundle by a wire wound round
the outside. There are usually seven strands, one forming a
core round which are placed the remaining six. It was found
impossible to employ a seven-strand cable in the construction
of the East River Bridge, New York, as the individual strands
would have been far too bulky to manipulate. The same ob-
jection held against a thirteen-strand cable (thirteen is the next
number giving an approximately cylindrical shape), and it was
finally decided to make the cable with nineteen strands. Seven
of these are pressed together so as to form a centre core, around
which are placed the remaining twelve, the whole being con-
tinuously wrapped with wire.
703
704
THEORY OF STRUCTURES.
In laying up a cable great care is required to distribute the
tension uniformly amongst the wires. This may be effected
either by giving each wire the same deflection or by using
straight wire, i.e., wire which when unrolled upon the floor
from a coil remains straight and shows no tendency to spring
back. The distribution of stress is practically uniform in un-
twisted wire ropes. Such ropes are spun from the wires and
strands without giving any twist to individual wires.
The back-stay is the portion of the cable extending from an
anchorage to the nearest pier.
The elevation of the cables should be sufficient to allow for
settling, which chiefly arises from the deflection due to the load
and from changes of temperature.
The cables may be protected from atmospheric influence
by giving them a thorough coating of paint, oil, or varnish, but
wherever they are subject to saline influence, zinc seems to be
the only certain safeguard.
2. Anchorage, Anchorage Chains, Saddles. The an-
chorage, or abutment, is a heavy mass of masonry or natural
rock to which the end of a cable is made fast, and which re-
sists by its dead weight the pull upon the cable.
FIG. 456. FIG. 457. FIG. 458.
The cable traverses the anchorage as in Figs. 456 to 458,
and passes through a strong, heavy cast-iron anchor-plate, and,
if made of wire rope, has its end effectively secured by turning
it round a dead-eye and splicing it to itself. Much care, how-
ever, is required to prevent a wire-rope cable from rusting on
account of the great extent of its surface, and it is considered
advisable that the wire portion of the cable should always ter-
minate at the entrance to the anchorage and there be attached
to a massive chain of bars, which is continued to the anchor-
plate or plates and secured by bolts, wedges, or keys.
ANCHORAGE, ANCHORAGE CHAINS, SADDLES.
In order to reduce as much as possible the depth to which
it is necessary to sink the anchor-plates, the anchor-chains are
frequently curved as in Fig. 458. This gives rise to an oblique
force, and the masonry in the part of the abutment subjected
to such force should be laid with its beds perpendicular to the
line of thrust.
The anchor-chains are made of compound links consisting
alternately of an odd and an even number of bars. The friction
of the link-heads on the knuckle-plates considerably lessens
the stress in a chain, and it is therefore usual to diminish its
sectional area gradually from the entrance E to the anchor.
This is effected in the Niagara Suspension Bridge by varying
the section of the bars, and in the East River Bridge by vary-
ing both the section and the number of the bars.
The necessity of preserving the anchor-chains from rust is
of such importance that many engineers consider it most
essential that the passages and channels containing the chains
and fastenings should be accessible for periodical examination,
painting, and repairs. This is unnecessary if the chains are
first chemically cleaned and then embedded in good hydraulic
cement, as they will thus be perfectly protected from all at-
mospheric influence.
The direction of an anchor-chain is changed by means of a
saddle or knuckle-plate, which should be capable of sliding to
an extent sufficient to allow for the expansion and contraction
of the chain. This may be accomplished without the aid of
rollers by bedding the saddle upon a four- or five-inch thickness
of asphalted felt.
The chain, where it passes over the piers, rests on saddles,
the object of which is to furnish
bearings with easy vertical curves.
Either the saddle may be constructed
as in Fig. 459, so as to allow the
cable to slip over it with compara-
tively little friction, or the chain may be secured to the saddle,
and the saddle supported upon rollers which work over a per-
fectly true and horizontal bed formed by a saddle-plate fixed
to the pier.
7 o6
THEORY OF STRUCTURES.
3. Suspenders. The suspenders are the vertical or in-
clined rods which carry the platform.
FIG. 460.
FIG. 461.
FIG. 462.
FIG. 463.
In Fig. 460 the suspender rests in the groove of a cast-
iron yoke which straddles the cable. Fig. 461 shows the
suspender bolted to a wrought-iron or steel ring which em-
braces the cable. When there are more than two cables in
the same vertical plane, various methods are adopted to insure
the uniform distribution of the load amongst the set. In Fig.
462, for example, the suspender is fastened to the centre of a
small wrougnt-iron lever PQ, and the ends of the lever are
connected with the cables by the equally strained rods PR
and QS. In the Chelsea bridge the distribution is made by
means of an irregularly shaped plate (Fig. 463), one angle of
which is supported by a joint-pin, while a pin also passes
through another angle and rests upon one of the chains.
The suspenders carry the ends of the cross-girders (floor-
beams), and are spaced from 5 to 20 ft. apart. They should
be provided with wrought-iron screw-boxes for purposes of
adjustment.
4. Curve of Cable. CASE A. An arbitrarily loaded flexible
cable takes the shape of one of the catenaries, but the true
catenary is the curve in which a cable of uniform section and
material hangs under its own weight only.
Let A be the lowest point of the cable, and take the ver-
tical through A as the axis of y.
Take the horizontal through O as the axis of x, the origin
O being chosen so that
Hmp, ...... (i)
/ being the weight of a unit of length of the cable, and H the
horizontal pull at A.
CURVE OF CABLE.
707
m or AO is the parameter, or modulus, of the catenary, and
OG is the directrix.
Let x, y be the co-ordinates of any point P, the length of
the arc AP being s.
Draw the tangent PT and the ordinate PN.
The triangle PNT is evidently a triangle of forces for the
portion AP, PN representing the weight of AP (viz., ps), PT
\
O N
FIG. 464.
\
the tangential pull T at P, and NT the horizontal pull H at
A.
dy n ~ AT PN PS s
* -7-= tan PTN= -=jr =-=-, ... (2)
dx TN H m
which gives the differential equation to the catenary.
It may be easily integrated as follows :
ds
or
dy^ l~ ~~? i
v I + * = *
f. (s)
ds
dx
. ? tog(*
in
c being a constant of integration.
When x = o, s = o, and therefore log m = c.
Hence,
log M^+^ =
708 THEORY OF STRUCTURES.
or
or
m
- \ / \
-e -). ...... (4)
Again, ^ ji
dy s i* *
-y- = - = -(e e );
V
dx m 2
and hence,
tn -
(5)
The constant of integration is zero, since y = m when
x o.
The last equation is the equation to the catenary, while eq.
(4) gives the length of the arc A P.
By equations (4) and (5),
(6)
Draw NM perpendicular to PT, and let the angle PTN =
PNM=e. Then
PM = PN sin = y = s , . . . (7)
<fV + m 9
and
MN = PN cos 0=y =??==: m, . . . (8)
since tan = -f- = .
or 7
Thus, the triangle /W7V possesses the property that the
side PM is equal to the length of the arc AP t and the side
MN is equal to the modulus w (= AO).
The area ^/W6>
/* x m* *
=y 7</4r = (^ ^"w) ms = 2 X triangle PMN.
CURVE OF CABLE.
The radius of curvature, p, at P
709
y
m*
(9)
PG being perpendicular to PT.
At A, y = m, and the radius of curvature is also m. (id)
Again,
Z
ps
PT
_
y
H m
(II)
(12)
p c being the radius of curvature at A.
These catenary formulae are of little if any use in the design
and construction of suspension-bridges, as they are based upon
the assumption of a purely theoretical load which never occurs
in practice, viz., the weight of a chain of uniform section and
density.
CASE B. Let the platform be suspended from chains com-
posed of a number of links, and let W be the whole weight be-
tween the lowest point O of the chain and the upper end Pof
any given link. Let the direction of this link intersect that of
the horizontal pull (//) at O in E. Drop the perpendicular PN.
THEORY OF STRUCTURES.
The triangle PNE is evidently a triangle of forces ; and if the
angle PEN = 6,
PN W
and hence
tan oc W.
Thus, by treating each link separately, commencing with the
lowest, the exact curve of the chain may be easily traced.
Generally speaking, the distribution of the load may be
assumed to be approximately uniform per horizontal unit of
length, the load being suspended from a number of points
along each chain or cable by means of rods. The curve of the
cable will then be a parabola.
Let w be the intensity of the load per horizontal unit of
length.
Let x, y be the co-ordinates of any point P of the cable
with respect to the horizontal 6Uf and the vertical OY as axes
of x and y, respectively.
Let be the inclination of the tangent at P to the horizon-
tal. The portion OP of the cable is kept in equilibrium by
the horizontal pull H at O, by the tangential pull T at P, and
by the load wx upon OP, which acts vertically through the
middle point E of ON, PN being the ordinate at P.
Hence, the tangent at P must also pass through E, and
PEN is a triangle of forces. Hence,
x
H 2 2H .
= -, or *= -y, ..... (i)
the equation to a parabola with its vertex at O, its axis vertical,
2H
and its parameter equal to .
Again,
T _ PE _ i
H~ EN~ cos~0'
PARAMETER, ETC. 7 II
and hence
r CO s0 = ;7= ' (*
2y '
and the horizontal pull at every point of the cable is the same as
that at the lowest point.
Also,
Z/Jt" 2 /
/ / x*
F = W* \ I I + 5
x l V 4/
4
The radius of curvature at P
*
so that the radius at O is
H
or
5. Parameter, etc. Let ^,, ^ 2 be the elevations of A
and .#, respectively, above the horizontal line COD, Fig. 465.
Let OD = a lt OC = a,, and let a l + a t =a= CD.
By equation (i), Art. 4.
2ff a, a^ a,
y i//^ a v^, + ^ a y^ + Vk
Denote the parameter by P. Then
w
Also,
712 THEORY OF STRUCTURES.
If 0, , a be the values of at A and B, respectively.
tan 6>, = 2 A- and tan 3 = 2 Ai,
Note.li )&, = >&, = ,
and hence
tan ft-as= = tan ^,.
6. Length of Arc of Cable. Let OP = s, Fig. 465,
wx
Since tan = -==,
2
or
sec a OdB = -^^r = ~</j cos ft
ds =
w cos 3
Hence,
TT /^ 7^a rr
^ = ^7J tan ^ sec .0 + log, (tan
cos
Again,
tan 6 = x,
and
sec v =
WEIGHT OF CABLE. 713
Note. An approximate value of the length of the arc may
be obtained as follows :
/ i wV\
.-. ds dx\\ + -- -J7r]> approximately.
Integrating between O and P,
.
* /
7. Weight of Cable. The ultimate tenacity of iron wire
is 90,000 Ibs. per square inch, while that of steel rises to
200,000 Ibs., and even more. The strength and gauge of cable
wire may be insured by specifying that the wire is to have a
certain ultimate tenacity and elastic limit, and that a given
number of lineal feet of wire is to weigh one pound. Each of
the wires for the cables of the East River Bridge was to have
an ultimate tenacity of 3400 Ibs., an elastic limit of 1600 Ibs.,
and 14 lineal feet of the wire were to weigh one pound. A very
uniform wire, having a coefficient of elasticity of 29,000,000 Ibs.,
has been the result, and the process of straightening has raised
the ultimate tenacity and elastic limit nearly 8 per cent.
Let W, be the weight of a length a, (= OD) of a cable of
sufficient sectional area to bear safely the horizontal tension H.
Let W^ be the weight of the length s,( = OA) of the cable
of a sectional area sufficient to bear safely the tension 7\ at A.
Let /be the safe inch-stress.
Let q be the specific weight of the cable material.
Then
and
7 1 4 THEORY OF STRUCTURES.
'. --.| -<-*+!<- +.4
or
+ V), nearly.
A saving may be effected by proportioning any given section
to the pull across that section.
At any point (x, y) the pull = H sec 0, and the correspond-
ing sectional area j . The weight per unit of length
= -7. g, and the total weight of the length s l (= OA) is
r sec ds
Hq
f
Hence,
and also
The weight of a cubic inch of steel averages .283 Ib.
The weight of a cubic inch of wrought-iron averages .278 Ib.
IT
The volume in inches of the cable of weight W 1 = \2a l p .
DEFLECTION OF CABLE. 715
W,
-- = .283 lb. or .278 lb.,
according as the cable is made of steel or iron.
Let the safe inch-stress of steel wire be taken at 33,960 Ibs.,
of the best cable-iron at 14,958 Ibs., and of the best chain-links
at 9972 Ibs. Then
W, = Ha, X .283 X = for steel cables ;
W t = Ha, X .278 X -~- g = for iron cables ;
t = Ha, X .278 X ~ = for link cables.
Note. About one-eighth may be added to the net weight of
a chain-cable for eyes and fastenings.
8. Deflection of a Cable due to an Elementary Change
in its Length.
By the corollary of Art. 6 the total length (S) of the cable
A OB is
Now a, and #, are constant ; h^ h 9 is also constant, and
therefore dh^ = dh v Hence,
If the alteration in length is due to a change of / in the
temperature,
dS = etS,
e being the coefficient of linear expansion and = ^ ^~
per degree Fahr. for wrought-iron.
THEORY OF STRUCTURES.
In England the effective range of temperature is about 60
Fahr., while in other countries it is usual to provide for a range
of from 100 to 150 F.
If the alteration is due. to a pull of intensity /per unit of
area,
dS = gS,
E being the coefficient of elasticity of the cable material.
If h, = h, = h,
a 16 h
# , = a t = , and dS = dh.
9. Curve of Cable from which the load is suspended by a
series of sloping rods.
r o E N
FIG. 466.
Let O be the lowest point of such a cable. Let the tangent
at O, and a line through O parallel to the suspenders, be the
axes of x and y, respectively.
Let w' be the intensity of the oblique load. Consider a
portion OP of the cable, and let the co-ordinates of P with
respect to OX, OY be x and y.
Draw the ordinate PN, and let the tangent at P meet ON
in E.
As before, PNE is a triangle of forces, and E is the middle
point of ON. Then
PN 2
_ _ ^ 2 _
~~' ~~ ~ y '
the equation to a parabola with its axis parallel to OY and its
focus at a point S, where ^SO ,- .
CURVE OF CABLE WITH OBLIQUE SUSPENDERS.
Cor. i. Let the axis meet the tangent at O in T f , and let
its inclination to OX be i.
Let A be the vertex, and ON' a perpendicular to the axis.
Then
SO = ST' = SA + A T f = SA + AN'.
But AS .AN' = ON" = N f T'*tan*i
.-. AS = AN' tan 2 i, and SO = AS(i + cot 2 *) =' -..
sm a z
Hence, the parameter 4^45 = ^SO sin 2 /.
Cor. 2. Let P be the oblique load upon the cable between
O and P.
Let Q be the total thrust upon the platform at E.
11 w " " load per horizontal unit of length.
" q " " rate of increase of thrust along platform.
" / " " length of PE.
Then
w
w . , and q = w cot z ;
sm i sin i
p = ff ^
x y
x*
t* y + + *y cos i.
4
Cor. 3. Let s be the length of OP, and let 6 be the inclina-
tion of PE to OY. Then
s = AP- AO
tan (9 - ^ sec(9 - ^
+ log,jtan (90 &) + sec (90 - 0)} - tan (90- *) sec (go *)
- log, { tan (90 - 1 ) + sec (90 - i) } j
//"sin* ( ;-. cot # + cosec <5 )
--- : -J cot cosec 6 cot / cosec i 4- log, - r^
2W ( cot z -f~ cosec z )
7i8
THEORY OF STRUCTURES.
It may be easily shown, as in the Note to Art. 6, that ap-
proximately
. .
s = x 4- y cos t -4-
sin 2 *
t . .
X -j-J/COSZ
10. Pressure upon Piers, etc.
Let 7 1 , be the tension in the main cable at A.
" T; " " " " " back-stay at A
" ?, /? be the inclinations to the horizontal of the tangents
at A to the main cable and back-stay, respectively.
The total vertical pressure upon the pier at A
7 1 , sin a -j- 7", sin y# /?.
The total resultant horizontal force at ^4
= 7i cos f ~ T z cos fi = Q.
If the cable is secured to a saddle which is free to move
horizontally on the top of the pier (Fig. 467),
Q the frictional resistance to the tendency to motion,
or Q = ^R t
fa being the corresponding coefficient of friction.
FIG. 467.
Let A Fig. 468, be the total height of the pier, and let W
its weight.
Let FG be the base o
of the centre of pressure.
4
be its weight.
Let FG be the base of the pier, and K the limiting position
, * f
AUXILIARY OR STIFFENING TRUSS.
Let /, q be the distance of P and W, respectively, from
Then
pp _L Wq
for stability of position Q = -- />""
and for stability of friction, when the pier is of masonry,
. the coefficient of friction of the masonry.
If /^ is sufficiently small to be disregarded, Q is approxi-
mately nil, and 7, cos a = 7", cos fi = H. The pressure upon
the pier is now wholly vertical and is //"(tan a -j- tan /?).
When the cable slides over smooth rounded saddles (Fig.
459), the tensions 7, and 7, are approximately the same.
Thus,
R = T, (sin a + sin ft) and g = T t (cos a cos ft).
It a = fi, Q = o, and the pressure upon the pier is wholly
vertical, its amount being 27, sin a.
The piers are made of timber, iron, steel, or masonry, and
allow of great scope in architectural design.
The cable should in no case be rigidly attached to the pier,
unless the lower end of the latter is free to revolve through a
small angle about a horizontal axis.
II. Auxiliary or Stiffening Truss. The object of a stiff-
ening truss (Fig. 469) is to distribute a passing load over the
cable in such a manner that it cannot be distorted. The pull 1
upon each suspender must therefore be the same, and this vir-
tually assumes that the effect of the extensibility of the cable
and suspenders upon the figure of the stiffening truss may be
disregarded.
7 20 THEORY OF STRUCTURES.
The ends O and A must be anchored, or held down by
pins, but should be free to move horizontally.
Let there be n suspenders dividing the span into (n -\- i)
equal segments of length a.
Let P be the total weight transmitted to the cable, and z
the distance of its centre of gravity from the vertical through O.
Let T be the pull upon each suspender.
Taking moments about O,
/ being the length of OA.
Also, if t is the intensity of pull per unit of span,
r
tl = nT, and hence Pz t .
Let there be a central suspender of length s. There will,
therefore, be - suspenders on each side of the centre.
r
The parameter of the parabola = =- .
Hence, the total length of all the suspenders
If there is no central suspender, i.e., if n is even,
the total length = (n i)(s + -" -jJ.
Denote the total length of suspenders by L. Then
the stress-length = TL = ~PL.
AUXILIARY OR STIFFENING TRUSS. ? '21
Let w be the uniform intensity of the dead load.
CASE I. The bridge partially loaded.
Let w' be the maximum uniform intensity of the live load,
and let this load advance from A and cover a length AB.
Let OB x, and let R l , R^ be the pressures at O and A,
respectively.
For equilibrium,
R^R^ + tl-wl- /(/ - *) = o ; (i)
J_z/-^(/-*) =0. ... (2)
Also, since the whole of the weight is to be transmitted
through the suspenders,
>(l x) ....... (3)
From eqs. (i), (2), and (3),
(4)
which shows that the reactions at O and A are equal in mag-
nitude but opposite in kind. They are evidently greatest when
x = -, i.e., when the live load covers half the bridge, and the
w'l
common value is then r- .
O
The shearing force at any point between O and B distant x'
from O
which becomes j(l x) R l = R^ when x' equal x.
Thus the shear at the head of the live load is equal in magni-
tude to the reaction at each end, and is an absolute maximum
722 THEORY OF STRUCTURES.
when the live load covers half the bridge. The web of the
rf
8
truss must therefore be designed to bear a shear of ^- at the
centre and ends.
Again, the bending moment at any point between O and B
distant x' from O
= R l x' + t x" = ~ l -=(x-xx'), . . (6)
which is greatest when *' =-, i.e., at the centre of OB, its
VOJ / _ JK
value then being f~ x *- Thus, the bending moment is
O /
d * 2
an absolute maximum when ~j~(tx* **) = o, i.e., when x =. -/,.
w'
and its value is then --- /*.
54
The bending moment at any point between B and A dis-
tant x' from O
" - - x? = j(x> - x)(l- x'\ (7)
d
which is greatest when ~r~ t {(^' ~~ *Yt ~ x '}\ = i- e -> when
14-x
x' = , or at the centre of AB, its value then being
w' x
(I x}*. Thus, the bending moment is an absolute niaxi-
/
3
mum when -j-\x(l x)*\ = o, i.e., when x = , and its value is
then + /'.
54
Hence, the maximum bending moments of the unloaded and
loaded divisions of the truss are equal in magnitude but opposite
in direction, and occur at the points of triscction (D, C) of OA
AUXILIARY OR STIFFENING TRUSS. 723
when the live load covers one-third (AC) and two-thirds (AD) of
the bridge, respectively.
Each chord must evidently be designed to resist both
tension and compression, and in order to avoid unnecessary
nicety of calculation, the section of the truss may be kept uni-
form throughout the middle half of its length.
CASE II. A single concentrated load W at any point B of
the truss. W now takes the place of the live load of intensity
<w'.
The remainder of the notation and the method of pro-
cedure being precisely the same as before, the corresponding
equations are
R 1 + R, + (t-w)l-W = o (i')
'-W(l-x) = o (2')
2
W
(3')
(4')
which shows that the reactions at O and A are equal in mag-
nitude but opposite in kind. They are greatest when x = o
and when x = /, i.e., when W is either at O or at A, and the
W
common value is then .
2
The shearing force at any point between O and B distant x'
from O
W
W
which is a maximum when x' = x, and its value is then .
2
The web must therefore be designed to bear a shear of
W
throughout the whole length of the truss.
724 THEORY OF STRUCTURES.
Again, the bending moment at any point between O and
B distant x' from O
First y let w < . The bending moment is positive and is a
maximum when x' =. x, its value then being
+ (/
21
Next,\&. x > . The bending moment is then negative va^
is a maximum when x' = x -, its value then being
Wi
The bending moment at any point between B and A dis
tant x' from O
= R,x' + (t- w )-- W( X ' -x) = -j(x' - /)- -
which is a maximum when
I W I l\
i.e., when x' = x -4- -, and its value is then -Ax } .
1 2 2/\ 21
Note. The stiffening truss is most effective in its action, but
adds considerably to the weight and cost of the whole struc-
ture. Provision has to be made both for the extra truss and
for the extra material required in the cable to carry this extra
load.
AUXILIARY OR STIFFENING TRUSS. 725
Stiffening Truss hinged at the Centre. Provision may be
made for counteracting the straining due to changes of tem-
perature by hinging the truss at the centre E.
Let a live load of intensity w advance from A.
First, let the live load cover a length AB = x f > ).
Let R^ , R^ be the pressures at O, A, respectively.
The equations of equilibrium are
l + R t + (t-iv)/-iv'x = ; .... (I)
--*-'=" W
(3)
Eqs. (2) and (3) being obtained by taking moments about E.
Hence,
w'
t-w=- -(/ 3 - 4** + 2*') ; ... (4)
; (5)
Next, iet the live load cover the length BO ( < -)
Let AB = x as before, and let JR/ t RJ, t' be the new values
of R lt R^ t, respectively.
The equations of equilibrium are now
'X-'V-~*)=o; . . (7)
726 THEORY OF STRUCTURES.
*,'- + (;'-Hg=o; (9)
and hence,
/'- W = 2-(/- *)'[=-(/-, -a-')]' (10)
I W
\i -*)>(= -R) (12)
Diagram of Maximum Shearing Force. The shear at any
point distant z from A in the unloaded portion BO when the
live load covers AB
^R^(t-w)(l-z) ......... (13)
>-, _/)(/_*)}
' - /)(/- z) - w\l- z)
= minus the shear at the same point when AB is
unloaded and the live load covers BO.
For a given value of z the maximum shear, positive or
negative, at any point of OB, is found by making (see eq. (13) )
or
3*) - (/-*K- 4/+ 4*) = O,
or
7 4# 2/
jr=z/ T- (H)
AUXILIARY OR STIFFENING TRUSS.
Hence, by eqs. (4), (5), (13), (14),
727
I 2X
the maximum shear = %w'x , . . (15)
/-*
and may be represented by the ordinate m
{positive or negative) of the curve mnpq.
For example, at the points defined by
*= /, v, */,
the shears are greatest when
*= K |/, i/,
and their values are, respectively,
o.
o
VI
n I
FIG. 470.
Again, the shear at any point distant z from A in the
loaded portion BE when the live load covers AB
= ^ + (/ _ ,)(/- ^) - W '(^ - ^) .... (16)
- w -
w\l x)
= minus the shear at the same point when AB is
unloaded and the live load covers BO.
Hence, by eqs. (4), (5), (16),
\w'
the shear = q= - -^(l - ^z](l - x)\ . (17)
increasing for a given value of z with / x, and, therefore, a
maximum when x = z. Thus,
the maximum shear = =f ~r(^ "~ 4 Jtr )(^~- ^t ( J 8)
and occurs immediately / front of the load when it covers
AB, and immediately behind the load when it covers BO. It
7 28 THEORY OF STRUCTURES.
may be represented by the ordinate (positive or negative] of
the curve orsq.
For example, at the points defined by
the maximum shears given by eq. (18) are, respectively,
o, yfrw7, T Ve>'/, AV*//, \w'l.
Diagram of Maximum Bending Moment. The bending
moment at any point in BO distant z from A when the live
load covers AB
(19)
*}
= minus the bending moment at the same point when
the live load covers BO.
Hence, by eqs. (4), (5), (19), the bending moment
I w' I w'
= - 2 -j(f - At* + 3**X/ -*) =F - -f(l* - Al* + z**)(l -*) 1 .
For a given value of z this is a maximum and equal to
W ! zl zl 2.Z 2lz
T (1-2,) when * =
Thus, the maximum bending moment may be represented
by the ordinate (positive or negative] of a curve.
For example, at the points defined by
* = /, v, */, t/, '-,
AUXILIARY OR STIFFENING TRUSS.
the bending moments are greatest when ^
v, T y, v, v, V-,
their values being, respectively,
The absolute maximum bending moment may be found as
follows :
For a given value of x the bending moment (see eq. (19)) is
a maximum when
or
/ ze;
Hence, the maximum bending moment
h " -
2 * w " - 8 /' 4/*r -f 2* 2 '
It will be an absolute maximum for a value of x found by put-
ting its differential with respect to x equal to nil.
This differential easily reduces to
x = f/ is an approximate solution of this equation, and the cor-
responding maximum bending moment = -^-^w'T.
The preceding calculations show that at every point in its
length the truss may be subjected to equal maximum shears and
equal maximum bending moments of opposite signs.
Again, it may be easily shown, in a similar manner, that
73 THEORY OF STRUCTURES.
when a single weight W travels over the truss,
the maximum positive shear at a distance z from A
W
= 7 r(2/' -5/2
the maximum negative shear
W
either = -(/ 2 - 5 /*+ 42*)
I W
or = - -j
and the maximum bending moment
W
= T*(/ -*)(/- 2,).
12. Suspension-bridge Loads. The heaviest distributed
load to which a highway bridge may be subjected is that due
to a dense crowd of people, and is fixed by modern French
practice at 82 Ibs. per square foot. Probably, however, it is
unsafe to estimate. the load at less than from 100 to 140 Ibs. per
square foot, while allowance has also to be made for the con-
centration upon a single wheel of as much as 36,000 Ibs., and
perhaps more.
A moderate force repeatedly applied will, if the interval
between the blows corresponds to the vibration interval of the
chain, rapidly produce an excessive oscillation (Chap. Ill,
Cor. 2, Art. 24). Thus, a procession marching in step across
a suspension-bridge may strain it far more intensely than a
dead load, and will set up a synchronous vibration which may
prove absolutely dangerous. For a like reason the wind
usually sets up a wave-motion from end to end of a bridge.
The factor of safety for the dead load of a suspension-bridge
should not be less than 2^ or 3, and for the live load it is
advisable to make it 6. With respect to this point it may be
remarked that the efficiency of a cable does not depend so
much upon its ultimate strength as upon its limit of elasticity,
MODIFICATIONS OF THE SIMPLE SUSPENSION-BRIDGE. 731
and so long as the latter is not exceeded the cable remains un-
injured. For example, the breaking weight of one of the 1 5-inch
cables of the East River Bridge is estimated to be 12,000 tons,
its limit of elasticity being 81 18 tons ; so that with i only as a
factor of safety, the stress would still fall below the elastic
limit and have no injurious effect. The continual application
of such a load would doubtless ultimately lead to the destruc-
tion of the bridge.
The dip of the cable of a suspension-bridge usually varies
from ^ to Y 1 ^ of the span, and is rarely as much as y 1 ^, except
for small spans. Although a greater ratio of dip to span would
give increased economy and an increased limiting span, the
passage of a live load would be accompanied by a greater dis-
tortion of the chains and a larger oscillatory movement.
Steadiness is therefore secured at the cost of economy by
adopting a comparatively flat curve for the chains.
13. Modifications of the Simple Suspension-bridge.
The disadvantages connected with suspension-bridges are very
great. The position of the platform is restricted, massive
anchorages and piers are generally required, and any change in
the distribution of the load produces a sensible deformation in
the structure. Owing to the want of rigidity, a considerable
vertical and horizontal oscillatory motion may be caused, and
many efforts have been made to modify the bridge in such a
manner as to neutralize the tendency to oscillation.
(a) The simplest improvement is that shown in Fig. 472,
where the point of the cable most liable to deformation is
attached to the piers by short straight chains AB.
FIG. 472.
(b) A series of inclined stays, or iron ropes, radiating from
the pier-saddles, may be made to support the platform at a
number of equidistant points (Fig. 473). Such ropes were used
in the Niagara Bridge, and still more recently in the East River
73 2 THEORY OF Sl^RUCTURES.
Bridge. The lower ends of the ropes are generally made fast
to the top or bottom chord of the bridge-truss, so that the cor-
responding chord stress is increased and the neutral axis pro-
portionately displaced. To remedy this, it has been proposed
to connect the ropes with a horizontal tie coincident in position
with the neutral axis. Again, the cables of the Niagara and
FIG. 473 .
East River bridges do not hang in vertical planes, but are in-
clined inwards, the distance between them being greatest at
the piers and least at the centre of the span. This drawing in
adds greatly to the lateral stability, which may be still further
increased by a series of horizontal ties.
(c) In Fig. 474 two cables in the same vertical plane are
diagonally braced together. In principle this method is similar
FIG. 474.
to that adopted in the stiff ening truss (discussed in Art. 1 1), but
is probably less efficient on account of the flexible character of
the cables, although a slight economy of material might doubt-
less be realized. The braces act both as struts and ties, and
the stresses to which they are subjected may be easily calcu-
lated.
(d) In Fig. 475 a single chain is diagonally braced to the
platform. The weight of the bridge must be sufficient to insure
FIG. 475.
that no suspender will be subjected to a thrust, or the efficiency
of the arrangement is destroyed. An objection to this as well
MODIFICATIONS OF THE SIMPLE SUSPENSION-BRIDGE. 733
as to the preceding method is that the variation in the curva-
ture of the chain under changes of temperature tends to loosen
and strain the joints.
The principle has been adopted (Fig. 476) with greater per-
fection in the construction of a foot-bridge at Frankfort. The
FIG. 476.
girder is cut at the centre, the chain is hinged, and the rigidity
is obtained by means of vertical and inclined braces which act
both as struts and ties.
(e) In Fig. 477 the girder is supported at several points by
FIG. 477.
straight chains running directly to the pier-saddles, and the
chains are kept in place by being hung from a curved chain by
vertical rods.
(/) It has been proposed to employ a stiff inverted arched
rib of wrought-iron instead of the flexible cable. All straining
action may be eliminated by hinging the rib at the centre and
piers, and the theory of the stresses developed in this tension
rib is precisely similar to that of the arched rib, except that
the stresses are reversed in kind.
(g) The platform of every suspension-bridge should be
braced horizontally. The floor-beams are sometimes laid on
the skew in order that the two ends of a beam may be sus-
pended from points which do not oscillate concordantly, and
also to distribute the load over a greater length of cable.
734 THEORY OF STRUCTURES.
EXAMPLES.
1. The span of a suspension-bridge is 200 ft., the dip of the chains is
80 ft., and the weight of the roadway is i ton per foot run. Find the ten-
sions at the middle and ends of each chain. Ans. 31 \ tons ; 58.94 tons.
2. Assuming that a steel rope (or a single wire) will bear a tension of
15 tons per square inch, show that it will safely bear its own weight over
a span of about one mile, the dip being one-fourteenth of the span.
Ans. Max. tension = 33,074 Ibs.
3. Show that a steel rope of the best quality, with a dip of one-seventh
of the span, will not break until the span exceeds 7 miles, the ultimate
strength of the rope being 60 tons per square inch.
Ans. Max. tension = 59.545 tons per square inch.
4. The river span of a suspension-bridge is 930 ft. and weighs 5976
tons, of which 1439 tons are borne by stays radiating from the summit
of each pier, while the remaining weight is distributed between four
I5~in. steel-wire cables, producing in each at the piers a tension of 2064
tons. Find the dip of the cables. Ans. 66.44 ft.
The estimated maximum traffic upon the river span is 1311 tons
uniformly distributed. Determine the increased stress in the cables.
Ans. 596.4 tons.
To what extent might the traffic be safely increased, the limit of
elasticity of a cable being 8116 tons, and its breaking stress 12,300 tons ?
Ans. To 13,303 tons uniformly distributed.
5. If the span = /, the total uniform load = W, and the dip = ,
show that the maximum tension = 1.58 W, the minimum tension
= 1.5 IV, the length of the chain = i.oiS/, and find the increase of dir>
corresponding to an elongation of i in. in the chain.
6. A cable weighing p Ibs. per lineal foot of lengthjs stretched be-
tween supports in the same horizontal line and 20 ft. apart. If the max-
imum deflection is ft., determine the greatest and least tensions.
Ans. Parameter m = 100 ft.; max. tension = ioo/; min. ten-
sion = ioo^.
7. A light suspension-bridge carries a foot-path 8 ft. wide over a
river 90 ft. wide by means of eight equidistant suspending rods, the dip
being 10 ft. Each cable consists of nine straight links. Find their several
lengths. If the load upon the platform is 120 Ibs. per square foot, and
EXAMPLES. 735
if one-fourth of the load is borne by the piers, find the sectional areas
of the several links, allowing 10,000 Ibs. per square inch.
Ans. Lengths in ft., 10 ; 10.049; 10.198; 10.44; 10.77.
Tensions in Ibs., 45000; 45004/101 ; 45004/104; 4500^109;
4500 4/1 1 6.
Areas in sq. in., 4.5 ; 4.522 ; 4.59 ; 4.698 ; 4.847.
8. A suspension-bridge of 200 ft. span and 20 ft. dip has 4^ sus-
penders on each side ; the dead weight = 3000 Ibs. per lineal foot ; the
live load = 2000 Ibs. per lineal foot. Find the maximum pull on a sus-
pender, the maximum bending moment and the maximum shear on the
stiffening truss. Also, findthe elongation in the chain due to the live load.
Ans. Max. pull = 12,500 Ibs. ; max. shear = 30,000 Ibs.; max.
B.M. = 1,066,666$ ft. -Ibs. ; elongation = 89,600,000 -f- EA, A
being sectional area of a cable, and E the coefficient of elas-
ticity.
9. A foot-path 8 ft. wide is to be carried over a river 100 ft. wide by
two cables of uniform sectional area and having a dip of 10 ft. Assum-
ing the load on the platform to be 112 Ibs. per square foot, find the
greatest pull on the cables, their sectional area, length, and weight.
(Safe stress = 8960 Ibs. per square inch ; specific weight of cable = 480
Ibs. per cubic foot.)
Ans. H = -=T= 56,000 Ibs.; area =6.73 sq. in.;
4/29
length = io2f ft.; weight = 2302.65 Ibs.
10. Find the depression in the cables in the last question due to an
increment of length under a change of 60 F. from the mean temperature.
(Coefficient of expansion =i -f- 144000.) Ans. .0802 ft.
n. Each side of the platform of a suspension-bridge for a span of 100
ft. is carried by nine equidistant suspenders. Design a stiffening truss for
a live load of 1000 Ibs. per lineal foot, and determine the pull upon the
suspenders due to the live load when the load produces (i) an absolute
maximum shear ; (2) an absolute maximum bending moment.
Ans. Max. shear = 6250 Ibs.; max. B.M. = 92,592^ ft.-lbs.; pull
on suspender = (i) 2777! Ibs., (2) = 185 iff Ibs. or 3703-1 rf Ibs.
12. In a suspension-bridge (recently blown down) each cable was de-
signed to carry a total load of 84 tons (including its own weight). The
distance between the piers = 1270 ft.; the deflection of the cable = 91 ft.
Find (a) the length of the cable ; (b) the pull on the cable at the piers
and at the lowest point ; (c) the amounts by which these pulls are changed
by a variation of 40 F. from the mean temperature ; (</) the tension in
the back-stays, assuming them to be approximately straight and inclined
to the vertical at the angle whose tangent is f.
THEORY OF STRUCTURES.
T
Ans. (a) 1287.4 ft.; () H = 146^ tons ; (<r) depression due
to change of temp. = .936 ft. and amount of change in H = ' H
= ii tons, in T =1.45 tons; (d) 394.55 tons, neglecting pier
friction.
13. The platform of the bridge in the preceding question was hung
from the cables by means of 480 suspenders (240 on each side). Find
the pull on each suspender and the total length of the suspenders, the
lowest point of a cable being 14 ft. above the platform.
Ans. .35 ton ; 10,565^!-! ft.
14. A suspension-bridge has a dip of 10 ft. and a span of 300 ft. Find
the increase of dip due to a change of 100 F. from the mean tempera-
ture, the coefficient of expansion being .00125 per 180 F.
Ans. 1.17 ft.
Also, find the corresponding flange stress in the stiffening truss,
which is 12^ ft. deep, the coefficient of elasticity being 8000 tons.
Ans. 6.24 tons.
15. The ends of a cable are attached to saddles free to move horizon-
tally. If Aa is the horizontal movement of each saddle due to the ex-
pansion of the cables in the side spans, and if AS is the extension of the
chain between the two saddles, show that the increment of the dip (fi) is
approximately
$h~a\
16. The platform of a suspension-bridge of 150 ft. span is suspended
from the two cables by 88 vertical rods (44 on each side) ; the dip of the
cables is 15 ft.; there are two stiffening trusses ; the dead weight is 2240
Ibs. per lineal foot, of which one-half is divided equally between the two
piers. Find the stresses at the middle and ends of the cables when a
uniformly distributed load of 78,750 Ibs. covers one half of the bridge.
Also, find the maximum shears and bending moments to which the stiff-
ening trusses are subjected when a live load of 1050 Ibs. per lineal foot
crosses the bridge.
Ans. Pull on suspender = 2741$ Ibs.; H= - - T= 203,437^^5.
^29
Max. shear on each truss at centre and due to 78,750 Ibs.
= 9843^ Ibs. = that due to 10*50 Ibs. per lineal foot.
Max. B. M. due to 78,750 Ibs. is at centre of loaded and un-
loaded halves and = 184,570^ ft. -Ibs.
Abs. max. B.M. due to 1050 Ibs. per lineal foot is at points
of trisection and == 218,750 ft. -Ibs.
EXAMPLES. 737
17. Solve the preceding question when the trusses are hinged at the
centre.
Ans. Pull on suspender = 2741! Ibs. ; H --= T = 1 54,2 1 Sf Ibs.
Max. shear due to 78,750 Ibs. = 9843! Ibs. at centre of
span and at end of loaded half of bridge ; max. shears
due to 1050 Ibs. per lineal 001 = 13,125, 5906^, 4921^,
&35TZ8> anc * 9843! Ibs. at ends of the half truss and at
the points dividing the half span into four equal seg-
ments.
Max. B. M. due to 78,750 Ibs. is at centre of half truss and
= 184,570^- ft. -Ibs. Max. B. M. due to 1050 Ibs. per lineal
foot =176, 1 80 ff|, 221,484!, and 153,808-^1 ft.-lbs. at points
dividing the half truss into four equal segments.
1 8. Show that the total extension of a cable of uniform sectional area
A under a uniformly distributed load of intensity iu is
wl
3 / 16 </ 3 \
- I + -TTl,
8.M 3
/ being the span and d the dip.
19. The dead weight of a suspension-bridge of 1600 ft. span is \ ton
per lineal foot; the dip = . Find the greatest and least pulls upon
one of the chains. The ends of the chains are attached to saddles on
rollers on the top of piers 50 ft. high, and the back-stays are anchored
50 ft. from the foot of each pier. Find the load upon the piers and the
pull upon the anchorage.
Ans. 255 tons ; 243! tons ; 637^ tons ; 344.6 tons.
20. A bridge 444 ft. long consists of a central span of 180 ft. and two
side spans each of 132 ft. ; each side of the platform is suspended by
vertical rods from two iron-wire cables ; each pair of cables passes over
two masonry abutments and two piers, the former being 24 ft. and the
atter 39 ft. above the surface of the ground ; the lowest point of the
cables in each span is 19 ft. above the ground surface ; at the abutments
the cables are connected with straight wrought-iron chains, by means of
which they are attached to anchorages at a horizontal distance of 66 ft.
from the foot of each abutment ; the dead weight of the bridge is 3500
Ibs. per lineal foot, and the bridge is covered with a proof load of 4500
Ibs. per lineal foot. Determine
(a) The stresses in the cables at the points of support and at the
lowest points.
(d} The dimensions and weights of the cables (i) if of uniform sec-
73 8 THEORY OF STRUCTURES.
tion throughout; (2) if each section is proportioned to the pull across it.
(Unit stress = 14.958 Ibs. per square inch.)
(c) The alteration in the length of the cables and the corresponding
depression of the platform at the centre of each span, due (i) to a change
of 60 F. from the mean temperature ; (2) to the total load, E being
30,000,000 Ibs.
(d) The pressure and bending moment at the foot of pier.
(e) The mass of masonry in the anchorage necessary to resist the
tendency to overturning and to horizontal displacement.
Data. Weight of masonry per cubic foot = 128 Ibs. ; safe compres-
sive stress per square foot = 12,000 Ibs. ; coefficient of friction =76;
deviation of centre of pressure in base of pier from centre of figure = f
x thickness of base.
Am. (a) Side span, Tiz= H l 387,200 Ibs. = T 3 -^= , T,
1/509 4/146
and 7a being the tensions in a cable at summits of
Q
low and high piers, respectively; centre span, T-=-. -
V97
= 405,000 Ibs. = H, T being tension at summit of high
pier.
() Side span : Length = 135/3 ft- ; sect, area at summit
of high pier = 28.43 sc l- m - 5 weight if of uniform section
= 12,834 Ibs., if proportioned to pull 11,710 Ibs.
Centre span : Length = i85ff ft. ; sect, area at summit
of pier = 29.6 sq. in.; weight if of uniform section
= 18,361 Ibs., if proportioned to pull = 17,267 Ibs.
(c) (i) .0594 ft. for side span and .0775 ft- f r centre span ;
(2) .0675 " " " " " -0927 " "
(d) High pier: Overturning moment = 694,200 ft. -Ibs. ;
bearing area at summit = uSf sq. ft.; thickness
= 8 ft. ; uniform width = 14! ft.; thickness of base
= 10.7 ft. ; weight of pier = 692,348.8 Ibs. ; total pres-
sure on base = 2,116,348.8 Ibs.
(<?) Weight to resist upward pull = 29,333^ Ibs. ; weight
to resist horizontal displacement = 509,474 Ibs.
21. In the preceding question, if the piers are wrought-iron oscillat-
ing columns, and if equilibrium, under an unequally distributed load, is
maintained by connecting the heads of the columns with each other and
with the abutments by iron-wire stays, determine the proper dimensions
of the stays, assuming them approximately straight. Assume that the
proof load covers (a) a side span ; (b) two side spans ; (c) the centre span.
Ans. (a) Pull on stays in centre span = 840,050 Ibs.
(b) " J' " " " " = double that in (a).
00 " " " " side span = 948,432 Ibs.
EXAMPLES. 739
22. A floating landing-stage is held in position by a number of 4^-
iti. steel-wire cables anchored to the shore, a shoreward movement being
prevented by rigid iron booms, pivoted at the ends and stretching from
shore to stage. The difference of level between the shore and stage at-
tachments of the cables is 50 ft., and the horizontal distance between
these points is 150 ft. The horizontal pull upon each cable is 1360 Ibs.
Find the length of the cable, and the tensions at the points of attach-
ment. (Weight of cable = 490 Ibs. per cubic foot ; form of cable a com-
mon catenary.) Am. 342.82ft.; 12,267.2 Ibs. and 10,132 Ibs.
CHAPTER XIII.
ARCHES AND ARCHED RIBS.
I. AN arch may be constructed of masonry, brickwork,
timber, or metal.
FIG. 478.
In the figure ABCD represents the profile of an arch. ' The
under surface AD is called the soffit or intrados. The upper
surface BC is sometimes improperly called the extrados. The
highest point A" of the soffit is the crown or key of the arch.
The springings or skewbacks are the surfaces y^.5, DC from
which the arch springs, and the haunches are the portions of
the arch half-way between the springings and the crown.
Upon each of the arch faces stands a spandril wall, and the
space between these two external spandrils may be occupied
by a series of internal spandrils spaced at definite distances
apart, or may be filled up to a certain level with masonry (i.e.,
backing] and above that with ordinary ballast or other rough
material (i.e., filling).
A masonry arch consists of courses of wedge-shaped blocks
with the bed-joints perpendicular, or nearly so, to the soffit.
740
EQUILIBRATED POLYGON AND LINE OF RESISTANCE. 74!
The blocks are called voussoirs, and the voussoirs at the crown
are the keystones of the arch.
A brick arch is usually built in a number of rings.
Consider the portion of the arch bounded by the vertical
plane KE at the key and by the plane AB.
It is kept in equilibrium by the reaction R at KE, the reac-
tion R l at AB, and the weight F, of the portion under con-
sideration and its superincumbent load.
Let S and T be the points of application of R^ and R,
respectively.
Let the directions of R, and R intersect in a point. The
direction of F, must also pass through the same point.
Taking moments about 5,
p l and jj/j being the perpendicular distances of the directions of
R and F, from S, respectively.
Similarly, the portion KECD of the arch gives the equation
F 9 being the weight to which it is subjected, and / a , y^ the
perpendicular distances of the directions of R and F 2 from the
point of application Fof the reaction at the plane DC.
If the arch and the loading are symmetrical with respect
to the plane KE,
Y l = Y 9t yi=y^y and therefore /, / 2 .
Hence the direction of R will be horizontal, which might have
been inferred by reason of the symmetry.
The magnitudes of the reactions are indeterminate, as the
positions of the points of application (5, T, F) are arbitrary,
and can only be fixed by a knowledge of the law of the varia-
tion of the stress in the material at the bounding planes AB,
KE.
2. Equilibrated Polygon and Line of Resistance.
Suppose an arch divided into a number of elementary portions
THEORY OF STRUCTURES.
ke' , k' e" . . . (e.g., the voussoirs of a masonry arch) by a series
of joints ke, k'e' . . .
FIG. 479.
FIG. 480.
Let W iy W^j . . be the loads directly supported by the
several portions. These loads generally consist of the weight
of a portion (e.g., ke') + the weight of the superincumbent
mass -\- the load upon the overlying roadway ; the lines of
action of the loads are, therefore, nearly always vertical.
Each elementary portion may be considered as acted upon
and kept in equilibrium by three forces, viz., the external load
and the pressures at the joints. If the pressure and its point
of application at any given joint have been determined, the
pressures and the corresponding points of application at the
other joints may also be found.
For, let i 2 34 ... be the line of loads, so that 12= W^
23= W,
Assume that the pressure P and its point of application r
at any given joint ke are known.
Draw o I to represent P in direction and magnitude.
Then 02 evidently represents the resultant of P and W l in
direction and magnitude, and this resultant must be equal and
opposite to the pressure P l at the joint k'e' .
Hence, a line n'n drawn through n, the intersection of P
and W lt parallel to 20, is the direction of the pressure P lt and
intersects k'e' in the point of application r' of P,
Again, o 3 represents the resultant of /\ and W^ in direc-
tion and magnitude, and this resultant must be equal and
opposite to the pressure P 9 at the joint k"e" .
The line n"ri drawn through ', the intersection of P^ and
EQUILIBRATED POLYGON AND LINE OF RESISTANCE. 743
W 2 , parallel to 30, is the direction of the pressure P 9 and inter-
sects k"e" in. the point of application r" of P t .
Proceeding in this manner, a series of points of application
or centres of resistance r', r", r'" , . . . may be found, the
corresponding pressures being represented by 02, 03^04, . . .
The polygon of pressures formed by the lines of action of
P, P lt P^, . . . is termed an equilibrated polygon, and is a funic-
ular polygon of the loads upon the several portions.
The polygon formed by joining the points r, r f , r" , . . .
successively, is called the line of resistance.
In the limit, when the joints are supposed indefinitely near,
these polygons become curves, the curve in the case of the
equilibrated polygon being known as a linear arch.
The two curves may, without sensible error, be supposed
identical, and they will exactly coincide if the joints (of course
imaginary in such a case) are made parallel to the lines of
action of the external loads. This may be easily proved as
follows :
Let the figure represent a portion of an arch bounded by
the joints (imaginary) KE, MN parallel to the lines of action
of the external loads, which will be assumed vertical.
Reduce the superincumbent loads to an equivalent mass of
arch material.
Let h, e.g., be the depth of material of specific weight w l ,
overlying the arch at any given
point, and let Q be the load per
unit of area of roadway.
Also, let iv be the specific
weight of the arch material.
Then x, the equivalent depth,
is given by
wx = wjt -\- Q.
FIG. 481
If the value of x is deter-
mined at different points along the arch, a profiU, en may be
obtained defining a mass ENne of arch material which may be
substituted for the superincumbent load. Denote the weight
of the mass MKen by W.
744 THEORY OF STRUCTURES.
Let the pressure P and its point of application O at the
joint KE be given.
Take O as the origin, the line OA in the direction of P as
the axis of x, and the vertical through O as the axis of jy, and
let be the angle between the two axes.
Let the lines of action of P and W intersect in G. The
line of action of their resultant will intersect MN m the centre
of resistance O r
Let X, Y be the co-ordinates of O,.
Let z be the depth of an elementary slice of thickness dx,
parallel to OK at any abscissa x. Its weight wzdx sin 0.
Then
W. OG = wzdx smB.x= W(X - AG\
But -vp= ~TQ ~~p~> since tne triangle AGO is evidently a
triangle of forces for the forces acting upon the mass under
consideration.
Also,
W = / wzdx . sin 0.
.'. Cwzxdx . sin = WX - W~Y= X ffvz sin Odx - PY.
t/o W t/o
This is the equation to the line of resistance.
Taking the differential of this equation,
wz'X sin BdX = Xwz' sin BdX + WdX - PdY 9
z' being the depth corresponding to the abscissa X.
dY W AO
Thus the tangents to the curve of pressures and to the
curve of centres of pressure at any given point coincide, and
the curves must therefore also coincide.
CONDITIONS OF EQUILIBRIUM. 745
3. Conditions of Equilibrium. Let the figure represent
a portion of an arch of thickness unity, between any two bed-
joints (real or imaginary] MN, PQ.
Let W be its weight together with that of the superincum-
bent load. Let the direction of
the reaction R' at the joint MN
intersect MN in m and the direc-
tion of J^in n. For equilibrium,
the reaction R" at the joint PQQ.
must also pass through n. Let its
direction intersect PQ in O. In O +W
order that the equilibrium may be
stable, three conditions must be HIG ' 48z<
fulfilled, viz. :
First. The point O must lie between P and Q, so that there
may be no tendency to turn about the edges Pand Q.
Second. There must be no sliding along PQ, and therefore
the angle between the direction of R" and the normal to PQ
must not exceed the angle of friction of the material of which
the arch is composed.
N.B. The angle of friction for stone upon stone is about
30.
Third. The maximum intensity of stress at any point in PQ
must not exceed the safe resistance of the material.
Further, the stress should not change in character, in the
case of masonry and brick arches, but should be a compression
at every point, as these materials are not suited to withstand
tensile forces.
The best position for O would be the middle point of PQ,
as the pressure would then be uniformly distributed over the
area PQ. It is, however, impracticable to insure such a dis-
tribution, and it has been sometimes assumed that the stress
varies uniformly,
With this assumption, let ^Vbe the normal component of R".
Let /be the maximum compressive stress, i.e., the stress at
the most compressed edge, e.g., P.
Let OS q . PQ, S being the middle point of PQ, and q a
coefficient whose value is to be determined.
746 THEORY OF STRUCTURES.
PO
Then if PO < ,
N- f ' PQ
and in the limit when PO = , i.e., when the intensity of
stress varies uniformly from /at Pto nil at Q,
?=* and &**
(See Art. 16, Chap. IV.)
Similarly, if Q is the most compressed edge, the limiting
position of O, the centre of resistance or pressure, is at a point
PQ
O' denned by QO' - ~.
Hence, as there should be no tendency on the part of the
joints to open at either edge, it is inferred that PO or QO'
PQ
should be > , i.e., that the point O should lie within the
middle third of the joint.
Experience, however, shows that the " middle-third "
theory cannot be accepted as a solution of the problem of
arch stability, and that its chief use is to indicate the proper
dimensions of the abutments. Joint cracks are to be found in
more than 90$ of the arches actually constructed, and cases
may be instanced in which the joints have opened so widely
that the whole of the thrust is transmitted through the edges.
In Telford's masonry arch over the Severn, of 150 ft. span,
Baker discovered that there had been a settlement (15 in.)
sufficient to induce a slight reverse curvature at the crown of
the soffit. Again, the position of the centre of pressure at a
joint is indeterminate, and it is therefore impossible as well as
useless to make any calculations as to the maximum intensity
of stress due to the pressure at the joint. What seems to
JOINT OF RUPTURE. 747
happen in practice is, that the straining at the joints generally
exceeds the limit of elasticity, and that the pressure is uni-
formly distributed for a certain distance on each side of the
curve of pressures. Thus, the proper dimensions of a stable
arch are usually determined by empirical rules which have
been deduced as the results of experience. For example,
Baker makes the following statement :
Let T be the thrust in tons or pounds per lineal foot of
width of arch.
Let f be the safe working stress in tons or pounds per
square foot.
An arch will be stable if an ideal arch, with its bounding
i T
surfaces at a minimum distance of from the curve of pres-
sures, can be traced so as to lie within the actual arch. An
advance would be made towards a more correct theory if it
were possible to introduce into the question, the elasticity and
compressibility of the materials of construction. These ele-
ments, however, vary between such wide limits that no
reliance can be placed upon the stresses derivable from their
values.
4. Joint of Rupture. Let I 2, 3 4 be the bounding surfaces
between which the curve of pressures must lie, and let 4 be
2
FIG. 483.
the centre of pressure at the crown. A series of curves of
pressure may be drawn for the same given load, but with
different values of the horizontal thrust h.
Let AfXy be that particular curve which for a value //of the
horizontal thrust is tangent to the surface I 2 at x ; the joint at
x is called the joint of rupture.
The angle which the joint of rupture makes with the
748
THEORY OF STRUCTURES.
horizontal is about 30 in semicircular and 45 in elliptic
arches.
The position of the joint in any given arch may be tenta-
tively found as follows :
Lety be any joint in the surface I 2.
Let Wbe the weight upon the arch between /"and I.
Let X be the horizontal distance between J and the centre
of gravity of W.
Let Y be the vertical distance between J and 4.
It will also be assumed that the thrust at 4 is horizontal.
If the curve of pressure be now supposed to pass through
J, the corresponding value of the horizontal thrust h is given
by
kY= WX.
By means of this equation, values of h may be calculated
for a number of joints in the neighborhood of the haunch, and
the greatest of these values will be the horizontal thrust H for
the joint x. This is evident, as the curve of pressure for a
smaller value of h must necessarily fall below ^xy.
When this happens, the joints will tend to open at the
lower edge of the joint I 4 and at the upper edges of the joints
at x and at 2 3, so that the arch may sink at the crown and
spread, unless the abutments and the lower portions of the
arch are massive enough to counteract this tendency.
If the curve of pressure fall above A^xy, an amount of back-
ing sufficient to transmit the thrust to the abutments must be
provided. The same result may be attained by a uniform in-
crease in the thickness of the arch ring, or by a gradual increase
from the crown to the abutments.
For example, the upper sur-
face (extrados) of the ring for an
arch with a semicircular soffit
A KB, having its centre at O, may
be delineated in the following
manner:
^ Let x define the joint of rup-
ture in the soffit ; then AOx 30.
\
MINIMUM THICKNESS OF ABUTMENT.
749
be
Y
the
In Ox produced take xx r = 2 X KD, KD being the thick-
ness at the crown.
The arc Dx' of a circle struck from a centre in DO pro-
duced may be taken as a part of the upper boundary of the
ring, and the remainder may be completed by the tangent at
x l to the arc Dx' .
5. Minimum Thickness of Abutment. Let
resultant thrust at the horizontal joint BC of a
rectangular abutment ABCD.
Let y be the distance of its point of applica- C
tion from B.
Let // and V be the horizontal and vertical
components of T.
Let w be the specific weight of the material
in the abutment.
Let h be the height AB of the abutment.
Let t be the width AD of the abutment. FIG. 485.
In order that there may be no tendency to turn about the
toe D, the moment of the weight of the abutment with respect
to D plus the moment of V with respect to D must be greater
than the moment of H with respect to D. Or,
or
- + V(t-y)> Hh,
H \ I 2 4- 2V 4- V *
h > V W~* Wk- ufk*
This relation must hold good whatever the height of the
abutment may be ; and if h is made equal to oo ,
which defines a minimum limit for the thickness of the abut-
ment.
750 THEORY OF STRUCTURES.
6. Empirical Formulae. In practice the thickness /at
the crown is often found in terms of s, the span, or in terms of
p, the radius of curvature at the crown, from the formulae
t = c Vs, or t = Vcp,
t, s, and p being all in feet, and c being a constant.
According to Dupuit, / = .36 Vs for a full arch ;
/ = .27 Vs for a segmental arch.
According to Rankine, = l/.i 2p for a single arch ;
/ = V.ijp for an arch of a series.
7. Examples of Linear Arches, or Curves of Pressure.
\(a) Linear Arch in the Form of a Parabola. Suppose that
the cable in Art. 4, Chap. XII, Case B, is exactly inverted,
and that it is stiffened in such a manner as to resist distortion.
Suppose also that the load still remains a uniformly distributed
weight of intensity w per horizontal unit of length. A thrust
will now be developed at every point of the inverted cable
equal to the tension at the corresponding point of the original
cable. Thus the inverted parabola is a linear arch suitable for
a real arch which has to support a load of intensity w per
horizontal unit of length.
The horizontal thrust at the crown = H = wp,
p being the radius of curvature at the crown.
(fr) Linear Arch in the Form of a Catenary. Transformed
Catenary. If the cable in Art. 4, Chap. XIT, Case A, is in-
N T o N verted and stiffened as before, a linear
arch is obtained suitable for" a real
arch which has to support a load dis-
tributed in such a manner that the
weight upon any portion AP is pro-
p , portional to the length of AP, and is
in fact =ps. The area OAPN ' ms.
Thus, a lamina of thickness unity
and specific weight w, bounded by the curve AP, the directrix
ON, and the verticals AO, PN, weighs wms, and may be taken
EXAMPLES OF LINEAR ARCHES. 75 r
to represent the load upon the arch if wms ps, i.e., if wm = J>,
i.e., if the weight of m units of the lamina is w.
The horizontal thrust at the crown H wm = wp,
the radius of curvature (p) at the crown being equal to m.
A disadvantage attached to a linear arch in the form of a
catenary lies in the fact that only one catenary can pass
through two given points, while, in practice, it is often neces-
sary that an arch shall pass through three ^points in order to
meet the requirements of a given rise and span. This difficulty
may be obviated by the use of the transformed catenary.
Upon the lamina PAPNN as base, erect a solid, with its
horizontal sections all the same, and, for simplicity, with its
generating line perpendicular to the base.
Cut this solid by a plane through NN inclined at any re-
quired angle to the base. The intersection of the plane and
solid will define a transformed catenary P'A'P', or a new linear
arch, and the shape of a new lamina P'A'P'NN, under which
the arch will be balanced. This is evident, as the new arch and
lamina are merely parallel projections of the original.
The projections of horizontal lines will remain the same in
length.
The projections of vertical lines will be c times the lengths
of the lines from which they are projected, c being the secant
of the angle made by the cutting plane with the base.
Let .*, Y be the co-ordinates of any point P' of the trans-
formed catenary.
Let x, y be the co-ordinates of the corresponding point P
in the catenary proper.
Then
YPW A'Q M
_ _
= ~
y ~ PN ~ AO
The equation to the catenary proper is
(2)
75 ^ THEORY OF STRUCTURES.
Substituting in the last equation the value of y given by eq. (i),
which is the equation to the transformed catenary.
With this form of Linear arch the depths M over the crown
and Y over the springings, for a span 2x, may be assumed, and
the corresponding value of m determined from eq. (3).
It is convenient, in calculating m, to write eq. (3) in the form
-'- ... (4)
The slope i' at P' is given by
dY Ml*- -c\ Ms
tarn = y = I* e 1 = ,
dx 2m\ I m
s being the length AP of the catenary proper, corresponding
to the length A'P' of the transformed catenary.
'P'N= C
<J o
The area OA'P'N= Ydx = -e - e m = Ms.
The triangle P' TN is a triangle of forces for the portion
A'P'.
The triangle PTN is a triangle of forces for the portion AP.
(The tangents at P and P' must evidently intersect ON in
the same point T.)
Let H' be the horizontal thrust at A', //being that at A.
Let P' be the weight upon A'P' t P being that upon AP.
Let R' be the thrust at P'.
Then
P' __ area OA'P'N _ Ms _M
~P' - area OAPN ~~ ~ms~~^n y
EXAMPLES OF LINEAR ARCHES.
and hence
pt _ p wms wMs ;
m m
H = P' cot i' = ivMs-jrf- = wm* = H\
/ J/V
R'= H' sec i' =wm\/ i -\ = w
V m
The radius of curvature p' at the crown = -=-= .
. '. H' = wMp' H wp,
and the radius of the " catenary proper " is M times the radius
of the transformed catenary.
The term " equilibrated arch " has generally been applied
to a linear arch with a horizontal extrados.
(c) Circular and Elliptic Linear Arches. A linear arch
which has to support an external
normal pressure of uniform inten-
sity should be circular.
Consider an indefinitely small
element CD, which may be as-
sumed to be approximately j-~
" KlG. 487.
Let the direction of the result-
ant pressure upon CD, viz.,/ . CD, make an angle B with OB.
Let CE, DE be the vertical and horizontal projections of
CD.
The angle DCE = 8.
The horizontal component of p . CD = p . CD cos 6 = p . CE.
This is distributed over the vertical projection CE.
i /. the .horizontal intensity of pressure = / . CE -j- CE = p.
Similarly, it may be shown that the vertical intensity of
pressure />.
754
THEORY OF STRUCTURES.
Thus, at any point of the arch,
the horizontal intensity of pressure
= vertical intensity = normal intensity =/.
Again, the total horizontal pressure on one-half of the arch
= 2(p. CE) = p2(CE) = pr = H,
and the total vertical pressure on one-half of the arch
= 2(p . DE) =
Hence, at any point of the arch the tangential thrust = pr.
Next, upon the semicircle as base, erect a semi-cylinder.
Cut the latter by an inclined plane drawn through a line in the
plane of the base parallel to OA. The intersection of the cut-
ting plane and the semi-cylinder is the semi-ellipse B'AB', in
which the vertical lines are unchanged in length, while the
lengths of the horizontal lines are c times the lengths of the
corresponding lines in the semicircle, c being the secant of the
angle made by the cutting plane with the base. A semi-
elliptic arch is thus obtained, and the forces to which it is sub-
jected are parallel projections of the forces acting upon the
semicircular arch.
These new forces are in equilibrium (see Corollary).
Let P' = the total vertical pressure upon one-half of the
arch ;
H' the total horizontal pressure upon one-half of the
arch ;
EXAMPLES OF LINEAR ARCHES, 755
P
py = vertical intensity of pressure = ^-57 ;
tft
p x ' = horizontal intensity of pressure = ~7=rr
Then
P f = P=H = pr; ........ (i)
P P
y '- OB'~ c.OB~~ cr ""
H f = cH = cP = cP'-, ....... (3)
H'
Hence, by eq. (3),
/_ ___
px ~- OA' ~ OA ~~ : r ~
*L OB '
~P~~ ~OA'
or, the total horizontal and vertical thrusts are in the ratio of
the axes to which they are respectively parallel, and, by eqs.
(2) and (4),
A.' -I OA "
p x '~S~ OB'* ;
or, the vertical and horizontal intensities of pressure are in the
ratio of the squares of the axes to which they are respectively
parallel.
Any two rectangular axes OG, OK in the circle will project
into a pair of conjugate radii OG' , OK' in the ellipse.
Let OG' = r lt OK' = r 2 ;
Q = total thrust along elliptic arch at K\
E> a /"
Then
H r H r
THEORY OF STRUCTURES.
or, the total thrusts along an elliptic arch at the extremities of
a pair of conjugate radii are in the ratio of the radii to which
they are respectively parallel.
The preceding results show that an elliptic linear arch is
suitable for a load distributed in such a manner that the vertical
and horizontal intensities (eqs. (2) and (4) ) at any point of the
arch are unequal, but are uniform in direction and magnitude.
Corollary. It can be easily shown that the projected forces
acting upon the elliptic arch are in equilibrium.
The equations of equilibrium for the forces acting upon the
circular arch may be written
T being the thrust along the arch at the point xy, and X, Y
the forces acting upon the arch parallel to the axes of x and
y, respectively.
If T', X', V be the corresponding projected forces,
~ = ~, Xds = cX'ds', Yds = Y'ds*.
Hence, the above equations may be written
d j^p cdx^ + cX'ds' = o,
and
or
and
d\
( Tid ] + *'***
Hence, the forces T 1 ', X', and Y' are also in equilibrium.
EXAMPLES OF LINEAR ARCHES.
(d) Hydrostatic Arch. Let the figure represent a portion
of a linear arch suited to support a load
which will induce in it a normal pressure at
every point. The pressure being normal
has no tangential component, and the
thrust (7") along the arch must therefore be
everywhere the same.
Consider any indefinitely small element
CD.
It is kept in equilibrium by the equal -FIG ' 489 '
thrusts (7") at the extremities C and D, and by the pressure
/ . CD. The intensity of pressure / being assumed uniform
for the element CD, the line of action of the pressure/. CD
bisects CD at right angles.
Let the normals at C and D meet in O l , the centre of
curvature.
Take Of O,D = p, and the angle CO,D = 2 AS.
Resolving along the bisector of the angle
6 =p. CD pp
or
2TA& = pp. 2 AS \
and hence,
T = pp = a constant. . . . (i)
Thus, a series of curves may be obtained in which p varies
inversely as/, and the hydrostatic arch is that curve for which
\\\e pressure p at any point is directly proportional to the depth
of the point below a given horizontal plane.
Denote the depth by y, and let w be the specific weight of
the substance to which the pressure/ is due. Then
P = wy, .......... . . (2)
and
T pp wyp = a constant. . . . (3)
The curve may be delineated by means of the equation
yp = const ........... (4)
75 8 THEORY OF STRUCTURES.
It may be shown, precisely as in Case (c), that the horizontal
intensity of pressure (p^)
=: the vertical intensity (p y ) =fl (5)
Take as the origin of co-ordinates the point O vertically
above the crown of the arch, in the given horizontal plane.
Let the horizontal line through O be the axis of x.
" " vertical " " " " " " " y.
Any portion AM of the arch is kept in equilibrium by the
O equal thrusts (T) at A and M,
j and by the resultant load P upon
AM, which must necessarily act
in a direction bisecting the angle
ANM.
FIG. 490. Complete the parallelogram
AM, and take SN NM to represent T.
The diagonal NL will therefore represent P.
Let be the inclination of the tangent at M to the hori-
zontal.
The vertical load upon AM vertical component of P
= LK T sin 6 = pp sin = wyp sin = wy p sin 0, . (6)
y ot p being the values of y, p, respectively, at A.
The horizontal load upon AM= horizontal component of P
= NK=SN-KS = T-
= 2pp (sin -) = 2wyp sin = 2wy p (sin -) . . (7)
\ 2i' '
Again, the vertical load upon AM
/ "* pdx w / ydx = wy p sin ; (8)
e/o vo
the horizontal load upon AM
f*y f*y iv / \
= J pdy = wj^ ydy = -(/ - y*) = 2wy p (s'm -j . (9)
EXAMPLES OF LINEAR ARCHES. 759
Equation (8) also shows that the area bounded by the curve
AM, the verticals through M and A, and the horizontal
through is equal to y p a sin #, and is therefore proportional
to sin 0. At the points defined by d = 90 the tangents to
the arch are vertical, and the portion of the arch between these
tangents is alone available for supporting a load. The vertical
and horizontal loads upon one-half the arch are each equal to
WJW
Corollary. The relation given in eq. (i) holds true in any
arch for elements upon which the pressure is wholly normal.
This has been already proved for the parabola and catenary,
in cases (a) and (b).
At the point A' of the elliptic arch,
_ OB'* _ c*r* _
= '-~ r ~~ ~
Hence, the horizontal thrust at A'
= PyP = ~P = P Cr = H -
(e) Geostatic Arch. The geostatic is a parallel projection of
the hydrostatic arch.
The vertical forces and the lengths of vertical lines are
unchanged.
The horizontal forces and lengths of hori-
zontal lines are changed in a given ratio
c to I.
Let B' A be the half-geostatic curve de- FIG. 491.
rived from the half-hydrostatic curve BA.
The vertical load on AB'
P f = P= thrust along arch at B'. ... (i)
The horizontal load on AB'
= H' = cH thrust along arch at A. . . . (2)
The new vertical intensity
-^'- -*-A2 ,
f > OB'
THEORY OF STRUCTURES.
The new horizontal intensity
H 1 cH
~ =C ^ = ^ .... (4)
Thus, the geostatic arch is suited to support a load so dis-
tributed as to produce at any point a pair of conjugate press-
ures ; pressures, in fact, similar to those developed according
to the theory of earthwork.
Let R l , R^ be the radii of curvature of the geostatic arch
at the points A, B' , respectively, and let r lt r y be the radii of
curvature at the corresponding points A, B of the hydrostatic
arch.
The load is wholly normal at A and B ' . Thus,
H' =p y f R l =^R l =:cH=cpr l . ... (5)
' R* = ^ ....... ..... (6)
Also,
cR, = r,. . ..... '. . . . . . (8)
(/) General Case. Let the figure represent any linear
p arch suited to support a load which is sym-
metrically distributed with respect to the
crown A, and which produces at every point
of the arch a pair of conjugate pressures,
the one horizontal and the other vertical.
Take as the axis of y the vertical through
the crown, and as the axis of x the hori-
FIG. 492. zontal through an origin O at a given dis-
tance from A.
Any portion A M of the arch is kept in equilibrium by the
horizontal thrust H at A, the tangential thrust T at M, and
the resultant load upon AM, which must necessarily act through
the point of intersection N of the lines of action of //and T.
Since the load at A is wholly vertical, H is given by
X.=P,P., - ...... (i)
EXAMPLES OF LINEAR ARCHES. jl
p and p being, respectively, the vertical intensity of pressure
and the radius of curvature at A.
Let MN = T, and take NS = H .
Complete the parallelogram SM\ the diagonal NL is the
resultant load upon AM "in direction and magnitude.
The vertical (KL) and the horizontal (KN) projections of
NL are, therefore, respectively, the vertical and horizontal
loads upon AM.
Denote the vertical load by V, the horizontal by//. Then
(2),
and
H=KN=SN-SK=H ~ Fcot 0, . . (3)
being the angle between MN and the horizon.
dV
p yy the vertical intensity of pressure, -j . ..... (4)
p x , the horizontal intensity of pressure
"> ..... (?)
EXAMPLE. A semicircular arch of radius r, with a hori-
zontal extrados at a vertical distance R from the centre.
The angle between the radius to J/and the vertical = 6.
.'. x r sin #, y = R r cos 0. . . . (i)
dx=r cos Ode, dy = r sin OdO ..... (2)
p y wy = w(R r cos 0), ...... (3)
w being the specific weight of the load. Hence,
V = wf\R - r cos 0)r cos BdB
I rtt r sin 2#\
= wr(R sine-- - ]. ... (4)
7^2 THEORY OF STRUCTURES.
Equations (3) and (4) give H '; for
p. = w(R-r), (5)
ind hence
HQ == wr(R -r) (6)
p x , the horizontal intensity of pressure,
d . ( n r B - sin cos 6 _\ , ,
= -7- (Fcotff) = w\R . -a rcosfl). (7)
dy^ \ 2 sin 9 /
Rankine gives the following method of determining whether
a linear arch may be adopted as the intrados of a real arch.
At the crown a of a linear arch ab measure on the normal a
length aCj so that c may fall within the limits required for
stability (e.g., within the middle third).
At c two equal and opposite forces, of the same magnitude
as the horizontal thrust H at a, and acting at right angles to
ac, may be introduced without altering the equilibrium.
Thus the thrust at a is replaced by an equal thrust at c, and
a right-handed couple of moment H . ac.
Similarly, the tangential thrust T at any point d of ab
may be replaced by an equal and parallel thrust at e, and a
couple of moment T . de.
The arch will be stable if the length of de, which is normal
to ab at dj is fixed by the condition T . de = H . ac, and if the
line which is the locus of e falls within a certain area (e.g.,
within the middle third of the arch ring.
8. Arched Ribs in Iron, Steel, or Timber. In the fol-
lowing articles, the term arched rib is applied to arches con-
structed of iron, steel, or timber. The coefficients of elasticity
are known quantities which are severally found to lie between
certain not very wide limits, and their values maybe introduced
into the calculations with the result of giving to them greater
accuracy. There are other considerations, however, involved
in the problem of the stability of arched ribs which still render
its solution more or less indeterminate.
It has been shown that the curve of pressure, or linear arch,
ARCHED RIB UNDER A VERTICAL LOAD. 763
is a funicular polygon of the extraneous forces which act upon
the real arch. It is, therefore, also the b ending-moment curve,
drawn to a definite scale, for a similarly loaded horizontal
girder of the same span, whose axis is the springing line.
When the arched rib carries a given symmetrically dis-
tributed load, it will be assumed that the linear arch coincides
with the axis of the rib, and that the thrust at any normal
cross-section is axial and uniformly distributed.
The total stress at any point is made up of a number of
subsidiary stresses, of which the most important are : (i) a
direct thrust ; (2) a stress due to flexure ; (3) a stress due to a
change of temperature. Each of these may be investigated
separately, and the results superposed.
9. Bending Moment (M) and Thrust (T) at any Point
of an Arched Rib under a Vertical Load. Let ABC be the
axis of the rib.
Let D and E be points on the same vertical line, E being
D-D
FIG. 493.
on the axis of the rib and D on the linear arch for any given
distribution of load.
Resolve the reaction at A into its vertical and horizontal
components, and denote the latter by H.
Since all the forces, excepting H, are vertical, the difference
between the moments at D and E = H . DE.
But moment at D o. Hence,
moment at E = M= H . DE.
Let the normal at E meet the linear arch in D'. Then, if
T is the thrust along the axis at E,
n r E
Tcos DED' = ff= 7 > approximately,
or
H .DE = T.D'E = M.
764 THEORY OF STRUCTURES.
10. Rib with Hinged Ends ; Invariability of Span.
Let ABC be the axis of a rib supported at the ends on pins or
FIG. 494.
on cylindrical bearings. The resultant thrusts at A and C
must necessarily pass through the centres of rotation. The
vertical components of the thrusts are equal to the corre-
sponding reactions at the ends of a girder of the same span
and similarly loaded, and H is given by the last equation in
the preceding article when DE has been found.
Let ADC be the linear arch for any arbitrary distribution
of the load, and let it intersect the axis of the rib at S. The
curvature of the more heavily loaded portion AES will be
flattened, while that of the remainder will be sharpened.
The bending moment at any point E of the axis tends to
change the inclination of the rib at that point.
Let the vertical through E intersect the linear arch in D
and the horizontal through A in F.
Let 8 be the inclination of the tangent at E to the hori-
zontal.
Let /be the moment of inertia of the section of the rib
at .
Let ds be an element of the axis at E.
_, Mds H.DE.ds
Change of inclination at E dv = ^ = - -=j -- .
If this change of curvature were effected by causing the
whole curve on the left of E to turn about E through an angle
dO, the horizontal displacement of A would be
ARCHED RIB WITH HINGED ENDS.
This is evidently equal to the horizontal displacement of
, and the algebraic sum of the horizontal displacements of all
points along the axis is
H.DE. EF. ds rH. DE . EF. ds
2- m - -=J- - - = Q, . . (i)
since the length AC is assumed to be invariable.
Thus, the actual linear arch must fulfil the condition ex-
pressed by eq. (i), which may be written
rDE.EF.ds
J- - -=0, (2)
since H and E are constant.
If the rib is of uniform section, /is also constant, and eq. (2)
becomes
CDE.EF.ds=o (3)
Also, since DE is the difference between DF and EF,
f(DF ~ EF)EF. ds=o =f^F. EF. ds-J*EF*ds (4)
Remark. Eq. i expresses the fact that the span remains
invariable when a series of bending moments, H . DE, act at
points along the rib. These, however, are accompanied by a
thrust along the arch, and the axis of the rib varies in length
with the variation of thrust.
Let H be the horizontal thrust for that symmetrical loading
which makes the linear arch coincide with the axis of the rib.
Let T be the corresponding thrust along the rib at E.
The shortening of the element ds at E of unit section
T ~ T
E
EXAMPLE I. Let the axis of a rib of uniform section and
hinged at both ends be a semicircle of radius r.
Let a single weight W be placed at a point upon the rib
whose horizontal distance from (9, the centre of the span, is a.
7 66
THEORY OF STRUCTURES.
The " linear arch " (or bending-moment curve) consists of
two straight lines DA, DC.
FIG. 495.
Draw any vertical line intersecting the axis, the linear
arch, and the springing line AC in E' , D', F', respectively.
Let OF' = x, and let dx be the horizontal projection upon
AC of the element ds at E'.
Then
-^ = cosec E'OF' = -== ,
dx E'F '
or
(i)
Applying condition (4),
f D'F'rdx + f D'F 'rdx = f E'F'rdx,
or
f D'F'dx + f D'F'dx = f E'F'dx,
or area of triangle ADC area of semicircle.
And if z be the vertical distance of D from AC,
zr =
ARCHED RIB WITH HINGED ENDS.
767
or
nr
z = = one-half of length of rib.
(2)
nr
(3)
Hence, if h be the horizontal thrust on the arch due to W,
= M = W
r* a'
2r
(4)
Similarly, if there are a number of weights W^ W^ PF 3 , . . ,
upon the rib, and if h^ h^, h^, . . . are the corresponding hori-
zontal thrusts, the total horizontal thrust //will be the sum of
these separate thrusts, i.e.,
(5)
It will be observed that the apices (D lt D^, D s , . . .) of the
several linear arches (triangles) lie in a horizontal line at the
nr
vertical distance from the springing line.
Ex. 2. An arched rib hinged at the ends and loaded with
weights W,, W^ W^ . . .
-. L__ _J i._.J__
\
FIG. 497.
Let i 2 3 4 ... be the line of loads, W^ being represented
by i 2, JF 2 by 2 3, W 3 by 3 4, etc., and let the segments \x,
THEORY. OF STRUCTURES.
iix, respectively, represent the vertical reactions at A and C.
Take the horizontal length xP to represent H, and draw the
radial lines Pi, P2, P$ t . . .
The equilibrium polygon Ag^g^ . . . must be the funicu-
lar polygon of the forces with respect to the pole P, and there-
fore the directions of the resultant thrusts from A to lt E l to
v , 9 to E 3 , . . . are respectively parallel to Pi, P2, ^3, ...
The tangential (axial) thrust and shear at any point p of
the rib, e.g., between E t and E s , may be easily found by draw-
ing Pt parallel to the tangent at/, and 3^ perpendicular to PL
The direct tangential thrust is evidently represented by Pt,
and the normal shear at the same point by 3/. The latter is
home by the web.
If/ is a point at which a weight is concentrated, e.g., t ,
draw Pt't" parallel to the tangent at , and 5/', 6t" perpen-
dicular to Pt't".
Pt' represents the axial thrust immediately on the left of
E+ , and 5/' the corresponding normal shear, while Pt" repre-
sents the axial thrust immediately on the right of E t , and 6t"
the corresponding normal shear.
A vertical line through P can only meet the line of loads
at infinity.
Thus, it would require the loads at A and C to be infinitely
great in order that the thrusts at these points might be vertical.
Practically, no linear arch will even approximately coincide
with the axis of a rib rising vertically at the springings, and
lience neither a semicircular nor a semi-elliptical axis is to be
recommended.
Ex. 3. Let the axis of the rib be a circular arc of span 21
and radius r, subtending an angle 2ot at the centre N.
Let the angles between the radii NE, NE' and the vertical
be ft and 0, respectively.
The element ds at E' = rd6.
Also, E'F' = r(cos 6 cos a) ; AF = / r sin 6 ;
D'F' = (l ~ r sin
ARCHED RIB WITH HINGED ENDS. 769
Applying condition (5),
/ r (cos 8 - cos 'a)*rdO
I rr~(t r sin 6>)r(cos 6 cos a)rdB
+ f j^ (lr sin 6>)r(cos 6 cos a)rdO,
Jt l ~- a
which easily reduces to
r\a(cos 20. -f- 2) f sin 2#J
T -j / a ( s i n ^ ^ cos ) -| (cos 2<af cos 2ft)
- a ( 4
rl cos <*(cos <* cos ft) la(sin ft ft cos a) v ,
an equation giving z or Z?/''. Also,
iD^r^/r-^?r
and the corresponding horizontal thrust may be found, as
before, by the equation
/ a <2 2
77 THEORY OF STRUCTURES.
Note\i a c/T,
7t 2z IF a\ nr
r -- -p_ ~\ /> or z as in Ex. I.
Ex. 4. Let the axis be a parabola of span 2/ and rise
(Fig. 498, Ex. 3). From the properties of the parabola,
la '
and
ds* = d
or, approximately,
ds dx\\ 2X '
Applying condition (5),
which easily reduces to
an equation giving z or DF.
Note. If the arch is very flat, so that ds may be considered
ARCHED RIB WITH ENDS ABSOLUTELY FIXED. 771
as approximately equal to dx, the term 2j-,^ 2 in the above
equation may be disregarded, and it may be easily shown that
16
or
2 =
32 k
II. Rib with Ends absolutely Fixed. Let ABC be the
axis of the rib. The fixture of the ends introduces two un-
'K L'
FIG. 499.
known moments at these points, and since H is also unknown,
three conditions must be satisfied before the strength of the
rib can be calculated.
Represent the linear arch by the dotted lines KL ; the
points K, L may fall above or below the points A, C.
Let a vertical line DEF intersect the linear arch in D, the
axis pf the rib in E, and the horizontal through A in F.
As in Art. 10, change of inclination at E, or dO, = ^- r .
\ El
But the total change of inclination of the rib between A and
C must be nil, as the ends are fixed.
*Mds
Mas
-7 = =
H.DE.ds
which may be written
(I)
(2)
since H and E are constant.
77 2 THE OR Y OF STRUCTURES.
If the section of the rib is uniform, / is constant and eq.
(2) becomes
Again, the total horizontal displacement between A and C
will be nil if the abutments are immovable. If they yield, the
amount of the yielding must be determined in each case, and
may be denoted by an expression of the form yw//, yu being
some coefficient.
As in Art. 10, the total horizontal displacement
p
~ J
H.DE.EF.ds
H.DE.EF.ds
~ET
But H and E are constant.
*DE . EF. ds
(5)
If the section of the rib is uniform, / is also constant, and
hence
fDE.EF.ds = o or =;... (6)
and since DE is the difference between DF and EF, this last
may be written
ds = o or = . ... (7)
Again, the total vertical displacement between A and C
must be nil.
The vertical displacement of E (see Art. 10)
ARCHED RIB WITH ENDS ABSOLUTELY FIXED.
Hence, the total vertical displacement
*H.DE.AF ,
= r
which may be written
.AF
/JJtL . Sif , , .
-f" *= fe>
since H and are constant. If the section of the rib is also
constant,
.AF.ds. (10)
Eqs. (2), (5), and (9) are the three equations of condition.
In eq. (9) AF must be measured from same abutment
throughout the summation.
The integration extends from A to C.
EXAMPLE I. Let the axis of the rib be a circular arc of
span 2/, subtending an angle 2a at the centre N.
Let a weight W be concentrated on the rib at a point E
whose horizontal distance from the middle point of the span
is a.
Let the radius NE make an angle /3 with the vertical.
The " linear arch " consists of two straight lines DA', DC' .
Let A A = r,, DF = z,CC' = y, .
774 THEORY OF STRUCTURES.
Draw any ordinate E'F' intersecting the linear arch in D f .
Let the radius NE' make an angle with the vertical.
Then
E'F' r(cos 6 cos ).
AF' = l rsintf, and D'F'=(lrs\n
if F f is on the left of F\
and D'F' = (l-rsm
if /*"' is on the right of F.
Also, ^y = rdO.
Applying condition (i),
os 6> cos *)<# ...... (i)
Applying condition (3), and assuming ^ = o,
/ B (cos ^ - cos ) | (/ - r sin ^q^' +^ 1 ^
3
+ Acos - cos a) I (/ r sin ^)y^ +7, I
os 6> - cos r)V0 ..... (2)
ARCHED RIB WITH ENDS ABSOLUTELY FIXED.
Applying condition (5),
+ A/+r sin 0) j (/- rsin tff^+f
= rC a (cos # cos )(/ r sin 0)</0
s cos )(/+ r sin 0)</0. (3)
Equations (i), (2), (3) may be easily integrated, and the re-
sulting equations will give the values of 7, , z, and y t .
The corresponding horizontal thrust, //, may now be ob-
tained from the equation h . DE = M = h(z EF).
Note. If the axis is a semicircle, and if Wis at the crown,
a = o, a = 90, ft = O,
and eqs. (i), (2), (3) reduce to
it 2 4+ 2n TT*
/. z = r -- , and y. = y t = r --- .
4 TT' 4 TT
Ex. 2. Let the axis be a parabola of span 2/ and rise k
(Fig. 500 in Ex. i).
As in Ex. 3, Art. 10,
THEORY OF STRUCTURES.
Also,
D'F' = y l + (/ *)TTT^- on the right of DF,
and
& _.---. <\t
= y t + (/ *}-7 r? on the left of DF.
J * ' '/ ^
The equations of condition become
{ ^ + c - -fi~ } - ?)* +
/
r j 7, + (/ - *)^5 }(/-*> (i + 2
+ r j^ + (/ - ^-yf~ }(/+*)(>+ 2
y
r *(i - ^) (/ - *) (i + 2^)^
/-a
// / r s
4* - r
.
EFFECT OF A CHANGE OF' TEMPERA TURE. 777
These equations may be at once integrated, and the result-
ing equations will give the values of y lt jj> 2 , z.
If the arch is very flat, so that ds may be taken to be ap-
proximately the same as dx, it may be easily shown that
2 /+ 50 2 I 5# 6
12. Effect of a Change of Temperature. The variation
in the span 2/ of an arch for a change of t from the mean
temperature is approximately = 2etl, e being the coefficient of
expansion.
Hence, if H t is the horizontal force induced by a change of
temperature, the condition that the length AC is invariable is
expressed by the equation
DE.EF.ds
__ 2et i _ a
If the rib is of uniform section, 7 is constant; and since
E is also constant, the equation may be written
. EF. ds 2etl = o.
EXAMPLE I. Let the axis AEC of a rib of uniform section
FIG.
be the arc of a circle of radius r subtending an angle 2a at the
centre.
First, let the rib be hinged at both ends.
7/8 THEORY OF STRUCTURES.
It is evident that the straight line AC is the "linear arch.'*
Then,
J*DE . EF. ds ^j^EF^ds = r* J*\cos B - cos afdB
r*{a(2 -f- cos 201) | sin 2a\.
Also, / r sin a.
H / 3
Note. If the axis is a semicircle, a = 90, and
!L?L
EI 2
2etl = o.
Second, let the rib \>z fixed zk both ends.
The " linear arch" is now a straight line A'C' at a distance
= Z?/*') from ^4 C given by the equation
CDE.ds^O.
.-. CDF. ds =fF. ds y
or
z Cds = r* /"(cos ^ cos a)dO,
or
a.3- = r(sin a a cos a).
Also,
. EF. ds =f(DF . EF^EF^ds = zj*EFds ~ C EF *ds
= 2.3T 2 (sin a a cos <*)~ r*{a(2-\- cos 2) f sin 2a\.
TT f -^
.'. -JT\ 2Jsr\sin a a cos a) r*{a(2 + cos 2a) f sin 2a\ (
2etl = o,
and / = r sin a.
Ex. 2. Let the axis ^7 of a rib of uniform section be a
parabola of span 2/ and rise k. (See Fig. 501 in Ex. I.)
EFFEC7" OF A CHANGE OF l^EMPERATURE. 779
First, let the rib be hinged at both ends.
The straight line AC is the linear arch. Then
CDE .EF.ds = f l F 2
and hence,
Second, let the rib be fixed at both ends.
The linear arch is the line A'C' at a distance z (= DF)
from A C given by the equation
or
E .ds = o = C(DF ~ EF)ds,
DFfds = fEF. ds.
*T(i + 7^)^ = A(i - f!)(i + 2^>)^.
or
2 2
Also,
.EF.ds =&F .EF.ds-^ (*
7^0 THEORY OF STRUCTURES.
Hence,
= o.
Remark. The coefficient of expansion per degree of Fah-
renheit is .0000062 and .0000067 for cast- and wrought-iron
beams, respectively. Hence, the corresponding total expansion
or contraction in a length of 100 ft., for a range of 60 F. from
the mean temperature, is .0372 ft. (= ^/') and .0402 ft. (= ").
In practice the actual variation of length rarely exceeds one-
half Q( these amounts, which is chiefly owing to structural con-
straint.
13. Deflection of an Arched Rib.
FIG. 502.
Let the abutments be immovable.
Let ABC be the axis of the rib in its normal position.
Let ADC represent the position of the axis when the rib is
loaded.
Let BDF be the ordinate at the centre of the span ; join
AB, AD.
Then
* = AD* - AF* = AB- AF\
arc AB
But
arc AB arc AD /
~~ ~ ~E '
/being the intensity of stress due to the change in the length
of the axis.
/. DF* = AB*i - - AF* =
ELEMENTARY DEFORMATION OF AN ARCHED RIB. /8l
AB* 2 -
f\*
= BF*- DF* = (BF- DF)(BF + DF)
= 2BF(BD\ approximately.
^rj is also sufficiently small to be disregarded. Hence,
h I
AB* f V + I* f
BD, the deflection, = -jr= -^ 7 -g , approximately.
14. Elementary Deformation of an Arched Rib.
FIG. 503.
The arched rib represented by Fig. 503 springs from two
abutments and is under a vertical load. The neutral axis PQ
is the locus of the centres of gravity of all the cross-sections, of
the rib, and may be regarded as a linear arch, to which the
conditions governing the equilibrium of the rib are equally ap-
plicable.
Let A A ' be any cross-section of the rib. The segment
AA'P is kept in equilibrium by the external forces which act
upon it, and by the molecular action at A A'.
The external forces are reducible to a single force at C and
to a couple of which the moment M is the algebraic sum of
the moments with respect to C of all the. forces on the right
of C.
The single force at C may be resolved into a component T
along the neutral axis, and a component Sin the plane A A'.
THEORY OF STRUCTURES.
The latter has very little effect upon the curvature of the neu-
tral axis, and may be disregarded as compared with M.
Before deformation let the consecutive cross-sections BE 1
and AA meet in R ; R is the centre of curvature of the arc
CC' of the neutral axis.
After deformation it may be assumed that the plane A A'
remains unchanged, but that the plane BB' takes the position
B"B'". Let AA' and B"B"' meet in R' ; R is the centre of
curvature of the arc CC' after deformation.
Let abc be any layer at a distance z from C.
Let CC = As, CR = R, CR' = R' y and let Aa be the sec-
tional area of the layer abc.
By similar figures,
ac __ R' + z ab _ R+z
~~A r>/ and ~ ~
As R As R
i i i\
The tensile stress in abc
- A be As.z
= E . Aa; = E . Aa -r
ab ab
z? A I I l \ i
= jfi . Aa . x\~pT jrjji very nearly.
The moment of this stress with respect to C
Hence, the moment of resistance at A A '
the integral extending over the whole of the section.
ELEMENTARY DEFORMATION OF AN ARCHED RIB. 783
Again, the effect of the force T is to lengthen or shorten
the element CC', so that the plane BB' will receive a motion
of translation, but the position of -R' is practically unaltered.
Corollary i. Let A be the area of the section AA r .
The total unit stress in the layer abc
T Ms
the sign being plus or minus according as M acts towards or
from the edge of the rib under consideration.
From this expression may be deduced (i) the position of
the point at which the intensity of the stress is a maximum for
any given distribution of the load; (2) the distribution of the
load that makes the intensity an absolute maximum ; (3) the
value of the intensity.
Cor. 2. Let w be the total intensity of the vertical load per
horizontal unit of length.
Let w, be the portion of w which produces only a. direct
compression.
Let //be the horizontal thrust of the arch.
Let P be the total load between the crown and AA' which
produces compression.
Refer the rib to the horizontal OX and the vertical OPY
as the axes of x and y, respectively.
Let x, y be the co-ordinates of C.
Then "
P=ff~-' t but dP = w,dx.
dx
....... (3)
MW
also, :
784 THEORY OF STRUCTURES.
15. General Equations.
Let / be the span of the arch.
Let x, y be the co-ordinates of the point C before deforma-
tion.
Let x\ y' be the co-ordinates of the point C after deforma-
tion.
Let be the angle between tangent at C and OX before
deformation.
Let 0' be the angle between tangent at C and OX after
deformation.
Let ds be the length of the element CC' before deforma-
tion.
Let ds be the length of the element CC' after deformation.
d(T i dO i
Effect of flexure. ~jp ~g> and ~fa ~ R'
Mi i dO' de dO' - dO
Let i be the change of slope at C. Then
Mds Mds
dt = dO dB' =
C
-0' = f.+ /
El ~ Eldx
**M ds
t being the change of slope at P, and a quantity whose value
has yet to be determined.
Again, the general equations of equilibrium at the plane
A A' are
d
for the portion w l , Cor. 2, Art. 14, produces compression only
and no shear.
GENERAL EQUATIONS. 785
S 9 being the still undetermined vertical component of the shear
dv
at P, and ~ the slope at P. Also,
y, ; - *, (8)
J/ being the still undetermined bending moment at P.
Equations (5), (6), (7), and (8) contain the four undeter-
mined constants //, 5 , M , / .
Let M l , 5 t , and z, be the values of M, 5, and z, respectively,
at Q.
Equations of Condition. In practice the ends of the rib are
either j&ritf? or free.
If they are fixed, z' = o ; if they are free, M = o. In either
case the number of undetermined constants reduces to three.
If the abutments are immovable, x l / = o. If the abut-
ments yield, x^ / must be found by experiment. Let ;r, /
= /<//, >w being some coefficient. T\\t first equation of condi-
tion is
x l l=o 1 or x,~l=^H. .... (9)
Again, Q is immovable in a vertical direction, and the
second equation of condition is
(10)
Again, if the end Q is fixed, i l = o ; and if free, M l = o ; and
the third equation of condition is
* t = o, or M l =o ...... (il)
Substituting in equations (7) and (8) the values of the three
constants as determined by these conditions, the shearing force
and bending moment may be found at any section of the rib.
Again,
cos Q' = cos (0 i) cos + i s ^ n ^ J
sin 0' = sin (0 i) sin 6 i cos 6.
786 THEORY OF STRUCTURES.
dx' dx ,dy dy' dy dx
" ~j~r ~r + * j and jr = j l ~r- ( I2
ds' ds ds ds ds ds v
Hence, approximately,
d . , .dy d .dx
jrt* x) t -j- and -j( y' y\ i .
ds^ ds ds^ ds
Thus, if JTand Fare respectively the horizontal and verti-
cal displacements,
dX .dy dY .dx
-j- = tr- and r =. tr,
as as ds ds
or
dX . dY
16. Effect of T and of a Change of t in the Temperature.
Also, if there is a change from the mean of t in the tem-
perature, the length ds\i T^J must be multiplied by
(i e/), e being the coefficient of linear expansion.
-p-j et), approximately. (14)
By equations (i 2),
<& = (dx + i . dyy^ = (dx + i . dy)(i -jj &)
and
dy' = (dy - i.dx)^ = (dy - i.dx](i - ~ et).
GENERAL EQUATIONS. 787
.-. dX = d(x' - x )^idy--
and
= d(y f y) = idx - ^ etdy, approximately,
Hence,
and
Note. A nearer approximation than is given by the pre-
ceding results may be obtained as follows:
Let x + dx, y -\- dy be the co-ordinates of a point very
near C before deformation.
Let x' + dx' , y' -f- dy' be the co-ordinates of a point very
near C after deformation.
Then
ds* = dx* + dy* and ds'* = dx'* + dy'\
.-. ds'* - ds* = dx'* - dx* + dy'* - dy'*,
or
(ds'-ds)(ds f + ds) = (dx'-dx)(dx f + dx) + (dy'-dy)(dy : + dy).
. (ds' ds)ds = (dx' dx)dx + (dy' dy)dy y approximately.
. , , .ds dy
.'. dx dx = (ds ds)-j- (dy dy)-~
ax 'ax
and
ds dx
dy' -dy = (ds'
Hence, by equations (12) and (14),
.dy J T(ds
dx' dx t-^j-dx -=r-.( I dx eti-- dx
dx EA\dx' ^dx
788 THEORY OF STRUCTURES.
and
T (ds\dx J lds\dx ,
dy > -dy=~ ^^-dz -
/. ay / 1 i ds\ r x tds\
i -j-dx I -rr-Ti j 1 ax et I I -j- 1 dx
dx / hA\dx I J \dxl
'
and
r IdsVdx , r*tds\*dx
These equations are to be used instead of equations (15)
and (16), the remainder of the calculations being computed
precisely as before.
The following problems are, in the main, the same as those
given in Art. 180 of Rankine's Civil Engineering, I3th edition-
17. Rib of Uniform Stiffness. Let \hzdepth and sectional
form of the rib be uniform, and let its breadth at each point
vary as the secant of the inclination of the tangent at the point
to the horizontal.
Let A), /j be the sectional area and moment of inertia at
the crown.
Let A, I be the sectional area and moment of inertia at any
point C, Fig. 503.
Then
(17)
Also, since the moments of inertia of similar figures vary
as the breadth and as the cube of the depth, and since the
depth in the present case is constant,
(18)
T HsecO H
Again, -j- = - - = -j-, and the intensity of the thrust
is constant throughout.
ARCHED RIB OF UNIFORM DEPTH. 789
Hence, equations (5), (15), and (16), respectively, become
/*
-*=J*
dy
H
(21)
Equation (19) shows that the deflection at each point of the
rib is the same as that at corresponding points of a straight
horizontal beam of a uniform section equal to that of the rib
at the crown, and acted upon by the same bending moments.
Ribs of uniform stiffness are not usual in practice, but the
formulae deduced in the present article may be applied without
sensible error to flat segmental ribs of uniform section.
18. Parabolic Rib of Uniform Depth and Stiffness, with
Rolling Load; the Ends fixed in Direction; the Abut-
ments immovable.
D E O
FIG.
504-
Let the axis of x be a tangent to the neutral curve at its
summit.
Let k be the rise of the curve.
Let x, y be the co-ordinates at any point C with respect
to O.
Then
and
d~x-~ -T\~2-*>'
79 THEORY OF STRUCTURES.
Let -w be the .dead load per horizontal unit of length.
2/ " " live " " " " " "
Let the live load cover a, length DE, = r/, of the span.
'Denote by (A) formulae relating to the unloaded division
OEj and by (B) formulae relating to the loaded division DE.
.Equations (7) and (8), respectively, become
(o Z, JLT \
-)*; ......... (24)
(B) S=S,+ ^f - w} X - w'\ x - (i - r)l\. . . (25)
;. ...... (26)
(B) M = M, + S.x+--w-\x-(i-r)l\>. (27)
Since the ends are fixed,
*'. =0 = *; ...... . . (28)
Hence, by equations (19) and (26),
I ' (UH
I ( x }
(A) ,= --^ ^Ms + S. +(-jr -w)^ J; . (29)
and;hy equations (19) and (27),
i . IZkH
. (30)
i= -
When x = /, i = *', = o, and therefore, by the last equation,
\- - /
) 9 -H-\/ 2 W )A & r l *- (30
ARCHED RIB OF UNIFORM DEPTH. 791
dv
Again, let i = -- . Then
/"' .dy J C l dv dy , .dy C l d*y ,
I ?*"= / Txix dx = l ^x- / v ^ dx '
*/ o y o */ o
But i, = o, and - = --.
r i dy , s& r i u r i r* mj
' / V 1 *^ ~ ~T / vdx =-~r \ 7 ^
ty o t/o t/o */o
By the conditions of the problem, x' x and j/ 7 are
each zero at Q. Hence, equations (20) and (21), respectively,
become
(33)
=-J MX. .......... (34)
Substitute in eqs. (33) and (34) the value of i given by eq.
(30), and integrate between the limits o and /. Then
i* r tun \ r r
o -- h
" '
c o o -- --- ---
EI l ( 6 ' "24 ' \ I* J 120 120
and
ss.tuff
i IMS
-17.1
which may be written
79 2 THEORY OF STRUCTURES.
and
Hence, by eqs. (31), (35), (36),
4 ' ' 3
When x = l,M=M lt and 5=5,.
Hence, by eqs. (25) and (27),
and
\/ a wW a
w \ .
/2 2
-AjH; (37)
: ('-Z')+f^ (38)
15 ,
H=- --- _-^_ *J. . . . (39)
Substituting in these equations the values of S , M , given
above, we have
, . . . (40)
A I l>
and
r = - - M//VI- - -r + -) + -kH. . (41)
12 \2 3 4/3
To find the greatest intensity of stress, etc. The intensity of
T f-f
the stress due to direct compression *=-*-
ARCHED RIB OF UNIFORM DEPTH.
793
The intensity of the stress in the outside layers of the rib
due to bending is the same as that in the outside layers of a
horizontal beam of uniform section A 1 acted upon by the same
moments as act on the rib, for the deflections of the beam and
rib are equal at every point (eq. (19) ). Also, since the rib is
fixed at both ends, the bending moment due to that portion of
the load which produces flexure is a maximum at the loaded
end, i.e., at Q. Hence, the maximum intensity of stress (/,)
occurs at Q, and/, = r- -A/, 7, z l being the distance of the
AI /,
layers from the neutral axis.
H and M l are both functions of r, and therefore /j is an ab-
solute maximum when
But
and
dp,
o
i
dH
8.
\- '
dM,
dr
dH
150
~A
// a
dr -
L A
dr
r}*
dr
4
k
i-t-
15.
7, '
dM.
V44)
Hence, /, is an absolute maximum when
The roots of this equation are
r = i
and
2 4
r = 7
53 A
1
I
(45)
794
THEORY OF STRUCTURES.
11 .
r I makes -j-r- zero, so that the maximum value of p.
dr*
corresponds to one of the remaining roots.
Thus,
thewwwr. thrust '
and
(4 6 )
the max. tension = - H+ -jMj = //', (47)
the values of H and M l being found by substituting in eqs.
(39) and (40
i+^^-
2 4 Aft
or
. , 45
_2_ ' 4
(48)
according as the stress is a thrust or a tension.
If eq. (47) gives a negative result, there is no tension at any
point of the rib.
Note. The moment of inertia may be expressed in the form
q being a coefficient depending upon the form of the section.
Hence,
the maximum intensity of stress = ( H '-|- l ) . . ( : tg)
Corollary I. If the depth of the rib is small as compared
with k, the fraction j will be a small quantity, and the maxi-
mum intensity of stress will approximately correspond to r = -J-.
ARCHED RIB OF UNIFORM STIFFNESS. 795
The denominator in eq. (39) may be taken to be k, and it may
be easily shown that the values of //,//' are
: - (50)
,_ i (,// i i S *,\ 5 rf/ 54 v/r\
' - T, I TV + 7^*4^1 + 3125^: 5' (5I)
. 2. If the numerator in eqs. (48) is greater than the
denominator, then r must be unity. Hence, by eq. (39) and
and by eqs. (38) and (41),
Thus, //,/," can be found by substituting these values of
H and M l in eqs. (46) and (47).
19. Parabolic Rib of Uniform Stiffness, hinged at the
Ends.
Let the rib be similar to that of the preceding article.
Since the ends are hinged, M 9 = o = M l , while i is an un-
determined constant.
The following equations apply :
(A) S=S.+f=_- W J:r; (54)
(B) S=S.+ K- w )*- v /\x-(i-rW; (55)
79^ THEORY OF STRUCTURES.
(A) M =Ss + ? -;. ....... (56)
(B) M =S,x + -w-{x-( l -r- } l\\ (57)
%kH
(58)
!-* - + -- - i*-(-^ (59)
Assume that the horizontal and vertical displacements of
the loaded end are nil.
Substitute in eqs. (20) and (21) the value of i given by eq.
(59). Integrate and reduce, neglecting the term involving the
temperature. Then
_//.J.l _^ JL. : (6o )
S6ff V s ,.
-
From (57), since M, = o,
\/ , 7 r 3
-^- -wj--ze;7-. . . . (62)
Equations (60), (61), and (62) are the equations of condi-
tion.
Subtract (61) from (60). Then
UH \r lr* r*\ H
"/ r -
which may be written
ARCHED RIB OF UNIFORM STIFFNESS. 797
Subtract (63) from (62). Then
Hence,
/'U + V-5r ; +2o}
L " (6S)
Eliminating 5 between (61) and (62),
UH \/ 3
Also, by (55),
r - / - ^V/ = - /> suppose. (67)
Eliminating 5 between (62) and (67),
_ / , =5i= (^_^_<4_. . . . (68)
Eqs. (62), (65), (66), and (68) give the values of H y S ,S lt
and i .
Again, the maximum bending moment M' occurs at a
dM
point given by j- = o in (57), i.e.,
-w'\x-(i-r)l\. .(69)
Subtract (69) from (67). Then
_/>, = 5, = (^?- ;)(/- *) - /(/-*).
79$ THEORY OF STRUCTURES.
Hence, the distance from the loaded end of the point at
which the bending moment is greatest is
w --
Substitute this value of x in (57), and, for convenience, put
w --w pr m>
Then
p
M , = / -
m
'-m)l- W 'rl\
P'/w' - m a
m"\ 2~ '
But by (62), o = S.
Hence, M' ', the maximum bending moment,
\
(71)
As before, the greatest stress (a thrust)
= -H+M= P ;, .... ( 72 )
and the value of r which makes // an absolute maximum is
given by ~- = o. But by (71), M' involves r 10 in the numera-
ARCHED RIB OF UNIFORM STIFFNESS. 799
dpi
tor and r 5 in the denominator, so that -~ = o will be an
ar
equation involving r 14 .
One of its roots is r = i, which generally gives a minimum
value of //. Dividing by r i, the equation reduces to one
of the thirteenth order, but is still far too complex for use. It
is found, however, that r =. % gives a close approximation to the
absolute maximum thrust.
With this value of r, and, for convenience, putting
15 I, i
" "'
By (65),
By (62),
By (68),
By (66),
By (70),
n i w'
(74)
- I W'
(75)
(77)
r^TJ-^-+y
By (70.
. If the rib is merely supported at the ends but not
fixed, the. horizontal displacement of the loaded end may be
800 THEORY OF STRUCTURES.
represented by ^H (Art. 1 1). Thus the term ^H must be
added to the right-hand side of eq. (15).
20. Parabolic Rib of Uniform Stiffness, hinged at the
Crown and also at the Ends. In this case M=o at the
crown, which introduces a fourth equation of condition.
By (57).
L ( UH _ ^_^Tf_l V
~~ s 2 ~r" \~r~ ' w ls ~ 2 \ 2 "" r l '
which may be written
>r-r \ J f i\
'-r + -). (79)
Eliminating S between (79) and (62),
UH
w = w ( 2r -\- ^r i).
Hence,
H = -7}^ w'(2r* 4r+ i)}. . . . (80)
8 (
By (79)
S =*^-(3r* -4/-+I) (81)
By (68),
By (66),
By (70) and (82),
l-z=J =L. . (84)
By (71),
PARABOLIC RIB OF UNIFORM STIFFNESS. 8oi
When r = j,
w'r w f r \ ' (86)
t = - -77-7-, and M' = -?.
384 hi, 64 J
These results agree with those of (73) to (78), if n = i.
In general, when n I,
/
w-\- (5r a 5r 4 -f- 2r 6 ) =. w w\2r* ^r -f- i),
by (65) and (80). Hence,
and the roots are r j-, r = i, r = 4/2.
Hence, w = I only renders the expressions in (86) identical
with the corresponding expressions of the preceding article
when n = J or i.
Again, the intensity of thrust is greatest at the outer flange
of the loaded and the inner flange of the unloaded half of the
rib, and is
r uw a . w 1
The intensity of tension is greatest at the inner flange of the
loaded and the outer flange of the unloaded half of the rib,
and is
w' if w'
The greatest total horizontal thrust occurs when r = I, and
its value is
8O2 THEORY OF STRUCTURES.
21. Maximum Deflection of an Arched Rib. The deflec-
tion must necessarily be a maximum at a point given by i = o.
Solve for x and substitute in (16) to find the deflection y' y;
the deflection is an absolute maximum when -j\y' y) = o.
The resulting equation involves r to a high power, and is too
intricate to be of use. It has been found by trial, however,
that in all ordinary cases the absolute maximum deflection
occurs at the middle of the rib, when the live load covers its
v/hole length, i.e., when x -, and r = I.
CASE I. Rib of Art. 18. For convenience, put I + ~^ = s.
Then, by (39),
By (38) and (41),
-M.= ^ + < V 'f-^^^^ = -M,.. (88)
By (36) and (38),
S.= -6 ............. (89)
By (30), (38), (89),
i=-jTJ \ M X ~ 3 M J + 2M *J\ (9>
Hence, the maximum deflection
C" ., M, f*l x* , x \ MJ'
~ J ( ldx = ~ Tit (* ~ i-T + 2 r) dx = ~ F/32
r w + w' * i 5 e t r
=: - n: 7* T d. , suppose. . . (01 )
384 hi, s ~ 128 s k
MAXIMUM DEFLECTION OF AN ARCHED RIB. 803
The central deflection d^ of a uniform straight horizontal
beam of the same span, of the same section as the rib at the
crown, and with its ends fixed, is
Hence, neglecting the term involving the temperature,
4 = ^4 ......... (93)
CASE II. Rib of Art. 19.
By (65),
By (66) and (62),
By (30), (94), and (95),
Hence, the maximum deflection
n
^. (97)
If the ends of the beam in Case I are free, its central de-
flection
5 *l
El
_
(98)
Thus, the deflection of the arched rib in both cases is less
than that of the beam.
804
THEORY OF STRUCTURES.
22. Arched Rib of Uniform Stiffness fixed at the Ends
and connected at the Crown with a Horizontal Distribut-
ing Girder. The load is transmitted to the rib by vertical
struts so that the vertical displacements of corresponding
points of the rib and girder are the same. The horizontal
thrust in the loaded is not necessarily equal to that in the un-
loaded division of the rib, but the excess of the thrust in the
loaded division will be borne by the distributing girder, if the
rib and girder are connected in such a manner that the hori-
zontal displacement of each at the crown is the same.
The formulae of Art. 18 are applicable in the present case
with the modification that /, is to include the moment of
inertia of the girder.
The maximum thrust and tension in the rib are given by
equations (64) and (65).
Let z' be the depth of the girder, A' its sectional area.
The greatest thrust in the girder = ; -] --- ~. (99)
MJ_
~2EL
The greatest tension in the girder =
2EL
h~- ( I0 )
H and M l being given by equations (66) and (67), respectively.
The girder must have its ends so supported as to be capable
of transmitting a thrust.
23. Stresses in Spandril Posts and Diagonals. Fig. 505
represents an arch in which the spandril consists of a series of
vertical posts and diagonal braces.
1
n n+l
FIG. 505.
Let the axis of the curved rib be a parabola. The arch is
then equilibrated under a uniformly distributed load, and the
diagonals will be only called into play under a passing load.
STRESSES IN SPANDRIL POSTS AND DIAGONALS. 805
Let x, y be the co-ordinates of any point F of the parabola
with respect to the vertex C. Then
4&
y = TT*
Let the tangent at Fmeet CB in Z, and the horizontal BE
in G.
Let BC = k'. Then
BL^BC- CL = BC- CN=k' -y.
Let Nbe the total number of panels.
Consider any diagonal ED between the nth and (n -f- i)th
posts.
Let w be the greatest panel live load.
The greatest compression in ED occurs when the passing
load is concentrated at the first n I panel points.
Imagine a vertical section a little on the left of EF.
The portion of the frame on the right of this section is
kept in equilibrium by the reaction R at P, and by the stresses
in the three members met by the secant plane.
Taking moments about G,
D.GE cos B = R.AG,
D being the stress in DE, and the angle DEP.
Now,
w'n(n i)
~2 7v^'
Also,
" k' - y ' 2y
and hence,
^ z? r* D i k'x -\- xy A r A I > k'x x
GE GB -4- x = - --, and GA = ---
2J 22^
Hence,
2 A^ IT j: + xy
The stresses in the counter-braces (shown by dotted lines in
the figure) may be obtained in the same manner.
806 THEORY OF STRUCTURES.
The greatest thrust in EF ~ w' -|- w.
The greatest tension in EF = ZJcos w, w being the
dead load upon EF.
If the last expression is negative, EF is never in tension.
24. Clerk Maxwell's Method of determining the Re-
sultant Thrusts at the Supports of a Framed Arch. Let
As be the change in the length s of any member of the frame
under the action of a force P, and let a be the sectional area of
the member. Then
.
Ea
the sign depending upon the character of the stress.
Assume that all the members except the one under con-
sideration are perfectly rigid, and let Al be the alteration in
the span / corresponding to As. The ratio is equal to a
constant m, which depends only upon the geometrical form of
the frame.
.. Al = m . As mP-^- .
Ea
Again, P may be supposed to consist of two parts, viz.,/,
due to a horizontal force H between the springings, and / 2 due
to a vertical force V applied at one springing, while the other
is firmly secured to keep the frame from turning.
By the principle of virtual velocities,
/, M
Similarly, - is equal to some constant n, which depends
only upon the form of the frame.
=- (m*H+- mnV\~.
CLERK MAXWELL'S METHOD.
Hence, the total change in / for all the members is
If the abutments yield, let ^Al = }*H, ^ being some co-
efficient to be determined by experiment. Then
If the abutments are immovable, 241 is zero, and
(D)
2 (m*~
the same as the corresponding reaction at the end of a
girder of the same span and similarly loaded. The required
thrust is the resultant of H and F, and the stress in each
member may be computed graphically or by the method of
moments. In any particular case proceed as follows :
(1) Prepare tables of the values of m and n for each member.
(2) Assume a cross-section for each member, based on a
probable assumed value for the resultant of V and H.
(3) Prepare a table of the value of w?-- for each member,
and form the sum '2{m' i ^~
\ Ea
(4) Determine, separately, the horizontal thrust between
the springings due to the loads at the different joints. Thus,
let v l , v^ be the vertical reactions at the right and left supports
due to any one of these loads. Form the sum
using v l for all the members on the right of the load and v^ for
all those on its left. The corresponding thrust may then be
808 THEORY OF STRUCTURES.
found by eq. (C) or eq. (D), and the total thrust H is the sum
of the thrusts due to all the weights taken separately.
(5) Repeat the process for each combination of live and
dead load so as to find the maximum stresses to which any
member may be subjected.
(6) If the assumed cross-sections are not suited to thes^
maximum stresses, make fresh assumptions and repeat the
whole calculation.
The same method may be applied to determine the result-
ant tensions at the supports of a framed suspension-bridge.
Note. The formulae for a parabolic rib may be applied
without material error to a rib in the form of a segment of a
circle. More exact formulae may be obtained for the latter in
a manner precisely similar to that described in Arts. 18-22,
but the integrations will be much simplified by using polar co-
ordinates, the centre of the circle being the pole.
EXAMPLES. 809
EXAMPLES.
1. Assuming that an arch may be divided into elementary portions
by imaginary joint planes parallel to the direction of the load upon the
arch, find the limiting span of an arch with a horizontal upper surface
and a parabolic soffit (latus rectum = 40 ft.), the depth over the crown
being 6 ft. and the specific weight of the load 120 Ibs. per cubic foot;
the thrust at the crown is horizontal (= P) and 4 ft. above the soffit.
2. A masonry arch of 90 ft. span and 30 ft. rise, with a parabolic in-
trados and a horizontal extrados, springs from abutments with vertical
faces and 10 ft. thick, the outside faces being carried up to meet the
extrados. The depth of the keystone is 3 ft. The centre of resistance
at the springing is the middle of the joint, and at the crown 12 in. below
the extrados. The specific weight of the masonry may be taken at 150
Ibs. per cubic foot. Determine (a) the resultant pressure in the vertical
joint at the crown ; (&} the resultant pressure in the horizontal joint at
the springing ; (c) the maximum stress in the vertical joint aligning with
the inside of an abutment.
3. The intrados of an arch of 100 ft, span and 20 ft. rise is the segment
of a circle. The arch ring has a uniform thickness of 3 ft. and weighs
140 Ibs. per cubic foot ; the superincumbent load may be taken at 480
Ibs. per lineal foot of the ring. Determine the mutual pressures at the
key and springing, their points of application being 2 ft. and i ft., re-
spectively, from the intrados. Also find the curve of the centres of pres-
sure.
4. The soffit of an arch of 30 ft. span and 10 ft. rise is a transformed
catenary. The masonry rises 10 ft. over the crown, and the specific
weight of the load upon the arch may be taken at 120 Ibs. per cubic foot.
Determine the direction and amount of the thrust at the springing.
5. A concrete arch has a clear spring of 75 ft. and a rise of 7^ ft. ; the
height of masonry over crown = 5 ft. ; the weight of the concrete = 144
Ibs. per cubic foot. Determine the transformed catenary, the amount
and direction of the thrust at the springing, and the curvatures at the
crown and springing.
Ans. m 23.9 ; thrust = 91,354 Ibs. ; slope at springing = 25! ;
radius of curvature = 114.2 ft. at crown and =248.7 ft. at
springing.
6. Determine the transformed catenary for an arch of 60 ft. span and
15 ft. rise, the masonry rising 6 ft. over the crown and weighing 120 Ibs.
per cubic foot. Also find the amount and direction of the thrust at the
abutments.
8 10 THEORY OF STRUCTURES.
7. Determine the transformed catenary for an arch of 30 ft. span and
7^ ft. rise, the height of masonry over the crown being 4^ ft. ; weight of
the masonry = 125 Ibs. per cubic foot. Also find the thrust at the spring-
ing and the curvature at the crown and the springing.
8. In a parabolic arch of 50 ft. span and 10 ft. rise, hinged at both
ends, a weight of i ton is concentrated at a point whose horizontal dis-
tance from the crown is 10 ft. Find the total thrust along the axis oi
tUe rib on each side of the given point, allowing for a change of 60 from
the mean temperature (e = .0000694).
9. A parabolic arched rib of 100 ft. span and 20 ft. rise is fixed at the
springings. The uniformly distributed load upon one-half of the arch
is loo tons, and upon the other 200 tons. Find the bending moment
and shearing force at 25 ft. from each end.
10. An arched rib with parabolic axis, of 100 ft. span and 12^ ft. rise,
is loaded with I ton at the centre and i ton at 20 ft. from the centre,
measured horizontally. Determine the thrusts and shears along the rib
at the latter point, and show how they will be affected by a change of
100 F. from the mean ; the coefficient of linear expansion being .00125
for 1 80 F.
u. A parabolic arched rib hinged at the ends, of 64 ft. span and 16
ft. rise, is loaded with i ton at each of the points of division of eight
equal horizontal divisions. Find the horizontal thrust on the rib, allowing
for a change of 60 F. from the mean temperature. Also find the maxi-
mum flange stresses, the rib being of double-tee section and 12 in. deep
throughout. (Coefficient of linear expansion per i F. / -f- 144000.)
12. The axis of an arched rib of 50 ft. span, 10 ft. rise, and ninged at
both ends is a parabola. Draw the linear arch when the rib is loaded
with two weights each equal to 2 tons concentrated at two points 10 ft.
from the centre of the span. If the rib is of double-tee section and 24
in. deep, find the maximum flange stresses.
If the arch is loaded so as to produce a stress of 10,000 Ibs. per square
inch in the metal, show that the rib will deflect .029 ft., E being 25,000,000
Ibs.
13. A steel parabolic arched rib of 50 ft. span and 10 ft. rise is hinged
at both ends and loaded at the centre with a weight of 12 tons. Find
the horizontal thrust on the rib when the temperature varies 60 F. from
the mean, and also find the maximum flange stresses, the rib being of
double-tee section and 12 in. deep.
14. A semicircular rib, pivoted at the crown and springings, is loaded
uniformly per horizontal unit of length. Determine the position and
magnitude of the maximum bending moments, and show that the hori-
zontal thrust on the rib is one-fourth of the total load.
15. Draw the linear arch for a semicircular rib of uniform section
EXAMPLES. 8ll
under a load uniformly distributed per horizontal unit of length (a) when
hinged at both ends ; (b) when hinged at both ends and at the centre; (c)
when fixed at both ends.
16. A semi-elliptic rib (axes ia and 2b) is pivoted at the springing.
Find the position and magnitude of the maximum bending moment, the
load being uniformly distributed per horizontal unit of length.
How will the result be affected if the rib is also pivoted at the crown ?
17. Draw the equilibrium polygon for a parabolic arch of 100 ft.
span and 20 ft. rise when loaded with weights of 3, 2, 4, and 2 tons, re-
spectively, at the end of the third, sixth, eighth, and ninth division from
the left support, of ten equal horizontal divisions. (Neglect the weight
of the rib.)
If the rib consist of a web and of two flanges 2^ ft. from centre to
centre, determine the maximum flange stress.
18. Find the flange stresses at the ends of the rib, in the preceding
question, and also at the points at which the weights are concentrated,
when both ends are absolutely fixed.
19. A semicircular rib of 28 ft. span carries a weight of ton at 4 ft.
(measured horizontally) from the centre. Find the thrust and shear at
the centre of the rib and at the point at which the weight is concen-
trated.
20. The axis of an arched rib hinged at both ends, for a span of 50 ft.
and a rise of 10 ft., is a parabola. Draw the equilibrium polygon when
the arch is loaded with two equal weights of 2 tons concentrated at two
points 10 ft. from the centre of the span. Also determine the maximum
flange stress in the rib, which is a double-tee section 2 ft. deep.
21. The load upon a parabolic rib of 50 ft. span and 15 ft. rise, hinged
at both ends, consists of weights of I, 2, and 3 tons at points 15, 25, and
40 ft., respectively, from one end. Find the axial thrusts and the shears
at these points.
Ans. Horizontal thrust = 9.6 tons.
Axial thrusts : above i ton = 9.3 tons ;
below i " =97 "
above 3 tons = 8.3 "
below 3 " = io.r "
Shears : above i ton = 3. i tons ;
below i " = 2.2 "
above 3 tons = 5 "
below 3 " = 2.6 "
22. Draw the linear arch and determine the maximum flange stresses
for an arched rib of 80 ft. span, 16 ft. rise, and loaded with five weights
each of 2 tons at the end of the first, second, third, fourth, and fifth
division, of eight equal horizontal divisions. The rib is of double-tee
8l2 THEORY OF STRUCTURES.
section and 30 in. deep. Also find the shears and the axial thrusts at
the fifth point of division.
23. A wrought-iron parabolic lib of 96 ft. span and 16 ft. rise is
hinged at the two abutments; it is of a double-tee section uniform
throughout, and 24 in. deep from centre to centre of the flanges. Deter-
mine the compression at the centre, and also the position and amount
of the maximum bending moment (a) when a load of 48 tons is concen-
trated at the centre ; (b) when a load of 96 tons is uniformly distributed
per horizontal unit of length.
Determine the deflection of the rib in each case.
24. Design a parabolic arched rib of 100 ft. span and 20 It. rise, hinged
at both ends and at the middle joint ; dead load =40 tons uniformly
distributed per horizontal unit of length, and live load = i ton per hori-
zontal foot.
25. Show how the calculations in the preceding question are affected
when both ends are absolutely fixed.
26. In the framed arch represented by the figure, the span is 120 ft.,
t * ie r ' Se I2 * L> l ^ e c ^ e P t ^ 1 ^ tne truss at the
crown 5 ft., the fixed load at each top joint
FIG. 5 o6. 10 tons, and the moving load 10 tons. De-
termine the maximum stress in each member with any distribution of
load. Show that, approximately, the amount of metal required for the
arch : the amount required for a bowstring lattice-girder of the same
span and 17 ft. deep at the centre : the amount required for a girder of
the same span and 12 ft. deep :: 100 : 155 : 175.
27. The steel parabolic ribs for one of the Harlem River bridges has a
clear opening of 510 ft., a rise of 90 ft., a depth of 13 ft., and are spaced
14 ft. centre to centre. The dead weight per lineal foot is estimated at
33,000 Ibs. and the live load at 8000 Ibs. ; a variation in temperature of
75 F. from the mean is also to be allowed for. Determine the maxi-
mum bending moment (assuming /constant), and the maximum deflec-
tion. R = 26,000,000 Ibs. Show how to deduce the play at the hinges.
28. A cast-iron arch (see figure) whose cross-sections are rectangular
O'g'o' and uniformly 3 in. wide, has a straight horizon-
tal extrados, and is hinged at the centre and at
the abutments. Calculate the normal intensity of
stress at the top 'and bottom edges D, of the
FIG. 507. vertical section, distant 5 ft. from the centre of
the span, due to a vertical load of 20 tons concentrated at a point dis-
tant 5 ft. 4 in. horizontally from B. Also find the maximum intensity
of the shearing stress on the same section, and state the point at which
it occurs. (AB 21 ft. 4 in.).
INDEX.
Allowance for the weight of a beam, 403.
Alternating stresses, 152.
American iron columns, 532.
Anchorage, 704.
Angle of repose, 237.
" torsion, 568.
Angular momentum, 177.
Anti-friction curve, 320.
" pivots, 320.
Arch, 470.
Arch abutment, maximum thickness of,
649.
Arch, conditions of equilibrium of, 745.
" formulae for thickness of, 750.
11 linear, 743, 750, 760.
Arched ribs, 740, 762.
" " deflection of, 780, 802.
Arched ribs, effect of change of tem-
perature on, 770, 786.
Arched ribs, elementary deformation of,
781.
Arched ribs, general equations of equi-
librium of, 784.
Arched ribs, graphical determination of
stresses in, 677.
Arched rib of uniform stiffness, 788,
789, 795, 800, 804.
Arched ribs with fixed ends, 771.
'* " with hinged ends, 764.
Arched ribs with axis in form of circu-
lar arc, 769, 773.
Arched ribs with parabolic axis, 760,
775-
Arched ribs with semicircular axis, 765,
775-
Arches, middle-third theory of, 746.
Auxiliary truss, 719.
Back-stays, 16, 704.
Baker's formulae for strength of pillars,
549-
Balancing, 198.
Beam acted upon by oblique forces, 396.
Beam, transverse strength of, 340, 429.
Beam, transverse vibration of loaded,
461.
Beams, equilibrium of, 93.
" of uniform strength, 358-365.
Bearing surface, 314, 315.
Belts, 321.
" effect of high speed in, 325.
" effective tension of, 324.
" slip of, 326.
" stiffness of, 327.
Bending moment, 96, 118, 434.
Bending moment in plane which is not
a principal plane, 354.
Bending moment, relation between, and
shearing stress, 108.
Bevel-wheels, 335.
Boilers, 586.
Bollman truss, 56, 618.
Bowstring truss, 61, 618.
Brace, i, 25.
Brakes, 323.
Breaking-down point, 149.
Breaking stress, 147.
weights, 348, 399.
Breaking weights of iron girders, 369,
370..
Breaking weights, tables of, 212.
Brickwork, 1.49.
Bridge, bowsinng suspension, 626.
" loads, 600.
" trusses, 17, 52.
chords of, 625.
" depth of, 597.
Bridge trusses, maximum allowable
stress in, 657.
Bridge trusses, stiffness of, 598.
" stringers of, 656.
Bridges, 597.
position of platform of, 598.
Buckling of pillars, 513, 515.
813
8i4
INDEX.
Cable with sloping suspenders, 717.
Cables, 703.
' curves of, 706.
' deflection of, 714.
' length of arc of, 712.
' parameter of, 711.
weight of, 713.
Camber, 388, 659.
Cantilever, 365.
" curve of boom of, 634.
deflection of, 638.
" depth of, 637.
" weight of, 632.
Cast-iron, 147.
Catenary, 34, 706, 750.
Cement, 150.
Centres of gravity, n.
Centre of resistance, I, 743.
Centrifugal force, 181.
Centripetal force, 182.
Clapeyron's theorem, 292.
Coefficient of cubic elasticity, 255.
elasticity, 141, 143.
fluidity, 162.
hardness, 164.
lateral elasticity, 144.
rigidity, 254, 285.
rupture, 248.
torsional rupture, 574.
transverse elasticity, 285.
Collar-beams, 25.
Columns, see Pillars, 513, 538.
flexure of, 554, 5*57.
Compound strain, 236.
Compression, 141.
Conjugate stresses, 247.
Continuous girders, 463.
Continuous girders, advantages and dis-
advantages of, 486.
Continuous girders, maximum bending
moment in, 465.
Coulomb's laws, 568.
Counterbrace, 60.
Counter-efficiency, 328.
Counterforts, 270.
Covers of riveted joints, 665.
Cranes, 13.
bent, 31.
" derrick, 16.
jib, 13-
pit, IS-
Crank effort, 207.
Cubic elasticity, 255.
" strain, 283.
D<.ad load, 143, 600.
Deflection, curve of, 434.
of girders, 384-386, 638.
Deformation, 140, 251, 254.
Dock walls, 270.
Dynamometer, Prony's, 327.
Earth foundations, 258.
Earthwork, 255.
pressure of, 257.
Earthwork, Rankine's theory applied to
retaining walls, 264.
Efficiency of mechanisms, 335.
" of riveted joints, 666.
Elastic curve, 355.
" moment, 96, 340.
Elasticity, 140.
coefficient of, 141, 143.
" cubic, 255.
" lateral, 144.
" transverse, 285,
limit of, 145.
Ellipse of stress, 241.
Ellipsoid of stress, 281.
Empirical rules for wind-pressure, 663.
Encastr6 girders, 458.
Energy, 207.
" curves of, 207.
" fluctuation of, 207.
" kinetic, 167, 169, 170.
" potential, 167.
Envelope of moments, 121.
Equalization of stress, 349.
Equalizer, 629.
Equilibrated polygon, 740.
Equilibrium of beams, 428.
Equilibrium of beams, general equations
of, 428.
Equilibrium of flanged girders, 366.
Euler's theory of the strength of pillars,
537-
Examples, 69-92, 132-139, 216-234,
294-298, 337-339* 407-427, 490-512,
563-567, 580-585, 594~59 6 > 689-702,
734-739, 809-812.
Expansion of solids, 215.
Extension of prismatic bar, 289.
Extrados, 740.
Eyebars, 661.
steel, 665.
Factor of safety, 150.
Fatigue, 152.
Fink truss, 54.
Flanged girders, 365.
" " equilibrium of, 366.
" stiffness of, 384.
Flanges, 365, 597.
" curved, 366.
" horizontal. 366.
Flexure of columns, see Pillars, 554,
557-
Flow of solids, 162.
INDEX.
815
Fluctuation of stress, 151.
Fluid pressure, 162.
Force polygon, 3, 7, 119.
Foundations, earth, 258.
limiting depth of, 258.
of walls, 270.
Fracture, 141.
Framed arch, stresses in, 804.
Framed arch, Clerk Maxwell's method
of determining stresses in a, 806.
Frames, i, 2.
incomplete, 27, 61.
Friction, 300.
angle of, 237.
coefficient of, 300, 313.
journal, 310.
rolling, 310.
Funicular curve, 10.
polygon, 3, 7, 117, II9 .
Gins, 17.
Girder of uniform strength, 381.
Gordon's formulae for pillars, 522.
Hinged girders, 127-131.
Hodgkinson's formulae for the strength
of pillars, 513. 517-521.
Hooke's law, 142.
Howe truss, 58, 611.
Impact, 184.
Impulse, 176.
Incomplete frames, 27.
Inertia, 198.
moment of, 12, 342.
pressure due to, 200.
Inflection, point of, 453-463.
Internal stress, 235.
Isotropic bodies, 283.
Joint of rupture, 747.
Keystone, 741.
Lateral bracing, 654.
Lattice girder, 600.
Launhardt's formula, 153.
Lenticular truss, 626.
Limit of elasticity, 145.
Line of loads. 5.
resistance, 273-276, 741, 750.
" rupture, 265.
Linear arch, 743, 753-760.
Loads, live, in, 115, IIQ> 600,639,641,
Loads, stationary (dead), 118, 600.
Long pillars, 535.
Mansard roof, 6.
Masonry, 149.
Mechanical advantage, 294.
Middle third theory of arches, 746,
Modulus of elasticity, 141.
" rupture, 348.
" transverse elasticity, 254
Moment of forces, 116.
" inertia, 12, 342.
" examples of, 371-81.
Moment of inertia, variable section of
455-
Moment of inflexibility, 96.
" resistance, 96.
Momentum, 176.
Mortar, 150.
Neutral axis, 340.
of a loaded beam, 435-454.
surface, 340.
Oblique resistance, 169.
OsciLatory motion, 190, 195.
Panel points, 52.
r^anels, 54.
D iers, 65.
513.
Euler's formulas for, 527.
failure of, 515.
flexure of, 515.
formulae for American, 527-532.
Gordon's formulae for, 522
Pillars, Hodgkinson's formulae for '517-
521.
Pillars, Rankine's formula for, 526.
Pillars with stress uniformly distributed
516.
Pillars with uniformly varying stress,
Pins, 661.
Piston velocity, curves of, 205
Pivots, 316.
conical, 319
cylindrica., 316.
Schiele's (anti-friction), 320
Plasticity, 141.
Poisson's ratio, 142
Pole, 7.
Polygon of forces, 3, 7.
" pressure, 743.
Pratt truss, 60.
Primitive strength, 153.
Principals, 33, 34.
Prony's dynamometer, 327.
Proof strain, 171.
stress. 171.
Purchase 304.
Purlins, 33, 34.
Radius of gyration, 174, 528-531
p "' of twist, 289.
8i6
INDEX.
Redundant bars, 48.
Reservoir walls, 271.
Resilience, 171.
Retaining walls, 260.
Retaining walls, conditions of equiib-
rium of. 260.
Retaining walls, Rankine's theory ap-
plied to, 264.
Rivet connection between flange and
web, 660.
Riveted joints, 668.
covers of, 675.
design of, 678.
efficiency of, 679.
failure of, 670.
stresses in, 670.
theory of, 671-675.
Riveting, 666.
Rivets, 666.
" dimensions of, 667.
Rocker-link, 629.
Rollers, 35, 639.
Roof trusses, 17.
" distribution of loads on, 39.
types of, 33.
weights of, 37.
Ropes, 321.
Saddles, 704.
Schiele's pivots, 320.
Screws, 306.
" endless, 309.
Sections, method of, 62
Set, 145. -,. ' *~~**v
Shafting, distance. EetvySfen bearings of,
575- ' /'* * 1
Shafting, efficiency of, 577.
intprrfal-streiB&iii, 237.
stiffness of, 573.
" ' torsional strength of, 288-291.
Shear, 141.
" maximum, 121, 237.
Shearing force, 95.
" '. ^ et examples of, 97, 108.
Shearing force and bending moment,
relation between, 108.
Shearing stress, 198.
. " distribution of, 391.
Shear-legs, 17.
Similar girders, 401-404.
Skew-backs, 34, 74-
Soffit, 740.
Spandril, 740.
Specific weight, 143.
Spherical shells, 591.
Spritigings, 740.
Springs, 355, 456-458.
" simple rectangular, 456.
Springs, spiral, 477.
Springs of constant depth, but triangu-
lar in plan, 4.57.
Springs of constant width, but parabolic
in elevation, 457.
Statical breaking strength, 153.
Steel, 148.
Stiffening truss, 719.
" hinged at centre, 725.
Stiffness, 190, 387, 389.
Strain, 140, 251.
Straining cill, 18.
Stress, 141, 251
" and strain, relation between, 281.
" general equations of, 277.
principal, 241.
planes of, 237.
Stresses, conjugate, 248, 250.
Stress-strain curves, 147-149.
line, 144.
Strut, i.
St. Venant's torsion results, 572.
Surface loading, 350.
Suspenders, 706.
Suspension-bridges, 703.
" loads on, 730.
Suspension-bridges, modifications of,
73i.
Suspension-bridges, pressure upon piers
of, 718.
Swing-bridges, 470-472.
Tables of breaking weights and coeffi-
cients of bending strength of timber,
212, 213.
Table of coefficients of axle friction,
336.
Table of coefficients in Gordon's for-
mula, 524.
Table of coefficients in Rankine's mod-
ification of Gordon's formula, 526.
Tables of diagonal and chord stresses,
644-650.
Tables of efficiencies, 587.
" " elliptic integrals, 562.
" " expansion of solids, 215.
" " eyebar dimensions, 665.
" " factors of safety, 214.
Tables of loads for highway bridges,
687.
Tables of strengths, elasticities, and
weights of iron and steel, 210.
Table of strengths, elasticities, and
weights of various alloys, 211.
Tables of weights of modern bridges,
682-687.
Table of weights and crushing strength
of rocks, 214
Table of weights of roof coverings, 67.
INDEX.
8I 7
Table of weights of roof frames, 67.
Tension, 141.
Theorem of three moments, 463-470.
Thick hollow cylinder, 588.
Timber, 149.
Torsion, 141, 568.
" St. Venant's results, 572.
Torsional coefficient of elasticity, 145.
resilience, 574.
Torsional strength of shafts, 288, 289,
571, 572.
Transverse strength, 141.
Transverse vibration of a loaded beam,
461.
Trellis girder, 600.
Tresca's theory of flow of solids, 162.
Tripods, 17.
Truss, 2.
' composite, 31.
king-post, 21.
' queen-post, 25.
' roof, 32.
' triangular, 19.
Trussed beams, 53.
Twist, 141.
Unwin's formula, 159.
Values of k*, 174, 528-531.
Vibration strength, 153.
Voussoir, 741.
Warren truss, 57.
Web thickness, 382.
Wedge, 303
Weights of roof coverings, 67.
" frames, 67.
Weyrauch's formula, 153.
Weyrauch's theory of buckling of pillars,
550.
Wheel and axle, 329.
Whipple truss, 618.
Wind pressure, 38, 67, 629, 651, 653.
Wind-pressure, American specifications
of, 652.
Wind-Pressure Commission rules, 653.
Wind-pressure, empirical regulations,
653.
Wohler's law, 150.
Work, 167.
effective, 178.
external, 168.
internal, 168.
useful, 178.
waste, 178.
Work done in bending a beam, 460.
Work done in small deformation of a
body, 292.
Work of journal friction, 314.
Working load, 150.
" strength, 150.
" stress, 150.
Wrought-iron, 148.
Yield-point, 149.
ERRATA.
Page 12, 1 5th line from top, for A\ , A*, As, read a\ , a*, a*.
23, Fig. 34, for BO read AO, and for AO read BO.
16562 16562
47, 2d line, for , read -=-.
52, 3d line from bottom, for W* read W*.
55, 8th line from top, for 2 7i read 2 7\.
63, 7th line from bottom, for "revolving" read " resolving."
3 V 13 3 V's
64, 3d line from top, for - read
2V 13
127, Fig. 181, line x* should lie between W* and B instead of
between E and W*.
215, i st line, for ACTORS read FACTORS.
219, 4th line from top, for 17,828 Ibs. read 17,328 Ibs.
223, 6th line from top, for 8625 ft.-lbs. read 15,750 ft.-lbs.
239, nth line from top, in 2d equation, for = I read = i.
274, 5th line from top, for CD read QP.
276, ist line (numerator), for ABQP read OBQ.
294, 2d line from bottom, for "thrust" read "tension," and for
"tension" read "thrust."
295, I2th line from bottom, for 11 38' read 3 29'; nth line from
bottom, for 3 26' read 63 26'.
297, loth line from top, for 9.8 ft. read 12.2 ft.; I3th line from
bottom, for 13.19 ft. read y - ft.
298, 7th line from bottom, for 20,720 Ibs. read i4,o49 T Jt Ibs.
326, 3d line from top, for $T*(<? a i) read \T^ a i)v.
344, 5th line from bottom, for 1000 Ibs. read 100 Ibs.
363, for equation at bottom of page substitute
WJ WJ
364, 8th Iin6 from top, for read -
(Over)
Page 371, Fig. 292*2, 7/ t and h* should be interchanged.
413, I4th line from top, for 230 ft. read 129.6 ft.
417, 8th line from top, for 42^ tons read 42 tons.
422, Add to last line: The tensile and compressive working
strengths being 2000 and 5000 Ibs. per square inch,
respectively. Assume that the thickness of the web is a
fraction of its depth.
" 423, ist line, for 26^ in. read 2o^ G in., and for 28.8 sq. in. read
25^ sq. in. ; 2d line, for 5.76 sq. in. read 6|f sq. in. ; and
3d line, for 93 sq. read 53 sq.
" 455, 4th line from top, equation, for read 7.
474, i ith line from top, for " the wheels of a locomotive passing "
read " five wheels passing."
485, 1 3th line from top, for/ 2 read y*.
" 501, 3d line from top, for ^i , fa , fa , read p\ , pz , p<>.
530, ist line under (-), for S = 26/1 I? read S = 2.bh &* ; and
in denominator of next equation, for // 2 read ^.
555, 7th line from bottom, equation, for dQ read d(f>.
565, 3d line of Example 27, for "maximum" read "minimum."
571, bottom line, for i.57/) 2 Tread 1.57 D*Tf.
578, ist line of Example, for 50 read 95.
. " 580, Examples 7 and 8, for m read G.
581, Examples 10, u, and 15, for m read G.
582, Examples 18 and 19, for m read G.
583, Example 30, for m read G.
647, Table of Maximum Stresses in the Verticals, change 3d and
4th lines to read
7/3 = 42625 + 16200 = 58,825 Ibs.
7/4 = 28700 + 5400 = 34,100 "
649, 9th line from top, for 6250 read 7600, and for 1525 read 2875.
650, ist line of Compression Chord Table, for di read d\.
707, Equation (2), for PS read ps.
711, 6th line from bottom, after Art. 4 add " Case B."
714, 9th line from top, for H\ sec 6 read H sec 6.
715, 9th line from bottom, equation, for ^ read .
724, 4th line from top, for iv read x.
758, Equation 7, for 2iuyp sin read 2wyp\s\t\ ) .
u I* I
772, Equation (5), for read juE; Equations (6) and (7), for ^
read nEI, //
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