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r
■ITT
I
APPLIED MECHANICS.
'i
BY
HENRY T. BOVEY, M.A.,
ASS. Mem. Inst. C.E. ; Mem. Inst. Mech. E. ; Mem. Am. In.st. M.E.
Professor of Civil En.ineerinr, and Applied Mechanics, MrGill University, Montreal.
tMomoj Queens' Colkye, Camhridye, (Eng.).
MONTREAL:
PRLNTED BY JOHxN LOVELL & SON.
1882.
6^ /-?^
207 a
/3^
I
:;
' , ^. ^A. in thp vear One Thousand Eight
.• »trv Aot of Parliament of Canada, m tne year v»i.c
Knterkd according to Act of rariia , . c„;„ j„ tho Office of the Minister of
Hundn.d and Eighty-two. by John Lovkll & Son, m
Agriculture.
\
CONTENTS.
r of
CHAPTER I.
FRAMES.
PAfJE
Definitions ,
Frame of two Members ^^
Frame of three or more Members 2
Method of Sections <
^'■^"^s !ll!V.''"l*l°!!'l ib.
Bridge and Roof Trusses g
Compound Roofs 25
Gothic Roof Truss Ig
CHAPTER II.
BOOFS.
Types of Truss I g
Principals, Purlins, etc 19
Roof Weights 21
Wind Pressure 22
Methods of Determining the Total Stresses in Roof Members 24
Tables 32
Examples ^ i
CHAPTER III.
BRIDGES.
Classification 42
Depth of Truss ' j^^
Position of Platform i^
Number of Main Girders 44
Bridge Loads 4-
Chord and Web Stresses 4,;
Friction Rollers ^^
i
IV
CONTENTS.
Trellis Girder
Warren Girder
Howe Tru.MS
Pratt TruHH
Fink TrusH
Bollinan Truss •
Quadrangular and Post Systems
Bowstring Girder
Bowstring Girder with Isosceles Bracing ^^
Lenticular Truss '
70
PAOB
47
49
55
59
CI
62
63
ib.
ber.
<jani
Tests .
Tables
72
73
CHAPTER IV.
SUSPENSION HRIDGES.
Cables.
Anchorage ; Saddles
Suspenders
Curve of Cable ; Vertical Suspenders
Pressure upon Piers
Length of Cable
Weight of Cable
Deflection of Cable
Curve of Cable ; Inclined Suspenders
Suspension Bridge Loads
Modifications of Suspension Bridge
Auxiliary, or Stitlening, Truss ^'^
CHAPTER V.
75
76
77
78
79
80
81
82
83
84
85
ARCHED RIBS.
Linear Arches
Arclied Ribs • •
Bending Moment at any point of an Arched Rib.
Rib with Hinged Ends ; Invariable Span
Value ofH
Process of Designing a Rib
Rib with Ends absolutely fixed
Deflection of an Arched Rib
Elementary Deformation of an Arched Rib. ...
General Equations
Rib of Uniform Stiff"nes8 ■
Parabolic Rib of Uniform Depth and Stiffness . .
Parabolic Rib of Uniform Stiffness
91
92
94
95
96
97
ib.
100
101
103
105
106
112
91
92
94
9y
96
97
ib.
100
101
103
105
106
112
r
CONTENTS. V
Maximum Deflection of an Arched Kib hq
Arched Kib of Uniform Stiffness connected at the Crown witli a IlorV-
zontal Distributing Girder jig
Semi-circular Timber Rib ' ib
Stresses in Spandril Posts and Diagonals 119
Maxwell's Method of obtaining the Resultant Thrusts at th'e'suuporis of
a Framed Arch " 22O
CHAPTER VI.
bbi'AILS OF rONSTRUCTION.
Main Chords ^.o
Platform ,24.
!*'»« ....!.!!.........*.....!..... 125
Bending Action upon a Pin '/_ 127
General 'iemarks ] _ _ j 28
Kjvets ...*....!..!.......... 129
Dimensions of Rivets " j.^q
Strength of Punched and Drilled Plates '.. .','.'.'.'.",'.'.*.*.".'. ib.
Riveted Joints joi*
Strength of Riveted Joints '...*."'..'.',,.. 132
Single Riveted Lap and Single Cover Joints 135
Single Riveted Double Cover Joints V.'. V.'.V. ".'.'.". '.V. ib.
Chain Riveted Joints.. ' * ib
'Ig-zag and other Riveted Joints ...'.'.'...... 136
Covers r^7
Rivet connexion between Flanges and Web .."...'"...*. 138
Examples .oa
APPLIED MECHANICS.
CHAPTER I.
Frames.
(1). Dcfantions. — Frames are rigid structures composed of straight
struts and ties, jointed together by means of bolts, straps, mortises and
tenons, &c. Struts are members in compression, 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 lines of action of these resultants intersect the joints
at or near the centre in points called the centres of resistance. The
figure formed by joining the centres of resistance in order is usually a
polygon, and is designated the line of resistance of the ' <^me.
The position of the centres should on no account be allowed to
vary. It is assumed, and is practically true, that the joint.^ of a frame
are flexible, and that the frame under a given load does not sensibly
change in form. Thus, an individual member 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 tioo mcnbers.
OA, OB are two bars jointed at and supported at the ends A, B.
The frame in Fig. 1 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 direc-
tion 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.
FUNICULAU POLYGON.
Let the an,i,'b AOB= a, and *-he an<?lc COf>=fi.
Let iS',, St, be the stresses along OA, OB, respectively.
■'• P~ 00 ~ nina *" /' ~ 0'0~ siiui
(3.) Franie of three or more inemhers.
Let .'li^j^'lj ... be a polyf:;oiial franie jointed ut vl,, .ij, J3...
Let ]\, 1\ 1\ ... be the resultant furce.s at the joints J,, A^, A3 ...
respectively. Lit >S'„ jS'j, »S', be the forces alon<f Jjvl^, ^l^ A^^ ...
respectively.
Consider the joint A,.
The lines of action of three forces, i', , »S', and Pg, int^^-rscct in this joint,
and the forces, beinjr in equi'.ibrium, may be represented in direction and
magnitude by the sides of the triangle Os^s^, in which .s, Sg is parallel to
i* , Os^ to >S'i, and Os^ to S^.
Similarly, F^, <S'i, jS'j, may be represented by the sides of the triangle
OsiS^ which has one side Os^ common to the triangle Os^s^, and so
on.
Thus, every joint furnishes a triangle having a side common to each
of the two adjacent triangles, and all the triangles together i'orm a closfd
polygon s, SjjSj... The sides of this polygon represent in magnitude and
direction the resultant forces at the joints, and the radii from the
pole to the angles s,V3 ••• represent in magnitude, direction and
character, the forces along the several sides of the frame AiA^A^ ...
The polygon SjSjSj ... is the line of resistance of the f"ame, and is
called the funiculaT polygon of the forces /*,, P., 1\, ... with respect
to the pole 0.
FUNICULAR POLYGON.
I to each
la closxi
ide and
lorn the
kon and
tUAi ...
and is
respect
Corollary 1. — The converse of the preceding is evidently true.
For if a system of forces is in equilibrium, the polygon of forces s, ^.^st^...
must close, and therefore the polygon which has its .sides respectively
parallel to the radii from a pole to the angles .s,,»a,8.,, ... and which
has its angles upon the lines of action of the forces, must also close.
Corollary 2. — Let the resultant forces at the joints be parallel.
The polygon of forces becomes the straight line s,Sj, which is often
lermed the line of loads.. Thus, the forces 1\, P^, ...1\ are represented
by the sides s, s^, s, Sj, ... s^ Sj, which are in one straight line closed by
s, Sg and Sj Sg, representing the remaining forces /^, and F^
Draw OH perpendicular to s,S5, the line of loads ; OH represents
in magnitude and direction the stress which is the same for each
member of the frame.
Let oi, aj, 03... be the inclinations of the members yl, A^, A^ ^3,.. ,
respectively, to the line of loads,
. • . 0H= Hsi. tan oi and also, 011= Hsi. tan oj
.'. OE. (cot a + cot n) = Hs, + Hs^-s s. = P + P, + P + P = P. -h P.
Pi+ ..^ P, i\+^, A • ,\. 4.
. nn — -= — r~ i — - ^"a is the stress common to
. . ua _ ^^i ^^ ^ ^y^ ,^^ cot a, + coi a^
each member.
OFT
Again, the stress in any member, e.g., At A^, is Os, = -: — -
Corollary 3. — Let the resultant forces at the joints ^,, A^, be in-
clined to the common direction of the remaining forces, and act in the
METHOD OF SECTIONS.
directioiiS shewn by the dotted lines (Fig. 4'). Let P\, P r, be the
magnitudes of the new forces ; draw Sj .s'g parallel to the direction of P,
so as to meet Os,-, in s'f„ (Fig- 5) ; join s^ .s'e- i^'^nce theie is equili-
brium, s'fi Sr, must be parallel to the line of action of P',,.
Thus .s, s'e Sr, is the force polygon.
Corollary 4. — The forces, or loads, Pj, P3...P;,,
are generally vertical, while P„ Pq, are the vertical
reactions of two supports. In this case, s^ and II
coincide, and II is the horizontal stress common
to each member of the frame.
The stress in any bar may be found as in Corol-
lary 2.
Corollary 5. — If the number of the forces P„
P^, ... is increased indefinitely, and the distance be-
tween the Hues of action of the forces indefinitely
diminished, the funicular polygon becomes a curve
and is called t\iQ fvnicular curve.
Corollary 6. — If more than two members meet at a joint, or if the
joint is subjected to more than one load, the resulting force diafrram
will be a quadrilateral, pentagon, hexagon ... according as the number of
members is .S 4, 5 ... or the number of loads is 2, 3. 4...
(4) Method of Sections. It often happens tliat the stresses in the
members of a frame may be easily obtained by the method of section?.
This method is precisely similar to that employed in discussing the
equilibrium of beams, and depends upon the following principle : —
If a frame is divided by a plane section into tico 2>arts, and if each
part is considered separately, ,.ne stresses in the hars (or memhers')
intersected by the secant plane must balance the external forces upon
the part in question.
Hence, the algebraic sums of the horizontal components, of the vertical
components, and of the moments of the forces with respect to any point,
are severally zero, i.e., analytically,
S(X)=o, I,(V)=o, and M=0
These equations are solvable if they do not involve more than three
unknown quantities, i.e., if the secant plane does not cut more than
three members, the corresponding stresses are determinate.
(5) Jib-Crane. Fig. 7 is a skeleton diagram of an ordinary jib-
crane.
OA is the post fixed in the ground at 0.
CRANES.
OB is the jib, AB is the tie.
The jib, tie aud gearing are sus-
pended from the top of the post by a
cross head, which admits of a free ro-
tation round the axis of the post.
Let the crane raise a weight W.
Three forces in equilibrium meet
at B, viz., W, the tension T in the
tie, and the thrust C along the jib.
They may be represented by the sides
of the triangle AOB to which they
are respectively parallel.
.*. T = W. i^andC= W 12
AO AG
The stress S in the chain tends to produce a compressive stress of
equal magnitude in any member along which it passes. Thus, if it
pass along AB the resultant stress in thj tie is T-S, while if it pass
along BO, the resultant stress in the jib is C + S.
W
Also, neglecting friction, S= — , n being the number of falls of chain
from B.
Again, the post is a semi-girder fixed at one end and acted upon at
the other by a force T.
Let Mj. be the bending moment at any point E of the post distant x
from A.
M^ =T.CosBAD.x=W.
AB
^^\.x^wM.x
AG' AB AO
The bending moment at is evidently W.AD, and is independent
of the height of the pf^at.
The vertical component of T, viz., T. —
W. — , is transmitted
AO'
through the post. Hence, the total unit-stress in the layers of the
transverse section at E, distant v from the axis, is — ^ ■^~ ± -7 -'"''
' ^ 'A AO J
A being the area of the section, and / its moment of inertia with
respect to the axis. The total resultant pressure along the post at
rp • T, An^n ' onn- w BD , .jr BO BD + AO
= — T. sm BAD + C. sm BOF = — ^- -tt{-\- ^V. -_- — - — :
AU AU MU
w
If the post revolves about its axis, the jib and gearing aie bolted to
i, and the whole turns on a pivot at the toe G, In this case the frame,
G
CUANES.
as a whole, is kept '"n equilibiiuin by the weifiht M\ the horizontal re-
action //of the curb-plate at 0, and the reaction /£ at G. The first
two forces meet in /', and therefore the reaction at G must also pass
throuirh F.
Hence, since OFG is a trianjjjle of forces,
//= W. ^, and R = Tr. ^.
OG GG
(CO Derrick-
Grave. Fij^. 8,
shews a combination
of a derrick and
crane called a der-
rick-crane. It is
distiniiuished from
the jib-crane, by
having two bac/i-
sf,(ii/K, which are
u-iUiilly situated in planes at ritiht-aiigles to each other. 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 recjuired.
The derrick-crane is gevierally of wood, is simple in construction, is
easily erected, has a vertical as well as a lateral motion, and a range
ccjual to a circle of from 10 to G(» ft. in radius. It is, therefore, very
useful for temporary works, setting masonry, &c.
The stresses in the jib and
tie are obtained as in ^(5),
and those in the back-stays
and post may be calculated
as follows : —
Let Orr OK, OL. be the
horizontal traces of the tie
and back-stays.
In //O produced, take ON^
to represent the horizontal
component of the stress in
the tie, i.^ ., T, sin f/,« being
the inclin.ition of the tie to
the vertical. Complete the
parallelogram X Y.
CRANES.
very
OX and 01' are the horizontal components of the stresses in the back-
stays. Let the angle NOX = 6.
.•. OX = ON. cos d = T. sin a. cos 6, and is a maximum when — 0"
OY=OX. sin d= T. sin a. sin f), '• " '' t)=9{\o
Hence, the stress in the back-stay is <rreat<^st, when the ttlane of the
back-stay and post coincides with that of the jib and tie.
Again, let /? be the inclination of the back-stays to the vertical. The
vertical components of the back-stiiy .stresses arc T. sin a. cos o. cot ft and
T. sin a. sin 0. cot /g, so that the corresponding stress along the post is
T. sin (I. cot ft. (cos + sin e), which is a maximum when ^=45".
(7). Jient Crane.
Fig. 10. shews a convenient form of crane when much hoad-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 bo
braced together as in the Fig. for hiads of less than 10 tons. The
flanges may be kept at the same distance apart throughout, or the
distance may be gradually diminished from the base t<iwards the
peak.
Let the letters in Fig. 10 denote the stresses in the corresponding
members. Three forces *S^,, C.^, and W. act through the point (1), so
that Si and C^ may be obtained in terms of \V ; three forces *S'|,»S'.„7'„
act through (2), so that >S'a and 7\ ni;iy be obtained in terms of »S'i and
therefore of W; four forces >S'j„ f„ /S,, C,, act through (3), and the
values of iS",, 6V being known, those of aS'^, T^, may be determined.
Proceeding in this way, it is found that of the forces at each succeeding
joint, only two are unknown, and the values of these are consequently
determinate.
il
8
TIMBER TRUSSES.
The calculations may be checked by the method of moments and by
the stress diaj^ram, (Fig, 11).
e.(j., let W= 10 tons.
Take moments about the point (7).
.'. Tr. (y7)==10. (x?) or T, =
3. 25
1. 25'
10 = 26 tons.
No other forces enter into the e(|uation o^ moments, as the portion
of the crane above a plane intersecting 08 and passing through (7) is
kept in ecjuilibrium by the weight of 10 tons and the stresses Tj, S^, Ce)
the moments of S,-, and C^ abou [7) are evidently zero.
.3. 9
Again, from the stress diagram, 7\ = QR-= ,-^- 10 = 26 to>ns.
(8) Timhcr bridge <nid roof
trusses of small span.
(a). — A single girder is the
simplest kind of bridge, but is
only suitable for very short spans.
When the spans are wider, the
centre of the girder may be sup-
ported by struts OC, OD, Fig. 12, through which a portion of the weight
is transmitted to the abutments.
Let the weight at be P, and let a be the angle AOC.
P OJl _^P 1
2 sin a
The thrust along OC = „ . -,-^
Z At
The horizontal thrust at = ^. -,4y =-6- ^^^ «
i At/ ^
The horizontal and vertical thrusts upon the masonry at C (or />),
respectively.
P P
are — . co< a, and _
2 2 sin a
If the girder is uniformly loaded, Pis one-half of tho whole load.
{It). — In Fig. 18 a straining ciU, E F, is introduced, and the girder
is supported at two points.
Let P be the weight at each
of the points E and F.
The thrust in EC (or FD)
= P- 1,^ and the horizontal thrust
At
in the straining piece = 7'. iiii-
"* 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).
i
TIMBER TuUSSES.
By means of straining cillsj
the jjirdcrs may be supported
at several points 1, 2, ... and the|
weight concentrated at each may
be assumed to be one-half of the I
load between the two adjacent
points of support. The calculations for the stresses in the struts, etc...
are made precisely as above.
If the struts are very long they are liable to bend, and countcrbraces
AM, iiiV, are added to counteract this tendency.
(<!). — 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. 14 represents the
simplest form of roof-truss,
the dotted lines being the
lines of resistance of the
bars.
AC, BC are rafters of
equal length inclined to
the horizontal at an angle a,
and each carries a uni-
formly distributed load 11'.
The rafters react horizontally upon each other at C, and their feet
are kept in position by the tie-beam AB.
Consider the raftt'r AC
The resultant of the load upon AC, i.e., W, acts through the middle
point D.
Let it meet the horizontal thrust II oi BC upon AC in F.
For ccjuilibrium, the resultant thrust at A must also act through F.
The sides of the triangle AFE, evidently represent the three forces.
AE W AE _W
2
AF .__ lAErrHF'^^^^
Hence, H= W. -=,.:,= _ = .^. vat a.
' EF 2 JJ"
AF /
R^w.-^E'^^w.yl'
FE'
'1 +
1 ^'
4 • DE'
cot* a.
The thrust R produces a tension // in the tic-beam, and a vertical
pressure IV^upon the support.
10
KING-POSTS.
Also, if y Is the angle FAE,
tan a.
ion y=^^: = 2,211 = 2
AE AE
If the rafters .4^, BC. are
iiiu'<jual, }et or,, a.^, be their
inclination to AB, respectively.
}j<!t W^ be the nniforuily
distributed load npon A(-, W.^
that itfion BC.
Let the direction of the nintual thrust 7* at C make an an^le /3 with
the vortical so that if CO is drawn per|x>ndicnlar to FC, the angle COB
= /3 ; the angle A CF = W—ACO = 00"— 3 — "i
Draw AM pcrpcndicnlar to the direction of /', and consider the
rafter A(\ As before the thrust /j*, at .1. the resultant weight >r,
at the middle point D of JC, and the thrust P at C, meet iu the point
F. Take luouients about A,
.*. P AM = ir,. AE
But AM = AC. sinACM= AC. cnsT^a^ and AE= ^. cos. ax,
W\ co.<i a ,
•. P =
"2"' cos (/3-a,j
Similarly, by considering the rafter BC,
W2 co.'i di Wi cn.f at
2 ■ cos (;8 + rto)
IF,
P =
2 • i,'?» (/3 + a-i - 90)
Hence, -r-. — ;„ — ^ ~ -^ — ~— tt-- —
' 2 con ()3-rti) 2 cos
IT, + W2
rn.*i a
(/3 + at)
and
<an /3 =
W\. tan cii- tVi. tan ay
The horizontal thrust of each rafter = P. sh> $
The vertical thrust upon the support A =-- W^ - P. cos 3
B =. W, + P. ens $
(d). — Kfng-pnsf truss. The
simple triangular truss may be
modified by introducing a king-
post CO, (Fig. IG), which car
ries a portion of the weight of
the beam AB, and transfers it
through the rafters so as to act
upou the tic in the form of a tensile stress.
d
KING-POSTS.
11
*(
r the
It >r,
point
Let P bo the wvij^lit borne by tlu- kin<,'-po,4 ; represent it ))y (JO.
Dr?,w OB parallel to BC and l)E parallel to AB.
CF I' 1
DC= . '' ■ = TV — 1 is the thrust in CA due to /*, and is of course
sm a I sin a,
equal to DO, i. e., the thrust alonu; CB.
P
DE — CE.cot a = .-r . cot n, is the horizontal thrust on each rafter, and
is also the tension in the tic due to /*.
Let IF be the uniformly distributed load upon each rafter.
The total horizontal thrust upon each rafter = ( ir f /*)
r
rot a
IT
The total vertical pressure upon each support = ^'+-:7
If the ape.x; (J is not vertically
over the centre of the tie-beam,
(Fi<c. 17), take CO, as before, to
ri'present the weijzht P borne by
the kintr-post ; draw OD parallel
to BO, and DI'J parallel to AB.
The weiulit J* produces a
thrust CD along CA, DO along CB, and a horizontal thrust DE upon
each raftt'r
CE is the portion of /' supported at ^1 and EO that supported 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 coluniii or wall (Fig.
18), the king-post being a pillar
against which the heads of the
rafters rest.
Consider the rafter ^4^'.
The normal reaction A" of CO
upon AC, the resultant weight
irat the middle point D, and the
thrust Rat A meet in the point /'.
Take moments about A,
• •• li . AC—W.AE, or R = _. cos a
Thus the total thrust transmitted through CO to the support at ia
, W ITT
2. — cos a, cos a = W. cosi a
w
12
KING-POSTS.
The horizontal thrust upon each rafter = —. cos a. sin a = ^
(<•). — If the raflers arc incon-
veniently lontr, or if they are in
(lunger of bending and breaking
across, the centres may be sup-
ported by struts 0/), OE, (Fig.
19). A portion of the weight
upon the rafters is transmitted
through the struts to the king-post (or rod), 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.
Indeed, the main duty of struts and ties is to transform transverse
into longitudinal stresses.
The load upon the rafters {2.W) may be assumed to be concentrated
at the points A, D, C, E,and B, in the proportions Jf, il. 1 , i i and
4 2 2 2
) respectively. Drop the vertical DF to represent the load __ at /) ;
4 ' 2
draw FG parallel to OZ), and GK parallel to AB.
GF ( = DG) is the thrust along DO, and may be resolved at into
its vertical (= DA") and horizontal ( = GK) components ; the latter is
counteracted by an equal horizontal thrust from the strut OE. The
vertical component DK(—— )is borne by the king-post, so that the
W
total pull upon the king-post is — + P, P including the weight of the
struts and post.
Take CJ/to represent this pullpZw-s the weight at C, i.e., CM= W+ P.
Draw ML parallel to BC, and ZiV parallel to AB.
CLz=
W+ P 1
Sin a
is the thrust along the rafter from C to D.
CL + DG = C"^i— 4- -^V ~ — , is the thrust along the rafter from
\ 4 2 / sm a '
Dto^.
LNjf GK= P-i^+^\ cot ci, is the tension in the tie.
Note.— The king-post truss is the simplest and most economical
frame for spans of less than 30 ft. In larger spans, two or more sus-
penders may be introduced, or the truss otherwise modified.
■;
ROOF AND BRIDGE TRUSSES.
13
Collar-beams {^OK), queen-posts {DF^ J^O), braces, &c..., (Figs.
20, 21), may be employed to pre-
vent the defleetion of the rafters,
and the complexity of the truss
necessarily increases with the
span and with the weight to be
borne.
The stresses in the several
members are calculated in the
manner already described.
In I?ig. 21, e. g., it may
assumed that one-half of the
weight upon AC=z(lL\, is con-
centrated at D, and that the
component of this weight per-
pendicular to the rafter viz.,
W ...
Y • cos a, IS distributed be-
tween the collar-beam DE and the queen-post DF.
(9) Trusses for bridges and roofs (}f large span.
In bridges and roofs
of large span the rafters
or braces, AC, BC, are
very long, and the unsup-
ported distance between
A and 0, or B and 0, is
too great. The rafters
are therefore cut off at
D and E, (Fig. 22), a
straining piece DE is introduced, and tlie tic-btani AB is supported
at two points by queen-rods (or posts), DF and Ed.
Let W be the load upon each of the braces AD, BE,
Let P be the weight borne by each queen-rod.
The thrust along DE, the resultsmt weight upon EB, and the resul-
tant thrust at B evidently meet in the point A'.
Take KL to represent W, and draw LM parallel to EK.
LM is the horizontal thrust in DE, or the tension in Ali, due to W.
Take EX to represent /■*, and draw iVT' parallel to EB.
ET is the thrust in ED, or the tension in AB, due to P.
14 ROOF AND BRIDGE TKUSSES.
Hence, the total thrust in DE, or tension in AB, =:LM+ET
= (7w).co,„.
The values of P and W dopend upon tl»e character of the loadinpj.
If the load, includint; the wei};ht of the truss, is uniformly distributed
over the tie-beam, as in the case of a bridge, it may bo assumed to be
concentrated at the points A, F. G and li, in the proportions onr-hol/
of the load upon AF, one-half of tlie luad upon A(i, om-half (tf the load
upon BF, and oije-lalf of the load upon Bd', rospeetively.
Suppose the load to be un-
equally distributed, and that
a single weight W' is suspend-
ed from E. This produces an
upwaid pull at D, e(iual in
D^.agnitude to the downwa' d
pull at E.
The latter force may be neu-
tralized by an equal and opposite
force at E, or by a downward
W'
pull of at i>, together with
z
W'
an \ipward pull of — at E,
it
which is the manner in which
the beam AB neutralizes the
efiFect of the weight W .
The beam AB is therefore acted upon by two opposite forces, the one
at F^ the other at 6?, ejich being equal Xa Z- m. magnitude.
Let R be the reaction at A ; let AB ■=. I and FG — c.
. p, W \l^c W' l-e ' W , „ W c
• • R.l == — -o~ "T^" + — •' — 1-= — — • c, and R — — — . z.
2. I 2 2 2 2 1
The magnitude of the bending moment at F and G, at which pointe
. . . 1 , . . W' c I r
it 18 evidently a maximum, is . —
^ '212
The ordinates of the dotted lines, Fig. 24, are proportional to the
bending moments due to the two equal and opposite forces — FF being
the bending moment due to the upward pull at B, and GG' that due to
the downward pull at E.
Again, the rigidity of the truss may be increased by introducing
diagonal braces BG and EF, as shewn by the dotted lines in Fig. 22.
H" l-e
:
■
COMPOUKD ROOFS,
15
El £
2 'I
points
The shoaring force for points between /' and (i is ^ mAxinium whf n
^r> _ yy. _ II'. ^_^
the load is at /\)r G, itn vahie being ± — c -t- — = -+- y"" ~7~" ^ '^'^
force has to be transmitted to the horizontiil b^'ams throufjrh the uitidiuui
of a diajj^onal, which is iJli when the load is at /' and KF when the
load is at U. Thus, the nnijrnitude of the greati^'st thrust iu either
of the diai^oiials is, —r ' — r-,
^ 2 I (I '
s bein^ the K'nirth of a diagonal, and d the depth of the truss.
(10) (li'iicnil ruu'trks. — In the trasH«.» described in (</) and (t)
§ (8), and in i^ (1>), th^; vortical members are ties, i.e., are in tension,
and the inclined members are struts, /.»■., are in compression, liy
invertint^ the respective fiLrures another type of truss is obtained in
which the verticals are struts while the inclined members are ties,
lioth systems are widely ustd. and the mtiithod of calculatiuj,' the stresses
is precisely the same in each.
In desitniinir any particular member, allowance must be made for
every kind of stress to which it nuiy be subjected. The collar-beam
J)J'J in ri_u'. -1, for example, must be treated as a pillar subjected to
a thrust in the direction of its length at er^ch end ; if it carry tv
transverse load, its streiiirthas a beam, support<-Hi at the points DdwdJJ,
must also be determined. Similarly, the rafters AC, BC, Fig. m,
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 there-
fore practically sufBcient to assume that
the rai'ters are divided into a number of
short lengths, each of which carries 07ie-
/laZ/oi' the load betweeu the two adjacent
supports.
When a tie-beam is so long as to re-
quire to be spliced, allowance must be
made for the weakening effect of the splice.
(11) Compound Roofs. — In mansard
rov^fs and many other framings a number
of rafters are made to abut against each
other as in Fig 25.
The relative position of the rafters is
fixed by the condition that the horizontal
pressure upon each must be the same,
■ ■ ,•
/_;j;Vll.
-^K,
*
■■■"-'
'v>
..
Vk
' ^A
^
/I/
^-|f
.^A/
^^■^^■S^^E^H
1
■ " •/
/
/
'i ,
— In
riozs
16
GOTHIC ROOF TRUSS.
in order that the prcflsurc may be completely tranamittod from one
rafter to the next.
Let IK„ ir„ ir, ..., a„ a„ a,... be the wei}?ht3 and inclination.s to
the horizontal of the rafters AJi, B,C, CD... respectively.
Assume that the weights arc concentrr.ted at the points A, B, G
in the proportions ^, -^-^-^\ ^^^^\ , respectively.
The triangles BFA, BAG, CGII,... represent the forces at these
points.
Hence the horizontal pressure upon AB = -y- -^j, = -^-cotai = //»
The resultant thrust along BC= — L_^ — ?. --- = — I- — . -^-- —
and its horizontal component = ; . — ■. — — = Ha.
2 sin ia-i - «i)
Similarly the horizontal component of the thrust along CD
W'i + Wi ens a.2 cos a,
)
2
sill (fla -Ui)
= //j, and so on.
But III == Hi == //j . ...
W Wi + T^ cos tti. cos a-i
''2. '"' "' - 2 • sin (a, - a,)
TT,.
or —7-" coi a\ =
Wi -y W,
2
tan a-i — tan Uj
W^ + W3 COS 052 COS flj
2 * COS (aa ~ 02)
_ \V^ +_^,. 1_
2 tan 03 — <«'i <h
Hence, tan a-i == ( 2 + -jp-J.tan a,
/ 2 JF2 TTA
<fl?i ai) == ( 2 + — jTT + -rr^ J.tan a\
<a« 04 = ( 2 + vp= 1 . fan ai
&c. &c.
If F;= W.,= TF3..., .'. tanai^l-- tan n^=\-tana^ = - tan a,=
(12) Co^/iic Roof Truss.~T\\\» class of truss exerts an oblique
pressure upon the supporting walls, and the framing should be so
designed as to reduce the horizontal component of chis pressure to a
minimum.
The primary truss K D E L supports the weights at D, C,and E,
(Fig. 26).
The weight If upon AC, or BC, may be assumed to be concentrated
. W W W W
at the points A, £>, C, E, and B, in the proportions -j-, -^, -^-j "2",
d
GOTHIC KOOF TKUSS.
17
w
and -J-' rcspcctivoly. Hence
the total weij^ht uj)on the pri-
uiary truHs == -W. Trace the
stress (liairraui ; the fine oflomh
MN is evidently vertical ; take
MN » -^ . W i draw OM, ON,
parallel to DK, EL, respect-
ively.
The horizontal line Oil is the
thrust aldiijj; L)E, and OJ/is the
thrust aloni:; DK.
Draw or parallel to AD;
the sides of the trianule OTIl
represent the stresses in the
members AD, DF, and FA.
Also, since DF is equal and
parallel to AK, T is the middle
point of Mil.
The thrust along ^lA" =^ the
thrust along DF + portion of
weight concentrated at A
•trr
The secondary truss DCE carries a weight -^ at C
W
Take IIS ^= ^ and draw SO' parallel to CD ; SO' II is a triangle
of forces for one-half of the secondary truss.
W
Also,
OH -
Oil .
OH
an
IIS
4 2
: .J — = lu and the resultant thrust aloag DE is
1
OH.
1
CHAPTER II.
Roofs.
(1). — Ti/pcs of Truss. — A roof tru.«s may bo constructed of timber,
of irctii, or of timber and iron combined. Timber is almo.'it invariably
employed for small spans, but in tlie longer spans it has been larirely
superseded by iron, in consequence of the couibined liuhtness, 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 imprac-
ticable t<t make any further distinction than that which may be drawn
between those in which the reactions of the supports are vertical and
those in which they are inclined.
Fig. 1 is the simplest form of truss for spans of less than 30-ft.
Fig. 2 is a superior framing for spans of from iJO U) 4()-ft. ; it may
be still further strengthened by the introduct'on of struts, Figs. 3 and 4,
and with such modification has been employed to span opi'nings of 90
ft. It is safer, however, to limit the use of the type shewn by Fig. 3 to
spans of less than 60-fi,. Figs. 5, 6, 7, 8 and 9 are forms of truss suit-
able for spans of from 60 to 100-ft. and upwards.
Arched roofs (Figs. 8 and 9) admit of a great variety of treatment.
PRINCIPALS, PUHLINS, &C.
19
^
Thoy have a pieasiiiij^ appoaranco, and cover wide spans witliout inter-
mediate supports. The flatness of th<* areh is limited hy the reijuireuunt
of a uiininiuni thrust at the abutnu'nts. The tlirust may be resisttid
either by tliiekoniiijr the abutments or by intnulueintr a tie. If th(( only
load upon a roof-truss were its own weight, an arch in the form of au
inverted catenary, with a shallow rib, might be u.sed. But the action of
the wind induces oblicjue and transverse stresses, so tliat a considerable
depth of rib is generally needed. If the depth exceed 12-ins., it is better
to connect tho two flanties by braces than bv a solid web.
Koofs of wide span are occasionally carried by ordinary lattice girders.
( 2) J'riiicijxt/if, J'lir/i'ns, dr.
treatment.
The principal rafters in Figs. 1 to 7 are straight, abut against each
other at the peak, and are prevented by tie-rods Irom spreading at the
lieels. When made of iron, tee (T), rail, and channel (both single > — "
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 aniformly from heel
to ptiak.
Tlie heel is fitted intti a suitable cast-iron skew-back, or is fixed between
wroughtriron angle brackets, (F'igs. 10, 11, 12), and rests either directly
upon the wall or upon a wall plate.
When the span exceeds GO-ft., allowance should be made for alterations
i
20
PRIN'CIPALS, PURLINS, &C.
of leniith due to chariijesof tempcraturo. This may be effoctctl by inter-
posing'' a set of rollers between the skew-back and wall-plate at one heel,
or by fixing one heel to the wall and allowinj;- the ojjposite skew-back to
slide freely over a wall-jiiate.
The junction at the peak is made by means of a casting or wrought
iron plates, (Fig«. 13, 14, 15).
Light iron and timber beams as well as angle-irons are employed as
purlins. They are fixed to the top or sides of the rafters by brackets,
or lie between them in cast-iron shoes, (Figs. 1(J to 21), and are usually
held in place by rows of tie-rods, spaced at G or 8 ft. intervals between
peak and heel, running the whole length of the roof.
The sheathing boards and final metal or .^late covering is fa.stened upon
the purlins. The nature of the covering regulates the spacing of the
purlins, and the size of the purlins is governed by the distance between
the main rafters, which may vary from 4-ft. t^) ui)wards of 25-ft. Hut
when the interval between the rafters is so great as to cause an undue
deflection of the purlins, the latter should be trus^ d. Each purlin may
be 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. 22.
Struts are made of timber or iron. Timber struts are rectangular in
section. Wrought iron struts may consist of L irons, T-bars, or light
columns, while cast-iron may be emjtloyed for work of a more orncmental
character. The strut-heads :;re attached to the rafters by means of cast
caps, wrougbt-iron straps, brackets, &c., (Figs. 23 to 26), and the strut-
I
or ligbt
■
ROOF WEIGHTS.
21
iVet are easily desigiu'd both for pin and screw connections, (Fius. 27
to 30).
Ties may be of flat or roniid bars attached either by eyes and pins or
by !<crtw ends, and occasionally by rivets. (Fiirs. .11. ;-i2). Tlu' greatest
care is nectssary in projK'rly pro))ortioninii' the dinunKions ol' the (.yes
and ])ins to the stresses that eonie upon them.
To obtain Lireater security, cacli of the end paiu'lsol'a roof may be
provided with lateral ))raees. and wind-ties are often made to run the
whole length of the structure through the feet ol'tlie main struts.
Due allowance must hi' made in all eases for eliaiiu'es of temperature.
(;{ ) Riiiif )iiliihtx. — In ealeulatini!' the stres.>^es in tlie different mem-
bers of a HMif truss two kinds of load have to be dealt with, the one
jn'niiiniriif and the other <i<-ct<liiit<il.
The permanent load consists of the cnrrrlny, the j'nini in y, and arcu-
■))ii(fftfiinis of siiiiir.
Tabh'S at the end of the chapter sliew the \vei<ilits of various coverings
and frauiinirs.
The weight of freshly fallen snow may vary from 5 to 2()-lbs. per
22
WIND PRESSURE.
cubic ft. English and European engineers consider an allowance of 6
lbs. per gq. ft. suflScient for snow, but in cold climates, similar to that of
North America, it is pr<)bably unsafe to estimate this weight at less than
12-lbs. per sq. ft. .
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-lbs.
per S(j. ft. of surface perpendicular to the direction of blow. Ordinary
gales blow with a force of from 20 to 25-lbs., which may sometimes
rise to 84 or 35-lbs., and even to upwards of 50-lbs. during storms of
great severity. Pressures much greater than 50-lbs. 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 groat differences that exist
in all recorded wind pressures are primarily due to the unphilosophic,
unscientific, and unpractical character of the anemometers which give no
correct information either as to pressure or velocity. The inertia of
the moving parts, the transformation of velocities into pressures, and
the injudicious placing of the anemometer, w'aich renders it subject to
local currents, all tend to vitiate the results.
It would be practically absurd to base calculations upon the vio-
lence of a wind-gust, a tornado, or other similar phenomena, as it is
almost absolutely certain that a structure would not lie within its range
In foct, it may be assumed that a wind pressure of 40 lbs. per sq. ft.
upon a surface perpendicular to the direction of bbw 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.
(4) Wind prei^)inre upon inclined surfaces. — The pressure upon an
inclined surface may be obtained from the following formula, which was
experimentally deduced by Ilutton, viz :
!.«» .11/1 (I
P„ = P. ,s//( a
(A),
P being the intt'nsity of the wind pressure in lbs. per sq. ft. upon a
surface perpendicular to the direction of blow, and
J\ being the normal intensity upon a surface iuclined at an angle a
to the direction of blow.
Let P,„ P„ be the components of P,„ parallel and perpendicular ,
respectively, to the direction of blow.
•*• A == P„. sin a, and P„ == P., cos a.
i
DISTRIBUTION OF LOADS.
23
upon a
I
i
Hence, if the inclined surface is a roof, and if the wind blows hori-
zontally, a is the roofs pitch.
A};ain, let /' be the velocity of a fluid current in ft. per second, and
be that due to a head of /t-ft.
Let w be the weijrht of the fluid in lbs. per cubic ft.
lictj) be the pressure of the current in lbs. per stj. ft. upon a surface
perpendicular to its direction.
If the fluid, after strikinjr the surface, is free to escape at right
angles to its original direction.
w
.'. ;j = 2. /(. ?t' = —
g
Hence, for ordinary atmospheric air, since ir=:.08 lbs., approximately,
When the wind impinges upon a surface oblique to its direction, the
inU'nsity of the pressure is ( -^ — ] , v being the absolute impinging
velocity, and /s being the angle between the direction of blow and the
surface impinged upon.
Tables prepared from formula; A and B are given at the end of the
chapter.
(5) J)isfn'hii(!(in of Looih. — Engineers have been accustomed to
assume that the a^-eidental load is unif(trmly distributed over the wliftle
of the rot»f, and that it varies from HO to JlS-lbs. per sq.ft. of covered
surface for short spans, and from H.^ to 4(l-lbs. for spans of more than
60-ft. But the wind may blow on one side only, and although its direc-
tion is usually horizontal, it may occasionally be inclined at a consider-
able angle, and be even normal to a roof of high pitch. It is therefore
evident that the horizontal component (^j^) of the normal pressure (yj„)
should not be neglected, and it may cause a complete nitrsnf of stress in
members of the truss, especially if it is of the arched or bracecl type.
In the following examples it is a.ssumed that the wind blows on one
side only, and that the total load is concentrated at the panel jKtints (or
points of support) of the principal rafters.
Let «' be the permanent load per sq. ft. of roof surface.
„ j>„ „ normal wind pressure per sq. ft. of roof surface.
d „ distance in feet from centre to centre of principals.
/, and /jj be the lengths in ft. of two adjacent panels.
The total ^>iT7««Ht'«Moad at the panel point on the windward side
9
'sr.W. d.
1'
24
EXAMPLES.
The total accuhuttil load at the panel point on tlio windward side
^^ /'-
<l.
/, + /.
The total Inad at the e()rrts})iiii(linL: jtaiu'l point on the leeward wide
= W. t(. r—
2
T(» dt'tcrniine the total .stresses in the various uieud)ors ol' tlu' truss,
two methods may be pursued, viz. ; —
(</). The normal ))ressuri' ( />„) maybe resolved into its vertieal (/>,)
and horizontal (/'/,) eomponeiits ; j>,. may be eombined with the perman-
ent load and />,, dealt with independently ; the respeetive efleets are to be
added toiictiier.
(I)). The )»ernianent and aceidental loads may be treated separately,
and the respeetive effi'ets added toiiether.
It' / is the Uiii:th (d'a ratter, and it' A" is tlie reaction at the windward
sup|iort (hie to the horizontal jiressure of the wind, by takinu' moments
about tile leeward sujiport,
h .1 . /. i'ii>y II • y*„. .s/// (I. I. il . — =<>,orA = — /'„ — 7-' '.
Hence. A'' is inifnttrr and acts downwards ; it must be le.ss than the
reaction at the same su]>port due to the ju'ruuinent load and the vertieal
wiiul pressure, or the root' will blow over.
AV. 1. — Method ((/) aj»plietl to the roof tru.ss repre.sented in Fiir. 38-
EXAMPLES.
25
(1 side
itoly,
Fi<^. 34 is the stress iliap:ram for the vertical load upon the roof.
/> (j is the vertical reaction at H = -^- (>"«• '"«" + 2. cO-
yz', />/•, /•«. yw, ifjjrcscnt tlu! stresses in the nienihers to \vhi<'li they are
respectively parallel, viz., the rafter J /A and the ties HI), /)A. l>h\
Fi<;. 3,1 is the stress diai-rani for the load due to the liorizontal com-
ponent oi'tlu^ wind jiressure when there are no rollers under the heel ol'
either of the rafters.
The horizonta! reaction at each heel may be assumed to be one-half
of the total horizontal force of the wind, and e(jual tt) -■ "' ' .sin a.
, , . , , , . ,, }},, I. (I sill'' a
J) q IS the dowmimrd reaction at Ji= — .
4 ro.s ii
The stres.se8 in the ineud)ers AB, BD, DA, I)E, arc represented by
the lines q'r',j//, r's\ y/.s'. to which they are respectively parallel, and
are evidently reversed in kind.
Thus the total resultant compression in Ali = (jr - ij'r',
the total resultant tension in HI) — ^n- p'r', in DA = rs - r's',
in IJI'J = i>s-j,'s'.
Fi^. 30 is the stress diagram for the load due to the horizontal com-
ponent of the wind pressure, when rollers are placed at B.
The total horizontal reaction is now at f, and equal to y)„. h <1. sin <i.
, , • , , • ,. /'»• ^- </ •'*'"^ "
I) t \^{\w ilinmiriird W'AC on 7> = —
4 ''".s' 1 1
t' y' is the Iidrizniital force of the wind at Ii.
q' /•'. // /■', /•' s\ p s', are the stresses in AH, HP. DA, Dh\ respec-
tively, and the n'sultant stres.seH in the dift'erent members are q r-q' /
p i—j> /•', r s-r' s'. and /ts~p' s\ as ))efore. ,
Fig. 37 is tlie stress diagram for the load due to the horizontal com-
ponent of the wind pressure, when rollers are placed at <\
The total horizontal reaction of the wiiul is now at B, and equal to
^>„. f. (h sin ii.
The horizontal force of the wind at Bis . shni, so that the re.
sultant horizontal force at B is ".,-" i^ln >' = t' q'.
2* t IS the duicnicnfd reaction at B - — r— .
CUS u
26
EXAMPLES.
The resultant stresses are obtained as above.
The stress in I)E equilibrates the stresses in DA. BD, and also thoso
in A\l, EC, so that the stresses in DA, DB arc respectively e(iual to those
in EA, EC.
The dotted lines in the diagrams represent the stresses in the mem-
bers .1 C, E (\ E A, to which they are respeetivcly parallel.
Ex. 2. — Method (</) applied to the roof truss represented in Fij;. 33.
Fig 89 is the stress diat^rani for the vertical load upon the roof.
l.d
})q is the vertical reaction at i? = -— . (2, jy„. ma a + 3. ?/').
qo is the weight at F = --— . (2 />„. '•'>.s " + 2. ?'•)•
qi\ OS, sr, pr. sf, pf. represent the stresses in the members to wliicli
they are respectively parallel, viz.. BE. FA, DF, DB. DA, 1)E.
Fig. 4(» is the stress diagram for the load due to the horizontal component
of the wind pressure, when no rollers are placed either at B or C.
p„. I. d .fin- a
j>'/)i' is the downward reaction sit B = — ;j^ — cos^'
pn. I. d
m'(j' is the resultant horizontal force at B == . s*'«. a.
The line througli </' parallel to the rafter BA, passes through ;/, so
that there is no stress in BD from the horizontal wind pressure.
The strisses in the members BE, FA, DF, DA, DE, are represented
J
'
EXAMPLES.
27
I
i
by the lines, q'r\ s'n', p's', ,sV, p't', to which they are respectively
parallel, and are evidently reversed in kind.
Thus, the total resultant compression in IiF= </r-q'r', in FA^os -
a'li', in 1)F - sr - s'r , the total resultant tension in IW = ju- - o, in
IJA - sf - s'f\ in I)F -. pf - p't'.
Fijr 41 is the stress diagram for the load due to the horizontal compo-
nent of the wind pressure when rollers are placed at /i.
, , . . . , . ,. p„.l.'/ sin'' a
It t)t IS the vertical reaction at Ji = - — : — -.
' 4 cos a
m' a' is the resultant horizontal force at li - - " • P»- I- ''• •'"'" f'-
4
The total resultant stresses in HF, FA, DF, Dli, DA, I)F, are,
V ~ '/'■') "•*'■ ~ ■'*'"') ■''■'' ~ •''''■'j /"■ ~ /'''■'» •'"■' - ■"*'''? J'^ ~ y''' respectively.
The dotted lines in the diairranis represent the stresses in the ineiubers
CG, GA, GE, E(\ FA. to which they are respectively parallel.
]t(j', in Fig. 39, is the reaction at C --—.-• (/'„. (''>•">' '< + '^. "")•
g^'o' ia the weight at G =- -^. I. d. w.
Ex. 3. — Metliod {(i) applied to the roof truss represented in Fig. 42.
Fig. 43 is the stress diagram due to the vertical load upon the roof.
pq is tlic vertical reaction at B
r- P"-''"" " . (3. BIl + All), d + -.(4. BH + 2. AH ). d.
28
KXAMPLES.
qm is the weight ^i F - IliiiSH^^JLl^ . BII --= the wi-i-lit at //-- mn
Fiir. 41 i« the Htrcss cliaiinim duo to the hnrizoiital e<.»iui»onent of tlie
wiiul pressure, when rollers are placi'd at li.
/>„• I- 'f •*•'"'' "
//</ is the downward reaction at li -- — , — • •
■• 4 I'ns (I
o'q' is the horizontal f'oreo of tlii' wind at /» -- Jjll — '.'. /)/•"'.
'L
q'm' ~-- m'ii\ is the horizontal force of the wind at For //. and iscijnal
to Ji^f'jii!!_ii, nil.
2
The total resultant stresses in the menihers BF, FIT. 11 A. DF. I) IT.
DB, T)A. Die. are, y/' - '/'/•'. iiix - Ill's', lit - ii't' . sr -■ .s'/-', sf - .s'/',
jjt' —j/r', tr — f'r'. pr- />'i''. rcsjuctivdy.
E.C. 4. — Mt'tliod i^ii) ap|(li('d to the roof truss represented in Fit:'. 45.
Fig. 46 is the stress diagram due to the vertical load upon the roof.
jiq is the vertical reaction at7i= — -. (10. p„. cos a + 14. n-).
qm = mil = no, is the weight at 0, F, or Q, and is e(i[ual to
— . [j),,. cost! + n-).
Tf />„. rnn II =- y. tlie ;ioint o coincides with ^).
To avoid ambiguity it is gem-rally assumed that the stresses in IIO,
JIF, are ecjual to those in LQ, LF, respectively.
KXAMI'LES.
29
Fi«r. 47 is the stress diaurain duo to tho horizontal couiponoiit of the
wind pressure, wlieii rollers are placed at ('.
, „ . , , , . „ />,, /. '/ «///* '/
J) I IS tlie downward reaction at xj = , — , .
4 ra.v II
/' q' is the resultant horizontal force at B == -. j)„. I. <l. sin it.
•15.
I
q' III' == ni' )i' r-z ii' o'. is the horizontal force of the wind at (). /•'. or
Q, and is e«|ual to —^ — . sin it,
4
Ex. G. — Method ((/) applied to the roof truss represented in Fi^. 48,
Fijr. 41) is the stress diayraui for the vertical load upr^n the mof.
pq is the vertical reaction at B == -!,-^. (7. j>„ cos a + 1(». ir),
gm = 7)1 n = --— . (/>„. ros n + w), is the weight at F or //.
o
Fig. 50 is the stress diagram due to the horizont;il component of the
wind pressure when rollers are placed at B.
'£) I IS tlie downward reaction at li =- — - —
V q' is the horizontal force of the wind at B ==
cos a
p„. I. d .
, sin (I.
Ex. 7. — Method [h) applied to the roof truss represented in Fig. 51.
Assuming the direction of the wind to be horizontal, th>' load at any
joint of the bowstring roof in Fig. 51 due to wind may be taken to be
30
EXAMPLES.
the pressure upon a surface o(jual in area to the portion of the roof
supporttid by one panel or bay, and inclined at the same anf];le to the
horizontal as the tanjirent to the curve of the roof at the joint in question.
Let It be the resultant of the load /*, , 1\, I\, I\, at the joints B, d, c,
/, and let it make an anj^'le tt with the horizontal.
Fig. 5li is the stress diagram due to the normal wind pressure, when
rollers are placed at C.
The whole of the horizontal reaction (= K. cos f>) is at B, and is
represented by mn.
nt is the vertical reaction at B.
ts, sr, rq, qp^ represent the loads />,, j9j, ^^3, p^^ respectively.
«(?', rV §2', p3', are the stresses in Bd, de, c/,/A
md',me',m/\ma', " '' 51,12,23,30
<fl', e'2',/3', aV, " " d\,e2,/S,A0 "
The stress diagrams for the normal wind pressure when rollers are
placed at B, and also for the permanent load, may be easily drawn as
already indicated.
(6). — Anali/sis. The results in the preceding examples may be easily
verified analytically.
To determine, for example, the stresses in the members of the truss
in Ex. 2 due to the vertical load.
-^ . (2 jo„ cos a + 3. w). (1)
Let R be the vertical reaction at B
W
weight at F = ~. (2 p„. cos a + 2. to).
4
(2)
' C,, C„ C3, be the compressive stresses iu BF, FA, DF, respectively.
' r„ T„ r,, " " tensile " " DB, DA, DE " "
' the angle ABD == /3 == the angle DAB ; .-. ADE = a + /3.
EXAMPLES.
31
Resolve tlie forces at li porpendicular to the rafter,
,' , R, ms a = 7',. sin $.
Resolve the forces at li perpendicular to the tie DB,
.'. Ji. cos « - ^ = (\. sin /8
Resolve the forces at F parallel to the rafter,
, ■ . ^^1 — \y. sin (I = t\
Resolve the forces at F perpendicular to the rafter,
.-. W. i-osn = (\
Resolve the forces at D perpendicular to the tie DE^
••• 2\. sin a - )8 + 6',. cos a = T.^. sin « + /8
= R. "]'^'^ . sin a - )3+ (\. cos <t. by (3) i
silt & ^ J V / f
(3)
(1)
(7)
Resolve the forces at D perpendicular to the tie DA,
•*• T^. sin 2 i3 — C;,. cos j8 == 'J\. sin ITT^
= 2. R. cos a, cos /3 - ^3. cos d. by (3) (8)
From these equations, C,, C.^, C^, T,, 7'j,, 1\, maybe found in terms
ofji„ and u\
The value of y, may be verified by considering a section of the roof
made by a plane a little on one side of A.
Let h be the distance of BE from .1 ; take moments about ^4.
•*• T.^. h = R. h cos a — II'. -. cos a
„ I sin (I + fi I
or 7,. -. -— =Jt.Lcos a — IV. -. cos a
^ 2 cos /3 2
. • . jT,. sin (d + P) =: 2. R, cos a. cos /3 - W. cos a. cos j8, which agrees
with 8, for C^ = W. cos a by (6)
^1
32
TABLES.
WEIGHTS OF VARIOUS ROOF FRAMINGS.
Description of Roof.
Pont
Coiniiion Tnipsi
liu
(lu
do
Coiiinion Truss
do
do
^^—.j. Timber
|RSnrnfteri< ami
JSg5|»'truts, iron
ties.
Location.
Cover-
ing.
Ijiverpoo!
Docks
Felt
Liverpool
Docks
do
do
do
Zinc
Zinc
Zinc
Slates
Manchester..
Litne Street..
Birtninghain
Strasburg. ..
Paris
Dublin
Derby
Sydenham.. .
do
St. Pan eras.
Cremorne.,..
Span.
lAVelf^UiTi Til's]
Width i i'<'''«i. n.(,f
,)£ I covered .-irea.
Bays. 7^ 7, 1
•' il" ram- Cover-!
Pitch.
mg.
4.(1
WA\"\ 12M)"
50^0"j lO^U" ;^.o
53^3"! 11^0" 2.085
o4^o"i i4^o"i !).r)i
uo^o"i (/.(i" II (ji
72^0"|2U^()" 7.0'
ing.
o.OC 30
c2^o"
7(i-'.0"
79^0"
80^8"
90^2"
84^0"
1 00^0"
130^0"
50^0"
154^0"
211^0"
97^0"
153^0"
4F.0"
8F.6"
120^0"
72^0"
240^0"
45^0"
12^0" 3.01.3; 5. GO
25^0"
13^0"
IF. 8"
20^0"
9^
14^
20 ^
W
2r/.
24^
13^
20^
16^
24^
29^
I-/,
30°
2.01 7.72 .SO"
3.80 5.42 30°
-^•'2' 12.1 20^34
13. G
8.D
7.0
0.4
9.0
4.9
ll.O
12.0
15.0
10.7
16.8
11.8
U.3
24.5
11.5
TABLES.
33
WEIGHTS OF VARIOUS ROOF COVERINGS IN LBS. PER SQ. FT.
Thatch 6.5
Flit, a.sphalted 3 to .4
Tin 7 to 1.25
Hheet lead 5 to 8
Sheet zinc 1.25 to 2.(1
Copper 8 to 1.25
Sheet iron, (1-16 in.
thick) 3.i»
Sheet in)n, (corruijated) 3.4
Cast-iron plates, (f-in.
thick) 15.0
Sheet iron, (16 W.G.),
and laths 5.0
iron
and
Corruirated
lath.s 5.5
Slates, (ordinary) 5 to 9
Slates, (large) 9 to 11
Tiles
Pantiles
Slates and iron laths
Boardinir, (.f-in. thick)
Boarding and ^heet
iron, (20 W. G.)
Timbering of tiled and
slated roof's, addi-
7 to 20
6 to 10
10.0
2.5
6.5
tional 5.5 to 6.5
Table of the values of P,,, F,. J\ in lbs. per scj. ft. of surface, when
i^=40, as determined by the formula P„ = P.sln ^'S*"'""-!
Pitch of roof.
P.
1\
Pn
5"
5.0
4.9
.4
10"
9.7
9.6
1.7
20"
18.1
17.0
6.2
30"
26.4
22.8
13.2
40"
33.3
25.5
21.4
5<l"
38.1
24.5
29.2
60"
40.0
2(1.0
34.0
70"
41.0
14.0
38.5
80"
40.4
7.0
39.8
90"
40.0
0.0
40.0
Tabic prepared from the formula, ^j =
20
Vt'lDcitic in feet
Velocities in miles
Pressure in lbs. per
per tecond.
per hour.
sq. ft.
10
6.8
.25
20
13.6
1.00
40
27.2
4.00
60
40.8
9.00
70
47.6
12.25
80
54.4
16.00
90
61.2
20.25
100
68.0
25.00
110
74.8
30.25
120
81.6
36.00
130
88.4
42.25
150
102.0
56.25
1'^
I
EXAMPLES.
(1). If the pole of a funicular polygon describe a straight line, shew
that the corresponding sides of successive funicular polygons with respect
to successive positions of the pole will intersect in a straight line which
is parallel to the locus of the pole.
(2). 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, shew that the
tangents of the angles which successive bars make with the horizontal
are in arithmetic progression.
(3). If an even number of bars of equal length and weight rest in
equilibrium in the form of an arcli, and if a,, (i.i,...a„, be the respective
angles of inclination to the horizon of the 1st, 2nd, n"* bars count-
ing from the top, shew that tcui a„_^.^ = .tl ■ tan a„.
2 /i - 1
(4). Four burs of equal weight and length, freely articulated at the
extremities, form a scjuare ABCJJ. 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 /i and I). If H^be
the weight of each bar, shew ((/).-that the stress at C is horizontal and
=~2~j ('>)-that the stress on BC Sit B is ir.-y and makes with the
, (r).-that the stress on ^^ at B
vertical an angle t<in'^
18
Vi '•{ . . 3
W. — :j- and makes with the vertical an angle ^nr'-s", (rf). -that the
5
stressupon jliiat J. is-— .ir, (<?).-that the tension of the string is 2. W
(5). 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 H and E. If the weight of each bar be IK, shew that the
W
tension of the strmg is ——.(jan 54° + 3. tan. 18"), and find the magni-
tudes and directions of the stresses at the joints.
(6). Six bars of etjual length and weight (= W), freely articulated at
the extremities, form a regular hexagon.
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 J)E, shew
that, (</).-the tension of the string is 3 If, (6) .-the stress at C is ..
2.\/3
EXAMPLES.
35
and horizontal, (r).-the !,trcP8 at D is ir.V- and niakos with the vertical
an angle tan'^ t4'h-
Secoinl, if the system rest in a vertical plane, the bar DEXw'wv^ fixed
in a horizontal position, and the form of the hexairon being preserved by
means of a string connecting the joints C and F, shew that, (f^.-the
t(!nsion of the string is W. va, (i),-the stress at C is W.yJ'-[\&\\A u akes
with CBan angle *'" v ji^;, (f)--the stress at £ is W.yJ !_ and makes
with CD an angle sin-^ ^/^
Thinly if the system hang in a vertical plane, tlie joint A being fixed,
and the furm of tiie hexagon being preserved by means of strings
connecting A with the joints E, JJ, and (7, shew that. (r/).-the tension of
each of tlit; strings AE and AC is \V. VJ, (/>).-the tension of the string
AD is 2 ir, and determine the magnitudes and directions of the stresses
at the joints, assuming that the strings are connected with pins distinct
from the bars.
(7). Shew that the stresses at C and F in the first case of Ex. 6
remain horizontal when the bars AF, FE, IU\ CI), are replaced by any
others whieh are all ecjually inclined to the horizon.
(8). An ordinary jib-crane (Kig. 7, Chap. I) is required to lift a
weight of lO-tons at a horizontal distance of (i-ft. from the axis of the
post. The post is a hollow cast-iron cylinder of 10-ins. external diar. ;
find its thickne><s, assuming the safe tensile and compressive stress to be
H-tons per scj. in.
The hanging j)art of the chain is '\\\ four falls ; the jib is 15-ft. long,
and the t^ip of the post is lU-ft. above ground ; find the stresses in the
jib and tie when the chain passes (l).-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.
(9). If the post in the preceding question were replaced by a solid
cylindrical wrought-iron post, what should its diar.be; the safe inch-
stress being 3 tons as before ?
(10). The horizontal traces of the two back-stays of a dvirrick-crane
are* and y ft. in length, and the angle between them is a ; shew that
'•OS. (0-0) X
being the
the stress in the post is a maximum when
"^ cos I) y
angle between the trace ;r and the plane of the jib and tie.
('11). The Fig. represents a crane employed in the construction of the
staging for the Holyhead break- water. \t>
principal members are two wrought-iron I
rods ABC, supported at H upon two timber
uprights BD, and connected at A and Cwithj
two timber longitudinals ADE. The crane rests upon a truck, and the
36
EXAMPLES.
centre of pressure is maiiitainecl at /> by iiieans of ;v movalile balanee-
l)ox V)etweeii I) aial A'. 'I'lie crane is recjuired to lift a weight of 2-tons.
deteruiine the stresses in the different members. (Data: — AK—M
ft.. AI)^'.V,)-{\., D('='l-2-\\., BD^Mt., vertical distance between J
and/;-7ift0
Point out a method by which the members may be efiectively braced
together.
(12). The inner flantre of a bent crane. (Fig. 10. chap. T), forms a
(jnadrant of a circle of 2(>-ft. radius, and is divided into/oKr e(|ual bays.
The iHifir flange forms the segment of a circle of 2.")-ft. radius. The
two flanges are o-ft. apart at the foot, and are stru(,'k from centres in
the same liorizuntal line. The bracing consists of a st'ries of isosceles
triangles, of which the bases are the eipial l)ays of the inner flange. The
crane is required to lift a weight of lO-tons, determine the stresses in
all the members.
(];}). A braced semi-arch is 10-ft. dei>p at the wall and ]>rojects 40-ft.
The upper flange is horizontal, is divided into /'**//• eijual bays and
carries a uniformly distributed load of 4n-tons. 'I'he lower flange form-!
the segment of a circle of lU4-ft. radius. The bracing consists of a
series of isosceles triangles of whica the bases are the e(|ual bays of the
upper flang(>. |)etermine the stresses in all the meud).'rs.
(14). Three bars, freely articulatt'd. form an e(piilati'ral trianglt!
AJi(\ Tile system rests in a vertical plane upon suj»port< at /i ami (^
in the same horizontal line, and a weight IT is suspended from .1,
determine the stress in Ii(\ neglecting the weight of the bars.
(15). Three bars, freely articulated, iorm a triangle . I /^<^. and the
system is kept in c((uilibrium 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
the forces meet. ('*).-in the centroid. (A).-in the orthoeentre of the
triangle ?
(16). A triangular truss of white pine consi.-^tsof two c(iual raftiTS ..J li,
AC, and a tie-beam B(^ ; the span of the truss is l{l>-ft. and its rise is
7^-ft. ; the uniformly distribute I load upm each raft.M* is S4<K)-lbs , and
•A weight of 1(1.0(1(1 lbs. is suspenth'd from the centre of the tie-beam ;
det^irmine the dimensions of the raiters and tie-beam, assuming the safe
tensile and compressive inch stresses to be 3'{0U and 2700-lbs., respec-
tively.
(17). The beam J /? is sup-
ported at /i ami strengthened
by a tie or strut ( '/) ; a
weight ir is suspended from
^1 ; .shew how to determine
the dimeusious of the beam
and brace.
FA'AMPLES.
37
(18). A triangular truss consists of two equal raftors aP, AC, and
a tit'-bcam B(\ all of white pine ; the centre I) of the tit'-beam is
supported from ^1 by a wrcu^rht-iroii rod A/); tln' uiiifurnily distributed
load upon each rafter is 84()()-lbs. and upon the tie-beam is 36,(MM)-lbs ;
di'tcrniinc the dimensions of the different members, BC being 40-ft. and
A/) 20-ft.
What will be the effect upon the several members if the centre of the
tie-beam be supported upon a wall, and if for ihc rod a post be .substituted
against which the lieads of the rafters can rest ?
(19). A triangular truss of white jiitie consists of a rafter A(\ a
vertical post vl/i, and a horizontal tic-beam AC; the load upon the
rafttr is 300-lbs. per lineal ft.; J6' == 30-ft., .1/^ == 6-ft. ; find the
resultant pressure at (^.
How nuich .strength will be gained if the centre of the raft r be
supported by a strut from /i ?
(20). The rafterH of a roof are 20-ft. long, and inclined to the vertical
at IJO" ; the feet of the rafters are tied by two rods which meet under
the vertex and an; tied to it by a rod 5-ft. long ; the ro(d' is loaded with
a weight of 35U0-lbs. at the vertex, determine the stres.ses in all the
members.
(21). The feet of the equal roof rafters .1/^, .1^', are tied by rods BI),
CI), which meet under the vertex and are joined to it by a rod AD.
If li'and W are the distributed loads in lbs. upon the rafters, and if
•s is the span of the roof in it., shew that the weight of metal in the ties
5 ir+ w
in lbs. is -. -. . s. c(tt /s. / being the safe inch-stress in lbs., and fi
the angle ABD.
(22). TheraftersylT?, .ir. in the Fig. are20-ft.,
long, and are sui)ported at the centres by the struts
l)K, l)F ; the centre I) of the tie-beam B(^ is sup-
ported by a tie-rod AD, 10-ft. long; the uniformly
distributed load upon AB is80()0-lbs.. and ui)on AC\
is 2400-lbs. ; determine tlu' .^tresses in all the members.
What will be the effect upon the several mend>ers if ^ifl be subjected
to a horizontal pres: ure of 156-lbs. per lineal ft. ?
(23). Determine the dimensions of all the members of the truss in
the preceding question, assuming the tie-beam to be also loaded with
a weight of 12tM)-lbs. per lineal ft.
(24). A roof truss consists of two equal rafters inclined at 60° to the
vertical, of a horizontal tie-beam of length /, of a collar-beam of length
n, and of a queen-post at each end of the collar-beam ; the truss is loaded
with a weight of 2600-lbs. at tlie vertex, a weight of 4000-lbs. at one
collar-beam joint, a weight of 1200-lb8. at the other, and a weight of
38
EXAMPLES.
13,500-lbs. at the foot of each quci'ii ; doteruiiiie tlic stress in each
inoiuber.
(25). Tho platform of a bridj^o for a cloar span of I
60-ft. is carried by two tiuibcr trusses 15-ft. deep,
and of the form shewn in the Fij;. ; the load upon the I
bridge is 5n-lbs. per sq. fi. of platform, which is 12- 1
ft. wide ; what should be the diuieiisions of tho ditiereut members,
assumiiifz; the workiuij; stress to be 4('(l-lbs. per sq. in. ?
If a single load of tiOdO-lbs. pass over the bridjje, and if its effect be
equally divided between the trusses, find the greatest stress produced in
the diagonals ^1.1, BB.
(26). A frame is composed of a horizontal top beam 40-ft. long,
two vertical struts 8-ft. long, and three tie-rods of which the middle one
is horizontal and 15 ft. long; liiid the greatest stress produced in the
several members when a single load of 12.(l0<)-lbs. passes ov.t th e truss.
(27). The truss represented by the Fig. is conqioscd
of two e(|ual ratters, A B.AC fu'.r equal tie-rods A I).
BIJ, AE, C/'J, and the horizontal tie DE.
If ir be the distributed load upon AB, IP that
upon AC, s the span, $ the angle ABD, and/ the safe ineli-stre.-s in the
5 \V + W
metal, shew that the total weight of the tie-rods is —.
/
s. cot $.
GO" - 1
(28). If the rafters in the preceding Question are inclined to tho
horizontal at an angle a, and if the tension on BD or EC is equal
to that on DE, shew that /3 =- «, (t. e., D and E must fall in the hori-
zontal line BC), or ^ ■
Petermine the stress in ^D corresponding to the latter value of/?.
(29 1. The trusses for a roof of the same type as that in Question 27
are 12-ft. centre to centre, the span is 40 ft., the horizontal tie is 16-ft.
long, the rafters are inclined at OlC'to the vertical, the deadweight of the
roof including snow, is estimated at 10-lbs. per sq. ft. of roof surface,
there are no rollers either at B or C ; determine the stress in each mem-
ber wlien a wind blows horizontally o.i one side with a force of 40-lbs.
per sq. ft. of vertical surface.
(30). The accompanying truss is similar t(> that
in question 27 with the addition of the struts l)F,
EG.
(a). — With the same data as in question 27 shew |
f> a W+(W^ W). cos 'a
that the weight of the metal in the tie rods is -• - • . ,
b y • s(« /3- c"s j3
and is a minimum when tan'' )3 == 2 +
W
EXAMPLES.
39
an
d is a miuimuui when tan'^fi^
(h). If there arc no rollers either at B or at C, and if a wind blows
horizontally upon .1 /i, tshew that the horizontal component of it^ pres-
sure upon AB produces no stress in the rod Bl).
(c). In a given roof, the rafters are of pitch-pine, the tie rods of
wroughtr-irou, the sjiun is tiO-ft., the trusses are 12-i't. ecntre to centre,
DF== 5-ft. == J'JG, tlie anj;le.4/i6' =- 30", the dead vvei-ht of the roof,
including snow, is 9-lhs. per sq. ft. of roof surface, rollers are jilaced at
C, a single weight of HllOO-lbs. is suspended from /', :ind the rool" is also
designed to resist a normal wind pressure of U(i.4 lbs. per sq. it. of roof
surface on one side; determine the dimensions of the several members.
(31). The accompanying truss is similar to that
in Question 27 with the addition of the/owr equal
struts I)F. nil. AY/, EK.
(((). — With th(; same data as in question 27 shew |
that the weight of the metal in the tie-rods is
5_s_ 4 W+'d. (W+ W). cos'^
18 /' cos fi. sin fi '
7_ 3 W'
4 ^i"W'
(6). — In a truss of this type adopted for the roof of a shed at Birk-
enhead, the riifteri are of piich pine, are rectangular in section, have a
constant breadtli of U|^-ins., and a depth which gradually diminishes from
13-ins. at the heel to It^-ins. at the peak ; the struts, of which 1>F and
EG are vertical, are also of pitch pine, are rectangular in section, and
have a scantlini: of Si-ins. by Gi-ins. ; the tit-rods are of wroui;ht-iron
and AD == BJJ -= AE == CE -= 23-ft. ; BD and EC are round 'l}i-\n.
bars, AD9.\\i\ AE are round H-in. bars, DE is a round 1^-in. bar;
the angle ABC =- 30" ; the span == 7!)-ft. ; the trusses are 13 ft. centre
to centre ; the heel B is free to slide on a smooth wall-plate ; the dead
weight of the roof, including snow, is 8-lbs. per sq. ft. of roof surface ;
determine the stress per unit of area to which each member is subjected
when the wind blows horizontally witi a force of 40-lb8. per sq. ft. of
verti'.'al surface, (1 — upon the side AB, (2) — upon the side^lf.
(32). The Fig. represents the form of truss recent-
ly ado})ted for the roof of a shed at Liverpool. The
rafters are of pitch pine, are rectangular in section,
have a depth of 14-ins., a breadth of 8-ins., are I
inclined at GO" to the vertical, and are each supported
at 5 equidistant points ; the six struts are also of pitch pine and reetau
gular in section ; 1)F ViW A EC are each 10-ft. G-ins. long, and liave a
scantling of 12-ins. by 8-ins. ; the remaining four struts have a .scant-
lings of 8-ins. by 8-ins. ; the tie-rods are of wrough^iron ; BX and
CO are round 2§-in. bars, ND and OE are round 2|-in. bars> Dl*
and EQ are round Ig-in. bars, FA and QA are round 2;^ in. bars,
FN, FF, GQ, and GO are round 1^-in. bars, DE is a round l.f in.
bar ; the trusses are 20-11. centre to centre, and the weight of a bay of
40
EXAMPLES.
the roof is 24,41 (i-lbf^ ; tho span is 90 ft.4-ins. ; the heel li is free to
slide upon a sinonth wail-plat(^ ; dcteniiiiie the stress per unit of area in
eaeli uunilur when the wind prodiiees a normal pressure of 2'>.4-lbs.
pcrsq. ft. of roof surface, (l)-upon the side .1/^, (2)-upon the side AC.
(.33). Make an alUsmative desi<rn for the truss in the preceding
question, usin^i iron instead of timber for the rafters and struts.
(34). Tlie trusses for a roof are to be 2r)-ft. centre
to centre, and of the tvpe represented l)y the Fijr. ; the i
span />V'=- l(l()-ft., tiie an.ule A/U'--:W<\ a sinjrle
weijiht of lOOO-lbs. is to be suspended from //. and one I
of 2(HM(-lbs. from 0^ rollers are placed at (', the roof is|
to be designed to resist a horizontal wind pressure on one side i>f40-lbs.
per sq. ft. of vertical surface, the ties are to be of wrouudit-iron. the
rafters and struts of timber or wrought-iron ; determine the dimensions
of the different members.
(35). The domiul roof of a ga^
holder ibr a clear span of HO ft. is I
strengthened by secondary and primary
trussing as in the Fig. The points /i\
and 6' are connected by the tie /^/V passing beneatli the central strut
AP, 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 secondarv truss ABEF,
consists of the e(|ual bays All, HG, (Hi, the tven BE, Et\ FA, of which
BE is horizontal and the struts d'E, FIf, which are each 2 ft. (5 in.
long, and are parallel to the radius to the centre of Gil; the secondary
truss ACLK'is similar to ABEF; when the holder is empty the weight
support^'d by the truss is 3fi,fl00-lb>;., which may be asumed to be con-
centrated at G, II, A, JA N, in the proportions.'HOOO, 4(100. 1000, 4000
and 8000-lbs., respectively; determine the stresses in the different mem-
bers of the tru.ss.
Should the braces HE and ML be introduced ? Why ?
(3G). Dcisign the bowstring principals for the following roofs —
(a). Span ■= 50-ft., rise == 10-ft., depth of truss at centre == 5-ft.,
distance between principals =r U-ft.
(/>). Span=n 150-ft., rise =:, 30-ft., depth of truss at centre == 12-ft.,
distance between principals -= 20-ft.
(c). Span == 200-ft., rise == 40-ft , depth of truss at centre == 20-ft.,
distance between principals == 20-ft.
{<!). Span == 100-ft., rise==25-ft.
(37). — A funicular polygon is drawn for a series of parallel loads at
given distances ; shew that, {<i).-hy properly drawing the closing line of
the polygon a bending moment curve is obtained which corresponds to
any position of the series of loads on at»y given beam, (i).-so long as
the closing line lies on the same two polygon sides, its positions for any
given beam form the envelope of a parabola, (c),-the centre of the
i' r
EXAMPLES.
41
i; I
hoam cnrrcpponflinf: to a jrivon olnsin.tr line bisects the distance between
tlie verticals through the intersection of the polygon sitles and the point
where the closing line touches the parabola.
(38). Vertical loads of 4, 3. 7 and 'J-tonc are concentrated upon a hori-
zontal beam of2(l-ft. sj.an at distances of .5, 7. 12. and ir)-ft.. respectively,
from the left su]>]iort ; prove uenerally that tlie vertical ordinate inter-
cepted between a point on the corresponding I'unicular polygon and a dos-
ing lino wlio.se hori'/ontal projection is the span of the beam, represents on
a certain determined scale the bending moment of a section at the ]ioint ;
find its value by scale measurement f(«r a section at !>-it. from the left
support, using the following .scales :-for hiKjths, ^-inch^ 1-foot ; \'oy forces,
^-inch == 1-ton ; the jxilar distance == 5 tons.
Determine grapliieally by means of the same diagram the greatest bend-
ing moment that can be ])roduced on the same section by the same series
of loads travelling over the span at the stated distances apart.
(89). The inclined bars of the trapezoidal trus.s repre-
sented by the Fig. make angles of 45" with the horizontal ; a
load of 10 tons is applied at the top joint of the left rafter in
a direction of 45" with the vertical ; draw a frame diagram
and determine graphically the two reactions, assuming' the
one at the right to be vertical ; aKo, by means of a recipro-
cal figure, find the stresses in all the pieces of the frame.
Ex))lain in what manner the right hand reaction may be made approx-
imately vertical.
CHAPTER III.
Bridges.
(1). — CliisHijicatiun. — Bri(lij;i!a may be dividi'd intti tliroe general
classes, viz. ; (I). — Bridge.s with horizontal girders, (2). — Suspension
bridges, (3). — Arched bridges.
The present chapter will treat of bridges of the 1st class oidy. In
these the platforiu is supported by two or more main girders resting upon
two or more supports, and exerting thereon pressures which are vertical
or very nearly so.
The neutral line (or surface) divides the girders into two parts, of
which one is stretched and the other compressed.
(2). — Comparative advuntages of curved and horizontal J(anges. —
Jiittle. if any, advantage is gained by varying the depth of a girder, ana
the practice is open to many grave objections. The curved or parabolic
forui is not well suited to plate construction, and a diminution in depth
lessens the resistance of the girder to distortion. Again, if the Vjottom
flange is curved, the bracing lor 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.
The depth is sometimes varied for the sake of appearance, but, gene-
rally speaking, the best and most economical form of girder is that in
which the depth is uniform througlunit, and in which the necessary
thickness of flange at any point is obtained by increasing the number of
plates.
(3). — Depth of girder, or trnas. — The depth may vary from ^'^th to
Tjth of the span, and even more, the tendency at present being to approx-
imate to the higlier limit. If the depth fall below -j-'ath of the span,
the deflection of the girder bec(unes a serious consideration.
The depth should not be mon; than about Ih times the width of the
bridge, and is therefore limited to ^-i-ft. for a single and to 40-ft, for a
double track bridge.
(4). — Position of platform. — The platform may be supported either
at the top or bottom flanges, or in some intermediate position. It is
POSITION OF PLATFORM.
43
cliiirncd in favour of the last tliat tho main ^jinlers may be braced to-
gether below the iilatform ( Fii;. 1), while the upper portions serve a.s
parapets or {guards, and also that tiie vibration comumuicated by a
passing train is diminished. The position, hoW(!ver, is not conducive to
rij^idity, and a large amount of metal is required to form the connections.
The method of supporting the platform by the top flanges, (Fig. 2),
renders the wh(>le depth of the girder available for braeing, and is best
adapU'd to girders of shallow depth. Heavy cross girders njay be
entirely dispensed with in the case of a single track bridge, and tlie load
most effectively distributed, by laying the rails directly upon the flanges
and vertically above the neutral line. Provision may be made for side
spaces by employing sufliciently long cross-girders, or by means of sliort
cantilevers fixed U) the flanges, the advantage of the former arrangement
being that it increases the resistance to lateral flexure, and gives the
platform more elasticity.
Figs. 3, 4, 5, shew the cross girders attached to the bottom flanges,
and the desirability of this mode of support increases with ths depth of
the main girders, of which the centres of gravity should be as low as pos-
sible. If the cross-girders are suspended by hangers or bolts below the
flanges, (Fig. 5), the depth, and therefore the resistance to flexure, is
increased.
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 cros-t-girders may be
44
NUMBER OF MAIN GIRDEUR.
prf)lon^'e(l on cacli side and the end jc»iiK'd to the top flanircs by suitable
barH.
When the depth of the main girders is more than about 5-ft., the top
flanges should b • braced together. But the minimum clear head-way
over the rails is 10-ft., so that some other method slumld be adopted for
the support of the platform, when the depth of the main girders is more
than 5-ft. and less tlian IG-ft.
Assunu' that the depth of the platform below th(^ flanges is 2-ft., and
that the dejith of the transverse bracing at the top is 1-ft. ; the total
flmiti>it/ depths are 7-ft. andl9-ft., and if J is taken as a mean ratio of
the depth to the span, the corresponding limiting spans are 5G-ft. and
152-ft.
(5). — Comparative advantagps of two, three and four main girders.
— A bridge is generally constructed with two main t^irders, but if it is
crossed by a double track a third is occasionally added, and sometimes
each track is carried by tiro independent girders.
The tiro-girder system, however, is to be preferred, as the rails, by
such an arrangement, may be continued over the bridge without devia-
tion at the approaches, and a large amount of material is economized, even
taking into consideration the increased weight of long cross-girders.
The employment of four indepesident girders possesses the one great
advantage of facilitating the maintenance of the bridge, as one-lialf may
be closed for repairs withoi,t interrupting the trafiic. On the other liand,
the rails at the approaclios umst deviate from the main lines in order to
enter the bridge, the 'vidth of the bridge is much increased, and far more
material is required in its construction.
nillDGE LOADS.
45
Fow, if any, reasons can bo uiLri'tl in favour of tho introduction of
ii third intonucdiato j^nrdcr, since it presents all the object ionahle features
of the last system without any corresponding ncouunendatinn.
(0). — Bridge hmds. — In order to deteruiino 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 wliieh the bridjio
may be subjected. The load is partly tlrml^ partly Hn\ and depends
upon the type of truss, the span, the number of tracks, and a varii'ty of
other conditions.
The dead load increases with the s| an, and embraces the weight of
the main girders (or trusses), cross-girders, platform, rails, ballast, and
accumulations of snow.
A table at the end of the chapter gives the dead load which may be
adopted as a first approximation in calculating the strength and
designing the parts of a bridge.
(7). — Lirr load. — The live load due to a passing train is by no
means unifnmdy distributed, but consists of a nund>er of diH'erent
weights concentrated at definite points. This variable distribution is of
little importsnce for spans of more than 100-ft., as the ratio of the dead
to the live load then becomes very large, but it is nf vital moment when
the span is short.
The extreme weight of a locomotive upon its three drivers, with a
wheel-base rarely exceeding 12-ft., variiis from 72,000 to ^^4,tl0t( lbs.,
and the maximum weight u]»on one pair of drivers may be taken at
:-iO,()00-lbs.
Thus, 30,000-lbs. may bo concentrated at the centre of spans of 12-ft.
and under. This is etpiivalent to (50.(<00-lbs., uniformly distributed, /.*'.,
to 12,000-lbs. for a 5-11. span, and to 6000-lbs for a 10-ft. sj.an, per
lineal ft.
If the locomotive is placed centrally upon spans of 15, 20, 25, and
30-ft.. and if it is assumed that its weight ( == 84,000-lhs.) is equally
distributed between the drivers, the bending moments at the centre are,
147,000, 252,000, 357,000, and 402,000 ft.-lbs., respectively. The
g
equivalent uniformly distributed loads per lineal ft. are, 147,000 X —^^
(==52205-lbs.), 252,000 x _(- 5040-lbs.), 357,000 x A.^(..4569a
lbs), and 462,000 X -- ( == 4106§-lbs.)
iifl
46
CHORD AND WEC STRESSES.
Two locomotives at the head of a train will cover a span of GO-ft., and
the load is practically uniform, or 2HO0-lbs. per lineal it.
Two locomotives and tenders will cover a span of lOO-ft., and impose
a practically uniform load of 2750 lbs per lineal ft.
The couplinfjj of more than two locomotives is sometimes necessary
upon snow-roads, but rarely occurs els«. where, and the case should not be
included in a general rule for bridge design. It is a common practice,
however, of the English Board of Trade, to prove a bridge by covering
each track with as many locomotives and tenders as possible, and mea-
suring the resulting deflection — an unreasonably severe proof for spans
of more than lOii-ft.
The weight of the heaviest train, exclusive of the engine, may be taken
at 2250-lbs. per lineal ft. When a train crosses a bridge the conse(jueut
strains are far more intense than if the train were at rest, and in France
the government ri'gulations require a bridge to be proved under a moving
as well as under a dead load. This additional intensity rapidly dimin-
ishes as the span increases, and may be disregarded lor spans of more
than 75-ft.
For spans of less than 75-ft. the live load may be reduced to an equi-
valent dead load by adding to tlie former certain arbitrary percentages,
which may be taken at 10 for 75-i't., 20 for GO ft., 30 for 50-ft., 40 for
40ft., 50 for 30-ft., GO for 20-ft., and 70 for 10-ft.
For spans of more than 75-ft. the total load -- ihad load + lice load.
Provisit n may be made against accidents from high winds by the intro-
duction of lateral bracing sufficiently strong to withstand a pressure of
40-lbs. per sq. ft. of truss and train.
Tables at the end of the chapter give the uniformly distributed live
loads upon railway and highw.iy bridges of diff"erent sjtans.
(■g), — Chord and web stresses. — It may be assumed that the load upon
a bridge is concentrated at the panel points of the main trusses.
The stresses produced in the chords (flanges) by the passage of
a train are to be calculated on the basis of a uniformly distributed load.
In determining the strength of the web members allowance may
be made for the irregularity in the dist'-ibution of the live load by sub-
stituting for the nearest standard panel weight a weight equal in amount
to that upon an independent sj)an of the same length as a panel.
(9). — B'Mviiig, Friction (^also called expansion and hearirg) rollers,
(.((•_ — The area of the bearing surface at the supports (abutments) must
be sufficient to prevent any undue crushing. The length of the bearing
for a cast iron girder is often made one-fifteenth of its central depth.
BKAKING, FRICTION ROLLEUS.
In order to provide for the lon<i;itadinal contraction and expansion of a
girder from changes of temperature, a smooth metallic sliding surface
(bed-plate) is placed under one end, and, if the span is more than 60-ft.,
friction rollers are interposed between the girder and plate. The crush-
ing strength of a cylindrical roller is proportional to the product ol'
its length and diameter, and that of a sphere to the square of its diameter,
while the relative strengths of a cube, the inscribed cylinder upon its base,
the inscribed cylinder upon its side, and the inscribed sphere, are
as, 100, 80, 32, 26. It is not desirable to subject cast-iron rollers to
a greater pressure than that of 4000 to fiOOO-lbs. per lineal inch, r?
340-lbs. per sq. in. of diametral section, and in practice, when there ^re
several rollers, the working pressure should not exceed 185-lb8. per sq. in.
of diametral section, as an iillowance must be made for a variation in
the j)Osition of the centre of pressure at the bearing, owing to the
impossilnity of loading the rollers equally.
The elastic strength of the rollers should remain unimpaired, and
therefore the bed-plates should be fixed with the utmost care. Imper-
fect fixing cannot be remedied by increasing the number of the rollers.
It is uncertain whether cast-iron bed-plates continue long in working
order, and hard-wood wall -plates are often substituted i'orthem in t^pans
of even 150-ft.
Let W be the total weight upon a bridge truss in lbs.
B " breadth of the bearing at the roller end in inches.
L " length " " " " "
2 X 185 ■ ^ 370*
B for moderate spans usually exceeds the width of the flange (chord)
by f) or Gin., and maybe still further increased, in the case of large
bridges.
Note. — In the construction of the Yougltioglieny bridge, l*a., the
specification requires that the pressure per lineal inch of roller is not to
exceed the product of 5<)0-lbs. by the square root of the diameter in
inches (500.Vd).
(10). — Trrllin, 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. tl). The system of tnllis is said
to be single, double, treble, — according to the number of diagdiials met
by the same vertical section.
II
T
48
TRELLIS, Oil LATTICE GIUDEUS.
Vortical stifFenors. united to the chords mid diagonals, may be intro-
duced at retrular intervals.
Figs. 7, 8, 9, 10 shew appropriate sections for the top chord ; the
bottom chord may be formed of fished and riveted plates, or of links
and [lins.
The verticals and diagonals may bo of an L, T. I. H, i^ 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 cf girder is the number of the joints.
The stresses in the duigonals are determined on the assumption that
the shearing force at any vertical section is ecjually distributed between
the diagonals met by that section, which is e(|uivalent to the substitution
of a lui'dn stress for the different stresses in the several bars.
r. (J., let »' be the ;>(7v;('A»/'/(Moad concentrated at each apex in Fig. 6.
Let be the inclination of the diagonals to the vertical.
The reaction at .1 == 7^. v, and the shearing force at the section MX
== 7J,. u- - 4. >r - U. ir.
This shearing force must be transmitted through the diagonals.
Hence, the stress in ah due to the permanent load -=
U. w
sec ==
-. u\ sec 0,
Again, let w' be the flee load concentrated at an apex.
The live load produces in ah the yrniffst stress of the same kind as
that due to the permanent load, when it covers the longer segment
up to and including b.
WARUEN GIRDP:R.
49
55
Tlic sheariivj; force at J/.V ( == tlie n-actidu at .1) i,s thou '~.w', and
55
the correspond! II ii: stress in ah is ' '-.w'.sirO.
' ' ('.4
•'• the maxiniuiu stress in nh due to botli kinds of loads =
/7 55 A
The sheariiiiT t'oree ]irnduces its (jrmto^t eff, ct in rcrirsiiKj the strrss
in (ih due to the piTuiaiii iii load, when the live IikmI covers the shurffr
sejriueiit up to and iiieludlMi:' n.
5
The sheariiiij: force at AJX ( r- the rtaetioii at B) is then -. w', and
5
the corresponding stress in ah is —j-.ir'.sfc H;
•*• the resultant stress in (i/> when the live load covers the sliorter sej:ment
I .SVC IK
= (^•'" - .To'"')-^'
This stress may he. noirative. a»id must be provided ior by introducing
a counter-brace or by proportioning the bar to bear (jitfli the greatest
tensile and the greatest conij)ressive stress to which it may be subjected.
The stress in any other bar may bi- obt.iinetl as above.
The (7(o/-(/ stresses are greatest when the live load covers the whole
of the girder, and may be obtaini'd by the method of moments, or in
the manner described in the succeeding articles.
(11^. — Witnrn Girder. — The Warren girder consists of two hori-
zontal chords and a series of diagonal br 'es forming a single triangula-
tiou, or zig-zag, Fig. 11.
The principles which regulate the construction of trellis girders are
equally applicable to tho.oc of the Warren type..
The cro.ss-girders ('loor-beams) are spaced so as to occur at the apex
of each triangle.
When the platform is supported at the top chords, the resistance of
50
WARREN GIRDER.
the Btructure 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 at the bottom chords, additional cross-
girders may be suspended from the apices in the upper chords, which also
have the eflfect of adding to the rigidity of the main-girders.
Let w be the dead load concentrated at an apex, or joint.
w' " " live
I *' " span of the girder.
k " " depth " "
s " ** length of each diagonal brace.
" to the vertical
ti " " inclination "
N " " number of joints
Two cases will be considered.
Case I. — All the joints loaded.
Chord Stresses. — These stresses are greatest when the live load covers
the whole of the girder.
Let *S^„ be the shearing force at a vertical section between the joints
n and n + 1.
Let H„ be the horizontal chord stress between the joints n-l and n + 1.
The total load due to both dead and live loads = (w + w'). N- 1.
w + w'
The reaction at each abutment due to this total load =
The shearing forces in the different bays are,
10 + w'
2
N-l.
So =
w + w'
Si = — 5 . ^^3,
S,=
2
w + w'
. iV-1, between and 1,
a 1 » 2,
iV-5.
S^ =
w -f- w
and S„ =
. iV-7,
w + w
" 3 " 4.
. (iV-2n-l)
•T
he corresponding diagonal stresses are,
So-see 1^, Si.sec ^ , S^.$ec 9.
WARREN GIRDER.
31
The last stresses multiplied by sin d give the increments of the chord
stresses at each joint. Thus,
I
N.k
H, = tension in 2 = ^o- <«« ^ = 9 — . iV-1,
_ w + w' l_ N-\
2 k' N '
II2 ~ compression in 13 = S^. tan d + Si. tan d
_ to + w' I N- 1 w + w' l^ iV-3 _ tp-^w ' l_ 2.(iV-2)
" 2 Ic N "*" 2 k' N ~ "Ik' N
„ „ „ w^-xo' I 3.(.V-3)
iZ, = tension in 2.^ = Hy\-Sy. tand + Sy tand=: ___._•
w + w' I 4.(iV-4)
B^ = comp'n in 36 = //, + S,. tan e + S3, tan e = ^ -r j^
and j^„ = horizontal stress in chord, between the joints n-1 and n + l
- ^ —.— . "^ — 7= — , being a tension for a bay in the bottom
chord, and a compression for a bay in the top chord.
JVb^e. — The same results may be obtained by the method of momenta.
e. g., find the chord stress between the joints n-1 and n + 1.
Let a vertical plane divide the girder a little on the right 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 n, 7i + !•
Take moments about the joint n,
■'• //„. k
w
+
to'
2
*
w
+
to'
I
n. I
iV-l.n.^ -n-l.w-^w'.^^
.1. n.
N-n.
2 N
.'. //„ = &c.
Diagonal Stresses due to dead load.
Let d^ be the stress in the diagonal n, n + \ , due to the dead
load.
The shearing forces in the different bays due to the dead load are,
-^.iV- 1, between and 1, ^.iV-3, between 1 and 2
Z-N-^ " 2 " 3,|.iNr-7, " 3 " 4
2 ^
and — .(iV-2 n-1), between n and n + l.
52
WARREN GIRDER.
The corresponding diagonal stresses are,
w
"2
w
70
a compression ^.^- 1- «ec. = -^.iV- 1. — — (f, in 01,
= (Z, in 12,
c?j in 23,
a tension --.iv- 3. «fc. 0= -c^Jy - o. ,
1^ u li
W
W
a compression —
— .JV- 5. sec. =— .iV- 5.-.- =
A;
and the stress in the n-th diagonal between n and n + I is.
?tf
/,.= -:5-. (.V-2»-l).
/^
being a tension or compression according as the brace slopes down or
up towards the centre.
Diagonal stresses due to live load.
The live load produces the greatest stress in any diagonal, («, n + 1),
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; represent
this maximum stress by Z)„.
The live load produces the greatest stress in any diagonal {n, n + 1),
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',,.
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 cor-
responding diagonal stress is the product of this shearing force by sec 6,
or -.
K
The
/>» =
values of the different diagonal stresses are,
= compression in ^ when all the joints are loaded
s w' N.N- 1
A =
N '
tension in 1 2 when all the joints except one are loaded
s w' N-l. N-2
~k 2" N
Dj = compression in 2 3
_ s ro' N-2.N-3
^T2' N
Di = tension in 3 4
_ s w^ N- 3.^-4 :
"IT N
" 1 and 2
<' 1,2, and 3
WARREN GIRDER.
63
D^ = stress in ?», n + 1 when all the joints except 1, 2, 3 . . . and n loaded
_s w'. N-n. N- n-\
~ /? 2 N '
//,= stress in 1 before the load comes upon the girder =
D\ z= compression in 1 2 when the joint 1 is loaded =!_,—. 1
S Hi'
i>'2 = tension "23 " joints 1 and 2 " =-I —.3
C 'If J
D'a = compression " 34 " " 1, 2,&3 " = /•• "y'^
/)'„ = stress in 7i, n + 1,
" l,2,...&?t" = -7
s w' H. n + I
fc iV 2
The total maximum stress in the ?t-th diagonal of the sa7/if kind as
that due to tlie dead load = </„ + J)„.
The resultant stress in the n-ih diagonal when the load covers the
shorter segment "= d„ - />'„.
This resultant stress is of the same kind as that due to the dead load
so long as </„ > Z>'„, and need not be considered since </„ + A, is the
maximum stress of that kind.
If D'„ > d„. it is necessary to provide for a stress in the given diago-
nal of a kind opposite to that due to d„ + JJ„, and equal in amount to
D'. - 'L-
This is effected by counterbracing or by proportioning the bar to bear
both the stresses r/, + /)„ and D'„ - d„.
Case II. — Ordi/ joints denoted hy even numbers haded.
Chord Stresses. — The stresses are greatest when the live load covers
the whole of the girder.
»'■ !
64
WARREN GIRDER.
iV-2
The total load due to both dead and live loads = (w> + «?'). — r — .
w + w' -7— —
The reaction at each abutment due to this total load= — -, .iV- 2.
To find H,. take moments about 1, .'• H,.k = — j — -V- 2. ~
2 Hk- '^•^"' iv^r9 o ^
I
l_
N
4,
- (to + w?'). ^.
4 iV
-(«^ + ^«').2.^.
" //«
n, and^r*^ let n be even ;
5-.A; =
w + ?o'
4 •^-2-]^
-(w +w'). j^ln-'I + 7t-4 + ... + 6 + 4 + 2|
n 71. 71 - 2 1
4~ J
iV- / 4
and, Zr„ ==
i«
+ 10' I 71. iV- n
Next let 71 be odd.
4 • fc' ^
. . J?„.A; = — - — . N- 2. n. ^.
4 JV
~(w + w). -^.\n-2 ^ 71-4 + ... + 5 + 3 + 1|
= (wj + w'VJl J N-2.n 71-1
">4-{
and, i7„= -^— . ^
4 4
W + 70' ? iV- 2.7i -71-1*
i\r
-AT,
iVb^e. — If — be even,
.'. /ijv^, the stress in the middle bay = — zrpi — T.iv
r
16 ■ k
HOWE TRCSS.
55
If ^ be odd,
:.IIk, the Btress in the middle bay^
r »
k
16 'k' N '
Diagonal stresses due to the dead load.
The shearing forces in the difftrent bays due to the dead load arc,
w
10
■ic
.N-2 between and 2,- . JV- 6 between 2 and 4, — . A"- 10
4 4 4
between 4 and 6, etc.
The corresponding diagonal stresses are,
rfo in 1= r-
w
. jV - 2 = <i, in 1 :i
w
d, in 2 3 = -r. -,. A^- 4 = u\ in 3 4
- k 4
rf, in 4 5 = j^ .-^. iV- 10 = rfj in 5 6
etc.,
etc..
etc.
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
D, = />, D', = D\
D, = D, D\ = //,
A = A ^'. = ^'»
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 = d^ + Dj = dj + Dj,
and the maximum resultant stress in 3 4 when the shorter segment
is loaded = d^ - D\ = d.. - D\.
j^ote. — e is generally GO^, in which case s = ^~]^'
(12). — Howe Truss.
Fig. 14 is a skeleton diagram of a Howe truss.
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
56
HOWE THLSS.
nionibcrs, pliipcd a little tlistaiuH! apart so as to allow iron susponders
with sercwt'd ends to pass bctwtvn tlirin (Kii;s. 15, 1<; ).
Each meuibor is uiade up of a number of lengths scarfed or fished
together (Figs. 17, 18).
The main-braces, shewn by the full diagonal lines in Fig. 14 are
composed of two or metre members.
The counter-braees. which are introduced to withstand the v<^ffect of
a live load, and are shewn by the dotted diagonal lines in Fig. 14, 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 tlie points of
intersection.
The main braces and counters abut against solid hard wood or
hollow cast-iron angle-blocks (Fig. 10). They are designed to with-
stand compressive forces only, and arc kept in place by tightening up
the nuts at the heads of the suspendi-rs.
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 cliord 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. 19, 20).
HOWE TRUSS.
57
In Nucli a case tlie lower annK'-bloeks should have grooves to receive the
burs. HO as to provoiit lateral flt-xure.
It'tlio truss is iiiaile cntinly of iron, the top chord uiay be foruiod of
K'li^ths of cast-iron provided with suital)le 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 same data are
assumed as i"or the Warren girder, except that N is now the number of
panels.
(.'Iiiird strrssi's. — These stresses are greatest when the live load covers
tlie whole of the girder.
Jict II„ be the chord stress in the 7j-th panel.
The total load due to both dead and live loads == w + ?//. iV- 1.
The reaction at each abutment due to this total load == — ^ — . A'-l.
•■J
Let a plane MM' divide the truss as in Fig. 14. The portion of the
truss on the left of the secant plane is kept in e<|uilibrium by the load
upon that portion, the reaction at the left abutment, the chord stresses
in the ?i-th panels, and the tension in the Ji-th suspender.
Jursf, let the load be on the top chord and take moments about the
foot of the n-th suspender ;
7/1 4- 1/1 /
I 71 . H - 1
Ar-"' + ''-N'—^-
W + w' , 11. N'— 11
-. /,
or //„ =
jw + w' I n, N— n
Next let the load be on the bottom chord and take moments about the
head of the H-th suspender.
.'. //„. k =
IV + iv' , //. X- n
I.
, as before.
2 ■ X
Thus, 7/„ is the same for corresponding panels, whether the load is on
the top or bottom chord.
DiagojKil stresses due to the dead load : —
Let V'„ be the shearing force in the «-th panel, or the tension on the
?/-tli suspender due to the dead load.
First, let the load be on the top chord,
TV ^^' \ /^- 1 \
.-. y ,= -^.X- I - n. w = ^'''\— — »^ )•
Next, let the load be on the bottom chord,
••• v'„= ^ . ix- i)-:n.w --^v. (^ -u).
58
HOWE TRUSS.
The corresponding diagonal stresses are,
Diagonal stresses due to the live load.
Let F"„ be the shearing force in the «-th pan^l, or tension on the «-th
suspender, when the live load covers the longer segment.
First, let the load be on the top chord.
The greatest stress in the H-th brace of the same kind as that produced
by the dead load occurs when all the panful points on the right of MM'
are loaded.
.*. V"„, the shearing force on the left of MM' -■= the reaction at
w' M— ?i
H"' ■^~ n -1. — rj— , and the corresponding diagonal stress
" k " k' 2 ^' N
Hence, the resultant tension in the n-th suspender due to both dead
and live loads = F„ = F„ + F',
/iV-1 X ir ._
j\r-n
iV
T f
and the resultant maximum compression on the n-th brace due to both
dead and live loads
_ * T7 * i /iV-l \ w'
.V-H-l.
N-n
N
}-
d„+D„
The live load tends to produce the greatest stress in the 7i-th counter
when it covers the shorter segment up to and including 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 + 1 )-th brace.
The shearing force on the right of MM' = the reaction at N'
I)'n the corresponding diagonal stress = — — ■
k 2 N
and the resultant stress in the counter = !)'„ - (/„.^,
s ( w' n.n + l /N-3 \ )
'=lc'l2-N--"'\-li V}
Next let the load be on the bottom chord.
... F' =!!^7V-„^^±1
„ s w' ,_ N- 11 + 1
_ w' n.n 4- 1
\'~ir'
PRATT, OR WHIPPLE TRUSS.
Hence V„= ?',+ F'„ = t«. (^^ - n) + ~.N-n.
and, < f /;„ - y| ^- (-^ - ") + y ^- "• —jr~ \
59
N
Also the fltresfl in the n-th counter is
71+1
JVb<c. — A common value of « is 45", vrhen sec e= -^ = \ .414, and
I
tan e
N.k
1.
The end panels and posts, shewn by the dotted lines in Fig. 14, may
be omitted when the platform is suspended from the lower chords.
(13). — Pratt, or Whipple truss.
Fig. 21 is a skeleton diagram of the simplest from of Pratt truss.
The verticals are now in compression, and the braces in tension, so that
the angle-blocks are placed above the top and below the bottom chord.
Counter-braces, shewn by the dotted diagonals in the Fig,, are to be
introduced if necessary, to withstand the effect of a live load.
The principles which regulate the construction of Howe trusses are
equally applicable to those of the Pratt type.
The stresses in the different members are found precisely as in the
preceding article.
Fig. 22 shews a modification of the Pratt truss, the braces forming
two independent systems, which may be treated separately.
60
PRATT, OR WHIPPLE TRUSS.
Let 6' be the inclination of AB to the vertical.
" ^ ■' " Al, CD, ''
Chord sfressrs.— These stresses are greatest when the live load covers
the whole of tlie girder.
The reaction at A from the system ABCD.., == 4.(w + w').
« 1-3 ==^_(^w + ic')
The shearing forces in the different bays are,
4. (^v + w') in AC, from the system ABCD...
r,-(u- + w') In AC
^.(w + ic') in C2,
^.(if + jf') in 2E
^12 3...
ABCD...
^123...
ABCD...
^123...
ABCD,,,
^123...
/in
2.(iv + w') in Ei, " "
2^.(?« + «;') in iG " "
].(w + w') in G^6, " "
-.(?« + ?«') in 6 1 " "
The corresponding diagonal stresses are,
4. (w-{-w').sec O'in AB,3l(w + w').sec 6 in A l,S.(w + w'). sec On
CD, 2^. (iv + 7r'),sec in 23, etc.
Hence the top chord stresses are,
C, in AC = 4.(w + w').tan 6' + 3h(w + w').tan fl, (7, in C2 = C, +
3. (w + w'). tan e = 4.(w + w')MmO' + 6^.(w + w').ia7ie, C, m2E =
G'l + 2\. {w + xc').tan 6 == 4.(w + w').tan 0' + 9.(w + w'). tan 6, etc.
The bottom chord stresses are,
T,mBl=z4.{w + w').tane', T, in 1 D= T, + 3^. (w + w').tan 8
= 4.(w + w').tan 6' + 3i.(w + w').tan e= C,
So, T; = (7,, T, = q,, etc." etc.
Again the stress in any diagonal 4 5 of the system A 1 2.... due to the
dead load = l^.w.sec e.
The live load produces the greatest stress in 45 of the saoie kind as
that due to the dead load, when it is concentrated at all the panel points
of the system A 12 3... on the right of 4.
1 K
The reaction at A is then -^ .w\
o
15
and the corresponding diagonal stress = — .w.' sec 0,
o
FINK TRUSS.
61
•*• Tbe maximum resultant stress in 4 5 = ( — av + ~-.i''' ) .sec ff
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^ 12 3
on the left of 8.
•J
The reaction at the right abutment is then '-.w',
■i
and the corresponding stress in the counter = -jv' seed;
'8 . 1
*« the resultant stress in the coun
•■■■=(4-"''-2-")-
sec e.
~. w.sec being the stress in 6 7 due to the dead load.
Similarly the stresses in any other diagonal and counter may be
found.
{U).—Fmk Truss.
Fig. 23 represents a Fiuk truss. The platform may be either at the
top or bottom.
The verticals are always struts and the obliques are always ties.
The stresses in the different members are to be calculated for a
maximum distributed load.
Let weight upon post IC, i. e., ^ of weight between A and 2. be TF,
" 2D, " " A " 4, be W.,
" 3F, " " 2 " 4, be W^
" 4F, " " A " li, be W^
and let TFj, Wf, Wj, be the weights upon the posts 5(7, Gil, and 71,
respectively.
Let the inclinations of A.C, AD, AF, to the vertical be a, $, y, res-
pectively.
Let Ti, Ti, T^, be the tensions in the ties as in the Hg.
The tensions in the ties meeting at the foot of a post are evidcnily
equal.
62
BOLLMAN TRUSS.
••• 2.7',. cos a = TT,
2.7,co. y= ^\ + {T,+ T,).cos^+(^T,^T,).cosa
2.T,.cosa= W,
2.T,.cos(i= W, + (T,+ Tr).co8a
2.Tj.cosa= Wj
■•■ r, = l.W,.sec a., T, = \.(w, +i!i-tZ»).„e,J
^a = 5. ^^3- sec a
^
^-M*"'
+
^i±^^w^±^^w^j^^^^\+w,
''-]■
sec y
n = l m. sec a, r. = 1. (r. + J!^).„,*.,
and 2^7 = —. TF;. sec a.
Again the compre sion in A2 = r, sm a + T,. sin + T,. sin y
at 2= T,. sin + T,. sin Y
^" 24 =- T,. sin 0+T,. sin y + T, sin a
" at 4=- T,.siny.
etc. etc.
the compression in 1<7 = 2. 7; .cos a == W^
" 2D= 2.T,.cos
" 3E=: 2.T,.cos a= W,
" iG = 2.r,.cos y
etc. etc.
e. g., let the load upon the struss be a uniformly distributed load, W.
.-. W,= W,= ^. = r,= f, Tf,= >re=|:,and IF.= f. '
I, -i3 _ij- /; = ___. 5ec a, T^ T,= —■ sec 0, and T,=~-secy.
If the truss represented by Fi^. 23 is inverted, another system is
obtained m which the obliques are struts and the vertical ties.
(l5).—J]oUman Truss.
Fig. 24 represents a Bollman truss. The platform may be either at
the top or bottom.
The verticals are always struts and the obliques are always ties
Each pair of tics, as AC, BC, is independent, and supports the load
QUADRANGULAR AND POST SYSTEMS.
63
upon a vertical, {. e,, the weight of panel, or one-half of the weight be-
tween the panel points A and E.
Let T , Tj be the stresses in AC, BC, respectively.
Let W\ be the weight upon the post DC
Let a„ ttj, be the inclinations of AC, BC, respectively, to the vertical.
sin ttj _j /T7 Ttr '^'^ ^1
f,= W.
and 'R = W.
sin (a, + ajj)' ' ' ' sin (a, + a.^)
The stress in any other tie may be obtained in the same manner.
The compression in the top chord is the sum of the horizontal compo-
nents of the stresses in all the ties which meet at one end,
(16). — Quadrangular and Post si/stems. —
Fig. 25 represents the Quadrangular truss in which tiic bottom chord
is supported at intermediate points half way between the verticals.
Fig. 26 represents the Post system. In reality ;}- is the Pratt
(Linville) truss shewn in Fig. 22, modified by inclining the compression
members at an economical angle.
The stresses in the different members are to be calculated in the
manner described in the preceding articles.
(17). — Bowstring girder, or truss.
The bowstring girder in its simplest form is represented by Fig. 27,
and is an excellent 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 cylindrical tubes,
but any suitable form may be adopted, and the inverted trough offers
special facilities for the attachment of verticals and diagonals.
64
BOWSTRING GIRDEH, OR TRUSS.
The tie is constructed on the same principles as those employed for
other iron jrirders, 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 vertical bars
which arc 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 verfieals and diagonals may be lattice-work.
If the load upon the girder is uniformly distributi'd and stationary,
verticals only are required for its suspension, and the neutral axis of the
bow should be a parabola. An irregularly distributed load, such as tliat
due to a passing train, tends to change the shape of the bow, and di-
agonals are introduced to resist this tendency.
A circular arc is often used instead of a parabola.
To determine the stresses in the different members, assuming that the
axis /lZ^(7of the top chord is a parabola : —
Let w bo the dead load per lineal ft.
" I " span of the girder.
'' Je " greatest di'pth BD of the gird;T.
Chord strcssfs. — These stresses are greatest when the live load covers
the whole of the girder.
The total load due to both dead and live loads = (ir + w'). I
u- + ?r'
The reaction at each abutment due to this total load
-. I
Let // be the horizontal thrust at the crown.
Let T " " tension in the tie.
Imagine the girder to be cut by a vertical plane a little on the right
of Bl>. The portion ABD is kept in equilibrium by the reaction ut A
the weight upon Al), and the forces // and 1\
BOWSTRING GIRDER, OR TRUSS. 65
Take moments about B and D ;
8
and • ' J =- — — . -j-==Jl
Lot //' be the thrust alonj^ the cliord at any point /*.
Let X be the liorizoiital distanee of 7' from li.
The portion J'Ji is kept in i'(|uilibriuni ])y the thrust // at B, the
thrust //' at /* and the weinlit {w + tr').x between /' and Ji ;
••• /r.siv' (•- //'•-'=. IP + (»• + »')■■../•-,
i being tlie inelination of the tanjxent at P to the horizontal.
Hence, the thrust at .1 = C'~r~-l) ( 777-7^ + 1 j
Diiigrntnl strcssri^ diw to Urc fo'uf.
Assume that the hiad is concentrated at the panel points, and let it
move from .1 towards C
l^m^m
ri6.2L8
If the diau'onals sliipr as in the Fig. 28, thi'y are all ties, and the live
load produces the greatest stress in any one ol'them. as QS, wlu'ii all the
panel points from A up to and including (^ are loaded.
Let ,r, y. be the horizontal and vertical co-ordinates, respectively, of
any point on the parabola witii respect to B as origin.
Tlu' e((uati(»n of the parabola is. // — - " ..r-
(1)
Li't tlu! tangent at the ap'X /* meet />>/) produced in L, and /-^'' jiro-
duced in A'.
Draw the horizontal line I'M.
From the properties of the parabola, LM ^- 2. BM.
Let I'M --= X, and BM ■■= >/.
From the similar triangles L MP and LDl^,
M r~ D E '"' ~ ~ xTt^E'
66
BOWSTRING GIRDER, OR TRUSS.
2.y 8.x
. CE l-2x
(i-2xy
Sx
'QE^l + 2x (2)
Draw EF perpendicular to QS produced, and imagine the girder to
be cut by a vertical plane a little on the right oi' FQ.
The portion of the girder between FQ and C is kept in equilibrium
by the reaction li at C, the thrust in the bow at F, the tension in the
tie at Q, and the stress in the diagonal QS.
Denote the stress by Z>„. and let the panel OQ be the u-th.
Let be the inclination of QS to the horizontal.
Take moments about E,
:. £>,, EF -- Jl CE
orD„ = R0 cosec .
Let N be the total number of panels,
:. j^is & panel length, and w^ is a panel weight.
(^)
Also X = 71. ^r^ - ^^ , and . . , _
^V 2' lr2x QE
I -2.x CE N-n
H
R, the reaction at C when the // panel points preceding Tare loaded
_?f^ , n.n T 1
Hence, equation (3) becomes.
iV
J^n = -^-f.n + 1. .coscc
4.k
w
Again, by equation (1), /.• - ST=~. DT'= 4./.-. ( ^Lll _ A'
>ST ST
Thus, finally,
...^/ iv^-„[^^+^-K^-'^-i)(«+oi^]*
^"~ 8 ^ -A^
This formula evidently applies to all the diagonals between B and C.
(5)
BOAVSTRIXG GIRDER WITH ISOSCELES BRACING.
G7
(2)
' to
um
the
;^)
'i)
r>)
Similarly, it may bo easily shewn that the stress in any din<j;onal
between B ami A is given by an expression of precisely the same t'orm.
Hence, the value of I), in ecjuation (5) is general for the whole girder.
A load moving from C towards A requires diagonals inclined in an
opposite direction to those shewn in Fig. 28.
JSt n'ss<)i in the verticals due to the live load.
Let V„ be the stress in the »i-tli 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 are loaded.
Imagine the girder to be cut by a i)lane *S" .S"' very near PO, Fig. 2S.
The portion of the girder between »S" S" and C is kept in equilibrium
by the reaction li' at C, the thrust in the bow at P, the tension in the
tie at 0, and the compressittn V„ in the vertical.
Take moments about E.
V,, QE =-- R'.CE., or T; = R'. — ,
and R\ the reaction at <' when the ('(-1) panel points from ^-1 up to
1 • I J- r\ 111 "■' ,"-"-1
and includniir O are loaded = — ./.-
■z '
Ilenc, V
(6)
a general formula for all the verticals.
Let r„ be the tension in the ii -th vertical due to the dead load. The
resultant stress in it when the live load covers AO is i\-]\, and if
negative, is the maximum compression to which J^Q is subjected.
If v,-V„ is positive, the vertical PQ is never in conq)res.sion.
The maximum tension in a vertical occurs when the live load covers
the whole of the girder and == !«''.-+ the tension due to the dead load.
Note. — The same results are obtained when N is odd.
(18). — Buicst ring girder with isoscelrti Itrariug.
Diagon(d stresses due to the dead load. — Undera dead load the bow is
equilibrated, and the tie subjected to a uniform tensile stress eijual in
amount to thii horizontal thrust at the crown. The braces merely
serve to transmit the load to the bow and are all ties.
Let y,, T.^ be the tensile stresses in the two diagonals meeting at any
panel point Q. Let", f^j, be the inclinutions of the diagonals to the hori-
zontal.
Lot irb-' the panel weight susr^ended from Q.
68
BOWSTRING GIRDER WITH ISOSCKLES BRACING.
Tiic stress in the tie on each side of Q is the same, and thuieiore
T*,, 7j, and TFare necessarily in equilibrium.
. — — :, and y, "- >K -r -.
«<« ('^ + ",)' ' sin (y, + W,)
Hence, T, - >r. -;-
Diagonal stresses due to the live lorn/.
Let N be the number of half panels.
2.1 . 2.1
The lentrth of a panel == —rr-', the weight at a panel point = "•'. -~^-
Let the load move from A towards C. All the braces indiiKil like
07' are ties, and all those inolinrd like (Jl' dw struts,
girdir from A up to and including ().
Tlie live load produces the greatest stress in OT when it covers the
Denote this stress by />„ ; ()G is the /;-th half-panel.
As before, I),. EF - A*. CE. ( 1 ) .
rpv- 1 1 4/» ' ' ] • n '"'•"•*' 'i + 2
The load upon AO = n.w , — , and • • Ji = — -— . _^
The ratio of CE to iJF is denoted by the same expression as in the
preceding article.
.„_»_: 2 ,^ x-» [^ t5'^!i" -')"'^"iil' (2)
■■ " 8'//7i + l' iV ■ N-,>- 1
The live load produces the greatest stress iti OM when it covers the
girder up to and including D.
Denoti this stress by D'„ ; DG is now the 7i-i\i half panel.
Let W be the reaction at (\
As before, D\r A". l]^p. cosecO, (3)
e being the angle MOD.
The weight upon AD= (h - l),!o'.— ,, and .'.li' = -^.l. -ttj—
BO^VSTKING SUSPENSION lUUIMJE.
69
10
be
It may be easily aliown, as in the precediiijr article, that,
i
oe'
n+ 1
, and co8ccd = N.
[/■+ !^fi(.v-„).<]
• D' =
/ H- 1
4n.K'. (N-)i)
(4)
8 k n iV ■ N-n
Hence, when the load moves from A towards C, equation (2) gives
the diagonal stress when n is even, and equation (4) gives the stress
when u is odd.
If the load moves from C towards A, the stresses are reversed in
kind, so that the braces have to be designed to act both as struts and
ties.
Note. — By inverting Fig. 29, a bowstring girder is obtained with the
horizontal chord in compression and the bow in tension,
(19). — Jiou-striiig Suspension Bridge, {Lentie.ular Truss). — This
bridge is a combination of the ordinary and inverted bowstrings. The
most important example is that erected at Sultash, Cornwall, which
has a clear span of 445— ft. 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.
In Fig. aO, let BD^k. B'I) = 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.
//=
w + w
P
8 k + k'
r^.= T.
First, let k = k'.
ponding stress in a bow-string girder of span I and depth k.
•. 11= T— — :,— T) or one-half of the corres-
lo k
70
CAMIiKU.
Ono-hiilf of the total loud is .supportod by the bow and one-lialf is
trausmittod through tlie verticals to the tie.
.".The stress in each vortical = .^.0''' + "'"),
A ^ '
w" being the portion of the dead weight per lineal ft. borne by the ver-
ticals, and jY the number of panels.
The diagonals are strained nnly under a passing load.
Let /'/*' be a vertical through K, tlie point of intersection of any two
diagonals in the same panel, and let the load move from A towards C.
By drawing the tangent at i^and proceeding as in § (18), the expres-
sion for the diagonal stress in i^S, becomes as before,
,. xv' . H.(//-l) /-2x
^^"^T-'— A^'TTii^-^"*""'- (1)
Similarly the stress in the vertical QQ is
„ I «•' , n.(n-\\l-2x
\ = }(' — ^ — / — -^ '-• •
\V 2 ^'^ l+'lx
(^)
'.under these loads,--- ■ — — II .
o k
ir=
(3).
Next, let A* and A*' be unequal.
Let W be the weight of the bow, IF ' the weight of the tie.
F' ""^ 7r " A;'"
The verticals are not strained unless the platform is attached to them
along the common chord ADC. In such a case, the weight of the plat-
form is to be included in W '.
The tangents at P and F* evidently meet AC produced in the same
point 0', for EO' is independent of A or k'. Bence, the stresses in the
verticals and diagonals due to the passing load may be obtained as
before.
(20). — Camber. — Owing to the play at the joints, a bridge truss,
when first erected, will deflect to a much greater extent than is indi-
cated by theory, and the material of the truss 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 made straight they would curve downwards, and,
although it does not necessarily follow that the strength of the truss
would be sensibly impaired, the appearance 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
CAMBER.
71
(
shortening the lower chord, and the difference o? length should beoqual-
ly divided amongst all the panels.
The lengths of the wt-b uifuiber.s in a cambered truss are not the
same as if the chords were horizontal, and must be carefully calculatt-d,
otherwise the several parts will not tit accurately together.
Tojindnn opjiroximate value for the Camber, etc.
Let d be the depth of the truss.
Let .s-,. .v^, be the lengths of the upper and lower chords, respectively.
J^<-'t/i,/i, be the unit stresses in " " « u
Let <?,. </j, be the distances of the neutral axis from the upper and
lower chords respectively.
Let ^ be the radius of curvature of the neutral axis.
Let I be the span of the truss,
</i s, - / J\ (f_ I- s, f
'•~R~ l~ - E^ ^^^lt~ ~T~ ir' "PP^^'^'^mately,
the chords being assumed to be circular arcs.
Hence, the excess in length of the upper over the lower chord
Let .T„ X,, be the cambers of the upper and lower chords, respectively.
^+(Z, and /^- r/ji are the radii '' " <' «
By similar triangles,
the horizontal distance between the ends of the upper chord =^^^.l
R
, R-d.,
lower
Heuce,(-^.^jy= x,.2.{R+dd, approximately.
2
2. n = x,:2.(R- (4), approximately.
R
^J
\2 R
■ ^^^ ^*"^0'' ^ ~
d^
R
72
TABLES.
TESTS FOR THE MATERIALS OF A BRIDGE.
I. WROUGHT IRON.
Tension Tests. — (1). All wrouj;ht-iron to have a limit of elasticity of
not \em than 2G,000-lb8. per sq. in,
(2), Full sized bars of flat, round, or wjuare iron, not over i^-aq. ins.
in sectional area to have an ultimate strength of 50,000-lbs. per aq.
in., and to stretch 12^ per cent, of the whole length.
(3). Bars i)f a larger sectional area than 4h sq. ins, to be allowed a
reduction of 1,000-lbs, per sq. in. for each additional sq. in. of section
down to a minimum of 46,000-lbs. per sq. in.
(4). Specimens of a uniform section of at least one sq. in., taken from
bars of 4i— sq. ins. vsection and under, to have an ultimate strength of
52,00Q-lbs. per sq. in., and to stretch 18 per cent, in 8-inches.
(5). Similar specimens from bars of a larger section than 4^sq. ins.
to be allowed a reduction of 500-lbs. per sq. in. for each additional
sq. in. of section, down to a minimum of 50,000-lbs, per sq. in.
(6). Similar specimens from angle and other shaped iron to have an
ultimate strength of 50,000-lbs. per sq. in., and to stretch 15 per cent,
in 8-inches.
(7). Similar specimens from plate-iron to have an ultimate strength
of 48,000-lbs. per sq. in., and to stretch 15 per cent, in 8-inches.
Bending Tests. — (8). All iron for tension members to bend cold,
without cracking, through an angle of 90° to a curvs, of which the
diameter is not more than twice the thickness of the piece, and at least
one sample in three to bend 180'^ to this curve without cracking.
(9). Specimens from plate, angle, and other shaped iron, to bend cold
without cracking, through an angle of 90° to a curve, of which the
diameter is not more than three times the thickness of the specimen.
II. STEEL (mild).
(1). The steel to have an elastic limit of 40,000-lbs. per sq. in., and
a co-efficient of elasticity varying from 26,000,000 to 30,000,000-lb8.
(2). Full-sized bars to have an ultimate tensile strength of 70,000
lbs. per sq. in., and, when marked oflf in 12-in. lengths, to shew a
stretch of 10 per cent, for the whole length, and of 15 per cent, at least
for the length including the fracture.
(3). Bars drawn down to a suitable size from the crop ends of each
bloom slab, and not more than one inch in thickness, to bend cold, with-
out cracking, through an angle of 180° to a curves, of which the diam-
eter is not more than the thickness of the bar.
I
TABLES.
73
rth
III. CAST-IRON.
A specimen bar of the iron, 5-ft. long, 1-in. wiuare, and resting
upon supports 4-t't. G-inn. apart, to bear, without breaking, a weight of
550-lbs. .suspended at the centre.
Tables of the safe working loads upon conjprossion members
WROUOIIT-IRON.
Rntlo of IruKth to
le«»t tran»v«n(e
(lliufiiHtftn.
Sdfe w<irklii(f Imiil lu Ibpi. per «(|. In.
Katin of Icntfth to
least triuiffv<>rse
dlnit'imlou.
Sufe wcrklnB lo»il In Ibn. per iu\. In-
Flat MidH.
Hound i-udfi.
Klftt unU.
Uoiuicl (inila.
4,000
3,500
2.500
2,000
10
10 to 15
15 " 20
20 " 25
25 " 30
10,0000
9,000
8,000
7,500
6,800
7,000
6,500
6,000
5,500
5,o;-o
30 " 35
35 " 40
40 " 50
50 " 60
6,000
5,000
3,S(I0
3,000
Cast-iron compression members should not exceed 22-diam8.
in length and should be subject to the same stresses as those prescribed
for wrought-iron.
TIMBER.
Ratio of length to least
transverse dimension.
Safe load in lbs. per sq. in.
OAK.
PINE.
10
10 to 20
20 " 30
30 '' 40
1,000
800
600
400
9U0
700
500
300
RONDELET S RULE FOR OAK AND PINE.
Batio of length to least
transverse dimension.
Extrt'uie load in lbs. per sq.
in. borne by pillar without
lateral tlexuri!.
Safe Vforking load in lbs. per
uq.in. with "as factor
ot safety.
1
12
24
36
48
60
72
5,974
4,978
2,987
1,991
996
498
249
853
711
426
23i
112
71
35
In flexure, the greatest allowable stress per sq. in. may be
taken at 1,200-lbs for oak, and 1,000-lbs. for pine.
iV. J5. — The figures in the above tests and tables are by no means
absolutely fixed, and are merely given to shew the standard practice of
engineers.
74
TABLES.
Table of loads for Highway Bridges.
Sp.in in feet.
City and Suburban
Bridges liable to heavy
t.rattic.
Bridges in Manufac-
turing Districts.
Ballasted Koads.
Bridges in Country
Districts, rnballast-
ed Roads.
100 and under.
100 ()200 ....
100 lbs. per sq. ft.
HO " "
90 lbs. per sq. ft.
60 •' •'
70 lbs. per sq. ft.
60 " "
200 to 300 ....
70 "
50 '* "
50 " "
300 to 400 ....
(JO " "
.■)0 " "
45 "
400 and over...
50 " ''
50 "
45 " "
i
Table of loads for Railway Bridges.
Dead load in
lbs. per
lineal ft.
Span in ft.
"!0
15
20
25
30
40
50
75
451)
500
550
625
625
65;>
700
750
800
Live load in lbs.
Span in
per lineal ft.
ft.
12,000
100
5,500
125
5,000
150
4.750
175
4,500
200
4,000
250
3.500
3o0
3,250
350
3,000
400
Dead load in lbs
per lineal ft.
900
1135
1225
1300
1500
2000
2400
3000
4000
Live load in lbs.
per lineal ft.
2750
2600
2500
2500
2400
2400
2250
2250
2250
"try
.Hast-
.ft.
a
(I
a
u
lbs,
t.
CHAPTER IV.
Suspension Bridges.
(1). CahJcs. — The modern suspension bridge consists of two or more
cables, from whi^h the platform is suspended by iron or steel rods. The
cables pass over lofty supports (piers), u' d 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 liar, hut 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 cahh,. have been used in a few cases, but the practice is
now abandoned, on account of the diifieulty attending the manufacture
of endless hoop-iron.
Win -rope cablis 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 cylin-
drical bundle by a wire wound round the outside. There are usually
seven strands, one forming a core round which are placed the remain-
ing six. It was found impossible to employ a serm-strnud cable in the
construction of the East River Bridge, a.« tht; individual strands would
have been far too bulky to manipulate. The same objection held against
a thirteen-strand cable (13 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 continuously wrapped with wire.
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 .s7ra/(j|/t^ 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 untwisted wire ropes. Such ropes are spun from the wires and strands
without giving any twist to individual wires.
76
ANCHORAGE, ANCHORAGE CHAINS, SADDLES.
The htchstay is the portion of the cable extending from an anchoraj^e
to the nearest pier.
The elevation of the cables should be sufTioient to allow for settling,
wliich chiefly arises from the deflection due to the load and from changes
of temperature.
The cables may be prot<!cted from atmospheric influence by giving
them a thorough coating of paint, oil or varnish, but wherever they are
subject to saline influence, zii-.o seem.stobe the only certain safe-guard.
(2). Anchorage, anchorage chains, sadffhs.
The ancl.ovage, or abutment, is a heavy mass of masonry or natural
rock to which the end of a cable is made fast, and which resists by its
dead weight the pull upon the cable.
The cable traverses the anchorage, as in Figs. 1 , 2 and 3, 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, however, is required to prevent a wire-rope cable
from rusting on account of the great extxint of its surface, and it is con-
sidered 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.
In order to reduce as much as possible the depth to which it is neces-
sary to sink the anchor plates, the anchor chains are frequently curved
as in Fig. 3. This gives rise to a force in a direction shewn by the
arrows, and the masonry in the part of the abutment subjccU^d 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 even number of bars. The friction of the link-heads on
the knuckle plates lessens to a considerable extent the stress in a chain,
and it is therefore usual to dimini.sh its sectional area gradually from the
entrance E to the anchor. This is effected iu the Niagara Suspcusiou
SUSPENDERS.
77
Brldfje by varyin<r the K'ction of the bars, and in the East River bridge
by varyinjjboth tlie scetion and tin; number (if'the bars.
The neecssity fit' presirvinu- the anelior ehains frnni rust is of'sueh im-
portance, that many entrineers consider it most essential that tlie passages
and channels containing the chains and i'astcnings should be accessible
for periodical examination, painting and npairs. This is unnecessary
if the chains are lirst eluniically elea.i-tl and then cuibcddrd in godd
hydraulic cement, as they will thus be perfectly jirotectid from all atmos-
pheric influence.
'I'lie direction of an anclior 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 exjiansion and contraction of the chain. This may be
accomplished without the aid of rollers by bedding the saddle upon a 4 or
5-in. thickness of asphalted felt.
'file chain, where it passes over the
piers, rests on saddles, the object of I
which is to furnish bearings with easy
vertical curve-s. Either tht' saddle may !
be constructed as in Fig. 4, so as to |
allow the cable to slip over it with comparatively little friction, or the
cliain may be secured to the saddle and the saddle supported ujmhi rollers
which work over a perfectly true and horizontal bed formed by a saddle-
plate fixed to the pier.
(3). Siispi'ii(/<rs. — The suspenders are the vertical or inclined rods
■which carry the platform.
In Fig. f). the suspender rests in the groove of a cast-iron yoke which
straddles the cable. Fig. ») shews the suspender bolted to a wrought-
iron or steel ring which embraces tlic cable. When there are more than
two cables in the same vertical plane, various methods are adopted to
ensure the uniforni distribution of the load amongst the set. In Fig. 7,
for example, the suspender is fastened to the centre of a small wrought-
78
CrUVE OF CABLE ; VERTICAL SUSPENDERS.
iron lever PQ, and the ends of the lever are connected with the cables by
the equally sstrainod rods PR and QS. In the Chelsea bridge, the dis-
tribution is made by means of an irregularly shaped plate (Fig. 8), one
angle (if 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). — Ciiriw of Cable; vertical suHpenders. — An arbitrarily loaded
flexible cable takes the shape of one of the catenaries, but the trii<' cate-
nary is the curve in which a cable of uniform section and material hangs
under its own weight only. If the load is uniformly distributed per
horizontal unit of length, the cable forms a parabola. In most examples
of suspension bridges the load is supported by rods from a certain num-
ber of points along each cable, and its distribution may be assumed to
be approximately uniform per horizontal unit of length. Hence, the
curve usually adopted for the cables of a suspensiivi bridge is a parabnla •
Let Obe the lowest point of the cable AOB passing over piers at .1
and J3.
JiCt the load supported from the cable by vertical rods be w j^er hori-
zontal unit of length.
Let X ;i be the coordinates of any point /' of the cable with respect to
the horizontal OX a'ld the vertical 6'Fas axes of x and //, respectively.
Let ^ be the inclir ation of the tangent at iF*to the horizontal.
The portion 0/*of the cable is kept in equilibrium by the horizontal
pull //at 0, by the tangential pull 7' at /*, and by the load wx upon OP
which acts vertically through the middle point E of OX, /W being the
ordinate at P.
Hence, the tangint at P must also pass through E, and PEX is a
triangle of forces.
X
~9~
JJ_^^ ,2.//"
XC X 1/ ' w ' '
0)
'■ti'Mm^^-
PARAMETER, &C.
79
the equation to a parabola with its vertex at 0, its axis vertical, and its
2 //
parameter equal to — — *
w
, . T PE 1
^^''""^ rr EN- 7^^ ■■■J^-=T.coso, ^^^^
and the horizontal pull at every point of the cable is the same as that at
the lowest point.
T PE V/+ ;
Also,-
W.X PJ\
y
(3)
or, T=u\x. V 1 + — —
4 >/-
(5). —Parameter, (f-c— Let /j„ /(,. be the elevations of A and B, re
spcctively, above the horizontal line COl).
Let OD rrz o„ 0C= oTjj, and let ((, + o.^ = u = CD.
by equation (1), a\= ""./(„ and a;^= '^~ .h.
w
But a, -f- a.^ — a,
The Parameter =
(-t)
«.V/'i
(t.'^h.
= -= , and (f.^ --
2 //^ «r <r
Let ^„ #^, be the values of'^at .1 and B, respectively.
.-. fan 0, = 2.^T/Ii±^, and tan 6.^ = 2.VT,^^+^'^~ C^)
(5)
(7)
CoroUari/. — If /(, = h.. = /;, .-. „, = f/^= - .
the parameter^—, by equation (G),
4./(
(10)
and tan f^-tan '^,,= -— , by equation (8).
(6). — Pressure upon jners, dr.
Let r, be the tension in the main cable at .1
'* T' " " backstay "
" ''' be the inclination to the horizontal of the tangent to the back
stay at A.
80
LENGTH OF ARC OF CABLE.
The total vertical pressure upon the pier = 7\.sin ^, + T'.ainif — P
The resultant horizontal force at ^l - 7', roa fl, - 7". con ii' = Q
When the cable slide.s over .smooth rnunded saddles (Fi^. 4), the
tensions T^ and T' are approximately the same;
.-. F=-- 1\. (sin 0^ + sin ()'), and Q= 7\.{cofi «, - cos tl').
In order that the rcsulttint pressure upon the pier may be wholly ver-
tical (^ must be zero, or 0\ = ()', and the corresponding pressure is,
F=i2.1\.si)t,i\
Again, the resultant pressure may be
made to act vertically, by securing the
cable to a saddle wliich is free to move
horizontally on the top of the pier,
(Fig. 10).
In this case, T,.c.nsO,= T\ cos^y .=^ H
and .•. P- II. {tan l\ + tan ft').
Su])p(ise the friction at the saddle too great to be disregarded.
Let I) be tlie total height of the pier, and
let \y be its weight.
Let FG he tliu base of the pier, and A' the
limiting position of the centre of pressure.
Let /;, (/ be the distance of P and 11', re-
spectively, from K;
.". ftr sftdx'lifi/ fif position, Q '^ —d. '1' \
and for stdhiliti/ of frirtion when the pier is!
of urasonry. p — r.-._i the co-efficient of friction.
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 Olid of the latter is free to revolve through a small angle about a
horizontal axis.
(7). — LiiKjth of arc of cable. — Let OP = s.
tan 6= -jr-, by equation (7),
.'.scc\n.d0 = J .,Jx = ^.Js.cos
or ds=:
II do
w cosO
WEIGHT OF CABLE.
81
Hence, s= . / — 5—= -r — .\tane. secd + logJtanG \-secd)\ (11)
w J cos 2.W ' ^ / > V /
Again, fail 0:
w
.X, and sec 6
=v
1 + -^.x'
CoroUuri/. — The length of any arc of the parabola measured from
i8 approximately the same as that of the osculatory circle at measured
from and having the same abscissa.
Let R be the radius of this circle.
R=
Parameter x*
2
2.3/
"^^^= b-'(^)y= (^ + -7eO* = ^ + ^- approximately.
-.x +
X
H./e*"
2 y^
--X + - . •—
O X
(13)
(8\ — Weight of Cable. — The ultimate tenacity of iron wire is 90,000
lbs. per sq. in , while that of steel rises to 160,000-lbs, and even more.
The strength and gauge of cable wire may be ensured by specifying that
the wire is to have a certain ultimate teracity and elastic limit, and that
a given number of lineal ft. of the wire is to wpigh o?ic pound. Each of
the wires for the cables of the East River bridge was to have an ultimate
tenacity of 3,400-lbs., an elastic limit of 1600-lbs., aad 14 lineal ft. of
the wire were to weigh oiie pound. A very uniform wire, having a co-
efficient of elasticity of 29,000,000-lbs., 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 ^i (^ = 0D) of a cable of sufficient
sectional area to bear safely the horizontal tension //.
Let W2 be the weight of the length s, (=0-4) of the cable of a sec-
tional area sufficient to bear safely the tension T^ at A.
Let/ be the safe inch stress.
Let p be the specific weight of the cable material.
Wy= —.(ii-p, and W.i= — '—, — '-^i-P
f
f
^^\^ ir,.-^.sec<
:0 = _±.( a, + -.-')( l+—-i + ...)
or Tf,= >r,(l+|^i-), nearly,
6
(14)
82
DEFLECTION OF CABLE.
A saving may be effected by proportioning any given section to the
pull across that section.
At any point («,//), the pull=//. sec0, and the corresponding sec-
, Jf.secff ^, . , .^11 Il.sccd , ,
tional area= — ;; — . 1 he weight per unit oi lcngth= -, — .p, and the
total weight of the length s, (^=0A) is,
/
W.
/ U.secB ds H.p/ ,., H.p I {. 2.y
^....y.^
Butx'-T^.y
.■.F.=f!./-(U2.|.^....):..=f<(..i.'^),nea.,
4 Ai
Hence, W^=^^-(} + ^^
The weight of a cubic inch of steel averages .283-lb.
" " wrought iron " .277-lb.
The volume in inches of the cable of weight Tr, = 12.a,.
(15;
IV'
.', ^— 77=. 283-lb. or .277-lb., according as the cable is made of
12. a,.:^
f
steel or iron.
Let the safe inch-stress of steel be taken at 33,960-lbs., of the best
cable-iron at 14,958-lbs., and of the best chain-links at 9,972-lbs.
•'• ^^=^' «'• ^^ ■'-^'^''3im= i^m ^'' ''''^ ''^^'''
r,=JT,«,x.277x
W,=ff.a,x.277x
12
14,958 "" 4,500
for iron cables.
12
H.a,
, for link cables.
(16)
(17)
(18)
9,972 3,000'
Note. — About |^th may be added to the net weight of a chain cable
for eyes and fastenings.
(9). — Deflection of a cable due to an elementary change in its length.
By the Corollary of § (7), the total length (S) of the cable AGB is,
2 hi 2 hi
Now a, and a,, are constant; A, - /i, is also constant, and .'.dhi^dh.^.
Hence, dS=t(^-^+^'A.dh,
6' \a. a J
_^__. ■ k
Cl'ltVK OF CABI.K.
If /,,Z=/,,= /i.
(i,~(i.,=
10 /.
and (/aS'=^ o • ~- <^^*
(V.))
rosp
Draw tlu! ordinate /■'.V, and let the tangent at y-* uijct (J.X in JJ.
As before, I*.VE is a triangh' of forces, and ^is the middle point of
ON.
•" - (20)
the cquatiotj to a [larabola with its axis paralKl to OY, and its focus at
a point aS' where 4. <SV> ^ "" , ■
CurnUir\j 1.— L't tlie axi'. meet the tangent at in 7", and h't its
inclination to OX be ('.
Let .1 be the vertex, and ON' a perpendicular to the axis.
.•..SY>=,S'7"=*SM 4- AT'z=ii\ 4- AN'
But, \.AS.A N'=ON"=N'T'\ tan^l =\. AS \ Vurl
\ S
■ AS:=AN'. tat^i and SO- AS. (1 + C»t'i)= -,~Tv
Hence, the parameter =4. AS=i. SO. .sin'l. (21)
Corollari/2. — Let F be the obli({ueload upon tlie cable between &i F
'• Q " " total thrust upon the plarform at E.
" w " '* load per horizontal unit of length.
'• q " " rate of increase of thrust along platform.
'' t " " length of i'i;.
w
w '= - — -I and q=-w. cot i
sin I ^
(22)
84
SUSPENSION BRIDGE LOADS.
' -»
)/• ,r
-' //
H=",:L=2>r'.SO=2AS.
2.( w'.f.x
sin^i
=2. A S. -."-.- (23)
r=iir^'= " •'- (24)
.c //
f=,r + ^ + x.t/.Cosi (25)
CnroUiin/ 3.— Let s ha the lengtli of ()l\ and let f) bo the inclination
.'.s=Ar—AO
- ^•^^'l'l[tan (90-/?) ..wc (90-/0 t- Ai(7,.('"» 90^ + s^>c. 9(rr«)
2.W L
-tnn (90-0.«''C (90-(")-/o(;, ( ^ni !KM + scr 0(Pi) |
— , Acot iKcosec H — cot i.oscr I -^ dx]. ,
(20)
It may b.' (>iisily shown as in tlio Cornllary of v^(7) that aj)proxiuiatoly,
•> .y-. ^Z//- '• ^27)
,s=:.l' + y.<'".s' * H — .-
3 .'•
//. ''ON i
(11). — Siisjuuninii hriilijc lodds. — Tho heaviest distributtMl load to
which a hiLihway bridire may be subject'd is that due to a dense crowd
of people, and is fixed by modern French practice at S2-lbs. per sq. ft.
Probably, however, it is unsafe to estimate tlie loud at less than iVoni
100 to 140-lbs. per s(|. ft., while allowance has also to be made for the
concentration upon a single wheel of as nnich as HO.OOO-lbs., and per-
haps more.
A procession marching in step across a suspension bridge may strain
it far more intensely than a dead load, and will wt up a synchronous
vibration which may prove absolutely dangerous.
The fucfor of sa/ctj/ for tlie 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 uuiy be remarked that the efficiency
of a cable does not depjnd so niueh u[)oii its ultimate strength as
upon its limit (>f elasticity, and .so long as the latter is not exceeded the
cable remains uninjured. For example, the hnak'ng weujht of one of
the 15-inch cables of the East River Bridge is estimated to be 12,000
tons, its limit of elasticity being 8,118-tons, so that with H only as a
factor of safety, the stress would still fall below the ela.stic limit and
MODIFICATIONS OK TIIK SIMPLE RUSPKNSION lUIIDOE.
8.')
,1
it
us
•:>
liiive no iiijuriouH lifft-ct. Tlio riDitlinui/ applieation of such ii loail
would doubtlt'SH iiltiiiiatcly K-ad to thi; dostria'tioii of the bridiio.
The dip of tho cable of a suspoii.sioii hridLTC ttsiiully varies fi-oiu
l-15th to 1-I2th of the span, and i.^ rarely as much as much as 1-lOth,
except for small spans.
(12). — Minlljifiifiotis of the simple sitxprusioii hr!ihji\ —
The disadvantages connected with suspension bridt,'es are very <;reat.
The po.sition of the platform is restrieted. massive anclioraijes and
})iers are jrenerally re(|uired. and any change in the distribution of the
load produces a sensible deformation in the struct\ire. Owinir to the
want of rigidity, a considerable vertical and lutrizontal 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 >s that
shewn in Fig. 13,
where the point of the
cable mo.st liable toi
deformation is attach-
ed to tlu' piers by short
straight chains A B.
(b). - A series of in-
clined stays, or iron!
ropes, radiating from
the pier saddles, niay
be made to support the
platform at a number
of equidistant points,
(Fig. 14.) Such ropes were used in the Niagara Britlge, and still more
recently in the East River Bridge. The lower ends of the ropes are
generally made fast to the top or bottom chord of the briilge truss, so
that the corresponding chord stress is increased and the neutral axis
proportionati'ly displaced. To remedy this, it has been proposed to
connect the rojH?s with a horizontal tie coincident in position with the
neutral axis. Again, the cables of the Niagara and East Kiver bridges
do not hang in vertical planes, but are inclined inwards, the distance
between them being greatest at the j)iers and least at the centre ot the
span. This draicing in adds greatly to the lateral stability, which may
be still further increased by a series of horizontal ties.
IMAGE EVALUATION
TEST TARGET (MT-3)
10 ^
I.I
1.25
IIM 112.5
I: iiS
1.4
liiiU
II 2.0
1.6
%
^3
^"
/J
c^.
(Tl
^'/ ^'^#
:^ >
V
Qx
w-
\
86
MODIFICATIONS OF THE SMPLE SUSPENSION BRIDGE.
(c). -In F'v^. 15 two
CJibles in the f^anio ver-
tical [tlane are diagonal
ly braced t<tgether. In
principli' this method i-
similar to that adopted
in the sfijfrniiig truss,
(discussed in § IJJ), but is probably less efficient on account of the
flexible cliaraeter of the cables, althouLrh a slight economy of mattjrial
might doubtless bo realized. The braces act both as struts and ties,
and the stresses to which they are subjected may be easily calculated.
(<0.-In Fig. Itj a sin-
gle chain is diagonally
braced to the platform .
The weight of the bridge
must be sufficient to en-
sure tliat no suspender will
be rubjected t. a thrust, or'
the efficiency of the arrangement is destroyed. An objection to this as
well as to the preceding method is that the variation in tlie curvature of
the chain under changes of temperature tends to Ioc^-mi and strain the
joints.
The Drinciple has
been adopted with
greater prefection i n
the construction of a
foot-bridge at Frank-
fort. The 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 (Fig. 17.)
(O.-In Fig. 18, the I
girder is supjiorted at sev-
eral points hy 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
AUXILIAKY OR STIFFENING TRUSS,
87
ical and
nrought-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 hori-
zontally. The floor beams are sometimes laid on the skew in order that
the two ends of a beam might be suspended from points which do not
oscillate '^oncordantly, and also to distribute the load over a greater
length of cable.
(13). — Auxiliary or\
stiffening truss. — The
object of a stifi'ening
truss (Fig. 19) is to
distribute a passing load
over the cable in such a
manner that it cannot be distorted. The pull upon each suspender
must therefore be the same, and this virtually assumes that the effect
of the extensibility of the cable and suspenders upon the figure of the
stiffening truss may be disregarded.
The ends and A must be anchored, or held down by pins, but
should be free to move horizontally.
Let there be n suspenders, and let T be the pull upon each,
. ^ I T
:. t, the mtensity of the pull per unit of span = T-^ r = n + 1 -y
I being the spa,n.
Let w be the uniform intensity of the dead load.
Let R■^, i?2 , be the reactions at and A, respectively.
Case I. — Th'; bridge partially loaded.
Let w' be the greatest uniform intensity of the live load, and let it
advance from A, and cover a length AB.
Let OB=x
For equilibrium,
R, + R^ + (^ - «'). I -^''- G - x)=0 (1)
&nd,R^.l + l~^.P-'^.{l-.xy =
(2)
Also, since the whole of the weight is to be transmitted through the
suspenders, (<-«?).? — ?«'".(?- x)=0
to
or, t~ic =—.(l-x)
(3)
Wsl
88
AUXILIARY OR STIFFENING TRUSS.
From equations (1), (2), and (3),
-R,=^.j.(l-x)=.R,
(4)
which shews that the reactions at and A are equal in magnitude but
opposite in kind. They are evidently greatest when x=-, i.e., when
w'.l
the live load covers half the bridge, and the common value is then -—
o
The shearing force at any point hetioeen and B distant x' from 0,
=R, + {t- w).x'r=w'!-^. (x' - -|) (5)
which becomes —.-.(I- x)=.-R^=Ri, \s'hen x' equal x. Thus, the shear
at the head of the live load is equal in magnitude to the reaction at each
end, and is an absolute maximum when the live load covers half the
bridge. The web of the truss must therefore be designed to bear a
shear of
w'.l
8
at the centre and ends.
Again, the bending moment at any point between and B, distant
x' from 0,
=i?,x'+i:^> =|'.^il.(^'^_,,.x'), (6)
which is greatest when x'= „, i.e., at the centre of OB, its value then
w Z — cc
being — ^. — — , x^ Thus, the bending moment is an absolute maxi-
mum when , Cl.x^ - x^") = 0, i.e., when x = ^.l, and its value is then
dx X ; ^ ^
w
54"
V
The bending moment at any point between B and A distant x' from (?,
=/?l.x + —^ .X " - — .(x - x)* =. -r-. .-.X - x.l - x'
(7)
d
l + x
which is greatest when — -, (x' -x. Z-x')=0, i.e., when .x'=— ^, or at
dx:
w x
the centre of AB, its value then being q-.j-.(^ - x)*. Thus, the bend-
ing moment is an absolute maximum when (x.l - x ) = 0, i.e., when
dx
X = _, and its value is then + — P.
3' 54
AUXILIARY OR STIFFENING TRUSS.
89
!
Hence, the maximum bending moments of the unloaded and loaded
divisions of the truss are equal in viagnitude hut opposite in direction,
and occur at the points of trisection ( D, C) of OA, when Ihe live load
covers oners one-third (AC ) and tuw-thirds (AD) 0/ the bridge, respec-
tively.
Each chord must evidently be designed to resist both tension and
compression, and in order to avoid unnecessary nicety of calcuhition, the
section of the truss may be kept uniform throughout the middle half of
its length.
In order that the truss
might act most efficiently,
it should be made in two
halves, (Fig. 20), con-
nected at the centre by a
bolt strong enough to re-
sist the shear of
u/.l
The trusses can then deflect freely and are not
sensibly strained by a rise or fall of the cable under a change of tem-
perature.
Case II. — A single concentrated load W at an 1/ point B 0/ the truss.
W now takes the place of the live load of intensity w'.
The remainder of the notation and the method of procedure being
precisely the same as before, the corresponding equations are,
Jii + Jii + {t-w).l- W={) (1')
^-^.P-W(l-x)=0 (2')
RJ +
P- W(l-x)=0
W
t-^w=
I
-^=f(-4)=^'
(3')
(4')
which shews that the reactions at and A are equal in magnitude but
opposite in kind. They are greatest when x— and when x=l, i e.,
W
when W is either at or at A, and the common value is then
2
The shearing force at any point between and B distant x' from 0,
=R, + {t-w).x'=^.(x'-x+^-y (5')
W
which is a maximum when x' = x, and its value is then -^'
90
AUXILIARY OR STIFFENING TRUSS.
w
The web must theicftire be desijiiicd to bear a shear of throui^hout
the whole leriL:th f)f the truss.
Ajraiii. the bendinj; moment at any p(»int hifirccii (ind B distant x'
from O,
First le x< -• The bondinir moment is positive and is a maximum
when x'=x, its value then being +-—- (?.x — x^j.
iVcx/ let x>-^. The bending moment is then negative and is a maxi-
^ • , t- ,. . "' / /\'
mum when x =x — -— , its value then being - • / x - -y ) .
The bending moment at any point hetwecn B and A distance x' from
= .S,.x' -. (< - ,.)/^ - ir.(x' - x)= -.(x' - ( ;^- - ^^ (7')
which is a maximum when -, i x'- ^Y o- - -^ ) [ =0, /.f-., when
2 . W / l\^
— , and its value is then — 5— •( a; — :, ) .
X =eX +
Ahte. — The stiffening truss is most effective in its action, but adds
considerably to the weight and cost of the whole structure. Provision
has to be made both for the extra truss and for the extra material re-
quired in the cable to carry this extra load.
CHAPTER y.
A R C II K D K I B S .
(1). — Lincdr arrhcs. — When a cord han^rs from two points of sup-
port, and is loaded with a iiunibcr of weights, it tends to assume the
form of a /iinini/iir, or vquiUhriinn polyfron, and the horizontal pull
at every point of the cord is the same, (^ (3). Chap. I).
Let and .1 l)e the two
points of support in the same
horizontal plane.
Let ./•. ij, be the co-ordin-
ates of any point Q of the
cord witli respect to 0.
Let P be the resultant of
the weights bt'tween and Q.
Let X, bo the horizontal distance of /'from Q.
Let Fand //, respectively, be the vertical aud horizontal components
of the resultant force at 0.
Take moments about Q.
ILy-V.x + P.x, (1)
Suppose the cord to be exactly invcrtal and to be stayed in such a
manner that it retains its form unchanged. Also, let the weights be the
same in magnitude and distribution. The stresses at different points of
the cord are now reversed in /^lud but not in mapnitude, and equation
(1) becomes,
n.,/=V.x ~ P.x.^M (2)
M being the bending moment at a distance x from \n sl horizontiil
girder of the same span and similarly loaded.
But JI is constant and .'. // oc M.
'Tence, the vertical ordinate^ at points of the inverted cord measured
from the horizontal axis A are proportional to the bending moments
for corresponding sectioits (fa girder of the same span and similarly
loaded.
92
ARCHED RIBS.
If the numbei" of the weights is increased indefinitely, and the
distance between the points of application indefinit<.'ly diminished, the
inverted cord becomes a curve, and forms a lincur arch, or cquilUrrated
rib.
An infinite number of linear arches, or curves of equilibrium, may be
drawn through the points of support, as they are n.erely the bending
moment curve plotted to different scales. These arches transmit a
thrust only, and can never be constructed in practice, as the equilibrium
is destroyed by the slightest variation in the distribution of the load
from that for which the arch is designed, but the principles upon which
they depend may be applied in the dt. termination of the lines <>/ resist-
ance for arches and arched ribs.
Corollitry 1. — If 7/ be unity, the ordinates (j/) are ecjuul to the bend-
ing moments.
Corolhry 2. — The linear arch for a load uniformly distributed along
the horizontal is 2i, parabola, (§ (4), Chap. IV).
(2). — Arched Ribs. — An arched rib is any truss in which both the
chords are concave or convex in a vertical plane, and which gives rise to
oblique reactions at the supports.
Arches and suspension bridges are the two extremes in which the
horizontal stresses developed by the load are met by horizontal forces
applied at the points of support, instead of being made to neutralize each
other within the structure itself, and are therefore the only systems in
which the material should or might be subjected to the same kind of
stress throughout.
Wrought-iron and mild steel are very suitable for small arches in
which parts of the rib are liable to tensile tresses. Cast-iron and cast-
steel, having a greater compressive strength, might seem better adapted
to large spans, as they admit of greater economy, but they are ill-suited
to withstand shocks, and have been almost entirely superseded by the
more elastic materials.
The rib may consist of a number of tubes bolted together, as in the
St. Louis steel bridge, but is usually of an /-section. The depth of the
rib is not necessarily uniform from end to end, but may be advantageously
increased in parts subjected to an excessive bending action.
The space between the roadway and rib, i.e., the spandril, may be
filled up with some kind of lattice work, which should be very carefully
designed, as the artistic appearance of the arch is largely dependent
thereon.
ARCHED RIBS.
93
Generally, the ends of the rib are
securely bolt<!d to iron skcwbaeks built in-
to the abutments. It iss far preforublc to
support the ends on pins or on cylindrical
bearinjis,
Fij;.
2, which allow of a free
rotsitional movement at the sprin^injr.",
so that the straining action in fhosi'
parts due to chaiifres of temperature or
other causes are eliminated. Besides, the h<,rizontal thrust, which is
indeterminate when the ends are fixed, now becomes approximately
axial and determinate, and a consideration of the actual deformation of
different portions of the rib will render its strength completely ascertain-
able. Similar results have been ob'ained in roofs by .suspending the ribs
from single links attached to the supports.
The betiding action at any point of the rib may be made to vanish by
the intn»duetion of a hinge, and the lint ar arch mu^t pass through all
such points. With a live load, however, there cannot be more than
three hinges, or the equilibrium will be unstable and the arch will fail.
The practice of constructing timber arches with bent planks, bolted
and scarfe^l ,ether, is not to be recommended. When wood is em-
ployed, the a -hould bo a frame of which the several members are
straight bal'.s.
Timber arches are usually in the form of a .segment of a circle, and a
semi-circle is often employed for the sake of architectural appearance.
Fig. 3 represents a good
form of metal arch for mod-
erate spans, the chords bt^ing
braced together, so as to pre-
vent distortion. If the .^span
is Inrge, die depth of the rib
may bo increased and kept unitbrm throughout. The cliords uro then
parallel, and the necessity for long braces at the abutmencs is obviated.
A horizontal distributing girder introduced at the top of an arched
rib gives an increased depth, and is also intended to share the variable
stresses induced by a pa.ssing load, so that the maximum stress at anj
point of the rib may never exceed the greatest stress produced when the
bridge is completely covered by the passing load. Much uncertainty,
however, exists as to the mutual relation of the girder and rib and it
seems better to concentrate the additional metal in the rib.
< i
94
GENERAL PUOrOSITION.
With a passing Utad the thrusts upon a pier supporting the onds of
adjacent arches in a viaduct are often unequal, and if the conditions of
stability will allow the pier to bear only a part of the resultant thrust,
the ri^mainder must be converted into an artificial thrust along tlu; via-
duct.
Note. -When the ends of a rib are hinged, // is indeterminate withiu
limits dependent upon the frictional bending moment at the bearings.
(!J). — General proposition. — It is now proposed to describe the uuthod
of calculating tlic /)>v>Ar(/j/« maximum stresses at the diiferent sections
of an arched rib. The stresses cannot be found with mathematical
accuracy, as they depend upon a variety of conditions which are altogether
indeterminate.
The total stress at any point is made up of a number of subsidiary
stresses, of which the most important are : (1) a direci thrust, (2; a stress
due to flexure, (3) a stress due to a change of temperature. Euoh of
these may be iiivo^stigated separately.
(4). — Bmclivg moment {M) at any jwint of an arched rth.
Let A B c, the axis of an
arched rib fiiiiged at the ends,
coincide with a linear arch for
a given load. The thrust at
any cross-section of the rib is
necessarily axial and uniformly
distributed.
For a diflferent load the linear arch assumes a new form, as a d c, and
thf^ etrcBs is no longer axial.
Let V and //, respectively, be the vertical and horizontal components
of the thrust at A ,
Let P, the resultant of the load between A and any ordinate n E F,
meet A c in G.
''•M at E,= V.af-P.qf-ILef
But, = F. A p - P. a p - //. D p
.'. i/= /f ( DF - EF) = //. D E (1),
Hence, ihe bending moment at any point of the axis is proportional to
the vertical distance of that point from the linear arch.
The same is true if the ends of the rib are absolutely ^icec?.
Let Ml be the moment due to the fixture at A.
.". M= V. AP - P. GP - B. EP - Ml
& = F. AP - P. GP - //. DP - Ml
& M=H. DK, as befor3. (2).
I' y I
RIB WITH HINOED ENDS.
95
iV is zero at II, the point of intersection of the axis and linear arch, and
is ncjrative, i.e., clianj^ca in direction, at points on the right of ll.
Corollary. — Let T be the thrust along the axis at E; draw d'e a nor-
mal at £.
.-. r=//.s^cDED' = //.^
D E
and •'*M= ZT. D e= T. d'e
(5). Rih with hinged ends ; iuvnriability of span.
Let ABC bo the axis of a
rib supported at the ends on
pins or on cylindrical bear,
ings. The resultant thrusts
at A and c must necessarily
pass through the centres of
rotation. The vertical components of the thrusts are e(jual to the cor-
responding reactions at the ends of a girder of the same span and
similarly loaded ; H remains to be found.
Let ADC be a linear arch for any given distribution of 'he load, and
let it intersect the axis of the rib at ll. The curvature of the mor*
heavily loaded portion AEH will he flattened, while that of the remainder
will be sharpened.
The bending action at E tends to change the inclination of the rib
at that point, and if the end a were free to move, the line E/ would
take the position EL, al being of course very small ; draw the ver-
tical LO.
AO represents the horizontal displacement of the point E.
Since LAE is approximately a right-angle,
(1)
,AO = AL.
AE
M
But the angle Ael = the change of curvature=-Tr-7 nearly,
/being the moment of inertia of the section of the rib at E.
.-. al_^,^
and .'. A0 = -
AE, nearly,
M „ DE.EF
(2)
But the length AC is assumed to be invariable, so that the total dis-
placement between A and c is nil.
'Zr.DE.EP\ _
(■
E. I J
96
VALUE OF II.
or, since 11 and E are constant,
L\ ■;!!
m
1
1!!
(i^)=o
C^)
The actual linear arch may now be ascertained by drawinj; several
linear arches and selecting the one which most nearly satisfies equation (3).
Corollary 1. — Equation (li) may be written,
/DP-EP\ /l»K. KK\ /EF'\ - ...
l(^— p-j.EF = 0,ors(— ^-j-2(-y)=0 (4)
Hence, the ordinates of the required arch may be directly calculated
from the bending moments at the several sections of a girder of equal
span and similarly loaded.
Corollary 2. — If the section of the rib is uniform, / is constant, and
equations (3) and (4) respectively become,
2. (de.ep) = and 2 (df.ep)-5; (ep*)—
Corollary 3. — Let the
axis of a rib of uniform
section of span / and rise k,
be a parabola, and lot a
single weight be placed up-
on the rib at a point of
which the horizontal dis-
tance from the centre of the span is x.
The linear arch now consists of the two straight lines da, do, and it
32 k P
is found, by applying the condition (4), that D P=— • - „'. ./
O 0.' ^TC.X
Corollary 4. — If the axis of the rib is semi-circular, instead of a
Ml
parabola, it is found that dp is equal to . , i.e., is half the length of the
rib.
No linear arch, however, will even approximately coincide with the
axis of a rib rising vertically at the springings, so that a semi-circular
or semi-elliptical axis is not to be roconiniendcd.
(6). — Value of II. — Let
ADC be the actual linear arch
of a rib of which the axis is
ASC.
Let DP, the maximum or-
dinate of the arch,-=y
Let x, be the horizontal
distance from DP of P, the resultant of the load between A and dp.
PROCESS OF DESIGNING A RIB.
97
Let AF = X
Take uiomonts about D,
v.x—p.x,=rr.
V and V", th
il
npoiicntfi of tlic thrusts at A and c, are
cfjual to tlio corresponding reactions at the ends of a fj;irdt'r of the eame
span and simihirly h".adod, Tho resultant thrusts at A aud C are there-
fore V V' + IP ^"'^ V Y'^ + II -, respectively.
Tho resultant thrust at any other point of the rib is now easily com-
puted, and is res,»lvahle into two coT»ij)onents, the one normal to tiie sec-
tion of the rib at the u;iven point and the other in tho plane of the section.
The latter i;ives rise to a shearing stress and the former (unless axial)
to a uniformly varying stress,
(7). — I'l'ofi'ss of ib'signing
a rih. — For a fixed distribu-
tion of the load the axis of the
rib should be mado to coin"
cide with a linear arch for that
load.
(generally, the rib lias to be designed to carry a live., in addition to a
deatf load, so that provision must be made for the stresses which may
arise from every possible distribution of the former.
Let MN be a numbi',r of equi-distant crosfl-^ections of a given rib.
Draw a linear arch for each of the following cases : —
(l).-When the live load covers the whole of the bridge.
(2),-When the live load covers the middle hal/ofiho bridge.
(3). -When the live load covers one-half of the bridge from one end.
(4). -When the live load covers three-fourths of the bridge from one end.
Calculate the flange stresses in the diflferent sections from the thrust in
the nearest of the four linear arches, and add or remove metal according as
the stress is excessive or falls below the safe limit. If the form of the rib
is much changed by this process, a second approximation must be made.
(8). — ]{ib with ends ah-
solutely fixed. — Let ABC be
the axis of tlie rib. The
fixture of the ends introduces
two unknown moments, and
since H is also unknown,
three conditions must be
satisfied before the strength
of the rib can be determined.
98
RIB WITH ENDS ABSOLUTELY FIXED.
:t
The linear arch, as shewn by the dotted lines, may fall above or be-
low the points A and C ; draw any ordinate df.
Since the ends of the rib are absolutely fixed, the total change of in-
clination of the rib between A and o must be zero.
M
But the change of inclination at any point Eis ~, approximately.
■■•'{in)='=<iu)
or,3.(";)=0 (1)
If the abutments are immovable the span A c is invariable; if they
yield, the total horizontal displacement between A and c may be found
by experiment ; iet \t=/'M.
:.hyp,i{^-^ )=0,or=f,M (2)
The total vertical displacement between A and c nmst also be zero
Referring to § 5 the vertical displacement of any point E is LO.
ButL0=AL.^=^^-^^-^-^*^-^^
AE KI
^DE.AP
A\I
• 2(^^)
(3)
Equations (1), (2) and (3), are the three equations of condition.
CoroUari/ 1. — Let the
axis of the rib be a parabola
of span I and rise k, and let
the section of the rib be uni-
form.
Let a single weight be
placed upon the rib at a
point of which the horizon-
tal distance from the centre of the span is x. The linear arch becomes
the two straight lines da', dc'.
Let aa'=^„Cc'=^j, and DP = y.
Condition (1) gives,
I . \ . /I \ 4,
y.i + (^ + ^) -yi + (^— ^) -y^ = g-'-^
Condition (2) gives,
^5 « .A y . /I
8
yii^-^y^ (^ + .).(^.?-x).y.., (,'-x).(^./.x).y.= «/c.i.
les
EFFECT OF A CHANGE OF TEMPERATURE. 99
Condition (3) gives,
y-(|'-.r)./. (/ «) Q./-.).,, . (-;-.):,.=2./V.
Solving those three equations.
(i , 2 /+10..C, , 2 /-lO.r. ;
-^ f) "' 15 /+ 2,c '^- 15 l~l..c
(9). — /v/ftr^ of (I change of fimprrnfiirc. — The variation in the span
?ofan arch for a change of f from the mean temperature is ± a.t.l,
approximately, a being the co-efficient of expansion.
Hence, if //, is the horizontal force (thrust or tension) induced by
the change of tenipi-rature, the condition that the length AC is invariable
is expressed by tiie e([uati(»n,
"'■K-irf) -"■'■'=' w
Cornllan/ 1. — Let a rib of uniform section of span I tud rise k be
hinged at the ends. Fig. G. The linear arch is the line A c, and equa-
tion (4) becomes.
The greatest bending action is at the crown, and is equal to I^^.k
Coi'olhirti 2. — If the axis of the rib in the previous Corollary is a
semi-circle, instead of a parabola,
.-. ^^'.2(EF-') -\-a.t.l=0=JL.l.P^,;,fJ
E.I "- ^ - E.l IG -
The greatest bending action is at the crown and is equal to //, -
CnroUdri/ )]. — Let the ends of
the rib in Corollary 1 be abso-
lutely iSxed.
The linear arch is a straight
line MN which miust satisfy con-
dition 1, in § 8.
.•. the areas ams and cnt are
together equal to the area SUT,
and .•. MA — NC = ^/r:.
3
100 dp:fllction of an auchkd hiij.
Henco, equation (4) becomes,
^ . l{ izv'-ik.-EF^-^ a.t.l=0 = 3. .( Ijr.l-'^ Jcr\ +<(.f.l
, jj _45 ,, jo.t.
The niaxiumin bondiu'c action= //,• -. A;= T — .K.I. --
3 '1 k
Coral/dii/ -1. — The co-efficiont of expansion per (le^iTce of Fahren-
heit is .(J0()(HIG2 and .0(10001)7 for cast and wruuirht iron beuuis, res-
pectively. Hence, the corresponding total exj)an.si(in or contraction in a
length of 100 ft., for a range of GO" F. from the mean temperature is
.037J ft. ( = .-;*'') and .0402 ft.( = L'y
In practice the actual variation of length rarely exceed.s onehdf of
these amounts, which is chiefly owing to structural constraint.
(10^. — Drjlc-tion •>/ (1)1
orchid n'h. — Li t the abut-
ments be immovable.
Let AHC be the axis of the
rib in its normul ])nsition.
Let ADC represent the posi-
tion of the axis when the rib is loaded.
Let BDK be the ordinat.^ at the centre of the span ; join ah, ad.
... I)F"-AD''- af'=.\ij-.( ) -af^
\(trc. \Ji/
But,
"/•C.AB-'/rr.AD
/being the intensity of stress due to the change in the length of the
axis.
.•.DF^=AB^ A-ZV- AF'=BF'''-An^ | 2. -^ -J lY ]
.-. atjI I 2. j,-(C)' I =BF=' - i)f' = ( hf-dp) («f + df)
— 2.nF.(Bj:))., approximately.
( -_, ) is also sufficiently <mall to be disregarded.
BD, the deflection, =
AB^/
BF E'
k'
k E
n . ^
•en-
res-
II a
! is
•of
tlie
ELEMENTARY DEFORMATION OF AN ARCHED RIB.
(11). — Elementary deformation of an arched n'h. —
101
The arched rib, roprosented by FIl,'. 13, sprintis fiom two abutments
and is under a vortical load. The neutral axis P(^ is the locus of tin;
centres of ijravity of all the eross-seetions of the rib, and may be
reirarded as a linear arcli, to which the conditions L^overiiiuL:; the equili-
briuui of the rib are equally applicable.
Let .LI' bi^ any cros.s-section of the rib. The sepnent .Li'/* is kept
in ecjuilibriuni by the external forces which act upon it. and by tho
molecular action at .LI'.
The external forces are reducible to a siiiirle force at C and to a
couple of which (he moment J/ is the al<rebraic sum of the moments
with respect to C of all the forces on the riirht of (\
The sinj>le force at C may be resolved into a component T alonj»
the neutral axis, and a component S in the plane; A A'. Tlif latter haa
very little effect upon the curvature of the neutral axis, and may be dis-
rejrarded as compared with M.
Before deformation let the consecutive cross-sections lili' and AA'
meet in R; R is the centre of curvature of the arc (.'(/" of the neutral
axis.
After deformation it may be assumed that the plane .LI' romain.s
unchantrod. but that the jilane Bli' takes the position B"B"'. Let AA'
and B"B"' meet in R' ; R is the centre of curvature of the arc GC
after deformation.
i
I
102
ELEMENTARY DEFORMATION OF AN ARCIICD RIB.
Let nJx'. be any layer at a distance z from C
Let CC' = ^>i, CK=R, CR' r= li\ and let la be the seeiioual area of
the layer tihc.
T> • ., £ "c R' + z . iih R+z
By siuular figures, = — „— , and - = — j^-
Aii R Js R
mi M .1 ,, f"' ., A.*.' /I 1 \
iho tensile Stress iiwf6r = 7:<.Aa.^-=:/i'.A^/.—- — )
(lb ah y R R/
= E.\<, -z. (^j^- - — ^ , very nearly.
The moment of this stress with respect to C=E.l((.;.^.\ — j^ \
Hence, the moment of resistance at AA'
.jA-..„..^(±-i)=i'.(;^- ;>jw,
the intc'Tal extendin"- over the wliole of the section.
•"=^-'(-/r- /t)
(1)
Airain, the effect of the force T is to lonjithen or shorten the element
CC, so that the plane BB' will receive a motion of translation, but the
position of R' is practically unaltered.
Curulliirij 1. — ]jet A be the area of the section ^l.l'.
T M.z
The total unit stress in the; layer (thr=p=r -i____, (^2)
the sign being 7>^^s■ or iiuiiks according as J/ acts towards or from the
edge of the rib under consideration.
From this expression may l)e deduced. (1). — the position of the point
at which the intensity of the stress is a maxinium for any given distri-
bution of the load, (-). — the distribution (if the load that makes the
tensity an absolute maximum, (!>). — the value of the intensity.
Corolla n/ 2. — Let ir be the total intensity of the vertic il load per
horizontal unit of length.
Let »", be the portion of w which produces only a direct compression.
Let 7/ be the horizontal thrust of the arch.
Let y^ be the total load between the crown and .LI' which jtroduces
compression.
Refer the rib to the horizontal OX and the vertical OPV as the axes
of X and y, respectively.
)
1)
i
GENERAL EQUATIONS,
Lot X, 1/, be the co-ordinates of C.
:.P=Ij:^;hvLtdF=w,.dx
d.c '
103
also T=ill.
d?
ds
dx
(12). — Gcncnd equations
Let X, y, be tlie co-ordiuates of the point C before deformation
\ ^V/'- " " " " ^iftrr
'' e be an-le between tr.M-ent at G and OX before deformation
"'^'(='^-^^0 " <• " after '. ■
" ds be the length of tlie element CC before deformation.
(3)
(■i)
ds'
after
<(
Effect of flexure. ^L.^.^,,n^f='
ds W ds R
■ ■ £.1 R' R ~ ds' di-
''•AO:=:e-e'=^Ae„-r-^/t-dx,
J L.J dx
ds
ve
ry nearly.
(5)
^(^o being the alteration of slope at F, which has yet to be determined.
Again, cos e'=zeos (ft - so)=eos ft + Afi.sinO
and siti 0'=zsi)iQ) - Aft) = sin ft - Aft.eos ft
. dx' dx di/ di/' dij
ds'
ds
'ds
ds' ds
■ AO.
dx
17
(fi)
Effect of T and of a change of t" in the temperature. —
ds is compressed by T, and . • . its new length due to the compression
In order to make alhiwunce for the change in temperature, this last
expression must be multiplied by {l±a.t), a being the co-efficient of
linear dilatation of the material of the rib.
nt
.■.ds' = ds.(l-^--^y(l±a.t). (7)
Hence, by equations (G),
dx'==(dx + A ft.di/). (j. - -Ly^i j.„.,)
_ T
— dx + A d.di/ - -p—-dx±^a.t.dx, approximately.
104
GENERAL EQUATIONS.
and dy'={ihj - a fl.^/x). (l - -^^ (1 ± ,,.0
T
= d\j- i^O.dx ~ p A 'dy ± a.t.dy, approximately
dy T
or, dx' - //x=d e,-d.Jx - -- .dx ± a.t.ilx
(tx Jli.A
and dy' -dy= -^ e.dx - -—'-■L^dr.-hn.t.'-^.dx
E. A dx dx
O O II
and J, -y= - j ^".^x- J^_.^,,fa ±j ,,. (. -,f & (9)
o
Let / be the span of the arch.
Let x„( = 0),y„, be tlie values oi'x.y, at 1\
Let A Ox, .''i,. .'/u be the values of Aft, x\ y\ respectively, at Q.
C' M ds ,
..A.. = Aft„-J^-^.--.<fe (10)
o
a-,- /= r Aft.|£-</x - r^^-'^'^i r«-'-^- (11)
o o o
y,-y„=) A^.?x-j -^^-^^-.ci.± j ...-^.^, (12)
Again, the general equations of equilibrium at the plane AA' are,
^M dS ^ ._ / ,,</^/x (13)
for the portion «',, Cor. 3, §(11), produces compression only and no shear,
.■.5=5„-J-...^x.//,(|-|^) („)
aS*,, being the still undetermined vertical component of the shear at P,
and '-— the slope at P,
UXg
and, iff = Mg-^-So-x - r A7(/x» + II. (y-y^- x.^'^ (15)
o o
Mg being the still undetermined bending moment at P.
Equations (8), (9), (14) and (15) contain the four undetermined
ds
constants H, S^, ^o,A^„, T being H.j-
11
RIB OF UNIFORM STIFFNESS.
105
LctJ/,, S„ bo the values of M awl S, rospectivLly. at Q.
Equations of condition.-In practice the eiuis of the rib arc cither
Jixrd or free.
If they are fixed A ^,=0. if they arc free 3/;=0. I„ either case the
number of undotermiued constants reduces to f/iree.
If the abutments are immovable, .r, - /=(). If the abutments yield
x,-l must be found by experiment ; let .r, - I =,. //, . beinir ,c.me co-
eincient.
The^/-A'^ equation of condition is a-, - /— 0, or .r - (=ru If
A,<;ain, Q is immovable in a vertical direction.'
The second equation of condition is, //, - y^—Q
Again, if the end Q is fixed, A«,=0, and if free, .V, =
The third equation of condition is ^^, = 0, or il/, =
Substituting ,a equations (14) and (15) 'the values of the three 'con-
stants as detennmed by these conditions, the shearing force and bendin^.
moment may be found at any section of tlse rib.
(i;]) /p of uniform stiffness. -hot the depth anA sectional form
of the r.b be_ un.form, and let its breadth at each point vary as the
semnt of the .nchnation of the tangent at the point to the horizontal.
Let .4,, /„ be the sectional area and moment of inertia at the crown
A I, " " « . • ^
at any point C.
:.A=A.sece=A,.^ (19)
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,
(16)
(17)
(18)
• r r T ds
.1= I\. seiid=I,.-~
. . T II, seed
Again, -= _
A Ai. sec e
dx
(20)
//
-2, and the intensity of the thrust is con-
stant throughout.
Hence, equations (5), (8), and (9), respectively beeom,,,
II
X' -X —
f<
-,— dx
dx
EA
x± a.t.x
(21)
(22)
y'-y~-j ^^''^^~^;j{y~y.)±a.t.{y-y,) (23)
fi
106
rARABOLIC RIB OF UNIFOIIM DEPTH Ax\D STIFFNESS.
Equation (21 ) shews tliat the dofloction at each point of the rib is
the same as that at correspond in;^ points of a straijj;ht liorizontal bi'aiu
of a uniform section ((jual to tliat of the rib at the crown, and acted
upon by the same bendini:; moments.
Ribs of uniform stiffness are not usual in practice, but the formulne
deduced in the present article may be applied without sensible error to
flat sejimental ribs of uniform section.
(13). — I'linif/olu: rlh of uniform dcpfh and stiffness, with rolling
load; the endu jiM'd in dircctioi: ; tlic idnifrntnts immovahlc.
9
i
D
V ■
J
^ .
- ,— — — rm. _Z^<— _ ==^— — — — —. _^ : ^ • 1
/■ -
.__yL__
Fic.l^
1 '//y'^/yyy-. -
Let the axis off bo a t:iii,ii*'nt to tlie neutral curve at its summit.
Let h be the ris(i of the curve.
Let X, ^, be the co-ordinat^^s at any point C with respect to 0.
A.kfi y
(24)
d Sf: /I di,, 4.k dji^ 'i.kd-/ HA;
and, ; - = ( _ _ x), -^ = — r' ~r — r 'ti = i^ C-5)
dx r V 2 (/.'•„ / '/'i I dx' /- ^ '
Let «> be the dead loail per horizontal unit of length.
Letwj' " live " " "
Let the live load ccver a lenutli DIJ, - /'. /, of the span.
Denote by (.1) formula) rclatinu' to the unloaded division OE, and
by (7i) formuhv! relating to the loaded division DJJ
Equations (14) and (1.5) respectively become,
(A) s^-S,+ (^^-ir).x (26)
(B) S=>%, -f (^--AZ^ - »•) . X - w'. ]x -(1 - r).l\ (27)
(A) M^- M + .%,.x + (^dL _ ^o) -^ (28)
(B) J/- ^f, + S..X + (^-^ - «) .^l^'ix-(i-o./r (29)
PAUABOLIC RIB OF UNIFORM DEPTH AND STIFFNESS.
Since the ends are fixed, _so„ = = Aff
Hence, by wiuatiuiis (21) and (28),
and by erjuations (21) and (29),
When .r=A A^^rr^^, =. (». a,.,] therefore by tlie hist e(,uatinn,
107
(30)
(31)
o=^i/„ 4- >y
Again, let :^fl^ ,
dx
(32)
(33)
But A ff. = 0, and'i^^-= ^■^'
By the conditions of the problem, x' - x and / - y are each zero at (J.
Hence, equations (22) and (23) respectively becoue,
O
Substitute the value of A ^ in equation (32). and integrate between
the limits;
J -^(l.tj.
which may be written,
o=j/„ +§•.,+ (^/_ „.).f._ :::,.o,.+?./a_ -'.A'v,
^^T3i
u
108
rAR\noiJc Rin of uniform dfpth and stiffness.
Ecjuations (133), (37), ami (3S), arc the equations of condition.
Stihtract (37) from (33),
.Siil)tract (37) from (3S). uml iiiiilti]iK the result by 3,
Sul)tr:ict (3!0 from (40).
■•^•==H-7^-'')-.io-'^'-'-U-i.r(r)-2\i.:A-^ 4 ' /r ^^^^
lleiico, 7/= (42)
V 4 K.A-.^^
Elimiiiatin<,' .S'„ bctwct'ii (33) and (38),
" \ p / 12 V 3 4-^ ^ ^
By equation (20), 3/ at Q is,
.,,,,.,A,..v„./..C^-)f'p-'' W
Eliniiiuvtinjz S^ between (33) and (44).
••• - ^^^' == - ^^^" + V-7- - "" ) 6 - "■ •'"• (2-3; ^^5)
Substitute in this equation the value oi' M„ in (43),
' V r^ / 12 V2 3 4/ ^ ^
To find the greatest intensity 0/ stress, etc. —
T II
The intensity of the stres.s due to dire-^t compression — - = —
The intejisity 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^ acU'd upon by the same moments as act on the rib,
for the deflections of the beam and rib are equal at every point, (Equa-
tion 21). Also, since the rib is fixed at both ends, the bondinj:^
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 inten-
sity of stress (p,) occurs at Q, and pi= — ± 3/,.-' (47)
Zi being the distance of the layers from the neutral axis.
rAUAnoLic Kin of unifohm depth and stiffness. 109
// and J/, aro both functions of ,-, and thcrofun. p, is an uhmlnte
aximum when '^' = 0= A . '''^^- ^' 'Z''^'
But, }.»y wjuation (42),
and by equation (40),
k.(\ l^^.\'^ \
" di:
dr \i"-dr
1 , n-'./^llV. (,_,,.
■'^■/•.(i-f:*^.A \ A'
^ 4 ,l„/r7
(49)
(50)
(51)
which may b(! written,
The ro(jt.s of tliia equation are,
1 , ^^i /,
•■— 1 and r=:-+-.2_ ^ vl|./.-'
(52)
(5:5)
r= 1 makes Ul zero, so that the maximum value of p corresponds to
one of thj remaining roots.
llence, the maximum thnist= 1 ( II +^ .M^) ^p\ (54)
the values of II and M^ being obtained by substituting iu Equations
(42) and (4G), ^ i ^
(55)
r = ~
I ^--^^ A
^ 1 -_^ A
2 1-/-
and the maximum tension = —. ^ - H + ^^[ J^/^ =^y/' /'gg\
the values of // and M^ being obtained by substituting in Equations
(42) and (4i;),
no
p\ii\noLic lun OF unifgum depth and stiffness.
»• ==-
1 , 45 A
5,3 A
(57)
1 + X-
2 vl,.2,.A'
Efjuation (54) will give the jn-opi-r Hcctioruil area wlii'ii the depth
ami i'lirui (if tlu- rib have been fixctl.
If Eiinatioii (.')()) gives a negative result there is no tension at any
point of tiK! rib.
Curolliin/ 1. — The moment of inertia liiay bo expressed in the form,
/, = g.2'f./l„
q being a eo-effic'eiit whieh depends upon the. figure of the section.
.'. the vi'i.n'minn thrust = -•(II -f — ' ) (58)
JiV q.z\/
and the corresponding value of /• is,
1 -1 45 3?
1 + _ .n. _' ,
;} z, y (^
— O.J
o ■< /.
— -■-, suppose.
(59)
By (42;, 7/ =
v'.l* ,15 .EI. t n '5 T 15 4 ,3 5\
8 - H k V4 8 • 4 /
k. b
(GO)
,.2 9
By(4n),.,/,= -r.,.//,!^„„,,.(^l.-.,.,_;^)
lb ^ A-'' 4 k Vb 4 'If, ,., //-^ 2 , r*\
= 1 +"-'^-(2-3'"4)
•• '^ q.z, S.h \k 2 /tV"*" 4q.z^\ 2/^7
^^(^,5 15 3,x /5 5 rS) ,^^
^5..~,./>( /c W S-'^U- J Vb' 4' '^2Ji'^y..^, V-i 3 "^4/
8.6 VA- 2 A-V 4 q.zyk h q.z, jb* 4 2) ( h j
-^^;(2-3-''^r)
But 1 - ;=• V=c and r=--
2 a; c
. T1 a. '^^> - "--^ /I 15 2, \ _ 1 a.^J?./, «,'.;» /r* ,.' r« x ^^^^
PAUABOLIC RIB OF UNIFORM
l»i:i"TII AND STIKFXESS. HI
Hence, ;/„ tho muxitnum thrust = ^
.V,
7 -f
15
r-k > I
'> i.i
1 +
45 :
rn
-^\at.E.L ir',p
Similarly./,",, the m;xhunn teminn = -^ ( - n,\- ^'^' )
15 .:, 1 ^ '
4 /i"
3
v^- /•=•+".
(C3)
t^>;W/..^ 2.-Lct tho depth of the rib be small compared with k.
.-. ;■ is a small fraction, and the maximum value of ;>. both for tension
and compression corresponds to r = |, approximately.
Hence, equations (02) and (G3), respectively become
P
P
A U U -^ 2 kO ^4 7^-+ aT25- ,T ! (64)
^..0..-, 3.-If 2.>5.. then . mustt'lv, '"^ ''^'^
^^Lquations (42), (40), (C.2) and (G3), in thi^' case respectively be
I'w + w' 15 a. t E.I,
H^
1 15 2,
aA 8 '"-r-^VTrx +20-^.1
i'. = -r
_ 1 15^,
^,( 8
(68)
(69)
^ Lorof/,n-y 4.— A nearer approximation than is given by the preced-
ing results may be obtained as follows :- « J^ « Pieced
deLmatio '^^' ^ ^ ^^' ^^ ^^"^ co-ordinates of a point very near C before
112
PARABOLIC Rin OF UNIFORM STIFFNESS.
Let x' + dx', y' + dy', be the co-ordinates of a point very near C after
deforuiatlon.
.-. ds"" = dx^ + dy-" and f/.s'^ = dx"' + dy'""
.-. ds'^ - di = dx!^ - dx? + dy" - dy"
or, (ds' - ds). (ds' + ds) = (dx' - dx). {dx' + dx) + {dy' - dy)(dy' + dy)
.• (ds - ds).ds:=(dx' - dx).dx + {dy' - dy)dy, approximately.
ds , -, , , ^ dy
.-. dx' - dx = ids' - ds). "- - (dy' - dy)/li
dx 'ix
da
and dy' - dy - (ds' - ds)'!l /if - (rfx' - dx) . ^
dx dy
Hence, by Equations (6) and (7),
y%.. T /ds^^
dx
dx'-dx=.M'^,dx - -£. Y f y,c7;c +«.<. C^y,
E.A \dxj - \dxJ
Jx
T /ds\'dx J , /dsVdx
.dx ^3«.^ .
dy \dxJ dy
and dy' - dy= - SO.dx - J. (% \ !^^,dx ±a.t.{'ll] ."^^.dx
XL. A \dx/ df/
■ < f^'^y
:.x' -X— /Mr' '.dx- / __. _ . ((x-h<i.t. I (—-\ .dx
J dx J E.A\dx) - J ydx)
J J EA\dx)\hj ~ J \dJ-dy-
O o o '
These equations are to be used instead of equations (8) and (9), the
remainder of the caicuhitions being computed pre isely as before.
(14). Parabolic Rib of uniform stiffness, supported at the ends, but
not fi.t(<l. —
Let the rib be simihir to that of the preceding article, but let its ends
be free. In this case J/„=ilf, = 0, while a is an undetermined con-
stant.
'J'he following equations apply : —
(A) S=S,-\-(^-^- w^x
(B) S=So+i^-^- n^x-w'.\x-(l-r).l\
(A) M= S,.x + (^— - - w) •-
, T»x i# cy /S.k.II \ X* to' .
(B) M= S^.x + (-^- - «, j ._ - _.|x - (1 - r).ll^
(26')
(270
^28')
(29')
(31')
(32')
113
PARABOLIC RIB OF UNIFORM STIFFNESS.
Assume that the horizontal and vertical displacements of the loaded
end are ml Substitute the value of a 6 from (32') m (35) and (3G) •
integrate and reduce, neglecting the term involving the temperature '
-''=^''-U<-(^^"--)L-'^'^ (380
From (29*), :ince J/,=0,
Equations (37'), (38'), and (70), are the equations of condition
Subtract (38') from (37'),
''•'=m}'^
/> 1
which may be written,
Subtract (71) from (70),
o)--- w'.l. ( V - - -r ~\ + 3 7/ ^' . ^
/5 V2 2"5;+^-^^x^
.-.0=/^
7V
Hence, Ii=
PAio + ~^(5y-5.r* + 2.r')l
(71)
(72)
(73)
8Jc.(l-,l^A.L)
\ 8 A./c^J
Eliminating *S'„ between (38') and (70),
Also, by (27'),
^i = ^o+ {^—ji wj .1 - w'.rj= - P, suppose. (75)
Eliminating ,S; between (70) and (75)
••■-^=^-('#'-)f»'K'-2-) W
(70), (73). (74) and (76) give the values of //, ^„, ^„ and
Agam, the maximum bending moment M' occurs at a point given by
^ = 0.n(29').
8
(7V)
M
114
PARABOLIC RIB OF UNIFORM STIFFNESS.
Subtract (77) from (75),
Hence, the distance from the loaded end of the point at which tlie
bending moment is greatest is,
F
(78)
l — x=-
to + V)
S.k:H_
Substitute this value of .r in (29'), and. for convenience, put
8./.-.//
ii; + }C —
P
== 711.
^ ?/•' -- m , w' , ,
But by (70), 0=S„+-^,^—l-,yi^.l
P' (79)
Hence, M\ the maximum bendlug m<)nuiif. = -
-~7 , S.I:.IJ\
As before the greatest stress (a fhriisf) = -j' ( H + ^j^M'j -7''i-(^^)
and the value of r which niakrs //, an ohsnhitr niaxiniuui is given by
^=0. But by (79), M' ijivolvos /•'" in the numoraior, and r' in the
dr
denominator, so that 'p = <l. will bo an ((lUiitimi involving /•".
(ir
One of its roots is r=l. which g.iuTixlly -ives a niiinwinn value of
//,. Dividing by r- 1, the o(iuiitinn reduces to one of the thirtroith
order, but is still far too complex lor use. It is found, howeve;, that
r=i "ives a riosr (i/iprnj-iinnfimi to the dhsolufi' maximum thrust.
. , 15 7, 1 _
With this value of /•, and, i(.>r oonvenience, jiutting 1 4 - -j^ j^^
= n.
By (73), -^=8:/;r7\""'-2.)'
By (70), S.= ^-^{>r-,-^y~^-^\,
(HI)
(82)
•8)
PATIABOLIC RIB OF UNIFORM STIFFNESS.
.,(74), -.=,^,{0-i;>'i^'-^'}
B, (76), -S, = P=U(„:^'^)^^"'
Ej (78),
By (70)
( "-'N n-\ w'
,_8-{("'^ 2) — ^-|
/ w'\ n-\ w'
M=
115
(83)
(84)
(85)
(80)
(15). — PuriihoUc Rib of uni/onu stiffness hinged at the crown and
also at the ends.-h^ this case J/= at the crown, which introduces a
fourth equation of condition.
By(2n o=.„.^.(?^_.).(:.:^.(_i,,y
which may be written,
Eliminating S^ between (85) and (70),
8./.-.//
.-. -J- - w =. w'.( - 2r' + 4r - 1)
Hence, ^i=^^- \w- io'.(2r^ _ 4^ + 1 W
By (87),
By (76),
By (74),
S^=~{(3r'~4r+1)
i.=s,=i^--',(,.-i)-
A 6.,=
7C'.P
" 24. E.I
■(l-4.r + 4.r'-r*)
--(r-1)
By (78) and (^QO),l-x=;^ _
tj.w (^r —- L )
By (79), ^f='!^Xr-ly
_ _/
'~ 4
(88)
(8;))
(90)
(91)
(92)
(93)
116 MAXIMUM DEFLECTION OF AN ARCHED RIB.
When r=\,
1 U'\P . ,„ ,r'./^
I
(94)
These results agree with those of (81) to (80), if?( = l.
In gi'mnil, when » = 1,
w + ^ (5.,-^ - b.r* + 1.r') = w - U-' (2./-^ - 4./- + 1), by (73) and (88)
.•.2./-''- 5.r« + 9./-' + 8./- + 2 = 0= (2./- - !).(/• - 1 y.{r' - 2),
1 , -
and the roots are, r=-, r=l, r- ± v2.
Hf'iico, ?( = 1 only renders the expressions in (94) identical with the
corresponding expressions of the preceding article when h=\ or 1.
Again, the intensity of thrust is greatest at the outer flange of the
loaded and the inner flange of the unloaded half (if the rib. and is
=8-i;-i7,H-n-("'-^2J i
2,h/ 1 / V'\ .
The intensity of tension is greatest at the inner flange of the loaded
and the outer flange of the unloaded half »tf the rib, and is
_ P (2,
■ O,
The greatest total horizontal thrust occurs when r=l, and its value is
-^.("■ + «0
(16). — Maximum (leJl"cti'on of an archrd rib The deflection must
Dccessarily be a niaxiuiuui at a point given by A fl={). Solve for x and
substitute in (9) to find the deflection y' ~ y ; the deflection is an abso
hue maximum when — .(_y' - y)=i{). The resulting ecjuation 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 deflec-
tion occurs at the middle of the rib, when the live load covers its whole
length, I.e., when Jt^ = —y, and r=l.
45 /,
=s
Case I. — Rib o/ §(13). -For convenience, put 1 •♦"~T'~r7.2=
/» w + w ' IbaAE.I^
8jk r -¥« "1^
.-. By (42),
U=
(95)
MAXIMUM DEFLECTION OF AN ARCHED lUB. II7
By (48) ana (4«),-,V„=^^.(„. ,. ,,ylzJ^l'^J,^ _ ,,^ (,,,,
liy (3S; and (4:i), ,V„= - (1.''^" ^gj^
By m, (4:i,, (,,7,, .„= - ^!-^. (,,„.. _ ,,,,,;^, ,,.,,;j;') ^...^
IK'uoc, the niaxiimitii dofloctidn
384/;./,' s "^l^w'T'Z-"'^" ^"PP"'^^- (99)
The central d.-flectmn ,/, of a uniform straight b.rizontal beam of
same span, ofsamo section as the rib at the crown, and with its ends
nxed, IS,
I* w + w'
(100)
'^^"r84i^
Henco, neglecting the term involving the temperature,
^h=~^.L
Case U.—jm, of^{U).~
By (78), //=
P w + w'
(101)
(102)
By (74) and (70), ., = ^-^,.. .o/^=^,(103)
By(32),(102)and(103),..= ^.(;:.|!,i;) ^,,,^
Hence, the maximum deflection
A^/■:/ V12 2 -3T/;-'^-^=3^47r/<'''-*-'^')-— ='^'. (1^5)
If the ends of the beam in Case I are free, its central defection
384 EJ "
2)
, . « , d
n
(106)
Thus, the deflection of the arched rib in both cases is less than that
of the beam.
118
HORIZONTAL DISTUIBUTING GIRDER.
(17). — Archfd rlh of uniform slijf'm'ssfxid at the. ends and cnnnerted
at the crown viith n horizonful distrlhufing girder (Sac ^ (2). — The
load is transiiiitU'd U) the rib by vertical struts so that the vertical dis-
plaeemeiits (if correspoiidiiijj; puiiits df the rib and ^irdiT are the same.
The lutrizoiital thrust in tiie load^il is not lu'cossarily erjual to that in
the uiiloadi'd division of the rib, but the excess of the tlirust in the
loadt'd division will be borne by the distributinir trirdei, if the rib and
frirder are connected in such a manner that the horizontal displacement
of each at tlu' crown is the same.
Till- forniulii3 of )^ (18) are applicable in the present case with tlie
modification that /, is to include the uioment of tlie inertia of the girder.
Tilt' maximum thrust and tension in the rib arc given by Equations
((11) and (05).
Let ,: ' be the depth of the girder, A ' its sectional area,
11 M,.z'
The ureatest thrust in the girder ==
((
tension
.'l, + .l'"^ 2.A\/,
(107)
(lOfi)
7/ and J/, being given by Etjuations ((iO) and (t)7), respectively.
The girder must luive its ends so supported as to be capable of trans-
mitting a thrust.
(18). — Semi-circulnr timber rlh. hinged at the ends.
Let F C Q be the axis of such
a rib. Consider a portion (X'"
bounded by vertical planes at C
and C.
Let ;r, i/ be the co-ordinates of
r' with resjM'ct to (\
Let /• be the radius of the semi-
circle.
Let w 1)0 the intensity of the
load which is assumed to be uni-
form/i/ distributed horizontalli/.
The moment of resistance of the rib at 6" = the algebraic sum of the
moments of the thrust at C, and of the weight upon CC, with respect
to (7'.
M=-U.ij +
ir. x^
STRESSES IN SPANDRIL POSTS AND DIAGONALS.
119
v\ r
>>
At either end, x — i/-r. and Jf= 0. .", //=
alsox* = ^. (2. >•--//)
?r
J/ is a maximum when // = -^, its value then being — ^
In the horiz(tntal tan^-nt at (\ take CG = — ; Q G will be the direction
of the resultant thrust at Q.
(19). — Sfrcsst's in spiDufri/ j>(>stx itnd duujnnuh.
Fip:. 1(5 ri'prosents
an arch in which the
spaiulril consists (»t' a
series of vertical posts
and diau'onal braces.
Let the axis of the
curved rib be a para-
bola. The arch is tiuiii equilibrated under a unil'uruily distributed
load, and the diagonals will be only called into play under a passing
load.
Letx, ^, be the co-ordinates of any point i^ of the parabola with re-
^f' 2
spect to the verte^ o. .'. y = -jy' -'^
Let the tangent at F meet C B in />, and the horizontal B E in G.
Let BC=k' :. BL = BC-CL = BG-CN=k'-y.
Let i.Vbe the total number of panels.
Consider any diagonal E I) between the /<-th and (h 4- l)-th posts.
Let u''be the greatest panel (ivc load.
Tlie greatest compression in ED occurs when the passsng load is con-
centrated at the first h-1 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 wjuili-
brium by the reaction R at I* and by the stresses in the three members
met by the secant plane.
Take moments about G.
:. D.GE.cos e=R.AG,
D being the stress in DE, and 6 the angle DEF.
120
METHOD OF DETERMININC RESULTANT THRUSTS.
The greatest thru.st in EF = w' +
" " tension " = D.coxd-
Now,;;=";'.'^
, x+C B k' + y ,, „ k'.x - .T.y
and • aE=GH^.c J''^;^^'!' ^ «,.d GA =U^L±Z^
•J,y 2 'l.y
,, T^ "'' « " - 1 /'/ + k'.x -xy
Hence, Z>=—. r^ — .-^^ -.strd
2 N k ..I ^x.ij
The stresses in the counter braces, (shewn by clotted lines in the Fig.),
may be obtained in the same manner.
w
xus f - ?p,
%o being the dead load upon EF,
If the last expression is negative, EF is never in tension.
(20) — CUrk MuxwvU s method 0/ determining the resultavf thrustn
at the anjiporis of a framed arch.
Lei .^ .s be the change in the length s of any meu-bcT of the frame
under the action of a force P, and let u be tlie sectional area of the
member.
P
,•. ±-rr--*>' = •^•'*j the sign depending upon the character of the stress.
E.a
Assume that all the members, except the one under consideration, arc
perfectly rigid, and let Sf be the alteration in the span / corresponding
a/
to As. The ratio — is equal to a constant m, which depends only upon
the geometrical form of the frame.
.'.a/ =w. as = rh ni.P. ,\—
E.a
Again, F may be supposed to consist of two parts, viz., /, due to a
horizontal force // between the springings, and /, 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,
f_A/ ^^^^
II AS
A:.
Similarly, ! is equal to some constant n, which depends only upon
the form of the frame,
P==f+A==m.JI+n.V
.: Al=± (m\II +m.7i.V)-r^
METHOD OF DETERMINING RESULTANT THRUSTS. 121
Hence, the total clian^o in I for all the members is,
I.sl=±^(in- /I. ■' ) 4- :l( m.ii.V. ,' ]
If the abutments yield, let ^:s/=u.//, n being some co-offieient to be
determined by experiment.
If the abutments are iuunovable, 2..^/ is zero.
(A)
and 11= -
(B)
T^ is 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 77 and T', and the stress in each member may be computed
grapliically or by the method of moments. In any particular case pro-
ceed as follows : —
(1 ).-Prepare tables of the values of /u and n for each menibcr.
(2).-Assunie a cross-section for each member, based on a probable
assumed value for the resultant of Fand //.
(3) .-Prepare a table of the value of ;>r.^— for each member, and
h.ii
form the sum j; ( »r. 1,- )
V E.n I •
(4). -Determine, separately, the horizontal thrust between the spring-
ings due to the loads at the different joints. Thus, let c,, r.^ be the
vertical reactions at the right and left supports due to any onv of these
loads. Form the sum zim.n.V. — ^ using c, for all the members on
the right of the load and r^ for all those on its left. The corresponding
thrust may then be found by E<iuation (A) or Equation (B), and the
total thrust 11 is the sum of the thrusts due to all the weights taken
separately.
(5).-Ref eat the process for each combination of live and dead load
so as to find the maximuiu stresses to which any member may be sub-
jected. (§7).
122 MKTIIOD OF DETEKMINING KESULTANT THRUSTS.
(C).-Ifthe assuuifd cross-soctions arc not suitiulto these maximum
stresses, make fresh assumpticms, and rcjxat the wlide ealcnhition.
i\^o/r -The same method may he applicl to determine the resultant
tensions at tlie supports of a framed suspension bridge.
Notr.— The fcrmuhe for a paraholie rib may be applied without
mate-rial error to a rib in the form of a se-ment ofacirele. More
exaet formulio may he obtained for the latter in a manner precisely
similar to that described in Articles ll} to 17, but the integrations will
be n.ueh simplified by using polar coordinates, the centre of the circle
beiiig the hole.
!■ ^ I
CTTAPTER VI.
DETAILS OF CONSTRUCTION.
(1). — Chonh (tfmit'iu tnissrs (itiJ Jfoor hffnvs. — Tlio diordsof a main
truss are to be dcsiu'iu'd fur a uiiii'orm distributiDii of the live load over
the whole of the bridire.
First, let the depth of the truss be uniform throuirhout. Draw the
curve of bending moments A c E G K M o Q s to a given scale. Any ordin-
ate of the same curve may, by altering ihe scale, be made to represent
tlie thickness of the cliord at its foot.
Erect ordinates 1 C, 2 E, 'A (i...at each panel point, and complete the
rectangles BS, DQ, FO, and IIM. The broken outline a B C D E F G u i. m
o P Q R S is a delineation of the least admissable thickness of the chord at
different points in its length.
The floor-beams of a highway bridge are subjected to a distribution of
load which is ajtproximately uniform, and may be designed in the same
manner,
IVicorctioiUi/, the thickness of the cliord at the ends is zero, but in
practice it is made equal to that at the nearest panel point.
Next, let the depth of the truss or beam vary. Draw the bending
moment curve to such a scale that its greatest ordinate coincides with
the central depth. If this curve be taken as the axis of the chord, the
thickness of the latter is constant.
124
PLATFORM.
Let A B be
the betidiiif; iiio-
liiciit curve for
tilt' U'lisidii fliingc!
of 11 fldor-bt'iiiu
I'ur ;i hiirliway
bridp', the ci;ii-
tral depth of the
beam being II
(Fig. 2). AO B\)i the uxia of the curved flange, but. in order to avoid
the difficulty of curving, the flange is oft^in constructed iu three straight
pieces AG, CD, Dli, tangential to the curve,
Iu r<tl/)r(ii/ brid-
ges, the live load is
concentrattid at sin-
gle points of the
floor beams. Thus,
for a si.'xife track,
the bending mo-
ment carve is the
broken lin^ ACDB
(Fig. 3). the load
being applied at the
two points E and
F, while for a
(hmhie track the curve is the broken line AG C D II B, the load being
applied at the four points A', E. F, and L, (Fig. 4).
As before, if // is the central li'pth of the beam, the broken line
in each Fig. is the axis of the corresponding tension flange, whi-ih may
therefore be made of fhne .'Straight pieces for the single, and of Jive for
the double track. In the latter case, however, the flange is generally
made of the three straight pieces, AM, J/.V, NB.
The depth of a floor-beam may vary from i-th to J-th of its clear span,
and should be made as great as possible.
(•^)- Fhifform. — The floor of a higJnrni/ bridge is laid upon a series
of stringers, which run parallel to the centre line of the bridge, are
spaced 2-ft. centre U^ centre, and are supported by the floor-beams. The
latt<}r are consequently loaded at several equi-distant points, and the dis-
tribution of the load may be assumed to be approximately uniform. In
PINS.
125
a railway bridp;e a strinper is usually placod uiulornoath each line of
rails, and is supported by the floor-beams, wliich may, thori'f'ore, bo loadi'd
at two ox four points accordinj^ to the numb;.^ of the tracks.
The entire platform may be constructed of timber wherever such
material is abundant. The floor of a narrow bri(l};e, for examiijc. may
consist of strong? planks stretehing between the main trussi's, or. better
sti'.l, the planks may be laid upon timber floor-beams. These fl(»or
beams, for a sinjjjle-track bridj^e, are often 12-ins. widt; by !)-ins. deep,
and are spaced about 3-ft. centre to centre. The rails are laid upon
longitudinal strinjrers which are about 12 ins. wide by 9-ins. deep and
are notched and spiked to the floor-beams. In all such ca.ses the main
trusses should be braced together with iron tie-rods.
A timber platform may be .set on fire by the cinders from a locomo-
tive, and it is sometimes neces.sary to protect the surface with iron
plates, gravel, etc. If there is any traffie underneath the bridge the
planks should be tongued, and the joints caulked, so as to prevent leak-
age.
The entire roadway is rarely made to depend upon timber only,
except in the case of small bridges. The longitudinals and floor-beams
may be of wrought-iron or steel, and are nearly always /-shaped in
section. Rolled joists, from 4-ins. deep and 18-lbs. per lineal yd. to 15-
ins. deep and 2()0-lbs. per yd., may be obtained in the market at a
reasonable cost, and are recommended as floor-beams. In railway
bridges, however, built plate-beams are more common, as they can be
connected to the main trusses with greater ease and efiect. In design-
ing these plate-beams, the dimensions of the plates should be kept as
uniform as is practicable, and should be such as are obtainable in the
market, special sizes invariably adding to the cost.
The floor is often made of timber for the sake of economy and light-
ness, but a floor of curved, corrugated, buckled, or flat plates stiffened
with angles or tees, will add to the strength and durability of the whole
bridge.
Cast-iron floor-beams have been rarely adopted except in cast-iron
bridges.
(3). — Pins. — One of the most important members of a bridge con-
structed on the pin-system is the eye-bar, or link, which is a flat bar with
enlarged ends, having eyes for the insertion of bolts or pins. The eyes
are forged in the ends under hydraulic pressure, with suitably shaped
dies. Careful mathematical and experimental investigations have been
undertaken to determine the proper dimensions of the link-head and pin.
126
PINS.
In consequence of the complex character of the stresses devclopoJ, an
exdct mathematical solution is impossible, but approximate results may
be obtained whicli differ very little from the results of experiment.
Let d be the
depth, and t ^hc
thickness of the
sh<nik of the cyo-
bar represented in
Fi,L'. 5. Let S be
the width of the
meLal at the sides
of the I'ye, and JI
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 em-
braced, and failure may arise from any one of the filldwing causes: —
{(i). -The pin mnij he shorn thr'mgh.
Hence, if the pin is in double shear, its sectional area should be at
least one-hdf {hat of the shank.
It may happen that the pin is bent, but that fracture is prevented by
the closing up of the pit ces between the pin-head and nut ; the efficiency,
howi'ver, of the connection is destroved, as the bars are no longer free to
turn on the pin.
2 4
In practice, I) iovji<it bars, varies from ^-d to - d but usually lies
between - -a and —-il.
4 5
The diar. of the pin for the end of a round bar is generally made equal
to Irjr-times the diar. of thi; bar.
Tiie pin should be turned so as to fit the eye accurately, but tlie best
practice allows a difference of from ^L to j J^ in the diars, of the pin and
eye.
(J)). -The link may tear acroasMN.
Hence, the sectional area of the metal across ^VA^'must be at least equal
to that of the shank, and in practice is always greater.
11 5
S varies from ^q'^^ to -^
EFFECT OF BENDING ACTION UPON A PIN.
1.27
Tlic sectional aroa throiiuh tho Hides of the eye in the head of a round
bar varies from li-timcs to ticicr thatof tlio bar.
(c).-The pin may he torn through the head.
Theoreticallij, the sectional area of the metal across /'(^should be one-
half that of the shank. The metal in front of the pin. however, may be
likened to a uniformly loadeil girder with both ends tixed, and is subjected
to a bending as well as to a shearing action. Hence, the tain iiniiin value of
11 has been fixed at - •(/, and if // is made equal to cl, both kinds of action
will be amply provided for.
(d).— The bearing s)ir/ace niai/ he instiffielint.
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.
In practice, adequate bearing surface is obtained, by thickt'ning th<i
head so as to confine the maximum intensity of tin; pressure within a
given limit.
(^e).— The head may he torn through the shou/drr at XV.
ii !!■ , A'l'' is made CfpUil to (/.
The radius of curvature Ji of the shoulder varies frmn Ih-d to 1 .Gd
d 1
Note. — The thickness of the shank should be - or „•</ at least,
4 /
(4). — Efect of bending action ujion pin.
Figs, li and 7 represent groups of cy mjuis a.s tliey often oeeii in prac-
tice.
Let two sets of 2.» bars pull upon the pin in opposite directions, and
assume that the distribution of pressure between the pin and the eye-
bar heads is uniform.
Let P be the stress in each bar.
128
GENERAL REMARKS.
Let;) ue the distance between the centre lines of two consecutive bars.
All the forces on each half of the pin are evidently equivalent to a
couple of which the moment is n.P. j).
Hence, at any section of the pin between the innermost bars,
c
/bcinir the stress in the material of the pin at a distance c from the
neutral axis, and / the moment of inertia.
If there are 2« - 2 bars in one set, the bending action at the centre of
the pin is nil.
In general, the bending action upon a pin connecting a number of
vertical, horizontal, and inclined bars, may be determined as follows : —
Consider one-half of the pin
only.
Let V, Fig. 8, be the resultant
stress in the vertical bars.
It is necessarily equal in magni-
tude but opposite in direction to
the vertical component of the re-
sultant of the stresses in the inclined bars. Let v be the distance
between the lines of action of these two resultants. The correspond-
ing bending action upon the pin is that due to a couple of which the
moment is V. v.
Let h be the distance between the lines of action of the equal resul-
tants //of the horizontal stresses upon each side of the pin. The cor-
responding bending action upon the pin is that dug to a couple of which
the moment is //. h.
Hence, the maximum bending action is that due to a couple of which
the moment is the resultant of the two moments V.v and Il.h., viz.
(5). — General remarks. — Tension chord pins are most severely strain-
ed when the live load covers the whole of the bridge.
The advantages of pin-connected structures are facility of erection
and transport, and the interchangeability of many of the parts. On the
other hand, pins are often badly designed, the effective strength of a bar
with hinged ends as a strut is considerably less than if the ends were
securely fixed, and the hammering under a live load causes the eyes to
elongate and i\vi pins to work loose.
A properly riveted bridge is certainly far more rigid than one con-
Btructed on the pin system.
Rm:TS.
129
to
(6). — 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 p<in-shaj)(d
head (Fig. 10). 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 the hole prepared for it. and the tail is made
into a hut ton, or point. The hollow cuji-tool gives to the point a nearly
heniispherieal shape, and forms what is called a smiji-riret. (Fig. 2).
Snap-rivets, partly for the sake of appearance, are commonly used in
girder-work, but they are not so tight as con lca^pointed rivets {staff-
rivet^). wliieh are hammered into shape until almost cold, (Fig H).
When a sunMith .-iirfaee is required, the rivets are counter-sunk
(Fig. 4). The C(»unter-sinkirig is drilled and may extend throiujh the
plate, or a shoulder may be left at the inner edge.
Cold riveting is adopU^d for the small rivets in boilor work and also
wherever heating is imprac'ticable. but tightly-driven turned bolts are
sometimes substituU^l fur the rivets. In all sueh eases the material of
the rivets, or bolts, should be superior in (juality.
Loose rivets are easily discovered by tapping, and, if very loose, should
be at once replaced. It must be borne in mind, however, that expan
sions and contractions of a complicated character invariably accompany
hot-riveting, and it cannot be supposed that the rivets will be perfcftly
tight. Indeed, it is extremely 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 eflfective. A machine will squeeze a rivet, at almo.st any
t>'mperature, into a most irregular hole, but the exigencies of practical
conditions often prevent its use, except for ordinary work, and its advan-
tages cannot be obtained where they would be most appreciated as,
e.g., in the riveting up oi' connexions.
130
STRENGTH OF PUNCHED OR DRILLED PLATER.
(7.) — Dlinrnslnns of rivets — The diam(-ter {d) of a rivet in ordinary
girder-work varies from ^-in, to 1-inch, and rarely exceeds l^in.
The thickness (t) of a plate in ordinary girder-work should never be
5
less than i-in., and a thickness of fin., or even -r-;-in., is preferable.
lb
Let T be the total thickness through which a rivet passes.
According to Fairbairn, when t < ^-in., d should be about 2.t
" " •' " < > i-in. " " " \\.t
** '' Unwin, when t varies from ---in. to 1-in. and passes
through two thicknesses of plate, d lies
V . 3,5 ,7,8
between -t + 7-7 and 7:-t + ~.
•i lb 8 8
T 5
" " " when the rivets join several plates, <Z= • -f- —
u
<(
(I
" French practice, diar. of he;\d = l^-'A
'• " " length of rivet from head=r+l§.(^
'• Rankine, lenijth of rivet from head = T+2.}f.d
The rise of the head —d
•J
The diar. of the rivet hole is made larger than that of the shank by
from .-7-;in. to -^-.in., so as to allow for the expansion of the latter when
32 8
hot.
There seems to be no objection to the use of long rivets, provided they
are properly heated and secured.
(8). — Strength of Punched and Drilled Plates. — Generally speaking
experiments indicate that the tenacity of iron and steel plates is con-
siderably diminished by punchuuj and that the effect of drilling is
similar, though less in degree. It would seem as if the deterioration in
tenacity is due, not to the cracking caused by the punch, nor to any
reduetinn (if tenacity in the neighbourhood of the hole, b)it rather to an
alteration in the elasticity which destroys the power of elongation in a
narrow annulus around the liole. The removal of the annulus (about
— in. thick^ neutralizes the effect of punching, and the plan is therefore
sometimes adopted of punching the holes -.in. less in diar. than the riv-
ets, the holes being subsequently enlarged and counter-sunk.
RIVETED JOINTS.
131
oil-
is
in
my
an
n a
out
ore
nv-
Punching docs not sensibly aflFect the 8trenf!;th of Landore 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, how-
ever, 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 diar. from the edge than when it is so far that
there is no bulging at the edge.
All experiments go to prove that the original strength of a punched
plate may be restored by annealing.
Whenever the metal is of an inferior quality, the holes should be
drilled. J^rilling is a necessity for first-class work when the diar. of the
holes is less than the thickness of ihe plate and also when several plates
are piled. It is impossible to punch plates, bars, angles, etc., in spite
of all expedients, in such a manner that the holes in any two exactly
correspond, and the irregularity bec^.ues intensified in a pile, the pass-
age of the rivet often being completely blocked. Recourse is then had
to the barbarous drift, or rimrr, which is driven through the hole by
main force, cracking and bending the plates in its passage, and separa-
ting them one from another, (Fig. 5).
The holes may be punched for ordinary work, and in plates of which
the thickness is less than the diar. of the rivets.
If the punch is helical or spiral inform the injury to the plate is less,
and may be diminished still more by giving the hole in the die-block a
diar. greater than that of the punch.
Hence, the loss of tenacity in punched plates depends upon : —
(l).-The character of the material operated upon.
(2).-The amount of the metal between the holes, or, probably, the
ratio of the diar. of the punch to tlie pitch or the punched holes.
(3). -The diar, of the punch, or, probably, the ratio of the diar, of the
punch to the thickness of the; plate,
(4:).-The ratio of the diar. of the punched hole to the diar. of the
hole in the die-block.
(9). — Riveted Joints. — Riveted joints may be classified as Lup,
Fish, and Butt ynntfi.
In Lap joints, the plates over-lap, (Figs. 15 and 18), and are riveted
together by one or more rows of rivets.
132
STRENGTH OF RIVETED JOINTS.
[3^
mmm^W
PI
I
tesuKMme^A
^^^^■^H
^s^
FiS
16
Pig. 16
r
Fig. 17
Fic.18
In Fi&\i joints, the ends of the plates meet, and the plates are riveted
tea 8injj;le, (Fig. 16), or to two, (Fij^. 17), covers by means of one or
more mws of rivets on each side of the joint.
A Fish joint is termed a Rntt joint when the plates are in compres-
sion. The plates should butt evenly against one another, althougli they
seldom do ^!0 in practice. Indeed, the mere process of riveting draws
the plates slightly apart, leaving a gap which is often concealed by caulk-
ing. A much better method is to fill up the space with some such liard
Bub-tance 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
platiis carefully, and then bring them into close comtact, 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, ibnihle, triple , according as
the joint is secured by one, two, three, , rows of rivets.
Double, triple, , riveting may be chain, (Fig. 19), or zig-zag,
(Fig. 20). In the former case the rivets form straight lines longitu-
dinally and transversely, while in the latter the rivets in each row
divide the space between the rivets in adjacent rows.
(10). — Strength of riveted joints. — The strength of a joint, as derived
from the rivets alone, evidently depends upon the number of shears to
which they are liable. Thus, a rivet in double shear, (Figs. 17 and
18), is twice as strong as a rivet ia single shear, (Figs. 15 and 16).
According to Fairbairn, if the strength of an unpunched plate be
taken at 100, the comparative strength of single and double riveted
joints are as 56 and 70, respectively, but this estimate is probably too
high. At least it does not seem to be founded upon a satisfactory basis,
STRENGTH OF RIVETED JOINTS.
133
nor to give satisfactory results, for the ratio of the tensile to tlie shear-
ing stress, as deduced from Fairbairn's experiments, varies with different
thicknesses of plates, which appears to be contrary to reason.
Fairbairn proposed to make the joint and unpuncheJ plate equally
strong, by increasing the thickness of the punched portion of the plate,
but it is difficult to carry this out in practice.
Very few, if any, trustworthy data have been supplied respecting the
resistance of plates to indentation by rivets >r bolts. The resistance
must of course vary with the quality of the iron or steel, but the degree
in which the tenacity of the plat«'s is affected by the indentation is most
uncertain. The mean crushing intrnsiti/ may be defined to be the ratio
of the total stress to the bearing area (the j)roduct of the diar. of the
hole by tht- thicftness nf the plate), and has been fixed at 5 tons per
stj. in. for icrought-iron, but on theoretical rather than practical grounds.
If it is also assumed that the bearing and tearing strengths of a joint are
the same, it is found that the crushing pressure rapidly increases with
the ratio of the diar. of the rivet to the thickness of the plate. When
the ratio exceeds 3, the crushing pressure in a double-cover joint is more
than 40,000-lbs. per sq. in., i.e., it has reached its ultimate limit, and the
joint will fail, although the tension in the plate is less than SGOO-lbs. per
.sq. in. Hence, the diar. of the rivet should not be greater than three
times the thickness of the plat«. In all probability the failure of the
joint at such a low tensile stress is partly due to the deformation of the
metal around the hole, which causes an unequal distribution of stress.
Until further experiments give more definite information, the safe
bearing intensity for wrought iron may be taken at 5-tons per sq. in. in
ordinary work, and at 7^-tons per sq. in. in chain rivetted joints with
well supportt>d rivets. The safe bearing intensity for steel may be taken
at the same proportion of the ultimate stress as in the case of wrought-
iron.
Let S be the total stresses at a riveted joint.
" /I'/s'./'a-/*! be the safe tensile, shearing, compressive, and bear-
ing unit stresses, respectively.
" t be the thickness of a plate, and w its width.
" A' be the total number of rivets on one side of a joint.
" n '' '' in one row.
" p be the pitch of the rivets i.e., t!\e distance centre to centre.
" (/ be the diar. of the rivets.
" X be the distance between the centre line of the nearest row of
rivets and the edge of the plate.
134
STRENGTH OF RIVETED JOINTS.
It is impossible to apply the ordinary mathemafical rules to the
determination of the ultimate .<trengthof a riveted joint, as the distribu-
tion of the stress which produces rupture may prove to be very
irregular.
This irregularity may rise, (1). -Because the stress does not coin-
cide in direction with the centre line, (2) -Because of local action,
(3).-Because of the existence of t^traius within the metal before
punching.
The joint may fail in any one of the following ways : —
(1). — The riveta may shear.
(2). — The rivets may be forced into and crush the plate.
(3). — Tlie 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.
Vulw of X. -It has been found that the minimum safe value of x is
d, and this in most eases gives a sufficient overlap (=:2..c), while
3
x=--d is a maximum limit which amply provides for the bending and
u
shearing to which the joint may be subjected. Thus the overlap will
vary from l.d to 3.(^.
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 x-.^, 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 excepted) may be likened to a uniformly loaded
beam of length c?, depth x^ - -.~, and breadth <, with both ends^et/. Its
moment of resistance is therefore ^-.^ (x-i — o) , /being the max. unit
stress due to the bending. Also, if P is the load upon the rivet, the
. P
mean of the bonding moments at the end and centre is -d.
Hence, ajiproxini'iteli/, —a = - •^ ( x, - - j
-, 4/.</ </\^
It will be assumed that the shearing strength of a rivet is equal to the
strength of a beam to resist cross-breaking.
STRENGTH OF KIVETEI) JOINTS.
135
-J\ = {p-d).t.f,= d.t.J\
(11). — Singlc-ruettd lap and single cover Joints {Figs. 15 invl Ul)
(1)
2..,.,/,=-^,/,,
■K <
/«
3 (/
(12). — Single-riveted dt)ubic cover Joints, (Fig. 17).
■A=(p-d).t./,.^d.t./,
TT d^
2..r,.^/,=2. --—/„
.T,= —
d^
(4)
(5)
(<5)
3 ./ v^'-ii^-r 4-/'' •••'■'^^n- v:;-!.^
Ar„^^.__The joiiits in § (llj are weakened by the bending action,
possibly also by the concentration of the stress towards the inner faces of
the ^''ates, and oxperinjtiit shows that they are not nearly so strong as
the joints in i^( 12 ) of the same theoretic strength. The second cover in
the latter must necessarily diminish the effect of the bending and pre-
Bcrvo a more uniform distribution of stresa. In all probability, the
equality is restored by riveting a stiffeuer (e.g., an angle-iron) to the
plates in the lap or single cover joint, and in any case the difference will
wholly disappear if the joint is double riveted.
Again, equations (1) and (4) shew that the bearing unit-stress is
twice as great in the joints of § (12) as in the joints of ^ (U), so that
rivets of a larger diar. may be employed in the latter than is possible in
the former, for corresponding values of — ••
(13). — Chdin riveted Joints, (Fig. 19).
Mw-H.d).t = S=J\.N:d.t
^ = ^'—r--fii when there is one cover only.
n.d^
^ = ^' -ij- -fi, when there are two covers.
(7)
(8)
(9)
This Class of joint is employed for the flanges of bridge girders, the
plates being piled as in Figs. 21, 22, 23, and n being usually 3, 4 oi .5.
In Fig. 22. the plates are grouped so as to break Joint, and opinions
differ as to whether this arrangement is superior to they«// butt shcwu
136
ZIG-ZAG AND OTHER RIVETED JOINTS.
, _• », r\r\r\ r< r\ r\ /^,r^. ^ ^ . ' *
r • ■ ' f ' — ' ■ 1 J . « ■
^ ' : —^ .
. I, '' v^ v^t^ <^ VJ \J \^ wj ' I
J - I • , • ■ '
<..".. • , Fig. 21 . r.
^ r^ ^ ^ r^r-. O o r-i o' Oi O , ,r^ ^ r.. ^ ^ ■ ^ ^ ^
C^ ' ' , =^\ n — ^
Fig. 22
>
'' v-" v-f •-0 — — »=» — «=> — «::> — ^13 — iw)i ^_^ v.^ 1
FiG.23
in Fij;. 23. The advantiif^ea of the latter 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
accompanying l/roken joints are dispensed with.
A good practical rule, and one saving much labour and expense, is
to make the lengths of the plates, bars, etc., multiples of the pitch, and
to design the covers, connectiona, etc., so as to interfere with the pitch
as little as possible.
The distance between two con.secutive joints of a group, (Fig. 22), is
generally made equal to twice the pitch.
(14). — Zig-zag and other riveted joints (Fig. 20), —
A few experiments indicate that chain is a little stronger than zig-
zng riveting. An excellent plan for lap and single cover joints is to
arrange the rivets as shewn in Figs. 24 to 28. In reality, the strength
COVEHS.
137
, 13
ind
tch
is
'9-
to
th
nf the plate at the joint is (>iily Wfukonod by nnr rivet Imlc. fur the
pliii cannot tear at its weakest section, i.r., alonjr the central row of
rivets {ii<t). until the rivets between it and theedi^e are shorn in two.
Let there be in rows of rivets, 11, 22, HiJ (^^'J-'- 2G).
The total number of rivi'ts is evidentlv w".
Ii(t/,, 7„. 7„ y, bo the unit tensile stresses in the plate alon<'
the lines, 11, 22, 3'^, , respectively.
.-.S^ (u-->I).(j\~^'^.vi\/,, for the line 11
(H'-'J.l/)^7,= ^ .(m^-l).y'. '• " L'2
{w - 3d).tq, = .(/n» - :j ) /;. " •• 3.i
4
{w - 4.d).t.(j, = ^- .(m^ - t; )./;, ••
4-t
ur - 1
Assume that /,=7„ .-. »•=(?«'+ 1).^/.
Hence, by substituting this value of w in the first of the above
I *• ^^ '^ /i •
relations, ^ =j-.'--- since q,^,q^, are each less than /„ so that the as-
sumption is justifiable.
(1.5). — Covers. — In tension joints the strength of the covers must not
be less than that of the plates to be united. Henco, a single cover should
be at loa.st 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 borne by the nearest cover. It is. tluTcfore,
considered advi.^^able to make its thickness |-ths. that of the plate.
The .lumber 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 dispen.sed with in per/ret y«w/* joints, and a
great ecommiy of material effected, if the difficulty of forming such joints
and the increased cost did not render them impracticable. Hence, it
may be said that covers are required for all eomprrssion 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
138
KIVET CONNEXION 15KTWKKN FLANOKS AND WEFl.
of the best exiiniploH of bridyc construction, the tension and couipnssion
joints are i(k'ntical.
(IG). — liio(t ('i)nnexion hvtween Jl(ni(j< s and web.
The web ia usually riveted to the angle-irons which form part of the
fla" «
.1/ be the bending niomont. and .S' the shearing force, at the jioint
of >vnicli tlie abscissa is x, x being measured along the girder from any
given origin.
Let /( be the deptli of the girder,
liCt/; be the safe shearing stress per unit of area.
Two cases may be considered.
Otisc I.-l'i'(te \Vt;h. — The shearing area of the rivets connecting the
■web with tlie angles must be e({ual to the theoretic horizontal section of
the web, i.r., to .y, and the corresponding number of rivets per unit
Jl-d[_ 11 nJ'/^
1^ .'be the liorizontiil stress transmitted to the angles through the
web, and tending to make them sUde over the flange face.
,'.F.h, the increment of the bending moment, = -^ = S,ot F=—.
dx
n
s
The requisite shearing area = -r-.- and number of rivets is therefore
the same as before.
Cuse II -Open M'eh. — One of the
braces which meet at each apex is a
strut, and the other a tie.
The sectional area of the rivets con-
necting each brace with the an<rle-irt>n
must be ecjual to the net section of
that brace.
Let /' be the horizontal component
of the diagonal stresses.
.•. F.h the increment of Uie bending moment, = aS'
.". F =-j-, is the force which tends to make the angle slide over the
flange face.
Hence, the number of rivets between two consecutive apices.
_ 14 1 S
EXAMPLES.
139
iton
the
lint
my
ihe
(.f
lit
he
re
le
Examples
(1)- — A trellis girder rests on two supports without exorting any
lateral pressure upon them. A vertical section of the girder uieet^ two
sets of H bars inclined in opposite directions, and it is assumed that in
each set a mean stress maybe substituted for the different stresses in the
bars. If />,/>', are the mean stresses and 7'C, the horizontal forces in
the flanges at the section, shew that «.(/)- D') sin 0+ T- C must be
zero at all points of the span, independently of any additional hypothesis,
being the common inclination of the bars to the vertical.
Note. — If T = C, D must necessarily be equal to D'.
(2). — The tensile and compressive unit stresses in the chords of a bridge-
truss of span I and uniform depth (/, are nowhere to exceed /', and/j,
respectively ; shew that the greatest central deflection is approximately
(/i+/2)- g ^, , • Hence, if the sp:in is eight times the depth, and if
the grade line is to be truly horizontal when the bridge is loaded, shew
that the length of the top chord should exceed that of the bottom chord
by an amount etjual to the camber.
(H) — A lattice girder 200-ft. long and 20-ft. deep, with two systems
of right-angled triangles, carries a dead load of 800-lbs. per lineal ft. ;
determine the greatest stresses in the diagonals of the fourth bay from
one end, when a live load of 12()0-lbs. per lineal ft. passes over the
girder.
(4). — A lattice girder 80-ft. long and 8-ft. deep carries a uniformly
distributed load of U-ijOOO-lbs. ; find the flange inch-stresses at the
centre, the sectional area of the top flange being SG^sq. ins. gross, and
of the bottom flange 45-sq. ins. 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 ?
(5). — A lattice girder 80-ft. long, and 10-ft. deep, with four systems
of right-angled triangles carries a dead load of 1000 lbs. per lineal it. ;
determine the greatest stresses in the diagonals met by a vertical plane
ju the seventh bay from one end, when a live load of 2500-lbs. per lineal
ft. passes over the girder. Design the flanges, which are to consist of
plates riveted together.
140
EXAMPLIS.
The lattice bars are riveted to angle irons ; find the number of ^ -in.
o
rivets required to connect the angle-irons with the flanges in the first
bay, 10,000 lbs. per sq. in. being the safe shearing strength of the
rivets.
(6). — The bracing of a lattice girder consists of a single system of
triangles in which one of the sides is a strut and the other a tie inclined
to the horizontal at angles of a and ^, respectively ; in order to give the
strut sufficient rigidity its section is made /;-times that indicated by
theory, t e co-efficient /•; being > unity ; shew that the amount of material
in the struts and ties is a minimum when - — = «.
(6). — A bridge of n equal spans crosses a space of 7/-ft. ; w ^ and w,
are respectively the live load and weight of the platform, permanent
wry, etc., in tons per lineal ft.; shew that the weight ?" , of the main
girders in tons per lineal ft. may be suppressed in the form
L. Wj. {p. k 4- r) -f- w^.n
n - L. (p-fc + q)
k being the ratio of the span to the depth, and p, q, r, numerical co-
efficients. Hence, determine the limiting span of a girder.
If X is the cost of each pier, Y the cost per ton of the superstruc-
ture, determine the value of ?i which will make the total coat per lineal
ft., i.e.,- ■■{■w[.Y, a minimum, and prove that this is approxi-
mately the case where the spans are so arranged that the cost of a span is
equal to that of a pier.
(8). — A warren girder with its bracing l
formed of nine equilateral triangles, and
with every joint loaded, is 90-ft. long; its
dead weigl' 5001bs. per lineal ft.; prepare
a table shewing the maximum stress in each bar and bay when a live-
load of 1350-lbs.per lineal ft. crosses the girder, allowing for an engine-
excess of 4,(l001bs. per panel, extending over two consecutive panels.
The diagonals and verticals are riveted to angle-iront forming part of
the flanges ; how many f in rivets are required for the connection of AB,
AC, and AD, at A ? Also, how many are required between A and E to
resist the tendency of the angle-irons to slip longitudinally ?
EXAMPLES.
141
(9). — If a force of SOOOlbs. strike the bottom chord of the <:irdor in
the prccedinfr quoi^tion at 20-ft. from one ' nd, and in a direction inclined
at 30° to the horizontal, determine its effect upon the several members.
(10.) — The platform of asinirle-track
bridtre is supported upon the top chord>
of two warren girders ; each girder i.^
100-ft. loni.', and its bracing is formed of tin L'cjuilateral triangles; the
dead weight of the bridge is 900-lbs. per lineal ft.; the greatest total
stress in A A when a train crosses the bridge is 41,;-{94.8-lbs.; determine
the weight of the live load per lineal ft. Prepare a table shewing the
greatest stress in each bar and bay when a single load of 15,000-lbs.
crosses the girder.
(11). — The compression and tension bars in certain two bays of a
lattice girder are required to bear stresses of 90-cwts. and 130-cwts
respectively ; design the bars and specify the size and number of the
rivets at the attachments, allowing 100-ewts. per sq. in. of net section
in tension, 65-cwts. per sq. in. of gross section in compression, and a
sliearing stress on the rivets of 100-cwts. per sq. in.
(12).-The two trusses for a 16-ft. road-
way are each 100-ft. in the clear, 17-ft. 3-in.
deep, and of the type represented in the
Fig.; under a live load of 1120-lbs. per
lineal ft. the greatest total stress in Ali is 35,400-lbs., det<;rmine the
permanent load.
The diagonals and verticals are riveted to angle-irons forming part
of the flanges ; how many |-in. rivets are required for the connection
of AB and BC at B ? also, how many are required between B and D to
resist the tendency of the angle-irons to slip longitudinally?
(13). — Design a double flanged plate girder, 8-ft. deep and 80-ft.
long, to carry per lineal ft. a dead load of 500-lbs. and a live load of
1600-lbs., the safe tensile and compressive iuch-stress being 10,000-lb8.
and 8,000-lbs., respectively.
Determine the number of rivets required per lineal unit of length at
the end and centre of the girder to coanect the flange angle-irons with
the web.
(14). — Find the points in each chord of the girder in the preceding
questioa at which the chord stress is equal to the vertical shearing
stress.
142
EXAMPLES.
(15). — A Howe truss, 80-ft. long and 8-ft. deep, has a single set of
diagonals at 45"; find the chord and diagonal stresses in the ^rd hay :
(a)-Whcn the truss carries a uniformly distributed load of 40-tons, (b)-
Wheo the 40-tons are concentrated at the middle of the span. (^Neglect
the weight of the truss).
(16). — Design a 16-panelled Howe truss for a clear span of 200-ft.,
and of such a depth that '.,he chord stress in the 8th and 9th panels is
one-half of the total distributed load (dead load per lineal ft.=:800-lbs.
live load per lineal ft. = 1200-ibs.).
(17). — What should be the depth of the truss in the preceding ques-
tion 30 that the chord and diagonal stresses in the second panel may be
ecjual ?
(18). — A Howe truss with A^-panels and a span of ?ft. is designed for
a dead load of ?{'-lbs. and a live load of i«'-lbs. at each of the panel points
in the bottom chord ; assuming that the sectional nreas of the diflferent
members are directly proportional to the stresses to which they are sub-
jected, determine the depth of truss which will secure the greatest
economy of material,
(19). — The platform of a double-track
railway bridge at Thamesville is supported
from the bottom chords of two trusses bracid
and counter-braced as in the Fig. The leng<'. of each truss is 184-ft.
2-in.. its depth is 34-ft., and it is designed to carry per lineal ft. a live
load of 2250-lb3. and a dead load of 1 1 00-lbs. ; prepare a table shewino-
the max, stress in each member.
(20).— Design a cross-tie for the bridge in the presiding question,
the live load for the floor-system being 800il-lbs. per lineal ft.
(21).— A 16-panelled Muiphy-Whipple truss 200ft. long, and 18-ft.
9-ins. deep, is designed to carry per lineal ft. a dead load of 750-lbs. and
a live load of 1200-lbs., the panel excess weight due to the engine being
4000-lbs., and that due to the tender 3000-lbs. ; prepare a table shewin*'-
the max. stress in each member, and specify the bars which require to
be counter-braced.
(22). — Determine the greater weiglit that may be constructed at 50-
ft. from one end of the truss iu the preceding question.
(23). — Design a quadrangular truss (See ^^^^
Fig.) with square panels for a span of 91-ft., ^^^SlMMMMS
the chords and posts being of timber and the
ties of wrouKht-iron.
EXAMI'LK-!.
143
(24). — Design a Fink truss for a 242-ft. span from the followini^
data : — dopth of truss = 30-ft. ; kntrth of panel = lo-ft H ins. ; dead
load per lineal ft. of truss = 1282 lbs. live load per lineal ft. of truss
= 1300-lbs. ; the en_i:;ine excess ■weij:^ht upon each of the end posts =
17U4-lbs.
(25). — A^, A.,, and a,, a.^, are respectively the sectional areas and
inclinations to the vertical of the two ties which meet at tlie foot of a
post in a Bolhnan truss; if the sectional area of each tie is proportioned
to the stress to which it is subjected, shew that
>1,. cos'n.siti a = A .i.cos^aj.sin n^.
(26). — Design an cight-panolled Bolhnan truss. 100-ft. long and 12iV-
ft. deep, to carry a uniformly distributed load of 200-tons, together with
a single load of 10-tons concentrated at 25-ft. from one end.
(27). — In a Post truss for a 200-ft. span there are IH-panels, the posts
incline towards the centre, and have a run of half a bay, the ties cross
two panels, and are inclined at 45" to the vertical, the counter-ties cross
one panel, and have the same inclination, the panel weight of engine =
17,600-lbs., of tender = 16,16(1 lbs., end of cars = 12,500-lbs.; the maxi-
mum stress in the seventh tie from one end = 56,000-lbs.; determine
the panel weight of truss, and specify the bars which require V, be
counter-braced.
(28). — A weight is placed at the centre of a truss of which the web
consists of a single system of struts and ties; shew how to de.Vrmine
the inclination of the web members to the vertical, so that the amount of
material in them may be a minimum.
(20). — Design a bowstring girder, (^^;)-with vertical posts and diagon-
al bracing, (6)-with isosceles bracing, from the following data: —
The span is 80-ft., the depth at the centre is 10 ft., the dead load per ft.
run is ^-ton, the live load per ft. run is 1-ton.
(80). — The bowstring girder repre-
sented by the Fig. is designed to sup-
port a load of 840-lbs. per ft. run ; its
span is 100-ft., its central depth 14-ft.,
its depth at each end 5-ft., and its
upper chord is a circular arc ; determine the stresses in all the members.
(31). — The span of a suspension bridge is 200-ft.. the dip of the chains
is 80-ft., and the weight of the roadway is 1-ton per ft. run ; find the
tensions at the middle and ends of each chain.
144
EXAMI'LES.
(Ii2). — Assuinin<r ' .vt a rope (or sitiirlo wirt;) will safely bear a tiMi-
sioii of ir)-ton.s per sij. in., show that it will safoly b.'ar its own weight
over a span of about one uiile. the dip being j'^-th of the span.
(;{;}). — Siiew that a steel wire rope of the best (|Uality. with a dip nf
■|-th of the span, will not break until the span execds 7 miles.
(34). — The platform of a suspension bridgt.' of 150-ft. span is sus-
pended by 90 vertical rods (45 on each side) from the cables, which
have a dip of 15 ft. ; the weight of the platform is 224(J-lbs. per lineal
ft. ; find the stresses at the middle and ends of the cables, and in the
suspenders, when a uniformly distributed load of 78. 750-lbs. covers half
the bridge.
(35). — Design a pair of auxiliary girders for the bridge in the pre-
ceding (juestion, to counteract the effect of a live load of l()50-lbs. per
lineal ft., and examine the effect of hinging the girders at the centre.
(3G). — Abridge 4-t2-ft. long consists of a central span of 180-ft. and
two side spans each of 131-ft. ; each side of the platform is suspended by
vertical rods from two iron wire cables ; each pair of cables passes over
two ma.sonry abutments and two piers, the former being 24 ft. and the
latter 40-ft. above the surface of the ground ; the lowest point of the
cables in each spaa is 19-ft. above the ground surface ; at the abut-
ments the cables are conuectt^d with straight wrought-iron chains, by
means of which they are attached to anchorages at a horizontal dis-
tance of GG-ft. from the foot of each abutment ; the dead weight of the
bridge is 3500-lbs per lineal ft., and the bridge is covered with a proof
load of 4500-lbs. per lineal ft. ; determine,
Qi). — The stresses in the cables at the points of support and at the
centre of each span.
(/>). — The dimensions and weight of the cables, (1) — if of uniform
section throughout, (2) — if each section is proportioned to the pull
across it.
(c). — The alteration in the length of the cables and the corresponding
depression of the platform at the centre of each span, due (1) — to a
change of G0° F. from the mean temperature, (2) — to the proof load.
(rf). — The crushing and bending efforts at the f«ot of a pier.
(<'). — The bearing area at the top of the abutments, and the mass of
masonry necessary to resist the tendency to overturning and to hori-
zontal displacement. {Weight of masonry per cubic ft. — Vl'A lbs., Its
safe comjjressive strength per sq.ft.~2ii^-lbs., its cuejficient uf friction
= .76).
EXAMPLES.
145
the
as of
lori-
Its
•tiun
(37). — Solve the preceding question when the cahlcs for the several
spans are independent, and have tlieir ends securely attached to the
points of support. The piers are wrouirht-iron oscill'.lng columns, and
in order to maintain the equilibrium of the structure under an unequally
distributed loads the heads of the columns are connected with each
other and with the abutments by iron wire stays ; determine the proper
dimensions of the stays, assuming then to be approximately straiglit.
(38). — The river span of a suspension bridge is 930-ft., and weigh.s
5976-tons, of which 1430-tons are borne by stays radiating from the
summit of each pier, while the remaining weight is distributed between
/our 15-in. steel wire cables producing in each at the piers a tension of
20G4-tons ; find the dip of the cables.
The estimated maximum traffic upon the river span is 1311-tons,
uniformly distributed ; determine the increased stress in the cables.
To what extent might the traffic be safely increased, the limit of
elasticity of a cable being 8116-tons, and its breaking stress 1^3,300-
tons?
(39).— A semi-circular 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 shew that the
horizontal thrust on the rib is one-fourth of the total load.
(40). — Draw the equilibrium polygon for an arch of 100-ft. span and
20 ft. rise, when loaded with weights of 3, 2, 4, and 2 tons respectively,
at the end of the 3rd, 6th, 8th, and 9th division from the left support,
often e(|ual horizontal divisions. (Xeijlect 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.
(41), — A parabolic rib of span / and rise /.•, having its ends A and B
fixed, supports a given load at a liorizontal distance x from the middle
of the span ; the ecjuilibrium polygon consists of two straight lines CD
CE, meeting the verticals through .1 and li in D an E respectively ;
if ^•IZ>=//,, BE=ij.i, and if rj is the vertical distance of (J from AB
shew that,
2 /+10..V, 2 /-10.;i- ^^ 6 ,
(42). — A parabolic double-flangod rib 2i-ft, deep, of 100-ft. span
and 20-ft. rise, is fixed at both ends and loaded in the same maimer as
the rib in Question 40 ; draw the equilibrium polygon, and det<!rmiuo
the flange stresses at the ends and at the division points.
10
146
EXAMPLES.
(48). — A ril) ill the form of tin- scixuii-nt of a cirelo. of sp;in I and
rise /,-, is hiiitri'tl at its oihIh AJi. and supports a W(.'i;:ht M' at a liori-
zojital (listanco .r from tho uiiddlo of the span: the t'(|uilibrium polygon
consists of two straight lines CA, CB, tho vertical distance of ^' from
AB beiiii: 1' ,• determine the value of 1', and sliew that the horizontal
thrust on the rib is ir.-
4. /. 1'
(44). — Xf a rib in the form of the seiiiiient of a circle be substituted
for the rib in Question (41), shew how to tind y.., y^, Y, and apply to
the case of semi-circular rib.
(45). — Draw the curve of o(iuilibrium for a semi-circular rib of uni-
form Mction under a load uniformly distributed per horizontal unit of
length, (") — when hinjred at tlie ends, (/>; — when hinged at the ends
and centre, (c) — when fixed at ihe ends.
(4(3). — Det^^-rmine the horizontal thru>t on a rib in tlie form of the
scirment of a circle due to change of t" from the mean temperature.
(47). — In CliMpfi-r X, determine the changes to be made in the
analysis, (a) — of JJ i;}.14, 15, when the e(|uations deduced in Corollary 4,
§(13), are employed instead of equations Id, 11 and 12.
(fj). — Of ^ 13 and 14, when the abutments yield to the thrust ^o as
to t'ohirge the sjian by an amount //.-//., // being a co-efficient to be
determined by experiment.
(r). — Of ^5 14 and 15, when the effect of a change of temperature is
to be taken into account.
(48). — Illustrate J^ 13, 14, 15, by the example of a rib in which
h = ~=i\S.z, and q =■-;•
h u
(49). — A wrought-irou parabolic rib, of 90-ft. span and 10-ft. rise, is
lunged at the two abutments ; it is of a double y-section, uniform
throughout, and 24-ins. 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, {<() — when a load of 48-tons is concen-
trated at the centre, (/>). — when a load of 9G-tons is uniformly distributed
per horizontal unit of length.
IX'termine the deflection of the rib in each case.
(50). — In the framed arch repre-
sented by the Fig., the span is 120-
ft.. the ri.<e 12-ft., the depth of the
truss at the crowu 5-ft., the fixed load
EX.\MPLES.
147
ibuted
at each top joint 10-totis, anrl the moving load 10-tons; doteruiino the
maximum stniss in oadi mrmbiT with any distriltiition of load.
Shew that, a))|iriiximati'ly. the amount oi' metal rc((uir(.'d for the arch :
the amount required for a bowstrinir lattice girder of the same s])an and
17-ft. deep at tlu! centre: the amount recjuired for a girder of the same
Fpan and 12-ft. deep ;: 100 ; If)') : 175.
(51). — Design an arched rib of 100-ft. span and 20-ft. rise, the fixed
load being 40 tons uniformly distributed per horizontal unit of length,
and the moving load 1-ton per horizontal ft.
(52). — A cast iron arch, (."^eeFig. ), who.'ie
cross-.HH'tiiMis are rectangular, and uni-
formly !^-ins. wide, has a straight horizontal
cxtrailos, and i.s hinged at the centre and at
the abutments. Calculati' the nm'mal intensity of stress at the top and
bottom edges d, E, of the vi'rtieal section, distant 5-ft. from the centre
of the si)an, due to a vertical load of 20-tons concentrated at a point
distant 5 ft. 4-ins., horizdutally from B ; also find the maximum intensity
of the shearing .stress on the same section, and .state the point at which
it occurs. (A15=: 21 ft. 4 ins.)
(5|}) The gket-ch .shews the
truss of a counterbalanced swing-
bridire. supposed to be wholly
supported by the turntable at A
and B ; the dead weight is ()50-lbs. per lineal ft. of the bridge, and the
couuterpoi.se consists of concrete hung in a box from the points c and
D ; ro(iuired the weight of the counterpoise a.ssuming, ///•.</, that it is all
carried on B, and, sirmid, that a portion of it is transmitted to A by
the dotted bar, so that the reaction at A may be e((ual to tliat at H. and
the whole wt>ight of the bridge be equally distributed over the turntable.
Shew how to find the stresses in the different members of the truss.
(54).— The truss for
a counterbalarced swing-
bridge is of the I'orm and
dimensions shewn by the
Fig., the counterpoise
consisting of concrete
hung in a box from the points c and D ; the dead weight is 050-lbs.
per lineal ft. of the bridge ; determine the stresses in the different mem-
148
EXAMI'LES,
bern, (</). — when tlic bridge is opon iiiid is wholly supportrtl l)y tlu^ turn-
table at A and B, (h) — wlu-n the liridgt! is closed, and is subjected to
a passing load of 30(IO-lbs. per lineal ft.
Design tlie joint at E so that, however unecjually the arms may be
loadi'd, the whole of the load may be transmitted to the main posts EA
and EB and evenly distributed over the turntable.
(55). — A girder bridge of 100-ft. span deflects 3-ins. below the hori-
zontal line under a certain litad IT. at rest ; tind the increased pressure
due to centrifugal force if ITcros-ses the bridge at GU-miles an hour.
(5()). — A .stiffening girder for a suspension bridge has its ends fixed .so
as U) be incapable of vertienl motion.
If a load is concentrated at the centre of the girder, shew that th^
ratio of tlie vertical distance between the highest and lowest points of the
central axis to the vertical distance between the end and lowest point
IS
9
If a uniformly distributed load covers /itv;7A('rc7s of the girder from
one end, shew that the ratio of the vertical distance between the ends
and the lowest point of the neutral axis to the vertical distance between
the ends and the highest point is 4.
Determine the depression of the lowest point below the highest when
one-third of the girder from one end is uniformly loaded.
(57). — The plate in Fig. is 7^-ins. wide
and ^-in. thick ; B and c are |-in. rivet holes ;
AB= IJ-ins. OF = 3-ins. ; find CE so that the
efficiency on the line abce may be equal to
that on the live ABD.
(58). — Determine the strength of a joint of
the type represeat<id by Fig. 28, Chap. VI., from the following data : —
widthofplate = 4 r--ius., thickness of plates= thickness of cover =— in.,
16 2
13 . 7
diar. of rivets = -—-ins , pitch of rivets on line aa = 2— -^-ins., on line
BB=3J-ins., distance between aa and bb =l^-ius., between aa and
cc= 3|^-ins.
(59). — A ^-in. flat diagonal bar at an angle of 45° is riveted to a
i-in. vertical flange plate ; if the admissible stress per sq. in. on the
shearing area of rivet be 5-tons, and 7^ tons on the bearing area, shew
that the strength of the joint may be always made greater than that of
the bar with two rows of rivets.
EXAMPLES.
UO
in.
(♦HO. — A riveted joint may fail l»y orushin<r, the bearinir area being
insufficient; assuming that tlie rivil and rivet hole retain tlieir cylin-
drieal forms when the joint is straini'd. and that the pressure at any
point of the hearing is normal to tlio surfaces in contact and eipial in
intensity to /. cos n., a being the angle between the normal and the
direction of the resultant stress, shew that -y, =-,-:• J' and / being re-
spectively the maximum and mean intensities of the crushing pressure in
the direction of the resultant stress.
((51). — A chord-pin joint is of the type represented by Fig. 7, Chap.
VI. ; the eight bars are each S-ins. deep by 1-in. wide; tlie two inner-
most bars are 1-in. apart, and the spaces between the remaining bars
are closed up ; find the diar. of the pin, assuming that the ultimate unit
stress in a bar is -T--ths. of the actual stress in the pin at its cireum-
27 '
ference.
(G2).-The curvature of a girder at a vertical section of depth h on
one side of a chord pin joint is increased '/-times by the deformation
of the web; r is the radius of gyration of the .section of the chord; if /'
be the coefficient of sliding friction, shew that the diameter of the pin
which will just prevent rotation is a[)proximately - . —
State the practical effi.'Ct of this result.
(G3).-Two 1^ ins. lateral rods are coupled on a piu ; the outer is
subjected to a .stress of 2^-tons and the lever arm of its longitudinal
component is H-ins., the inner is subjected to a stress of 9i-t(rtis and
the lever arm of its longitudinal component is 1-in.; the stressi's in the
web members produce a coujile having its axis inclined at 4^)" to the
vertical, and a moment of 20-inch tons; determine the proper diameter
of the pin,
(()4)-I)(if(t. — A deck bridge consists of two
independent spans of 200-ft., supported at one
end by an iron braced pier 50-ft. high. The
trusses ;ire 80-iY, centre to centre of chords ;
the distance between the centre lines of the
tru.s.scs is 17-ft. ; the planes of the pier faces
batter 2-ivis. in 1-ft.; a horizontal section of the
pier is square. The \vind pressure is 401bs.
per sq. ft.; 10-sq. ft. per lineal ft. of bridge
and lO-.sq. ft. per lineal ft. of train, are subject
to wind pressure; the dead load per lineal ft. is
150
KX.VMl'LKS.
1000-lbs. and the 11\.> load 280()-lbs.; tlio uhsymod wiM^dit of the pior
is IHOO-llw. ])er ft. in heij:;ht; the cyliiidiT of tiw. locomotive is iniiis.
in diani., tlm pri'ssurt; on the piston is l;{()-lbs. per sc^ in.: —
(<i)- — Find tli(! stresses in the wind b''acini:.
{/)). — It a train of empty cars is on the l)rid;j;t^ and are on the point of
overturning, ascertain whether the inelined posts will be subjected to
tensile stn-sses. The cars weigh DOO-lbs. per lineal ft., and the centre
of priissure is 7-ft. above the rails.
(o). — Ascertain whether the wind pressure of 40-lbs. per s(|. ft. upon
the empty bridge will produce tension anywhere in the inclined posts,
the centre of pressure being li-ft. above the middle of the truss.
('/). — A loaded train rests upon one span, the engine being directly
over the inclined ])osts ; find the stresses in the traction bracing when
the whole pressure oi'130-lb-. per .s(j. in. is suddenly applied upon the
piston.
(t). — A loaded train, 200-ft. in length, is running at a speed of 30
uiiles an hour. Sufficient power is applied to the car-brake to bring
the train to rest in 300-1't.; find the stresses in the traction bracing
when the engine is over a pair of inclined posts.
(/), — Find the max. stresses in inclined posts in each of the three
sections, by the method of moments, neglecting the effect of traction.
(r/). — Will the shoe at the foot of a post slide under any distribution
of il'o load ? (Cocff. of frictioa=/)
I