KINETIC THEORY

ENGINEERING STRUCTURES

DEALING WITH

STRESSES, DEFORMATIONS AND WORK

FOR THE USE OF

STUDENTS AND PRACTITIONERS IN CIVIL ENGINEERING

DAVID A. MOLITOR, B.C.E., C.E.

AUTHOR "HYDRAULICS OF RIVERS, WEIRS AND SLUICES," ETC.; MEM. AM. soc. c. E.; MEM. DETROIT

ENGINEERING SOC.; MEM. SOC. FOR THE PROM. ENG. EDU.J MEM. A. A. A. S., ETC.;

FORMERLY DESIGNING ENGINEER ISTHMIAN CANAL COMMISSION;

FROFESSOr, IN CIVIL ENGINEERING, CORNELL UNIVERSITY.

McGRAW-HILL BOOK COMPANY

239 WEST 39TH STREET, NEW YORK

6 BOUVERIE STREET, LONDON, E.G.

1911

BY
DAVID A. MOLITOR

THE SCIENTIFIC PHES9
ROBERT ORUMMONO AND COMP
BROOKLYN. N. V.

PREFACE

" ALL that mankind has done, thought, gained, or been; it is lying in magic preserva-
tion in the pages of books"; and so in presenting this work for publication, the author
hopes to preserve the results of many years of painstaking labors, study and exped-
ience, to civil engineering, his chosen profession.

The field of usefulness which it is proposed to supply is that of an advanced treatise
on stresses and deformations in engineering structures, for the use of students and prac-
titioners in civil engineering and particularly for specialists engaged in the design of
so-called " higher structures " requiring more than the ordinary methods of statics for
their analysis.

The reader is supposed to possess a thorough knowledge of higher mathematics,
including the calculus; also, of the elements of statics, embracing the composition
and resolution of forces, force and equilibrium polygons, etc. All these are usually
given in the two first years of a Civil Engineering course in connection with mathema-
tics and mechanics.

Details pertaining to the design of bridge members and connections are not dealt
with.

The present volume might advantageously be employed as a text-book in bridge
stresses with the aim of giving a more thorough training in the analysis of stresses and
deformations by economizing somewhat on the time at present devoted to detailing
and shop practice, which is not really justified in a four-year course of theoretical
preparation.

In the opinion of the author this would be highly desirable, because details and
shop practice can best be learned in the shop, where experience is the teacher. It
devolves on the college to give to its students that thorough theoretical training which
they cannot readily acquire in practice, a feature which must ultimately distinguish
the college-bred engineer from the purely practical man.

The present work, which is the gradual evolution of years of labor, thus represents
an effort to place before the profession a treatise on the analysis of engineering struc-
tures which is based on the most advanced theories and researches of the present time.

To enhance the educational value of the book, each subject is prefaced by a few
brief historical remarks and all important theorems bear the names of their originators,
a practice which has been shamefully neglected by many modern writers.

The author has made free use of any and all literature bearing on the various sub-
jects treated and hereby expresses his grateful acknowledgment to all who have contrib-
uted to the sum total of our present knowledge. Prominent among these should be

vi PREFACE

mentioned Hooke, Bernoulli, Lagrange, Coulomb, Navier, Lame, de Saint- Venant,
Clapeyron, and Menabrea, as the founders of the theories of elasticity and virtual work.

To Clerk Maxwell, Castigliano, Mohr, Fraenkel, Culmann, Winkler, Mueller-Breslau,
Engesser, Weyrauch, Manderla, Asimont, Landsberg, Melan, Zimmerman, Ritter, Land,
and Mehrtens, we are indebted for advancing these theories to their present state of per-
fection and usefulness. Professor Otto Mohr enjoys the high distinction of being the
foremost originator of novel principles in the analysis of engineering structures.

A bibliography of the works used by the author will be found at the end of this
volume.

The author has avoided a sharp differentiation between analytic and graphic methods
and uses both without discrimination, aiming always to secure the most practical solu-
tion for any problem in hand.

While this work was produced with very great care and diligence, hoping to eliminate
typographical and other errors, it is certain that some have escaped detection. The
author, therefore, invites the kind indulgence of the reader and will be truly thankful
for having his or the publishers' attention called to any errors that may be discovered.

Finally, as the love liberated in our work is a true manifestation of character, so
may the present production reflect the character of, the man, the engineer, the author.

ITHACA, N. Y., September 9, 1910.

DEFINITIONS OF TERMS USED THROUGHOUT THIS WORK

FREQUENTLY some of the letters are employed in a special connection other than here
defined, but in such cases the definitions locally given will prevent misunderstandings.

ART. 3. p = number of pin points of any frame.
m = number of members of any frame.
n = number of redundant conditions in any structure.
n' = number of external redundants in any structure.
n" = number of internal redundants in any structure.

e = number of elements or simple frames in a structure.
Sr = number of necessary reaction conditions for any structure.
2 = the summation sign.
ART. 4. S = total stress in any member of a frame resulting from any loads P or other

causes and + for tension.

& = the stress produced in any member by applied loads P.
/= the unit stress in any member, + for tension.

Z = length of any member or span of a structure for the condition of no stress.
JZ = change in length of a member due to any cause, positive for elongation.
F cross-section of any member, prismatic in form.
t=& uniform change in temperature in degrees, + for rise.
= coefficient of expansion per degree temperature.
E= Young's modulus of elasticity for direct stress.
a = the angle which a member makes with the x axis of coordinates.
/? = the angle which a member makes with the y axis of coordinates.
m = any point of a structure chosen for illustration.
p = l/EF = the extensibility of a member employed for brevity.
.R = any reaction force, or condition, of a structure resulting from loads P.
.ft = any reaction force, or condition, of a structure resulting from loads P.
Jr = any displacement of the point of application of a force R in the direction

of this force.

P = any external load or force applied to a structure.
P = a load or force identical in position and point of application with the force

P but having a different intensity.
ART. 5. J = any displacement of a pin point.

<? = the displacement of the point of application of any force P in the direction
of this force.

viii DEFINITIONS OF TERMS USED THROUGHOUT THIS WORK

A = actual work = force times distance through which the force moves.
A e = ihe positive or externally applied work of deformation of any structure.
.A t - = the negative or internally overcome work of deformation of any structure.
W = a virtual work of deformation.

W a virtual work of deformation produced by a conventional case of loading.
ART. 7. X a , Xb, X c , etc., are the stresses in redundant members or reactions A, B, C, etc.
<, = the stress in any member of a principal system, due entirely to loads P
when all redundant conditions X are removed. This is known as con-
dition X = 0.

S a = ihe stress in any member of a principal system, due entirely to the conven-
tional loading X a =l, applied to the principal system. Condition X n =l.
Sb. S c , etc., are similarly defined for conventional loadings Xb=l; X c =l, etc.
St = the stress in any member of a principal system produced by a uniform change

in temperature.

R t = a reaction produced by a uniform change in temperature.
M t = & moment produced by a uniform change in temperature.
R , R a , R b , etc., have definitions analogous to those given for S , S a , Sb, etc.
M 0f M a , M b , etc., are moments similarly defined for the conventional loadii^s.
da, $b, $c are changes in the lengths of redundant members or supports X a , X bf

X e , respectively.

L represents the length of span of a composite structure.

ART. 8. dma = the displacement of the point of application m of any load P m , in the
direction of this load, when the principal system is loaded only with
X a = 1 . Condition X a = 1 .
Smb = a similar displacement of the point m for the conventional loading X b = 1 .

Condition Xb=l.

<5 TOC = the same for condition X c = 1.

daa = the change in length of the member a for condition X a l.
(5 j> = the change in length of the member a for condition Xb=l.
d&a = the change in length of the member b for condition X a =l, etc.
d at = the change in length of the member a resulting from a change in temperature
of t, when the principal system is not otherwise loaded, hence X = 0, P = 0.
dbt, $ct are similarly defined for members b and c respectively,
ART. 13. f x , f y , / 2 = the unit normal stresses in the directions X, Y, and Z, respectively ,

for any isotropic body.

T XV and r x:s = the unit tangential stresses in the YZ plane.
r vx and r vz = the unit tangential stresses in the XZ plane.
T ZX and T zv = the unit tangential stresses in the XY plane.
G = the modulus of tangential stress.

m = the Poisson number or ratio of lateral to longitudinal deformation.
ART. 14. N = & normal thrust on any section.
R = any oblique thrust.
Q = a tangential force or shear.
I = the moment of inertia of anv cross-section.

DEFINITIONS OF TERMS USED THROUGHOUT THIS WORK ix

t/ = an ordinate usually the distance from the neutral axis to the extreme fiber

of a cross-section.
h = height of a section.
6 = base or breadth of a section.

/? = coefficient of shearing strain, Eq. (ML) and Eq. (14M).
ART. 16. w = weight of a moving body.

H = height of fall. Later used to represent the horizontal thrust of an arch,

also, the pole distance of a force polygon.
v = velocity at instant of impact.
g = acceleration due to gravity.
ART. 17. jj = any ordinate to an influence line.
d = a panel length.

p = uniform moving load per unit of length.
ART. 18. m = a load point, being any one of the many possible points of application of

a certain moving load.
n = the point for which art influence line is drawn. Also, the location of a section

under consideration.
ART. 20. i = the load divide for positive and negative effects. For arches the load

divide is sometimes designated by d.
ART. 21. r = the lever arm of a member in the equation S = M/r.

S a = ihe stress in any member S due to a reaction unity at A, when the load

producing this reaction is to the right of the section.
Si, = the stress in the same member due to a reaction unity at B, when the load

producing this reaction is to the left of the section.

ART. 28. e = the kernel point for the extreme fiber of the extrados of a solid web arch section.

i = the kernel point for the extreme fiber of the intrados of the same arch section.

ART. 35. w = an elastic load, representing an abstract number without any particular unit

of measure.

ART. 45. <S = the stress in any member of a statically indeterminate frame for the con-
dition J = and P=l.
M = a moment for the same conditions.
A = the reaction at A for the same conditions.

/x = an influence line factor by. which all ordinates must be multiplied to obtain
the actual influences.

CONTENTS

PAGE

PREFACE v

DEFINITIONS OF TERMS vii

INTRODUCTION 1

CHAPTER I
REACTIONS AND REDUNDANT CONDITIONS

1. Reaction Conditions 5

2. Structural Redundancy 6

3. Tests for Structural Redundancy 7

CHAPTER II
THEOREMS, LAWS, AND FORMULAE FOR FRAMED STRUCTURES

4. Elastic Deformations, Fundamental Equations 11

5. General Work Equations for any Frame 14

6. Displacements of Points. Statically Determinate Structures 18

7. Indeterminate Structures, by Mohr's Work Equation 19

8. Indeterminate Structures, by Maxwell's Law 24

9. Prof. Maxwell s Theorem (1864) 27

10. Theorems Relating to Work of Deformation 29

(a) Menabrea's law (1858). Theorem of least work 29

(b) Castigliano's law (1879). Derivative of the work equation 30

11. Temperature Stresses follow Menabrea's and Castigliano's Laws 31

12. Stresses Due to Abutment Displacements 32

CHAPTER III

THEOREMS, LAWS, AND FORMULAE FOR ISOTROPIC BODIES

1 3- General Work Equations 34

14. Work of Deformation due to Shearing Stress 38

15. Work of Deformation for any Indeterminate Straight Beam 41

16. Work of Deformation due to Dynamic Impact 43

xii CONTENTS

CHAPTER IV
INFLUENCE LINES AND AREAS FOR STATICALLY DETERMINATE STRUCTURES

PAGE

1 7. Introductory 46

18. Influence Lines for Direct Loading 4<S

19. Influence Lines for Indirect Loading 50

20. The Load Divide for a Truss 51

21. Stress Influence Lines for Truss Members 52

22. Reaction Summation Influence Lines 54

23. Positions of a Moving Train for Maximum and Minimum Moments 57

24. Positions of a Moving Train for Maximum and Minimum Shears 62

CHAPTER V
SPECIAL APPLICATIONS OF INFLUENCE LINES TO STATICALLY DETERMINATE STRUCTURES

25. Double Intersection Trusses 65

26. Cantilever Bridges 67

27. Three-hinged Framed Arches 72

28. Three-hinged, Solid Web and Masonry Arches 78

29. Skew Plate Girder for Curved Double Track 80

30. Trusses with Sub-divided Panels S3

CHAPTER VI
DISTORTION OF A STATICALLY DETERMINATE FRAME BY GRAPHICS

31 . Introductory 87

32. Distortions due to Changes in the Lengths of the Members 87

33. Rotation of a Rigid Frame about a Fixed Point 89

34. Special Applications of Williot-Mohr Diagrams 91

CHAPTER VII
DEFLECTION POLYGONS OF STATICALLY DETERMINATE STRUCTURES BY ANALYTICS AND GRAPHICS

35. Introductory 9!)

36. First Method, according to Prof. Mohr 100

(a) Influence on the deflections due to chord members 100

(6) Influence on the deflections due to web members . 102

37. Second Method, according to Prof. Land , 107

(a) The w loads in terms of the unit stresses in the members . . 107

(6) To find the changes in the angles of a triangle 1 07

38. Horizontal Displacements 114

39. Deflections of Solid Web Beams .. . 117

CONTENTS xiii

CHAPTER VIII
DEFLECTION INFLUENCE LINES FOR STATICALLY DETERMINATE STRUCTURES

PAGE

40. Influence Lines for Elastic Displacements 122

41. Special Applications of Displacement Influence Lines 124

(a) Cantilever beam, deflection polygon 124

(b) Simple truss, displacement of any pin point in any direction 125

CHAPTER IX
INFLUENCE LINES FOR STATICALLY INDETERMINATE STRUCTURES

42. Influence Lines for One- Redundant Condition 127

(a) When the redundant condition is internal 128

(b) When the redundant condition is external 128

43. Influence Lines for Two Redundant Conditions 130

44. Simplification of Influence Lines for Multiple Redundancy 132

(a) Structures having two redundant conditions 133

(b) Structures having three redundant conditions 135

45. Stress Influence Lines for Structures Involving Redundancy 137

(a) Multiple redundancy 137

(6) One redundant condition 139

CHAPTER X
SPECIAL APPLICATIONS OF INFLUENCE LINES TO STATICALLY INDETERMINATE STRUCTURES

46. Simple Beam with One End Fixed and Other End Supported 140

47. Plate Girder on Three Supports 143

48. Truss on Three Supports 148

49. Two-hinged Solid Web Arch or Arched Rib 153

50. Two-hinged Spandrel Braced Arch 166

51. Two-hinged Arch with Cantilever Side Spans 176

52. Fixed Framed Arches 178

CHAFFER XI
DESIGN OF STATICALLY INDETERMINATE STRUCTURES

53. Methods for Preliminary Designing 194

54. On the Choice of the Redundant Conditions 202

55. Solution of the General Case of Redundancy 203

56. Effect of Temperature on Indeterminate Structures 206

57. Effect of Shop Lengths and Abutment Displacements 208

xiv CONTENTS

CHAPTER XII

STRESSES IN STATICALLY DETERMINATE STRUCTURES

PAGE

58. Dead Load Stresses 210

(a) General considerations 210

(6) Aug. Hitter's methods of moments (1860) 211

59. Live Load Stresses 218

(a) The critical positions of a train of moving loads 218

(b) The method of influence lines 218

(d) The author's method 210

7 CHAPTER XIII

SECONDARY STRESSES

60. The Nature of Secondary Stresses -.-. . . 226

61. Secondary Stresses in the Plane of a Truss due to Riveted Connections 227

62. Secondary Stresses in Riveted Cross Frames of Trusses 244

(a) Dead and live load effects 244

(6) Wind effects 247

63. Secondary Stresses due to Various Causes 251

(a) Bending stresses in the members due to their own weight 251

(6) Secondary stresses in pin-connected structures 253

(d) Effect of temperature and erroneous shop lengths 256

64. Additional Stresses due to Dynamic Influences 256

(a) Forces acting longitudinally on a structure 256

(6) Forces acting transversely to a structure 258

(d) Fatigue of the material 265

(e) Unusual load effects 266

65. Concluding Remarks 207

(a) Features in design intended to diminish secondary and additional stresses 267

(b) Final combination of stresses as a basis for designing 268

CHAPTER XIV
MITERING LOCK GATES

66. The Nature of the Problem 271

67. The Theory of the Analysis 273

68. Example : Upper Gate, Erie Canal Lock No. 22 279

CONTENTS xv

CHAPTER XV

FIXED MASONRY ARCHES

PAGE

69. General Considerations 298

70. Modern Methods of Construction 301

71. Preliminary Designs 304

72. Determination of the Redundant Conditions by the Theory of Elasticity 309

73. Co-ordinate Axes and Elastic Loads 314

74. Influence Lines for X a , Xb, X c , and Mm 317

75. Temperature Stresses 318

76. Stresses on any Normal Arch Section 320

77. Maximum Stresses and Critical Sections 322

78. Resultant Polygons 324

79. Example : 150-foot Concrete Arch 326

(a) Given data arid preliminary design 326

(6) Analysis of the preliminary design by the theory of elasticity, treating the structure

as symmetric 331

BIBLIOGRAPHY

Theory of Elasticity 359

Graphostatics 359

General Treatises .' , 360

Secondary and Additional Stresses 360

Special Treatises on Arches 361

KINETIC THEORY

ENGINEERING STRUCTURES

INTRODUCTION

The work of deformation constitutes the basis for the solution of all those problems
which involve the elastic properties of the material and which are not susceptible to
analysis by the methods of pure statics.

Work is the product of a constant force into the displacement of its point of applica-
tion in the direction of the force.

When a prismatic body of length I and cross-section F is either lengthened or
shortened a small amount M within the elastic limit, by a force S gradually produced
without impact, then a certain amount of work or kinetic energy is stored up in the
body. In the instant when the force ceases to alter the length a condition of static
equilibrium is established and the total kinetic energy stored then assumes the form
of potential energy or energy of position. W T hen the force is gradually released and
the body is allowed to resume its original length, the potential energy is liberated.

The work of deformation or applied kinetic energy may be thus expressed:

The principle of virtual work. When a rigid body is set in motion by any group
of forces, then the kinetic energy imparted is equal to the algebraic sum of all the work
performed by the several applied forces.

A real displacement can never result in zero work,' because motion cannot exist
without the presence of a force, and the combination of force and motion constitutes
work.

However, for a state of equilibrium wherein the applied forces do not actually alter
the motion, a certain equilibrium proposition may be formulated by considering possible
displacements that might be produced but may or may not be real. Such displacements
are called virtual, and any work resulting therefrom is called virtual work.

2 KINETIC THEORY OF ENGINEERING STRUCTURES

The fundamental principle of statics may, therefore, be stated thus: A body at
rest remains at rest when for any infinitesimally small but possible displacement the
summation of the work of the applied forces is zero.

Letting P represent any force and d any possible virtual displacement of its point
of application in the direction of the force, then, according to Lagrange (1788), the law
of virtual work is represented by the following equation:

............. (2)

The law of virtual velocities follows from the law of virtual work by introducing the
element of time, giving

........ ..... (3)

The relations thus existing between the possible conditions of equilibrium of a rigid
body and the virtual displacements resulting from any cause, lead to important geometric
relations which are generally applicable to all branches of technical mechanics.

Clapeyron's law (1833), which covers the entire work of the external forces and
internal stresses acting on any frame or body in equilibrium, is derived, from Eq. (1),
by summing all the partial effects for an entire structure. Then, by the " doctrine of the
conservation of energy," the applied work A e must equal A; the work overcome, giving

......... (4)

wherein the loads P, stresses S, displacements d projected onto the directions of the
loads, and the changes Al in the lengths of the members, all result from a simultaneous
condition of loading for a case in equilibrium.

Proceeding from Clapeyron's law. Professor Clerk Maxwell, in 1864, derives his
important law of reciprocal displacements and Professor Otto Mohr, in 1874, evolved his
two work equations which together with Maxwell's law, afford a complete solution for
all problems relating to frames and isotropic bodies.

Chapter I deals with the general relations existing between a frame or beam, arid
its supports, furnishing the definitions and necessary criteria for distinguishing between
statically determinate and statically indeterminate structures or those involving redundant
members or redundant reaction conditions.

Chapter II then presents the derivation and discussion of all the general laws per-
taining to framed structures and these are applied to isotropic bodies in Chapter III.

Chapters IV, V, and VIII are devoted to influence lines for determinate structures,
while Chapters VI and VII treat of the various problems relating to distortions and
deflection polygons of frames and solid web structures.

Chapters IX and X apply the various theories and methods to the analysis of
statically indeterminate structures chiefly by the use of influence lines.

In Chapter XI, the general question of designing a statically indeterminate structure
is then taken up and Chapter XII deals with stresses in statically determinate structures,

INTRODUCTION 3

taking advantage of some of the more advanced ideas presented in the earlier chapters.
Special attention is directed to Art. 59, dealing with stresses due to concentrated live
loads.

Chapter XIII gives a brief treatment of secondary stresses and the two last chapters
are devoted to the problems of mitering lock gates and fixed masonry arches.

It is more than likely that the use of the word kinetic in the title of the present
volume will be questioned by some. However, after long and careful consideration
the author concluded that he was justified in employing the term without really depart-
ing from the best modern dictionary definitions. It is true that in hydraulics the term
kinetic energy is exclusively confined to represent the static equivalent h=v 2 /2g of a
dynamic head. In mechanics generally, the expression is almost limited to the term
Wv 2 /2g=mv 2 /2, representing the potential or kinetic energy of amoving mass. Other-
wise the term is not met with.

While the author employs the term kinetic in connection with engineering structures
which are, generally considered, at rest, it is precisely with the same idea above expressed
in the mechanical sense. That is, a bridge or other structure composed of elastic mate-
rial ceases to be at rest the moment it is acted upon by external forces. The distinction
between the previous and present uses of the term is, therefore, only relative, being
here applied in a different field and to a special class of bodies, where the displacements
are small and follow certain empiric laws depending on the elastic properties of the
material. Strictly speaking, then, we have no license to regard an elastic structure as
statically at rest only when it is neither stressed nor distorted.

The equation of work is thus applied to a structure in the sense that the latter is
a mechanical contrivance or machine in motion, instead of an inelastic body at rest.
It is true that this motion prevails only while a change in the static balance is taking
place, as when loads are added to, or removed from, the structure, but this may apply to
any mechanism of interrupted activity.

Therefore, while we speak of a structure as in static equilibrium, 'we may also speak
of it as in a state of dynamic equilibrium, a state which the structure assumes in the instant
that the superimposed loads do not produce any further elastic deformations. The
same would be true when all the loads are entirely removed, in which case the dynamic
equilibrium returns to the special state of static equilibrium. This is merely a broader
viewpoint of the subject, embracing at once all of the conditions as they really exist
in a frame sustaining loads. The term kinetic thus appears quite appropriate to the
use and application here made.

For the same structure the magnitude of the deformation of the deflection is a
direct function of the loads, the unit stresses which they produce, and the elasticity of
the material.

On the other hand, the stiffness of a given structure is independent of the magnitudes
of the unit stresses, but depends entirely on the ability of the structure to resist stress
and this in turn is a function of the elasticity of the material and the geometric shape
of the structure.

Theoretically speaking, and disregarding economy, a structure of maximum stiff-
ness is one of infinite height compared to its span, resulting in zero stress; and con-

4 KINETIC THEORY OF ENGINEERING STRUCTURES

versely, a structure of minimum stiffness is one of zero height involving infinite
stress.

Maximum stress never occurs simultaneously in the chords and web of any structure.
The maximum total load produces maximum chord stresses, but only a small fraction
of the maximum web stress. Hence, the influence of the web on the total deflection
of a bridge is always small as compared with that produced by the chords and is fre-
quently negligible.

Throughout the present treatise, exepting Chapter XIII, all members of a frame
are supposedly connected at their ends by frictionless pins, and no account is taken of
the so-called secondary stresses occasioned in the members by friction on these pins or
by the rigidity of riveted connections. Chapter XIII deals separately with the secondary
stresses induced by riveted connections and friction on pins, etc.

Unless otherwise specifically stated, a clockwise moment of forces on the left of a
section is regarded as positive, so also an upward shear on the left side of a section.
Tensile stress and elongation are positive, while compressive stress and contraction are
negative. Work is positive when the direction of a force coincides with the direction
of the displacement of its point of application, independently of the algebraic signs of
the forces or stresses and the displacements.

The nomenclature is uniform throughout, though in a few instances slight departures
from American custom became necessary owing to conflicts encountered in combining
so many subjects in one book. The definitions of terms are always given in appropriate
places and also in summary form following the table of contents.

Equations and figures are designated by letters and numbers facilitating the matter
of locating cross-references. The number refers to the article and the letter to the
position of the equation, figure or table in that article. The article number appears in
the heading of every page.

ART. 1. REACTION CONDITIONS

The general purpose of engineering structures, of the class here considered, is to
carry loads over otherwise impassable distances to certain fixed points.

The distances are called the spans, and the fixed points, to which the loads are
finally transmitted, are called the supports or reactions.

It is thus clear that the sum total of the superimposed loads, together with the
weight of the structure, must exactly equal the sum of the vertical reaction forces when
equilibrium exists.

The nature and purpose of a structure determine the manner in which it must be
supported to insure stability under all circumstances that are likely to occur.

This can always be accomplished by one of three types of supports known, respec-
tively, as movable, hinged, and fixed. They are illustrated in Figs. IA, IB, and Ic.

FIG. IA.

FIG. IB.

FIG. Ic.

Fig. IA shows the type of support known as movable, which can exert a single
upward reaction only. Any lateral force applied to such a support would result in
horizontal displacement provided the roller is frictionless.

Fig. IB shows a hinged support and can resist horizontal and vertical forces or
any resultant of these.

Fig. Ic represents a fixed support which may be made to resist any direct forces
and a bending moment.

Statically considered, the first type satisfies one stability condition, the second
type offers two such conditions, and the third type supplies three stability conditions.
These conditions may be represented by forces as shown in Figs. ID, IE, and IF.

Let r represent a single vertical or horizontal force to be known as a reaction con-
dition. Then the three types of supports will involve, respectively, one, two, and three
such conditions. Each reaction condition may be represented by a short link, or member,
which may transmit direct stress only; that is pure tension or compression.

KINETIC THEORY OF ENGINEERING STRUCTURES

CHAP. I

Hence, for a single truss on two supports, with one end movable and the other
end hinged, the number of reaction conditions is Sr=3. Also, for a two-hinged arch,
with both ends on hinged supports, 2r=4; and for an arch without hinged ends, com-
monly called a fixed arch, 2r=6.

FIG. ID.

FIG. IE.

r
FIG. IF.

ART. 2. STRUCTURAL REDUNDANCY

Any structure, statically considered, may be classed as determinate or indeterminate,
depending on the manner in which it is supported, and the arrangement of the members
employed to form the structure.

A structure is externally indeterminate when it involves more than three reaction
conditions for a single truss element or frame. This implies that the reactions are
counted with the external forces.

A structure is internally indeterminate when it includes more members within its
frame than are required for internal stability according to the conditions for static equilib-
rium, regardless of the reactions.

Hence external and internal redundancy may exist simultaneously. The terms
redundant and indeterminate, as here used, are synonymous with impossible, only in so
far as the laws of pure statics are concerned. However, the modifying term static may
sometimes be omitted for the sake of brevity. Thus indeterminate will always mean
statically indeterminate.

The conditions for static equilibrium applied to any structure as a whole are repre-
sented by the following equations :

27=0, and 2M=0, ...... (2A)

wherein 2# is the sum of all the horizontal components of the external forces, includ-
ing the reactions; 2F is the sum of all the vertical components of these forces and reac-
tions; and SM is the sum of the moments of these forces and reactions about any point
in the plane of the structure. The sums are always algebraic sums.

These condition equations apply equally to all forces (externally applied and internal
stresses) active about any pin point of any structure.

When the external forces are not thus balanced among themselves and with the
internal stresses, then a structure is said to be statically under determinate or to be unstable.

A structure is, therefore, externally determinate when the three condition equations
for static equilibrium suffice to determine the reactions.

ART. 3 REACTIONS AND REDUNDANT CONDITIONS 7

A structure is internally determinate when its members are arranged to form triangles
in such manner that the successive removal of pairs of members uniting in a point,
finally reduces the structure to a single triangle. The triangle is thus seen to be the
primary element of all determinate systems.

A structure is in every sense determinate when all reactions and stresses can be expressed
purely as functions of the externally applied loads, unaffected by all temperature changes
or temperature inequalities, and small reaction displacements.

ART. 3. TESTS FOR STRUCTURAL REDUNDANCY

Let p= number of pin points of any given frame.
m= number of members in the frame.
n= total number of redundant conditions.
ri = number of external redundant conditions.
n" = number of internal redundant conditions.

e= number of elements or simple frames in a structure.
2r= number of necessary reaction conditions.
Then for any statically determinate structure

which condition must be satisfied in any event, and when satisfied, the structure is
always statically determinate but not always stable. This exception will be discussed
below.

When 2p>m + Zr, ........... (SB)

then the structure is not in stable equilibrium and may be called statically insufficient.
When 2p<w + Sr then the structure involves redundant conditions or members.
the total number of which is given by the equation,

n=m + Zr 2p, ........... (3c)

wherein a negative n would indicate static insufficiency.

The number of necessary reaction conditions r for any structure depends on the
number of frame elements or simple frames of which the structure may be composed.
Hence in order to distinguish whether the n redundant conditions are external, internal,
or both, it will be necessary to determine one or the other of these by a separate
criterion. This criterion is readily established for the reactions. Thus in Art. 1, Sr=3
was found to express the number of required reaction conditions for a simple truss on
two supports.

A composite structure, which may be composed of any number e of simple truss
frames, each one of which is directly supported by piers at one or two points, will always
have more than three reaction conditions. The requisite number of reaction conditions
for any statically determinate structure is

r=3+e-l=e+2. . . (3D)

KINETIC THEORY OF ENGINEERING STRUCTURES

CHAP. I

The number of external redundant conditions n', for any given structure of e elements,
is thus:

w' = 2r-e 2, (3E)

and hence the number of internal redundant conditions must be

n =nn.

OF)

The number of elements entering into the makeup of any structure is thus an important
consideration in establishing whether the redundant conditions, indicated by Eq. (3c),
are external or internal. In general, a truss element may be denned as a frame or girder
which in itself constitutes a simple truss or beam. The illustrations given below will
serve to amplify this definition.

The exception to Eq. (3A), when a structure is apparently determinate but unstable,
will now be discussed.

FIG. SA.

FIG. 3s.

The three-hinged arch, in Fig. SA, is in every sense a determinate, stable structure
for which the horizontal thrust H =Pl/2f. All the reactions A, B, and H are finite
quantities and the stresses in the several members are also finite. This will always
remain true except when the middle ordinate /=0, whence // = sc .

This special case is illustrated in Fig. SB, where ACB is a straight line and the load
P cannot produce stress without causing a slight rotation of the two elements I and II
about the points A and B. Hence equilibrium does not exist until rotation through
a small arc ds has taken place and the stress along AC and CB is then infinite. Struc-
tures involving this peculiar feature of the well-known " toggle joint " are regarded tem-

porarily unstable, and equilibrium in them
can be established only after certain
infinitesimally small displacements have
taken place.

Another striking example of this kind
is shown in Fig. 3c, where the supports A
and C are hinged, and the support B is
movable, and hence 2r=5. Also w = 19
and p = \2, making n=w + Hr 2p=0.

This structure is made up of three
elements I, II and III, each one of which
may be regarded as a simple frame in direct contact with a support. Hence e=3 and

This proves the structure to be internally as well as externally statically determinate,

ART. 3

REACTIONS AND REDUNDANT CONDITIONS

though a condition of temporary instability exists because the point k is common to
the paths of the three elements in the first instant of rotation about their respective
centers of rotation at A, B, and C.

FIG. 3o.

FIG. 3K.

FIG. SQ

FIG. SE.

FIG. 3F.

FIG. 3n.

FIG. 3j.

FIG. SL.

FIG. 3n.

The point k may thus be regarded as an imaginary crown hinge for a three-hinged
arch AkC. Hence a load P, acting at D, will cause a small rotation of the three elements
before equilibrium is established and then the stresses become infinite.

KINETIC THEORY OF ENGINEERING STRUCTURES

CHAP. I

The only loads which will not cause rotation are those passing through k, but for
any of these loads, there will result an infinite number of conditions of equilibrium and
as many different systems of stresses.

The following illustrations are presented for the purpose of showing various applica-
tions of Eqs. (3c) and (SE). Figs. 3D to 3M are all examples of simple truss elements
for which e = l. Fig. SN has six such elements counting the two suspenders cd, which
must always be included when the elements alone are regarded. For Fig. 3o, e=2 and
for Fig. 3p, e=8.

It should also be noted that Eq. (3c) is applicable to any structure as a whole,
whether composed of one or many elements. This equation will apply when the individual
members and pin points are counted and also when the elements and their connecting
points only are considered.

For example, in Fig. SN, there are 56 members, 32 pin points, and 8 reaction con-
ditions, making n = Sr+ra 2p=8+56 2X32=0, whence the structure is statically
determinate. Now when the elements alone are considered, then e=m--=Q, p=7 and
2r=8, and Eq. (3c) again gives ri = 5>+ra-2/>=8+6-2X7=0. Eq. (3E) gives
n' = 2r e2=86 2=0. Hence, both equations may serve to test the external redun-
dancy, when the elements alone are considered, but internal redundancy requires count-
ing all members and pin points in the entire structure. It is thus important to understand
the exact interpretation of these two test equations.

As a general rule it is best to count intersecting web members as four members
instead of two, though the result is usually identical. Thus in Fig. SK, n = 2r+m 2p =
34-1720=3+2124=0. But in this example the structure is not composed of
a succession of triangles and when pairs of members, meeting in a point, are removed,
the structure will not reduce to a single triangle, showing that it is not stable.

Fig. 3cj is another example of the toggle joint when the members at c are not
connected.

The following table was arranged to illustrate all these points with reference to the
several cases represented by Figs. 3D to 3n.

Fig.

n = ~ r + -m 2p

n 1 = r-e-2

Remarks.

3+ 1- 4 =

3-1-2=0

Determinate.

6+ 1- 4 = 3

6-1-2=3

3 times ext. ind.

6+ 1- 4 = 3

6-1-2=3

3 times ext. ind.

3+ 8-10=1

3-1-2=0

Once int. ind.

4+21-24 = 1

4-1-2=1

Once ext. ind.

6+17-20 = 3

6-1-2=3

Thrice ext. ind.

3+21-24 =

3-1-2=0

Not stable.

5+45-48 = 2

5-1-2=2

Twice ext. ind.

3+33-36 =

3-1-2=0

Determinate

8+56-64 =

8-6-2=0

Determinate.

4+10-14 =

4-2-2=0

Determinate.

10+16-26 =

10-8-2 =

Determinate.

3<>

3+ 9-12 =

3-1-2=0

Case of infinite stress.

8+ 3- 8 = 3

8-3-2=3

Thrice ext. ind.

CHAPTER II

THEOREMS, LAWS, AND FORMULAE FOR FRAMED STRUCTURES

ART. 4. ELASTIC DEFORMATIONS, FUNDAMENTAL EQUATIONS

Elastic Deformations

Let S = total stress in any member of a frame resulting from any cause, or causes,

designating tension by + .
I = length of any member when its S = 0.
Al= change in I due to stress S, + for elongation.
F = cross-section of any member, prismatic in form.
t = a uniform change in temperature in degrees, + for rise.
= coefficient of expansion per degree of temperature.
E = modulus of elasticity.
fS/Fumt stress in any member.
JZ/Z=relative elongation.
l/FE =|0=the extensibility, frequently employed for brevity.

Then according to Hooke's law and within the elastic limit of the material,

from which

(4A)

This equation represents the elastic deformation
for any member of any frame and is a fundamental
elasticity condition.

Referring now to Fig. 4A, let AB be any member
of any frame which, as a result of elastic distortions of
the frame, is made to undergo displacements Ji at A
and J 2 at B and a change in its original length of Al.
The new length of the member is then I + M and its new
position may be shown as A\Bi.

Since the displacements Ji and A% are very small
compared with the length AB, the two lines AB and
A\BI are assumed parallel for the present purpose.
The member and its displacements are referred to rectangular axes in Fig. 4A. Then

FIG. 4A.

12 KINETIC THEORY OF ENGINEERING STRUCTURES CHAP. II

which, differentiated, gives

2ldl=2(x b x n ) (dx b -dx a ) +2(y b -y a ) (dy b -dy a ) .

Also from the figure x b x a =l cos a. and y b y a ^l cos /9, which values substituted
in the previous equation, dividing through by 21, and replacing the differentials by
small finite increments A, gives

Al = (Ax b - Jx a ) cosa + (4y b - Ay a } cosp = -- + etl, ..... (4s)

tir

which is a fundamental elasticity condition for any frame.

Hence, every frame will involve twice as many of the unknown, subscript-bear-
ing displacements Ax and Ay, as there are pin points, while there will be as many equa-
tions of the form Eq. (4e), as there are members in the frame.

It will now be shown that any frame, whether involving redundancy or not, is capable
of anatysis by proving the following: For any frame in stable equilibrium, there are as
many possible condition equations as there are unknown quantities, provided the elasticity
conditions Eq. (4s) are included. This will be true of a statically determinate frame
without including the elasticity conditions.

The ultimate analysis of any frame includes the determination of the reaction forces
or conditions; the stresses in all the members; and the deformation of the frame as a
result of these stresses. The deformation is considered solved when the displacements
Ax and Ay, of all the pin points, are found.

It is assumed that for any structure under consideration, the externally applied loads
P, the changes in temperature t and the abutment displacements Ar (if any exist) are
all known and that wherever movable connections or roller bearings occur, these are
frictioriless.

Let S\, 82, 83, etc., be the stresses in the several members meeting in some particular

pin point m.
ct\, (Xz, 3, etc., be the angles which these members make with the x axis of

coordiantes.

/?i, /?2, /?3, etc., be such angles with the y axis.
P x =the sum of the components parallel to the x axis, of all the external forces

P acting on the point m.

P v =the sum of the components parallel to the y axis, of all these forces P.
p, m and 2r as previously defined in Art. 3. Then

(4c)
cos ,5=0

because for a state of equilibrium, thje sum of the horizontal and vertical components of
all the forces acting on any one pin point must respectively equal zero.

Hence, for every pin point of a frame, two equations of the form of Eqs. (4c) may
be written, expressing equilibrium of the internal and external forces for that pr'-

ART. 4 THEOREMS, LAWS, AND FORMULAE FOR FRAMED STRUCTURES 13

Also, one equation of the form of Eq. (4s) may be written for each member of a frame.
Besides these, there will be one equilibrium equation for each reaction condition.

Therefore, every stable frame affords 2r+m+2p condition equations of the first
degree, involving as many unknowns as there are equations, thus:

2p equations of the form (4c) involving ra unknowns S.
m equations of the form (4B) involving 2p unknowns Ax and Ay.
Zr equations for the reactions involving Zr unknowns R.

Total 2p-fra + Zr equations, involving 2p+m + r unknowns.

It follows then that the stresses S, the reactions R of known direction, and the Ax
and Ay projections of the pin-point displacements, may all be represented as linear func-
tions of the horizontal and vertical components of all the external forces P, plus similar
functions of assumed temperature changes t, plus linear functions of the abutment
displacements Ar, Jr 2 , Ar 3 , etc.

Thus any unknown function of any frame may be expressed by an equation of the

form

Z=f(P l ,P 2 , P 3 , etc.)+/i(0+/ 2 (M,M, ^ 3 , etc.), ..... (4o)

in which the coefficients are independent of the values P, t and Ar, but depend on the lengths
and directions of the members also on E, e and the manner in which the frame is supported.
The law of the summation of similar partial effects. In Eq. (4o), every set of causes
or conditions P, t and Ar, produces a partial value Z r for Z and the ultimate total value
of Z=Z' +Z" +Z'", etc., is the sum of all the partial values or effects resulting from the
respective sets of independent causes or conditions. Thus each effect, such as a stress,
whether due to loads, temperature or abutment displacements, may be ascertained or invest-
igated by itself and the sum total effect Z will then be the sum of the several similar partial
effects. This is the law of the summation of similar partial effects resulting from various
causes and is fundamental to the analysis of all structures involving redundancy.

The law of proportionality between cause and effect. Eq. (4o) , being true for any set
of effects, would remain true for any multiple of these effects. Hence, if the loads are
doubled the resulting stresses will likewise be doubled, etc. Therefore, the law of pro-
portionality holds true between the causes and their effects. Thus, if a set of loads P
produces stresses S in the members of a frame then another_set of parallel loads P,
acting at the same points, would produce stresses S which are P/P times as great as^the
stresses S. Or, if a single load unity, acting on a point m of any frame produces reactions
Ri, and stresses Si, then a load P m acting at the same point will produce reactions
R = P m Ri, and stresses S =P m Si.

14 KINETIC THEORY OF ENGINEERING STRUCTURES CHAP. II

ART. 5. GENERAL WORK EQUATIONS FOR ANY FRAME

The general work equation, Clapeyron's law. A loaded frame is a machine in equilibrium
and within the limits of proportional elasticity its deformation varies directly with the
magnitude of the superimposed loads.

The question of deformation does not enter into statics and hence a general and
comprehensive treatment of the elastic frame must necessarily involve the principles
of mechanics. ,

For example, given a frame with definite loading and supports, constituting a sys-
tem in external equilibrium. The frame in turn will undergo certain deformations which
will steadily increase in proportion to the internal stresses created in the members by
the external forces as they are gradually applied. The final deformation will occur
in the instant when the external forces are exactly balanced by the internal stresses, pass-
ing into the condition known as static equilibrium.

While this deformation is taking place the applied loads travel through certain
distances which are the paths or displacements of the points of application of the loads.
Thus a certain quantity of positive work of deformation is performed by the external forces.

The internal stresses in the members must accommodate themselves to the deformed
condition of the frame and in resisting this action must produce negative work of deformation.

In the instant when static equilibrium is established between the loads and stresses
the positive and negative work of deformation produced in the same interval of time
must exactly balance.

Let A e =the positive or externally applied work of deformation.
.4;= the negative or internally overcome work of deformation.
P=any externally applied loads including the reactions.
$=the stress in. any member due to the loads P.
M ihe change in length of any member due to the stress S.
=the displacement of the point of application of any force P measured in
the direction of this force.

A positive amount of work is always produced when the force and its displacement act
in the same direction.

The product %Pd represents the actual work produced by a force gradually applied
and increasing from its initial zero value to a certain end value P, thus exerting only
its average intensity during the entire time of traversing the path to perform this work.
Hence, for all the external forces acting on any frame, the total positive external work
of deformation would be A e =^Pd.

Similarly %SAl represents the actual work in any member subjected to a gradually
increasing stress of end value S, with an average intensity S/2 during the entire time
while producing a change in its length of JZ. Hence for all internal stresses in any frame
the total negative work of deformation would be A t -=^2$JZ.

By the " doctrine of the conservation of energy," the applied work must equal the
work overcome, hence

(5A)
or

ART. 5 THEOREMS, LAWS, AND FORMULAE FOR FRAMED STRUCTURES 15

which is Clapeyron's law (1833), and may be stated as follows: For any frame of constant
temperature, and acted on by loads which are gradually applied, the actual work produced
during deformation is independent of the manner in which these loads are created and is
always half as great as the work otherwise produced by forces retaining their full end values
during the entire act of deformation.

By regarding any elastic body as composed of an infinite multiplicity of members,
it is readily seen that Clapeyron's law applies equally to frames and solid web elastic
structures when properly supported. Clapeyron's law finds extensive application in
problems dealing with deformations of framed structures.

Cases involving dynamic impact would imply a certain amount of kinetic energy
in excess of the applied work of deformation. That is, the applied forces have some
initial value, greater than zero, while the stresses are still zero. This would not be repre-
sented by the work Eq. (OA), wherein A e Ai=Q, but would give rise to the following
equation :

Ae-Ai-jPf, ..... . ..... (OB)

indicating a state of accelerated motion instead of one of equilibrium. For problems
under this heading see Art. 16.

The law of virtual work will now be considered. It is a general law, permitting of
more varied application than does Clapeyron's law.

The law of virtual work was first enunciated by Galilei and Stevin, and in its more
general form by Joh. Bernoulli. Lagrange (1788) reduced the law to an algebraic expres-
sion for which the following derivation may be given.

According to the general law expressed by Eqs. (4c). all forces (and stresses) acting
on a pin point of any structure in equilibrium will have components parallel to any
axis and the sum of such a set of parallel components must be zero. Calling o: the angle
which any force or stress P makes with the axis chosen, then for any pin point

2P cosa=0.

Now if a displacement J, parallel to the axis, be arbitrarily assigned to this pin point
and assuming equilibrium to continue, then the product of J with the sum of the com-
ponents must still be zero. This is equivalent to multiplying the above equation by
J to obtain

2(Pcosa)J=0.

But the displacements A cos a=d are the projections of the displacements J on the
directions of the several forces, hence

0, ............. (5c)

wherein P signifies that the forces are in every sense independent of the displacements d.
This is the law of virtual work, and is applicable to any point or group of points and
hence to any frame or group of bodies. The displacements d may be any possible dis-
placements whether or not subject to the law of elastic deformation, provided equilibrium
exists.

KINETIC THEORY OF ENGINEERING STRUCTURES

CHAP. II

Stated in words this law implies that for any point, frame or body acted on by loads
P in an established state of equilibrium, the sum total work performed by these forces, in
moving over any small arbitrary but possible displacements o, must always be equal to zero.

If the time element is introduced into Eq. (oc) giving SPo/f =0, the law of virtual
velocities is obtained.

Professor Otto Mohr's work equations. Proceeding from Eq. (5c), Professor Mohr, in
1874, developed two equations which furnish the most comprehensive laws for framed struc-
tures. These equations are now derived.

Given any frame carrying the arbitrarily assumed loads PI, P 2 , P 3 , etc., which
in combination with temperature changes produce stresses Si, *S 2 , 83, etc., in the several
members of the frame and thus constituting a system of forces in static equilibrium.

All the pin points of the frame are now supposed to be subjected to small, arbitrarily
assigned displacements A\, J 2 , J 3 , etc., which may have been produced by some other
system of actual loads PI, P 2 , P 3 , etc., entirely independent of the loads P. The J's
will naturally vary in amount and direction for each pin point.

According to the la\vs for static equilibrium, the sum of the components, taken in
any fixed direction, of all forces acting on any pin point of a frame must equal zero.

Choosing for the fixed direction the displacement J lr
then for any particular pin point, Fig. SA, the follow-
ing equation may be written:

cos

which multiplied through by
JiP x cos 61 + Ji

5 cos 0=0;
i gives
cos 0=0.

But from the figure, A\ cos #1 =<?i, which value sub-
stituted in the above equation gives

0=0.

(on)

FIG. SA.

Eq. (on) represents the virtual work of all forces
acting on the pin point 1, in accordance with Eq. (5c).
One such Eq. (5n) may be written for each pin
point and the sum of all these equations would furnish
the total virtual work for the whole frame.

In this sum equation, each load P will occur only once, while each stress S will occu
twice, being involved once for each end of the same member.
The terms containing the forces P will give the sum

The sum terms involving the same stress S may be found for each member as

cos

AHT. 5 THEOREMS, LAWS, AND FORMULA FOR FRAMED STRUCTURES 17

since the two displacements J cos </> for the two ends of the same member give the sum
Jl, or actual change in the length of such member.

Hence, the terms containing the stresses S will give the sum

where the minus sign is due to the fact that the stresses applied to pin points as external
forces are always opposite in direction with their respective JZ's.
The final sum equation for all pin points then becomes

2P3 = 2SM, ............ (5s)

which may be said to represent a condition of elastic equilibrium as distinguished from
static equilibrium. The law thus expresses the equality between the external and internal
virtual work of deformation for real displacements and arbitrary cases of loading, provided
elastic equilibrium exists.

The external forces P necessarily include the reactions due to the applied loads P.
However, the work of the reactions is always zero when the abutments are immovable.
In the general case involving abutment displacements Jr in the directions of the reac-
tions R produced by loads P, Eq. (OE) may be written

.... (OF)

It should be repeated that the displacements d, Ar and Al are actual and mutually
dependent on the same causes, such as the actual loads P and temperature changes.
The arbitrary loads P, reactions R and stresses S form a system in elastic equilibrium
which is independent of the actual loads P, reactions R and stresses S, and hence inde-
pendent of the actual displacements.

In the special case where the forces P, S and R become identical with the forces
P, S and R, Eq. (5r) becomes

v _ v - y S2 *
EF'

which is in accordance with Clapeyron's law.

Eq. (5r) is the fundamental law of framed structures and includes all conditions
of equilibrium of the external forces; of the external forces and internal stresses; and
of all relations existing between the stresses and the distortions of any frame.

Professor Mohr's second work equation serves the purpose of determining any displace-
ment d m produced by any case of loading. It follows from Eq. (5r) by allowing all the
arbitrary loads to vanish and substituting therefor a single load unity and the stresses
Si and reactions RI resulting from such unit load. The new work equation then becomes

KINETIC THEORY OF ENGINEERING STRUCTURES

CHAP. II

wherein the unit loading, henceforth called the conventional loading, may be a single
force unity or a moment equal to unity. In the latter case the o will represent a rotation
measured in arc. By dividing Eq. (5n) by unity the stresses Si become abstract num-
bers and the equation will read

The applications of this law, universally known as Mohr's icork equation, will be shown
in the following articles 6 and 7. The work represented by the conventional loading will
generally be designated by W.

ART. 6. DISPLACEMENTS OF POINTS. STATICALLY DETERMINATE STRUCTURES
BY PROFESSOR MOHR'S WORK EQUATION

The relative displacement 8 m between any two pin points mi and m, of any frame Fig.
GA, when Al is given for each member of the frame, may be found by applying Eq. (5i) .

If the abutments undergo known displacements Ar, as a result of the given actual
loading, this effect on the required d m must be included.

Likewise temperature displacements may be considered, but the problem will first
be treated by neglecting this effect.

Now assume two unit forces, applied at the points mi and m respectively, and, acting
in opposite directions along the line tn\ < m. The directions of these unit forces should be so

chosen as to make the conventional work l-o m a
positive quantity. This means that if d m is an
elongation, then the unit forces must be so
applied as to elongate the distance m\m. This
rule will be universally applied to determinate
structures as well as to indeterminate for
redundant members or conditions. A negative
result will indicate an erroneous assumption
in the direction of the unit conventional
load.

The stresses Si and abutment reactions R i
resulting from the two unit loads acting at mi
and m are now determined from a Maxwell dia-
gram or by computation.

The work Eq. (5n) applied to this assumed or conventional loading then gives, after
substitution of values Al=Sl/EF, excluding temperature effect,

FIG. GA.

(GA)

The stresses S are those produced in the structure by any real case of loading, as loads
P, temperature changes or abutment displacements, giving rise to the actual changes
Al in the lengths of the members.

ART. 7 THEOREMS, LAWS, AND FORMULAE FOR FRAMED STRUCTURES 19

Eq. (6A) then offers a solution to the above problem, because all the quantities except
o, n are known or can be found from the given data.

If temperature effects are to be included then Eq. (4A) gives the value for JZ, thus
furnishing the final equation

~+eU)-ZR l Jr ..... \ . . (6 B )
&V /

In a similar manner the relative displacement of any pair of points, the deflection
of any point, or the angular change between any pair of lines of any frame, resulting from
given changes M of all the members of the frame, may be found by assigning such con-
ventional loads as will produce the conventional work 1 d on the frame at the point or
points in question. See also Art. 9 for other examples of this class.

ART. 7. INDETERMINATE STRUCTURES BY PROFESSOR MOHR'S WORK EQUATION

A truss, according to Chapter I, may include more than the statically necessary
number of members or reaction conditions, and is then called statically indeterminate.
It is now proposed to show the manner in which Mohr's work equation may be employed
to find the stresses and reactions in such a structure, loaded by any system of loads
P concentrated at the several pin points of the frame.

The structure must be so constituted that when all redundant members and reaction
forces are removed, the remaining frame will represent a statically determinate structure,
including the necessary conditions for proper support. This determinate frame will
always be called the principal frame or system..

Now let X a , Xi, X c , etc., represent the stresses produced by the applied loads in
any redundant members or supports, as the case may demand. When these stresses
are applied to the principal system, together with the external loads, the resulting stresses
in all the principal members will be identical with those produced in these same mem-
bers by the original loading of the whole indeterminate frame. This will also be true
of the deformations.

It follows then that the loads P applied to the indeterminate frame, and the loads
P and X applied to the principal frame produce identical stresses and deformations in the
principal members. Also, the deformations are in each case definitely fixed by the elastic
changes Al of the necessary members of the principal system. Hence, for applied loads
only (without temperature effects) the work of the stresses in the principal frame is
always greater when a redundant member is omitted. The work which a redundant
member can do must necessarily lessen the work otherwise required of the principal
system.

If, then, the elastic displacements of the several pin points representing the points
of application of the redundant forces X, are found for the principal system, these dis-
placements will suffice to solve one elasticity equation for each unknown X. This may
be done in two ways: (1) By applying Mohr's work equation to the indeterminate sys-
tem, and (2) by finding the requisite elastic displacements of the principal system from
Maxwell's law. The solution by Mohr's work equation will now be given.

20 KINETIC THEORY OF ENGINEERING STRUCTURES CHAP. II

The following definitions of terms will be strictly adhered to in all succeeding
discussions.

Let S =the stress in any member of the principal system.

>S> =the stress in this member due to the loads P when the several redundants

X and t are all zero, to be known as condition X=Q.
a =the stress in this member of the principal system when no load other than

X a = 1 is active. Condition X a = 1 .

Sb=ihe same when no load other than Xi = l is active on the principal system.
/S c =the same when no load other than X c = l is active on the principal system.
j=the stress in the same member caused by a uniform change in temperature.
R t =& reaction produced by a uniform change in temperature.
M t = a moment produced by a uniform change in temperature.
X a , Xb, X CJ etc., the stresses in the redundant members or reactions.
R , R a , Rb R c , etc., are denned like the S's with like subscripts, but represent

reaction forces instead of stresses.
da, <>b, $c, are changes in the lengths of the redundant members X a , X b , X c ,

respectively.

The forces X a , Xb, X c , etc., may be the stresses in redundant members or they may
be redundant reactions as in the case of fixed arches or continuous girders, etc.

Each of the above cases of conventional loading will be known as conditions. Thus
condition X =0 will mean that all the redundant conditions are removed, while condi-
tion X a = l will signify that this force alone is applied to the principal system and all
other X's, S t , and P's are removed.

The stress S, in any member of a frame involving redundant conditions, is a linear
function of the loads P, X a , Xb, X c , etc., all treated as external forces applied to the
principal system. This follows because all conditions of equilibrium are represented
by linear equations.

Hence, according to the law of the summation of effects expressed by Eq. (4o) , the
general equation for stress in any member of a truss involving redundancy, would have
the form

wherein ax =the stress in any member due to the external force X a acting alone on the
principal system. Similarly 5 6j .=the stress in the same member due to the external
force Xb acting alone on the principal system, etc.

But by the law of proportionality S ax =S a X a , Sb x =SbXb and S cx =S c X Ci etc. Hence,
the general equation for any case of redundancy may be written :
for the stress in any principal member

/S=*So S a X a SbXb S C X C , etc., +/S<

for any reaction of the principal system

R=R R a X a Rb Xb R c X c , etc., -\-Rt ...... (7 A)

and for any moment on the principal system

M=M -M a X a -M h X b -M c X c , etc.,+M

ART. 7 THEOREMS, LAWS, AND FORMULAE FOR FRAMED STRUCTURES 21

In these equations the quantities S , R and M are all linear functions of the
externally applied loads P, while all the stresses, reactions, and moments bearing
subscripts a, b, c, etc., are constants due to conventional loadings and are absolutely
independent of all external loads P and X.

The work of any member or reaction is obtained from Eqs. (7 A) by multiplying both
sides of these equations by the deformation which each sustains. Thus, neglecting
temperature effects, the work equations are

SM=[S -S a X a -S b X b -S c X c , etc.]JZ
RJr=[R -R a X a -R b X b -R c X c , etc.]Jr

The negative signs in Eqs. (7 A) and (7e) , indicate that the quantities involving the
redundants X must always be of opposite sign to the stress S resulting from the loads
P, for reasons above stated in this article.

Hence the increment of work performed by the redundants X, as shown by Eqs. (7s),
is always negative with respect to the work performed by the stress S because the forces
X are classed with the external forces.

The conventional unit loadings are always applied in the opposite direction to the
forces they replace, and this in turn puts these (negative) unit forces in the same direc-
tion as the displacements d which they accompany. Hence the work 1 o is always
positive.

This is exactly the same as the case illustrated in Art. 2, where the unit load was
independent of all redundant conditions.

Therefore, whenever in the following a conventional unit load represents a redundant
condition, the direction of this unit load must be taken in a direction opposite to the force
representing the stress in such member. This is equivalent to saying that when a
member X elongates under stress, then the unit load X = l must be so applied as to move
the adjacent pin points apart, and the converse when the member shortens under
stress.

Whenever the direction of a redundant force cannot be correctly foreseen, then some
assumption is made which, if it be erroneous, will result in a negative value for such
redundant X.

The following illustration is given to make these points clear and to avoid confu-
sion of ideas in all future problems.

Figures 7A to 7E represent an indeterminate structure involving three redundant
conditions as may be verified by applying Eq. (3c) to the problem in Fig. 7A.

The truss is supported by hinged bearings at A and D and by columns at B and C,
all of which may be subjected to certain displacements which may be estimated from
the conditions of the problem.

Figure 7s, then, represents the principal system carrying loads P, when all redundant
conditions are removed and shows condition X =0. The other figures represent, respec-
tively, the conventional loadings X a = 1 , X b = 1 and X c = 1 , the three X's being the redundant
conditions. Note here the directions which the unit forces take. They are negative

KINETIC THEORY OF ENGINEERING STRUCTURES

CHAP. II

with respect to the forces they replace, but in the same directions as the displacements
which their points of application undergo as a result of the loads P.

The reaction displacements are assumed to be downward by amounts A A, AB, AC,
and AD for these supports respectively, and it is also assumed that A and D move apart
by an amount AL. Temperature effects will be taken up later being omitted for the
present.

GIVEN CONDITIOM

CONDITION X=O.

FIG. 7A.

CONDITION Xf I.

FIG. 7c.

FIG. 7o.

CONDITION X c -l.

FIG. ?E.

The various reactions in the following table, given for the four conventional load-
ings, are now found for the principal system, which in this case is a three-hinged arch.

For Condition.

V f\

2P(L-e)

D (\

2Pe

A n L-P(L-2e)

Ao L

.DO U

o vJ

D L

H 2h

,, Uh-J:}

A-a

V, 1

^o =

l(L-d)

B a = v
R, n

a=0
c', n

L> a \J

D h - ld

fla-
il

H h ld -

Afc 1

V 1

A b L
A 1 ' d

Dfe U

<-& i

L
D KL-rf)

Hb 2h
l.d

A c 1

Ac ~ L

c (J

c U

Dc L

Hc " 2h

* Obtained by taking moment equation about the middle hinge O, in a clockwise direction.

ART. 7 THEOREMS, LAWS, AND FORMULA FOR FRAMED STRUCTURES 23

Hence from Eq. (7A) for reactions, these values give:

-e L-d d

A=A -

C=0

and

h k d d

H = H H a X a H b X b H C X C =H X a ylXb FJT

S & & a X a &bXi) S C X C

. . . (7c)

The work equation (GA), when applied to each of the three redundant forces X a , X b
and X c and observing that the conventional work 1-d in each case is positive, gives

= + 1 AL
- + 1 AB

(7D)

The values for ^R Ar, in each of the Eqs. (?D) , may also be written out from the above
tabulated quantities for the three conventional loadings X a = 1 , Xb = l and X c = I . They
are:

A b AA-D b AD-H b AL
L d . . d .^ d

-A C AA-D C AD-H C AL

d . L d . d .

(7B)

It should be observed that when the direction of a force is opposite to the direc-
tion of the displacement over which the force travels, then the product which represents
work is always negative. This is the case with all the work quantities in Eqs. (7E) . Hence,
when these are substituted into Eqs. (7o) they become positive.

The quantities 2>S a Al, HSbAl and ^S C AI still remain to be evaluated. Eq. (4A) gives
M=Sp + ett, where S=S S a X a -S b X b -S c X c from Eq. (7 A). Then by substitution

M=[S -S a X a -S b X b -S c X c ]p+dl,

which gives the final value

. . . (7c)

24 KINETIC THEORY OF ENGINEERING STRUCTURES CHAP. IT

Similar expressions follow for 2S&JZ and 2$ C JZ, and these substituted in Eqs. (7o)
give for the values d:

d a =

. . (7n)
d c =

wherein the summations, except those of the reactions, include only the members of
the principal system.

Now d a , 5, and d c being the changes in the lengths of the three redundants, these may
be evaluated from Eq. (4A) in terms of the lengths, areas and stresses of these members
(or reactions) and become

-A- , , , .

(73)

Hence all the terms in Eq. (7n) are now known except the three redundant forces
X a , X b and X c , and having three elasticity equations involving only these unknowns,
the latter may be found by solving Eqs. (7n) for simultaneous values of the X's, with
the aid of Eqs. (?E) and (7j) .

It is thus seen that the abutment displacements are of vital importance in determin-
ing the magnitudes of the stresses in any statically indeterminate structure. When-
ever these displacements cannot be determined with any degree of certainty, or when
small displacements indicate large resulting stresses, then such structures should not
be built. This applies particularly to fixed arches and continuous garders.

Eqs. (7n), based on Professor Mohr's work equation, thus furnish a means for
the analysis of any statically indeterminate structure.

ART. 8. INDETERMINATE STRUCTURES BY MAXWELL'S LAW

The application of Maxwell's law to the same analysis will now be given by express-
ing the summations in Eqs. (7n) in terms of the elastic deformations of the structure
instead of the stresses in the members.

Leto ma =the displacement of the point of application m of any load P, n , in the
direction of this load, when the principal svstem is loaded with only

Xa = l.

d mb =fi similar displacement of the same point m for the conventional load-
ing X 6 = l.

d nlc =ihe same for condition X c = l.

daa =the change in length of the member a for condition X a = l.
d^ =the change in length of the same member a for condition X b = l.
doc =the similar change for condition X c = \.
dba = the change in length of the member b for condition X a = 1 .
dbb =the change in length of the member 6 for condition Xb = l.

ART. 8 THEOREMS, LAWS, AND FORMULA FOR FRAMED STRUCTURES 25

dbc = the change in length of the member 6 for condition X c = l.
d ca =the change in length of the member c for condition X a = l.
d c t, =the change in length of the member c for condition X& = 1.
d cc =the change in length of the member c for condition X c l.
o al =the change in length of the member a resulting from a change in tempera-
ture t, when the principal system is otherwise not loaded, hence

x=o, p=o.

Ob<=the same for the member b.
d ct =the same for the member c.

For the sake of simplicity, the abutments will be assumed immovable, thus making
all Jr=0. This part of the previous problem remains unchanged.

The work Eq. (OE) for the condition X 0, (when the loads P only are acting on
the principal system and producing the stresses S ) now becomes for the three cases of
displacements

Similarly applying the work Eq. (GA) to the conditions X a = l, Xb = l and X e = l,
the various displacements become

1 d aa = 2S a M u ; l-3 ab = 2S a M b and 1 3 ac = 2S a M c ;
and by inserting values for Al aj 4l b and Al c from Eq. (4A) then

and similarly for conditions X b = 1 and X c = 1

2 S c S a p =d c a] S S c Si,p =d c b', 2 S c S c p = d c

(SB)

Finally the work Eq. (6B), for temperature displacements for each of the conven-
tional unit loadings, gives

SS etf - 1 3 at ; 2S b dl = l-3 bt and SS c stf = 1 d ct ...... (8c)

Substituting all these values from Eqs. (SA) , (SB) , and (8c) into Eqs. (?H) , the follow-
ing important equations are obtained :

Xcdac S# a Jr +d at

Ob = ZPmdmb X a dba X b dbbX c db c T>Rb
d c = 2P m 3 mc -X a d m -X h d cb -X c 3 cc - 2R c

(SD)

It should be noted that for the quantities on the right-hand side of Eqs. (So), the
single subscripts and the second one of the double subscripts always refer to the con-

26 KINETIC THEORY OF ENGINEERING STRUCTURES CHAP. II

ventional loadings X a = l, X b = l and X c = l, while the first of the double subscripts always
refers to a point or member of the structure.

According to Maxwell's law (proven in Art. 9) the following equalities exist in Eqs. (So) :

$ab = dba', dac=^ca and d bc = O cb .

This means that the order of the subscripts may be interchanged at will without
altering the equations, thus greatly simplifying the solution of problems.

All of the displacements d, in Eqs. (So), may now be determined by four Williot-
Mohr displacement diagrams (see Chapter VI), drawn respectively for the conditions
X a = l, X b = l, X c = l and t = l acting on the principal system. Hence, the three redun-
dant conditions X are again found by solving Eqs. (SD) for simultaneous values.

It is thus seen that any indeterminate structure may be analyzed in either of two
ways, by Eqs. (7n) or (SD), according to the choice of the designer or the nature of the
problem.

The redundants X being found by either of the above methods, all the stresses S
and reactions R may now be determined from Eqs. (7 A) . In these the X's are the redun-
dant forces from Eqs. (7n) or (So) and the stresses S , S a , S b and S c are those found for
the conventional loadings on the principal system and are independent of the values
X and of each other.

The summations in all the previous equations include only the members of the
principal frame. However, an equation of the form (7n) may be written to cover all
members, including the redundant, and this form is frequently very useful.

The work Eqs. (7o) when made to cover all members of an indeterminate frame
become

S# c Jr = SS c JZ ..... . (SE)

These values inserted into Eq. (7o), and others of that form, give

+ 2S a ea 1

(8r)

2# c Ar = 2S c S p -X a ^S c Saf) -X b ZS c S bf ) - X c ^S

wherein the summations extend over all the members including the redundant, and all
the terms retain their previous significance. Therefore, when X a = l, then S =0,
S a = l=X a , S b =0, and S c =0.

The first terms of the right-hand side of each of the Eqs. (8r), according to Eqs.
(SA), may be expressed in terms of the external loads as follows:

2iS S a p= 2P m $ mo ; ^S S b f>~^P m d mb ', 2)S *ST C jO=lIP m ^ 7nc , . . . (SG)

and may be evaluated from Williot-Mohr displacement diagrams, Chapter VI. All the
other summations may be determined once for all either from Maxwell stress diagrams,
or in terms of displacements by employing Eqs. (SB) .

Problems of the kind treated in Art. 6 can now be solved for any indeterminate
structure by following precisely the same method there indicated except that the redundants

ART. o THEOREMS, LAWS, AND FORMULA FOR FRAMED STRUCTURES 27

must first be found by one of the two methods just given, finally employing Eq. (?A),
to ascertain the stresses. These stresses being found, the elastic changes in all
members are computed. Then by applying a conventional unit load to the point for
which the displacement is desired, determine the stresses in the principal system for
this unit load and proceed as before by using Eq. (6A).

It may be well to mention in closing this subject that there is usually a wide latitude
n the choice of the redundant conditions. Thus, in Fig. ?A, one abutment might have
>een placed on rollers, thus making the principal system a simple truss on two supports^
Then X a would have become the horizontal thrust instead of the stress in a member

ART. 9. PROF. MAXWELL'S THEOREM (1864)

This theorem, known as the law of reciprocal displacements, and previously men-
tioned in discussing Eqs. (8 D ), establishes the mutual relation between the elastic dis-
placements of a pair of points, or a pair of lines, whenever these displacements result
from simultaneous conditions of loading ;_ provided that the arrangement of the membe
remains unchanged and the supports are immovable.

For the sake of simplicity it is assumed that the frame is in a condition of no stress
and that there are no temperature changes.

Clapeyron's law then applies and the equation for actual work becomes

wherein the P's are concentrated loads, and the 3's are the deflections of the points on
which the loads act in the respective directions of these loads.

FIG. QA.

FIG. OB.

Each of the products \P m d m may now be regarded as a summation of work pro-
duced by some group of loads such that the work of
the group is exactly identical with the work rep-
resented by %Pmd m . Figs. QA, 9s, and 9c will
illustrate just what is meant by such groups of loads.

If in Fig. OA, two equal and opposite . forces
P m are applied at the points m l and m, then the
resulting o m represents the relative displacement
between these two points. We call this case
the loading of a pair of points corresponding
to the case illustrated in Fig. GA, where the loads P are each equal to unity.

FIG. 9c.

28 KINETIC THEORY OF ENGINEERING STRUCTURES CHAP. IT

Fig. OB shows a pair of equal and opposite forces perpendicular to a line ik and
producing a couple of moment 1-Pm. This represents the loading of a line. The result-
ing d m represents the angular change in the line ik expressed in arc. This case may be
designated as the loading of a line m and, whenP = l, it becomes the unit loading of this
line.

Fig. 9c represents the loading of a pair of lines m^ and m. The d m of this group
of loads is the relative angular change between the two lines, or it is the change in the
angle 6, which results from the conventional loading.

The three groups of loadings are typical, and any particular load applied to a frame
may become one load of such a group, but not of two or more groups simultaneously.

For brevity, any such group of loads will be known as a loading and the corresponding
o will be the path or elastic displacement of the loading.

The several d's are always linear functions of the applied loads P and can be expressed
as follows:

d bm P m ; \ (9B>

O m ~ I>M r a ~r rnb* b ~r O mln L n

wherein the d's with the double subscripts are independent of the loads P, and for
example o am represents the special value of d a when P m = 1 and all the other loads P
are zero, all in accordance with the nomenclature employed in Art. 8.

Now apply loads P m to any frame, producing stresses S m in the several members,
and changes in their lengths Jl m =S m l/EF.

Likewise for a system of loads P n , producing stresses S n , and changes Al n = S n l/EF.

Let d' mn = the value of d m , for the point m, when certain loads P n only are active.
Also, let d' nm = the special value of d n for the point nj when a certain set of loads P m only
are active. Then note that d' mn = d mn when 2P n =l and o' nm = d nm when 2P TO =1.

The work equation for the loads P m , stresses S m and displacements d' mn and Al n due
the loads P n only, is according to Eq. (OE)

S I

2jPmO mn = ^o wz Jt n == Lo^-pp.

Similarly

S I

TiP n d' nm = 2}S n dl m ~ ^^n^W>

therefore,

(9c)

which is Betti's law (1872) and which extends Maxwell's law to the summation of all
members of a structure.

If instead of systems of loads P m and P n as above, only the unit loads P m = 1 and
P n = 1 are successively applied to the frame, then Eq. (9c) becomes

d mn = 8 nm , ............. (9D)

which is Maxwell's law (1864) and may be stated thus:

ART. 10 THEOREMS, LAWS, AND FORMULA FOR FRAMED STRUCTURES 29

1. The relative displacement d mn of a pair of points TOI and TO, resulting from a
unit loading of another pair of points n\ and n, is equal to the relative displacement o nm
of the pair of points n\ and n, caused by a unit loading of the pair of points m\ and TO.

2. The same equality exists between the relative angular changes of two pairs of
lines, successively loaded with a unit loading.

3. The same equality also exists between the linear change in a pair of points result-
ing from a unit loading of a pair of lines, and the angular change (expressed in arc) of
the pair of lines, resulting from a unit loading of the pair of points.

The practical value of Maxwell's law when applied to redundant conditions was shown
in Art. 8. Its application to displacement influence lines is given in Chapter VIII.

ART. 10. THEOREMS RELATING TO WORK OF DEFORMATION

(a) Menabrea's law (1858), or theorem of least work. Given a statically indeterminate
framework in an initial condition of no stress for which the temperature is known and
remains constant; also, assuming that the supports are either rigidly fixed or permit
frictionless movements, such that 2.RJr = for the entire structure, then the changes
in the lengths of the members are expressed by the formula M = Sl/EF, and the work
equations for the several statically indeterminate quantities X a , Xf,, X c , etc., are, accord-
ing to Eqs. (SB) ,

etc.,

the value S in which is given by Eq. (7A.) as

S = S S a X a SbXf, S C X C , etc.
The partial differentiation of S with respect to X a gives

likewise for the other X's,

(10B)

(lOc)

The actual work of deformation for the entire frame, including redundant members
as external forces, becomes by Clapeyron's law or Eq. (5A)

(10D)

30 KINETIC THEORY OF ENGINEERING STRUCTURES CHAP. II

which when differentiated gives

and by dividing through by ~dX a this becomes, after substitution of values from Eqs. (lOc)

EF EF '

which by Eq. (10A) must equal zero. Hence,

, . .. . dA . 9 A , .

= and similarlv ^-^-=0 and ^^=0, .... (10E)

which proves that the redundant or indeterminate conditions, reduce the actual work of defor-
mation of the frame to a minimum. This is the theorem of least work.

(b) Castigliano's law (1879), or derivative of the work equation. This law deals
with the displacement of the point of application of a force.

Any external load P m acting on a framework will, by Clapeyron's law, produce the
actual applied work of deformation

A=$P m d m , ......... . . (10F)

when d m is the displacement of the point of application of P m in the direction of P m
and within range of proportionality, or within the elastic limit of the material.

Also, the actual internal work of deformation, for the entire frame, becomes, accord-
ing to Eq. (10D)

The partial differential derivative of A with respect to any certain external load
P m is

'S 2 l\ .^ s dS_ J_

3P m EF'

wherein 3*S/3P OT is the derivative of the stress S in any member of a determinate or
indeterminate structure.

The general expression for S, from Eq. (lOs), by substituting for S its equivalent
value in terms of external loads becomes

, etc., +S m P m -S X a -S b X b -S c X c , etc., . . . (10j)

in which Si is the stress in the member S for PI =1, while S a is that stress when X a = l,
etc.; finally S is the stress in any particular member caused by the combined effects
of all P's and Z's.

The several values Si to S m are thus independent of the loads P and X. Also, the
loads P and X are independent of each other. Hence, S may be partially differentiated
with respect to P m or X and Eq. (10j) when so treated gives

--=S m and '' == ~ s > etc ........ ( 10K )

ART. 11 THEOREMS, LAWS, AND FORMULA FOR FRAMED STRUCTURES 31
Substituting this value in Eq. (lOn) then

but, according to Mohr's law, Eq. (GA), when abutment displacements are zero, then

(10M)

In Eqs. (IOL) and (10M) the summation extends to all the members, including the
redundant conditions, and hence

7)4

Ld m =-, ....... ........ (ION)

which, expressed in words, means that the path o m of a load P m is equal to the partial
differential derivitave of the actual work of the frame with respect to P m .

Castigliano's law thus expressed is equivalent to Mohr's work equation, that is,
leading to the identical result by a more circuitous process.

In similar manner the same law may be deduced for girders with solid webs.

Mohr's law thus offers the most direct method of finding any displacement which
may result from any specific cause. Castigliano's law will lead to the same conclusions
by a somewhat less direct method.

ART. 11. TEMPERATURE STRESSES FOLLOW MENABREA'S AND CASTIGLIANO'S

LAWS

When a structure is subjected to a uniform change in temperature t then, from
Eq. (4A),

This value of M when substituted into Eq. (10D) for Sl/EF, gives

; ........ (llA)

and the differential of A with respect to S is

(llB)

Similarly when the temperature effect is introduced into Eqs. (10A) then for immovable

abutments as before

y c cf/

2S^l==+8 a <&-Q ......... (He)

32 KINETIC THEORY OF ENGINEERING STRUCTURES CHAP. II

Also, Eqs. (lOc) apply equally when temperature effects are included, hence by divid-
ing Eq. (lie) by 3A" a , and substituting values from Eqs. (lOc), then Eq. (11s) becomes

9 A

a ~ EF
But, by Eq. (lie) this expression is equal to zero, hence, as before

"dA 3A

o^r- = and similarly ~^- =0 etc., ...... (HE)

oX a oXb

which proves Menabrea's law applicable to the general case of stress from temperature
and applied loads, provided the abutments are rigid and immovable.

The introduction of the temperature factor SefcS/ from Eq. (!!A) into Eqs. (10o),
to (ION) will suffice to prove that Castigliano's law also applies to the general case includ-
ing temperature effects. It is not deemed necessary to repeat the transformations here.

ART. 12. STRESSES DUE TO ABUTMENT DISPLACEMENTS

Abutment displacements produce stresses which follow Menabrea's and Castigliano's
laws.

This is readily seen when it is considered that the supporting elements may always
be replaced by linked members which take up these displacements and undergo elastic
deformations. The immovable supports are then outside of these connecting links as
illustrated in Figs. ID, IE and IF, and the links are counted with the structural members.
If then the abutment displacements are defined as distortions Jr in these several
links, which are in every way equal to the former displacements, the resulting effect on
the structure remains unchanged.

Therefore, when the distortions Jr must be considered in the computation of the
X's and d's, it is only necessary to extend the work equations to include these links. It
is also clear that these links may have any desired sections, lengths or values of E, as may
be required for metal or masonry supports.

Thus, let r be the length of a supporting link;
R the load which the link carries;
Ar the change in length of this link;
F r the cross-section of this link;
e r the coefficient of expansion ;
t the change in temperature from the normal;
E r the modulus of elasticity for any material.
then from Eq. (4 A)

The general work Eq. (11 A) then becomes

V027

ART. 12 THEOREMS, LAWS, AND FORMULA FOR FRAMED STRUCTURES 33

wherein Ar may be regarded as a constant and Menabrea's and Castigliano's laws again
apply.

For the case of one external indeterminate, R becomes X a and Ar becomes o a , then
from Eq. (12n)

By differentiating for 8 with respect to X a and imposing the condition for minimum
this becomes

s 3S 3 n9 v

:t + da ~

Also, S=S S a X a and its differential is

Substituting the value from Eq. (12o) into Eq. (12c) and solving for X a then the
latter becomes

2S S a ~ + 2dlS a -3 a
Xa= ~' ........ (12E)

which might be used as the expression for the horizontal thrust of a two-hinged arch
where X a is the redundant thrust.

CHAPTER III

THEOREMS, LAWS, AND FORMULA FOR ISOTROPIC SOLIDS

ART. 13. GENERAL WORK EQUATIONS

In the previous chapters no consideration was given to solid web structures or other
isotropic solid bodies only in so far as was necessary in demonstrating the general laws
and theorems of framed structures.

While it is true that the foregoing discussion is generally applicable to all isotropic
solid bodies which are supported in any of the ways given in Chapter I, and stressed
Avithin the elastic limit by externally applied loads, yet the formula? previously given
will require some modifications to better adapt them to solid web and other massive
and homogeneous structures. Also, the introduction of shearing stress now enters as
a further complication.

As was previously mentioned, all structures which are purely isotropic an involve
only external redundancy. It is, therefore, desirable to take up this subject to the

extent, at least, of giving the special formu-
lae applicable to any solid in which the
physical properties of the material are pre-
sumably uniform in any and all directions.
Such bodies are called isotropic solids.

It will scarcely be necessary to prove
at length all of the theorems and laws pre-
viously given for frames, but a passing refer-
ence at the proper time will be deemed suf-
ficient proof of their general acceptance.

The elastic deformation resulting from
given stresses in an infinitesimally small
particle of a body is a certain and definitely
measurable quantity and is dependent only
on the magnitude of the given stresses and
the physical properties of the material.

If then Fig. 13A represents a small parallelepiped of mass, referred to axes X, Y
and Z, and having dimensions dx, dy and dz, coincident with these axes respectively,
then the stress acting at the corner m is determined in magnitude and direction by
nine components, viz., three normal stresses and six tangential stresses acting in the three
coordinate planes and positive in the directions of the arrows.

FIG. ISA.

ART. 13 THEOREMS, LAWS, AND FORMULA FOR ISOTROPIC SOLIDS 35

Let f x , fy and f z be the unit normal stresses (tension or compression) in the direc-
tions X, Y and Z, respectively. Also, let - xy and r xz be the unit tangential stresses in
the YZ plane; r yx and r yz those in the XZ plane; and - zx and ~ zy those in the XY plane.
The first subscript indicating the normal stress to which they belong, and the second
subscript referring to the axis.

The two unit tangential stresses intersecting in any one point are equal, in each
case, otherwise the body would rotate about its center of gravity, which is the intersec-
tion of the three normal unit stresses. Also, because the three normal stresses and
four tangential stresses, ~, fz , r xz , r, y and T ZX , cannot have any moment about such a gravity
axis OZ', hence, the moment of the two remaining stresses -c yj . and r xy must equal zero.
But, the latter having equal lever arms must be equal to each other.
Therefore,

The resulting stress at the point m is then determined by the three normal stresses
f x ,fy,fz and the three tangential stresses r x , T U , and T Z in Eqs. (13A), and these in turn
determine the deformation of the parallelepiped.

Let Adx, M,y and Adz represent small elastic changes which the lengths dx, dy and
dz undergo, then the forces represented by the unit normal stresses / will produce the
following virtual work :

=f x dydzJdx+f y dzdxJdy+f z dxdyJdz = f x +

Also, let j- x , fy and f z be the tangents of small angular distortions of the
parallelepiped such that +f x is an increase in the angle between the Y and Z axes; f v
the change in the angle between the X and Z axes, etc. Then the virtual work of the
tangential forces is found thus : For the distortion of the YZ plane, the path is f x dy and
the force is r x dxdz, and since the vertical force r^jjan do no work in this direction then
the work of the tangential stress for the plane XZ becomes r x dxdz^r x dy, and similarly
for the other two planes, whence the total work becomes

x ...... (13c)

The expressions Adxjdx, 4dy/dy, and 4dz/dz in Eqs. (13B) are rates of elongation,
or coefficients, for which the values a x , a v and a z may be substituted. Then by making
dxdydz =dV, and combining Eqs. (13s) and (13c), the total internal virtual work is obtained.
Also, since by Clapeyron's law, this must be equal to the virtual work of the externally
applied forces, then the fundamental work equation for isotropic bodies becomes:

W =

This equation is applicable to any case of related displacements and elastic deforma-
tions o dr, a, and 7-, so long as these are small in comparison with the dimensions of the
structure. The external forces must include dead loads and all frictional resistances
which may be active at points of support.

36.

KIXETIC THEORY OF ENGINEERING STRUCTURES CHAP. Ill

As was previously shown for indeterminate frames, so also the various elements
in Eq. (13D) may now be expressed MS linear functions of the applied loads and any
redundant conditions, thus:

R=R R a X a R b X b , etc.

J x = J xo J xa-A-a J xb-^-bj CtC.
fy = fyofyaX a ~fyb^b, etc.
fz =^fzo fiaX a fzbXb, etc.
?x = T ro ~raX a T xb X. b , etc.

etc., etc.

Also, the virtual work of the reactions for each of the conventional loadings becomes,
(see Eqs. (?E)):

(13E)

xa f

(13P)

2R b Ar =J [f x b(

etc.
Then by writing Eq. (13D) for a load P a = l the following is obtained:

-'ZR a <lr. . . (13c)

from which problems of the kind described in Art. 6 may be solved.

Since the redundants X in Eqs. (13E) may be treated as independent variables,
their differentiation furnishes

_ f .- t

a '

"dX a ' 3Y

which by substitution into Eqs. (13r) give

etc., and

~ ~ r

etc.

' ' (13H >

Similar expressions result for 2/S^Jr, etc.

The general work equation is now found by inserting for the actual distortions a
and f their values in terms of stresses and the moduli of elasticity for direct and tan-
gential stress.

The length dx subjected to the unit stress f x and a rise in temperature of t degrees
will be changed by an amount

ART. 13 THEOREMS, LAWS, AND FORMULA FOR ISOTROPIC SOLIDS 37

The two other unit stresses/, and f z will diminish the effect on a x by an amount

/*+/.

mK '

wherein m is the " Poisson number " (1829) which is given as 3.33 for structural steel
and 3 5 for hi-h steel. Professor F. E. Turneaure gives 0.1 to 0.125 for concrete,
quantity m is also denned as the ratio of lateral to longitudinal deformation an
determined by experiment.

Hence, a x and similarly a y and a,, also, T *> Y y and r* may be evaluated as follows.

and

(13J)

and

wherein G is the modulus of tangential stress or shear and has the value G 2(jn + 1) '

while m is the " Poisson number "just given.

Inserting the values given by Eqs. (18,) into Eq. (13D), the fo.lowmg equat.on
obtained for the actual work of an isotropic body :

. . (13K)

\lso by substititingthe values from Eqs. (13i) into Eq. (13n), reducing and integrat-
in , tPS^/AT., etc., the following simple form is obtamed by msertmg the
value from Eq. (13KJ, when temperature effect is negled

VR^^- and similarly S^Jr = ^, etc. - - (13L)

OA a

When the abutments are immovable and no temperature effects exist then

-0 and -0, etc.,

MCT ab<)ve fed
P m , or redundant X m : ^

KINETIC THEORY OF ENGINEERING STRUCTURES

CHAP. Ill

The value of A in all the Eqs. (13L) to (13N) is that given by Eq. (13K).

\Yhen the abutments are immovable the last term of Eq. (13N), becomes zero and
Castigliano's law is again established.

Proceeding from Eq. (13o), and writing same for two related conditions of unit
loading, it is easily shown that Maxwell's law likewise applies here.

ART. 14. WORK OF DEFORMATION DUE TO SHEARING STRESS

For a beam of any section and any loading applied in the vertical plane YY, Figs.

and 14n, the resultant of the external forces on one side of any section A A may be
represented by a force R which may in turn be resolved into a normal force N and a
tangential force Q acting at the point of application of R on the section A A.

T,,

FIG. 14A.

Let M be the static moment of the normal force N about the Z axis and I z the moment
of inertia of the section about this same axis.

Then the unit stress/ at any point of the section is by Navier's law:

N My
f = J+~r (14.A)

* Z

If now this stress undergoes a small change df due to the differential change in R
for a neighboring section, distant dx from the first, then the shear on the area 2zidx
becomes

2T J * l dx=f(f+df)dF-ffdF=JdfdF.

where r x is the unit shear on the differential area d-F.

Assuming N constant in Eq. (HA) , and treating M as a variable, then by differentiation

(14c)

AKT. 14 THEOREMS, LAWS, AND FORMULAE FOR ISOTROPIC SOLIDS 39

Also, since the shear is the differential of the bending moment, then

dM=Qdx (14D)

Combining Eqs. (14B), (14c) and (14o), then

CydMdF Qdx C ,

1- x z-idx= I I ydF. 04E)

J L z l z J

and since J ydF is the static moment M s of the cross-section, this integration may be
considered performed to obtain

Return now to Fig. 14s, take any section pq perpendicular to the ?/ axis and let Q
produce a unit shear r in any point i of this section. The line of this shear intersects the
?/ axis in a point H, which is determined by the tangent pH. This is a pure assumption,
but from the nature of the case it is the most rational assumption to make.

The shear r may now be resolved into components T U and T Z , while r x , for this sectional
plane, must be zero. However, r y now has the same value as previously found for - x ,
and hence

*-!&->' <->

Again, from Fig. MB, tan0 = = - from which r g = -r y . But tan d = , therefore,

O7 i f) P P

/ * \ j. n ft A \

T a = Tj,( itan (14H)

\ z i/

For 0=0, T 2 =0, which is the case for any surface point, the tangent to which is parallel
to the y axis.

From Eqs. (14r) and (14o) it follows that for a given loading and section, the shear-
ing stresses depend only on M 8 and attain a maximum when the section coincides with
a gravity axis parallel to the neutral axis. In practical cases it is usually sufficient to
consider the shearing stress of a section as extending only over a unit length of the
beam as given by Eq. (14F). The actual work produced by shearing stress alone is
then found from Eq. (13K) as

By substituting the value of r, from Eq. (14n), into Eq. (14i), the latter becomes

(14K)

KINETIC THEORY OF ENGINEERING STRUCTURES CHAP. Ill

and by substituting for Ty its value from Eq. (14c), then Eq. (14K) may be solved for any
special section. This gives in general for any section

A=>

and

(14L)

In Eqs. (14L) ft is an involved function of the shape of the section, called by German
authors the " distribution number " of the section, or " coefficient of shearing strain."

This number @ is 1.2 for any solid square or rectangular section, and 1.111 for a solid
circular or elliptic section. Other values will be given after illustrating the computation
of this number by the use of Eq. (14L).

bh 3
For a rectangle of height h and breadth b = 2zi, then 1,. = -^; F = bh; and dF = bdy,

hence,

and

For any I section, Professor Mehrtens finds the general formula for /? as follows:
{3 = M\T 2 dF = j2\--(8b 3 l2b 2 t t 3 + 6bt 2 )+(-^-\ -+t' 2 a) , . . (14ivi)

wherein i= - and the special dimensions here used are shown in Fig. 14c.

Professor L. von Tetmajer, in his " Elastizi-
taets und Festigkeitslehre," 1905, p. 49, gives a
table of values for /? for I beams ranging from
3i"-4 Ibs. to 19"-95 Ibs., 'for which /3=2.39
to 2.03 respectively. Also for riveted girders
made of f" webs, 4 L's 3&"X3&"Xf" each,
and 2 plates 8f" xf" on each chord. Then for

_!.. __, ,

-Y

2.71

23$"

2.49

27 A"
2.35

The value is independent of the unit of length and is, therefore, the same for
metric and U. S. measures.

ART. 1.) THEOREMS, LAWS, AND FORMULAE FOR ISOTROPIC SOLIDS 41

ART. 16. WORK OF DEFORMATION FOR ANY INDETERMINATE STRAIGHT

BEAM

Direct ^tress. For axial or direct stress only, Eq. (13K) gives

(ISA)
Also, by observing that a x =f x /E, then Eq. (13n) becomes for direct stress alone,

- ........ < 15B >

Now ( f x dF=N, the normal direct stress, and, as dV =dFdx, then from Eq. (15A)

C&HV- C N f* dx - C N f* F dx_ CN*dx
J 2E a[ 'I ~2E~~J 2EF J 2EF '

and similarly from Eq. (15B)

ff x df x dV_ CN-df x dx_^ (N-df x Fdx_ fNZNdx (

J E-dX a J EdX a J EF3X a J EFdXa

The temperature effects in Eqs. (15A) and (15B) will now be determined on the
supposition that the effect is not uniform, but as shown on Fig. 14A, where

fo=the change in temperature from normal at the gravity axis of the section;
J=the difference in temperature of the two extreme fibers;

h = height of the section;

t/=any ordinate;

t=the temperature above normal, of any point of the section then

(15B)

Substituting this value for t into the temperature element of Eq. (15A), the latter
becomes

But ff x dF=N and r/ x ?/dF=M=moment of resistance of the section, therefore,

dx ......... (15F)

By differentiating Eq. (15?), and dividing through by 3X a , the value of the tem-
perature element for Eq. (15u), is obtained thus:

n . .

(15G)

42 KINETIC THEORY OF ENGINEERING STRUCTURES CHAP, in

The bending moment. From Eq. (HA), f x =My/I g and I z ==fy 2 dF. With these
values and neglecting the temperature element, Eq. (!OA) becomes

By differentiating Eq. (Ion) and dividing through by "dX a then

The general work equation, including all effects, may now be written out by collect-
ing the several Eqs. (14L), (15A), (15c), (15n), and (lor) into the following, represent-
ing the total actual work for any isotropic body by

r r M

Similarly the following differential expression of Eq. (!OK) is obtained from Eqs.
(14L), (15s), (15D), (15.i) and (15o) or directly by differentiating Eq. (!OK) and divid-
ing through by ~dX a , thus:

- (loL)

Equation (15L) being true for any force X a is, of course, true for a force P m , hence
by substitution of the latter value and then inserting the value 3A/3P TO . thus obtained
from Eq. (15L) into Eq. (13N) the following equation for elastic deflection is obtained:

c Q ^Q , C t ^N

j GFZP^ dx+ J t 3p-

The last three equations (15K), (15i.) and (15M) are the three fundamentals from which
all cases of redundancy for isotropic bodies can be solved. There are always as many
of these equations as there are redundant conditions X.

In each of these the first term expresses the effect due to direct or normal stress; the
second term that due to pure bending; the third term that due to shearing stress; the fourth
term gives the effect due to a uniform rise in temperature to; and the fifth term that due to a
difference At in the temperature of opposite extreme fibers.

ART. 16 THEOREMS, LAWS, AND FORMULA FOR ISOTROPIC SOLIDS 43

The object in presenting these rather long fundamental equations is not to make
the subject appear complicated, but rather with a view to showing once for all
combined effect from all causes, thus permitting the easy choice in combining any effects
or in omitting such as may seem negligible in any specific problem.

Equation (15K) may also be derived from Eq. (5n) for virtual work and offers very
useful applications.

Thus assuming a column subjected to a thrust N ttt and a bending moment M m
the virtual work on the column from Eq. (OH) becomes

where A r rt and M a are due to the usual conventional loadings. But

A^te nd . MJx

lioncG

W . C l N m N a dx , r lM rnM a dx

W = \-d m

r l M m M a dx
J El

When the virtual work becomes the actual work A then Eq. (15N) becomes identical
with Eq. (15K) term for term.

ART. 16. WORK OF DEFORMATION DUE TO DYNAMIC IMPACT

Problems involving the deflection or strength of a structure subjected to impact,
are frequently met with, and their treatment is here considered as properly belonging
to the subject of the present chapter.
Let LO = weight of a moving body;
H = height of a fall;
v= velocity at instant of impact;
g = acceleration due to gravity ;
a=any elastic displacement produced by the moving body in some structure.

Then the work expended by the moving mass is represented, either in terms of
velocity or height of fall as follows:

wherein v*/2g represents the velocity height or height through which a body falls in
acquiring a velocity v. When the body moves with a velocity v along a horizontal path,
thenv 2 /2g will be the height to which the body would raise itself in order to expend its
energy and come to a state of rest.

44 KINETIC THEORY OF ENGINEERING STRUCTURES CHAP. Ill

The work thus produced by kinetic energy, when it is instantly taken up by a
quiescent body or structure, is twice as great as the work of the same moving body
gradually applied, and hence produces twice the internal work in the body struck as
would result from an equivalent static load.

Problem 1. The weight w falls on top of a column, stressing the column normally,
what sectional area is required in order that the unit compressive stress does not exceed /?

From Eq. (!OK) the work due to direct stress is given by the first term, where N
is the total static load, and to balance the dynamic energy, twice this amount is taken;
hence

(16B)

and since o = ~ this gives for F,

Problem 2. A weight w falls from some height striking a beam resting on two sup-
ports. The weight strikes the center of the beam with a velocity v; what will be the
stress / in the extreme fiber for a given beam section? Shear and bending resistances
are to be considered.

From Eq. (!OK) the actual internal work is

Let P=an equivalent static load producing the same stress / in the given beam.
Then, for any point of the beam of depth h and span I,

P My Plh 8/7

-x; also /=--* = __ or P = -^- ..... (16s)

Hence

+ JM 2 dx_ P 2 rt 2 _ P2/3 _2f 2 ll

i_ ~2ET~4EI Jo X WEI~3Eh 2

(16r)

Also, for Q = the second term of Eq. (16D) becomes

=dx = _
2GF 4GFjo SGF GFlh 2 .......

The sum of Eqs. (16p) and (16c) gives A, according to Eq. (16D), and this must
equal a}V*/2g. Therefore,

AKT. 16 THEOREMS, LAWS, AND FORMULA FOR ISOTROPIC SOLIDS 45

Her e G=modulus of shear = 1 =0.385 E for structural steel (see Eqs. (13j).

The coefficient of shearing strain /? is given by Eqs. (14L) and (14M).

By applying Eq. (16n) to a steel beam 2.5X2.5 inches by 78 inches long, it is found
that when shear is included, the stress / is about f per cent smaller than when this term
is omitted, showing that the internal resistance due to shear is very small and usually
negligible. ' This is also true when computing deflections.

' In the same manner all problems involving impact may be solved.

Suppose a ship weighing w tons, and moving with a velocity of v feet per second
were to collide with a fixed structure, then the work which the ship is capable of expend-
in- is represented by a>v*/2g ft.tons. This work may be expended in injuring the
structure or the ship itself, a condition depending on the relative strength of the two
bodies, which must be ascertained form the design and construction of each.

CHAPTER IV
INFLUENCE LINES AND AREAS FOR STATICALLY DETERMINATE STRUCTURES

ART. 17. INTRODUCTORY

Professors MOHR and WINKLER, in 1868, published simultaneously the first treatises
on influence lines describing, at that early date, practically all the uses and applications
of these lines known at the present time. Professor J. Weyrauch, in 1873, introduces the
name influence line not used by Mohr and Winkler in their earlier work. Professor Mohr,
in a series of articles published from 1870 to 1877, was the first to apply influence lines
to deflections and to redundant conditions.

Professor Geo. F. Swain, in 1887, gave the first treatise on the subject in English.

An influence line is the graphic representation of some particular effect produced,
at a certain point of a structure, by a single moving load occupying, successively, all possible
positions over the entire span. The effect may be the shear, the bending moment or
the deflection at any certain point of the structure; it may also represent any reaction
force or the stress in any member. The single moving load is usually taken equal to
unity, though in certain special cases it may be desirable to use any load P.

An influence line represents a certain effect for a certain point or member of a
structure and for any position of a moving load, while a shear or moment diagram
represents effects due to a single position of the load or loads for all points of a structure.

The ordinate to any influence line is thus an influence number or factor, usually
designated by TJ when stresses are dealt with and sometimes by o when deflections are
under consideration.

As a matter of convention, all positive influence line ordinates will be laid off down-
ward from the axis of abscissae.

A load point is any particular one of the many possible positions of the moving
load.

A summation influence line is one which gives the total effect at some particular point
due to a train of concentrated loads. The actual loads are here employed and each
influence ordinate is made to represent the summation of influences of the same kind
for a certain position of the train of loads. Usually this position is taken so that the first
load is over the point for which the influence line is constructed.

An influence area is the area included between the influence line, the axis of abseissce
and the two end ordinates.

The maximum effect is always produced when the load point coincides with the
maximum ordinate of the influence area. Hence influence lines are eminently suited

ART . 17 INFLUENCE LINES AND AREAS 47

to the solution of all problems relating to position of moving loads for maximum and
minimum effects and stresses.

The load divide is such a load point, the ordinates to either side of which have
opposite signs. Hence, the influence ordinate at a load divide passes through zero and
the influence line intersects the axis of abscissae at such load divide.

It is assumed that if a unit load produces some effect 77 at a certain point of a given
structure, then a load P will produce the effect PTJ at this same point, so long as the material
is not stressed beyond the elastic limit. This follows from the law of proportionality
stated at the end of Art. 4.

Hence, the total effect Z produced by a system of moving loads, PI, P 2 , etc., will
be the sum of the effects Pi] of all the loads and the maximum and minimum value of Z
will be determined by the position of the moving loads.

The total effect is thus expressed by the following equation when the case of loading
is simultaneous and within working limits :

(17A)

Therefore, having given the influence line for a certain effect on some structure, then
the total effect, due to any system of loading, is easily found by a summation of the prod-
ucts of loads into corresponding influence ordinates or numbers for any desired positions

of the loads.

For a uniform moving load p per foot of length, Eq. (!?A) becomes

wherein the integral represents the area of the influence polygon between the end ordi-
nates of the uniform load.

The equation of any influence line may be written out by expressing the desired
function for a particular point in question in terms of a moving load unity acting at any
variable distance x from one end of the structure taken as origin.

Since influence lines represent all possible effects it is readily seen that maximum
and minimum stresses may be found from the same lines. Therefore, only one half of
a symmetric structure requires analysis. The left half is usually treated, as a matter

of conventional uniformity.

Direct and indirect loading. In the above it was assumed that the loads were directly
applied to the beam, which is rather the excep-
tion. Usually they are taken up by the floor
system and then transferred to the panel points
as load concentrations. The former case of
loading will be known as direct loading and the
latter as indirect loading.

The influence line between two successive panel

points is always a straight line, regardless of the riG 1?A

system of loading or the particular influence.

Let Fig. 17A represent two successive floor beams of any truss and the loadP =

IS KINETIC THEORY OF ENGINEERING STRUCTURES CHAP. IV

is transferred to the points a and b by the floor stringer ab. The load effects PI and
Po, which are carried to the panel points a and b, are

- and

Now the total influence of P l and P 2 , on the structure as a whole, must be exactly
equal to that produced by the resultant P.

But the influence line a'b' '', extending over the panel ab, must give, by Eq. (!?A),

, 17 x

from which

fd-:

For any influence line, >?i, r) 2 and d are constants, hence Eq. (17c) represents a
straight line, which was to be proven.

ART. 18. INFLUENCE LINES FOR DIRECT LOADING

The several influence lines for reactions, shears, moments and deflections for a beam
with direct loading will now be given. They will always be known by the names indicated
in Fig. ISA. Thus the A line is the influence line for the reaction A.

The influence line for a reaction is of the first importance, because the other lines
are generally derived from the reaction influence line. This is really seen from the
circumstance that for the unloaded portion of a beam or truss, the shear is always
equal to the end reaction and the moment is equal to this reaction into the distance from
the section in question to the end of the beam.

Hence, in drawing influence lines it is always best to consider the unloaded portion
of the span to the right or left of the section as the case may be, beca-use the only external
force on that side of the section will then be the reaction.

In the following the moving load will always be assumed as coming on the span from
the right end and the effect produced on the left half of the span only, is considered.

(a) Reaction influence lines A and B. Using the dimensions indicated in Fig. ISA,
the two reactions become

Px' Px

A= f- and B= ~T ..........

For x' and x variable, both expressions represent equations of straight lines which
are easily plotted. When P = l and x' =1 then A=l and for x' =0, A =0. Hence, the
A line is drawn by laying off a distance unity down from A and joining this point with B.
The reaction A for any load P acting at the load point m is then A m =Prj m .

In like manner the B line is found, and the corresponding reaction B m for a load
P at m becomes B m =Pj)' m , and i? m + fy' = l for every point of the span.

ART. IS

INFLUENCE LINES AND AREAS

(b) Shear influence lines. The shear Q n for a certain point n to the left of the load
point m is always equal to the reaction A. But for a load point to the left of n (not
shown) the shear is Q n = A P = B, because P =A +B.

The influence line for Q n is thus derived from the influence lines for A and B as
indicated in Fig. ISA, and consists of the polygon An'n"R.

The point n becomes the load divide for shears at n and hence there will always be
a maximum and a minimum value for Q n depending on whether the positive or the
negative influence area is fully loaded. The shear due
to any single load P must change sign in passing the
point n.

(c) Moment influence lines. The moment for
the point n is now found when the moving load is
to the right or left of the section at n. Then for

P.I

x>a, M n =Aa; and for

M n =B(la).

Thus the moment influence line is also derived
from the reaction lines because the ordinates of the
latter, when multiplied by a or (I a) as the case may
be, give the ordinates to the moment line. Hence
the bounding lines of the moment influence line are
easily found.

Since A and B are both unity, then the ordinate
A A' --=a and the ordinate BB'=la and the lines
AB' and A'B inclose the required moment influence
line. Also, the ordinate at n is the ordinate of the
intersection n' between the two bounding lines. Hence,
if BE' should fall off the drawing, then the line ~AB'
may be drawn from A to n' without finding B' '.

In either case the moment influence line is then
the polygon An'B with all ordinates positive so long
as the point n is not outside the span, a case which
will be given later. The maximum ordinate is always
under n and has the value l-a(l a)/l, which offers

still another construction for this influence line. The middle ordinate of the AB'
\me=(l-a) which furnishes a convenient construction for this line when B' falls off
the drawing.

It is clearly seen that if a single load is placed at n over the maximum ordinate,
then a maximum moment is produced.

(d) Deflection influence lines. The deflection for a point n, produced by a single
load at any point m, is given in terms of the moment of inertia of the beam section and
the modulus of elasticity, as follows:

FIG. ISA

QEIl

a*) P.

(18s)

KINETIC THEORY OF ENGINEERING STRUCTURES CHAP. IV

This is a cubic equation and when plotted for all values of x and making P = l, the
influence line for deflections is obtained.

In general, any deflection polygon, drawn for a load unity acting on a fixed point
n of any structure, is the deflection influence line for the point n of that structure. This
follows from Maxwell's law.

As the subject of deflection influence lines is fully treated in Chapter VIII, no
further consideration is given to it now.

ART. 19. INFLUENCE LINES FOR INDIRECT LOADING

In the present case the loads are applied to a stringer and floor-beam system and are
thus transferred to the beam at certain panel points as load concentrations. The

beam is then indirectly loaded and since an influence
line was shown to be a straight line between panel
points the present case will require some slight
modification of the influence lines just found for
direct loading. See Fig. 19A.

However, all that was said for cases of direct
loading applies here and the modifications made
necessary by the indirect loading occur only in the
panel wherein the point n is located. As before, n is
the point for which the influence line is drawn and m
is any one of the possible load points for the moving
loadP = l.

(a) Reaction influence lines remain the same
whether the loading is direct or indirect, since the point
for which the influence is sought is alwavs at A or B,

- 1 /

which are also panel points.

Furthermore, the reaction influence lines being
straight over the length of the span will always be
straight between successive panel points.

(b) Shear influence lines. The Q line, outside of
the panel ce and containing the point n, remains the
same as for direct loading. But since the influence
line within a panel must be a straight line, therefore,
the points c' and e' must determine the influence line
for the panel ce regardless of the location of the

point n so long as this point is within the panel ce.

Hence, the polygon Ac'e'B is the shear influence line for the panel ~ce and the point
i is the load divide for this panel. The limiting values for shear are thus the same for
any point n of the same panel.

(c) Moment influence lines. Here again the influence line within the panel ~ce is
all that requires modification in the case of indirect loading.

FIG. 19x.

ART. 20

INFLUENCE LINES AND AREAS

Hence for the same reasons just given to determine the modified Q line, the M line
for the point n is now the polygon Ac'e'B, instead of the triangle An'B for direct loading.
The point n is the center of moments.

ART. 20. THE LOAD DIVIDE FOR A TRUSS

In constructing influence lines for truss web members it frequently happens that
one or the other end ordinate falls outside the limits of the drawing.

There is a very simple way of locating the load divide i and thus facilitating the
construction of any influence line. The method is given first and the proof follows:

Let Fig. 20A represent a truss arranged for bottom chord loading, but no loads
are shown. Required to find the load divide i for the panel fg necessary to determine
the stress in the diagonal eg cut by the section tt.

FIG. 20A.

Construction. Prolong the unloaded chord member ek in the panel fg until it inter-
sects the verticals through the end reactions in a and 6. Then the intersection i of
the two lines af and bg will be the required load divide.

Proof. Suppose the unit moving load is now located in the vertical through i.
Then if i is the desired load divide, the unit load for the load point ii will produce zero
stress in the member eg and the influence line ordinate for the point i must be zero.

Let F and G be the panel concentrations in the points / and g due to the load P = 1
at i.

Then the polygon abfg may be regarded as the equilibrium polygon for the forces
A, B, F and G. Also, the resultant R of all forces on one side of the section tt must pass
through the intersection n of the two included sides of the equilibrium polygon.

But, the point n is the intersection of the two chords fg and ek, which point is also
the center of moments for the diagonal eg.

Hence the moment of this resultant R about n must be zero and cannot produce stress
in the member eg, thus proving that the load P = 1 is actually located in the load divide.

When the top chord is the loaded chord, a similar construction furnishes the point
i', as the load divide for the member eg, according to the same proof above given.

This construction applies to any structure even when one or both chords are straight.

When the center of moments n falls inside the span then there is no load divide and
all loads on the entire span will produce the same kind of stress in any particular web
member.

52 KINETIC THEORY OF ENGINEERING STRUCTURES CHAP. IV

ART. 21. STRESS INFLUENCE LINES FOR TRUSS MEMBERS

According to Professor Ritter's well-known method of sections given in Art. 58,
the stress in any member of a determinate truss may be expressed thus :

(21 A)

where M is the moment of all the external forces, including the reactions, on one side
of the section and r is the lever arm, of the member in question, measured from a center
of moments determined by the intersection of the two other members cut by the section.

Hence, the influence line for the stress in an}^ member is the influence line for the
corresponding moment for this member, the ordinates of which are divided by r.

For any moment influence line, the ordinate at any point represents the moment due
to a unit load at that point and, therefore, from Eq. (2lA), i)=M=rS.

But the end ordinates for a moment influence line were formerly found to be a and
a', therefore, the end ordinates for a stress influence line must be 1/r times as great,
whence

AA'=-, and BB"=- (21u)

r' r

Now l-a/r=$ =the stress in a member S due to a reaction unity at A, when the
load producing this reaction is to the right of the section through S.

Also I a' /r =${,= the stress in this same member S due to a reaction unity at B
when the load producing this reaction is to the left of the section.

Hence, the end ordinates for any stress influence line are equal to the unit stresses
S a and Sb produced in the member by reactions unity at A and B respectively.

Since these stresses are readily determined either by Ritter's method or by a Max-
well diagram, all stress influence lines for the members in any determinate truss are
easily found.

In Fig. 2lA, the stress influence lines are drawn for a general case of truss design.
The top chord is the loaded chord and the influence lines are drawn for the three mem-
bers L, U and D of the panel cut by the section it.

te =the stress in the member L for a reaction unity at A. This is evaluated by
Ritter's method in terms of the lever arms ai and r t .

Sib=ihe stress in the member L for a reaction unity at B and is expressed in terms
of the lever arms a'i and r t .

The stress influence line for the bottom chord L, is drawn in accordance with the
method illustrated in Fig. 19A, observing, however, that the end ordinates now become
the stresses S ta and Sn, respectively.

The center of moments for this member being at E, which is vertically over E',
the stress influence line for the member L becomes a triangle AE'B, determined by the
end ordinates A A' = to and BB' =#.

The influence line for the top chord, J7 is drawn in exactly the same manner, but in
this instance the resulting triangle AF'B does not offer a straight line over the panel

ART. 21

INFLUENCE LINES AND AREAS

Here F is

PJG and hence the line E'G' must be drawn to complete the influence line,
the center of moments for the member U.

The influence line for the diagonal D is similarly drawn and the load divide %' ', found
in the lower diagram, is seen to coincide with the point i found on the truss diagram
by the method previously given. Also the intersection 0', between the limiting rays

is verticallv under the center of moments for the

AE' and G'B of the D
member D.

line.

ISTRESS INFLUENCE LINE FOR L.

STRESS |NFJLUE!NCE LINE FOR. u."

' ' "

STRESS INKLUEJNCE LINE FOR o

It should be observed that in all three influence lines, Fig. 21 A, the limiting rays intersect
on the vertical through the center of moments f or the particular member. This then serves
as a check on the diagram.

Regarding the sign of the influence area the following rule should be observed:
When the center of moments is located between the supports A and B then all ordinates of
that particular stress influence line will be of the same sign; when the center of moments
is off the span, then there exists a load divide and there will be positive and negative influence
areas giving rise to stresses of opposite signs.

54 KINETIC THEORY OF ENGINEERING STRUCTURES CHAP, iv

The criterion for position of a moving load for maximum and minimum stresses in
any member is thus clearly shown by influence lines and the maximum effect of any load
is always produced when that load is situated over the maximum influence ordinate.
This subject is more fully treated in another article..

It is sometimes necessary to construct influence lines without the divisor r, and
dividing the final stress thus found by r. Either method is employed as circumstances
may demand, though it is always better to avoid the divisor and follow the method given
in Fig. 21 A, unless there is some very good reason for doing otherwise.

Attention is also called to the fact that moment and shear influence lines are not
required when the stress influence lines are used. The former were given to render the
treatment complete and may be employed to find stresses, but it is usually preferable
to draw the stress influence lines directly. The final result of an analysis will be more
satisfactory when this is done.

Influence lines are seldom used for finding dead load stresses, though such a procedure
may sometimes be warranted, and then the following points must be observed : For
direct loading and for beams, it makes no difference whether the influence lines be
used for dead or live loads, but in dealing with framed structures an error might be
committed because then the influence lines are always drawn for a certain loaded
chord while the dead loads act along both chords.

This is easily remedied by drawing the influence lines for both chords loaded and
applying the dead loads to the proper lines. These influence lines are identical except
in the panel containing the member in question. Hence by observing this circum-
stance, dead load stresses may be found from the same influence lines without difficulty.

ART. 22. REACTION SUMMATION INFLUENCE LINES

Most summation influence lines become rather complicated and, therefore, their
use is practically restricted to a few cases for which the construction is simple.

In general, any ordinate to a summation influence line is expressed by Eq. (17A), as

and it is readily seen that when P and ij are both variables, such a line would ordinarily
require much labor for its determination.

However, the summation influence line for an end reaction of a simple truss is
very easily constructed and serves a most valuable purpose in finding the end shears
for a system of moving loads. This influence line will be called simply the sum A line
to distinguish it from the ordinary A line in Figs. ISA and 19A.

The general usefulness, of the sum A line originated by Professor Winkler, will be shown
later; suffice it to say here that it affords a ready means of finding stresses in the web
members of any truss, because, as previously shown, the shears and moments for any
point of a truss are easily found when the end reaction for the particular loading is known.

The sum A line for a system of concentrated wheel loads will now be demonstrated
using but five loads for simplicity, though the method applies to any number of loads.

ART. 22

INFLUENCE LINES AND AREAS

Given the train of moving loads PI, P 2 , etc., in kips of 1000 Ibs., and spaced as
shown in Fig. 22A, to construct the sum A line for a span of 25 ft.

Since a reaction influence line is the same both for direct and indirect loading, the
sum A line is independent of the number and location of panel points.

For standard position of loads on a truss, the train is always assumed as coming
on the 'span from the right end and moving toward the left, but the above loads are
placed in exactly the reverse order, with the first load PI over the support B. The
reason for this will appear later.

The loads are applied consecutively from PI up on the vertical through A, using
any convenient scale.

POSITJON OF LOADS roft CONST *GcT ION or SUM A unr

FIG. 22A.

The several rays drawn from B to the respective load points c, d, e, etc., form
a force polygon with pole distance H =1. The sum A line is then drawn as the equilibrium
polygon for this force polygon and the loads P, by making 1 -2 \\Bd, 2 -3 \\Bc, 3 -4 \\Bf,
etc.

It will now be shown that the equilibrium polygon so obtained is really the sum-
mation influence line for the end reaction A .

Referring again to Fig. 22A, the line Be is easily recognized as the influence line
for A due to a moving load PI. Alsp^the influence line for A when the moving load is
P 2 , may be represented by the line Bd when Be is the axis of the absciss. Similarly
~Be represents the A line for P 3 , and so on for any other loads. The summation of
all odinrates should then represent the desired summation influence line.

For the train coming on the span from right to left, the ordinate TJ, under the
forward load PI, always represents the reaction A. Thus when the train has advanced

56 KINETIC THEORY OF ENGINEERING STRUCTURES CHAP. IV

to the point n, Fig. 22A, the ordinate v? represents the sum of the ordinates t, and
T}=A n =U+t 3 +t 2 +ti\vhkb. is readily seen by inspection. The ordinates t are the con-
tributions to A from the several loads on the span. It is now seen why the train
was first placed in reverse position for the construction of the equilibrium
polygon.

Hence the sum A line for direct or indirect loading is an equilibrium polygon drawn
for a pole distance equal to the length of span and any system of train loads placed in
reverse order on the span with the forward load at B.

The reaction A for any position of loading is then the ordinate to the sum A line
measured under the forward load when the train approaches from right to left. The
greatest possible reaction A will always be the end ordinate A5, Fig. 22A.

When extraordinary accuracy is required the end ordinate A 5 can be readily checked
by computation.

The sum A line for uniformly distributed loads may be constructed in precisely the
same manner as above illustrated for concentrated loads, merely by laying off the load
line Ag to represent the total load on the span =pl and dividing this into some con-
venient number of equal parts. The greater this number the more accurate will be the
result, and the polygon finally becomes a parabola.

The application of the sum A line, to finding the shear at any point of a beam or truss,
will now be presented.

The shear at any point n of a beam, for a case of direct loading, is equal to the
reaction A minus the loads between A and n. Hence, the sum A line gives a complete
solution for this case. The subtraction of any loads to the left of n can be performed
graphically on the diagram, Fig. 22 A, by taking^ off the proper ordinate less the loads
between A and n.

For indirect loading, the shear at n is equal to the end reaction A, provided the
forward load is exactly over the panel point n. But when the load extends over into
the panel, then the shear is equal to the end reaction A, minus the panel reaction a at
the left hand pin point of the loaded panel. See Fig. 22ra.

The values of o for all possible positions of loads in a single panel may be found
by drawing a sum a line noe for one panel, using as many of the loads from the forward
end of the train as may be placed into one panel. If this auxiliary sum a line is drawn
on tracing paper, it will also serve_ a ready means for finding the position of the train
for maximum shear in any panel mn.

The method of finding the sum a line is exactly the same as for the sum A line only
the pole distance for the former is the panel length d.

When the forward wheel of the train is at n, then the shear is the ordinate rj n .
When the forward wheel is at r the shear in the panel mn is Q r =rcre=ec where re = -n a
the panel reaction a for loads PI and P 2 in the panel and load P 3 at n. Similarlv when
the forward wheel is at s and the second wheel is at n the shear Q 8 =s/" os.

The position of the train for max.Q in the panel mn is then easily found by selecting
such a point s for which the ordinate O/=T? max. This maximum jj is easily found with
a pair of dividers and will always occur when some one of the forward wheels is at the
panel n.

INFLUENCE LINES AND AREAS

If the sum a line is drawn on tracing cloth, it may be superimposed on any panel
of the sum A line and TJ max. can be quickly determined.

It sometimes happens that two positions of the train give the same maximum shear,
in which case either may be used. In Fig. 22e this would be true if oe\\cf then ~ec=of
and both Q r and Q s would thus be equal.

Load potitiori for Qm<. in panel m-n

FIG. 22s.

ART. 23. POSITIONS OF A MOVING TRAIN FOR MAXIMUM AND MINIMUM

MOMENTS

(a) Point of greatest moment for direct loading. Given a span of certain length
I, loaded with a train of loads, PI, Pa, PS, etc., to find the point n which is subjected
to the greatest bending moment. It is assumed that all loads remain on the span.
See Fig. 23 A.

Let R = resultant of all loads P on the span.
RI = resultant of all loads to the left of n.
R 2 = resultant of all loads to the right of n.
A and B are the end reactions for R.
n =the point of maximum moments to be found.

Then the moment about n is

(22A)

58 KINETIC THEORY OF ENGINEERING STRUCTURES CHAP. IV

By differentiating and equating to zero, the abscissa for the point of maximum
moments is found thus :

dM R n a I r9 ., x

-=- r (l-2x-a) =0 or x+-=-^ ....... UOB)

ax t z z

Eq. (23B) signifies that the maximum moment under any system of loads occurs for
a point n when the center of gravity of the system and the point n are equidistant f rom
the center of the beam.

Introducing the value of x from Eq. (23 B) into Eq. (23A) the value of max. M is
obtained as

max.M^R^-R^. . . ........ (23c)

In any special case it is necessary to find the particular load P n which must act
at the point n to produce a real maximum. The moment influence line clearly indicates
that one of the loads must fall at n to obtain the maximum moment.

FIG. 23A.

Usually the point n falls very near the center of the beam and in most practical
problems it will suffice to place that load at n which falls nearest the center of gravity
of the sj'stem of loads.

When dealing with floor stringers it frequently happens that the stringer is long
enough to carry three wheel loads but that the max. M so found is less than when only
two of the loads are on the stringer. A few of these' special cases are here given.

Case I. When there is only one load P on the span, then x=l/2 and a=0.
Hence Eq. (23c) gives

M Pl

max.M= (23D )

Case H. When there are two equal loads P on the span, distant e from each other,
then R=2P and |=|, hence z=--|=_- 1 which gives, from Eq. (23c),

2P/Z
max. M =:-[

Equating Eqs. (23o) and (23c) to find when one or two loads give equal maxima,
this results in the condition e=0.5858Z. Hence, when e>0.5858Z, then one load gives
a greater moment than two loads, all loads being of same magnitude.

ART. 23

INFLUENCE LINES AND AREAS

Case III. When there are three equal loads P on the span, distant e from each
other, then R =3P and x =1/2 making a =-0. Also Ri=P and a t =e, hence

,, 3PI n n x

max. M=. Pe=P( e}.

4 \4

(23F)

From Eqs. (23E) and (23r) it is found that when e>0.4494Z, then two loads give
a greater max. M than do three loads, all loads being equal.

Case IV. For four equal loads on the span may. M occurs under the second load
and Eq. (23c) gives

max.M=P(l+-2e] ........ (23o)

Here again it is found that when e>0.2679Z, then three loads give a greater max-
imum than four loads.

(b) Critical loading for maximum moments. Direct loading. In this case the
moment influence line for any point n is a triangle and the discussion is, therefore,
applicable to any beam or truss for which the moment influence line is a triangle. See
Fig. 23B.

n-~lD..O 01 n h n

FIG. 23B.

Let R = 2P =the total load, or resultant of loads on the span.

RI =the resultant of all loads between A and n for a certain position of a train

of loads.
RZ= the resultant of all loads between n and B for the same position of the

train of loads.

Calling rji and rj 2 the influence ordinates for the load points of RI and R 2 respec-
tively, then the moment for the point n may be written as

2RiXi tan ai +R 2 x 2 tan 2 . . . . . . (23n)

Suppose now that the train is moved to the right by a small distance dx, producing
the change +dx in Xi and the change dx in x 2 , then the change in M n is dM n and the
differential coefficient thus obtained is equated to zero for maximum and minumum,
thus:

dM n

=Ri tan ai R% tan a 2 =0.

(23j)

60 KINETIC THEORY OF ENGINEERING STRUCTURES CHAP. IV

But since tan i =^-. and tan a 2 = , Eq. (23j) gives, by substitution,
a i a-2

^1=??. (23K)

ai a 2 '
By observing that a l +a 2 =Z and ^i +R 2 =R, then by composition, Eq. (23K) gives:

a\ -f 02 a\ I

Either Eq. (23K) or Eq. (23L) may serve as a criterion for the critical position of
the train load on the span. The load P n , falling at n, is divided equally between 1^
and R-Z.

In the first instance the criterion would be satisfied when the average unit loads on
both sides of the point n are equal, a condition which is absolutely fulfilled by a uniformly
distributed load.

According to Eq. (23L) the criterion for max. M would be that the average unit load
to the left of the point n must equal the average unit load on the whole span.

Also, since the maximum ordinate y is under the load point n, no position of loads
can give a real maximum unless one of the loads is directly over n. Hence it is readily
understood that there are usually two or more maxima for any point n, depending on
which load is placed over this point. Therefore, so long as the criterion is fulfilled, all
positions giving such a max. M n must be used to determine the real maximum.

In general for any moment influence line, Fig. 23s, the moment will be maximum
for a point n when a load P is applied at n and when the angle ^ of the influence line
exceeds 180. The moment at n will be minimum only when there is a load at some
point B for which 0'<180, that is where the angle between two successive elements
of the influence line is convex upward. Hence all angles in a moment influence line
represent critical load points and correspond to maxima or minima accordingly as these
angles are convex downward or upward, respectively.

(c) Critical loading for maximum moments. Indirect loading. For any point w,
other than a panel point, the moment influence line is a quadrilateral, hence all cases
for which this is true are included here.

Let .RI the resultant of all loads acting between A and c.
R% =the resultant of all loads acting in the panel ce.
R s =the resultant of all loads acting between e and B.

Other notation, as shown in Fig. 23c, which represnts a beam or girder with four
panels, each of length d.

From Eq. (23H) the general expression for an influence line of any number of sides
may be written as

M = 2Rx tan a, ........... (23 M )

and the differential coefficient, equated to zero for maximum and minimum, becomes

-- = 2R tan a =0 ...... ... . . (23u)

ART. 23

INFLUENCE LINES AND AREAS

The various values of R tan a are now found from the special case in hand, using
the notation and illustration in Fig. 23c. Three such terms are here required and these
are evaluated as follows:

jj V Tj c a\ d\ fje a 2 d 2

tana-i=--; tanas-; also ~~^~'

a\ a 2 TJ a\ T} a 2

from which tana 2 =

With #1, /2 2 and # 3 and these values of on, 2 and 3 , and observing that for dx
to the right the value R 3 tan 3 must be negative, then Eq. (23N) becomes

_
dx

_
d

(23o)

FIG. 23c.
which may be reduced to give the criterion

(23p)

Observing that R=Ri+R 2 +Rs as before; also that d=d l +d 2 and that l=(
then by composition of Eq. (23?) and substitution of these values the second form of

criterion is obtained as

_!l = :? (23Q)

a\ I

Eq (23r) expresses the criterion for max. M n when the average unit loads on both
sides of the point n are equal, provided the resultant R 2 is distributed between R, and R* in

the proportion di : d 2 .

Eq (23q) stipulates that the average unit load over the distance 01 must be equal to
the average unit load over the whole span, provided the portion R^/d is added to the

resultant R\. . .

The question of finding which particular load must be placed at either c <
fulfill either of the above criteria may be solved, by trial, as for the case of direct loading.

62 KINETIC THEORY OF ENGINEERING STRUCTURES CHAP. IV

ART. 24. POSITIONS OF A MOVING TRAIN FOR MAXIMUM AND MINIMUM

SHEARS

(a) Critical loading for maximum and minimum shear. Direct loading. The shear
influence line is then as shown in Fig. ISA and the max. +Q n is obviously produced by
a full loading of the entire span between B and the load divide n. The min.Q n is
produced by a full loading over the distance An, which corresponds to the negative
influence area. This is absolutely correct for uniformly distributed loads.

For a concentrated load system, the train must cover the portion of span between
n and B, with the first load just to the right of n so that its influence ordinate is nn" without
any negative effect from the ordinate nn' . However, if the first load is small compared
with the second load, it may necessitate placing the second wheel at n to attain the
real max. +Q n =A Pi.

Hence, it is clear that either PI or P 2 must be just to the right of n to obtain max. +Q,
and whichever gives the greater value is then the real maximum.

(b) Critical loading for maximum and minimum shear. Indirect loading. The
influence line for shear, Fig. 19A, or the stress influence line for a web member, Fig. 21 A,
are both included here. Hence the criterion for position of loads for a max. Q will also
serve for maximum stress in a web member.

For uniformly distributed loads the load divide, as found in Fig. 20A, will always
establish the point at which the head end of the load must be located for max. +Q or
min. Q.

When dealing with concentrated load systems the position for maximum and
minimum shear is most advantageously found by graphics, as illustrated in Fig. 22e,
where the magnitude of these shears is at once determined, together with the critical
position of loads.

When analytic methods are employed, the following criterion may serve a useful
purpose.

Let Fig. 24A represent a truss with bottom chord loaded and i is the load divide
for the diagonal D. The train of loads covers the span from B to i, making RI the resultant
of the loads in the panel CF and R 2 the resultant of the loads between B and F. The
line A'C'F'B' is the stress influence line for the member >.

Then the stress in the member D is

(24A)

By shifting the train dx to the right, both Xi and x are diminished by this amount
dx. Hence the stress now becomes

ART. 24 INFLUENCE LINES AND AREAS 63

Subtracting Eq. (24s) from Eq. (24A) the differential change in S D is obtained

dS =S D -S D ' =r t \ -^, +^===. | . (24c)

thus

From Eq. (24 c) the value dS/dx is found and equated to zero for maximum value

of S D . This gives

RI R 2 Ri+R 2 Ri R

T,=l^^ or by composition __=_=_, . . . (24D )

where R is the resultant of all loads between B and i. This furnishes the criterion,
provided no additional loads have entered the panel CF and no other loads have come
on the span in making the shift.

FIG. 24A.

It is evident from the influence line, Fig. 24A, that any loads to the left of i would
produce a negative influence on SD, so that the minimum value of SD may be found
by loading the left portion Ai of the span, and the criterion then becomes

RI R

This criterion expresses exactly the same condition as previously found for moments
in Eq. (23L), but the span now becomes the length a\ or a 2 , as the case may be, cover-
ing only that portion of the stress influence line D, which has the same sign.

As previously found, by the graphic method, it is always necessary to place a
load at the panel point F for max. D and one at C for min. D, allowing as many loads
inside the panel CF as may be required to satisfy the criterion.

64 KINETIC THEORY OF ENGINEERING STRUCTURES CHAP. IV

The above discussion of the various criteria for position of moving loads to produce
maximum and minimum stresses in any member of a statically determinate structure
is ample demonstration of the high value and practical use of influence lines.

It was not deemed advisable to extend this discussion to cover the various types
of special and composite structures, as the influence lines themselves, when used for the
determination of stresses, will give complete and comprehensive answers in all cases
without special analytic treatment. However, when such is desirable, the methods
previously employed will indicate the procedure to be followed for any particular
influence line.

The subject of influence lines for statically indeterminate structures and for deflec-
tions will be treated after presenting the general subject of distortions and deflection
polygons of framed structures.

CHAPTER V

SPECIAL APPLICATIONS OF INFLUENCE LINES TO STATICALLY DETERMINATE

STRUCTURES

ART. 25. DOUBLE INTERSECTION TRUSSES

Usually trusses of this type are analyzed as two separate systems, dividing the
loads between them in such manner as may seem most probable.

The stresses found for the separate systems are then combined for the double system
wherever the same member forms part of each system, provided simultaneous cases of
loading were used.

The method of influence lines has advantages especially in the distribution of
simultaneous load effects, and the combined stress in a member belonging to both systems
is at once found for the same position of loads.

Double intersection trusses should always be designed with an even number of
panels, thus retaining symmetry of both systems with respect to the center of the span.
Otherwise the loading carried by one system in the left half span must go to the other
system in the right half span, which is not desirable.

Fig. 25A represents a double intersection truss for which stress influence lines will
now be drawn.

Bottom chord member DE. A section tt, passed through the panel DE, cuts the
diagonals FE and GK, hence the chord DE forms part of two panels for which there
are two centers of moments, F and G. Therefore, each system gives rise to a stress
influence line for the member DE, shown in Fig. 25A, by the two dotted triangles A'C'B'
and A'D'B'. The top chord members are regarded straight between alternate pin points
for each system in question.

Loads acting at panel points are supposed to be carried by the system to which
each such point belongs except at the panel where the load is divided equally between
the two systems.

Hence a load at D goes entirely to one system and its influence is represented by the
ordinate DD' ', of the influence line A'D'B'; while a load at E is carried by the other sys-
tem and its influence ordinate becomes EE' in the influence line A'C'B'. This same
reasoning applies to all panel points.

Since influence lines between . successive panel points must be straight lines, the
actual influence line for DE is represented by the shaded area A'O'C'D'E'K' to B' .
The point 0' must lie midway between the two ordinates for because at this point
the load is carried equally by both systems. This is not the case at M, where the entire
load goes to one system.

KINETIC THEORY OF ENGINEERING STRUCTURES

CHAP. V

Top chord member LF. The influence line for DE is like the influence line for LF,
except that the end ordinates are slightly different. Here the lever arms are measured
normally to the members LF and FG.

Diagonal member FE. The method of reasoning just applied to chord members
answers in every sense for the web members.

The influence line for FE is constructed as indicated in the figure. The stresses
in FE, due first to a unit load at A and then at B, are found for the system to which FE
belongs.

FIG. 25A.

Since loads at the panel points of the other system cannot cause stress in FE, the
influence line ordinates for these points must be zero. This means that the other influence
line is the base line A'B'. Observing again that the required influence line must be
straight between successive panel points the final line becomes A'O'C'D'E'K' to B'.
The point 0' , as before, is midway between the ordinates under 0.

The influence line indicates clearly that when loads are spaced 2d apart, then the
entire web stress goes into one system, making the design verj'- uneconomical. This
would easily happen for 50 ft. locomotives and 25 ft. panels.

ART. 26 INFLUENCE LINES FOR STATICALLY DETERMINATE STRUCTURES 07

ART. 26. CANTILEVER BRIDGES

Any statically determinate structure extending over more than two supports (piers
or abutments), when continuous over r2 of these supports, may be regarded as a
cantilever. If r is the total number of supports, then such a structure must be provided
with r2 intermediate hinges such that the bending moments at these hinged points
are zero for any system of loads.

The reaction and typical stress influence lines for such a cantilever will now be
drawn.

Reaction influence lines for all simple structures are independent of the panels and
may, therefore, be illustrated on a simple cantilever beam, Fig. 26A.

FIG. 26A.

The illustration represents an anchor arm A B with a downward reaction or anchorage
at A and an upward pier support at B. The span EF is a simple beam on two supports
E and F, with an overhanging cantilever arm at_each_ end. _A similar arm projects out
from the anchor arm, making three cantilevers, BC, DE and FG, in the whole system.

The two spans CI) and GH are known as intermediate spans and are supported on
hinged ends which permit of horizontal displacements. Hence, there is only one roller
bearing at E, and the anchorage at A also admits of slight horizontal motion.

The whole system is, therefore, determinate and can have no temperature stresses,
neither will slight displacements of the supports cause any internal stresses.

The reaction influence lines thus involve no principles other than those already
elucidated for simple beams on two supports.

The central span WF is a suitable beginning, as it rests on two supports R 3 and #4
with a roller bearing at E. Hence the influence lines for these two reactions are drawn
precisely as for a simple beam between the points E and F. But the two arms at the
ends of this span must also be considered.

Regarding first the influence line for R 3 , it is seen that a load at F has zero effect,

KINETIC THEORY OF ENGINEERING STRUCTURES

CHAP. V

and as the load proceeds to the left its effect on R% is uniformly increasing and this
continues to be the case until the load reaches the point D, where the effect becomes
a maximum.

The same is true for a load between F and G, hence the ordinate unity at E'
determines the RS influence line from D^_io G' . A load at C or at H produces zero
effect on R s and loads outside the span CH have no influence, hence the line C'D'G'H'
is the influence line for R s . Further proof will scarcely be needed, though the several
moment equations for the separate spans CD and GH will readily supply this.

Similarly the line C"D"G"H" represents the influence line for R$, based on the
unit ordinate at F".

The influence line for Rs is the same as for a simple beam GH.

Precisely the same reasoning applies to the reactions Ri&nd R-2, the influence lines
for which are shown to the left in Fig. 26A. It should be noted with regard to RI that
the negative influence area being greater than the positive, provision must be made
for a downward reaction. When the upward reaction RI, due to live load from A to
B, is greater than the downward reaction R'i, due to dead load from A to D, then the
fastening at A must be designed to take stress in both up and down directions.

Shear influence lines. These are readily deduced from the reaction influence lines,
as may be seen by comparing Fig. 26s with Fig. 26A. Four typical panels are assumed.
One each for the central span EF, the cantilever arms BC and FG, and the anchor
arm A B, respectively.

FIG. 26s.

For the panel ef the shear influence line within the span is precisely as for any
simple beam on two supports, while outside the span these lines are continued to the
intersections D' and G' with the verticals through the hinges D and G, respectively.
The shear influence line for the panel ef is thus found as C'D'e'f'G'H' .

For the panel ab the same construction is applied as for a span on two supports
A and B followed by a cantilever to the right of B exactly as for the previous case.
The shear influence line for the panel ab is the polygon A'a'b'B'C"D" '.

When the panel is located in one of the cantilever arms as for gh, then the circum-
stances are slightly different, because the shear in any cantilever arm is always equal
to the sum of the applied forces between the section and the outer end of the arm.

ART. 26 INFLUENCE LINES FOR STATICALLY DETERMINATE STRUCTURES 69

Accordingly the shear must be composed of the end reaction from the intermediate

Hence the influence line
It should be noted that

truss GH and the constant effect of loads between h and G.

for the panel gh is easily found to be the polygon g'h'G"H' .

loads to the left of g, even on the next span EF, can have no effect on the shear in the

panel gh.

The shear influence line for the panel cd is similarly found to be the polygon c'd'C"'D" '.

Any panel located within either of the intermediate spans CD and GH, is treated
precisely as for a span on two supports, since no load outside these spans can produce
any effect on these trusses.

Moment influence lines. These offer very little difficulty and follow directly from
the methods previously given.

In Fig. 26c the anchor arm was omitted, as the portion of the structure from C
to H affords all the illustration required.

Here, again, the central span EF is treated exactly like a simple beam and the
negative effects of the adjoining cantilever arms are found by prolonging the lines so

FIG. 26c.

obtained to the intersections D' and G' with the verticals through the hinged points.
The moment influence line for the point n, in the panel EF, thus becomes C'D'E'e'f'F'G'H' ,
and it is readily seen that a load on any portion of the structure CH affects the moment
at n.

For the section t, in the panel gh, the moment is the reaction at G into the lever
arm a 3 , hence for a unit load, the ordinate at G must be l-a 3 , and observing that the
influence line must be straight over the panel gh, the required moment influence line
for the section t becomes F"g'h'G"H".

Stress influence lines. The general method for stress influence lines, illustrated in
Fig. 21 A, serves every purpose in connection with cantilevers.

In every determinate structure the stress in any member may be expressed as
S=M/r and from this it follows that the stress influence line for any truss member
is similar to the moment influence line drawn for the center of moments pertaining to the
member. The difference between the stress and moment influence lines is only the
constant divisor r applied to the ordinates of the latter to obtain the ordinates for the
stress influence line.

For convenience, the central span CE, Fig. 26o, will be treated first, because this

KINETIC THEORY OF ENGINEERING STRUCTURES

CHAP. V

portion of the structure is exactly like a simple truss on two supports, and all the rules
previously given apply here.

Regarding the web members it should be observed that the center of moments may
fall inside or outside the limits of the span. In the former case there will be no load
divide within the span. Both cases are illustrated in Fig. 26D, where the stress influence
lines for members U, L, D, and D' are presented.

The designations regarding the stresses in these members for unit reactions at C
and E, will be as previously adopted in describing Fig. 2lA, viz., S uc =ihe stress in the
member U for a reaction unity acting at C.

FIG. 26o.

Since these stresses are always necessary in constructing influence lines, it is best
to find them for all members of the structure before proceeding to draw any influence
.lines. These stresses are very easily found either by slide rule, using Ritter's moment
method, or by drawing a Maxwell diagram.

Stress influence line for U, central span. This being a top chord member, and in
compression, its influence line is negative between C and E, which is also recognized by
the stresses S^, and S ue found for this member for the reactions unity, acting first
at C and then at E.

These stresses are laid off from A'B', Fig. 26D, upward and on the verticals through

ART. 20 INFLUENCE LINES FOR STATICALLY DETERMINATE STRUCTURES 71

C and E, respectively. The stress influence line is then easily drawn, observing that
the intersection n' must fall under the center of moments n for the member U. Hence
this is a check and may also be used as a shorter method of finding the U line, when
only S llc is known. Between the panel points m and o the influence line must be straight,
hence m'o' is drawn.

To complete the influence line over the side spans, the U line, just found, is continued
to intersections B" and F" with the verticals through the hinged points B and F and the
final U line is then found to be A'B"m'o'F"G' . The portions below the base line A'G'
are positive and the upper portion is negative. The positions of loads to produce max. U
and min. U are thus clearly indicated and no other criterion is required.

Stress influence line L, central span. This is constructed in precisely the same manner
as the U line, but L being in tension, as found for C = l, the stress +Si c is laid off down-
ward under C. The point m" . vertically under the center of moments m for the member
L, is thus found arid may serve to complete the L line without using the stress Si e .
The L line is then obtained as A'B'm"F'G' and is positive over the span CE and negative
over the two cantilever spans AC and EG.

Stress influence line D, central span. The center of moments for this member is at
i, which is within the supports, and hence this member has no load divide. The stress
influence line is constructed exactly as was done for the U line, by laying off the positive
stresses S dc and S de to obtain the two limiting rays, with intersection i' under the point i.

However, since the rate of increase in the stress D must be uniform for a load advanc-
ing from E to o, the ray E' r7 i' must be continued to o ly finally drawing the line m^i to
complete the polygon.

Over the two side spans the D line is drawn as for the two previous cases, giving
the complete line A'"B f "C f "mio l E'"F"'G"'.

Stress influence line D', central span. The center of moments for D' is outside the
span at k. hence the influence line must show a load divide. The stresses + S'd c and S'd e
serve to construct the stress influence line for D' which is in all other respects like the
D line. The limiting rays intersect in k' and the load divide is at q.

Stress influence line U, cantilever arm, Fig. 26B. The effect of a moving load, coming
on the span from left to right, begins when the load is just to the right of A and increases
uniformly until the load reaches B. From B toward m the effect is uniformly decreas-
ing and becomes zero when the load is over n, which is the center of momentsjor U.
The stress in U for a load P = l at B is-S ub , and laying this off from the_base A'o', the
negative U line is easily drawn. For the effect due to loads in the panel mo draw m'o' to
complete the influence line. Loads to the right of o or to the left of A have no influ-
ence on U.

Stress influence line L, cantilever arm. The center of moments for L is at o and
the stress in L for a load P = l at B is+S lb . Hence the influence line becomes A"B"o"

and is positive.

Stress influence line D', cantilever arm. The center of moments for this diagonal
is at i, which is inside the cantilever arm. The rule for load divide is now reversed
because loads on opposite sides of i produce opposite stresses in D', and a load at i cannot
produce any stress. Hence i is the load divide for D'.

KINETIC THEORY OF ENGINEERING STRUCTURES

CHAP. V

The stress in D' due to a unit load at F is S' d f, hence making the ordinate under
F equal to this stress the influence line s'i'F'G' is obtained.

Since loads to the left of h cannot affect D', and since the influence line is a straight
line over a panel, the influence line is completed by drawing h's'.

Stress influence line D, cantilever arm. The center of moments now falls off the
arm at k and hence all loads produce the same kind of stress in D. Lay off the stress
+S df on the vertical through F and complete the influence line as shown in Fig. 26E,
obtaining s"v'F"G" for the D line.

The intermediate spans A B and FG are simple trusses on two supports and require
no further consideration.

FIG. 26E.

ART. 27. THREE-HINGED FRAMED ARCHES

In taking up this subject it is necessary to show briefly the general relations between
the reactions and external loads.

Let Fig. 27A represent any three-hinged arch ACB, which may be framed or solid.
The single load P may be applied at any point m.

For all positions of P between A and C, the reaction R% will always have the
direction BC and the two reactions RI and R% must be components of P. Similarly
for all positions of the load P between C and B, these reactions will have the directions
AC and eB, respectively. Hence for any position of the load P, the reactions at A and
B are found by resolving P into two components as shown, and the points of application
of all possible positions of P will lie on the prolongation of either AC or BC.

If the reactions R and R 2 be resolved, each into a vertical component and one
along AB, the two last-named components H' will be equal and opposite, while the two
vertical components A and B will be precisely the same as for a simple beam on two
supports A and B, because the forces acting at C are in equilibrium among themselves.

Therefore,

A+B=P;

Pa 2
A -,

and B= = .

(27A)

AKT. 27 INFLUENCE LINES FOR STATICALLY DETERMINATE STRUCTURES 73

The horizontal component of H' is known as the horizontal thrust of the arch and
is found from

H=H'cosa (27B)

If the reactions R\ and R 2 had been decomposed into components _4/, B' and H,
then the former would no longer be the reactions for a simple beam AB. However,
when a =0 and the end supports are on a horizontal line, then the vertical end reactions
are always as for a beam on two supports.

FIG. 27A.

Taking moments about the hinge C, when P alone is acting, then for condition of

equilibrium at C

M=Al l -H'y=Al l -H'fcosa=Q,

and substituting values from Eqs. (2?A> and (27s) this gives

Hi or H= J-J-. . .

(27c)

When P = l is applied at C the value H c becomes the ordinate vertically under C
for the influence line for horizontal thrust.
Hence,

,-i, (27D)

-_^ and when Zi=Z 2 =o, then

making the H influence line a triangle with middle ordinate T? C under C, as given by

Eq. (27D).

The influence lines for the vertical end reactions are found exactly as for a simple

beam AB.

The stress influence line for any member of a three-hinged framed arch will now be

developed.

When the structure is symmetric, the half arch only requires analysis, but for
unsymmetric arches the entire structure must be treated, though the method remains
the same as would appear from the previous discussion.

KINETIC THEORY OF ENGINEERING STRUCTURES

CHAP. V

In Fig. 2?B, the half arch is shown with a load P producing reactions R\ and R 2 ,
and since the latter must be transmitted to the point C, the right half of the span may
be considered removed and its effect replaced by R 2 acting at C. The left half of the
span is then held in equilibrium by the two reactions R : and R 2) the latter applied at C.

The reaction R\ is now resolved into A and K and the reaction R 2 is similarly
replaced by a vertical reaction C and a reaction K which is equal arid opposite to the
first K.

FIG. 27s.

If the forces K were zero, then the half span would represent a simple truss on two
supports, with vertical reactions A and C and the influence line for any member could
be found by the method previously given. Thus, the influence line for any member-
is easily drawn when the stresses, first for A=l, and then for 5 = 1 are known. Such
an influence line is drawn for the diagonal D, and is shown as the polygon A'E'F'C' ',.
Fig. 2?B, where the stress D a , due to A = 1 is positive, and the stress D c , resulting from
C = l, is negative. In each case the half span is supposed to be held first at C and then
at A in order to find these stresses.

ART. 27 INFLUENCE LINES FOR STATICALLY DETERMINATE STRUCTURES 75

The influence line thus found, must now be corrected so as to include the effect of
the force K.

Since K=Hsec{3, therefore K is proportional to H and the stress Dk, produced
by the force K, must likewise be proportional to K and H. Hence, the influence line
for the stress D k , is similar to a horizontal thrust line and must be a triangle whose
middle ordinate under the hinge C is equal to the stress D k , produced by a unit load
applied to the whole span at the point C.

To find this stress D k for a unit load at C, substitute P--=l into Eqs. (27 A) and
(27c) and obtain

A=i and # = l'i-

If the stress in D is found first for A = 1/2 and then for // =l/4f } separately, then
the combined effect will be the stress D k =%D a +D H , which is much more convenient
than to find D k directly for the diagonal force K, The stress D H is the stress in D
for the above force H = l-l/4f applied horizontally at A and supposing the half arch
rigidly held at C.

The stress D a was previously found, and hence the required stress D^ is readily
obtained from

(27B)

In the present case D^ was found to be negative and the influence line for the
member D due to the effect K is drawn as the triangle A"C"B" ' .

The final D line is now obtained by combining the two influence lines just found-
This is done in the last figure and gives the polygon A'"E"F"C'"B" r . The stress +D a
is laid off down from A'" and the stresses D c and D^ being negative, are laid off
upward from the base line A'"B'" and being of the same sign they are added. The
polygon is then drawn as indicated in the figure and as a check, the point n" must fall
vertically under the 'center of moments n for the member D. Still another check is
found from the circumstance that a load at i produces zero stress in the member D and
hence the influence line must intersect the base line vertically under i in the point i'.
This last mentioned condition is always true even though the point i falls below C.

The point i is the intersection between BC and An, where n is the center of moments
for D. This offers a very valuable condition and obviates the necessity of finding the
stress D c , whenever the center of moments n can be located within limits of the drawing.

It is thus seen that when n and i are conveniently located, which is always the case
for chord members, then the influence line can be determined when D a alone is given.
However, as a check, it is always desirable to know D^, so that when the stresses in
all the members are to be determined by influence lines, it is advisable to find S a and
Sh for each member before proceeding to draw the lines. This can be done by drawing
two Maxwell diagrams, one for A = 1 and the other for H=l/4f. A slide-rule computa-
tion will also answer if the lever arms are carefully determined.

In Fig. 27s the loads were supposedly applied to the top chord. Had they been
applied to the bottom chord then the panel line E"F" would have become G"H".

KINETIC THEORY OF ENGINEERING STRUCTURES

CHAP. V

Typical stress influence lines will now be given for the three-hinged, framed arch,
Fig. 27c, treating the top chord as the loaded chord.

The example will be taken up in such manner as to indicate a complete solution
for all stresses in the several members by the method of influence lines.

~?B

I UU INFLUENCE LINE.

"^luiii]|p|i||]^

90 LIN&THV

The first step will be to determine the stresses S a and S k for all the members and
this will be done analytically by Ritter's method of moments, being very careful about
the sign of each stress.

The same stresses could also be found in a very convenient way by drawing two
Maxwell diagrams, one for A~l, supposing the half truss ~AC fixed at C; and the other

ART. 27 INFLUENCE LINES FOR STATICALLY DETERMINATE STRUCTURES 77

for a load P = l acting vertically downward at C and producing the horizontal thrust
// = l-Z/4/= 0.833, which is again applied at A, supposing the half truss fixed at C.

The following table contains the stresses S a and S a thus obtained and the stresses
S=S a +S are then easil found.

MEMBER.

STRESSES.

MEMBER.

STRESSES.

-Sa

S H

U,A

-1.00

+ 1.02

+ 0.52

-0.45

+ 0.52

+ 0.30

f7,L,

+ 0.93

-0.57

-0.11

-1.37

+ 1.08

+ 0.40

-1.29

+ 0.79

+ 0.15

U 3 U t

-3.11

+ 1.97

+ 0.42

U 2 L 2

+ 1.32

-0.85

-0.19

AL t

0.00

-1.44

U 3 L 2

-1.65

+ 0.77

-0.05

L,L.

+ 0.85

-1.87

-1.45

U 3 L 3

+ 2.19

-1.15

-0.06

L 2 L 3

+ 2.08

-2.38

-1.34

U t L 3

-2.45

+ 0.96

-0.26

L 3 C

+ 4.40

-3.35

-1.15

u t c

+ 3.05

-1.19

+ 0.33

+ for tension and for compression.

The values in this table will suffice for the construction of all the stress influence
lines, but only those for (7 2 t/ 3 , L 2 L 3 , U 2 L 2 and / 4 L 3 will be drawn, as shown in Fig.
27c.

Stress influence line U 2 U 3 . The stress S a = 1.37 is laid off, to a convenient scale,
upward from A' and the stress ^=+0.40 is laid off downward from A'B', vertically
under C. The lines EC' and C'B' are then drawn.

The center of moments for U 2 U 3 is at L 2 , hence find Z/ 2 vertically under L 2 and
Finally since the influence line must be straight over the loaded panel

As a check observe

draw A'L 2 '.
UjU~ 3 , draw U 2 'U S

to complete the influence line A'U 2 UzC'B'.

that the load divide i/ falls vertically under i', which latter is the intersection of AL 2
with BC produced. Still another check may be had by resolving a unit load at U 2
into components parallel to U 2 U S and U 2 L 2 as was done to the right of the truss diagram.
The stress thus found for U 2 U 3 must equal the ordinate U 2 'F. Lastly, the ordinate
S c is the stress in U 2 U 3 for a load unity at C when the half truss AC is fixed at A
and the half span CB is removed.

Stress influence line L 2 L 3 . The method is precisely as for the previous case, being
careful to observe the signs of the stresses S a and S k in laying them off from the base
line A 7 /?'. The point U' 3 , vertically under the center of moments for this member,
again serves to complete the stress influence line, with all the checks just mentioned
for the upper chord member.

Stress influence lines U 2 L 2 and U 4 L 3 . The same method again applies and the two
lines illustrate the effects of the signs of the stresses S a and S k . The center of moments
n, for the member U 2 L 2 now falls outside the span, but the same relations hold good
as before. Similarly for the center HI of the member U 4 L 3 . In this last case the load
divide t' 3 falls below the middle hinge C and is no longer a real load divide, as the influence
line for TT^Ls indicates, but all other relations continue to exist as before.

KINETIC THEORY OF ENGINEERING STRUCTURES CHAP. V

ART. 28. THREE-HINGED, SOLID-WEB AND MASONRY ARCHES

Fig. 28A illustrates a three-hinged, solid-web arch, which may be built of masonry
or steel, according to circumstances. If of masonry, then the section will be rectangular,
and if of steel, then the section will resemble that of a plate girder.

Whenever the depth of the section is small compared with the radius of the arch
center line, then the effects of curvature are very minute and the stresses in the arch
rib may always be found by the well known beam formula of Navier. This condition
always prevails in bridges and all larger arches, but when dealing with small arched
culverts and structures of that class, then the effects of curvature may require special
consideration.

For any girder section then, the stress on the extreme fiber is derived from

wherein M is the bending moment, A" the normal axial thrust, F the cross-section, /
the moment of inertia of the section and ?/ the distance from the center of gravity of
the section to the extreme fiber. M k is the total bending moment of the external
forces about one of the outer kernel points e or i (Fig. 28A) and W is the moment of
resistance of the section.

Calling r the radius of gyration of the section, then

7 Fr 2 r 2 I

W=-= =Fk where k=^-~ ...... (2SB)

y y y Fy

The quantity k is the distance from the gravity axis to a kernel point and the negative
sign indicates that it is measured in the opposite direction from y for all sections.

Hence, the influence lines for the kernel moments M e and Mi will serve to find the
stresses f e and /; for the extreme fibers of extrados and intrados of any particular section
tl, by using the multiplier /y. = l/W = l/Fk. For unsymmetric sections the two kernel
distances are not equal so that {i e = l/Fk e for extrados and fi.i = \/Fki for intrados.

For each section the stresses on the extreme fibers thus become

''-7S, and fi =~K ..... ' ' (28c)

where M e =K ac e and Mi=K ac i, and K ac is the end reaction at A for a load unity at C
found by resolving the vertical reaction A along AC and AB. See also Art. 49 on two-
hinged arches.

The moment influence lines M e and M; are now found by the method previously
outlined for framed arches.

Each of these moments becomes zero when the unit moving load passes through
a load divide d. Hence the load divides are easily located for each case.

Also since moments and not stresses are now considered, the end ordinate for the
influence line at A is equal to x e or Xi as the case may be.

ART. 28 INFLUENCE LINES FOR STATICALLY DETERMINATE STRUCTURES 79

Hence to draw the influence line Mi, lay off Xi at A and project the point d down
to d', draw Gd'C' and after projecting down the center of moments i to find i', finally
complete the influence line by drawing A'i' and C'B'. Now find as a check, that
~C'~D = x'i.

FIG. 28A.

The Af t - influence line becomes the/; influence line with a factor m=*

The influence line M e is found in precisely the same manner.

In both cases the ordinate C'C" is equal to M t or M e as given by Eq. (28c) for a

unit load at C. The M e influence line becomes the / influence line with a factor

80 KINETIC THEORY OF ENGINEERING STRUCTURES CHAP, v

The shear influence line Q OT is found by a similar process. Here the center of
moments, for a web member at m, would be practically at an infinite distance because
the chords are parallel or nearly so. Hence the line AdAj.t will intersect the chord
C/I in the load divide d, which in this case is only imaginary, but serves the purpose
as before.

The equation for shear along the section it is easily found to be

and this gives, for a unit reaction at A, when the half arch is fixed at C, the end ordinate
A'G = l-cos <f>. Also since the shear is constant between A and m, the limiting lines
A'm' and Gd' must be parallel.

Hence the shear influence line is easily constructed by drawing GC'd' , and then
completing the polygon by drawing A'm' || Gd' and joining C' with B' ' .

Since the loading is directly applied the line Wira', vertically under m, completes
the Q 7> , line.

Finally the ordinate C'C" =H[ - - ) wherein 7/ = l--^rfor a unit load at C.

\ cos a ) fl

When the loads are indirectly applied, then the condition that the influence line
must be straight between load or panel points must be carefully observed.

For masonry arches the section is always rectangular and 7 =/>D 3 /12, hence k = r' 2 /y =
^F^Z), where D is the thickness of the arch ring. The points e and i thus become the
middle third points of the arch section and Eqs. (28c) become

, GM e ,

f*~~TF and ^ =

from which the stresses in a masonry arch section are readily found in terms of the
moments about the kernel points of the section. The stress influence lines thus become
moment influence lines with a factor 6/D 2 for each section.

ART. 29. SKEW PLATE GIRDER FOR CURVED DOUBLE TRACK

In general, influence lines for skew bridges are drawn in all respects like those for
any simple truss or beam with the exception that at one end of the span the load always
leaves the span before it is entirely outside of the limits of the longer truss. This merely
calls for a slight correction in the end panel of the influence line as ordinarily drawn.

At the opposite end of such truss some load comes on the end floor beam before
the moving load is on the truss, but since the end floor beam transmits its load directly
to the bridge support and not to the truss, this last named circumstance does not affect
the influence line. These points will be brought out in connection with the following
problem.

It may seem inappropriate to apply influence lines to plate girders, but the practical
designer who has dealt with problems of skew plate girders on curved double track
will readily appreciate the advantages of the method there presented.

KT. 29 INFLUENCE LINES FOR STATICALLY DETERMINATE STRUCTURES 81

MOMENT INFLUENCE LINE " ~ --
POINT 7, GIRDER NO.I.TRACK NO.2.

FIG. 29x.

82 KINETIC THEORY OF ENGINEERING STRUCTURES CHAP.)

This method was first suggested and used by Mr. T. B. Moenniche, C. E., in designing
a skew double-track girder of 125 ft. span for the Virginian Ry. Co., in 1906. Thi
author is indebted to Mr. Moenniche for the solution which follows.

The principal advantage in the present application of influence lines consists ii
reducing the solution of this rather complicated problem to a method instead of a time
consuming process of experimentation.

The principle of the method may be outlined as follows: The floor-beam reaction!
are first determined for a unit load situated on track Xo. 1, and then for a unit load 01
track No. 2, which requires a careful computation of the coordinates or intersectioi
points of each track on each floor beam. These floor beam reactions are then used a!
influence or reduction factors for the ordinates to any particular influence line, becaus<
the factors represent the proportion of a total load which can reach either girder a
any panel point. This will give two influence lines for each panel point of a girder
one separately for each track.

In Fig. 29A, the upper diagram represents the structure in plan and the two sue
ceeding figures show the floor beam reactions graphically. The last two diagram:
are the moment influence lines for point 7, girder No. 1, for tracks No. 1 and 2, respectively

The two reaction diagrams were drawn merely as a convenient way of showing
the figures and in practice they would serve the same purpose, though they might b(
dispensed with entirely. The reactions appearing as ordinates are plotted to sonn
convenient scale making the sum of two reactions for each floor beam equal to unity
except at the ends where the floor beams do not connect with both girders.

The moment influence lines for point 7 are drawn in the usual manner by laying
off a 2 vertically down from B and drawing A'T and 7'B' '. Since the load on track \o
1, when coming on from the right, begins to affect girder No. 1 as soon as it passes the
abutment at E, therefore, the distance #10 must be treated as a panel arid the influence
line is drawn straight from 10 to B'.

The ordinates d of the influence line are now successively reduced by multiplying
them with the corresponding ordinates or factors m of the reaction diagram to obtaii
the real ordinates y of the required moment influence line. Thus ^6 =< Vn 6 and similar!}
for the influence line point 7, girder 1, track No. 2, the ordinate Jjg=0 8 w 8 .

With the use of these two influence lines, represented by shaded areas, the actual
moment for point 7 may be found for any case of loading for each track and the sun:
of the two gives the combined max. M 7 .

Shear influence lines may be drawn in a similar manner, though they would scarcely
serve a useful purpose except when dealing with a truss. In the above example onl}
the reaction influence lines would be useful to determine the end shears.

AKT. :-o INFLUENCE LINES FOR STATICALLY DETERMINATE STRUCTURES S3

ART. 30. TRUSSES WITH SUBDIVIDED PANELS

From a point of economy, large trusses must have long panels, but with the increase
in panel length the stringers of the floor system soon attain such proportions as to
impose a practical limitation on the panel length.

To overcome this difficulty, modern practice has developed a new type of truss
especially adapted to long spans, wherein the stringers are supported by floor beams
placed at the half-panel points. These intermediate floor beams are carried at the ends
by a secondary set of truss members, which latter serve to carry the loads, thus locally
applied, to the main truss system. This has given rise to the truss with subdivided
panels.

In the analysis of these truss types, it is not always a simple matter to determine
the criterion for maximum stresses in the members, and it is believed that the following
treatment by influence lines will serve to clear up the doubts usually encountered.

Fig. 30A shows a truss of 200 ft. span with four different forms of subdivided
panels. These are all combined in the one structure, though in practice only one form
should be used in the same structure excepting in the first panels ALiL 2 , where the
several compression members shown offer distinct advantages in stiffening the ends
of the truss. Otherwise the panel L 2 L 3 , Case I, is the best and most economical type
for adoption, as will appear later.

In general, all primary members in the structure will receive stresses which are
at least equal to those which would exist if the secondary members were all removed
and in addition to these primary stresses nearly all of the members, excepting the
verticals, will receive increased stresses due to local loads from the secondary members.
This local loading is transmitted to the truss through the vertical M Z K which never
acts as a primary member and gets no stress other than that due to a full panel load,
besides the dead weight of bridge floor and bottom chords.

It is the load, locally transferred to the points M , which enters as a disturbing
element, and the four cases here treated will illustrate about all the practical points.

To make the four cases directly comparable, the same panel L 2 L 3 was retained,
but the diagonals were successively rearranged. The stress influence line for the
primary stress in U 2 L S remains the same in each case, hence it is drawn only once
and is then modified according to local conditions in the panel L 2 L 3 for each case.

The Diagonal Member U ,L ,.

Case I. The stress influence line for U 2 L 3 , when acting only as a principal truss
member, is drawn by laying off the end ordinate A' A" = +0.568 which is the stress
U 2 L 3 for a reaction unity at A. The ordinate B'B" (not shown) is 2.790 and is
the stress in the same member for a unit reaction at B. By drawing A" ' B' and A'B"
the points U' 2 and L' 3 are found and the line t/ 2 'L 3 ' completes the influence line
A f U 2 L s 'B' with load divide at i.

This gives the maximum stress for M 3 L 3 but not for U 2 M Zl because a load at K
would produce additional stress in the latter member without affecting the former.

Hence to find the influence line for U 2 M 3 , it is necessary to determine the stress

KINETIC THEORY OF ENGINEERING STRUCTURES

i STRESS INFLUENCE LINE M 3 L 3 .

FIG. 30A.

ART. :so INFLUENCE LINES FOR STATICALLY DETERMINATE STRUCTURES 85

in this member due to a local load unity at M 3 which gives the total ordinate
under M 3 .

A load at M z will be carried entirely by U 2 M 3 and M^U~ 3 , hence the stress due to
a unit load is readily found by resolving the unit load into components parallel to these
two members as indicated by the triangle of forces in the figure for Case I, drawn to half
the scale of the influence line ordinates. The stress in U 2 M 3 is thus found to be
+0.5, which is laid off from m to complete the influence line A'U 2 'M 3 L 3 B' for the
member U 2 M 3 . It is seen that the point M' 3 does not fall on the line B'A" as usually
assumed by authors on this subject. The error from this assumption is more apparent
in Case II, and is not always on the safe side.

Case II. The influence line for the primary stress in U 2 L 3 is as before, and again
gives maximum for the lower portion M 3 L 3 . However, a unit load at M 3 is now carried
to the members U 2 M 3 and L 2 M 3 , and these stresses are again found by a triangle of
forces drawn in the figure for Case II.

The stress thus produced in U 2 M 3 is now +0.62, from which the new influence
line for U 2 M 3 is readily obtained as A'U 2 'M 3 L 3 B'.

Case III. Here the primary stress holds good for the upper portion U 2 M 3 and
the stress in the lower portion M 3 L 3 is diminished by the compression due to local
loading at M 3 . A unit load at M 3 is now carried by M 3 L 3 and M 3 U 3 and the triangle
of forces again gives the ordinate iM 3 , which is here negative.

Case IV. This is like Case III, but a local load at M 3 is now carried by L 2 M 3
and M 3 L S equally. Hence the stress in M 3 L 3 due to a unit load at M 3 is found as
0.62, which gives the ordinate iM 3 .

Members L 2 M 3 and M 3 U 3 , whenever they occur alone, can act only as secondary
members to carry the portion of load locally applied at M 3 . When the panel is subject
to counter stress as in panels L 4 L 5 and L 5 Le, then the local stress must be combined
with the counter stress when a counter U^MQ is employed. However, it is usually
preferable to omit the counter member and design the main diagonal to take both the
direct and counter stresses.

The verticals U 2 L 2 , U 3 L 3 , etc., may be treated entirely as primary members assuming
the panels to be L 2 L 3} L 3 L 4 , etc.

The vertical UjLi is merely a suspender for the double panel AL 2 and its maximum
stress will occur when the span is fully loaded from A to L 2 .

The member M 6 V, and others of similar character often employed in large bridges,
are entirely secondary and perform no work as truss members. The sole purpose of
these members is to stiffen certain long compression members.

Bottom chord member L 2 L 3 . The influence line for this member is easily drawn.
The center of moments is at U 2 and the end ordinate at A is the stress in the member
for a unit reaction at A. The influence line thus becomes a triangle A'L 2 'B'.

Case I. With the diagonals all as tension members the local loads at the middle
points cannot affect the bottom chord stresses and the influence line just described
gives maximum stress.

Case II. Here the bottom chord stress is increased by the local load transmitted
from M 3 to L 2 . This is shown in Fig. 30A. For a unit load at M 3 the stress in the

86 KINETIC THEORY OF ENGINEERING STRUCTURES CHAP. V

bottom chord is the horizontal component in L 2 M 3 found in the small force diagram
for Case II. This stress, equal to +0.36 i _must then be the additional plus ordinate
under M 3 and the stress influence line for L 2 L 3 thus becomes A'L 2 K'L 3 B'. as shown.

Case III is exactly like Case II, and the horizontal component of M 3 L 3 due to a
unit load at M 3 will be the positive ordinate to be added to the ordinary influence line
under K.

Case IV. Here the two members L 2 M 3 and M 3 L 3 will transmit the unit load
from M 3 after the manner of a three-hinged arch. The local stress in the oottom

L 2 L 3 20

chord then becomes ===== = = +0.358 for a unit load at M 3 .

4M 3 K 4X13.95

Hence the positive ordinate under K must be increased by 0.358 to determine the
proper influence ordinate under K.

Top chord member U 2 U 3 has four cases as just described for the bottom chord.

Case I. Here the local load at M 3 produces a compression in the top chord like
a suspension s} r stem. The amount of this compression for a unit load at M 3 is readily
found from the Eq. (27o) given for three-hinged arches and is H' =l\L 2 /fd cos a, where
li +I 2 =d=pa,nel length, see the truss diagram Fig. 30A.

This will give the negative stress ordinate to be added to the negative stress
ordinate under M 3 for the final influence line.

Cases II and III are similar and here the local effect in the top chord is the com-
ponent in U 2 M 3 or M 3 U 3 parallel to U 2 U 3 .

Case IV does not influence the top chord for local load at M 3 . It is similar to
Case I for bottom chord.

It is thus seen that the best criterion for trusses of this type is obtained from
influence lines and the ease and clearness with which any case is solvable speaks greatly
in favor of this method of analysis, where critical positions of loads and stresses are all
given from the one solution.

In conclusion it may be remarked that the type of panel illustrated in Case I is in
all respects preferable to the others. The type Case II deserves second choice, and
the others, especially Case IV, should be avoided whenever possible.

The panel L\L 2 is in a certain sense indeterminate and should never be used except
in end panels as here illustrated, where the several posts tend to steady the shock of
trains coming on the structure.

CHAPTER VI
DISTORTION OF A STATICALLY DETERMINATE FRAME BY GRAPHICS

ART. 31. INTRODUCTORY

The displacement between any two points or pairs of lines of any determinate
frame may be found analytically as shown in Arts. 6 and 9. However, many problems,
to be considered later, require a complete solution for every pin point of a structure,
and then the analytic method would become quite laborious.

The graphic solution here given includes the two following problems: (1) To find
the distortion, or change in form, of a frame resulting from changes in the lengths of its
members, first published in 1877 by the French engineer Williot; and (2) to find the
effect on this distortion produced by a yielding of the supports, or reaction displacements.
This latter contribution came from Professor Otto Mohr in 1887, and without it the Williot
diagram had little real value. Hence it is proper to call the complete diagram a Williot-
Mohr displacement diagram, contrary to the position taken by Professor Mueller-
Breslau, who erroneously credits the entire subject to Williot.

Having given the changes in the lengths of all the members of a structure, by Eq.
(4fi), when the stresses and cross-sections are known, let it be required to find the hori-
zontal and vertical displacements of all the pin points.

The Williot diagram, to be presented first, offers a partial solution of this problem,
which is completed by Mohr's rotation diagram.

ART. 32. DISTORTIONS DUE TO CHANGES IN THE LENGTHS OF THE
MEMBERS BY A WILLIOT DIAGRAM

Since any determinate frame is formed by a succession of triangles, this elementary
frame is first examined. The process may then be extended to include any number of
such triangles or a complete frame.

In Fig. 32A suppose the point c to be connected with the points a and 6 by members
1 and 2, and that the lengths of these members also undergo changes Al and + J2,
respectively, while the points a and b move to new positions a' and &'. It is now
required to find the direction and amount of displacement resulting for the point c.

Both members are first moved parallel to themselves until they occupy the positions
a'c 2 and b'c i} respectively. The length of member 1 is now shortened by the amount
Jl because this is a negative change; similarly the member 2 is lengthened by the
amount J2, giving the new lengths a'c 4 and b'c 3 , respectively.

8S KINETIC THEORY OF ENGINEERING STRUCTURES CHAP. VI

With a' and 6' as centers and corrected lengths a'c 4 and b'c 3 as radii, describe arcs
which may intersect in a point c' representing the new position of the point c. But
because Al and Al are always very minute compared with the lengths of the members
to which they belong, the arcs c 4 c' and c 3 c' are replaced by their tangents, drawn respec-
tively perpendicular to the original directions of the members. The diagram shows
these displacements several hundred^ times larger than they actually are compared
with the real lengths of the members ac and be.

It is clear that the point c' might have been determined in a separate figure,
without including the members themselves, by dealing exclusively with the directions
and changes in the lengths of these members. Thus, in Fig. 32B, the Jl's are plotted
on a very large scale, affording greater accuracy in the results.

If the point is regarded as a fixed point or pole, and if from this pole the dis-
placements of the points a and 6 are drawn, making Oa' =aa' and Ob' =66', both in
magnitude and direction; also, applying at a' the shortening Jl in the length of the
member 1; and at b' the lengthening +J2 of the member 2; then the perpendiculars

FIG. 32B.

erected at the extremities of A\ and J2 will intersect in c' , which is the new position
of c. The displacement of c is then represented in direction and amount by the ray Oc'.

In laying off the values of Al, in Fig. 32s, the following rule must be observed with
regard to signs: If a' is regarded as fixed, then Al representing the shortening of ac,
it follows that c moves in the direction from c toward a, and hence Jl must be drawn
from a' in the direction of c to a. Likewise, if 6' is regarded fixed, then since +J2
represents a lengthening of be, it follows that c moves away from 6 and hence +J2
must be drawn from 6' in the direction of 6 to c.

Any succession of triangles, as in Fig. 32c, may be treated in the same manner.
It is necessary, however, to assume that one of the members, as ab, retains its direction,
and that some point of ab, as a, remains fixed. The changes Al in the several members
being given, the construction is carried out as follows :

The point 0, Fig. 32c, is the assumed pole, and since the point a is considered
fixed it must coincide with 0. Also, since the member ab does not change in direction,
then Jl is drawn parallel to ab and, being negative, it must be applied in the direction
of 6 toward a. The displacement of the point 6 is thus given by Ob'.

ART. 33 DISTORTION OF A STATICALLY DETERMINATE FRAME 89

The point c recedes from a by an amount +J3, and approaches b by an amount
J2. If, therefore, J2 be drawn from b' in the direction of c toward b and parallel
to cb, and if J3 be drawn from or a', in the direction of a toward c and parallel to
ac, then the intersection c' of the perpendiculars, to the extremities of J2 and J3, will
determine the new position of c. The displacement of c is thus given, both indirection
and amount, by a ray Oc' not drawn in the figure.

The point d is joined to a and c by the members ad and dc, Fig. 32c. Its displace-
ment Od' is now found, in Fig. 32o, by drawing +J4 parallel to ad from and in the
direction a to d; also by drawing J5 parallel to erf from c' in the direction d to c.
The intersection d' , of the two perpendiculars to the extremities of J4 and J5, is the
new position of d. Finally the new position e' of the point e is similarly found by
applying the same method.

I
\

\
\
\
\

FIG. 32o.

Fig. 32o, whose polar rays Ob', Oc' , Od' , and Oe' give the displacements of the several
points b, c, d and e, both in direction and magnitude, is called the "Williot displace-
ment diagram," for the frame abcde, giving credit to the name of its originator.

ART. 33. ROTATION OF A RIGID FRAME ABOUT A FIXED POINT PROFESSOR

MOHR'S ROTATION DIAGRAM

In the Williot diagram it was assumed that the member ab did not alter its position,
but merely changed its length. However, this is not always the case and more frequently
this member ab is also subjected to a rotation about some instantaneous center. In
the latter event, the elastic displacements, given by the Williot diagram, will require
further changes to make them comply with the initial change in the position of the
member ab. It is supposed, for the present, that the frame has undergone its elastic
deformation and has passed into a rigid state.

The solution of this phase of the problem was first proposed by Professor Mohr and
is based on the following theorem in mechanics: The motion of a rigid body, at any
given instant, may be defined as a rotation about a certain point called the instantaneous
center of rotation, and the direction of motion of any point of this body, at the instant in

KINETIC THEORY OF ENGINEERING STRUCTURES

CHAP. VI

question, will be perpendicular to the line joining such point with the instantaneous center
of rotation.

Thus, if abcde, Fig. 33A, represents a rigid frame rotating about the instantaneous
center P, each arrow, perpendicular to its respective ray, will then represent the direction
of motion of the point to which the ray is drawn.

Now, if from any pole 0, Fig. 33B, a series of radial rays be drawn respectively
parallel to the arrows representing the motions of the several points, then these rays
Oa", Ob", Oc", Od", and Oe" will in turn represent, both in magnitude and direction,
the several displacements of the several points, a b, c, d and e. The figure a"b"c"d"e"
will be the new position of the rigid frame abcde.

For, a"0-LaP; b"0-LbP; c"0-LcP; etc., because the direction of motion of each
such point of the rigid frame was perpendicular to the ray joining this point with the
instantaneous center of rotation.

Also. a"0:b"0:c"0, eic.=aP:bP:cP, etc., because the displacements of the several
points a, b, c, etc., are respectively proportional to their velocities, and these in turn

FIG. 33A.

FIG. 33n.

are proportional to the distances of the several points from the instantaneous center
of rotation.

From these conditions it. follows that:

(a) If the points a", b", c" , etc., of the diagram Fig. 33B, are joined by straight
lines, so that for every member ab of the frame, there will be a corresponding line a"b"
in the displacement diagram, then these latter lines will produce a figure a"b"c"d"e"
which will be similar to the rigid frame.

(6) Also, the line joining any two points of the rigid frame, as ab, will be perpen-
dicular to the corresponding line a"b" of the displacement diagram.

Hence, if it is possible to determine the position of any two of the points a", b",
c", etc., of the displacement diagram, then the figure a"b"c"d"e" can be drawn by-
similarity with the figure abcde. The method of determining two such points a" ', b"
will be illustrated in the examples.

Therefore, the secondary displacements, due to rotation of the rigid frame, are
found by inserting in the Williot displacement diagram a figure a"b"c"d"e" similar
to the frame abcde, and having its sides respectively perpendicular to the sides of the

ART. M DISTORTION OF A STATICALLY DETERMINATE FRAME 91

latter frame. The original displacements Ob', Oc', etc., are then combined with the
displacements Ob", Oc", etc., due to rotation to produce the resultant displacements
V'J', c r '7, etc.

The complete solution for the displacement of the pin points of any framed
structure, definitely supported, may then be outlined in the three following steps:

(a) Assume any member and a point on the axis of this member as fixed and
construct a Williot displacement diagram.

(b) Then assume the frame as rigid and subjected to a rotation such as will
conform to the actual conditions of the supports.

(c) The displacements of the several pin points may then be found, in magnitude
and direction, by scaling the distances between the m" and m' points. The direction
of any displacement will always be from m" toward m' '.

The complete diagram, combining the Williot displacement diagram with the Mohr
rotation diagram, will hereafter be known as a Williot-Mohr diagram.

The analytic solution of deflection problems is conducted with the aid of Professor
Mohr's work, Eq. (GA), and may be advantageously employed when only one point of
the structure is to be treated, or when great accuracy is required in the solution.

This method is fully discussed in Art. 6 and, in combination with Maxwell's theorem
given in Art. 9, offers the most accurate analytic solution of a great variety of problems
dealing with displacements of points and lines.

However, all problems requiring a complete solution for all the pin points of a
structure, may be solved by a Williot-Mohr displacement diagram when extreme
accuracy is not necessary.

Attention is called to the fact that the value for E, in all deflection problems,
should be chosen low rather than high, because there is always a slight amount of lost
motion and permanent set, which follows the first loading of a new structure and which
is not truly an elastic deformation. This can be compensated for in the calculations
by assuming a small E.

As a general rule, the changes Al, in the lengths of the members, are best figured
on gross sections rather than on net sections. Experience indicates that this gives
results nearer the truth. Also the stresses in the members must be simultaneous and
not maximum.

ART. 34. SPECIAL APPLICATIONS OF WILLIOT-MOHR DIAGRAMS

(1) Distortion of a simple truss. Fig. 34A has been solved in three ways, for the
purpose of illustrating that the displacements are not affected by the choice of the
member and point which are regarded as fixed in position, and also in order to show
how to select the most convenient of the possible forms of solution.

The truss is supported on rollers at e and is fixed at a. The extensions (+) and
contractions ( ) of the members are assumed to be as marked upon them.

The solution given in diagram 6 is made under the supposition that the direction
of the member af and the point a remain fixed.

According to the method outlined in Art. 32, a displacement diagram is constructed

KINETIC THEORY OF ENGINEERING STRUCTURES CHAP. VI

x DOUBLE *C/UE. /

/(d)\

OF DISPLACE MCNTS.

i , , , i , . . , i ,,,,11

o i 2!

FIG. 34A.

ART. 34 DISTORTION OF A STATICALLY DETERMINATE FRAME J3

with pole at 0. Since a is now supposed to remain fixed, the point a' coincides with
0, and the displacements of the various points 6, c, d, e and / are obtained by scaling
the distances Ob 7 , Oc r , Od' . . .Of.

In reality, however, the direction of the member uf does not remain fixed, for the
member revolves about the point a. Therefore, the displacements just found must
be combined with those displacements 06", Oc", Od", etc., which are caused by
revolving the rigid frame, or truss, abge about the point a until the point e shows a
resulting displacement, e'e" ', parallel to the direction of motion of the roller bed at e.

In other words, the resultant movement of the point e will be horizontal when
the roller bed is horizontal. If the bed were inclined, as em, diagram a, then the point
e would move parallel to em.

The figure a"b"c"d"e"f"g"h", similar to the figure of the truss, can then be con-
structed, making the sides respectively perpendicular to the members of the truss.
This figure is definite and can be drawn when the points a" and e" are found. The
point a" will coincide with a' or 0, since it remains fixed and a"e" will be perpendicular
to ae. Also e'e" will be parallel to the roller bed, which is horizontal in the present
case, but may be in any direction, as e'm" ', for a skew-back.

The actual displacements of the points 6, c, d, etc., will then be given in direction
and in amount by the distances &"&', c"c', d"d f , etc., but the horizontal and vertical
projections of these displacements are more generally desired and may be scaled from
the drawing.

The deflection polygon of the bottom chord is found graphically by projecting
the points a', a", 6', b", etc., on to the verticals through the panel points a, 6, etc.
The points a", 6", c", etc., will be projected in a'", 6"', c'", etc., and will form a straight
line a'"e'" . The points 6', c' , d' , etc., will fall in 6 , c , d , etc., and the ordinates 6 6"',
C(,c'" , etc., will give the vertical deflections of the panel points 6, c, etc.

In diagram c the direction cb is assumed and the point c is fixed. All that has
been said regarding the first solution applies here. It will be seen that this solution
gives exactly the same displacements as previously found, while it occupies only about
half the space of the first diagram because the member be has a lesser angular motion
than the member af.

The third solution, diagram d, is the simplest of all, and its diagram covers the
least area; for the member eg, which is now assumed as fixed, j-eally has no angular
motion, but simply drops vertically. The relative displacement b'g' of any two points,
6 and g, may be scaled off directly. Although this displacement diagram was drawn
on the assumption that the line eg and the point c remained fixed, nevertheless any
two points may be directly compared. Hence, any point may be chosen as the origin
of coordinates from which to scale the horizontal and vertical displacements of all other
points relatively to this origin. Naturally the displacements desired would be those
with respect to the point a, for this is the fixed point, giving for the point e a horizontal
movement from a' to e'. All other points move to the right and down from their
original positions by amounts which may be scaled from the diagram, taking a' for the
origin.

Hence it may be concluded that if a truss contains a member which will move

KINETIC THEORY OF ENGINEERING STRUCTURES

CHAP. VI

parallel to itself, or which has no angular motion, then this member is the best one to
choose as the fixed member in constructing the displacement diagram. The rotation
displacements are thereby eliminated.

(2) A French roof truss. Fig. 34s. In this example the stresses were found
from a Maxwell diagram, and the corresponding changes in the lengths of the members
were obtained by means of the formula Jl=Sl/EF, taking E at 25,600,000 pounds per
square inch for wrought iron. Changes of length, due to temperature, are neglected.
The values of Al being very small, ten times these values were taken. All the data
necessary in the solution of this truss are given in the following table:

TABLE OF THE VALUES OF S, F, I, AND 10JZ IN FIG. 34n

Member.

Stress,
Lbs.

Area,
Sq. in.
F

Length,
Inches,
I

10UO,
Inches.

Member.

Stress,
Lbs.
S

Area,

Sq. in.
F

Length,
inches.
I

10(40,
Inches.

-21,800

6.82

83.53

-0.106

-20,970

6.82

83.53

-0.102

-20.040

6.82

83.53

-0.094

-19,210

6.82

83 . 53

-0.091

- 18,300

6.82

83.53

-0.087

-17,470

6.82

83.53

-0.083

-16.540

6.82

83.53

-0.079

-15,710

6.82

83.53

-0.075

+ 19,730

2.02

96.53

+ 0.370

+ 15,330

2.02

96.53

+ 0.288

+ 16,430

2.02

96.53

+ 0.307

+ 13,570

2.02

96.53

+ 0.256

+ 9,040

1.86

100 . 08

+ 0.189

+ 9,260

1.86

96.53

+ 0.189

+ 6,200

1.86

96.53

+ 0.126

+ 12,560

1.86

96.53

+ 0.256

+ 7,960

1.86

96 . 53

+ 0.161

- 3,300

1.40

48.07

-0.043

- 1,680

1.40

48.07

-0.024

+ 3,300

0.78

96.53

+ 0.162

+ 1,680

0.78

96.53

+ 0.087

- 6,600

2.48

96.53

-0.102

- 3,520

2.48

96.53

-0.055

+ 3,300

0.78

96.53

+ 0.162

+ 1,680

0.78

96.53

+ 0.087

- 3.300

1.40

48.07

-0.043

28^

- 1,680

1.40

48.07

-0.024

The truss is composed of the two frames aeh and sqh, designated as I and II.
These are connected by the member as and by the pin at h.

The displacement diagram of frame I is first drawn, assuming as fixed the direction
of any member, as ab, and the position of a point, as a, of this member. The point a'
then coincides with the pole 0, and the displacement Ob' , of the point b will be equal
to J12. The points c' , d', e*, f, g' and h' are then found as directed in Art. 32.

The displacement diagram of frame II is next drawn, assuming as fixed the direction
of the member sk and the position of the point s.

Having thus determined the points k' , I', m', q' , ri ', p' and h' in diagram II, the
relative changes in the positions of the points h, s and q, parallel to the straight lines
joiningjthe points h and s, h and q, and s and q, may be found. These changes are called
Ahs, Ahq and Asq.

Diagram I may now be completed by inserting the values Ahs, Ahq and Asq previously
found from diagram II. The point s' is found from Al and Ahs, and the point q' is
found from Ahq and Asq. Since q moves on a horizontal roller bed and since e is fixed,
the figure c"d"b"g"h"f"a"s"q" can be drawn as in the preceding problem. This figure
gives the displacements of all the points of frame I, and those of the points s and q.

ART. 34 DISTORTION OF A STATICALLY DETERMINATE FRAME

The displacement diagram of frame II may now be completed by transferring the
displacements of the points h and q from diagram I into diagram II, thus determiniaf

Lj_
loo

\ DIAGRAM I.

\ N

DISPLACEMENTS.

DIAGRAM H.

too zoo

FIG. 34s.

the points q" and h" from which the figure q"l"s"ri'h"m"k"p" can be drawn.

As a check, it should be remembered that the line q"h" in diagram II must be

KINETIC THEORY OF ENGINEERING STRUCTURES CHAP, vi

perpendicular to the line hq, and that the displacement s's" of the point s must be the
same, both in direction and in amount, in both diagrams.

Diagram II might have been dispensed with in the present case, as the values dhs t
Ahq and Asq might have been found directly by summing the J/'s. However, the use
of diagram II is general, and it becomes necessary when the points h, p, k, m and q,
or h n and s, or /, s and q, are not in straight lines, as in the case of a curved top chord.

Measurements from diagram I show that the point g undergoes the greatest
displacement, having a horizontal movement of 1.93/10=0.193 inch to the right,
a vertical downward movement of 2.33/10 =0.233 inch, or a resultant movement of
oV 7 =3.02/10 = 0.302 inch. The horizontal movement of the point q is = q'q" =2.61/10 =

FIG. 34c.

0.261 inch. The displacements given by the diagram are here divided by 10, because,
as already stated, the changes Al were originally taken ten times too large.

(3) A three-hinged arch. Fig. 34c. It is required to draw the displacement
diagram.

Independent diagrams for each of the elastic frames, I and II, are first drawn by
assuming as fixed the direction of, and a point on, some member of each frame, as was
done in the previous problem.

The frames I and II are then regarded as rigid, and each frame is supposed to
revolve in such a way as to satisfy the conditions imposed by the supports.

The displacement diagrams are omitted and only the second part of the problem
is solved.

ART. 34 DISTORTION OF A STATICALLY DETERMINATE YRA ME 97

Supposing that the points a' and c' in Fig. 34c, diagram I, and the points b' and
c', diagram II, have been found from the corresponding displacement diagrams (not
shown) , let it be required to complete the solution by adding Mohr's rotation diagram.

The following conditions must exist between the figures a"c" and b"c" , which are
to be similar to the frames I and II respectively:

a. The displacement of the point a is zero, hence a" will coincide with a'.

b. Similarly b" will coincide with b'.

c. The line a"c", in diagram I, must be perpendicular to the line ac.

d. The line b"c" , in diagram II, must be perpendicular to the line be.

e. Both diagrams, I and II, must give the same displacement c'c" for the point c.

Hence a"c" is drawn through a', perpendicular to ac, also b"c" through b' perpen-
dicular to be. Now, in diagram I, c'n is drawn parallel to ac, intersecting a"c" in n,
and the projection of the required displacement c'c", parallel to ac, is thus obtained.
Likewise, in diagram II, c'm is drawn parallel to cb, intersecting b"c" in m, giving c'm,
the projection of c'c" parallel to cb.

Diagram I may now be completed by transferring c'm from diagram II and erecting
a perpendicular to c'm at m. This perpendicular will intersect na"c" in c", and the
figure a"c" can now be drawn by similarity with the frame I, since the members in the
two figures are respectively perpendicular.

In like manner diagram II may be completed by transferring c'n from diagram I
and drawing nc" perpendicular to c'n, thus determining c". As a check, the displace-
ments c'c" must be equal and parallel in the two diagrams.

(4) A cantilever bridge, similar in principle to the Memphis bridge, is represented
in Fig. 34D.

It is assumed, as in the two preceding problems, that separate displacement
diagrams have been made for each of the elastic frames, I, II, III, and IV, and of these,
the diagrams for I and III can be completely solved, as was done with the example in
Fig. 34A, since each is a determinate framed structure on two supports.

It is here deemed necessary only to complete the rotation diagrams of the frames
II and IV. The displacements of these frames depend on those of the points c and m
and upon that of the point s, respectively.

Let it be assumed that the points d' and n' have been determined for the elastic
frame II as in diagram II, and let it be required to find the figure d"n" which shall be
similar to frame II.

The members cd and ran are elongated by Jl and J2 respectively, and their elonga-
tions must be applied to the^points d' and n' in diagram II, giving the points c' and m'.
Now if the displacements ~c'c" and m'm" be drawn as found by diagrams I and III (not
shown), the points c" and ra" are obtained. These represent the original positions of
the points c and ra. The point n" must have originally occupied the same relative
level with ra", also d" must have been_at the same relative level with c". Lastly, the
line <Fn" must be perpendicular to dn. Hence, if the distance x, or the horizontal
distance between d' and d", can be found, then the figure d"n" becomes determinable
and the diagram can be completed. If it is granted that the point w will always be
midway between the points c and ra, this problem becomes definite, and the value x

KIXETIC THEORY OF ENGINEERING STRUCTURES

CHAP. VI

may be found from the figure, thus: x=^(e~ g +f) , which is (+) when measured to
the right.

Now assuming the displacement diagram of the frame IV drawn, and the points
v' and t' determined, as shown in diagram IV, required to find the figure v"t" .

The point v is a fixed support, hence v" must coincide with v' ; also the direction of
v"t" must be perpendicular to vt. To find t", apply J3 from t' to s', and then transfer
the displacement sV from diagram III (not shown) into diagram IV, thus giving the

FIG. 34o.

original position of s". The point t" must have dropped from the height s", and must,
therefore, be on the same horizontal line with s". Hence, t" is at the intersection of
tf't" and the horizontal through s", and the figure v"t", similar to the frame IV, can
then be drawn.

NOTE. For other problems see paper on "The Graphical Solution of the Distortion of a Framed
Structure" by the Author in the Journal Ass'n. Eng. Societies, June, 1894.

CHAPTER VII

DEFLECTION POLYGONS OF STATICALLY DETERMINATE STRUCTURES BY

ANALYTICS AND GRAPHICS

ART. 35. INTRODUCTORY

The deflection polygon for any series of panel points, either of the top or bottom
.chord, may be found from a Williot-Mohr diagram as shown in the previous chapter.

However, when the deflections, so obtained, are intended to form the basis for stress
computations, then greater accuracy is actually required than that obtainable from the
above method of solution. This is especially true of large structures, because the error
committed by substituting the tangent of an arc for the arc itself is of a cumulative
nature, always giving excessive values. This fact has been overlooked by most authors
treating of indeterminate structures.

Professor Mohr, in 1S75, established a very important relation between deflection
polygons and equilibrium polygons drawn for simultaneous cases of loading. Accordingly
a deflection polygon may be derived from a moment diagram by dividing the ordinates
of the latter by a certain constant which depends on the geometric shape of the structure.

The elastic deformation in the angle included between two successive chord members
can always be expressed in terms of the coexisting changes in the lengths of the truss
members or in terms of the stresses in these members for any simultaneous case of
loading. Such an angular change may be regarded as an elastic load w, which designation
will always be used in the following chapters.

The several elastic loads w for all pin points of an entire chord will then suffice
to determine the deflection polygon for that chord precisely as the external loads P will
determine an equilibrium polygon from which moments and stresses in the chords may
be found.

Naturally such a deflection polygon will furnish displacements in one fixed direction
only, and this is usually all that is required. However, two such polygons, for horizontal
and vertical deflections, respectively, would, if drawn for the same pin points and same
loading, determine actual displacements.

Since the total deflection of a truss is almost entirely due to the deformation in the
chords, it is often admissible to neglect the web system. For arches with parallel chords
this is always true.

The deflection polygon is usually drawn for the loaded chord only. If the loaded
chord is straight, as is often the case, then the computations are greatly simplified.

In the following, two methods will be given for finding such deflection polygons.
In the first method the elastic loads w are derived from the changes Al of all members

100 KINETIC THEORY OF ENGINEERING STRUCTURES CHAP, vn

of the structure, according to Professor Mohr. The second method is a modification of
the first, given by Professor Land, in 1887. and expresses the w loads in terms of the unit
stresses of the several truss members.

ART. 36. FIRST METHOD, ACCORDING TO PROFESSOR MOHR

(a) Influence on the deflections due to chord members. For any particular case of
loading, the stresses and changes in the lengths of the several members are supposed
to be known. Neglecting the web members for the present, by regarding their changes
Jl all zero, the influence of a single chord member is first considered. See Fig. 36A.

The deflection y m of any lower chord point m, due to a shortening in the upper
chord member opposite the point m, may be found from Mohr's work equation. Let
a unit force act downward from m (if vertical deflections are desired) . Then from
Eq. (GA), l-ym^Sma^lm, when there are no abutment displacements.

Let M ma represent the static moment about the point m when only the unit load
at m is acting. This is easily found from either one of the reactions, resulting from
the unit load, multiplied by the distance from m to that reaction. Then the stress
S ma =M ma /r m as may be seen from Fig. 36A.

Now if \-y m be regarded as a moment M mw due to a load w hung at the point m,
then by proportion

W m M mw l-y

~ = Wm =

The load w, thus expressed, will be known as an elastic load.

Hence, a moment diagram drawn for the elastic load w m acting at m, will give the
ordinate M mw =y m under the point m. Laying off w m on the load line to the right and
with a pole distance H=-l, to scale of forces, drawing the moment diagram A'm'B',
then the ordinate y m represents the deflection of the point m due to a change M m in the
length of the member fg opposite the point m, because

M, = 1 * =SmM m .

mu ,

Now the polygon A'm'B' is also the influence line of deflections for the point m for
any moving load w m . Hence, the deflection y nm , at the point n, is that due to a load
w m at the point m, and so on for any pin point of the chords.

In like manner the deflection y n is obtained for a load w n =Al n /r n and the polygon
A'n'B' is then the influence line of deflections for the point n for a moving load w n .
The ordinate y mri} under the point m, is now the deflection at m for a load w n at n.
But, by Maxwell's law y^ n = y nm .

By drawing these influence lines successively for all the pin points of the "top and
bottom chords, the several partial effects of all the elastic loads w on any particular
point as m may be found, and the summation of these effects 2?/ mn will be the total
deflection d m for the point m resulting from the given case of actual loading.

The summation, however, is more easily performed by combining the several loads
w into a load line 21 w and with a pole #=!, drawing the force and equilibrium polygons.

ART. 36 DEFLECTION POLYGONS BY ANALYTICS AND GRAPHICS

101

The latter is a moment polygon and also a summation deflection polygon, any ordinate
d m of which represents the actual deflection of the corresponding pin point m for that
particular case of loading and is measured to the scale of lengths. This does not
include the effect of the web members.

If the external loads are applied to the lower chord pin points then the true deflec-
tion polygon for the lower chord is the inscribed dotted polygon A'c"m"n"B' ', because
the deflection polygon between the successive floor beams must be a straight line.

When the loads w are multiplied by E and the scale of lengths is chosen 1 : a, then
by making the pole distance H =E to the scale of forces, the deflections will have the
scale 1 : a. When the deflections are. desired TO times actual to the scale I: a, then the
pole should be chosen E/m to the scale of forces.

FIG. 36A.

As is seen from the above, the elastic loads w are ratios; that is, they have no
dimension but depend on the unit stresses in the members and the geometric shape of
the structure. From this and the real value w = M/r, it follows that these elastic loads
may also be defined as tangents of the angular changes in the angles 6, or simply
w m =J6 m from Fig. 36A.

For a top chord point, the angular change Ad is positive, while for a lower chord
point it is negative, but in either case the angular change produces downward deflection.

For a top chord point

M . Al _ . Mp

and

'= . hence

and for a bottom chord point
-M=S and

S=-

hence again

102

KINETIC THEORY OF ENGINEERING STRUCTURES CHAP.VII

Therefore, for any chord point of a simple truss

^Jl m = M m p m= M,

'' r m r 2 = +IF;

(36s)

which is the fundamental equation for the elastic loads w, due to chord members only.

(b) Influence on the deflections due to web members. In considering the effect of the
web members, the influence of a single diagonal is first analyzed for the case when the
center of moments falls outside the span, Fig. 36s. The stress S a changes sign while

to-

FIG. 36B.

the moving load passes through the panel un, hence it is necessary to consider whether
the top or bottom chord is loaded as this determines the load divide.

Let mn be the member in question and the lower chord carries the roadbed. Required
to find the deflections of all the lower pin points for any given system of loads. See
Fig. 36s.

The influence line for the deflection of a lower chord point is determined when the
deflection for that point is known. Having this influence line, all the other required
deflections are easily found. This will now be shown.

The deflection y c of any chord point c, as previously found for a unit load at c, is
l-y c =S a Jl for which is found an elastic load w c = Al/r, wherein I and r refer to the
member mn with center of moments at o for a section pq.

ART. 30 DEFLECTION POLYGONS BY ANALYTICS AND GRAPHICS 103

This elastic load produces reactions A a and B a and, as previously shown, the moment
M wo (f r a U tne f rces on one s id e f tne se tion ail d f r tne center of moments o) is
found to be M WO =S a M = l-y c -

Hence the influence line for the moment M wo , for_a moving load w = Al/r, may be
found. By laying off T/^cV 7 as an ordinate from a'b' under the point c to any con-
venient scale, and drawing a line oV 7 to intersect the vertical through o in the point o',
the two limiting lines of the influence area are determined. Thus, a'u'o' represents the
influence of B a and o r b r a/' f thejnfluence of A a . Between the panel points n and u the
influence line is a straight line n'u'.

Also, for any other position of this elastic load the same influence line will be found.
Hence the most suitable point of application for this elastic load will be the load divide
i' for the panel whose section is under consideration. This would make any line through
the load divide i' an influence line for the elastic load w. Such a line would intersect
the verticals through the two adjacent panel points in points n' and u' and the two
limiting lines ~rW and oV must be made to intersect in the point o' which is any con-
venient point in the vertical through o. These two limiting lines cut off the closing line
tfV on the verticals through A and B. Then any load w=M/r, applied at the load
divide i, will produce a zero deflection in the corresponding point c and a zero moment
M wo about the point o.

Now the elastic load w may be replaced by two such loads w n and w u , acting at the
points n and u, respectively. These may be determined graphically by drawing the
force polygon Fig. 36B, corresponding to the moment diagram a'u'n'V.

The analytic method for finding w n and w u is as follows: The position of w is
evidently immaterial, because the resulting loads w n and w u will be the same for any
position of w. If the load divide is chosen as the point of application, then the moment
M WO =Q and the sum of the moments produced by the loads w n and w u must likewise
belero. Hence, the resultant of the two components must pass through the center

of moments o.

Also, by drawing a line nh\\um, Fig. 36s, and remembering that the resultant o
w n and w u must equal w, then by taking moments about 71

Ai _ M no M , no r

w u Xun=--no or w u = -'==- because ==-
r r un ' un u

wo r .
Similarly by taking moments about u, and noting that j-- , t

jl . Al uo M mo _M

w n Xun=-uo or > = .-= = =---.
r r un r mh r

The two new elastic loads, which were substituted for w, may now be evaluated
from the following equations, without involving the lever arm r, thus:

Wu =* and w n (36c)

104 KINETIC THEORY OF ENGINEERING STRUCTURES CHAP. Yii

In most cases this simple relation proves of valuable assistance, because r is
frequently outside the limits of the drawing.

Regarding the signs of the two elastic loads iv u and w n it is discerned from the above
that the signs must always be opposite and it is necessary to distinguish which of the
two is the positive load and then call the other one negative. The sign of Al alone
determines this without regard to the sign of the work S a Al and Professor Mohr gives
the following simple rule, which will always identify the positive elastic load w.

Calling all top chord members negative and all bottom chord members positive and
giving the proper sign to Al for the web member in question, then the positive w is found on
that side of the section or panel where the sign of Al coincides with that of the adjacent chord.

The same result is obtained from the force polygon, Fig. 36s, drawn for the three
loads w. This is also seen when the distortion in the quadrilateral ingun, due to the
change Al in the member mn, is considered. Assuming the top chord mg immovable,
then Al in mn will cause the point u to drop, hence w u is positive when the bottom
chord is the loaded chord. Were the top chord the loaded chord, then nn would be
considered immovable in applying this reasoning.

When the center of moments o falls inside the span there is no load divide, whence
the deflection influence line for a web member can have only a positive area. The
maximum stress is then due to the total span loaded, the same as for maximum moments.

However, this in no wise affects the previous discussion, but some proof of the
identity of the effects of the substitute loads w u and w n with that of w must be furnished
for this case.

Thus the elastic load w=Al/r, acting at c, is in equilibrium with the reactions A
and B resulting from w, and it is also the resultant of the two substitute loads w u and
w n . Hence the sum of the moments of the forces, A, B, w u and w n about any point o
in the plane of forces, must be zero. But the resultant w of w u and w n , always passes
through the center of moments o, hence this resultant must be equal and opposite to
the resultant of A and B. The method is thus general and applies equally to cases of
o inside and outside the span.

The elastic loads w. u and w n may also be expressed in terms of the moments of the
external forces, as was previously done for the chords, and the following general expres-
sions are thus obtained:

Al Mp Al Mp

w u = = and w n = = =F--;

. . . . . . . (36n)

U ^ r n r n r '~ '*

also

Al , Al Al Mp . 1,11

w ~= -*- = -f> hence -=- T - ...... (36 E )

v 'n ' > it r n

This condition is easily seen directly from Fig. 36B.

(c) The deflection polygon for the loaded chord. This deflection polygon may be
found by applying the formulae (36B) and (36o) to any given frame. The elastic loads
w are now computed for each chord and each web member and all those acting at the
same pin point are algebraically added for total effects.

The deflection polygon is then found by combining the loads w into a force polygon
and drawing the moment or equilibrium polygon exactly as shown in Fig. 36A, following

ART. 36 DEFLECTION POLYGONS BY ANALYTICS AND GRAPHICS 105

the description relating thereto. These moments may also be computed to obtain the
deflections analytically.

Before concluding this subject a few special cases will be illustrated.

(1) When there is no end post, as in Fig. 36B, then Hitter's section cuts only two
members 'ad and ac and some doubt may exist regarding the application of the formula

for w.

To overcome this difficulty, add a member ak\\cd and of any_convement length.
Then the center of moments o l will be the_center for the member ad treated as a web
member. The two imaginary members ak and kd are assumed rigid and Eqs. (36D>
again apply. The load w thus found for the point a will have no effect, and the result
is the same as would be found by applying Eq. (36n) to the center of moments at c

with the lever r\.

(2) When there is a vertical end post and the deflection polygon for the lower chord
is required, then a load applied at a, Fig. 36c, will not produce stress in the verticals
nor end post. The end chord section, and these members, therefore, do not influence
the deflection polygon of the lower chord.

FIG. 36c.

If the load is applied at the upper point c, then it is best to construct the deflection
polvgon by regarding the Al=Q for the two end posts. Now since the whole top chord
is lowered by amounts -M and -M, of these vertical end posts, it is merely necessary
to raise the closing line of the deflection polygon by these amounts at the two respective
ends. The deflection ordinates are thus corrected by the addition of the end pos

(3) Frames with vertical posts, as in Figs. 36c and 36D.

When the center of moments cannot be ascertained by a section through three
members, as for the member w', Fig. 36D, the elastic load for the point a' cannot

directly found. ,,

To overcome this difficulty assume the disposition in Fig. 36E, where the small

chord dl is interposed at a' and the vertical aa' is split into two posts such that the

deflection remains the same as for the original case. Then the original solution is again

applicable by passing two sections as indicated and cutting three members by each section

The vertical post is now made up of two substitute posts for which the work mu*

be equal to that of the original post, thus:

SI \

4-2-1

^2l EF I

10G

KINETIC THEORY OF ENGINEERING STRUCTURES CHAP, vn

Also for each substitute there will be a load w active at the center of moments o
and the resultant of these two loads will be w a , Fig. 860.

Hence,

Jl Al

In the ordinary case, Fig. 36c, the deflection polygons for top and bottom chords
are alike when the web system is neglected. When the latter is included it is best to
construct the deflection polygon of the chord which ends in the supports (the lower
chord in this instance) and the deflection of the other chord is then found by simply
correcting the first polygon by the Al of the verticals, being careful to regard the sign.

(4) Trusses with parallel chords become very much simplified because the height
of the truss and the inclination of the web members are then uniform, Fig. 36F.

FIG. 36F.

Hence for all chord members, Eq. (36s) gives

(36F)

For the web members, the center of moments is at infinity, making the elastic loads
equal and opposite for the same section. Each pair thus constitutes a moment which
is balanced by an infinitesimally small force acting at an infinite distance. For this
and other reasons it is better to determine the elastic loads for the web members directly
from the shear. Calling the shear Q, Eq. (36o) then gives

(36o)

Here r u =r n for each panel and <f> is the angle of the diagonal with the vertical.
The figure shows the case where the w loads are desired for the bottom chord panel points.

When <j6 is constant for all web members then the factor l/Er becomes constant
and may be applied to the scale of the elastic loads w, which are then l/Er times natural
size. Thus by making the pole distance Er/l times larger, the resulting deflection
polygon for web members will be unchanged.

(5) Composite structures, like three-hinged arches and cantilever systems, may
likewise be treated by the above method.

111

Ql ]

r u

EFr u

EFr u cos (f)

T w n

r- n

EFr n

EFr n cos <

ART. 87 DEFLECTION POLYGONS BY ANALYTICS AND GRAPHICS

107

For any positon of the moving elastic load w and any panel, the same work equation
applies and also Eq. (36e) .

Hence, with due regard to signs, the influence area of a deflection y is again the
influence area for the moment M^ and also the influence area of the particular member
under consideration. It is only necessary to observe the signs of the loads w and the
corresponding position of the closing line of the deflection polygon.

The w loads which result from the chord members will have the same sign as the
moment influence line for that panel. When the moment changes sign in any par-
ticular panel, as might be the case in cantilever systems, then the w loads of the adjacent
pin points must be of opposite signs.

The w load resulting from the diagonals will always be of opposite signs according
to the rule above given, no matter whether the center of moments for such diagonals
falls inside or outside the span.

(6) For indeterminate systems the foregoing method is made applicable by removing
the redundant members or reaction conditions and replacing these by external forces
X acting on the principal system in the manner described in Art. 7.

The deflection of an indeterminate system under certain loads P is the same as the
deflection of the statically determinate frame subjected to the loads P and X and the
work Eq. (GA) is in like manner applicable to this principal system, so loaded.

An example showing the application of the above method may be found in Table 50s.

ART. 37. SECOND METHOD, ACCORDING TO PROFESSOR LAND

(a) The w loads in terms of the unit stresses in the members.

This method is sometimes preferable to the former when the deflection polygon of
any succession of members is desired without regard to the remaining members of the
structure. The first method is, however, preferable when a more general solution is
required.

Here, any succession of members or rods is treated purely as a kinematic chain "
involving only the angular changes occurring
in the angles included between successive
members and the changes Al in the lengths of these
members.

The problem resolves itself into two parts:
(a) to find the changes in the angles of any
triangle resulting from changes M in the lengths
of the sides, and (6) to evaluate the loads w in
terms of these changes and finally to construct the
deflection polygon.

(b) To find the changes in the angles of a
triangle.

Let a, /? and- 7- be the three angles of a triangle ABC and/ a ,/ 6 and/ c the unit stresses
respectively in the three members opposite these angles, as given in Fig. 37A. Also let

108 KINETIC THEORY OF ENGINEERING STRUCTURES CHAP. VII

Ja, AB and Af be the changes in the three angles resulting from the changes Jl a , Jl b
and Al c in the lengths of the three sides.

Then from the figure, sin a\ =x/l c and sin 2 =y/lb, and by differentiation and
treating all quantities as variables,

cos

l c AxxAl c l b Jy-yJl b

and cos a^Ja<>= -.

r * /.*

Adding these expressions and solving for Aa\ + J2 = J gives

Jxx

Jl b

/ 7

?- (37A)

But J c cos i ={, cos 2 =/", hence Eq. (3?A) becomes

.. (37B)

ry b
Now from Fig. 3?A,

Jx_Jy_f f Mc_fc. Mb_fb. ^_ cot/? . and y._ cotr
T~Y F J Z c ~' Z 6 ~' r~ *<*> r~ tr '

which values substituted into Eq. (3?B) give

^a = (/ a -/ 6 )cot r + (/ a -/ c )cot^. . _. , "."- \ . . (37c)

The values EJ{3 and EAj- can be derived in the same way to obtain the following:
E Aa - (f a -f b ) cot r - (f c -f a ) cot B 1

= (f b -f c ] cota-(/ a -/ 6 ) cot r \. . ."../. . (37n)
= (f e -fJ cot B-(f h -ft cot a

(3?E)

wherein tensile stresses / are positive and compressive stresses are negative.

Since the sum of the three angles of a triangle is always constant, therefore,

whence the third value is easily found after having computed any two from Eqs. (37n).

The above formula are extensively used in Art. 61 dealing with secondary stresses.

When temperature changes are to be included then the unit stresses / must be
corrected by dE. A positive t produces elongation in all members, hence dE must
then be positively applied to all values/. The contrary is true for t.

The change in any peripheral angle ^ of a frame is easily found by summing the
changes Aa occurring in the several angles a whose apices meet in that angle 0. Then
^ = Sa and J^ = 2Ja, making

EI>Aa ...... .... (37F)

ART. 37 DEFLECTION POLYGONS BY ANALYTICS AND GRAPHICS

109

using the minus sign when 0=360 Ha which is the case when the vertices of the
angles a are on the opposite side of the kinematic chain from the peripheral angle <p.
The values EJa are obtained from Eqs. (3?D).

The products of the form (f a fb) cot 7- in Eq. (37c) may be graphically found
from a large scale truss diagram asjndicated in Fig. 3?A. The quantity (f a fb) is laid
off perpendicular to either AC or BC so as to include the angle ?. The base, to scale
of forces /,' will then be the desired quantity. The sign is easily found by inspection
of the given data and depends on the sign of (f a f b ) and the sign of the cot f.

(c) To evaluate the elastic loads w in terms of the angle changes J and the changes
M in the lengths of the members, and finally to construct the deflection polygon for the
kinematic chain.

In Fig. 37 R such a chain of chord members is shown and the J</ and Al are now
supposed to be given. The system is referred to coordinate axes (x, y).

FIG. 37s.

Now let /l=the angle which any member makes with a line parallel to the x axis
and through the right-hand end of the member. See Fig. 37s regarding the sign of A
for different members.

OA, 81, 3 2 are the displacements of the points A, 1, 2, etc., respectively, measured
from the closing line A'B' and parallel to the given y axis.

w\, w 2 , w 3 are the elastic loads, the moment polygon for which represents the

' These loads are now to be found

The

deflection polygon A'l'2'3' with closing line A'B'.

and then the polygon is easily constructed as for the first method above given,
loads w, parallel to the y axis, are applied, as before, in the direction in which the
deflections are to be measured.

In the figure rF|| 'A T B 7 \\Wc, hence it follows that ce^(o 2 diK- and ac = d 2 d 3
and by addition of these equations, ae is obtained as

)-T + (^2 03)-
d 2

(37o)

1 10 KINETIC THEORY OF ENGINEERING STRUCTURES CHAP. VII

But since the sides of_the deflection polygon are respectively parallel to the rays

y-f f> nij j.

of the force polygon, then -r-=-fr. This value of ae inserted in Eq. (37c) gives

ds ti

Also from Fig. 37s,

.Vi 2/2 =fe sin 4 :.

The differential of Eq. (37j) gives

Ji/i Ay 2 =A1 2 sin X 2 +1 2 cos k 2 d& 2 =i d 2 (37iv)

which must equal o\ o 2 because 'both expressions represent the deflection of the point
2 with respect to the point 1.

Also from the figure d 2 =1 2 cos X 2 which divided into Eq. (37K) gives

and in like manner is found for + ^

-- -j =-;- tan
"3 h

Making H = 1 and substituting these values into Eqs. (37n) then

~ tan A 2 +-r-^ tan A 3 ....... (37L)

But 180->(2 + >l3 = 2 then -

Also -r^=Tf and -7-^=^, whence Eq. (37i.) becomes
1 2 & /a Jb

f 2 t&n A 2 +/ 3 tan 4 ....... (3?M)

The general expression for any pin point n is then

Ew n = E4 tyn -f n tan X n +/ + 1 tan 4 . + 1 ..... (3?N)

in which Ed<f> n is given by Eq. (37r) for top chords, using the sign for bottom chords.

For any negative angle ^ (see Fig. 37s) tan A also becomes negative. Similarly
the signs of the unit stresses / must be considered in substituting values in Eqs. (37r)
and (37N). The equations as they stand are written for positive values in all cases.

// the elastic loads w are multiplied by E and the force polygon be constructed to any

ART. 37 DEFLECTION POLYGONS BY ANALYTICS AND GRAPHICS 111

convenient scale of forces with a pole H=E to scale of forces, when the scale of the truss
diagram is l:a, then the deflections will be actual to the scale I: a. But if the deflections
should appear m times actual when measured with the scale 1 : a, then the pole distance must
be made E/m to the scale of forces.

Finally the closing line A 'B' is found from the reaction displacements. If these
are zero, then dA and d B are zero and the closing line must join the ends of the equilibrium
polygon. Otherwise $A and d B must be ascertained from the conditions of the supports.

When the web system does not include vertical members then it is preferable to
choose these web members for the kinematic chain rather than a chain of chords, because
the deflection polygon will then furnish the deflections of all the pin points instead of
only those belonging to one chord.

When the deflections in one direction only are required, then the above polygon
affords a complete solution, but for a general determination of displacements in the
plane of the truss, a second deflection polygon would be required, giving deflections
perpendicular to the direction chosen for the first polygon. In the above illustration
this would be parallel to the x axis and the resultant displacement of any point would
be the resultant of the horizontal and vertical displacements of that point.

The displacements in the x direction could be computed from the same Eq. (37isr)
by substituting cot X for +tan A in each case, but a better solution is given below.

The method fails when any angle ^=90, hence no member of the kinematic chain
should be parallel to the direction of the deflections. If this occurs then a pair of
substitute members must be employed as was done in Fig. 36B. For this reason also,
the w loads from Eq. (37x) cannot usually be employed for finding horizontal displace-
ments as per Art. 38.

For a straight chord, the Eqs. (37F) and (37*) give the following simple formula
for w n

......... (37o)

Example. The above method will now be illustrated by the following example
of a two-hinged framed arch as shown in Fig. 37c. The same problem is also solved
in Art. 50, using the method given in Art. 36.

The unit stresses and angle functions appear in the truss diagram of Fig. 37c arid
the nomenclature used in Eqs. (37o) as indicated on a small triangle, is successively
applied to all the triangles of the frame.

The deflection polygons are drawn first for the bottom chord and then for the top
chord, both for the conventional loading X a =l, and the computations of the Ew loads
are given in detail for the bottom chord only, because these represent the general case
necessitating the use of Eqs. (37D), Eq. (37r) and Eq. (3?N). For the top chord, the
computations do not require the use of Eq. (3?N) and the resulting values of the loads
Ew only are given without the details of the figuring.

In the present problem, the method would not apply to the kinematic chain
f/0^1^1^2^2, etc., because the verticals are parallel to the direction of the vertical
deflections. Hence, the chords are treated separately, thus avoiding all vertical members
in the chains.

112

KINETIC THEORY OF ENGINEERING STRUCTURES CHAP. VI]

15*

15'
U f g -O.OI07

3 The ^ ig" r "bo'v* +he members reprlcserrl unit SfreSSesfl

Kip.

BOTTOM CHORD DEFLECTION POLYGON

All deflections af*- times actual 'in f>et, when meaeUred +

scale of lengths . The figures in ft. 'are E time* actual

I I

I
I
I

LENGTHS'

DEFLECTIONS.

FIG. 37c.

AKT. 157 DEFLECTION POLYGONS BY ANALYTICS AND GRAPHICS 113

The elastic load Ew Q for the bottom chord has no effect on the deflection of this
chord and is, therefore, neglected.

Computation of the elastic loads Ew for the bottom chord panel points.

For panel point L i :

By Eqs. (3?D), #Ji =(-0.0257 -0.0503)3.059 - (0.0255+0.0257)0.454 = -0.2557

=0.0 (0.0503+0.0107)0.890 = -0.0543

= (0.0625-0.0347)0.1125 -(-0.0342-0.0625)1.394 ==+0.1379

ByEq. (37F), -EAfa =EI>Aa = -0.1721
By Eq. (37N), Ewi =0.1721 -0.0255x0.660+0.0347X0.5596= +0.175

For panel point L 2 :

By Eqs. (37D), EActi =(-0.0342-0.0625)1.394 - (0.0347+0.0342)0.5596= -0.1733

EAa% =0.0 (0.0625 +0.0352)0.7173 = -0.0701

EJci 3 = (0.0791-0.0493)0.4192 -(-0.0498-0.0791)0.963 ==+0.1366

By Eq. (37r), -EJfa =EI>Aa = -0.1068
By Eq. (37N), Ew 2 =0.1068 -0.0347X0.5596 +0.0493x0.4312= +0.109

For panel point L :; :

By Eqs. (37D>, EActi= (-0.0498 -0.0791)0.963 - (0.0493+0.0498)0.431 =-0.1668

#J 2 =0.0 (0.0791+0.0736)1.038 =-0.1585

= (0.1065-0.0687)0.821 -(-0.0657-0.1065)0.649 ==+0.1428

By Eq. (37r), -EAfa =EI>Aa = -0.1825
By Eq. (37N), Ew 3 =0.1825-0.0493 X0.4312 +0.0687 X0.3147 = +0.183

For panel point L 4 :

By Eqs. (37o), EAon =(-0.0657-0.1065)0.6487- (0.0687+0.0657)0.3147 = -0.1540
=0.0 (0.1065+0.1000)1.542 =-0.3184

= (0.0889-0.0990)1.409 -(-0.0579-0.0889)0.462 ==+0.0536

By Eq. (37r), -EA$*=ET>Aa = -0.4188
ByEq. (37N), Ew =0.4188 -0.0687X0.3147 +0.0990X0.1865= +0.416

For panel point L- :

By Eqs. (37D), EAa^ =(-0.0579-0.0889)0.462 - (0.0990+0.0579)0.1865= -0.0971

EJa 2 =0.0 (0.0889+0.1081)2.164 = -0.4263

= (0.0541-0.1207)2.111 -(-0.0270-0.0541)0.400 -0.1082

By Eq. (37r), -EA^=E^Aa = -0.6316
ByEq. (37ar), Ew 5 =0.6316 -0.0990X0. 1865 +0.1207 X0.0621 = +0.621

For panel point L, ; :

By Eqs. (37D),jBJi= (-0.0270 -0.0541)0.400 - (0.1207+0.0270)0.062 =-0.0416
JJa 2 ==0.0 (0.0541+0.1290)2.500 = -0.4578

By Eq. (37p), -i^J0 6 =i^SJa= -0.4994
ByEq. (37u), ^Ewe =0.4994 -0.1207 X0.0621 = +0.492

114 KINETIC THEORY OF ENGINEERING STRUCTURES CHAP. VII

The minus sign in Eq. (37r) applies here because the angles a are on the side of
the chord opposite to the angle <p.

The Ew elastic loads are now combined into a force polygon and the equilibrium
polygon drawn for these loads acting through the several lower chord panel points will
then represent the deflection polygon for the lower chord.

If the pole distance H were made equal to E then the deflections would be actual
to the scale of lengths according to the rule above given. However, this would be too
small a scale for accurate results and therefore, the pole distance was made equal to 4w
units, giving deflections E/4 times actual to the scale of lengths of the drawing. By
constructing a scale four times smaller than the scale of lengths, the deflection ordinates
E times actual may be scaled directly from the drawing, and these values are written
on the ordinates.

The same problem is solved by the graphic method in Fig. 50s and again by the
method of Art. 36 in Table 50B.

Using the Ew loads computed for the top chord panel points the top chord deflection
polygon is obtained in precisely the same manner and these loads and the deflections
found for the top chord are comparable with the values in Art. 50 just referred to.

The numerical values of the Ew loads for the top chord panel points are as follows:

#w = -0.139 Ew 2 = +0.117 Ew 4 = +0.415 %Ew 6 = +0.479

Ewi = +0.098 Ew 3 = +0.203 Ew 5 = +0.608

The details for the computation of the load Ew% are given for illustration. Thus :

By Eqs. (37o), EAcn=( 0.0625+0.0352)0.7173 -(-0.0498-0.0625)1.394= +0.2266
= ( 0.0493+0.0498)0.431 -( 0.0791 -0.0493)0.419= +0.0302
= (-0.0657 -0.0791)0.963 -0.0 -0.1394

By Eqs. (37r) and (37o) Ew 2 =EA$ 2 =#Z Jo: = +0.1174

which follows for a straight chord when the angles a and ^ are 'on the same side of the
chord.

ART. 38. HORIZONTAL DISPLACEMENTS

To find the displacements in a horizontal direction when the deflection polygon
for vertical deflections is given.

It is readily seen, from the previous description, that the elastic loads w are indepen-
dent of the direction of the deflections because in every instance their value is dependent
on the shape of the truss and the changes in the lengths of the members, regardless of
how these were produced. See Eqs. (36s) and (36c). Hence, the same elastic loads
may be employed to find deflections in any direction.

The force polygon, used for the deflection polygon for vertical deflections, will thus
be the same for any other direction of deflections when revolved through an angle
equal to the angle included between these two directions. When horizontal and vertical

ART. 38 DEFLECTION POLYGONS BY ANALYTICS AND GRAPHICS .

115

deflections are under consideration this angle is 90 and the same force polygon may
readily serve both purposes, since the rays for horizontal deflections are then perpen-
dicular to the rays drawn for the vertical deflections.

In Fig. 38A, let the succession of members A to B represent the bottom or tension
chord of some structure, fixed at A and supported by a roller bearing on a horizontal
plane at B. Also given the deflection polygon A'c'd'e'f'B' to find the deflection polygon
A"c"d"e"f"B" for horizontal deflections. The loads w are all multiplied by E and the
deflections are m times natural.

The left-hand end of the chord havirig no horizontal motion, it is most convenient
to call all horizontal deflections positive to the right.

The pin points are horizontally projected on to the closing line A"e , which latter
is perpendicular to the closing line A'B'. The polygon A"c"d"e"f"B" is then drawn

*>-dL^

FIG. 38A.

by making the sides perpendicular to the respective rays of the force polygon and
maintaining the same order of succession in the pin points formerly used in constructing
the deflection polygon for vertical deflections. The horizontal abscissa B Q B" is then
the horizontal displacement of the point B. Similarly the abscissa cfod" is the horizontal
displacement of the point d, etc.

For any horizontal member, as ef, the horizontal displacement of the point / with
respect to e must naturally be the M for that member. Hence since_the scale of deflec-
tions is m times actual to the scale of lengths the displacement e"f" =mAl to the scale
of lengths used.

If the point B is made to roll on some inclined plane ik instead of the horizontal,
then a new closing line A7% must be determined for the two deflection polygons, as
follows: draw a line 'B^' \\~ik and erect a perpendicular to AB prolonged, and passing
through B". The intersection of these two lines gives k" and the vertical ordinate 3 B
represents the vertical displacement of the support B.

116

KINETIC THEORY OF ENGINEERING STRUCTURES CHAP. VII

Hence, making B'k' =d B the closing line A'k' may be drawn. Also a line A"k J_ A'k'
becomes the required closing line for the horizontal deflection polygon.

The same solution will apply to deflections in any direction other than 90 with
the vertical. Also, when the deflection polygon for vertical deflections of all points of
a truss is given, as in Fig. 36A, then the horizontal deflections of all pin points are found
precisely as above by adhering strictly to the sequence in which these points occur on
the given deflection polygon.

The change d AB in the length of the chord AB, Fig. 38A, may easily be found by
taking the loads w parallel to AB. In this case the displacement d^B becomes the
intercept on AB produced and included between the extreme rays of the equilibrium
polygon. Hence this displacement is equal to

,ww

(38A)

wherein y is the ordinate of any pin point measured normally to A B.
This same result may also be found by computation from

(38u)

which follows for a pole distance of unity.

For the case given in Fig. 37fi, where the angle changes A$ and the changes in the
lengths of the members are given, the total effect on the length of span AB then becomes
for n members,

^ B = Si"~V0 + SrJZ cos (;-), .--.. ,>.;-. . (38c)

where a is the angle which the span AB makes with the horizontal.

Example. The lengthening d^B for the span AB due to the loading X a = l in Fig.
37o is computed from Eq. (38c) , using the values E Al as given in Fig. 50A. The value
\Ed AB was found to check the value determined graphically in Fig. 50s. The
values Ed<{> are those above computed for the bottom chord panel points and a =0
because the span is symmetric.

Point.

E4$

Feet.

EyJ<l>

E4l Fig. 50A.
Feet.

cos >l

KJl co* X

0.1721

6.188

1.0649

. 286

0.834

0.2385

0.1068

14.58

1.5571

0.597

0.873

0.5212

L 3

0.1825

21.05

3.8416

0.805

0.918

0.7390

0.4188

25.77

10.7925

1.080

0.954

1.0303

0.6316

28.568

18.0435

1.511

0.983

1.4853

= 0.4994

29.50

14 . 7323

1.814

0.998

1.8104

= 50.0319

5.8247ft.

These values substituted into Eq. (38c) give for a lengthening E times actual,
4B =iE27/^ + A#2J/cos >l= 50.03 19 +5.8247 =55.8566 ft., where -29000 kips per
sq. inch.

r. 39 DEFLECTION POLYGONS BY ANALYTICS AND GRAPHICS 117

ART. 39. DEFLECTION OF SOLID WEB BEAMS
(a) Deflection due to moments. The differential equation of the elastic curve is

dx2 = -E~r ' ' ' ( 39A >

| Eq. (36s) gives the elastic loads w, for only one chord, in terms of moment and
truss dimensions as

Mp Ml

7/1 = L =

r 2 EFr 2 '

Considering both chords, each of area F, and neglecting the web effect, which is
usually quite small, then for a unit length of girder

EFr 2 El '

Hence Eq. (39A) gives directly the elastic loads w per unit length of girder as

d 2 v M

(39s)

An equilibrium polygon drawn for these loads, with pole distance equal to unity,
gives deflections to the scale of lengths chosen for the drawing.

By treating the moments M per unit length of girder, as loads which now become
El times too large, and constructing an equilibrium polygon for these loads M with a
pole distance H=El, the same deflection polygon is again obtained, giving deflections
to the scale of lengths of the drawing. In other words the moment polygon for any
case of loading becomes the load line for the deflection polygon corresponding to such
case of loading.

Deflections are usually drawn several times actual size, in which case the pole
must be divided by as many times the scale of the drawing. Thus, if the scale of lengths
is 1 :n and the deflections should appear twice actual, then the pole H =EI/2n.

When the value / is variable, the pole distance is made to vary as a function of 7,
as illustrated in the example below. See also Art. 47.

(b) Deflection due to shear. From Eq. (15M) the deflection of a straight beam
when acted on by shearing forces only, is

sn ^f)

(39c)

Disregarding all unnecessary refinements, since shear deflections are small compared
to the total deflection, tire value /? may be taken as constant and Q will be assumed
uniformly distributed over the web plate of sectional area FI =^

118 KINETIC THEORY OF ENGINEERING STRUCTURES CHAP. VII

For a single concentrated load 'dQ/'dP m = 1 and Q =R =the end reaction, hence

3 / 30 , 8 /

Taking (? =0.3332? and Fi=F/3 and assuming an average value for ,3=2.5 for|
ordinary girder sections, then approximately

(c) Deflection due to combined shear and moments. Ordinarily it will suffice to
figure the deflection due to shear for a point at or near the point of maximum moments
and to determine the percentage which this d m is of the moment deflection ordinate
at the same point. All other moment deflection ordinates may then be increased in
the same proportion to obtain the deflection polygon for combined shear and moment
effects.

For a uniformly loaded beam or constant / and neglecting shear effect, the deflection
becomes

1 r
= EIJ

When the depth of a girder is constant, but the moment of inertia varies so as to
maintain a constant stress on the extreme fiber at all sections, then the ratio M/EI
becomes constant and the deflection for such a case would be

-if:

where I m is the moment of inertia at the section of maximum stress.

When the moment of inertia varies abruptly the deflection may be expected to
fall between the two above values, making the numerical coefficient close to 1/9.
Taking in the shear effect, the approximate formula for maximum deflection, of a beam
of variable /, becomes

% '"max^ | ""*max fon^\

"-*"' ' '

However, for general cases of loading and variable 7 the graphic solution, above
given under Art. 39A, should be employed.

(d) Example. Given a plate girder of variable section and uniformly varying water
load as shown in Fig. 39A, required to find its deflection polygon. The girder is designed
as a double web beam and normally occupies a vertical position so that there are no
dead load stresses.

These girders are spaced 9.175 ft. center to center so that each will carry a full
water load P=31.25X9.175(?-a) 2 giving rise to the end reactions A and B, and moments
M, all as indicated in the diagrams, Fig. 39A.

DEFLECTION POLYGONS BY ANALYTICS AND GRAPHICS

119

FIG. 39A.

120 KINETIC THEORY OF ENGINEERING STRUCTURES CHAP. VII

The moments of inertia were computed for the various sections on gross areas, and
are plotted in the upper diagram.

The bending moments were also computed, using the formula

(39H)

and the results were plotted in millions of foot-pounds in the second diagram.

The w loads were then taken as the moments M over certain convenient lengths
Ax, instead of unit lengths according to the formula (39s) , whence

The lengths Ax are usually chosen with respect to the girder sections, so that / is
constant over each length Ax. Where the depth of section is variable the mean value
of / is taken for each length Ax.

Since M and Ax are both expressed in feet, while E and 7 are both for inches, the
factor 12X12 = 144 is introduced into Eq. (39 j) to establish a true ratio between like
units. Also, M and E are both expressed in millions of pounds, making the unit
weight one million pounds. This furnishes the true values for w in terms of the given

data as

144MAx MAx MAx ,_. .

--' (39K)

144

wherein E =28 is chosen low rather than high.

Now the scale of lengths was taken as 1 : 120, and if the deflections are to appear

twice actual size then the pole must be made equal to

OO T

The pole distances are thus a constant function of 7 and may be found for all
values of 7.

The force polygon is now easily drawn to any convenient scale of forces, using the
same scale for the loads w and the pole distances 77. This scale has no influence on
the deflection polygon, but merely affects the size of the force polygon.

The deflection polygon is then constructed as the equilibrium polygon of the w
loads with the various pole distances, and the ordi nates included between this polygon
and the closing line A'B' will represent the deflections to twice natural scale. Had
the pole distances been taken twice as long, then the deflections would have been actual.

The deflection due to shear, at the point of maximum moments, is found from

5x537X12
Eq. (39n) as o m = = ; ^-p-=0.09S inch which is 10.5 per cent of the maxi-

mum deflection due to bending alone.

ART. 39 DEFLECTION POLYGONS BY ANALYTICS AND GRAPHICS 121

The total deflection of the girder is therefore 10.5 per cent greater than the
amounts given on the deflection polygon of Fig. 39A. The maximum deflection due
to combined shear and bending effect is thus 0.93 +0.098 = 1 .028 inches, for the point 7.

The approximate formula (39o) gives for this same point

A 5.37X12X56.5 2 X12 2 .
07 = 9X28X134000 ' +

See Art. 47 for another method of dealing with variable moment of inertia, by
treating the quantities MAx/I as elastic loads and making the pole distance equal to E.

CHAPTER VIII
DISPLACEMENT INFLUENCE LINES FOR STATICALLY DETERMINATE STRUCTURE!

ART. 40. INFLUENCE LINES FOR ELASTIC DISPLACEMENTS

Deflection influence lines could not receive adequate treatment in Chapter T\
because these depended on a knowledge of deflection polygons. The, latter were taker
up in Chapters VI and VII. The subject is now treated from the most general aspec
covering the influence line for any elastic displacement.

Proceeding from Maxwell's law, Professor Mohr, in 1875, first developed displacemen'
influence lines. The application results from a consideration of two conventiona
loadings, of unit work each, applied to any frame so that one of the loadings represent;
the desired influence at some point n, and the other loading is a vertical moving loac
P = l applied at any load point m of the loaded chord. This includes all cases of con-
ventional loadings given in Art. 9, besides the simpler problems pertaining to displace
ments of points.

As applied to any simple beam or truss, the following theorem is now established
A deflection polygon, drawn for a load unity acting in a fixed direction on some detiniti
point n of any frame, is the deflection influence line for the deflection of the point n, in thi
flxed direction for any system of parallel moving loads applied to the loaded chord. Wher
the system of moving loads does not consist of parallel loads, then, of course, no influence
line is possible.

To prove this theorem, the simple beam, Fig. 40A, is loaded with a vertical loac
P n = l applied at the point n. A deflection polygon, drawn for this case of loading
will then be the deflection influence line for the point n, according to the following
demonstration. The method of drawing the deflection polygon is given in Art. 39.

The moment diagram is first drawn for the conventional loading P n = l kip, appliec
at n and thus making the maximum ordinate under n equal to M n = l-aa'/l=4.8 kip
ft. The symbol M is used to indicate a moment due to a conventional unit load.

This diagram is divided into suitable sections (2 ft. apart) and the moment areas
MAx are computed and treated as elastic loads w, for which the deflection poly-
gon A "B" is finally drawn, all as shown in Fig. 40A.

Assume the modulus '=28000 kips per sq. inch, 7=2087, in inches, M in kip
feet, and Jo? in feet. Then the pole distance for the force polygon becomes H =7/144
when the w loads are made equal to MJx, by Eq. (39j), for deflections of actual size,
However, the scale of lengths of the drawing was made 1 :36 and the deflections should
appear 200 times actual, hence H must be made 7/200X36X144=56.35 w units.

122

DISPLACEMENT INFLUENCE LINES

123

With a pole H= 56.35 and the w loads MAx, construct the force and equilibrium
polygons as shown, and the deflection polygon so found represents the deflection influence
ine for the point n according to the following application of Maxwell's law.

The vertical deflection of any point m, for the given load P n = l, is represented by
:he vertical ordinate rj m of the deflection polygon A"B", vertically under the point m.
For, by Maxwell's law, rj m =d mn =d nm , or in words, T? OT equals the elastic displacement
:>f the point n for a unit load P m acting at the point m. Hence m being any position

|g.iKip.
yn

24-"

SOIb. I Beam , I =2O87.z in*

; I

, i- fcw

M DIAGRAM FOR P = I KIP.

Scale for len^thsland momenta.

DEFJ.ECTIO|^ INFLL|ENCE L|NE FOR; POINT n.

Deflections zoo times actual.

FIG. 40A.

Df the moving load point, it follows that all ordinates rj m represent deflections for the
point n due to a unit load at the variable point m. This makes the polygon A"B" the
iesired deflection influence line for the point n.

Therefore, for any train of moving loads, the deflection at the point n becomes

(40A)

Since Maxwell's law is generally applicable to any case of conventional loading,
whether for a single point, a pair of points or a pair of lines, as illustrated in Art. 9,

124

KINETIC THEORY OF ENGINEERING STRUCTURES CHAP. VIII

therefore, the above application affords a solution for any displacement influence line
other than those for simple vertical deflections. However, the train of loads must
consist of parallel forces P, as otherwise no influence line is possible.

ART. 41. SPECIAL APPLICATIONS OF DISPLACEMENT INFLUENCE LINES

(a) Required the deflection influence line for the point n of the simple cantilever
beam, Fig. 4lA. The deflection polygon is drawn according to the method of Art. 31)
and illustrated in Art. 40, for the conventional loading P n = l at the point n. This
then becomes the deflection influence line for the point n.

FIG. 4lA.

The beam is anchored at A and supported by a hinged bearing at R and a roller
bearing at C. The point n is the hinged connection at the end of the cantilever arm
Bn. The following moments M and reactions R, resulting from the conventional
loading P n = l, are then found:

A = --^, B=-

L\
M A =0, MK--

^ V

M n =0.

The M diagram is constructed by making the ordinate at B' equal to 1 2 , and after
dividing this diagram into suitable sections and computing the loads w\ to w 8 , the
deflection polygon is easily drawn, observing the method followed in Art. 40.

The closing line of the deflection polygon is then found to be A"B"C", which
completes the influence line. The point B" is the intersection of the deflection line

ART. 41

DISPLACEMENT INFLUENCE LINES

with the vertical through B and n"C" is a straight line. Upward deflections are negative.

The support at A is elastic and its displacement o Aa , due to the conventional
loading P n =-l, should be computed from Eq. (4A) including temperature effect if
desirable. This displacement is then applied at the point A" and the final closing
line Ai"B"C\" i-s thus made to include this effect. By applying any train of loads the
resulting actual deflection of the point n, according to Eq. (40A) becomes d n = HPi).

For any case of variable moment of inertia of the given beam the pole distance H
is made variable, as was done in Fig. 39A.

(b) Given a simple truss, Fig. 41 B, on two supports, to find the displacement influ-
ence line for any panel point n for displacements d n in the fixed direction rm 7 . The
loading is to consist of a system of vertical concentrated loads P acting on the bottom
chord.

FIG. 41s.

Apply the conventional loading P n = l in the given direction nn' and compute the
reactions H, A and B for this load, then determine t-he stress S\ and resulting change
Jl in length, for each member. Finally compute the w loads for the several panel points
and draw the deflection polygon for the loaded chord (here the bottom chord) using any
of the methods of Chapter VI, but preferably the one given in Art. 36 (c) . The direction
of the w loads is always taken parallel to the system of loads P, hence the displacement
influence ordinates jj will be vertical ordinates, measured vertically under the respective
loads P. This is necessary because the direction of the deflections is determined by
the direction assigned to the w loads.

The proof that the polygon A 'B' is the desired influence line again follows from
Maxwell's law, viz., for any ordinate rj m =d mn = d nm .

The displacement d n of the point n in the direction nn', caused by the system of
vertical loads P is then

(c) Given the three-hinged arch, Fig. 41c, to find the influence line ior the
angular rotation 3 n between the two lines An and Bn. The train loads are to be carried
bv the top chord.

126

KINETIC THEORY OF ENGINEERING STRUCTURES CHAP. VIII

The conventional loading now becomes one of loading a pair of lines with a positive
moment equal to unity applied to each line in the direction in which the angle J would
increase for vertical loads acting on the top chord.

The horizontal thrust for this conventional loading is obtained by taking moments
about the crown n and gives H = (2-{-AV)/ < 2f and the reactions A and :i are found by
taking the moments of the external forces about B and A respectively.

Then for this case of conventional loading determine the stresses S\ and changes
Al in the lengths of the members and compute the w loads for the half span, which is
all that is required for a systematic structure.

FIG. 41c.

The moment diagram, drawn for the w loads, will then be the displacement influence
line for the change d n in the angle J. The d n for any particular set of loads is then
d n =2Prj, measured in arc.

The above problems will suffice to show the perfectly general solution of displace-
ment problems afforded by the application of Maxwell's law. These problems can also
be solved analytically with the aid of Mohr's work equation as indicated in Art. G.

CHAPTER IX
INFLUENCE LINES FOR STATICALLY INDETERMINATE STRUCTURES

ART. 42. INFLUENCE LINES FOR ONE REDUNDANT CONDITION

The principle, illustrated in Chapter VIII, for drawing deflection influence lines, is
easily applied to the construction of the influence line for any external or internal
redundant condition.

Eqs. (7n) and (So) for one redundant condition and a single external force P = 1
become

a a0 aa f-

and < 42i >

"a 1 ' "ma ~ X a O aa = X a f) a

after substituting the value d a as obtained from Eq. (4A) and neglecting temperature
and abutment displacements.

This gives X a in terms of displacements or stresses, whichever is preferred, as

Y -

where the redundant may be external or internal.

If the single load P = l is vertical and applied to some panel point of the loaded
chord then Eq. (42s) represents the influence line for X a for a unit load applied at any
panel point m.

A deflection polygon, drawn for the conventional loading or condition X a = l, will
give the ordinates r} m =d ma =d am = 'ZS S a p for the displacement- of the point a, for a unit
load at any point m of the loaded chord. Hence the deflection polygon drawn for
condition X a = 1 is the X a influence line provided all the ordinates are divided by the
constant denominator d aa +p a = 'Sc?o + p a , The constant displacement d a a=2>S a 2 o is
found by computation or otherwise, and p a is a given constant depending on the length
and section of the redundant member. When X a is a reaction then ,o a =0 for the case
of immovable supports.

Accordingly the following theorem may be stated: The ordinates to the influence
line of any redundant X a are some constant factor p. 1 ~ (d^ +p a ) times the ordinates to
a deflection polygon drawn for the loaded chord, for the conventional loading X a = l (P=0),
applied to the principal system.

127

128

KINETIC THEORY OF ENGINEERING STRUCTURES

CHAP. IX

(a) When the redundant X<j is an internal condition, as illustrated in Fig. 42A, the
above theorem is applied as follows: If the direction of stress in the redundant can be
anticipated, then this should be done, otherwise an arbitrary assumption must be made
and, if it was erroneous, the resulting value of the redundant X a will turn out negative
with respect to S . In any case the conventional loading X a = l, being external work,
should be so applied as to produce positive work when the redundant member is removed.
In the illustration, the member X a will be assumed in tension so that J/ a is a positive
elongation. Hence, the unit forces X a = 1 must be applied so as to move the points n
and HI apart. The principal system is the entire frame exclusive of the member X a .

The stresses S a are computed or found from a Maxwell diagram and from these the
w loads and deflection polygon for the loaded chord (which was taken to be the bottom
chord in the illustration) are determined by one of the methods in Chapter VII.

The deflection polygon so obtained may be represented by the broken line A'n'rii'B',
with ordinates i) m =d ma =d a m, which, according to Eq. (42B) are (daa+pa) times the actual

influence line ordinates for the redundant stress X a .
system of moving loads, becomes

Hence the stress X a , for any

(42c)

(b) When the redundant Xa is an external condition, the quantity p a becomes zero
for the assumption d a =0 for immovable supports, and this gives rise to a slight simpli-
fication in the solution of problems of this class.

Eq. (42n) then becomes in general

Oaa

(42D)

The illustration of a simple beam on three supports is given, in Fig. 42B, as a case
of one external redundant condition X c . Any one of the three supports might be
treated as the redundant one, but the middle support C is here chosen.

The conventional loading X c = 1 is again chosen so as to produce a positive quantity
of applied work. The deflection polygon A"C"B" of the simple determinate beam AB,

ART. 42 INFLUENCE LINES FOR INDETERMINATE STRUCTURES

125)

which now becomes the principal system, is found for this load X c = l exactly as was
done in Art. 40, Fig. 40A. The unit load is applied at C in a downward direction
because the deflection of this point of the principal system is also downward when the
support is removed.

Any ordinate rj m of this deflection polygon is equal to the deflection d mc =3 cm , which
is cc times the X c influence line ordinate according to Eq. (42D). Also the special
ordinate f) c =d cc is given by the same deflection polygon and is a constant quantity.
Hence the factor /* for the influence line X c is equal to l/d cc and the ratio dmc/dcc
determines the value of the influence line ordinate for any point m.

FIG. 42s.

It is thus seen that for the case of a single external redundant X c the magnitude of
the moment M M =lilz/l and the pole distance H of the deflection polygon, do not affect
the influence line ordinates T im =d me /8 ee , though they do affect each of the ordinates $ mc
and S cc separately.

For any system of moving loads the required redundant reaction becomes

(42E)

130 KINETIC THEORY OF ENGINEERING STRUCTURES CHAP. IX

ART. 43. INFLUENCE LINES FOR TWO REDUNDANT CONDITIONS

When there are two redundants, then two deflection polygons may be drawn for
the loaded chord of the principal system ; one for condition X a 1 and one for condition
X\)-=\. These polygons are then combined into an influence area for X a the ordinates
to which, times a certain factor /*, will give the influence line ordinates for the X a
influence line. Another combination similarly made for X b will give ordinates which
when multiplied by a factor fib, will give the ordinates for the'Afc influence line. The
analytic solution for these influence lines is derived from Eqs. (SD).

Thus for two external redundants with d a and 05 both zero and neglecting temper-
ature and reaction displacements, Eqs. (80) give, for a single load P,

= X a O aa

........ . . (43A)

These equations solved for X a and A^ and observing that ^=^06 by Maxwell's
law, give

__
ma

* -p

_ _ bb p _i}ma r -p

a * {ta^ma^

* * Oab T)a

aa o6"F~
066

mb ma *

wherein dma and dmb are the variable ordinates to the two deflection polygons X a = 1
and Afe = l, respectively, for any panel point m. The other quantities are all constants
given by the same deflection polygons.

These equations for a load P l, are the equations for the influence lines X a and
Xb. Their numerators then give variable ordinates which may be called r) ma and i) m b
while the constant denominators ij a and y b may be used as influence factors fta = l/T) a
and fjt b = l/r)b for the two influence lines X a and Xb respectively.

For any train of moving loads the redundant reactions are expressed by

ma }

\ ........... (43c)

inb

The problem of a simple beam, Fig. 43A, continuous over four supports, is chosen
to illustrate this case.

The redundant supports are considered removed and the principal system then
becomes a simple beam on two supports. In the present case the supports X a and A"
are treated as redundants, although any two of the four reactions could be thus regarded.

ART. 43 INFLUENCE LINES FOR INDETERMINATE STRUCTURES

131

The moment diagrams for the two conventional loadings X a = l and X h =--l are now
drawn and from these the deflection polygons are constructed precisely as illustrated
in Fig. 40A. The two poles H are made equal and as a check on the drawing <? a6 'must
equal o& a . When the structure is a frame, then the deflection polygons for the loaded
chord are drawn as described in Art. 37.

The deflection polygons so obtained are the influence lines for the deflections d n

FIG. 43A.

and #6 of the two redundant supports. For immovable supports these are zero and
Eqs. (43A) apply to any case of loading.

To obtain the influence areas for the redundants, the two deflection polygons are
combined into two influence areas by computing the ordinates from Eqs. (43s) as follows:

b^
Obb

132 KINETIC THEORY OF ENGINEERING STRUCTURES CHAP. IX

The factors /*, and fj. b are the reciprocals of the denominators of Eqs. (43B) all the
values for which are given on the two deflection polygons. The ordinates rj ma give,
when plotted, the shaded area C'D', which is the influence area for X a . The ordinate
T) a , under the support X a , gives the influence factor // a = l/V In similar manner the
ordinates rj mb furnish the shaded area C"D", which is the influence area for Xb with
the factor /j. b = I/fa, obtained from the ordinate rj& under the support X b .

A purely graphic construction of the influence area X a is given by Mueller-Breslau as
follows: Draw the d a deflection polygon for the w a loads, using any convenient pole H.
Then for the w^ loads draw an equilibrium polygon passing through the three points
C', B' and D' '. This is perhaps the best and most convenient solution and is also
illustrated in Fig. 43A.

The d a and db deflection polygons are first drawn as above described with any
assumed pole distance which may be the same for each polygon. Then draw the d b
line through the three points C', B' and D', and also draw the d a line through the three
points C", A" and D". The shaded areas thus formed represent the X a and X b influence
areas.

To find the new poles 0' in the two force polygons, which are necessary in drawing
an equilibrium polygon through three given points, proceed as follows: In the d b force
polygon, draw OK\\C"D" and OKf^C^W 7 thus determining the points K and K'.
Then draw KO'\\C'D' and K'0'\\C'B' and the new point 0', thus found by the inter-
section of K'O' and KO' ', will be the required pole for drawing the db line through C',
B' and D'.

The pole 0' in the d a force polygon is found in a similar way as indicated in the
figure, using the points F and F'. The new pole 0' ' , found by the intersection of FO'
and F'O', serves to draw the d a line through the points C", A" and D".

This construction is based on the fact that the influence of a load at B on X a must
be zero, likewise for a load at A on X b .

The above methods are not practicable for more than two redundants, and the
following method of simplification is given for multiple redundancy.

ART. 44. SIMPLIFICATION OF INFLUENCE LINES FOR MULTIPLE REDUNDANCY

In many problems of this class it is possible to so choose the redundant forces
X that they will have a common point of application. Then for certain directions
of the X'8 the d coefficients bearing different subscripts and appearing as factors of
the X's in Eqs. (So) may be reduced to zero. Whenever this is possible then
v a o- ba=--0, dac=dca-0, ^bc=^^d>=0', etc. ; and the following simple equations are
obtained 1 .

9* X^ 7~> S ~V & \^ Z? Af* 1 J} 1

O b = Ztr'mOmb AfcOfcfc Ltti b ar + 0bt

(44A)

ART. 44 INFLUENCE LINES FOR INDETERMINATE STRUCTURES 133

involving only one redundant X in each equation. The same assumptions may be
applied to Eqs. (7n).

This method of analysis was first introduced by Professor Mueller-Breslau, and
serves a most valuable purpose, especially when applied to fixed arches.

As will be shown presently, this disposition of the redundants is easily made
when their number does not exceed three. Beyond this number the analysis leads
to many complications. Fortunately, the important cases of redundancy are nearly
always limited to from one to three redundants, and then there is no difficulty in
following the scheme here proposed.

The solution of Eqs. (So) for simultaneous values of the X's is thus avoided and
the other chances for serious errors are greatly lessened.

It is usually expedient to treat temperature stresses and abutment displacements
separately from the primary stresses, and then Eqs. (44A) become for load effects only
and immovable abutments,

Z2->imma -\r ^*m mb /it \

a = ^ ; x b = 5 ; etc (44s)

*!=!--; *6i=l-rJ etc (44c)

Oaa Obb

By placing 2.R a Jr=0, d a also becomes zero, likewise for S-fi^Jr and d b , a circum-
stance which should be noted in writing Eqs. (44s) and Eqs. (44c).

In all of the following investigations, the X's will represent either a single force or
a couple applied to a principal system. The d's in the first case will then represent
linear displacements and in the second case they will be angular distortions expressed
in arc. Thus, if a redundant X a is applied to the point a then ^ is the displacement
of the point a in the direction of X a for a force X a = 1. When X a is a moment or couple,
then daa becomes the angular rotation of a rigid principal system as produced by a
couple X a --=l. The displacements d a , d b , etc., and the conventional loadings X a = l',
X b = l, etc., are always positive in the opposite sense in which the redundants X a , X b ,
etc., are supposed to act.

It should also be noted that in all cases here considered, the points a and b are
coincident and these designations are retained merely to distinguish the particular forces
from each other.

The Eqs. (44A) to (44c) are equally applicable to graphic and analytic solutions,
but the latter method is useful only when there are not more than three redundant
conditions, and great accuracy is desired. The graphic method is less laborious and
leads to a more comprehensive presentation.

The above will now be applied to general cases of two and three redundants.

(a) Structures having two redundant conditions. Here the two redundant forces
can always be applied at the same point and the case can be solved by applying Eqs.
(44s) and (44c), provided the directions of X a and X b are so chosen that #&, =&=().

134

KINETIC THEORY OF ENGINEERING STRUCTURES

CHAP. IX

To accomplish this, assume the direction of X a as may seem most convenient, and
find the displacement a\a of the point a for a load X a = l, Fig. 44A. Now take X b
acting at the same point a and at right angles with aid.

Then ^=0, because it represents the displacement of the point 6 in the direction
of Xj, when only the force X a = l is active. Therefore, dab also becomes zero
by Maxwell's law. If then the displacement 6j6 of X b is found for the load Xb = l,
this in turn must be perpendicular to X a because d ab is zero. This always fur-
nishes a valuable check on the solu-
tion when the graphic method is
employed.

The example of a two-hinged
arch with a column support at the
center, Figs. 44B to 44D, is used to
illustrate the application of the
method to two redundants.

Fig. 44B shows the given struc-
ture with hinged supports at A, B
and C, thus involving two external
redundants. By removing the sup-
port at A the structure becomes
determinate, involving a simple truss
on supports B and C, and an over-
hanging cantilever AC'. The redun-

FIG. 44A.

dant forces X a and X b are then applied to the principal system at A and are treated
as external forces.

The first redundant X a is assumed to act horizontally and the second redundant X b
is applied at the same point and making the angle 6 with X a , such that d a b=dba=0.

Fig. 44c represents the conventional loading X a = l and a Williot-Mohr displace-
ment diagram, drawn for this loading, will furnish the displacement aa t for the
point A and the displacement mm a for the point m. Hence d ao becomes the pro-
jection of act! on the direction of X a and <?*, is the projection of mm a on the direction
of the force P m .

ART. 44 INFLUENCE LINES FOR INDETERMINATE STRUCTURES 135

Similarly in Fig. 44D, another displacement diagram _drawn for the conventional
loading X b = l and acting at 90 with the displacement aal, gives the displacements
661 and mm b from which the values du, and dmb are found by projecting the displace-
ments on the directions of their respective forces X b and P m .

The angle 6 is thus graphically determined, and as a check, the displacement 66 t
must be perpendicular to X a .

Thus having the constants d m and d bb and the displacements d ma and d mb for any
pin point m, the influence produced by any load P m , according to Eqs. (44s) becomes

Zi mUma j v PmOmb /A4 -.

a= and X b = ^ , ........ (44D)

d aa O bb

where dma is negative with respect to the force P m , as may be seen from Fig. 44c.
The values X a and X b for any set of loads become

Z^imflma , _j v" ^^mPmb / AA \

a = 5 - and X b = r - ......... (44E)

and the several values d ma and dmh are found from the two displacement diagrams
above described.

When the load P = 1 is vertical then the values in Eqs. (44D) represent influence
line ordinates for any point m for the redundants X a and X b . The same displacement
diagrams will furnish all values d ma and d mb for finding all the influence line ordinates
for both redundants.

For a uniform rise in temperature of t, the point A will move horizontally an
amount etl and the projections of this displacement on the directions of X a and X b
will then be respectively d at = etl and d bt = etl cos 0, whence

1 M 1 V 1 & COS

^l-r- and X bt = l

(b) Structures having three redundant conditions. The most prevalent case of
three redundants is a fixed arch, and hence this style of structure is chosen, as an
illustration, see Figs. 44E to 44j.

Fig. 44E represents any general arch with fixed supports. It is transformed into
a determinate structure by removing one fixed support and converting the frame
into a cantilever arm to be treated as the principal system.

If, now, this support is replaced by three redundant conditions, a moment l-X a
and forces X b and X c , all acting on the rigid disc AJBO, Fig. 44r, the stresses in the
structure will remain exactly the same as in the original fixed_condition.

Then d a will be the angular rotation of the rigid disc ABO about some pole 0; d b
will be the linear displacement of the point in the direction of X b ; and o c will be that

136

KINETIC THEORY OF ENGINEERING STRUCTURES

CHAP. IX

displacement in the direction of X c . These displacements are supposed to be so chosen
that the moment X a and the forces Xb and X c , all applied to the rigid disc ABO, will

exert precisely the same effect on the principal
system as was previously produced by the re-
dundancy.

The pole is taken as the instantaneous center
of rotation of the disc, when acted upon by a
moment X a = l. If this pole is chosen as the point
of application for the redundants Xb and X c , then
dba=Q and ca =0. From this it follows that the
angular displacements <5 a & and 3 ac produced by the
loadings X b = l and X c = l acting on the disc, must
likewise be zero if the drawing or computation is
correct. Finally, if Xb is arbitrarily chosen in a
vertical direction and X c is taken perpendicular to
the displacement which the point b (pole 0) under-
goes as a result of the loading .X& = 1, then

PRINCIPAL SYSTEM.

CONDITION Xl

(J)

FlGS. 44E, F, G, H, J.

Hence this disposition of the redundant con-
ditions will cause all the quantities d bearing double
subscripts of different letters, to become zero, and
the simplified Eqs. (44A) will now apply.

When the abutments are considered immov-
able, d a , db, and d c become zero and Eqs. (44B) and
(44c) will suffice to solve the fixed arch. For
vertical loads these same equations furnish the
ordinates for the X a , X b and X c influence lines
by inserting a single load P m = 1 .

In proceeding to a solution it is best to apply
a unit moment e-l/e, Fig. 44c, to the points A and
B, representing the conventional loading X a = l
acting on the principal system. A Williot-Mohr
displacement diagram, drawn for this condition, will
furnish the displacements A^A and B t B of the two
points A and B. The pole 0, being the instan-
taneous center of rotation of the rigid disc, is
found by the intersection of two lines

and OB-LBiB. Then X b is applied at in a verti-
cal direction and X c is made active at and in a direction perpendicular to the
displacement found for from the displacement diagram drawn ior the condition
X b =*l. See Figs. 44n and 44j for the conventional loadings Xj, = l and X e = l.

ART. 45 INFLUENCE LINES FOR INDETERMINATE STRUCTURES 137

ART. 45. STRESS INFLUENCE LINES FOR STRUCTURES INVOLVING

REDUNDANCY

(a) Multiple redundancy. The previous articles 42 to 44 show how influence lines
for redundant conditions may be determined. In the present article it is proposed to
present the methods of drawing stress influence lines for the members of any struc-
ture involving redundant conditions.

Th? general equation for the stress in any member of the principal system of a
statically indeterminate structure is Eq. (7 A). This equation, if written for a single
external load P m = l, will represent^ the stress influence line ordinate, under the point
ra, for the member S. Hence, if represents the stress in the member for condition
X =0 and P m = 1, then Eq. (7 A) gives the desired influence ordinate for any point ra as

i) m =S=S -S a X a -S b X b -S e X e ,etc., ...... (45A)

where rj m is the algebraic sum of the partial influences S , X a , X b , X c , etc., on the stress
of a certain member S, due to a load P m = 1 acting at the panel point ra. In other
words, a separate influence line may be drawn for each term of Eq. (4oA) and the
algebraic sum of the several influence ordinates under a certain load point m becomes
the stress influence ordinate y m for that particular load point. Such an influence line
may be regarded as a summation influence line of partial stress effects instead of a
summation load effect as originally implied by the definition in Art. 17.

Accordingly, for any system of external parallel loads, the total stress in any
member is represented by

....... (45u)

As may be seen from Eq. (45A) such a stress influence line will always necessitate
drawing as many influence lines as there are redundants, plus one for the S stress.
However, the influence lines for the redundants remain unchanged for all members of
the same structure and hence need be drawn only once. Also the stressesjS a , S b and
S c are constants for a given member but vary for different members. The S influence
line is the same as the stress influence line for any member of a determinate frame and
will be different for each member. See Chapter IV.

Therefore, it is advisable to draw the influence lines for the redundants without
the S a , S b , S c , etc., factors and then draw the ~So stress influence line for any particular
member S. Finally insert for the X's the values for any set of ordinates, as for point
k, and these combined with the stresses S a , S b , S c in Eq. (4oA) give the required ordinate
rj k for the stress influence line S. The several ordinates rj for all panel points furnish
the required influence line. However, much of this work is done graphically as illus-
trated in Fig. 45A.

The example is a two-hinged framed arch with a column supporting the crown, thus
involving two external redundant conditions X a and X b as indicated in Fig. 45A. The
loads are vertical and applied to the top chord. Required to find the stress influence
line for the chord ik.

138

KINETIC THEORY OF ENGINEERING STRUCTURES

CHAP. IX

The influence lines for X a and X b are drawn supposedly by the method given in
Art. 43. The ordinary stress influence line for the chord ik is then drawn as indicated
and called the S line, meaning that it is the influence line for condition X = with
P = l. This line A'i'k'B' incloses a negative area representing compressive stress in
the top chord of a simple truss AB.

FIG. 45A.

The influence lines X a and X b are both positive and the stresses S a and Sb, for
the member ik, are found to be negative. The ordinate TJ^ for the panel point k is
found from Eq. (45A) as

....". T . . ,. (45c)

giving a negative residual for y k .

In like manner all ordinates for the S influence line may be found and plotted to
give the shaded area A 'B' , which is the influence area for the chord ik.

The multiplications S a r) a , etc., can be performed graphically by laying off angles
corresponding to tana=S a and tan ft=S b as indicated in the figure. The products
S a rj a and S b y b are then taken off directly for any ordinate, added or subtracted graphically
and the sum S a y a +S b T) b is then applied to the ordinate S in the proper direction. The
angles a and /? are constant for the same member but vary for different members.

ART. 45 INFLUENCE LINES FOR INDETERMINATE STRUCTURES

139

The stresses S a =tan a and Sb =tan /? are found from Maxwell diagrams or by computation.

It is thus seen that all loads between A and C produce compression in the member
ik, while all loads between C and B will produce tension.

The stress resulting from any system of loads is then

etc.=2Pi?,

(45D)

observing the sign of rj for each particular point.

Since the large majority of practical problems of redundancy do not, and should
not, involve more than one redundant condition, the general case is here treated in
less comprehensive manner. For a special method of deriving the influence line for a
web member from those of two adjacent chords, see Art. 52.

(b) One redundant condition according to Eq. (45A) gives rise to the simple stress
equation

/"cT \

(45E)

*jm *^ ^o *^&^*-a *^a I ct -^-ct
V>a

which makes it possible to represent the S influence area as the area inclosed between
the X a influence line and the S /S a influence line, which latter is drawn with as much
ease as the simple S line. The resulting S
influence area will then require a factor S a
applied to all its ordinates.

These influence areas may be drawn in
either of two ways: First, by plotting both
the X a and the S /S a lines from a common
straight base, observing the rule that areas
above this base are negative, see Fig. 45u.
The S line ordinates TJ will then_ be measured
between the X a line and the S /S a line and
these ordinates will be positive when measured
down from the X a line. Second, by first
plotting the X a line from a straight base and

then constructing the S area by applying the +S /S a ordinates y' down from the X a
line, see Fig. 45c. In this case the +X a ordinates are best plotted above the base so
that the 4- TJ ordinates of the S area will appear below the base in the customary way.

The first method is more generally used, though the second leads to a very interesting
property by which the S line may be derived from the X a line when one point of the S
line and its zero points are known.

This property is shown in Fig. 45c, where it is seen that over any distance, as m'B',
for which the ~S~ /S a line is straight, the corresponding elements of the S line and the
X a line will intersect in points b', b" , &'"_which are in a straight line parallel to P and
passing through the end zero point of the S /S a line.

Further details and applications are not given here, as these will be illustrated in
connection with specific problems in Chapter X.

CHAPTER X

SPECIAL APPLICATIONS OF INFLUENCE LINES TO STATICALLY INDETERMINATE

STRUCTURES

ART. 46. SIMPLE BEAM WITH ONE END FIXED AND OTHER END SUPPORTED

The solution by influence lines is illustrated by Figs. 46A, using the general method
developed in Art. 45B.

Applying the criterion of Eq. (3c) to this structure, where m = l, 2r=4 and 2p=4,
gives ra + 2r-2p = l. Hence there is one redundant condition and by Eq. (SB) this
condition must be external.

Accordingly, the reaction X a , acting at the expansion end B, is chosen as the
redundant support, reducing the beam to a simple cantilever arm as the principal
determinate system.

The problem is considered solved when the shear and moment influence lines are
found for any point n of the beam. The case of direct loading is assumed.

The X a influence line is represented by Eq. (42o) as

i ^

1 . Ont, , .

where dma is the variable influence line ordinate at any point m with [1 = 1/8^ is the
constant factor, when d ma and daa may be acutal or measured to the same scale.

The equations for moment and shear at any point n and for the vertical reaction
A , are given by Eqs. (?A) written in the form of Eq. (45E) as

~J A Y -- 4 Y \

** A a-A-a ~ -^rtl ~r~ -A-a
\Aa

(46B)

wherein M is the moment about the point n produced in the principal system by a
load unity acting at any point m of the structure. M a is the moment at the point n
due to the conventional loading X a = l. Similar definitions follow for Q~ , Q*, also A^
and A a .

140

ART. 46 SPECIAL APPLICATIONS OF INFLUENCE LINES 141

The following solution by influence lines is based on these Eqs. (46B), when the
influence line for X a is known or found from Eq. (46A) .

The ordinates d ma are ordinates of a deflection polygon drawn for the conventional
loading X a = l kip, being a load unity acting downward at the point B when the beam
is treated as a cantilever. This deflection polygon is drawn as described in Art. 40,
and as shown in Fig. 46Aa.

The M moment diagram for the conventional loading X a = l, is drawn by making
the end ordinate A^C' =1 and drawing C^B'. This end ordinate may be measured to
any convenient scale and was drawn to half the scale of lengths, making all the moment
ordinates to half scale of lengths. The figured ordinates represent moments, in kip feet,
and are scaled at points Jz=2 ft. apart.

The areas w =Jfdx are now computed and treated as elastic loads for the construc-
tion of the force polygon (a) and the deflection polygon (b) when the pole distance H=EI
Takino- E =28000 kips and 7=2087 in. 4 and reducing Ax to inches and M to kip
inches then the pole would be . When the scale of lengths is 1:60 then,

, ,, . ,, TT 28000X2087 O7n ^ nnita

for deflections 25 times actual, this pole becomes #=744x60x25 =270 -

When / is variable the pole is made variable as was done in Fig. 39A, or the w loads may

be divided by / as is done in Art. 47.

The force and equilibrium polygons are now drawn, using any convement_scale for
the w forces and the same scale for H . This gives the deflection polygon A'C', Fig.
46Ab with ordinates 25 times actual. Thus the actual end ordinate ^ = 1.96/25 inches.
This 'same polygon is also the X a influence line with a factor p = l/i) B irrespective of
the scale of ordinates, since all ordinates rj m and TJ B are measured with the same scale.
To obtain the X a influence line with a factory 1, divide all the ordinates of Fig. 46Ab
by the end ordinate , Bf whence the curve A'C', in Fig. 46AC, is obtained with the end
ordinate WC' = l, drawn to any convenient scale of ordinates. The redundant react
X can thus be found for any case of concentrated loads P and is X a = ^Pr l /t) B .

To obtain the M n influence area for moments about the point n, the X a influence
line is now combined with the M /M a influence line, with a factor p = M a according t

q ' In Fio- b the line ^C 7 represents the W influence line provided the end ordinate
WC' is made equal to z, but this end ordinate is actually , fl> hence the linem question
must have a factor p=z/y B . But since M a =z, therefore, the same line n C becomes
the W /M a influence line with a factor p-lfa which is the same as the factor for
the X a line. Hence the area IVC 7 represents the M n influence area with a fad

/ ' = ^irnifarly / ?n 5 'Fi g . c, the area A^V represents the M n influence area with the factor
a=M a =z because ,' fl was here made equal to unity. This was done to show the two
solutions and to illustrate the fact that the second step of drawing the Z influence line
with a = 1 is really unnecessary and adds nothing except to simplify the final factor.

The algebraic sign of the M n area is derived from_Eq. (46B) since the larger _ X
area (which is positive) is subtracted from the positive M /M a area, leaving a negat
area as the remainder.

142

KINETIC THEORY OF ENGINEERING STRUCTURES

Oi-

o/Q, LINE.

FIG. 46A.

ART. 47 SPECIAL APPLICATIONS OF INFLUENCE LINES 143

The moment area M m , for moments about a point m, is indicated to show a case
where a portion of the area below the X a line is positive, creating a load divide at the
point i.

The moment area MA, for moments about the abutment A, is also shown. It is
the area between the line A'C' and the X a line, which is entirely a positive area.

In all of these moment influence areas, the moments obtained will be expressed in
the same units as the applied loads P and the ordinate 2. Thus for P in Ibs. and z in
feet the moments will be ft. Ibs. The scale of the ordinates y is immaterial and is
eliminated in the factor p, Thus if Fig. b is used, the moment about n for any train
of loads would be

ii;-- S

and using Fig. c it would be

and the units would depend solely on those chosen for z and the loads P.

To obtain the Q n influence area for shear at the point n, the X a influence line is
now combined with the Q /Q a line, with a factor /*=Q a = l, according to Eq. (46e),
see Fig. 46Ad.

For a load unity to the right of n, the shear in the principal system would be
Q) = l and for X a = l the shear Q a = l, hence Q /Q a = ^ so that the broken line A'n'DC'
represents the Q /Q a influence line with a factor unity, and the Q n influence area is the
shaded area with a factor /*=Q a = l, with the portion below the X a line positive, as
before. Were Fig. b used, the factor would be / w=Q a /'?B == l/'?s-

The point n is always a load divide for Q H .

The end reaction A for any system of loads is simply HP X a , from Eq. (46B),
because A = DP and A a = l.

A uniform live load covering the whole span would give X a =3pl/8, as may be found
from Eq. (15j). The graphic diagram, Fig. 46Ab, gave X a =7Al2p=0.372pl

ART. 47. PLATE GIRDER ON THREE SUPPORTS

The example chosen is a general case of unequal spans and variable moment of
inertia as illustrated in Figs. 4?A. All dimensions in inches, and the loads are directly
applied to the girder.

According to the criterion of Eq. (3c) this structure involves one external redundant
condition. Any one of the three supports may be taken as the redundant condition,
but it is generally most convenient to assume the middle support at C and thus obtain
the principal system as a simple beam on two supports A and B.

The Xc influence line for the redundant support is again given by Eq. (42n) as

(47A)

where S mc is the variable influence line ordinate at any point m with the constant factor
fi = \/d ect and d mc and d cc must be measured to the same scale. The ordinates d mc are

144 KINETIC THEORY OF ENGINEERING STRUCTURES CHAP, x

the ordinates of a deflection polygon drawn for the conventional loading X c = \ kip,
acting downward at C on the principal system or simple beam AB.

This deflection polygon is the equilibrium polygon drawn for a series of elastic loads*
w = MJx/I when the pole H=E. The variable moment of inertia is not taken care of]
in the manner previously given, by varying the pole distance, but the alternative
method of involving / in the w loads is here employed. The moments of inertia are
given for the various sections of the beam in Fig. 4?Aa. The M diagram for X c =
1000 Ibs. is drawn in Fig. 47Ab, with an ordinate under the point C equal to 1000 l^lo/l =
175,000 in. Ibs.

The several M ordinates are scaled from the diagram or computed, and these
values divided by the corresponding 7's furnish the figures for plotting the ~kZ/7 diagram^
Thus the ordinate at C is 1 75000 H- 34982 =5. 00. The w loads are then computed aa
the areas of the M/l diagram using the horizontal distances between the ordinates in
inches. These loads are applied at the centers of gravity of the respective areas.

The w loads are now combined into a force polygon, Fig. 47AC, with pola
H =El 100 XI 20 -=2333 w units, all drawn to the same scale, but this scale may be taken
as any convenient one without regard to the deflections. The particular pole will then]
give deflections one hundred times actual, because the scale of lengths for the girder
span was chosen 1:120 and the pole was made 100X120 = 12000 times too small.
I/ =28,000,000 Ibs. per sq. inch. The resulting equilibrium polygon 'A'C'B', Fig. 47ACJ
is the deflection polygon for the load X C = IQOO Ibs. with ordinates 100 times actual
and it is also the X c influence line with a factor 1/c where c is the ordinate under C.
measured to the same scale as the other influence ordinates.

It is clear from Eq. (4?A) that the scale of the influence ordinates is immaterial so;
long as the same scale is used for the ordinate c, but when actual deflections are sought
then a natural scale, as inches, must be employed.

The reaction X c for any case of concentrated loads may then be found from Fig.
47AC, by multiplying the loads by the d ordinates and dividing the sum of these products
by c, thus X c = ^Pd/c, where c and the d's are measured to the same convenient scale.

The A and B influence areas are easily found when the X c line is given. Thus in
Fig. 47 AC, the line B'C'A" represents the A /A a influence line with some factor, and
the shaded area is the A influence area with a factor I/TJA to be proven.

For this proof the end ordinate A' A" is evaluated according to Eq. (46s), and the
factor required for this ordinate will be the factor for all ordinates TJ of the A influence
area.

For the point A, A = l and A c = l-l 2 /l and X c has the factor 1/c. Hence ~A f A c =
1/1 2 and the end ordinate should be

i 2 \i x c -\ i 2 \d I
= ~ = '' ...... (7B)

where the Tactor is l 2 /cl.

But = - = , hence the factor is simplv \fr\A.

d C T) A '

ART. 47

SPECIAL APPLICATIONS OF INFLUENCE LINES

145

k 9i

rlW

P
n t 5

1= fZ.U
6

a 3

i
i

S.94

*u=

FIG. 47x.

146 KINETIC THEORY OF ENGINEERING STRUCTURES CHAP.X

.
Similarly \/r IB is the factor for the B influence area indicated in Fig. 47 AC as the

area included between the X c line and the line A'C'B".

Hence the reactions A and B, for any train of concentrated loads, become

A= 2Pr y and B= 2Pr i '. (47 c)

f)A "T iB .-.

The algebraic signs of the areas are determined as before. All areas below the X c
line are positive and those above this line are negative. The ordinates may be measured
with any scale so long as the same scale is used for all.

The M influence area is a portion of the A area for all points between A and (7,
while for points between C and B the M area is a portion of the B area. See Fig.
4?Ad.

Thus for the point 4 the M 4 influence area is shown as the shaded area between
the X c line and the broken line A'-i'B'. The factor is M C /C=X/TJ A , where x is the
ordinate of the point 4. When x is expressed in feet the resulting moment is
in ft. Ibs.

Similarly the M-j influence area is indicated by dotted lines and the factor is
M c /c=x/r) B .

The usual directions as to algebraic signs and scales apply as before.

The Q influence area for shear is shown, in Fig. 4?Ae, for the point 4 as the shaded
area with a factor Q C /C = \/T)A the same as for the A influence are.a. The line A'4 7 is
parallel to B'C'4".

Similarly the Q 7 influence area is indicated by dotted lines and has a factor Q c /c =
I/T)B' B'7' is parallel to A'C'7". The signs are again determined with reference to
the areas above and below the X c line.

All of these influence areas have a load divide at the point C' and the Q areas have
two load divides each. This determines the positions of train loads for positive and
negative effects, and to obtain maximum effects, the heaviest loads should be placed
over the maximum ordinates.

Temperature effects. A uniform change in temperature will expand the girder
equally in all directions and will produce a slight lifting of the ends A and B equal to
d(72- 12) =0.0105 inch, for =25 F. and =0.000007. When the three supports are
on the same level then no stress will be produced by uniform temperature changes.

However, when the sun shines down on such a structure experience teaches that
the top flange is heated much more than the bottom flange and this difference in tem-
perature may become quite considerable and will cause the girder to assume a curved
position, convex upward, when the top flange is warmer than the bottom.

The maximum difference in temperature between the two flanges is bound to remain
somewhat problematic, but observations indicate that differences of 30 F. are quite
common. In the present example Jf=+20 F. will be assumed and on this basis and
that the temperature varies uniformly between the two flanges, a set of w t elastic loads
is now computed from the formula w t = -sJUx/h = -0.00014 dx/h, where Ax is the
horizontal distance between sections and h is the depth of girder, both in inches.

AKT. 47 SPECIAL APPLICATIONS OF INFLUENCE LINES

The following table gives the w t loads:

147

Point.

dx
Inches.

Inches.

10,000u> (

Point.

dx
Inches.

h
Inches.

w t

10,000u>,

19.5

0.00043

4.3

0.000117

1.17

34.5

0.00024

2.4

0.000117

1.17

49.5

0.00017

1.7

64.5

0.000098

0.98

64.5

0.000115

1.15

49.5

0.000127

1.27

0.000175

1.75

34.5

0.000183

1.83

0.000175

1.75

19.5

0.000323

3.23

An equilibrium polygon drawn for these w t loads with a pole distance unity, would
represent the deflection curve of the beam due to M. However, for convenience it is
better to multiply these small loads w t by some factor as 10,000 and then make the
pole distance 10,000 instead of unity.

The scale of lengths for the girder was made 1 : 120 and if the ordinates are to appear
say 5 times actual, the pole distance should be made equal to H = 10,000/5X120 =
16.67, using the scale chosen for the w t loads.

Fig. 47Af shows the force and equilibrium polygons for the w t loads, with
ordinates five times actual. Since these w t loads are all negative for +At, the
3 ct deflection curve was drawn above the closing line A 'B'. The ordinate
under the point C represents 5d ct and this must be increased by the small
ordinate 5x0.0105, previously found for the uniform rise in temperature to obtain
total effect.

According to Eq. (44c), the redundant reaction produced by this temperature
effect is X ct =d ct /d ee . The deflection polygon for X e = l kip, gives 100 cc =0".93,
whence + cc =0".0093, and the d ct deflection polygon gives -5d c< =0".74+0.05=0".79,
making 3 ct = 0'M6.

Hence X ct =d et /d ce = -0.16/0.0093 = -17.2 kips, which indicates a downward reaction
at C to maintain the girder on the middle support. As the entire girder has an approxi-
mate weight of 7440 Ibs. the above temperature effect would lift the girder off the center
support and cause the beam to carry itself on two supports, producing a compression
of about 3100 Ibs. per sq.in. on the extreme fiber. This temperature stress would not,
however, be fully developed unless the span is loaded down in contact with the center
support.

The dead and live load stresses must finally be combined with the temperature
effect to obtain the real stress at any point.

The application of these influence areas is illustrated by placing a single load P n at
any point n of the span and showing the values of the several functions. The same
process is followed for each load of a train of loads and the total effect of the train is
the sum of the individual load effects.

The ordinates r? under the point n, in each of the influence areas, are all scaled to
the same scale to which 7j_ 4 and rj B are measured. This may be any convenient scale,
which is universally used for all ordinates of the drawing.

148 KINETIC THEORY OF ENGINEERING STRUCTURES CHAP. X

The values of the various functions for the load P n are

Y fnOn _ ' -p _

A c = -- /"

C 57.

B = _M = _-M- PB = - .207P n ;
ijjs 15.94

+-B=P, which checks to 1 percent.

n =49.4P n in.lbs. for P n Ibs.

X ct l 2 a 17200X300X194 '.

M 4t J 2 - = - 72Q =1,390,330 in.lbs.

ART. 48. TRUSS ON THREE SUPPORTS

Figs. 48A illustrate the analysis by influence lines of a truss on three supports.
The bottom chord is the loaded chord and the lengths and cross-sections of the members
are written over the members in Fig. a, while the stress SA, produced by a unit reaction
at A on the simple span AB, are written below the members. The stress diagram,
from which the SA stresses are found, is shown in Fig. b. The solution is similar to
that given in Art. 47.

As in the previous problem, Art. 47, the present structure involves one external
redundant condition, which is again taken as the middle support C, reducing the truss
to a principal system on two supports, A and B.

The X c influence line is derived from Eq. (42o) as in the previous problem and
becomes

(48A)

where the variable influence line ordinate d mc is the ordinate for a deflection polygon,
drawn for the loaded chord and for the conventional loading X c = 1 kip acting downward
at C on the prinicpal system or simple truss AB. The influence line factor ju = l/ cc .
The elastic loads for this deflection poylgon are found from Eqs. (36s) as :

M c I

- (48B)

for each pin point of the top and bottom chords. The effect of the web members is
usually neglected as being insignificant, but when it is desired to include their effect,
Eqs. (36c) will give the w loads due to the web members and these are added to the

ART. 48

SPECIAL APPLICATIONS OF INFLUENCE LINES

149

chord loads found from Eq. (48R). See Art. 36, for a complete discussion of this method,
and Art. 50 for a complete problem.

A Williot-Mohr diagram might also be employed to obtain this deflection polygon.
See also the example in Art. 50, Fig. 50B.

In the present example the web system will be_ neglected and the w loads arc found
for the chords, using Eq. (48 B). The moments M are those produced by the conven-
tional loading X c = 1000 Ibs. and are represented in the moment diagram, Fig. b, expressed
in inch-lbs. The ordinate at the point C will be P//4 =528,000 in. Ibs.

Using these moments and the values for /, F and r given, in Fig. a, for each member,
the following computation of the w forces is made:

Member.

A/c
Inch-lbs.

I
Inch.

F
Square Inches.

r
Inch.

Md
Fr*

L L 2

127,125

508.5

22.7

384

14.8

t/!^

254,250

509.0

42.0

384

20.8

L 2 L 4

381,375

508.5

22.7

384

57.7

C7 3 f7 5

528,000

299.3

44.7

434.9

2X18.7

L 4 L 6

381,375

508.5

22.7

384

57.7

7 5 t/ 7

254,250

509.0

42.0

384

20.8

L 6 L 8

127,125

508.5

22.7

384

14.8

224.0

The modulus #=29,000,000 /6s. per xq. inch, was not embodied in the tabulated
values, hence the w loads are E times too large and the deflections would be natural size
for a pole H=E. But the scale of lengths of the drawing was 1:300, and wishing to
make the deflections 400 times actual the pole must be made equal to E/3QQ X400 =241. 1 'w
units.

The force and equilibrium polygons, Fig. c, are then drawn, using any convenient
scale for w forces. The X c influence line is thus found as the polygon A CB, with a
factor ft = l/c.' The actual deflection of the point C is d ee = l". 07/400=0". 0027,
obtained by measuring the ordinate c in inches. All influence ordinates are measured
with the same scale, which may, however, be any convenient scale, because the factor
JJL makes the quotients d mc /d cc independent of an absolute scale. For this same reason
the M diagram might have been drawn with any middle ordinate as unity, though
when the actual deflection d cc is wanted for temperature investigations, the method
here given is less confusing.

The A influence area is now found by combining the X c influence line with the
ordinary A line for the span AB, in such manner as to make the ordinate at C equal
to zero and then finding the factor u which reduces the ordinate at A to unity.

The line A'CB, Fig. c, is the ordinary A influence line with a factor I/IJA, and
when this is combined with the X c influence line, the areas included between the two lines
will represent the A influence area with a factor ^ = !/TJA =0.0187.

Since it is more difficult to redraw the X c line so as to make the ordinate c = l,
than it is to apply the factor p, the latter method is decidedly preferable to the one
frequently given.

150

KINETIC THEORY OF ENGINEERING STRUCTURES

CHAP. X

t i ? 1

I A INFLUENCE AREA , |fX/rk=+0.|Ql87

INFLUECE AREA ,

Upper chord
I Lower chord

Tension is + , compreision is

ART. 48 SPECIAL APPLICATIONS OF INFLUENCE LINES 151

The point C is a load divide for both reactions A and B, and in the present case
the same influence area will serve for both these reactions, because l\ =1 2 . When the
spans are unequal, then the B influence area is included between the X c line and the
line AC prolonged to the vertical through B and the factor then becomes ,B = I/>?,B.

The influence area for a chord member L 2 L 4 is obtained by combining the ordinary
S stress influence line of the member for the principal system with the A" c influence
line, Fig. d.

A load at C can produce no stress in any member of the indeterminate truss, hence
the ordinate under C, for every stress influence line, must be zero. Therefore, the line
BCA' must be common to all stress influence lines of the members between A and C,
web members included. The point C must be a load divide for all members.

The center of moments for the chord L 2 L 4 is at C7 3 , hence the line A3, Fig. d,
completes the S stress influence line with a factor SA/T)A, where SA is the stress in the
chord L 2 L 4 for a unit reaction at A.

The area included between the X c polygon and the S influence line is thus the
required influence area for the chord L 2 L 4 with the factor JJ.=SA/^A- The area below
the X c polygon is positive as usual. The stress *SU = +1.99, hence the factor
/i =+1.99/53.6 =+0.0371.

If the vertical L 3 t/ 3 were absent, then the S line would have to be straight over
the panel L 2 Z/ 4 and a line 2C would be necessary to complete the S influence line.
With the member L 3 t/ 3 acting, the chord stress L 2 L 4 is, therefore, greater than when
the vertical is omitted.

The S influence lines and their factors for all chord members are shown by dotted
lines in Fig. d, and the final influence areas are always positive below the X c line. The
factors ft are negative for the top chords because SA is then negative.

The stress in the member L 2 L 4 , for any train of moving loads, is expressed by

(48c)

where each ij is the ordinate of the shaded influence area vertically under its lespective
load P and S A is given in the same units as the loads.

This influence area is also the M 3 influence area with a factor fi=x 3 /i)A and gives
moments in foot units when x 3 is measured in feet as indicated.

The scale of the influence ordinates is immaterial so long as all ordinates are
measured with the same scale. This is apparent from Eq. (48A).

The influence area for a web member L 2 U 3 is found from exactly similar considerations
as those shown to exist for the chords. The shaded area in Fig. e is the influence area
for the web member L 2 U 3 . The S lines for the other web members are indicated by
dotted 'lines.

The \me-BCA', Fig. e, is again one of the limiting lines of the S influence line for
each web member, and when the chords are parallel in the panel containing the particular
web member, then the other -limiting line A2' will be parallel to BC and the line 2'3
completes the S line for the member L 2 U S .

The load divide i for this member, considering the principal system, must fall

152 KINETIC THEORY OF ENGINEERING STRUCTURES CHAP, x

vertically over the point i' where the line 2'3 intersects the base line AH. This serves
as a check or when the chords are not parallel, as in the panel L 3 L 4 , then it may be
used to complete the S line as indicated for the member t/ 3 Z/ 4 with the load divide t.

In case the chords are not parallel and the center of moments for the diagonal
falls outside the drawing, then the ordinate 753 as for the member U S L 4 , may be com-
puted from the formula

r/ 3 x 3 +a 3 Jx 3 +a 3 \ /63.6+91\. Q

u-ir or ^ 3= t^> 4 = r^r-r 3 - 6 < ^

where x 3 is the distance L L 3 from A to the panel containing the web member, and a s
is the distance from A to the center of moments of the member, both in the same units,
as feet.

The factor {J.=SA/T)A again applies and the sign of SA determines the sign for //.

Stresses due to temperature. When the three supports A, B and C are on the same
level then a uniform change in temperature produces no stresses in the structure.
However, when the sun illuminates the bridge from above, there is usually a difference
in the temperature of the two chords. Assuming this difference as Jf=20 F. as in
Art. 47, then the bottom chord would be cooler than the top chord by this amount and
the changes in the lengths of the bottom chord members would be eMl and the w t
elastic loads would be

(48 E )

These loads being extremely small, it is well to multiply them by say 10,000 and
then using a pole distance of 10,000, the d cf deflections would be natural size. However.
the scale of lengths on the drawing was 1 : 300 and for deflections five times natural the
pole becomes

10000 rr _
= 5X300 * units.

The loads w t are figured for the panel points 1, 3, 5 and 7, taking =0.000007,
J*=20 F. and r=384 inches. The d ct deflection polygon is shown in Fig. d, and has
negative deflections, five times actual. Hence o ct =0".95/5 =0".19 and d cc was previously
found from Fig. e, as 0".0027, hence from Eq. (44c)

This value is in kips because d cc is the actual deflection for one kip, and
X ct being negative produces the same effect on the principal system as a load X ct hung
at the point C. The reaction at A, due to a load of 70.4 kips at C, would then
be A t =70.4Z 2 /(Zi +Z 2 ) = +35.2 kips.

The stresses in the members are then S t =S A A t =35.2 S A kips. The stresses
S A , due to a unit reaction at A, are already found in Fig. b, hence the tem-
perature stresses are readily determined. Thus for the member U 3 U 4 , S A = -2A2,
hence S t = -2.42 X35.2 = -85.2 kips. The negative sign indicates compression.

ART. 49

SPECIAL APPLICATIONS OF INFLUENCE LINES

153

ART. 49. TWO-HINGED SOLID WEB ARCH OR ARCHED RIB

This and the two-hinged framed arch are perhaps the most common structures
involving external redundancy which are met with in practice. They will receive
special attention here as deserving a prominent place among commendable structures.
The external redundancy may be said to offer less objection here than in any other
class of structures.

The present theory will be developed in its most general application to unsym-
metric arches and will be equally applicable to masonry, concrete or steel arched ribs.

Fig. 49A represents an unsymmetric two-hinged arched rib of any cross-section and
the lettered dimensions are in general the same as those previously employed for three-
hinged arches in Art. 28.

The arch thrust along AB is treated as the external redundant condition. When
it is removed, by replacing the hinged support at A by a roller bearing, the simple beam
AB (though curved) on two supports, then becomes the principal system.

FIG. 49A.

According to Eqs. (7 A), the following values for the reactions, thrusts and moment
for any point ra may be written:

A =A A a X a , where
B=B -B a X a , where

and A a =-l-sina

and

= l- since

M m =M -M a X a , where M =A x-P(a' -x') and M a = l-ycosa

N m = N - N a X a , where N = Q sin <f> ; N a = - 1 cos (<f> -a)

T m = T -T a X a , where T =Q cos<f> = (A - X p) cos<; 7 7 = l-sin(0-a)

(49A)

154 KINETIC THEORY OF ENGINEERING STRUCTURES CHAP. X

in which A and B are the vertical end reactions; M m is the moment of the external
forces about any axial point m; R m is the resultant thrust, on the normal section mm',
produced by all the external forces acting on one side of this section; N m is the com-
ponent of R m , normal to the section mm'; and T m is the tangential component of R m
in the section mm'. Q is the vertical shear at any point m produced by all the loads
P acting on a simple beam AB. The conventional loading X a = \ is applied as indicated
in the opposite direction of X a .

Eqs. (49A), with the special values inserted, thus become

SPa' .

A = , -- 1- X a sm a

v .
X a sm a

M m =M X a y cos a
N m =Q sin </> +X a cos (< -a)
T m =Q cos < X a sin (<f> -a) J

SPa

A,, sm a

(49s)

where M and Q have the values in Eqs. (49A).
From Fig. 49A,

M m =N m v; R m =VN m 2 + T n ?; H=X a cosa. ^ . . .

In the above equations, all terms except X a are derived from the static conditions
and the solution becomes possible when X a is determined.

The X a influence line for the redundant haunch thrust. When the haunches are on
the same level then this thrust X a becomes the horizontal thrust H.

The equation for X a is given by Eq. (42o) as

where d ma is the vertical deflection ordinate of any point m of a deflection polygon,
drawn for the loaded chord and for the conventional loading X a = l applied to the
principal system, and o^ is the change in the length AB, due to X a = \ acting on
the principal system.

Since dma and d^ are not parallel displacements, they cannot be obtained from
the same deflection polygon as in the previous examples, Arts. 46^8. This circum-
stance necessitates the construction of an extra displacement polygon for daa, if the
graphic solution is strictly followed, or of finding d^a by computation, which is usually
advisable.

The X a influence line thus becomes the d TOa deflection polygon with a factor p. = l/aa-
Should this deflection polygon be constructed for a pole distance H =daa then the factor

p-1.

According to Art. 40, the dma deflection polygon is the equilibrium polygon drawn
for a set of elastic loads w with a pole distance H = 1. These elastic loads represent
partial areas of a moment diagram drawn for the conventional loading X a = l.

ART. 49 SPECIAL APPLICATIONS OF INFLUENCE LINES 155

The moment produced by X a = 1 about any axial point m is

M, n = 1 y cos a .......... (49E)

and the elastic loads, by Eq. (39i) become

M Au Axy cos a

^-JT--J -- -r,

El El cos (j> '

(49F)

where Au is the width of a partial moment area measured along the axial line, and Ax
is the horizontal distance between the vertical moment ordinates, making Au=4x/cos<j>.

The deflection d aa may be obtained from a deflection polygon drawn for the same
set of w loads by allowing these loads to act in a direction parallel to X a , as per Art. 38.

According to Eq. (38B) the deflection of any point in any direction, may be expressed
as the sum of the moments of all w loads on one side of the point, when the direction of
the w loads is taken parallel to the required direction of the deflection.

Hence

a daa = ^T ^(J/ COS )

(49o)

which affords a purely analytic solution for computing the X a influence line ordinates
and also the d m displacement.

Since the w loads are best found by computation, it is a comparatively easy task
also to compute the values wy cos a and thus obtain d aa = I>wy cos a, which is the
required pole distance for the d ma deflection polygon. Hence according to Eq. (49n)

i
and the X a influence line is an equilibrium polygon drawn for the elastic loads w with

a pole distance H = Swn/ cos a.

Since E enters into w, and hence into each term of Eqs. (49o), as a constant factor,
which cancels in the numerator and denominator of Eq. (49n), it would be proper to
multiply Eq. (49r) by E and thus make the elastic loads equal to Ew without affecting
the ordinates r? a from Eq. (49n). Should it be desirable to obtain values of d ma from
the X a influence line, then the jj a ordinates must be divided by E and multiplied by

cos a.

For very flat arches, the effect of the axial thrust should be considered in the
determination of X a , by allowing for the quantity d f m produced by N a , Eq. (49A), on
the displacement daa-

Thus for N a = -I cos (< -a) and Au = Jar/cos <j>, then from Eq. (15.v),

2 /j

156 KINETIC THEORY OF ENGINEERING STRUCTURES CHAP. X

which is the effect due to N a only, and this added to the displacement previously found
for moments only, gives the total displacement

(49HH)

Hence, when axial thrust is to be considered in Eq. (49n), then the value 3 aa should
be computed from Eq. (49HH) and used as the pole distance H in drawing the equilibrium
polygon for the elastic loads w. When the loads w are taken E times actual, then the
E in the second term of Eq. (49HH) is omitted.

Temperature stress. For any change in temperature above or below the normal,
X at is expressed by Eq. (44c) as

X at =m^, ..... V . '.I/ .. . (49i)

Oaa

wherein d al is the change in the distance AB due to any change in temperature of
acting on the principal system.

When the structure retains a uniform temperature then the change t will be uniform
in all directions so that the shape of the structure will always be similar to its normal
geometric shape. Hence, the distance AB will change as for any case of linear expan-

sion and

dl 0.000488/

for e =0.0000065 per 1F., and Z = 75.

When the change in temperature is not uniform, as when the upper flange is +Jl c
warmer than the lower flange, then d at must be found from a deflection polygon drawn
for a set of w t elastic loads computed from the formula w t = edtJu/D, where D is the
depth of section and Au is the length between sections measured along the axis. The
pole distance, if made equal to unity, will give the actual o at , and if the pole be made
equal to d aa = 'Zwy cos a as for X a , then the displacement found will be X nt measured
parallel to AB. The w t loads must likewise be taken parallel to AB in drawing the force
and equilibrium polygons since d at is a displacement along AB.

X at will be positive for +t, while for + At the w t loads are negative and X at would
also be negative.

In any case, when X at is determined, then the quantities M mt , N mt , and T mt are
found from Eqs. (49fi) by omitting all the terms involving the effects of the loads P,
thus:

M mt = ~X at y cos a =N mt v

N m t= X at cos((f>-a)
T mt = -X at sin (< -a)

Abutment displacements. When the abutments undergo displacements Ar in the
directions of the reaction forces R, then, for the case of external redundancy where
d a =0, Eq. (8D) gives for P=0,

ART. 49 SPECIAL APPLICATIONS OF INFLUENCE LINES 157

where R a has the several values A a =-l-sina, B a = \- sin a, and X a = l, while Jr
represents the vertical displacements A A, AB and a change M in the length of span I.
Hence the redundant X ar , due only to abutment displacements, is

S/g Jr _ A a 4

Xar ix

JB sin a A A sin a + M sec a.

X-ar = 5~~' ' " "

When AA=AB, which is usually the case, then

. (49L)

showing the effect due to an elongation M, in the span I, to be precisely the same in
character as that due to a uniform fall in temperaturegiven by Eq. (49i).

For a two-hinged arch with a tension member AB, and one expansion bearing, the
problem becomes one involving an internal redundant member X a and Eq. (42s) is
then used vvith the term p a in the denominator. The solution would otherwise remain

unchanged. .

Stresses on any arch section. Neglecting curvature, which is always permissible,
the stress on the extreme fiber of any arch section mm' is given by Navier's law as

N My (4j)M)

J~F I '

where N is the normal thrust on the section, M=Nv is the bending moment about the
gravity axis F is the area of the section, y is the distance from the gravity axis to the
extreme fiber and I=Fr* is the moment of inertia of the section about the gravity axis.

Fig 49B gives a graphic representation of the distribution of stress on any arch
section mm' in accordance with Eq. (49M). The center of gravity is at c and the uniform
axial stress f m =N/F is the stress ordinate at c, while the stresses on the extreme fibers
/. and /,-, are applied as ordinates giving the line tt' as representing the uniform ,
tribution of stress over the section. .

The point e is the kernel point for the extrados while * is the kernel point for the mtrados.
These points are determined by the distances k.=i*/e and k^fi/i, where r is the radius
of gyration of the section and e and i are the respective distances to the extreme f
the section measured from the gravity axis c.

When N and * are given, the line tt' is easily constructed as indicated by prolong-
ing i?_to 6 to find t, and by prolonging ?ito a to find f. The stress /. is then the inter-
cept bn while the stress f t is the intercept an.

158

KINETIC THEORY OF ENGINEERING STRUCTURES CHAP. X

Using e and i as subscripts, referring respectively to extrados and intrados, then
for M=Nv and I^Fr 2 , Eq. (49M) gives

and

(4*0

The moments about the kernel points e and i are evaluated from the figure as

N(v+k e )=M e and N(v-kd=Mi ........ (49o)

FIG. 49B.
For r 2 /e=k e and r' 2 /i=k{, Eqs. (49N) become

- _i 4. = _ - _j?

F\ + k e ) F\ k e I Fk e

F \ ki/ ''F \ ki i Fki
Also solving Eqs. (49o) for N and v, then

M e Mi

N=, -r and v =

(49?)

Me -Mi

(49 Q )

Hence Eqs. (49p) and (49cj) furnish a complete solution for the stresses f e and /,
on the extreme fibers, the normal thrust N and its distance v from the gravity axis,
for any unsymmetric section tt' of an arch ring, in terms of M et Mi, k e and &,.

ART. 49

SPECIAL APPLICATIONS OF INFLUENCE LINES

159

When the section is symmetric about a horizontal gravity axis, which is usually the
case, then these equations give, for e=i=D/2, or the half depth between extreme fibers,

_- __ _
= ,-i _______

M e Mi

=s ~ and i==

Me-Mi

1. T "~ , Ctllvl

i) M e +Mi

(49R)

2k M e -Mi 2N

For a rectangular section of unit thickness and of depth D, making e=i=D/2.

fc _* x> F==1 . D} /=i^- 3

M e +Mi

N=-

and ?'= :

(49s)

In all the above formulae M e and Mi have opposite signs when N acts between the
two kernel points e and i. When v>ki, both moments are negative and when v is negative
and larger than k e , both moments are positive. The figure shows v to be positive
when measured from the gravity axis at c toward the extrados. The stresses f e and
fi take their signs from M e and Mi respectively, and compression is regarded as a negative
stress.

The stresses on any arch section mm' may, therefore, be found for any simultaneous
position of a moving train of loads when the influence areas for M et M t and T m have
been drawn.

The resultant polygon for any simultaneous case of loading may also be drawn by
plotting the offsets v from the gravity axis of each section examined, observing that
+v is toward the extrados and v is toward the intrados.

The method of combining the X a influence line with ordinary moment and shear
influence lines to obtain the M e , Mi and T m influence areas for any section mm' will
now be illustrated.

Kernel moment influence areas. The moment equations for the kernel points accord-
ing to Eq. (49B) are

,_ \M oe ! [ M oe 1 ,

=M ae \ -a-, X n \=y e cos a \- X a \

lM ae J [j/ecos a

(49r)

The intercept of the M oe influence line on the vertical through A is simply the
ordinate x e of the point e. Similarly for the M oi influence line this intercept is the ordinate

160 KINETIC THEORY OF ENGINEERING STRUCTURES CHAP. 3

Xi of the point i. Hence the positive intercepts on the vertical through A, of th
M oe /y e cos a and of the M^/yi cos a influence lines, are easily computed when the coordi-
nates of the kernel points e and i are given, provided the X a influence line was drawn
for a factor /* = !.

The coordinates of the kernel points are derived from the coordinates (x m y m ) of the
axial point for the same section, when the kernel distances k and the angle are known
for that section. From Fig. 49s,

,
x e = x m +k e sm<f>, y e = y m -k e cos(j> }

I, V 49u)

X{ =x m -ki sin 0, yi =y m +k { cos ^ J

which apply to all points from A to the crown, and from there to the abutment B the
signs of the last terms are reversed. By using the I x ordinates the intercepts on
the vertical through B are found.

One intercept only need be computed as the two limiting rays of the M /y cos a.
line must intersect on the vertical through the center of moments. See Fig. 49c.

By constructing the X a influence area for // = !, which may always be done by making
the pole distance H = 2wy cos a, and since both the X a and the M /y cos a lines are pos-
itive, then by applying them below the closing line A'W, the area included between the
two lines will represent the M m influence area. The portion below the X a line will
be the positive area because X a is subtractive in Eqs. (49-r). The entire M m influence
area has a factor p. =y cos a =M a .

The T m influence area for tangential force on any section mm'. From Eqs. (49u)
the following equation for T m is obtained.

From this the end ordinate at A, for the T /T a Yme, becomes 1- cos <^ sin (0 a)
because the end ordinate for the Q line is unity. In this case the end ordinate at B is
numerically the same but negative. For axial points to the right of the crown the end
ordinate at B is positive and the one at A is negative. Other details are illustrated
in connection with the example. The final T m influence area has a factor /i=sin (<-a).

Example. A two-hinged arched rib, modeled after the Chagrin River Bridge near
Bentleyville, O., was selected to illustrate the application of the previous theory.
The structure was made unsymmetric by shortening the span at the right-hand end as
shown in Figs. 49c. The clear span thus became 164 ft. and the rise 27.89 ft. This
bridge was designed to carry a live load of 24 kips per truss per panel. The arch section
is composed of a f-inch web plate and 4-6" X6" X 11/16" angles, with 2-14" X7/16" flange
plates on each flange. Besides these there is one 14" X|" plate on each flange extend-
ing from sections 2 to 5 and 7 to 10. The sections are all symmetric about the gravity
axis and all general dimensions are given on the drawing and in Table 49A.

The bridge consists of two steel arched ribs 27 ft. between centers and carrying a
total dead load of 1,058,000 pounds. The following values are assumed as a basis

ART. 49

SPECIAL APPLICATIONS OF INFLUENCE LINES

161

X ordinores are ?.a times
I toicole of Ittjfeth ,.

M e INFLUENCE AREA. >\=y g CoiqC-18.-l

FIG. 49c.

162

KINETIC THEORY OF ENGINEERING STRUCTURES

CHAP. X

for the analysis: #=29,000 kips per sq.in. =4,176,000 kips per sq.ft.; e -0.0000065
per 1F.; andZ = 75F.

The first step in the analysis is to compute the Ew loads from Eq. (49r) using the
tabulated dimensions in Table 49A. In the same table the values Ewy cos a are computed,
giving the required pole distance H=E2wycosa for constructing the X a influence line
in accordance with Eq. (49n).

TABLE 49A
Xa INFLUENCE LINE

Sec-
tion.

Coordinates of
Gravity Axis.

<*>

in. 4

ft.

6 =
1

ft.

0J.r =
Ax

Ew =
Axy cos a

Ewy cos a

ft.

y
ft.

cos <t>

cos <f>

ft,

1 COS (j>

ft.

I COS (j>

ft.

36 15'

1.240

74,180

3.578

0.3466

9.4
15

15
15
15

15
15

15
15
15

4.6

0.652

9.4

6.01

32 15

1.182

74,180

3.578

0.3260

4.954

29.8

179

24.4

14.12

26 00

1.113

64,710
76,530

3.121
3.691

0.3567
0.3017

4.939

69.7

984

39.4

20.30

20 00

1.064

66,170

3.191

0.3332

5.014

101.8

2065

54.4

24.73

14 15

1.032

56,750

2.737

0.3770

5.673

140.3

3468

69.4

27.24

7 30

1.009

48,160
40,420

2.323
1.949

0.4352
0.5176

7.198

196.0

5339

84.4

27.89

1.000

33,590

1.620

0.6173

9.011

251.3

7007

99.4

26.77

7 30

1.009

40,420
48,160

1.949
2.323

0.5176
0.4352

7.198

192.6

5156

114.4

23.58

14 15

1.032

56,750

2.737

0.3770

5.673

133.7

3153

129.4

18.58

20 00

1.064

66,170

3.191

0.3332

5.014

93.1

1730

144.4

11.82

26 00

1.113

76.530
64,710

3.691
3.121

0.3017
0.3567

4.939

58.4

690

159.4

3.14

32 15

1.182

74,180

3.578

0.3260

3.983

12.5

164.0

34 20

1.211

0.3260

0.0

EZwycos a = 29,810

NOTE. The x abscissae are measured horizontally from A.
The y ordinates are measured vertically from AB.
The Jj are measured horizontally between sections.
a=l 06', cos a = 0.9998, sin a = 0.0192.

The products QAx in Table 49A are summed by the prism oidal formula treating
Jx as the length of a prismoid whose middle area is 6 and whose end areas are the means
of the successive tabulated values of 6. This is necessary because 6 is a variable quantity.

AHT. 19 SPECIAL APPLICATIONS OF INFLUENCE LINES 163

Calling 0o-i the mean between and 0i, and 0]- 2 the mean between 6\ and 2
etc., then the values 6 Ax become:

Q j x =^[200 +0 _!] =^[0.6932 +0.3363] =0.652;
o o

0! Jz = ^[0o-i +40i +0i_ 2 ] =[0.3363 + 1.3040 +0.3413] =4.954.

6 2 Ax =^[0i -2 +20 2 +20 2 ' +0 2 - 3 ] =^?[0.3413 +0.7134 +0.6034 +0.3174] =4.939.

6 z Ax = ^[0 2 -3 +403+03-4] =y [0.3 174 + 1.3328 +0.3551] =5.014,
etc., etc.

After collecting the tabulated data from the drawing, Fig. 49c, the remaining
computations are quite simple and no further comment is necessary here, as the several
operations are indicated in Table 49A.

The X n influence line is now drawn by combining the Ew loads into a force polygon
and making the pole distance equal to E2wy cos a. The resulting equilibrium polygon
represents the X a influence line with ordinates to the scale of lengths and /* = !, in
accordance with Eq. (49n).

In the drawing, the pole distance was made 1/20 of the actual length so that the
ordinates ?? are twenty times too large. For this reason a special scale of ordinates,
twenty times the scale of lengths, was constructed and used for all influence ordinates.

The scale of forces for the force polygon may be any convenient scale, so long as the
pole distance is measured with the same scale as the forces.

The influence areas for the kernel moments and tangential force are now drawn by com-
bining the M /M a lines and the T /T a line each with the X a line. According to Eqs.
(49x), (49u), (49v), this will require computing the kernel distances k, and the coordi-
nates of the kernel points and finally the end ordinates at A of all influence lines.
The end ordinates at B may be obtained by substituting for the abscissa? x the values
I x. These computations are given in Table 49s, which is self-explanatory.

In Fig. 49c the influence areas M e and Mi are constructed for section 3. The X a
polygon is copied from the original one by transferring down the several j) a ordinates
from a horizontal base' ~A r W. The M /M a lines are constructed from the end ordinates
A'A"=x c /y e cosa and A f A 7r '=Xi/y i cosa, using the coordinates for the two kernel,
points as given in Table 49u. The lines WA" and B'A'" are thus determined.

The kernel points e and i are the centers of moments and have the abscissae x e and
x i} as given in the table, and from these the points e' and i' are located.

The two moment influence lines are completed by drawing the lines A'e' and A'i'.

The M oe /M ae influence line is thus found to be the broken line A'e'B' and the shaded area

included between it and the X a line is the M e influence area for section 3. The fac-

tor for the ordinates i) e , when measured to the same scale as the rj a ordinates, is

i = cos a =18.41.

164

KINETIC THEORY OF ENGINEERING STRUCTURES

CHAP. X

TABLE 49n
KERNEL POINTS AND END ORDINATES FOR Al me , M mi , AND T m AREAS

Sec-
tion.

Sq. in.

in.

i 2I

k sin <
ft.

k cos <j>
ft.

Coordinates of kernel points,
Eqs. (49u).

End Ordinatea.

\-2FD

Eq. (49R)

ft.

x e
ft,

Ve
ft.

x i
ft.

ft.

x e.

x i

COS <ji

y e cos a

y^ COS a

sin (<t> a)

1
2

4
5

8
9
10

11
12

81.6
79.0
89.5
87.8
86.3
85.7
75.2
73.6
75.2
85.7
86.3
87.8
89.5
79.0
81.6

36.0
35.5
33.4
33.8
31.6
29.5
27.4
27.0
24.9
27.0
27.4
29.5
31.6
33.8
33.4
35.5
35.7

2.13
2.04
2.11
1.99
1.86
1.71
1.64
1.53
1.64
1.71
1.86
1.99
2.11
2.04
2.13

36 15
32 15

26 00
20 00
14 15

7 30
00

7 30
14 15
20 00

26 00
32 15
34 20

1.14

0.91

0.68
0.46

0.23
0.00

0.23
0.46
0.68

0.91
1.14

1.80

1.86
1.87
1.79

1.66
1.53

1.66
1.79
1.87

1.86
1.80

10.54

25.31

40.08
54.86

69.63
84.40

99.17
113.94
128 . 72

143.49
158.26

4.21

12.26
18.43
22.94

25.58
26.36

25.11
21.79
16.71

9.96
1.34

8.26

23.49
38.72
53.94

69.17

84.40

99.63
114.86
130.08

145.31
160.54

7.81

15.98
22.17
26.52

28.90
29.42

28.43
25.37
20.45

13.68
4.94

2.504

2.064
2 . 174f

1.058

1.470
1.747t

1.633

2.135

2 . 895f
4.307

8.884
55.555

8.884*
4.307
2.895

2.135
1.633

3.202
2 . 582*

2 . 869
2.265*

4.284

0.700

NOTE. D is measured between extreme fibers. For all dimensions in inches, k is in inches.
The coordinates of the axial points are given in Table 49A.

* For sections 7 to 12 the ordinates are for the B end.
t Values used in Fig. 49c.

Similarly the dotted line A'i'B' represents the M oi /M ai influence line and the area
included between it and the X a line represents the Mi influence area with factor
Pi=yi cos a =22.13 for section 3.

Hence for any simultaneous position of moving loads the two kernel moments for
section 3 are represented by the expressions

and

(49w)

where the subscripts refer to the kernel points.

The T m influence area for section 3, is found by laying off the end ordinate A' A" =
cos </sin ((f> a) =2.895 and drawing the line A"B'. The negative end ray A' 3' is parallel
to A"B', and the T /T a line is thus the broken line A'3'3"B' and the shaded area is the
T 3 influence area with a factor /*=sin (< a).

It will be seen that for sections near the crown the end ordinate at A becomes very
large and when <=, this ordinate approaches infinity, while becomes zero. Hence
there will always be several sections near the crown for which the T m influence area
must be found by making n = l and reducing the X a line accordingly. This is done
by using the original form of Eq. (49s) which is

T m =Q cos < X a sin (<j> a).

ART. 49 SPECIAL APPLICATIONS OF INFLUENCE LINES 165

For the A ordinate Q = l, hence this ordinate is simply cos < and by reducing all
the T) U ordinates by multiplying them by sin (<-), the .T 5 area is found with a factor

1=1.

The rj a ordinates are easily reduced by graphics, laying off the line B'z such that its
deviation from the vertical is sin (0 -a) in a distance unity. The rj a ordinates pro-
jected over horizontally are then reduced to the small horizontal intercepts between
Wz and the vertical.

The T 5 influence area is thus constructed and is represented by the shaded area
included between the broken line A'5'5"B' and the reduced X a line. The factor p = l.

The tangential force for section 3, for any train of moving loads is expressed by

(49x)

and the tangential stress by T Z /F 3 , where rj t is any ordinate to the T 3 area under some
particular load P.

The stress due to a unifo
(49 J) whence for t = 75 F.,

particular load P.

The stress due to a uniform change in temperature is found from Eqs. (49i) and

l-.a*_ etf 0.000488 1

Xat 5 it ?,

d aa COS Oi d aa COS a

Eq (49o) gives d aa = Hwycosa and Table 49A furnishes EZwij cos < =29810, hence
L=29810/JS ft. for X a = l kip. Making =4,176,000 kips per sq.ft., Z = 164 ft. and cos
a a =0.999S, then 0^-0.00714 ft. and X at = ^mfxo.9998 = n - 21 ki P s -

* Eqs. (49K) , will furnish the means of finding the values M mt , N mt , T mt , and v=M mt /N mt
and with these^and Eqs. (49?) the temperature stresses f e and f t may be computed for

any section mm'.

Stress due to abutment displacements. Eq. (49L) furnishes the haunch thrust X ar
for any change Al in the length of span. The quantity Al must be estimated from the
elastic properties of the abutments and is always more or less problematic, though it
is well to investigate the probable stress which might be created by such a displacement.

In the present example it is assumed that for the maximum case of loading the
haunches will spread an amount JZ=0.03 ft, then for d aa =0.0071< ft, X a - kip,
and cos a =0.9998, Eq. (49L) gives

y & - 03 = -4.20 kips.

ar = ~<LTc^ 0.00714X0.9998

The stresses /, and /,- may then be found in precisely the same manner as was
just described for the case of temperature stresses.

The stresses due to temperature and abutment displacements should be separately
investigated for several typical sections, especially the crown section, so as to
thldesigner to judge of the relative importance which these may have in comparison
with the combined dead and live load stresses.

The live load stresses for section 3 will now be found for a single jloac P acting at 3,
tion 4, merely to illustrate the use of the influence areas shown in Pig. 4

166 KINETIC THEORY OF ENGINEERING STRUCTURES CHAP X

M e3 =Me4P = 18.41 X0.45P = 8.285P.
M =i*P =22.13 X0.17P =3.762P.

= -6.84P per sq.ft. = -0.0474P per sq.in.

M,- 3.762P

= LOOP; T 3 =p t i)t*P =0.324 X0.94P =0.305P.

The resultant K A , of ^1 and JT a , is found graphically to be 1.19 P, which, for a single
load to the right of section 3, should check R 3 , as it does.

ART. 50. TWO-HINGED SPANDREL BRACED ARCH OR FRAMED ARCH

The analysis of the two-hinged frame arch is accomplished in the same general manner
followed in the previous article, where the corresponding solid web arch was treated,
only that the frame is usually simpler.

The general Eqs. (49s) for the reactions A, B, and- moment M m apply equally to an
unsymmetric framed arch, and the influence line for the redundant haunch thrust X a
is determined precisely as in the previous problem except that the w loads are found by
the method given in Art. 36.

The stress in any member then becomes

S=S S a X a =S a \-^-X a \, ........ (50A>

according to Eq. (45E) for one external redundant condition. This equation is later
employed to construct the stress influence lines for the members.

The X u influence line for the haunch thrust is represented by Eq. (49n) as

Z* ' Oma _ ! ' O-ma f ~p. \

a = K """v* 'Ja? ........ (OUB)

daa 2jWy COS a

where d^a is the vertical deflection ordinate of any point m of a deflection polygon, drawn
for the loaded chord and for the conventional loading X a = l applied to the principal
system; and d aa = 'wy cos a is the change in the span AB, due to X a = l acting on the
principal system.

According to Art. 40, the d ma deflection polygon is the equilibrium polygon drawn for a
set of elastic loads w, with a pole distance H = l, according to the method given in Art. 36c.

ART. 50 SPECIAL APPLICATIONS OF INFLUENCE LINES 167

The elastic loads w are functions of the changes in the lengths of all the members
of the frame as given by Eqs. (36fi) and (36c). These are algebraically summed for all
the panel points to obtain the total loads w. Thus the elastic loads for the chords alone
are w c = Al/r and each web member contributes two elastic loads w u = M/r u and w n = M/r n ,
acting at the two adjacent panel points u and n of the loaded chord. Hence when the
loads P are to be applied to the top chord, then the w u and w n elastic loads are com-
puted for the top chord panel points. The lever arms r for the chords are measured
as shown in Fig. 36A and the lever arms r u and r n for the web members are measured as
explained in Figs. 36s and 36r. The details of the computation of the w loads are illus-
trated in Table oOu, and Fig. 50A, in connection with a complete example.

For any braced arch with parallel chords, the w loads for the web members may
always be neglected.

The displacement d aa = 'Swy cos a is computed for the same w loads by taking the
sum of their moments about the line A B joining the haunches. The lever arms for an
unsymmetric span thus become the vertical ordinates of the respective pin points
times cos a. See also Table 50A.

The X a influence line thus becomes the equilibrium polygon drawn for the elastic
loads w with a pole distance o aa = ^wy cos a.

Still another method of finding this influence line consists in drawing the deflection
polygon for the loaded chord by means of a Williot-Mohr displacement diagram, which
also furnishes the value d^ from which the influence line ordinates rj a may be computed
and plotted. This solution is illustrated in Fig. 50B.

Since the modulus E does not affect the X a influence line it is more convenient to
comput-e all displacements and the w loads E times too large, thus avoiding the small
quantities resulting in many decimals.

Temperature Stress- Calling Al t the change in the length of any member due to any
temperature effect, then from Mohr's work Eq. (5n) the change d at in the length of
the span AB becomes

l.d at =ZS a 4l t =2S a dl .......... (50c)

The redundant thrust X at , due to any temperature effect, is found from Eq. (44c) as

. at _ a _ g

at ~~d^~ daa 'EZwycosa

For a uniform change in temperature of t, above or below a certain normal, the
change o^ in the length of the span 'AB, is found from Eq. (49j) as

dAB

cos a

In any case the stresses in the members are best found from S t = ^SaXat for X at , or
from a Maxwell diagram drawn for the external loading X at , acting on the principal system.

For a uniform change of t from the normal temperature, the resulting stresses
become S t = ^S a X at . The effect on the final stresses S for full loading, will then
be additive or S max = (S +S t .)

168 KINETIC THEORY OF ENGINEERING STRUCTURES CHAP. X

Abutment displacements will affect the stresses in the members by producing a redundant
thrust X ar as found from Eq. (49L) when JZ is an increase in the length of span AB.
The stresses S r would be found from a Maxwell diagram drawn for X ar acting on the
principal system. They could also be found as S r = =F*S a X ar , for X ar , since *S a is the stress
produced in any member S by X a = 1 acting outward.

The stress influence area for any member $ is derived from the previous Eq. (50 A)
from which any stress influence ordinate TJ becomes

s=s a ^-x a ^=y). .:.; . . (SOP)

These influence areas are alike in principle for all members, whether chords or web
members, and represent the area inclosed between the X a influence line and the S /S a
influence line, times a factor S a .

The lSo/S a influence line is the ordinary stress influence line S for any determinate
frame, nultiplied by the factor l/S a . Hence the end ordinates for the S line may be
found in the usual way. Thus the stress SA in the member S due to the upward reaction
.4=1 is the end ordinate of the S line at the point A, and this ordinate divided by
S a becomes the required end ordinate rj A =S A /S a of the S /S a line at A. Similarly the
end ordinate of the S /S a line at B is -fjB=SB/S a , where SB is the stress in the member
due to the upward reaction B = \ acting on the principal system.

Hence if the stresses SA, SB and S a , for all the members of the principal system, are
found either by computation or from three Maxwell diagrams, then all the data for the
several stress influence lines are at hand provided the X a influence line is known.

In drawing the S /S a influence lines the following points should be observed: 1,
That the end bounding lines must always intersect in a point i' on the vertical through
the center of moments i of the particular member treated. 2, that this S /S a line
must be a straight line between adjacent panel points of the loaded chord. 3, that when
one of the end ordinates is too large to be conveniently measured, then half this ordinate
may be laid off at the center of the span. This is frequent!}' done as in the example
which follows, see Figs. oOc.

The signs of the end ordinates and of the factors fJi=S a all follow from the signs of
the stresses SA, SB and S a .

Deflection of any point m. Applying Mohr's work equation (Gfi) the deflection
d, n of any point m, becomes

(600)

wherein Si is the stress in any member due to a load P = l, applied at the point m in
the direction of the desired deflection ; and S is the corresponding stress due to any cause
or actual condition of simultaneous loading for which the deflection is sought, includ-
ing load effects as well as temperature changes and abutment displacements. The sum-
mation covers all the members of the principal system.

ART. 50

SPECIAL APPLICATIONS OF INFLUENCE LINES

169

For any redundant condition X a the stress S by Eq. (7 A) becomes

S=S -S a X a +S t , . (50H)

where S t = S a X at is the stress due to temperature effect.

Example. A two-hinged, riveted, spandrel-braced arch, taken from a thesis by
Mr. A. V. Saph, 1901, Cornell University, is used to illustrate the above method.

The arch has a span of 168.75 ft. between pin supports, a rise of 29.5 ft., and weighs
1,060 ,320 Ibs. without abutment shoes, making a uniform dead load of 43.2 kips per truss
per panel. The top chord is the loaded chord. The abutments are symmetric and,
therefore, a =0.

Fig. 50A shows the half span with the lengths of members in feet below the lines,
and the values EM, in feet, as found from Table 50A, above the lines. The various lever
arms used in the computation of the stresses S A , S B and S a , and of the w loads, are also
shown. The actual values Al in feet =EM + 29000, because the areas F were not reduced
to square feet. See also the example in Art. 52, where M is actual.

TABLE 50A
COMPUTATION OF EAl, j) A , TJ B AND E2S a 4l t .

Stresses in Kips.

Unit

End Ordinates,

Temperature Effect.

Area,

..enscth

Stress,

S a l

*ES a 4lt

Member

S A

S a

*EM t

for

-stlh

A = l

5=1

Sq.in.

ft.

Kips. sq. in.

ft.

/ Sa

T>B Sa

ft.

Kip.ft.

Kip. ft.

- 0.320

- 5.437

-0.211

19.80

15.000

-0.0107

-0.1605

1.517

25.770

197.9

41.8

u c/

- 1.165

- 6.902

-0.697

19.80

15.000

-0.0352

-0.5280

1.671

9.917

197.9

137.9

u.u,

- 2 725

- 8.953

-1.457

19.80

15.000

-0.0736

- 1 . 1040

1.870

6.145

197.9

288.3

- 5 588

-11.755

-2.649

26.48

15.000

-0.1000

- 1 . 5000

2.109

4.438

197.9

524.2

ifu*

-10.008

-14.336

-4.121

38.11

15.000

-0.1081

-1.6215

2.429

3.479

97 9

815.5

u*u*

-14.062

-4.917

38.11

15.000

-0.1290

-1.9350

2.859

197.9

973.1

T V "

01 79

+ 1 085

42 50

11 228

+ 0.0255

0.2863

-0.159

88.9

96.5

366

6.231

1.388

40.00

17.190

0.0347

0.5965

0.263

4.481

136.1

188.9

LI*

1 269

7.516

1.848

37.50

16.335

0.0493

0.8053

0.687

4.067

129.4

239.1

L L

2 857

9 . 386

2.575

37 . 50

15.725

0.0687

1.0803

1.109

3.645

124.5

320.6

5 685

11.958

3.712

37.50

15.258

0.0990

1.5105

1.532

3.221

120.8

448.4

J* L

10 027

14 . 363

5.131

42.50

15.029

0.1207

1.8140

1.954

2.799

119.0

610.6

T? L

917

-0.605

23 . 52

35.942

-0.0257

-0.9237

1.516

379 . 5

229.6

1 179

- 2.043

-0.678

19.80

29.312

-0.0342

-1.0025

1.739

3.014

309.5

209.8

r/ 1 /" 1

1 502

- 1.976

-0.732

14.70

20.918

-0.0498

-1.0417

2.052

2.700

220.9

161.7

U>L 3

- 1.857
2 042

- 1.817
- 1 193

-0.773
-0.681

11.76
11.76

14.450 -0.0657
9.730-0.0579

-0.9494
-0.5634

2.402
2.999

2.351
1.752

152.6
102.7

118.0
70.0

U L

- 1.622

+ 0.109

-0.318

11.76

6.932

-0.0270

-0.1872

5.101

-0.343

73.2

23.3

77 V 5

0.0

11.76

6.000

0.0

63.4

0.0

rr r

If) 71

13 34

32 928

. 0503

1.6563

1.515

347.7

233.3

T7T

.UI/
1 451

2 514

0.834

13.34

25.741

0.0625

1.6088

1.740

3.015

271.8

226.7

2.166
3.413
4.869
4.367

2.848
3.339
2.843
- 0.294

1.055
1.421
1.622
0.857

13.34
13.34
18.25
15.84

20.828
17.871
16 . 524
16.156

0.0791
0.1065
0.0889
0.0541

1.6475
1.9033
1.4690
0.8640

2.053
2.402
3.002
5.096

2.700
2.350
1.753
-0.343

220.0
188.7
174.4
170.6
Tot'k

232.1
268.1
282.9
146.2
3293.4

3593 . 2

* Where =29,000 kips instead of 29,000X144. lE^S a M t - 299.8 kip. ft.

170

KINETIC THEORY OF ENGINEERING STRUCTURES

CHAP. X

U '

The f i^ure] obove-the members| represent EA]
other dijnen&ions are lengths in ft.

DEF-LEcflONS ARE \E TIlllES ACTUAL IN F*ET.

E= 29OOto KIPS PER SQ.IN. S
\ .XI

" " : r "

AKT. 50 SPECIAL APPLICATIONS OF INFLUENCE LINES 171

Table 50A gives these stresses, in kips, as found by computation, also the areas F
of the members, being gross for compression and net for tension members. The lengths
I of the members, unit stresses/ and quantities EAl (for shortening) are also included
in this table, and finally the end ordinates of the S /S a influence lines are obtained.

The X a influence line is now found by two different methods to illustrate the applica-
tions frequently referred to elsewhere.

The first method is by constructing a deflection polygon of the top chord by means
of a Williot-Mohr displacement diagram, Fig. 50B, using the quantities EM, in feet, as
the changes in the lengths of the members for the conventional loading X a = l, producing
stresses S a . The second method is by finding this deflection polygon from the w loads.

The first method requires little description other than to say that the displacement
diagram Fig. 50B is drawn for displacements EAl, Table 50A, on the assumption that the
point LQ and the direction of the member L 6 U 6 remain fixed and since the span is symme-
tric about this member and the point L 6 , therefore no rotation diagram is necessary.
The deflection polygon of the top chord is then found by projecting the vertical deflec-
tions of the top chord panel points onto the verticals through these points, furnishing
the polygon ~A rT lT y with a closing line A'A", horizontally through A' '. These deflec-
tions are E times actual, in feet, measured to the scale of displacements.

The horizontal displacement \Ed aa between A and L 6 is also obtained from the same
diagram as the horizontal distance between A' and L' 6 .

The vertical ordinates of the deflection polygon are values of dma, and hence the
X a influence line ordinates y a are found by dividing the several deflection ordinates
Eo ma by the constant Ed m according to Eq. (50s), giving

This shows that the factor E and the scale of the ordinates do not affect X a . The
ordinates r] f! , for all panel points, are plotted to any convenient scale to obtain the X a
influence line.

Second method. The same displacements EM, in feet, are here employed to com-
pute the elastic loads Ew, using the lever arms, also in feet, as given on Fig. 50A. See
Table 50s.

In the present example all the members are included and the table indicates exactly
how much each member contributes to the several total Ew loads. The method of
Art. 35 is rigidly followed. By using displacements EM which are E times too large,
the w loads are also multiplied by E.

Each chord member furnishes one w load which acts at the center of moments
for that chord while each web member contributes two loads w u and w n acting at the two
adjacent panel points u and n of the loaded chord or the chord for which the deflection
polygon is to be drawn.

" The w c loads resulting from the chord members are always positive as found by

Eq. (36s), thus:

M EM ro ,

uv = or Ew c = ......... (DUK;

172 KINETIC THEORY OF ENGINEERING STRUCTURES CHAP, x

The negative and positive loads for a web member are

Al Al EM EAl

w u =- and w n = or Ew u = and Ew n = -- . . (50L)

TU r n TU T n

as given by Eq. (36o).

The lever arm r for a chord, is the distance from the center of the moments for
such chord and perpendicular to the chord. The lever arms r u and r n are perpendicular
distances onto a web member as described in Figs. 36s, and 36F. They may be identified
by comparing the values in Table 50s with the dimensioned lever arms on Fig. 50A.

The rule for the signs of the two loads for any web member was stated in Art. 36,
and is as follows: Calling all top chord members negative and all bottom chord members
positive, and giving the proper sign to Al for the web member in question, then the positive
w load is found on that side of the panel where the sign of Al coincides mth the sign of the
adjacent chord.

For the panel point f/ the total load Ew (Table O()B) is made up of the positive
load produced by the chord L Li and one of the loads from each of the web members
U Li and U\L\, the signs of which are negative according to the above rule.

The w load for point A produces zero effect on the deflection and need not be
considered.

For the panel point U l the total load Ew\ becomes

ALL JUpLi AVjLz . JE/nZq J7 2 L 2

" ""' ~~ =auby

25.7 13.4 17.15 "' 11.7

and similarly for the other panel points of the loaded (top) chord.

The bottom chord points receive only the loads produced by the top chord members
except at LI where the member U L , being the end post, contributes a load Jt/ L /10'.2.

The total Ew loads acting in the same vertical are then summed for the seven panel
points of the half span and used in constructing the deflection polygon with a pole dis-
tance H =E2yw.

The y ordinates are measured vertically from the line A~B to the points of application
of the respective Ew loads, distinguishing between the loads acting at the top and
bottom panel points.

In order that the X a influence line ordinates may appear to a scale twenty times
as large as the scale of lengths, the pole distance was made equal to E2yw/20, see Fig.
50c. Note the agreement of the ordinates here found with those obtained in Fig. 50fi.

Stress influence areas. The end ordinates i) A and r) B of the S^/Sa lines are com-
puted in Table 50A and these serve to construct all stress influence areas for the several
members.

Fig. 50c shows stress influence areas for six typical members, and each one has a
factor S a as per Eq. (50A). The end ordinates are applied down from the closing line
when positive and up when negative. The signs of the influence areas are uniformly +
for areas below the .Y a influence line and the factor p takes the sign of S a .

SPECIAL APPLICATIONS OF INFLUENCE LINES

173

f^.

,_!

a; '

~ "3 *** s

& II - -i

<*-

T 1

"8 1 1

"o"2 c

TZ "t^

i-H

3 s2

-t-> i*"!

"S-2"

^ 'M O

1 *x ei

O O

_j_

2 II

UJ C ^^

5 CD 03 c

S~i

.5 aj

S >

tit;

OJ Sl*^

" -i

ai "Sto

* *
w

Tl ,-

fiq

BJJ

SjJl

o3 ^

= 2^^

.2 ~

o Sji

J-

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2ls

sl o

ItB^l

g g

;

CC C H-.

g5s c

^ -

r+K

T-l

CD
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rf

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t-_

i-C

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i-C

rf
i I

00
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O5
(N

-3-

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o

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IN

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4* .

b" ii

b ||

^ B

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*" ,c

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- 1

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1-4

eo d

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b ii

b II

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^ ii

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&

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rf

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'" (N

"i <=>

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^ ii

S, d
r^\ ii

S, d
^ 11

b (|

*~i d

b 11

""p

KJ !l

KJ ii

""rf V

j ^8 ^

| CD rl

H CO CC

CD ff

= c3

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3 2 "

i 3"=

*4 ^5

rf
CO

- 2

^ d

1C
(N

co
b d

u? II

4* d

>-3 d

*3 d

*s. ^

^(* II

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b* II

b" II

b ii

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<N ?

4 ii

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: ^ o ?
^ 2 e

J 00 C

3 2 '

*-3 11
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2 S c

- I'
J5 C,

^1 t-

!> ^

4 *"1 00 IN
5 (N O-
i ^ C

1 O C^

"^ rf u:

O rf

'"J rf

1 t T

N S

CO CT
1-H tf

S c

i-! tf

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3 d c

O i-

b"*

1 4*

174

KINETIC THEORY OF ENGINEERING STRUCTURES CHAP. X

FIG. 50c.

ART. 50 SPECIAL APPLICATIONS OF INFLUENCE LINES 175

The method of computing stresses for any train of loads need not be repeated here
except to call attention to the load divides, which must be carefully observed for each
member in placing the loads on the span.

Temperature effects. For a general case of unequal temperatures in the several mem-
bers the following assumptions are made:

Let the normal . temperature be 65 F., for which the structure has no temperature
stresses. Then assume a case where the top chord is heated to 130 F.; the bottom
chord to 104 F., and the web system to 117 F. For s=0.000007, and #=29000
kips per sq.in. the values of Ed become

Top chord Z = 130-65 =65, Eet = 13.20;
WebSystem t = 117 -65 -52, #rf = 10.56;
Bottom chord = 104 -65 -39, Eet= 7.92.

The values E Al t = dlE and ES a Al t are computed in Table 50A, using I in feet and
E in kip feet. Finally the half sum E2S a 4l t is found to be 3293.4 -3593.2 = -299.8
kip feet. Table 50fi gives ^E^lyw=54.7Q and from Eq. (50o) for cosa = l,

- *. *. * . i *

' == ~ = ~ 5 ' 7 PS '

which is a thrust acting outward the same as the conventional loading X a =
Hence the temperature stress in any member becomes

wherein S t has the same sign as S a .

A more severe stress would be produced when the top chord is colder than the
bottom chord, or when the top chord has a temperature of +10 F. at the same time
that the bottom chord has a temperature of -16 F, a case which might occur on a clear,
cold day. The stresses S t produced by such a condition would have the opposite sign
of S a when found from +X at .

For a uniform rise of 65 above the normal, the elongation for the whole span
becomes, by Eq. (50E),

_.! A D 771

= 2226.6 ft.,

cos a
and from Eq. (SOo),

Edat 2226.6

giving stresses S t = S a X ai .

For a uniform fall of 65 below the normal ^ == -2226.6 ft. and X at = -20.33 kips,

giving stresses S t =S a X at ,

Compare these results with those of the example in Art. 59, for a solid web arch o

about the same dimensions.

176 KINETIC THEORY OF ENGINEERING STRUCTURES CHAP. X

Abutment displacements. Assuming that as a result of yielding abutments the length
of span is increased an amount J/=0.03 ft. then from Eq. (49L)

0.03 X 29 ,000 _ _ . e
~ 109.53 '' lpS '

Comparing this result with the one obtained for the solid web arch in Art. 49, it
is clear that the framed arch is almost twice as stiff.

Deflection of the crown due to the temperature effect producing X at = 5.48 kips,
Table 50A, and Eq. (50M). In this case the abutments are assumed rigid making Jr=0 ?
and the stresses S are not included, since temperature effects alone are desired. This
reduces Eq. (50c) to

i /St i ,

where S t = -S a X at and = - = -EAlX at ; also S l = for a load unit Y at the

r r Zi

center of a symmetric span, and dlE =EAlt. Hence

which can easily be computed for any X at using the values S A) EM and EAl t given
in Table 50A.

ART. 51. TWO-HINGED ARCH WITH CANTILEVER SIDE SPANS

Occasionally a structure of this type is peculiarly adapted to certain sites as the
one at High Bridge, Ky. Its application has, however, received adverse criticism because
the analysis of stresses was considered too complicated.

This objection might apply to any of the ordinary methods of analysis, but not when
the problem is solved by influence areas.

As a type of bridge it is commendable and deserves careful consideration whenever
a particular site offers a suitable profile and good foundations for the center span.

Fig. 5lA represents one of several forms which might be employed. The center
span is the same as the arch in the previous article with the exception of the end posts
which are here made vertical, thus increasing the span to 180 ft. Owing to this dif-
ference, the computations in Tables 50A and 50s no longer apply because the stresses
SA, SB and S a are now slightly changed. Otherwise the preliminary computations would
be precisely the same in both structures.

The difference between the two-hinged arch and the present structure with cantilever
side spans is entirely in the principal system which results when the redundant thrust
X a is removed. In the first case the principal system is a simple truss on two supports
A and B, while in Fig. 5lA the principal system becomes a cantilever when the hinged
support at A is made movable for the purpose of analysis. The supports at D and F
are roller bearings and the simple trusses DC and EF are hinged at C and E respectively.

ART. 51

SPECIAL APPLICATIONS OF INFLUENCE LINES

177

178 KINETIC THEORY OF ENGINEERING STRUCTURES CHAP. X

The two-hinged central span, as in any ordinary two-hinged arch, is made deter-
minate by substituting a roller bearing for the hinged bearing at A.

The XQ influence line is found as in the previous problem and the computation of the
Ew loads is not repeated here. However, the loads Ew and_EVt- 2 , must_now be included
because these loads affect the directions of the extreme rays C'A' and B'E' of the equilib-
rium polygon outside of the span AB while they have no bearing on 'the A" a influence
line within this span.

The complete X a influence line for the whole span DF is found by prolonging the
extreme rays of the equilibrium polygon A'B' to the points C" and E' and finally draw-
ing the lines D'C' and E'F'. The middle ordinate A may be computed from the force
polygon as follows:

X= " measured to scale of lengths,

4/2

measured to scale of ordinates.

The lines A'C' and B'E', found by laying off A, are respectively parallel to the
extreme rays of the force polygon.

Stress influence areas- These are found precisely as in Art.' 50, so far as the span
AB is concerned, by laying off end ordinates T,A =&A/Sa and fjB = ^B/Sa and the factor
becomes a=S a . In each case the two end rays so determined will intersect in a point
i r , which is vertically under the center of moments i of the member in q^stion. Also
the S /S a influence lines must be straight over the panel containing the member.

Outside of the span A'B the S /S a lines- are drawn as for any cantilever system as
per Art. 26.

ART. 52. FIXED FRAMED ARCHES

A framed arch with fixed ends has three external redundant conditions according
to Eq. (3c) and Fig. 3j, and hence requires for its analysis three elasticity equations either
of the form of Eqs. (7n) or Eqs. (8D) .

Temperature changes and abutment displacements come into prominence here.
These effects are bound to remain more or less problematic because the actual circum-
stances attending the construction and later life of the structure cannot be foretold
with any high degree of certainty. Therefore, it would seem unnecessary to attempt
the analysis of the load effects with any extraordinary refinement and some assumptions
may be made to simplify the work, provided they are on the side of safety.

It is nearly always permissible to neglect the effect of the web system in comput-
ing the elastic loads w. In preliminary work it is also admissible to choose a constant
cross-section for the chord members.

As a general criticism it might be added that fixed framed arches are not commendable
for the flat type, since the temperature and reaction effects produce very considerable

ART. 52

SPECIAL APPLICATIONS OF INFLUENCE LINES

179

stresses which increase as the rise diminishes. Good rock foundations must be available
under all circumstances, otherwise the fixed arch is prohibitive.

It is always advisable to separate the computation of the load stresses from the
temperature and reaction displacement stresses so as to determine the relative importance

of the latter.

General relations between the external forces and the principal system. There are several
wavs in which the three external redundant conditions may be applied, depending on
the choice of the principal system. One of these was illustrated in Figs. 44E to J, where
the principal system was a cantilever fixed at one end. Another assumption might
be made by cutting the arch at the crown and creating two cantilevers fixed at the
respective abutments of the arch.

However, the simplest determinate structure is always a truss on two supports,
and in the present case that disposition will be made, as it affords the most compre-
hensive solution.

FIG. 52A.

All three methods have been used by Professor Mueller-Breslau and others, and the
final results, are identical and involve about the same amount of labor. In each case
the external redundant conditions may be so chosen that the application of the
plified Eqs. (44A) becomes possible in accordance with the discussion in Art 44

Fi 52A illustrates the relation of the external forces for a single applied load I ,
producing reactions R, and R z , intersecting in the point C on this load. This fixes the
points ttl and b, on the verticals through the outer supports a and ftjnth span I. ine
triangle ^Cb[ thus becomes a resultant polygon with the closing line a^.
R l may be resolved into the vertical component A and the haunch thrust H along afr.
The reaction R 2 may similarly be resolved into the vertical reaction B and the
H' which latter is equal and opposite to the H' acting at ai.

' The vertical reactions A and B are the same as for a simple beam o* with deter-
minate supports A and B . Hence A =P(l-e)+l and B = Pe/l. Also the moment, f,
P intTof the simple beam, equals M^-KH, wh_ere_H is the honzonta componen
' and K is the vertical ordinate of the triangle a.Cb, through the point m. Hence

180

KINETIC THEORY OF ENGINEERING STRUCTURES

CHAP. X

K = M om /H=M om /H' cos a. Therefore, the resultant polygon a l Cb l is determined
when H' and the closing line a^bi are found.

The external redundant conditions are now removed by relieving the fixed ends
and resting the curved structure on two determinate supports at the extreme outer
points a and b, Fig. 52B. This simple truss of span /, provided with a hinged bearing at
a and a roller bearing at b, constitutes the principal system.

To each end of the principal system an imaginary rigid disk is attached as shown
in the figure by the two shaded triangles. The redundant conditions are now applied
to these disks as external forces or moments, thereby re-establishing conditions of stress
in the principal system which are identical with those produced in the original structure
while the redundant conditions were active. The two disks are not connected at 0, but
are free to transmit a set of forces independently to each abutment.

The equal and opposite forces Xb are chosen vertically, and the one acting upward
is supposed to act on the disk OA. The forces X c are equal and opposite, but of unknown
direction /? with the horizontal, the one acting to the right is applied to the disk OA. The

FIG. 52s.

two moments X a are also equal and opposite and act separately, one on each disk. The
pole is not yet determined, but will be fixed by certain geometric conditions to be
established later.

To determine the exact relations between these redundant forces and the principal
system it is preferable to discuss only the forces acting on one end of the span. In Fig.
52c, the left-hand abutment is shown with the redundant forces which are active at
that end only. A similar set, not shown, would be active at the right-hand abutment.

The structure is now referred to coordinate axes (x, y) with origin at 0. The y axis
is made vertical, and the x axis is coincident with the redundant X c making the angle
/? with the horizontal. The location of and the angle /? are still unknown. The ordi-
nate y m of any point m is measured vertically from the x axis, while the abscissa x m of
this point is measured horizontally from the y axis instead of parallel to the x axis. This
is more convenient Jnjihe considerations which follow.

The rigid disk aa'O connects the origin with the arch along aa' ', and to this origin
are applied two equal and opposite forces H' , which are equal and parallel to the original
haunch thrust acting at a^ The equilibrium of the principal system and of the fixed

ART. 52

SPECIAL APPLICATIONS OF INFLUENCE LINES

181

arch thus remains undisturbed. In the original fixed condition the force P produced
the reactions RI and R2 and these were resolved into components A and H' acting at
ai and B and H' acting at 61.

Suppose now that all the external forces to the left of a section tt act on the principal
system only, and that the three forces H' and the vertical reaction A are applied to the
rigid disk aa'O and are thence transmitted to the principal system. Then the force H'
at a\ and the force H' ', in opposite direction at 0, form a couple with lever arm z cos a
producing a moment X a =H'z cos a. The other force H', acting at and to the right,
may be resolved into two components X b and X c , where X b is vertical along the y axis,
and X c acts along the x axis. The external forces to the left of the section and acting
on the principal system, are then P, A , X b , X c and a moment X a =H'z cos a =Hz .
Of these the two forces Xb and X c and the moment X a constitute the redundant condi-
tions while the forces P and A are known and all are applied to the principal system

FIG. 52c.

to the left of the section tt. A similar set of external forces acts on the principal system to
the right of the section, but these are not shown in Fig. 52c.

The moment of all external forces about any point m of any frame, involving thre
redundant conditions, is expressed by Eq. (7 A) as

M m =M om -M a X a -M b X b -M c X c

(52A)

wherein M om =A (l i -x m ) -Pd-^ the moment about m due to the load P acting on the
principal system. This is condition X=0. . .

M a = l=the moment about m due to the moment X a = l applied to the principal

system. Condition X a =1. . . ,

M 6 = l.* m =the moment about m due to the force X b = l acting on the prmcipa

system. Condition X b = l.

M c = l.</ m cos/?=the moment about m due to the force X c = l acting on the prin-
cipal system. Condition X c = l.

182 KINETIC THEORY OF ENGINEERING STRUCTURES CHAP, x

Substituting these values into Eq. (52 A) gives the following fundamental moment
equation for fixed arches:

M m =M om l-X a x m X b y m cos8X c ...... (52s)

Before M m can be determined for any point m of the arch, the three redundants
X a , X b and X c must be evaluated from three simultaneous work equations of the form
of Eqs. (44n) , which equations may be made to apply to the present' problem by so
locating the (x, y} axes that d ab , d ac and d bc all become zero.

Since there are as many values for the redundants as there are panel points on the
arch, and for each point there will be as many values as there are positions of the mov-
ing loads requiring investigation, the only practical solution of the problem is by means
of influence lines, first for the three redundants and then for the moment M m for each
member of the arch.

The stress S in any member may always be found from the moment M m about
the center of moments m for that member and the lever arm r m measured from this moment
center perpendicular to the direction of the member. Hence

M m M om M a M b M t

o = - ; o = --- , ^a -- > &b , ana o c = , . . . (o2c)

I'm I'm ?"m ^m I'm

where the lever arm r m is constant for the same member.

Therefore, Eq. (52s) will furnish the stress in any member as

M 1
S = - = [M om -l-X a -x m X b -y m cos px e ] =S -S a X a -S b X b -S C X C , (o2c)

' m 'm

M a I M b x m M c y m cos 8

where S a = - = ; S b = ?= ; and S c = = if - E ....... (52E)

* * - ' 4* 4" V /

' m 'm 'm 'm 'm 'm

Location of the coordinate axes. According to the conditions imposed by the simplify-
ing process of Eqs. (8n), discussed in Art. 44, the coordinate axes must be so located
that the displacements d, bearing different double subscripts, must be made zero.

These displacements d as given by Eqs. (SB), then become

=0',

and substituting the values for the stresses from Eqs. (52E) then

^ xl v 1

^ 2j XW = (J

. ^xycosfll v

=04, = 1 -^f = cos P2xyw a =0

(52r)

ART. 52 SPECIAL APPLICATIONS OF INFLUENCE LINES 183

where w a = l-l/EFr 2 for a moment unity, according to Eqs. (36B) and (36o), giving
w a =4l/r=Sl/EFr=Ml/EFr 2 and representing a certain geometric function called an
elastic load for some particular pin point with coordinates (x, y).

Eqs. (52r) then represent the conditions which determine the location of the
coordinate axes such that the simple work Eqs. (44A) become applicable.

The first two conditions (52r) imply that the origin of the (ar, ?/) coordinate axes
is the center of gravity of the several elastic loads of all panel points of the principal sys-
tem. This must be so because the moments ^lxw a and %yw a could not be zero unless the
resultant Hw a passes through the origin. According to the third condition, the angle
between the axes must be such that the centrifugal moment 2>xyw a =Q, which is true
when the axes are related as conjugate axes.

The origin may then be located with respect to any assumed pair of axes as the
(z, v) axes in Fig. 52c, where the z axis is taken vertically through the point a and the v
axis is any horizontal axis conveniently located say through a'. The coordinates z, i\
of all the panel points, are then determined from the arch diagram and tabulated. The
w a elastic loads are computed from Eq. (36fl) (and Eqs. (36D) if the web members are
included) for each pin point. See the problem in Art. 51. Finally the moments zw a
and vw a are found and from these the coordinates li and z ' for the origin are obtained

from

, ^vw a , 2zw a ._ ,

''- and * ~ .....

This fixes the y axis, which is parallel to the vertical z axis through the center of
gravity 0. The x axis, while passing through 0, makes some angle ft with the horizontal
such that I,xyw a =0, according to the last of Eqs. (52r).

The (x, y) coordinates are derived from the (z, v) coordinates when the angle /? is
determined. Taking /? positive when measured to the left of the origin and below the
horizontal, or to the right and above the horizontal as shown in Fig. 52c, then

x=li v and y=z z ' +x tan /? ...... (52H)

The angle is found by substituting the value for y from Eqs. (52n) into the
condition equation, giving

z f +x tan fi]w a =0;

or I>xzw a -z f Zxw a + tan p2x 2 w a =0;

and noting that 2xw a =Q, then

(52j)

The abscissa? x being known from Eqs. (52n) the values Xxzw a and Xx 2 w a are readily
found and tan /? is then obtained from Eq. (52j). Finally, the y ordmates are computed
by the second Eq. (52n) and the new axes and (x, y) coordinates are thus determined.

For symmetric arches the x axis is horizontal and tan =0, thus greatly lessening
the foregoing computations.

184

KINETIC THEORY OF ENGINEERING STRUCTURES

CHAP. X

Influence lines for X a , X&, X c and M m . The coordinate axes (x, y) having been located
to fulfill the requirements making ^=^=0, d ac =d ca =0 and d bc =o cb =0, the simplified
Eqs. (44A) become applicable to the present problem.

Since it is desirable to investigate separately the effects due to loads, temperature
and abutment displacements, the load effect will be considered first for immovable abut-
ments. Eqs. (44A) then take the simple form of Eqs. (44B), as
YP ft

*^*- m.u tnn. -*T -. jji^ jnu i -\r *-* * m^ jnr. / ~c\ \

(,)2K)

and

Y p ^

*->* m u mc

r s

Eqs. (SB) give the displacements dm, w> and cc in terms of the stresses in the
members. Noting that p =l/EF, and substituting the values for S a , S b and S c from Eqs.
(52E) , then Eqs. (SB) give

EFr 2 '

d bb =

'EFr 2 '

and calling l/EFr 2 =w a ; x 2 l/EFr 2 =x 2 w a =xw b and y 2 l/EFr 2 =ij 2 w a =yw c then Eqs. (52x)
become for a single load unity at any point ra :

1 $mb 1 ' '

* ' "

' "

(52L)

Eqs. (52L) furnish the values X a , X b and X c for any position m of a single moving
load P TO = 1, as functions of the deflections d ma , d^ and d mc of the point m resulting from
the conventional loadings X a = ], X b = l, and X c = l.

The exact significance of the deflections 3 ina , dmb and d mc may be determined from
their values as given by Eqs. (8 A), noting that S = M om /r m for a load P^-l applied at
m from Eqs. (52c), also using the values given by Eqs. (52E). Hence

is> v^oo ' y "* om'

b lEF

M om xl

=cos

(52M)

Now ^ ma is the deflection of a point m due to X a = l and P TO =0, and by Eqs. (52M)
it is equal to the sum of the moments M om w a of the loads w a about m. Hence d ma must
represent the ordinate to a moment digram drawn for the loads w a with a pole unity.
Also, if this moment diagram be drawn with a pole H a = 2w a , then the resulting ordinates
i) a will represent ordinates of the X a influence line according to Eqs. (52L).

ART. 52 SPECIAL APPLICATIONS OF INFLUENCE LINES 185

Similarly the moment diagram drawn for the loads w b =xw a with a pole H b =
will furnish ordinates jj b for the X b influence line and a third moment diagram drawn
'or the loads w c =yw a with a pole H c = cos {32>yw c will give ordinates )j c for the X c influence
ine, noting that one of the cos /? factors cancels.

Hence the equations for the influence lines of the three redundant conditions from
Eqs. (52L), become

1 /^ <>

(52N)

dr,

TTvi TT

cos Lw c H

*-<

These influence lines remain the same for the same structure and hence are drawn

only once.

The moment influence line M m , for any point TO with coordinates x m and y m , is derived
from Eq. (52s) , giving any moment ordinate as

r, m =M m =l\l f > m -\X a +x m X b +y m cos8X c ]=rj om -[-n a +x m -f) b +y m cos fa], . (52o)

where T lom is any ordinate of the ordinary moment influence line M om , drawn for the point
TO of a simple beam on two determinate supports A and B , which is a different line for
each point TO. The ordinates r) a , y b and TJ C are those respectively of the X a , X b and X c
influence lines all under the same panel point of the truss.

Hence, the moment influence line M m for any point TO is drawn by computing the
ordinates -[rj a +x m rj b +y m cos fa] for _all panel points of the loaded cord and plotting
these ordinates negatively from the M om influence line. Positive areas thus correspond
to positive moments.

This M m influence line will serve to find the maximum and minimum bending
moments for the point TO due to any system of concentrated loads. The influence line
gives the load divides.

The stress influence line for any member may be derived from the moment influence
line drawn for the center of moments TO for that particular member. Thus if r m is the
lever arm for a certain member S with center of moments m, then S=M m /r m , and the
ordinates T?, for this stress influence line by Eq. (52o) become

(52P)

where the several 9 ordinates are as above defined and all measured under the same
panel point. Positive areas then give positive or tensile stress. The moment influence
line for the center of moments for any member may be directly used by applying the
factor l/r w to obtain the stress.

The rather lengthy operation of computing all these ordinates for the several stress
influence lines for all the members may be considerably shortened by employing the

1S6

KINETIC THEORY OF ENGINEERING STRUCTURES

CHAP. X

following suggestions: Thus the multiplications x m -fj b and y m cos /?j? c , may be performed
graphically by laying off angles whose tangents are respectively x m and y m cos /?, as illus-
trated in Fig. 52o, using the scale of the TJ ordinates.
Two such diagrams are required for each M m or S line.

If a proportional divider is at hand, then it may be
used to perform these multiplications by setting the divi-
ders first for the ratio 1 : x m and then for 1 : y m cos /?.

Aside from the above methods, the stress influence
line for any web member may be derived from the
influence lines of two adjacent chords by using the
following method given by Professor Mueller-Breslau.

When the top chord is the loaded chord, then the
FIG. 52o. adjacent members of the bottom chord are used, and

vice versa.

In Fig. 52E, assume a unit positive stress in L\ and resolve the same into com-
ponents 1 parallel to D\ and +2 parallel to D 2 .

FIG. 52E.

FIG. 52r.

Then assume a unit positive stress in L 2 and resolve this into components + n\
parallel to D\ and /i 2 parallel to D 2 . Then for the top chord loaded and no load at
tr>, the stresses D\ and D 2 become

D 2 =+e 2 L l -

+L 2

(52Q)

By substituting influence ordinates for the stresses D and L this furnishes a ready
means of deducing a D influence line from the two influence lines of the adjacent chords.
Thus the DI influence area is the area between the L 2 line and the z\L\/ // t line and the

ART. 52

SPECIAL APPLICATIONS OF INFLUENCE LINES

187

final Di area thus obtained will have a factor m. The ordinates to be used in the equa-
tions are always under the same load point.

When the bottom chord is the loaded chord and there is no load at the point n,
then using the stresses as indicated in Fig. 52F, the diagonals D 2 and D 3 become

(52 R)

The value of D 2 will be the same in both equations Q and R and the signs will come
alike by noting that the influence ordinates for the L and U lines have opposite signs and
hence the coefficients s and have opposite signs from those used in Eqs. (52 Q ).

The resultant polygon for any case of applied loads P may be located by finding the
corresponding values of X a , X b , and X c from the three influence lines for thes
having previously located the (a;, y) axes.

The redundants X e and X b , Fig. 52c ? were taken as the components of H , respec-
tively coincident with the x and y axes. The angle /?, which the x axis makes with t
horizontal is given by Eq. (52j) and the y axis was taken vertically,
izontal component H of X c is

H =X C cos /?, and from Fig. 52c,
tan a.=t&n (3+-jj

cos a

z ~

cos a

X a = Xg

H'cosa H

. . (52s)

X a =

COS =

These dimensions fix the location of the closing line a
of the abutments also the haunch thrust H' all in terms of

on the two end verticals
X and X

188 KINETIC THEORY OF ENGINEERING STRUCTURES CHAP. X

A force polygon is drawn by laying off all the loads P in proper succession, dividing
this load line into the parts A and B and at the dividing point draw a line parallel to
a[bi of length equal to H' '. This determines the pole of the force polygon from whicli
the reactions RI and R 2 and the resultant polygon through ai and 61 are easily drawn.
See also the force polygon in Fig. 52A, showing how the pole is located when H' , A
and B are given.

Temperature stresses. For the general case of temperature effects, each member may
be supposed to undergo some change 4l t =etl in its length. These changes will pro-
duce a deformation in the frame, giving rise to external redundant forces X at , X bt and X ct ,
which may be found from Eqs. (44c), where abutment displacements and load effects
are excluded.

Using the values of d a t, $bt and d ct as given by Eqs. (Sc) and then noting the sub-
stitutions for Saa = w a , dbb = 2>xw b and d cc = *>yw c made in Eqs. (52N) , the Eqs. (44c) become

Y _dat_'S a Jlt

-A at * vi

Oaa

_d bt _

J\.bt ^

COS

(52T)

where d at represents a rotation measured in arc, while d bt and d ct are the displacements
of the origin 0, due to the assumed temperature changes measured in the directions of the
y and x axes respectively.

These redundant conditions would then produce a moment M m t about any point
m as given from Eq. (52s), where M om =Q, thus

-M mt =X at +x m X b t+y m cos^X ct . ..... (52r)

Since the changes Al t in the lengths of the members represent a simultaneous condition
and the stresses S a , S b and S c were previously found in computing the w loads, it is best
to solve Eqs. (52T) analytically, using the denominators previously found for the pole
distances of the X influence lines. The moment Eq. (52u) can then be solved for any
moment point to obtain the stress in .a corresponding member.

For a uniform change in temperature of t from the normal, an approximate deter-
mination of temperature stresses may be obtained for arches of high rise. For flat arches
this assumption would not be permissible. When the effect of the redundants X a and
X b is neglected then the stresses may be found from Eqs. (52T) and (52u) as

v -- ........ (52v)

The effect of abutment displacements may be investigated in a similar manner, using
Eqs. (44A) and omitting the terms representing the loads P and the temperature. The

\BT. 52 SPECIAL APPLICATIONS OF INFLUENCE LINES 189

iisplacements Ar must then be assumed or estimated and o a , d b and d c become zero. Hence
the redundants and moments may be found in precisely the same manner above illustrated
for temperature effects, thus

Xar

**.

(52w)

where the d's in the denominators are those used in Eqs. (52T).

Example. Owing to the comparatively few fixed arches of the framed type in
existence, it was difficult to find a suitable example to use in illustrating the above method
of analysis. For this reason the present example, Figs. 52a, was taken with slight modi-
fication, from the one given by Professor Mehrtens in his " Statik der Baukonstruk-
tionen," Vol. Ill, p. 343.

The bridge has a clear span of 162 ft. and rise of 23.69 ft. The roadway is 30 ft.
wide and the estimated weight is about 400,000 Ibs., making 20 kips per truss per panel.
The uniform live load is taken at 110 Ibs. per sq.ft. of roadway or 30 kips per truss per
panel, for medium highway loading. The top chord is the loaded chord and the arch
is symmetric, making /? 0.

" The top chord panel points are on the arc of a circle whose radius is R u = 163.7 ft.
The chords are parallel and the radius for the bottom chord points is R t = 148.7 ft.

Ordinarily the computation might be carried out by neglecting the effect of the
web members, but for the sake of completeness all members will be included.

The cross-sections F, lengths I, and lever arms r of all the members, are tabulated
in Tables 52A and 52n, and the Ew a loads are computed by successive steps indicated by
the headings of the columns. The lever arms r u and r n are found as described in Figs.
36B and 36F In the present solution the areas were converted into sq.ft. in computing
Al so that the modulus E enters with its real value of 4,176,000 kips per sq.ft. instead of
29 000 kips per sq.in. as used in the problem of Art. 51. The stresses S a due to a moment
fc-1 kip ft., are easily found, as the reciprocals of the lever arms r, being careful 1
observe the signs of the stresses.

The arch being symmetric, the y axis is known to be the vertical axis of symmetry,
and the angle ft, which the x axis makes with the horizontal, becomes zero. Hence, the
(x y) axes are located by computing the ordinate z ' =E2zw a /EXw a -.21.927 *; thus
determining the distance from the v axis to the center of gravity of the w a loads, bee
Table 52c. The y ordinates then become y=z-z '. The (x, z) coordinates must of
course be determined from the arch dimensions or the equations of the chord curves.
Table 52c then gives the functions Ew b =Exw a , Ew c =Eyw a , Exw b and Eyw c all in
terms of the w a loads. The sums EXw a , EXxw b and E2yw e represent the pole distances
for the X a , X b and X e influence lines such that the influence ordinates will
when measured to the scale of lengths.

190

KINETIC THEORY OF ENGINEERING STRUCTURES

CHAP. X

X c INFLUENCE LIKI^-ORDINATES IOO TIMtSfO SCALE OF LENGTHS

NCE LINE-ORDIMATCSACTUALTOSCALEOFLfNGTHS. !
I

FIG. 52c.

SPECIAL APPLICATIONS OF INFLUENCE LINES

191

TABLE 52A
w a LOADS FOR CHORD MEMBERS

.S',,

S a l

Chord.

*EJl = jr

*Ew a = ^

Panel
Point.

Sq.in.

ft.

Kips.

Kips sq.ft.

ft.

100

20.73

14.92

-0.0670

-0.0965

-2.002

0.1341

2 4

110

19.51

14.84

-0.0674

-0.0883

-1.728

0.1165

4 6

120

18.72

14.79

-0.0676

-0.0811

-1.512

. 1022

fi 8

124

18.25

14.77

-0.0677

-0.0786

-1.440

0.0975

8 10

130

18.03

14.79

-0.0676

-0.0737

-1.325

0.0896

20.59

15.14

0.0661

0.1161

2.390

0.1580

Q K

19.33

15.21

0.0657

0.1008

1.944

0.1277

^ 7

100

18.55

15.25

0.0656

0.0945

1.757

0.1152

6 .

7 9

104

18 . 13

15.28

0.0654

0.0906

1.642

0.1074

1(9-9')

128

9.00

15.31

0.0653

0.0734

0.662

0.0432

* Above w a loads are for ' = 29,000X144=4,176,000 kipa per sq.ft.

TABLE 52s
w a LOADS FOR WEB MEMBERS

Member.

F
Sq.in.

I
ft.

ft.

Kips.

S a
f-p

ips. sq.ft.

.<s ;
tUl-^s

ft.

r u
and
r n

ft.

*Edl

* Ew a =

For loaded panel
Points.

At
Panel
Point.

0-1
1-2

2-3
3^

4-5
5-6
6-7
7-8
8-9
9-10

8
14

18
10
18
6
14
6
10
6

15.03
24.06

15.26
21.79
15.66
20.09
16.19
18.78
16.91
17.76

780.87
132.91
443.54
154.04
331.25
175.75
260.26
206.93
231.40

+ 0.00128
0.00752
0.00225
0.00649
0.00302
0.00569
0.00384
0.00483
0.00432

+ 0.0132
0.0602
0.0324
0.0519
0.0725
0.0585
0.0922
0.0696
0.1037

+ 0.3176
0.9186
0.7060
0.8128
1.4565
0.9471
1.7315
1 . 1769
1.8417

{12.81
\13.05
/21.90
\18.96
[13.38
\13.65
J19.98
\17.70
/ 13.80
1 14.40
/18.45
\16.62
/14.37
\15.18
/17.10
\15.84
J15.03
\16.05

0.0248

2
2
4
2
4
4
6
4
6
6
8
6
8
8
10
8
10

0.0243
0.0419

0.0485
0.0533

0.0517
0.0407

0.0459
0.1055

0.1011
0.0513

. 0570
0.1205

0.1141
0.0688

0.0743
. 1225

0.1148

Above w. loads are for = 29,000X144 = 4,176,000 kips per sq.it.

192

KINETIC THEORY OF ENGINEERING STRUCTURES

CHAP. X

TABLE 52c
COMPUTATION OF w LOADS AND POLE DISTANCES

T/it <1

Points.

own
Ew a
Loads.

Ezw a

y=z-z '

Exw a

Ew c =

Eyw a

Exwb

Eyw c

Pin
Points.

ft.

0.0248

12.03

0.2983

- 9.897

2.232

-0.245

200.88

2.429

0.1341

0.00

0.0

-21.927

10.862

-2.940

879 . 83

64.474

0.1451

22.31

3.2372

+ 0.383

10.447

+ 0.056

752 . 20

0.021

0.1165

9.99

1.1638

-11.937

7.340

-1.391

462.39

16.601

0.1893

29.83

5.6468

+ 7.903

10.222

+ 1.496

552 . 00

11.823

0.1022

17.03

1.7405

- 4.897

4.599

-0.501

211.55

2.451

0.1292

34.99

4 . 5207

+ 13.063

4.651

+ 1.688

167.44

22 . 046

0.0975

21.53

2.0992

- 0.397

2.633

-0.039

71.08

0.015

0.1040

38.01

3.9530

+ 16.083

1.872

+ 1.673

33.70

26.900

0.0896

23.69

2 . 1226

+ 1.763

0.806

+ 0.158

7.26

. 279

4(10)

0.0027

39.00

0.1053

+ 17.073

0.0

+ 0.046

0.00

0.787

4(10)

-2w? a =

-2zw a =

1 . 1350

24 . 8874

55.664

+ 5.117

3338.33

147.826

24 . 8874

-5.116

H o=2. 270

*o'~

= 21.92

7ft.

#b = 6676.6

H c = 295.65

1.135

TABLE 52o
ORDINATES FOR MOMENT INFLUENCE LINES

M 4 Influence Line,

M 5 Influence Line.

Panel

rim = rjom [r)a + 54 T/b +7.9)} c ]

Tjm = rjom [rja + 45)?6 4.897 JJr]

Point.

i a

/ c

ft,

ft.

TjO

54, 6

7.9, e

rjm

45, b

4.897>? c

8.2

0.078

0.100

14.4

4.21

0.79

1.20

13.5

3.51

0.49

2. 28

14.6

0.104

0.330

28.8

5.62

2.61

5.97

27.0

4.68

1.62

9.34

18.2

0.088

0.535

25.2

4.75

4.23

-1.98

31.5

3.96

2.62

11.96

20.0

0.048

0.647

21.6

2.59

5.11

-6.10

27.0

2.16

3.17

8.01

20.5

0.000

0.657

18.0

0.00

5.19

-7.69

22.5

0.00

3.22

5.22

20.0

-0.048

0.647

14.4

-2.59

5.11

-8.12

18.0

-2.16

3.17

3.33

18.2

-0.088

0.535

10.8

-4.75

4.23

-6.88

13.5

-3.96

2.62

1.8S

14.6

-0.104

330

7.2

-5.62

2.61

-4.39

9.0

-4.68

1.62

0.70

8.2

-0.078

0.100

3.6

-4.21

0.79

-1.18

4.5

-3.51

0.49

0.30

All ordinates are in feet.

ART. 52 SPECIAL APPLICATIONS OF INFLUENCE LINES 193

The three influence lines for the redundants are now drawn by constructing the
force polygons, using the w a , w b and w c forces in the order of the loaded panel points and
drawing the corresponding equilibrium polygons. The scales used for the w forces arc
any convenient scales, noting that the pole distance must of course be laid off to the same
scale as the forces. The poles H b and H c were divided by one hundred, thus making
the tjb and rj c ordinatcs hundred times actual when measured to the scale of lengths.
The w c forces being of both signs it is best to plat their algebraic sums from some initial
point of the load line and number the points in the order of the loads, thus 0, 1,2, 3, etc.,
to 10. The pole rays are drawn in the same order. Note the check by which the end
rays of the X a line intersect in a point on the y axis.

If the w a forces were made to act horizontally, then by the method given in Art.
38, the horizontal resultant of these forces would be obtained acting at the intersection
of the outer rays. This resultant EHw a would intersect the y axis in the center of gravity
0. However, the method of finding zj by computation is preferable, as the point
must be accurately located.

The moment influence line for any point as panel point 4, is now constructed by
computing the i? w ordinates from Eq. (52o) as illustrated in Table 52o. The coordi-
nates of point 4 are x =54 ft. and y=7.9 ft. and the rj a , T) b and rj c ordinates are scaled
from the three X influence lines. The rj ordinates are obtained from the M o4 influence
line which is the ordinary moment influence line for point 4 of a simple beam of span
Z = 180 ft. The table indicates the details of combining these several ordinates to obtain
the ordinates which are finally plotted (Figs. 52c) , giving the M 4 influence line. The
MS line is similarly found.

The M 4 influence line may be used to obtain the stress S 3 _ 5 in the chord 3-5 with
lever r = 15.21 ft. Since S=M/r, this same influence line becomes a stress influence
line with a factor l/r.

The M 5 influence line may thus be regarded as the stress influence line for the
chord 4-6 with a factor l/r = 1 -*- 14.79 =0.068.

Stress influence lines might also be developed from the formula

S =S S a X a S b X b S C X

C ,

but this would require computing the stresses S b and S c in addition to the S a stresses
already found, and the stresses S A and S B required to draw the ordinary S line for any
member in question.

The influence lines for the web members can best be found from the chord influence
lines as described in the theoretical portion of this article.

It is not considered necessary to go further into this problem, as the theory is fully
treated and the general method is illustrated by finding a single stress influence line.

The temperature stresses are determined precisely as previously outlined.

CHAPTER XI
DESIGN OF STATICALLY INDETERMINATE STRUCTURES

ART. 53. METHODS FOR PRELIMINARY DESIGNING

The term indeterminate, according to previous definitions given in Art. 2 and the
tests for identification discussed in Art. 3, is always used in the sense that the laws ot
pure statics fail to give a solution when applied to structures involving statically inde-
terminate or redundant conditions. In all structures of this class the analysis of stresses
can be accomplished only by resorting to the theory of elasticity. Hence, indeterminate
is not synonymous with impossible only in so far as the application of the laws ot statics
is concerned.

The complete analysis of a statically indeterminate structure, according to Eqs
(7 A) and (7n), involves the solution of three separate problems as follows:

1. The determination of all internal stresses So, resulting in the principal system
from the externally applied loads P.

2. The determination of all internal stresses St and Sr produced in the principal
system by the temperature changes and abutment displacements.

3. The determination of elastic deformations in the principal system produced by
the several stresses S a , Sb, S c , etc., S t and S r from which the redundant conditions X a ,
Xb, X c , etc., X t and X r may be evaluated.

Eqs. (7n) and (SD) , furnish the solution of these three problems for any indeterminate
structure by means of considerations which are of similar character in each problem.

A statically determinate structure can always be analyzed by the methods known
in statics, when the live loads and the dead loads of the structure are known. Experience
in shop practice supplies approximate data for the weights of structures in common use,
but new types of structures, departing from the ordinary forms, necessitate repeated
approximations to determine the dead loads before a final analysis for the given live
loads becomes possible.

A statically indeterminate structure is likewise incapable of analyis until its dead
weight is known with some degree of certainty and will require a preliminary design
which is much more difficult to approximate than in the case of determinate structures,
because this involves a knowledge of the magnitude of the redundant conditions. These
redundant conditions in turn require that the cross-sections of the members be also
known.

Fortunately, the influence of a system of loads can be made to depend on relative
cross-sections, thus rendering an approximate solution of the redundants X possible,

194

ART. 53 DESIGN OF STATICALLY INDETERMINATE STRUCTURES 195

before the actual cross-sections are definitely evaluated. This is done by making cer-
tain assumptions like adopting a uniform cross-section for all chord members and
neglecting the web members entirely, whence the variable F, involved in p=l/EF or
Eq. (42u) , is treated as a constant. The w loads required for constructing the deflection
polygon for the loaded chord for any condition X = l may then be made EF times the
values given by Eq. (36s) , using a pole distance of EF. The details of this process will
be discussed later; suffice it to say now that it becomes possible to construct an approx-
imate influence line for any redundant condition without knowing the final cross-sections
of the members.

With the aid of these approximate X influence lines the redundants may be evaluated
for any given live loads and some assumed dead loads. Then applying all these loads
simultaneously to the principal system, the first approximate values for the stresses
S may be determined from a Maxwell .diagram. The redundants X ar*e applied along
with the external forces.

These stresses S will now serve as a basis for a close approximation of the cross-
sections F from which our new influence lines for the redundants X may be found in the
usual manner as illustrated by the problems in Chapter X. A final analysis of the stresses
is now possible and from this the final sections are found. If the first assumption for
dead load was grossly in error then the first X influence lines might be used again, employ-
ing revised dead loads before the final analysis is made.

This process is on the whole similar to the one above cited for new types of deter-
minate structures, the dead loads for which are not accurately known. The additional
difficulty here encountered consists in evaluating the redundants, which depend both
on the sections and loads at the same time. Therefore, all such work is likely to be some-
what tedious.

The method of determining approximate values for the redundants in terms of the
mutual relations of the sections of the members among themselves, instead of the actual
sections, will now be described.

In some cases as for arches with parallel chords it is always permissible to assign
a constant section to all chord members and neglect the web system entirely. This
is then a very simple case and affords a ready solution for the X influence lines by com-
puting all w loads EF times too large.

When there is only one redundant condition or when there are two or three of
these so chosen as to comply with the conditions discussed in Art. 44 then for the above-
mentioned case of constant chord sections and disregarding the web system, the redundants
may be found analytically from Eq. (42o) whence

S , 53 ,

- . .

where the stresses S must be obtained from a Maxwell diagram drawn for a maximum
case of combined live and assumed dead loads for the principal system. The stresses

196 KINETIC THEORY OF ENGINEERING STRUCTURES CHAP. XI

S a are found for the conventional loading X a = 1 from a similar stress diagram, and these
are independent of any actual loads. Eq. (53 A) shows that whenever any of the variables
common to the numerator and denominator become constant, they cease to affect the
value of X a .

The stress in each member of the principal system may now be found from

S=S S a X a , ... ........ (53s)

and the first approximate sections F may be evaluated from these stresses.

The approximate effect due to changes in temperature may also be included b}
finding X at from Eq. (44c) and Eqs. (SB) and (Sc) as

o at ZS a tl

A a< =^ = -- . , ......... (o3c ;

Oaa *

using some assumed constant value for F.

If for any reason the lengths of the chord members are all equal, then I, in Eqs,
(53A) and (53c) , would likewise be eliminated.

A somewhat more comprehensive method, especially when the chord sections are
not assumed equal, is obtained by drawing the influence lines for the redundants. This
method will now be outlined.

When the redundants are so chosen that the simplified Eqs. (44s) become applicable
then the X influence lines may be approximated by computing w loads for relative
sections of the members. Thus for any redundant

where F c is some constant chosen about equal to a typical chord section as described
below.

The w a loads from Eqs. (36s) and (36c), for condition X a = l, then become

and multiplying each by EF C , then,

^. . (53 E )

ART. 53 DESIGN OF STATICALLY INDETERMINATE STRUCTURES 197

M a is the moment, with lever arm r, produced by the conventional loading Z a = l.

If a deflection polygon be drawn for these EF c w loads with pole distance unity, the
ordinates r? will represent values EF c d ma measured to the scale of lengths. Hence the
numerator of Eq. (53n) represents the sum of the products Prj, and the denominator is

2>S%pr = ^p/--jr=EF c 'ZM a w, according to Eq. (53E), thus making it possible to

construct any X influence line provided suitable assumptions for the values F C /F can
be made.

The equation of the X influence line for a load P = 1 then becomes

77T TJT ^ 77T TJ1 JJt

Zj2jr c O ma Hir c u ma
- = . f^F^

7^ FW^ 1\/T iii * ^uor^

2>S~l-pT

in which F c need not be numerically known, but the ratios F C /F, for different members
of the frame, must be approximated in order to compute the EF c w loads from Eqs.
(53E).

The value of F c should be chosen equal to that of a chord section of most frequent
occurrence, which would usually be nearest the crown of an arch or where the chord
approaches the horizontal. This will make F c /F = l for most chord members and
usually that assumption may be made for both top and bottom chord members. The
web system may be entirely neglected in the first approximation or if it is thought
desirable to include the web members then F C /F may be given some constant value
ranging from 2 to 10 for these members, depending on the character of the structure.

The F c should not be regarded as an average value of the chord sections because
the deflection polygon of a frame is governed largely by the chords near the center of the
span. Thus, if the depth of an arch is much greater at the springing than at the crown,
then the crown sections will govern.

From the approximate X influence lines the redundant conditions may be evaluated
for any case of total loading and the stresses S , S a , etc., can be found by Maxwell dia-
grams or otherwise. Then Eq. (53B) will furnish the stresses S from which the sections
of the several members are derived. The stresses S may also be obtained from a single
stress diagram by applying all redundants and external loads simultaneously to the prin-
cipal system. Temperature effect may be included in the redundants at the same time.

Another method of making a preliminary design would result from the use of a
Williot-Mohr displacement diagram as used in Art. 50. However, the deflection poly-
gons as found from the w loads are more easily obtained except for oblique loads P.

To illustrate the entire procedure of making a preliminary design for a structure
involving redundant conditions, the fixed arch in Art. 52 will now be investigated.

Example. The data for this arch are given in the example of Art. 52 and need not be
repeated here.

The EF c w a loads will be computed from Eq. (53E) making F C /F = 1 and neglecting
the web system. The moment M a = I and the lengths I, lever arms r and ordinates z
are taken from Tables 52A and 52c.

198

KINETIC THEORY OF ENGINEERING STRUCTURES CHAP. XI

TABLE 53A
COMPUTATION OF APPROXIMATE LOADS EF c w c

Point.

Chord.

I
Feet.

r
Feet.

Feet.

z
Feet,

EF c w a ^~

zEF c w n

0-2

20.73

14.92

. 00449

0.00

0.0931

0.000

2-4

19.51

14.84

0.00454

9.99

0.0886

0.885

4-6

18.72

14.79

0.00457

17.03

0.0856

1.458

6-8

18.25

14.77

0.00458

21.53

0.0836

1.800

8-10

18.03

14.79

0.00457

23 . 69

0.0824

1.952

1-3

20.59

14.14

0.00500

22.31

0.1030

2.300

3-5

19.33

15.21

0.00432

29 . 83

0.0835

2.491

5-7

18.55

15.25

0.00430

34 . 99

. 0798

2.792

7-9

18.33

15.28

0.00428

38.01

. 0784

2.980

4(9-9')

9.00

15.31

0.00427

39.00 0.0384

1.498

0.8164

18.156

This makes z ' =

2zEF c w a 18.156

= 22.24 ft. from which the origin can be located

ZEF c w a 0.8164

and the ordinates y=zz ' be computed. The remaining w loads and pole distances
are then easily found.

TABLE 53 B
APPROXIMATE w LOADS AND POLE DISTANCES

Coordinates.

Point.

EF c Wa

EF c w b =

xEF c W a

EF c w c =
yEF c w a

xEF c Wb

yEF c Wc

Feet.

0.0931

-22.24

7.541

-2.070

610.82

46: 04

0.1030

+ 0.07

7.416

+ 0.007

533.95

0.00

0.0886

-12.25

5.582

-1.085

351.67

13.29

0.0835

+ 7.59

4.509

+ 0.634

243.49

4.81

0.0856

- 5.21

3.852

-0.446

173.34

2.32

0.0798

+ 12.75

2.873

+ 1.017

103.43

12.97

0.0836

- 0.71

2.257

-0.059

60.94

0.04

0.0784

+ 15.77

1.411

+ 1.236

25.40

19.49

0.0824

+ 1.45

0.742

+ 0.120

6.68

0.17

4(10)

0.0384

+ 16.76

0.000

+ 0.644

0.00

10.79

Totals..

0.8164

36 183

3 660

2109 72

109 92

H a = 1.633

+ 3.658

H b = 4219. 4

tf c = 219.8

The three influence lines for X a , X b , and X c may now be drawn precisely as was done
in Figs. 52c, by using the values in Table 53B.

ART. 53 DESIGN OF STATICALLY INDETERMINATE STRUCTURES 19i)

Comparing the ordinates found from these influence lines, Fig. 53A, with those pre-
viously obtained in Art. 52, it will be seen that the former are fairly close to the correct
values, but inclined to be a little large.

X,, INFLUENCE LINE - ORDtNATES ACTUAtTOSCALEOF LENGTHS-

ISO'

FlG. 53A.

The live load per truss per panel was stipulated at 30 kips and for an assumed dead
load of 20 kips the total load would be 50 kips. .

Using the ordinates from the influence lines in Fig. 53A and the total load of 50 kips
per truss per panel, the redundants X a , X b and X c are computed in the following Table
53c. For a symmetric loading X b =0 and need not be considered.

200

KINETIC THEORY OF ENGINEERING STRUCTURES

CHAP. XI

TABLE 53c
REDUNDANTS A' a AND X c FOR TOTAL LOADING

Ordinates.

Redundants.

Panel

Point.

f]a

TjC

Feet.

Kips.

Kip- feet.

Kips.

8.4

0.086

420

4.3

X a = 2X3730 = 7460 k. ft.

14.5

0.298

725

14.9,

X c = 2X97.9 = 195.8 kips

19.0

0.520

950

26.0

A = iSF = 225 kips

21.5

0.684

1075

34.2

i(10)

22.4

0.740

560

18.5

Totals

225

3730

97.9

The redundants might be obtained for any position of the live load, but since the
chords are stressed to their maximum for total loading over the span, and since the web
system plays a rather unimportant part in structures of the class here considered, no
further investigation of stresses is warranted at this point.

The stresses for the total loading, including the redundants X a and X c applied as
external forces, are now found from a Maxwell stress diagram in Fig. 53 B, and from these
the preliminary cross-sections of the members are deduced.

From the sums in Table 53A, the value of z</=22.24 ft. This locates the pole
on the y axis. The_rigid disk, upon which the redundants act, may now be replaced by
two rigid members 10 and 00. For the symmetric total loading, X b =0, and hence the
external loads are X c applied at 0; and a moment X a applied anywhere to the disk; a
vertical end reaction A ; and the loads P acting at the several top chord panel points.

In order that the moment X a may be incorporated in the stress diagram, it must be
resolved into two equal and opposite forces acting on the rigid disk. Since these forces
may be applied anywhere on the disk, it is best to choose the lever arm 1/2, thus making
the forces V =2X a /l each, one acting upward at the support A and the other applied
downward at the pole 0. The forces V and X c , acting at 0, are then combined Jnto a
resultant Z and from this resultant the stresses in the rigid members 10 and 00 are
found, making it possible now to draw the stress diagram beginning at the reaction A.
The remainder of the work offers no particular difficulty.

It might be desirable to test out the resultant polygon with the aid of Eqs. (52s).
From these 2 =Z a /X c =38.10 ft. and the ordinate CD, Fig. 53s, determines the inter-
section of the resultants R, RI and R 2 in the point C. The closing line aibj. for symmetric
loading, is fully located when OD=z is given, hence the points a } and 61 are found on
the horizontal through D. The resultant RI is obtained by combining A and X e at a!
and R 2 is the resultant of B and X e at 6 t . The resultant R = 2P must be in equilibrium
with RI and R 2 (see Fig. 52A) , hence the three forces must intersect in a point C.

A further check is obtained by computing the ordinate CI>=Z#/4X C = 103.4 ft.

The stresses S are now tabulated and cross-sections F are determined in Table 53D,
using a rather low unit stress of say 9200 Ibs. per sq.in., instead of an allowable 15,000

ART. 53 DESIGN OF STATICALLY INDETERMINATE STRUCTURES 201

Ibs. This then covers extra metal for joints, splices, latticing, etc., and for reduction
to net sections for tension members, and for l/r for columns.

o BOO KIM.

FIG. 53s.

TABLE 53D
PRELIMINARY STRESSES AND SECTIONS

Top Chord.

Bottom Chord.

Diagonals.

Mem.

Kips.

Sq.in.

Mem.

S
Kips.

F
Sq.in.

Mem.

S
Kips.

F
Sq.in.

Mem.

S
Kips.

Sq.in.

0-2
2-4
4-6
6-8
8-10

- 920
- 1028
-1114
-1178
-1210

100
110
120
124
130

1-3
3-5
5-7

7-9
9-9'

748
846
930
970
988

94
100
104
128

0-1
2-3
4-5

6-7
8-9

- 70

+ 166
+ 160
+ 117
+ 95

8
18
18
14
10

1-2
3-1
5-6

7-8
9-10

140
- 90
- 55

+ 10

+ 48

14
10
6
6
6

202 KINETIC THEORY OF ENGINEERING STRUCTURES CHAP. XI

These cross-sections were used in the example of Art. o2, in making the complete
analysis.

Naturally a two-hinged arch or other structure involving one redundant only,
would be susceptible to the same method of design above outlined, though the labor would
be greatly lessened.

Foi an unsymmetric arch the angle /? would have to be determined- and the closing
line aibi would not be horizontal.

ART. 54. ON THE CHOICE OF THE REDUNDANT CONDITIONS

In the analysis of a structure involving redundancy, it becomes necessary to remove
the redundant conditions, whether external or internal, and thereby reduce the indeter-
minate system to a determinte principal system to which the redundant conditions are
applied along with the external loading.

The particular reactions or members best suited to represent the redundant con-
ditions designated by X a , Xj,, X c , etc., are those which reduce the given structure to the
simplest possible principal system. This will usually offer no difficulty.

However, many cases present themselves wherein it would be difficult to decide
which of several possible assumptions would represent the most judicious selection of the
redundants in the direction of simplifying the analysis. Therefore, a few suggestions
along these lines will not be out of place.

According to the general method just referred to, the elastic deformation of an
indeterminate structure subjected to loads P, will be exactly the same as that of its prin-
cipal system loaded with loads P, X a , X b , X c , etc. In each case the deformation is
entirely derived from the elastic changes Al in the lengths of the members of the principal
system.

The structural deformation is thus affected by the redundant conditions merely
to the extent of altering the stresses in the necessary members of the principal system.

Hence, for the same case of loading, the stresses S in the principal system are
always diminished by the redundant conditions provided temperature stresses and
abutment displacements are excluded.

Generally speaking, any member or reaction of an indeterminate structure may be
removed to produce the principal system so long as the latter still remains a stable,
determinate structure and does not become subject to infinitesimally small rotation,
a condition described in Art. 3, Fig. 3c.

The rule should be to select a principal system of the simplest possible form, always
avoiding composite structures, such as the three-hinged arch or a cantilever, whenever
a simple beam or truss could as well be used.

Solid web structures should always be transformed into externally determinate beams
by assigning the redundant conditions to the supports.

Framed structures, if externally indeterminate, should always be so transformed
as to remove the external conditions. The only exception to this rule might be a con-
tinuous girder wherein a top chord member over each intermediate pier might be treated
as a redundant member.

ART. 55 DESIGN OF STATICALLY INDETERMINATE STRUCTURES 203

In the case of the fixed arch, always involving three redimdants, several assumptions
may be made. First removing one fixed end and thus reducing the arch to a simple
cantilever arm acted upon by a moment X a , a vertical force X b and a horizontal force X c .
Second, by removing two members adjacent to one end support and one member adjacent
to the other end support, thus forming a girder on two determinate supports acted upon
by three external forces replacing the three members thus removed.

Also three members may be removed at the crown converting the arch into two
cantilever arms with three external forces X a , X b , and X c applied to the end of each arm
to replace the redundant members.

Frequently all the redundant conditions (not exceeding three) may be applied at
one point as described in Art. 44. Then for certain assumed directions of the X's, the
work equations become exceedingly simple and afford excellent graphic solutions. The
best of these possible solutions was chosen in Art. 52, where a complete fixed arch problem
is solved.

Indeterminate structures, having a vertical axis of symmetry, will always have two
d's bearing double subscripts of like letters, which become equal. Thus d aa =d bb , which
condition greatly simplifies the determination of X a and X b , as illustrated in the case of
a girder over four supports with the two outer spans equal. See Fig. 43A. In this and
similar problems, when the above mentioned symmetry exists, the two influence areas
for X a and X b will also be equivalent but symmetrically placed.

When the redundant conditions are internal then the only way of deriving the princi-
pal system is to remove such redundant members and replace them by external forces A'.

Composite structures, or those composed of several determinate frames combined
into an indeterminate system, are best treated by assigning the redundant conditions
to the reactions and treat as for external redundancy.

ART. 55. SOLUTION OF THE GENERAL CASE OF REDUNDANCY

In the following it will be assumed that the preliminary design is completed and it
is now desired to make the final analysis for stresses due to any causes such as loads,
temperature and abutment displacements.

Problems of the general type are usually quite complicated, and since it is important
to know the stresses due to loads, to temperature changes and to yielding supports sepa-
rately, these should always be dealt with in this manner. This is not done merely as a
matter of convenience, but it is necessary to know the relation between the load and
temperature stresses as a criterion in judging the merits of any particular structure
under consideration.

In the general discussion which follows here, only the load effects will be included,
while the other matters will be taken up later. Hence the quantities 2#Jr and d t , S t ,
etc., will all be neglected, assuming for the present that they are all zero.

To make the case perfectly general, no distinction will be made between external
and internal redundancy, hence any of the redundants may be assumed to belong to
either class.

For n redundant conditions, Eqs. (?A), offer a general solution for the stress S in any

204

KINETIC THEORY OF ENGINEERING STRUCTURES

CHAP. XI

member, the moment M about any point, the shear Q on any section or any of the
reactions R, all for the principal system only. The redundant conditions X, involved
in these equations, are treated as external forces applied to the principal system along
with the loads P and their reactions. These redundants may be found from Mohr's
work equations (?H) or (80) of which there will always be as many equations as there are
redundants, rendering the solution of the X's possible by solving for simultaneous values
of n unknowns X in n equations.

It is usually perferable to employ Eqs. (80) in terms of deflections which are
obtained from deflection polygons of the loaded chord, though in principle the solution
remains the same whether dealing with conventional stresses or conventional deflections.

In general then for n redundants, including effects due to temperature changes and
yielding supports, Eqs. (?A) give

(55A)

S=S S a X a SfrXb etc. S n X n
M=M -M a X a -M b X b - etc. -M n X n M t +M r
R=R - R a X a - R b X b - etc. -R n X n R t +R
Q=*Qo - Q a X a - Q b X b - etc. - Q n X n Q t +Q

wherein tne quantities S , M , R and Q are all linear functions of the externally applied
loads P acting on the principal system as a result of what is known as condition .X"=-0.
The quantities bearing subscripts a, b, c, etc., to n, are constants due to conventional
loadings X a = l, X b = l, X c = l, etc., to X n = l, while the subscript prefers to temperature
effects and the subscript r to effects produced by yielding supports.

The n redundants may be obtained from Eqs. (7n) in terms of the constant stresses
due to the conventional loadings or from Eqs. (So) in terms of certain deflection constants
obtained from the same conventional loadings. The latter equations are

d n =

B X a d na

etc. X^d an

e tc. X n d bn

etc. X n d cn

etc. X n d nn

+d c

(55B)

wherein the o's have the definitions given in Art. 8, and all those bearing double sub-
scripts of like letters are equal by Maxwell's law. Thus

=d c

=d etc. d =

etc.

meaning that the order of the subscripts is immaterial and that only half of the d's
need be determined, or if they are all determined, that they must check.

Hence, either one of the subscripts may be made to refer to the point of application
of a load, while the other subscript deals with the conventional loading or condition
X \. The terms RAr express the effect due to yielding supports and the d t quantities

ART. 55 DESIGN OF STATICALLY INDETERMINATE STRUCTURES 205

express the effect of temperature changes, both of which will be placed equal to zero
for the present, though they will be considered in following articles.

The redundants X may represent indeterminate reaction conditions or they may
be stresses in redundant members, and in either case the displacements d a , d b , S c , etc.,
o n are the displacements of the points of application of the respective redundants in the
directions of their lines of action. These values o may thus be expressed in terms of
the redundants themselves in accordance with Eqs. (7j). When the point of application
of a certain redundant is rigid, then its d =0.

For each of the n redundant conditions there will be one work equation of the form
of Eqs. (55B) involving deflections due to conventional loadings. These together with
the case of actual loading due to loads P, will constitute in all n + l cases of loading on
the principal system to solve for one position of a moving train of loads.

The aim, in all practical problems of this nature, consists in representing all the
required stresses or other functions by influence lines, thus requiring n -f 1 influence line-*
to determine each such stress or function. However, the n influence lines for the n redun-
dants will be the same for all cases and all stresses or functions of the same structure,
while the influence line for the load effect must be separately found for each stress,
moment, shear or reaction.

For influence lines the applied load becomes unity and the functions ^P m d m = 1 d m .
Each of the n redundants will furnish a deflection polygon for condition X = l, drawn
for the loaded chord of the principal system. The n deflection polygons will then
furnish all the conventional deflections in Eqs. (55s) for a load P = l. Also, since the
subscripts may be interchanged, one such deflection polgyon drawn for X A = l will fur-
nish all the double subscript bearing o's of the first equation.

Thus as a matter of convenience, all of the d coefficients in a single equation
can be determined from one deflection polygon drawn for the loaded chord of the prin-
cipal system. The same might also be accomplished by drawing a Williot-Mohr dis-
placement diagram for the principal system, though this would usually be more laborious.

Therefore, the n Eqs. (55s) can be solved successively with the aid of n deflection
polygons, in accordance with the following form with transposed subscripts and for a

single load P = l.

X I

y- 7

X b O hc -X c d cc . . . - X ^nc =

, . . . (5oc)

where the values of d a , 3 b , S c , etc., were obtained from Eqs. (7j) and may be chosen to
represent anv of the possible cases of redundancy, as changes in the lengths of redundant
members elastic displacements of redundant supports or angular changes between pairs

206 KINETIC THEORY OF ENGINEERING STRUCTURES CHAP. XI

of lines according to Art. 9. When the redundants are immovable reactions, then these
S's become zero.

The d coefficients in Eqs. (55c) can thus be obtained from n deflection polygons
drawn as above described and the n equations solved for simultaneous values of the X's
thus furnish the values for a complete solution of Eqs. (OOA).

Eqs. (55c) are written for a single load P = l intended specifically for use with influence
lines, which are most valuable when dealing with concentrated load systems.

Examples involving dead loads only, or when the live load cannot occupy more than
one or two positions, could be solved advantageously without resorting to influence lines
according to the very interesting example of a lock gate in Chapter XIV.

For solid web structures and masonry arches, all the above is applicable, remember-
ing that in these cases the redundants must all be external. As soon as the X's are found
the stresses on any section are readily ascertained.

The solution of the n simultaneous equations is a matter of considerable labor and
may offer some difficulty. The method best suited to problems of the kind here con-
sidered is given in connection with the lock gate problem referred to above. See Chap-
ter XIV, Art. 68.

ART. 66. EFFECT OF TEMPERATURE ON INDETERMINATE STRUCTURES

Determinate systems are not materially affected by temperature changes. But it
is not correct to say that the temperature stresses are zero, because any structure which,
in consequence of changes in temperature, undergoes such deformations as to change
the positions of the points of application of the external forces must thereby create some
stresses. These will usually be small and are entirely negligible except possibly in very
flat suspension systems or very flat three-hinged arches where the horizontal pull or thrust
is a function of the middle ordinate.

For the case of indeterminate structures, wherein the redundant conditions ares
direct functions of elastic deformations, the general rule implies that stresses always
accompany changes in the lengths of structural dimensions, no matter what external
cause may have produced such changes.

All temperature changes in externally indeterminate structures will affect the reac-
tions and these in turn set up temperature stresses in the members. The only exception
to this case is a continuous girder with supports on the same level and subjected to abso-
lutely uniform temperature throughout. When such a structure is unequally heated
then excessive temperature stresses may result, as shown in the examples of Arts. 47;
and 48.

On the other hand where the redundancy is entirely internal, no temperature stresses
are produced so long as the whole structure retains a uniform temperature throughout.
When this uniformity does not exist, then temperature stresses are created, though the
reactions may or may not be materially affected.

Hence, in all cases of redundancy, it will be necessary to investigate the question
of temperature stresses in a very thorough manner, as in many instances these may
assume dangerous proportions.

ART. so DESIGN OF STATICALLY INDETERMINATE STRUCTURES 207

For this reason also, statically determinate structures should always receive the
preference, other considerations being nearly equal.

Internal redundancy is less objectionable than external, and according to the best
modern engineering experience, no design should be made to include more than one
redundant condition of any kind. This is especially true of steel structures, though to
a lesser degree applicable to masonry arches of short spans where the poor conductivity
of masonry acts as a protective agency against excessive temperature deformations.

For long-span masonry arches this consideration assumes greater importance. Even
though monumental structures of this class have been built and are regarded with pride
and admiration from an esthetic standpoint, they cannot be accepted as representative
of the best practice when viewed from the point of modern and progressive engineering.

The general effect of temperature changes on indeterminate systems is thus to pro-
duce deformations and resultant stresses of the same kind as those created by externally
applied loads in accordance with Eqs. (55A).

Since the temperature effects may be plus or minus in character, depending on the
existing conditions, it is always possible to increase the stresses due to the loading by
choosing appropriate changes in temperature. For this reason the temperature effect
is applied with a plus sign in Eqs. (55A), meaning thereby that the function, whether posi-
tive or negative, suffers a numerical increase. Of course some assumptions will produce
a decrease, but these are not vital to the problem and require no consideration.

It is always desirable to determine the temperature stresses apart from all other
influences so as to furnish a clear conception of their relative importance to the load
stresses. This then offers a criterion by which to judge the feasibility of a structure
involving redundancy.

The temperature stresses are determined from Eqs. (7n) and (80) by dropping out
all terms depending on P m , S and R, thus reducing these equations to the following
forms, covering all members of the principal system:

d a =ea a = ^S a stt-X at ^S^p -X bt 2S a S b p . . . etc. 1

k ... (56A)
d b = eil b = '2S b etl-X at 2SJS b p-X bt Z,Szp . . . etc. J

or from Eqs. (80)

8a=X a tpa=8 at -X a td aa -X b t8 ab etc. ]

\ (o6B)

8 b =X bt p b =d bt -X a t8 btt -X bt 8 bb ... etc. J

In both Eqs. (56A) and (56s) the displacements d a and d b become zero for external
redundancy with immovable supports.

Accordingly the redundant conditions X at , X bt , etc., may be found from either set of
the above equations, depending on the method used in finding the load effects from Eqs.
(7n), or (80) since both these and Eqs. (56A) and (56u) involve the same constants.

The stresses S t are then easily found from a Maxwell stress diagram drawn for the
principal system with the forces X at , X bt , etc., applied as external forces. See also examples
in Arts. 47, 48, 49 and 50.

208 KINETIC THEORY OF ENGINEERING STRUCTURES CHAP. XI

Regarding the assumptions which may be made as a basis for computing temperature
stresses, the following conclusions are taken from some experiments on steel arches at
Lyons, France, and given in An. d. ponts et chaus., 1893, II, p. 438-444.

For an air temperature in the sun of 5 F. higher than the shade temperature, when
the latter was 90 F., the steel had an average temperature of 115 F., while the parts
exposed directly to the sun were heated to about 130 F. and the shaded portions indicated
about 104 F. At the same place the coldest winter temperature was about 15 F.

The steel was thus subjected to a mean range of 15 to +115 = 130 F. giving
an average of +65 F. Therefore, such a structure should be designed for a mean
temperature of 65 F., allowing for uniform changes of 65 F. from this mean.

The difference between maximum and minimum simultaneous temperatures in
the steel, amounting to 26 F. in these experiments, may cause very serious stresses in
certain structures like fixed arches and continuous girders over several supports. In
the latter case the entire top chord would elongate relatively to the bottom chord and
thus set up an arching effect, relieving the intermediate supports and increasing the
two end supports, while a uniform change in temperature would produce no stresses.

Structures over a single span would not be so seriously stressed for unequal temper-
atures in the two chords. In fact this assumption might work to advantage in the
case of arches, though similar conditions for a clear cold day might prove more severe.

The painting of steel structures of the indeterminate class thus assumes considerable
importance, since light colors will obviously keep down the temperature while black
paint will absorb heat.

Masonry structures are not so severely distorted by temperature changes, but the
elasticity of masonry being proportionately lower, the temperature stresses may prove
equally dangerous.

ART. 57. EFFECT OF SHOP LENGTHS AND ABUTMENT DISPLACEMENTS

Every structure is designed for a certain geometric figure for a condition of no
stress. This is the figure which the structure should present when at a mean temperature
and when carrying no loads of any kind. Naturally the unavoidable errors in the shop
lengths of the members as built and the small inaccuracies in the joints and in the loca-
tion of the supports during erection, preclude the practical possibility of accomplishing
this end.

For structures of the determinate class this will have no significance, but for indeter-
minate types, these errors introduce initial stresses which may assume momentous
proportions.

To avoid such stresses requires the utmost care during construction and erection,
and even then only a partially satisfactory outcome can be expected.

The difficulties attending the erection of a structure so that no member will be
stressed prior to the introduction of the closing link, at the proper temperature, are
known to every practical bridge man.

In any event the final member should be inserted at the calculated mean tem-
perature. If this cannot be accomplished, then the exact length which this member

ART. 57 DESIGN OF STATICALLY INDETERMINATE STRUCTURES 209

should have must be ascertained by measuring the space which is actually available
when the structure is all in place and under no stress. Then applying proper tem-
perature corrections to the entire structure, the required length of the final member must
be ascertained before attempting to force the latter into its place.

When the end supports are not correctly placed or the entire structure has an eror
in length, then all the values S , S a , etc., and the redundant conditions will differ from their
computed values. If such displacements are in the same direction with a redundant
X, then the stress X is affected thereby, otherwise the principal system only receives
the additional stress.

Should a redundant member possess an erroneous length, then all the members
will receive certain initial stresses, while errors in the lengths of principal members would
change somewhat the geometric figure, but could not effect the stresses S , S a , etc., of the
principal system.

In dealing with abutment displacements it is necessary to distinguish between elastic
changes which the- supports may undergo as a result of loading, and permanent settlement.

An example of elastic displacement is given in Fig. 45A, where the supporting column
CD will be shortened by an amount d b , which can be determined and then be included in
the computations as for any other member of the structure.

However, should the pier at C undergo permanent settlement, or should one abut-
ment, as B, be pushed out horizontally or settle vertically, then serious difficulties
would arise unless these displacements could be determined beforehand, which is usually
quite impossible.

In either case the stresses occasioned by such displacements only, may be found
as follows, provided the displacements are known or can be estimated with reasonable
accuracy.

If, in Eqs. (55s), the terms involving the temperature effects be dropped, then for
internal redundancy, the displacements d a , d b , etc., are evaluated from Eqs. (7j), and the
work of the reactions is found as shown in the derivation of Eqs. (?E).

When the redundancy is purely external then the Eqs. (55s) again apply by treat-
ing the reactions R as the reactions of the principal system and evaluating the elastic
displacements d a , d b , etc., and the Ar for each reaction R, using such considerations as
given in Art. 12, or estimating these displacements as elastic yielding in the masonry
supports, etc. See also examples in Articles 49 and 50.

In any case the combined stresses from all causes are given by Eqs. (55A) and no
further comment is necessary here except to emphasize the inadvisability of adopting
externally indeterminate structures whenever immovable supports are not available. Even
for bed rock foundations, this will depend largely on the depth to which the masonry
must be carried and also on the quality of masonry used.

When steel towers or pendulum piers are employed to support an indeterminate
structure, then the elastic deformation of such supports must be considered in comput-
ing the structural stresses.

CHAPTER XII
STRESSES IN STATICALLY DETERMINATE STRUCTURES

ART. 58. DEAD LOAD STRESSES

(a) General considerations. The purpose of taking up the question of stresses ir
statically determinate structures is not with any intention of covering this subject exhaust-
ively, but merely to present in the briefest possible space the methods which are besl
suited to the analysis of all ordinary structures.

It was not deemed advisable to take up the present chapter before having treated
influence lines and developing the general criteria for the position of loads for maximum
and minimum stresses in Chapter IV. The fundamental conceptions there presented
are very material to a broader understanding of determinate structures and hence the
present order of subjects was considered justifiable.

Since it is desired to treat methods of analysis and not types of structures, only
the general case of non-parallel chords will be considered. Whenever a structure is sim-
plified by introducing parallel chords and otherwise simple relations between its dimen-
sions, then naturally the analysis becomes less complicated, until it might be said that
the problem is reduced to a mere application of arithmetic. The so-called algebraic
methods are, therefore, passed over without further consideration.

Since the dead loads are fixed in position and magnitude, the stresses produced by
them in any structure must be absolutely invariable. The live loads, however, owing
to their shifting position, may produce a variety of conditions and stresses which may
tend to increase or to diminish dead load stresses.

Therefore, every member may be said to be subjected to a maximum and a minimum
stress corresponding to the peculiar or critical positions of the live loads. These critical
positions are always determined from certain tests or criteria, differing for different
members of the same structure according to the principles discussed in Articles 20, 23
and 24.

It was found there that for the chord members the maximum stresses are
produced for the case of maximum loading over the entire span, while the minimum
stresses in these members result from the minimum total load, which is the dead
load.

This condition is very different for the web members, where the dead load rarely
produces minimum stresses because different positions of partial live load may pro-
duce live load stresses of opposite signs. When these are combined with the dead load

210

ART. 58 STRESSES IX STATICALLY DETERMINATE STRUCTURES

211

stresses to obtain total stresses, then the minimum total stresses are less than the dead
load stresses, while the maximum total stresses exceed the dead load stresses.

When a web member is designed to take only one kind of stress like tension, and this
stress is reversed by the live loads, then the panel containing such member must be
supplied with a counter web member or the original web member must be differently
designed so as to resist both tension and compression. This latter is considered the
best modern practice.

Hence, in order that the relation between the dead and live load stresses may be
clearly brought out in the analysis, it is always necessary to determine these stresses
separately. Also, the methods which apply to the solution of dead load stresses are usually
not the best adapted to finding the live load stresses, which presents another reason for
dealing with the general subject under separate headings.

However, each method given will be applied directly to all the members of a struc-
ture, showing the application both to chords and web members before proceeding
further.

(b) Aug. Hitter's methods of moments (1860). This method is based on the gen-
eral theorem of moments. Accordingly for all external forces in the same plane and
constituting a system in equilibrium, the sum of the static moments, about any point
in this plane, must equal zero.

If now such a set of forces or loads be applied to a frame, Fig. 58A, and these loads,
together with the two end reactions A and B, form a system in equilibrium, then if the

FIG. 58A.

frame be cut by a section tt, the stresses in the members cut, if acting as external forces,
must maintain equilibrium of all the external forces on either side of the section. If,
further, such planes can be passed which do not cut more than three members of the
frame, then the stresses in the members cut may be determined one at a time, by taking
as the center of moments the intersection of two of the members. The moments of the
two intersecting members thus become zero.

The following designations are used:

For member U the center of moments is n and the ever arm is r n ;

For member L the center of moments is m and the lever arm is r m ;

For member D the center of moments is i and the lever arm is r ;

41so calling M the sum of the moments of the external forces A, P, and P 2 to the

212 KINETIC THEORY OF ENGINEERING STRUCTURES CHAP. XII

left of the section it and using subscripts n, m and i to designate the centers about which
the moments are taken, then for positive moments in a clockwise direction

Ur n +M n =Q or U = ~

-Lr m +M m =Q or

-Dri-Mi =0 or Z>=- -1

(5SA)

where a positive stress indicates tension and a negative stress indicates compression.
Thus a top chord member is always in compression and a bottom chord member is always
in tension, while a web member may have either stress depending on its inclination with
respect to its center of moments.

Hence, the stress S, in any member of a determinate frame, may always be expressed
in terms of a moment M and its related lever arm r, by

(58B)

which is Ritter's fundamental moment equation so extensively used in the previous
chapters.

The centers of moments for the chord members are thus seen to be located opposite
the chord in question, while for a web member the center of moments may be anywhere,
depending upon the relative inclinations of the two chords composing the panel. The
lever arms should be scaled from a large scale drawing or be computed.

When the three members cut by a section intersect in the same point, the method
fails. Also, when the chords are parallel the lever arm for the diagonal becomes infinite.
Hence the stress in such a diagonal cannot be found by a direct application of Eq.
1.58B) without some modification.

For the case of chords which are parallel or nearly so, the stress in a, web member
may be found by first computing the stresses in the chords and then finding the web
stresses. This is done by choosing for the center of moments for a web member, any
convenient point of one chord and then writing a moment equation, including the other
chord of the panel cut among the external forces.

However, there is a simpler method of finding the web stresses in structures with
parallel chords and that is from the shear in the panel cut.

Thus in the center panel mq the chords are parallel and the diagonal pq is cut by
a section ft'. The lever arm for pq would be infinite.

Since the sum of the moments of the external forces to the left of the section and of
the three members cut must be zero, therefore the sum of the vertical components of
these forces and stresses to the left of the section must likewise be equal to zero.

The sum of the vertical components to the left of the section Ft' is called the shear Q.

ART. 58 STRESSES IN STATICALLY DETERMINATE STRUCTURES 213

It is equal to the end reaction A minus the sum of the panel loads between A and the section.
Thus

Q=A-*lP, .......... (58c)

where the summation covers only the panel loads to the left of the section and the shear
is taken positive upward on the left of the section.

The chords, being horizontal, cannot have a vertical component, hence the only
stress in any of the three members cut, which can have a vertical component, is the stress
in the diagonal pq=D. The vertical component of D l is then DI cos 0, and the sum
of the vertical components of all forces and members cut on the left of the section W
becomes

When DI is vertical, then 0=0 and the stress becomes equal to the shear.

The above moment equation also admits of a graphic solution which will not be
considered here.

(c) The method of stress diagrams. The first description of these diagrams seems
to have come from Bow, and in 1864 Clerk-Maxwell published a paper " On Reciprocal
Figures and Diagrams of Forces " in which he presents a scientific treatment of the sub-
ject. Cremona, in 1872, discussed the geometric properties of stress diagrams, showing
their general usefulness in connection with graphostatics.

English and American writers, therefore, call such diagrams "Maxwell stress diagrams,"
while in Germany and France they bear Cremona's name.

The Maxwell stress diagram, so called in the present work, serves a most valuable
purpose in the graphic analysis of all determinate frames, and is generally applicable
to all cases which are susceptible to treatment by Ritter's moment method.

A peculiar relationship exists between a frame and its stress diagram by which each
member of the frame is parallel to a line of the diagram and each pin point of the
frame is represented by a force polygon in the diagram. It is thus equally possible to
construct a stress diagram for a given frame or to construct a frame from a given stress
diagram. Hence, the term " reciprocal figures " used by Maxwell.

The closed force or funicular polygon, which constitutes the basis for the Maxwell
diagram, was known to Stevin, 1608, and Varignon, 1725, and marks the beginning of
graphics. Such a polygon may be drawn for any set of forces in equilibrium.

Since all the forces meeting in a pin point must constitute a system in equilibrium, a
closed force polygon may be drawn for each such point. Hence a Maxwell diagram is
merely a succession of closed force polygons drawn for all the pin points of a frame.

For any given frame, the directions of all forces and members are given or must be
assumed before proceeding to an analysis of stresses. Also, if the frame constitutes a
system in equilibrium, then the externally applied loads must be in static equilibrium
with the supports or reactions, and all these in turn must be in equilibrium with the
internal stresses in the members.

KINETIC THEORY OF ENGINEERING STRUCTURES CHAP. XII

Any pin point in equilibrium and acted upon by any number of forces of known direc-
tions, may then be represented in the force polygon by a succession of forces and stresses
respectively equal and parallel to the forces and stresses meeting in that pin point.
Since all directions are known, two magnitudes may be found by inserting the unknown
members in such a way as to close the force polygon.

This is the fundamental principle of the Maxwell diagram. It is illustrated in
Fig. 58s, where the force P and stresses S\ and 82 are known and the stresses S 3 and S 4
are of unknown magnitudes, but of given positions. All the forces meet in the point
A and are supposed to be in equilibrium.

The method of drawing the stress diagram and nomenclature used in the figure is
precisely the same in all cases and affords an easy way of deciding the direction of
action of each unknown force.

The clockwise arrow indicates the order in which the forces are assembled in the
stress diagram. Passing around the point A in this direction the first known force reached
is P. Hence the letters a, b, c, d, e are supplied as shown in the angles between the

Forces Acting on A.

FlG. 58B.

Stress Diagram.

forces, such that the force P is included between a and 6. In speaking of the force ab
in the stress diagram, we mean a force equal and parallel to the force P and acting
in the given direction from a to 6 in the stress diagram.

Observing this designation, the other known forces are added in their proper order
in the stress diagram and made to act in the given directions a-b-c-d as found by going
around the point A in a clockwise direction. The stress diagram from a to d is thus
obtained and may now be closed by drawing a line de' \ S% through d and another line
ae" || 84 through a. The intersection e thus found completes the force polygon and
determines the stresses 3 and S both in magnitude and direction of action. Thus the
arrows around the force polygon must be in the same direction as indicated by the
initial given force P. Accordingly the force 83 acts in the direction from d to e and the
force S^ acts in the direction from e to a. All forces in the force polygon are laid off to
a certain scale by which the unknown forces are finally determined. A force or stress
acting away from the pin point exerts a positive or tensile stress.

It is thus seen that the unknown stresses in two of the members meeting in any
pin point may be determined by means of a closed force polygon. However, if a given
frame presents no pin point involving only two unknown forces, then the method can-

ART. 58 STRESSES IN STATICALLY DETERMINATE STRUCTURES

215

not be applied except by first finding the forces in excess of two, by some other means.
When there are three unknown forces acting on the point, and there are no redundant
conditions involved, then Ritter's moment method will always furnish the necessary
solution for one of the unknowns, prior to drawing the Maxwell diagram.

The forces in Fig. 58B were arbitrarily assembled in a clockwise direction. A
counter clockwise direction might have been chosen with equal right, though the stress
diagram would then have occupied a symmetric position with respect to the present

case.

To illustrate the method, a simple truss, Fig. 58c, is used,
are each 10 kips and the reactions A =B =40 kips.

-43.0

B-40K.

The panel loads P

FIG. 5Sc.

The reactions must always be computed in order to supply the forces which are
necessary to establish a system of external forces in absolute equihbrium. The
diagram is then lettered and the direction of assembling the forces is chosen so as to locate
the stress diagram properly on the paper.

A pin point embracing only two unknown members must be fleeted for a point
beginning and by inspection it is seen that either abutmentsaUsfies this condition
theTress diagram iscommenced by drawingjhe triangle * inwhich the vertica upward
reaction is known and the two forces cb and tajire found, lollo^ -ound
triangle in the direction a-c-b, it is seen that the force cb acts toward the

216

KINETIC THEORY OF ENGINEERING STRUCTURES CHAP. XII

The stress in ba is found

produces negative or compressive stress in the member cb.
to act away from A , and is thus positive or tensile.

The construction of the diagram is now continued by assembling the forces around
pin point 1, beginning with the first known member be, then adding the known force
cg = Pi and finally closing the figure by drawing gf and/6, which are the two unknowns
for point 1 .

The members ub and bf being_known_the closed force polygon for pin point 2 may
next be drawn to find the stresses fe and ed.

-*TTZ\__ \_LI_U. _v- ' -

FIG. 58o.

The process is continued, taking the pin points in the following order: A-l-2-3-4-5-6-
7-8. As a final check rp= B.

AH the stresses are then scaled from the stress diagram and written on the truss
diagram with proper signs, for compression and + for tension.

The present example shows a truss in which the chord stresses are nearly all equal
and the web stresses are small. It is similar to the Pauli truss and is economical in design.

A very interesting Maxwell diagram is presented in Fig. 53e, where some of the
external loading consists of a moment.

It frequently happens that loads are not applied directly to the pin points, in which

ART. 58 STRESSES IN STATICALLY DETERMINATE STRUCTURES

217

case certain load concentrations must be effected before proceeding to draw a stress-
diagram. The members so loaded will usually receive a combination of direct stress and
bending as illustrated by the member AC, Fig. 58D, representing a portion of a roof truss.
The parallel loads PI to P 5 , acting on the member AC, may be combined (graphically
or analytically) into a resultant R of kncfwn position and magnitude, and R may then
be resolved into the components R \ and R%, constituting the pin point concentrations
at A and C respectively.

FIG. 58e.

After all loads have been thus concentrated on the several pin points of the struc-
ture the total reactions may be computed in the usual manner, taking into account only
the loads R,, R 2 , etc., which now replace the loads P, and then the stress diagram is
drawn in the usual way and the direct stresses in the members are obtained.

The reactions R, and R 2 are the same as for a span d, taken perpendicular to the
direction of the forces, and the bending moment M, for any point k distant x from A ,
may be found from the equilibrium polygon or by computation.

For a direct stress - S, as obtained from the stress diagram, the total thrust m the
member becomes N=-S-(Ri -ft -ft -ft) cos a. and the moment M--

218 KINETIC THEORY OF ENGINEERING STRUCTURES CHAP, xil

r]H, where H is the pole distance measured to the scale of forces and T? is
the ordinate of the equilibrium polygon measured to the scale of lengths.

The unit stresses on the extreme fibers of the column A C then become by Eq. (49ivi)

f _,

J p j ,

where y is the distance of the extreme fiber from the gravity axis of the section.

The dead weight of inclined or horizontal members would be considered in this
manner, only that for uniform loads the treatment is simplified.

The example given in Fig. 58E belongs to the class previously pointed out, in which
not less than three unknown members meet in every pin point, and hence a Maxwell
diagram cannot be constructed without first computing one member by Ritter's method
of moments.

The resultant R is found as in the previous figure, and supposing the support at .4
to be a roller bearing, the reaction at R\ must be vertical and RZ must pass through B
and d. Hence the reactions of known directions may be found from the force polygon
by resolving R into R \ and R%.

The stress in the member AB is then found by passing a section tt', cutting only three
members. Then with C as center of moments Eq. (58s) gives

AB-H-(Rr-R, l Y-(Rr-R r V-- R (r- r '
C" Kl 2)Ji ( K 2]h J\ r 2

RI and H being now known the Maxwell diagram can be commenced for the pin point
A and continued throughout the truss.

For vertical loads the solution may be conducted analytically.

ART. 69. LIVE LOAD STRESSES

(a) The critical positions of a train of moving loads to produce maximum and minimum
stresses in any member of a given determinate structure must be known prior to applying
any method for finding the stresses themselves. This question was fully discussed in
Arts. 20, 23 and 24, and in the General Considerations of Art. 58.

All the methods for the analysis of stresses which follow here presuppose a knowl-
edge of the criteria for the positions of loads. The reader is referred to the articles just
mentioned without repeating this discussion here.

(b) The method of influence lines, fully treated in Chapters IV and V, may be men-
tioned as the most universal, answering as it does all questions relating to criteria for
positions of loads and magnitudes of stresses for any determinate structure.

The method requires no further explanation here except to point out the types of
structures which warrant its application.

In general, the more complicated the geometric figure of a structure, the greater
the advisability of employing the method of influence lines, since the geometric relations
of the truss dimensions are thereby expressed^, independently of the loads.

ART. 59 STRESSES IN STATICALLY DETERMINATE STRUCTURES 219

Hence it would scarcely be advisable to apply influence lines to any truss with parallel
chords except in cases of complicated cantilever systems. On the other hand a saving
of labor might be expected in analyzing structures in which either or both chords are
curved. The method certainly offers obvious advantages in all cases illustrated in
Chapter V, especially when concentrated loads are employed.

It may be added that any structure which does not warrant the accurate analysis
by concentrated load systems should not be dignified to the extent of being called a
modern bridge. Nor should anyone not familiar with these more exact methods be
intrusted with the design of bridges.

It should also be emphasized that all computations of stresses should err decidedly
on the side of safety, since the secondary stresses produced by the friction on pin-con-
nected joints are frequently a very considerable quantity, to say nothing of the failure
to analyze properly the effect of the riveted connections.

(c) Discussion of methods in common use. In general, all methods involve the
solution of the following distinct problems: To find certain moments, shears and end
reactions resulting from certain critical positions of the train of loads and then to deter-
mine the maximum and mimimum stresses in a particular member in terms of these moments
or shears.

Both analytic and graphic solutions have been proposed to solve all of these prob-
lems in their various phases. Some are applicable only to trusses with parallel chords while
others are more general. Still others are used when the position of the train is chosen
so as to avoid the complications arising from the exact loadings required by the criteria
for maximum and minimum stresses in a certain member.

It would be outside the province of the present chapter to present all of these
methods in detail, though many of them are exceedingly interesting and ingenious. The
author would offer the general criticism that the methods in common use are too special
in their application, and when a more general problem is encountered no one method
will suffice, but several methods must be combined, thus necessitating an intimate knowl-
edge of many limited methods to accomplish a complete analysis.

The following method is an attempt to determine the live load stresses in each of
the members of any simple truss by applying the same process throughout.

(d) The author's method of determining live load stresses is based on Ritter's com-
prehensive and universal method of moments. The moment of the live loads on one
side of the section taken about the center of moments for any particular member is
derived from the principles of the sum A line, see Fig. 22fi.

Thus the stress in any member of any determinate truss is given by Eq. (58s) as

except for the web members of a truss with parallel chords where the stress is derived
from the shear according to Eq. (58o) as

( 59B )

220

KINETIC THEORY OF ENGINEERING STRUCTURES CHAP. XII

The values of M and Q for any position of a live load will now be derived and
evaluated first from the sum A line, and then from a single tabulation of moments.

The critical position of the loads for any particular effect on a certain member is
supposed to have been determined in accordance with Arts. 20, 23 and 24.

M 1 P il 11 1

, *m -

1 lam - .,

... | - _,

FIG. 59A.

Referring to Fig. 59A and calling MB the sum of the moments of all loads on the span
about the reaction B and M m the sum of the moments about any point m; also calling
Q m the vertical shear at the point m; then for loads 1 to q } covering a distance xq=ljr (l

!*P; (59c)

(59D)

(59E)
(59r)

M m =l am (A-

lam

(59c)

wherein A am = 3. Pe/l am may be defined as the end reaction at A of the loads 1 to m
covering the distance x m of a simple span Z am .

The values A and ^4 aw may be obtained from a reaction summation influence line,
or sum A line, drawn for the reaction A and span I, according to Art. 22, Figs. 22A and
22s. The ordinates of the sum A line may be computed from Eq. (59o), or constructed
graphically by the method given in Art. 22.

Such a sum A line is shown in Fig. 59s, drawn for a span of 200 feet and using a train
of Cooper's EGO loading, consisting of two locomotives followed by a uniform load of 3000
Ibs. per linear foot per rail.

The ordinate y x under the first load PI at x, represents the end reaction A for
the position of loads indicated, when the span A B is loaded for a distance 61 from B.

The reaction A am , for the same train covering the distance x m from m on the
span l am> may be obtained from, the same sum A line as follows: Had the sum A line

ART. 59 STRESSES IN STATICALLY DETERMINATE STRUCTURES

221

been drawn for a span l am then the ordinate at x m from B' would be the required reaction
A am . However, this polygon was actually drawn for a span I, and hence the ordinate
i) xm distant x m from B' is l/l am times too small, therefore,

(59n)

This illustrates how a sum A line drawn for any span may be used to obtain end
reactions for any other span. However, in order to avoid the multiplication of all ordi-

SCALE OF LOADS.

300 KIPS.

222 KINETIC THEORY OF ENGINEERING STRUCTURES CHAP. XII

nates by I, it is best to draw the sum A line for a span of 1000 ft., and then derive all
ordinates according to Eq. (o9n) ? by which the ordinates of one A line are to those of
another A line in the inverse proportion of the spans for which the lines were drawn.

A portion of such a 1000 ft. sum A line is shown in Fig. 59B, from which the actual
reactions A and A am , for a span /=200 ft., are obtained as

and ^= ....... (59.)

where JJA is the ordinate distant 61 from B' and f) xm is the ordinate distant x m from B'.

Any moment M TO or chord stress is now readily found from Eq. (59o) , using the values
from Eq. (59i) as follows:

l a

, ...... (59j)

and the stress in any chord member having the point m for its center of moments with
a lever arm r m , becomes by Eqs. (59A) and (59j)

(59K)

The web stresses for trusses with non-parallel chords are derived from moments Mi
of the external forces about the center of moments i for any particular web member.

Calling the lever arm r t - for a web member with moment center i distant Z at - to the
left of A then for x m <d and A nm = '

Mi = Al a i A nm (L a i + l an ) ,
or M^lOOO'-^l^^+U, .... V '.. . - . (59D

when x m >d, a case which happens very rarely, indicating that some loads extend to the
left of n, then

*" -"-w -Anmv/ai -rl an ) ^ ^ r(l n i

**^ x

wherein the two last terms must be computed. It should be remembered that this last
case does not occur more than once or twice, if at all, for a whole analysis and hence
involves only a slight amount of work.

In any case Eq. (59A) gives the stress in a web member for non-parallel chords, as

(59x)

The web stresses for trusses with parallel chords are derived from the shear accord-
ing to Eq. (59E)

ART. 59 STRESSES IN STATICALLY DETERMINATE STRUCTURES 223

Thus the vertical shear to the left of a section through any panel when there
are no loads to the left of n, making x m <d, becomes

0-1 A 1000> M 1000^ m

y m ^A-A nm = ^ ~^~, (59o)

wherein A nm is the reaction at n for the loads in the panel ^m extending over a distance
x m on a span d.

When x m >d, which would be the rare case for one or two loads to the left of n, then

r -2, R -3~2,, P > <w

where the two last terms would have to be computed for the loads between x and m as
in the previous case.

The stress in any web member of a truss with parallel chords is then found from
Eq. (58o) as

S=Q m secd, . . (59 Q )

where Q m is given from Eqs. (59o) or (59p).

The above demonstration, necessitating the use of one sum A line drawn for a span
of 1000 ft., was used only for the development of the formula?. In practice, the sum
A line is dispensed with by tabulating the ordinates 1000r M for any case of standard
loading.

One such table (see Table 59A) , will then suffice for the computation of all chord and
web stresses in any simple determinate truss for the general case of moving train loads.

Table 59A thus represents the values of 1000A =1000^ =M B =the sum of the
moments of all loads on the 1000 ft. span about the reaction B, expressed in kip feet.
These moments are readily computed from Eqs. (59c), and (59i), thus

4/5 = 1000^4=-- VVp + / 3 VV, . (59R)

where l q is the distance the train is moved ahead whenever a new load P 4 is brought
on the span.

Eq. (59n) , holds until the uniform load reaches the point B, after which the moments
must be figured as for uniformly distributed loads. The table suggests the ease of
computation and might be extended over the whole span of 1000 feet, though for pur-
poses of illustration this was not considered necessary.

To facilitate interpolation, the differences in M B for one foot were added in a separate
column. Interpolations for distances between those given are thus readily performed
and these are strictly accurate up to the point where the uniform load is reached. After
that there is a slight error because the moment for uniformly distributed loads varies
as I 2 . However, the error from this source is entirely negligible.

Example. Required the maximum live load stresses in the chord U 2 U 3 , and in the
diagonal ~U^L Z of the 200 ft. truss shown in Fig. 59e, using Cooper's E 60 loading.

224

KINETIC THEORY OF ENGINEERING STRUCTURES CHAP. XII

TABLE 59A.
REACTIONS FOR COOPER'S E-60 LOADING, SPAN OF 1000 FEET.

Wheel
No.

Load
Lenj^th,

Feet.

Total Load

S?P

Kips.

M B =IQOQT) A

Kips.

Diff. for
1 Foot.

Kips.

Wheel
No.

Load
Length,

&1
Feet.

Total Load

S?P

Kips.

-1/5=1000^
Kips.

Diff. for
1 Foot,

Kips.

523.5

15.0

0.0

144

531

41,293.5

15.0

538.5

45.0

120.0

149

546

43,986.0

45.0

553.5

75.0

345.0

154

561

46,753.5

75.0

568.5

105.0

720.0

159

576

49,596.0

105.0

583.5

135.0

1,245.0

164

591

52,513.5

135.0

598.5

154.5

2,460.0

169

606

55,506 .

154.5

613.5

174.0

3,232.5

174

621

58,573.5

174.0

628.5

193.5

4,276.5

179

636

61,716.0

193.5

643.5

213.0

5,244.0

184

651

64,933.5

213.0

658.5

228.0

6,948.0

189

666

68,226.0

228.0

673.5

258.0

8,772.0

194

681

71,593.5

258.0

688.5

288.0

10,062.0

199

696

75,036.0

288.0

703.5

318.0

11,502.0

204

711

78,543.5

318.0

718.5

348.0

13,092.0

209

726

82,146.0

348.0

733.5

367.5

16,224.0

214

741

85,813.5

367.5

748.5

387.0

18,061.5

219

756

89,556.0

387.0

763.5

406.5

20,383 . 5

224

771

93,373.5

406.5

778.5

104

426.0

22,416.0

229

786

97,266.0

426.0

793.5

109

426.0

24,546.0

234

801

101,233.5

433.5

808.5

114

441.0

26,713.5

239

816

105,276.0

448.5

823.5

119

456.0

28,956.0

244

831

109,393.5

463.5

838.5

124

471.0

31,273.5

249

846

113,586.0

478.5

853.5

129

486.0

33,666.0

254

861

117,853.5

493.5

868.5

134

501.0

36,133.5

259

876

122,196.0

508.5

883.5

139

516.0

38,676.0

264

891

126,613.5

All loads in kips for one rail.

ART. 59 STRESSES IN STATICALLY DETERMINATE STRUCTURES 225

For the chord U 2 U 3 , the center of moments is at m and the maximum stress occurs
when the span is fully loaded, making 6 t =1 = 200 ft., and x m =l am =60 ft. Also r m =34.5 ft.

Then Table 59A gives for 200 ft., 1000yj A =75036 +703.5 =75739.5, and for x m =60 ft
1000^ =6948 +4X228 =7860, whence by Eq. (59K),

= J- [75739.5(1^ -7S60J =430.78 kips.

For the diagonal U 2 L 3 with load divide at z w = 10.5 ft., making 6 t = 140 + 10.5
= 150.5 ft., Z an =40 ft., Z at -=98 ft., d=20 ft., and r^ = 132 ft., and thus x m <d, Table 59A
gives 1000r? A =43986 + 1.5X553.5 =44816.3 and 1000^ = 120+2.5X45=232.5.

Hence by Eqs. (59L) and (59N).

The minimum stress in C/ 2 ^3 may be found by treating the symmetric member L 7 U 8
in the right half of the span.

Similarly the stress in any member of the truss may be determined for any desired
position of the train of loads, all by means of the one Table 59A.

It is believed that the ease and simplicity of applying the above method, together
with its universal applicability, should commend itself to the practical designer.

Certainly it does not seem warranted to employ approximate methods when an
exact solution is possible without additional labor.

CHAPTER XIII
SECONDARY STRESSES

ART. 60. THE NATURE OF SECONDARY STRESSES

In the previous chapters a framed structure was regarded as a system of individual
members linked together by frictionless pin connections. It was also supposed that the
neutral or gravity axes of all members meeting in a panel point actually intersected in
one point. The stresses computed on this basis are called primary stresses.

Owing to the friction, which always exists in pin-connected joints, and the rigidity
introduced by riveted connections, this condition is never fully realized in practice. Hence
certain moments or bending effects are always produced in the proximity of the connected
ends, which set up bending stresses of variable character and magnitude. These are
generally called secondary stresses as distinguished from the primary stresses.

All structures involving redundancy are subjected to elastic deformations caused
by the loads, temperature, and abutment displacements giving rise to what may be called
additional stresses. These also exist to a slight degree in determinate structures, especially
when their geometric shape is peculiarly subject to large load and temperature deflec-
tions as in the case of three-hinged arches. Still other causes, such as erroneous shop
lengths of members, wind pressure, and impact and brake effects of moving loads, pro-
duce stresses belonging to this class which do not admit of exact determination and can
at best be merely estimated.

The first theoretical discussions of secondary stresses were given by Asimont, 1877;
Manderla and Engesser, 1879, and Winkler, 1881. Graphic solutions were published by
Landsberg, 1885; W. Ritter, 1890; and Mohr, 1891-3. A resume of most of these is given
by C. R. Grimm, C. E., " Secondary Stresses in Bridge Trusses," 1908.

The facts which were revealed by these discussions gave rise to very material
improvements in the modern practices of designing structures. Thus, the details of con 1
nections at panel points and between floorbeams and truss members have been vastly
improved, and particular attention is now being given to a more judicious design of the
members themselves, especially those subjected to compressive stress.

It is not the purpose here to treat all the various methods of calculating secondary
stresses, but merely to present what appears to the author as the most understandable
and usable method. The nature of the problem is such as to preclude the possibility
of considering all the complexities involved and yet remain within the limits of practical
utility. Hence some approximations must be applied, and justly so, because many of
the influences are too small to warrant the labor involved in their analysis, and others
are beyond reasonable limits of estimation.

226

ART. Gl SECONDARY STRESSES 227

The three primary causes for secondary stresses in bridge members are: the weights
of the members themselves producing deflections in all except vertical members and giving
rise to bending stresses; the absence of frictionless panel point connections; eccentric
connections between members meeting in a common panel point.

Riveted connections are here treated like fixed ends so that the elastic deformations
of the structure are supposedly taken up by flexure in the members themselves with-
out producing any changes in the angles at the panel points. While this assumption is
not absolutely true, because the bending moments cannot be resisted without producing
some elastic angular distortions, yet it is on the side of safety and tends to compensate
some other factors which are entirely neglected.

Pin-connected members, according to more recent experience, cannot be regarded
as free from bending moments on account of the rather excessive friction which always
prevails on the pins. In the case of eye-bars this may add considerable bending stress
owing to the slenderness of such members. This should not be construed to mean
that pin connections are inferior to riveted joints, but that the former do not possess
all the advantages usually attributed to them. The distinct superiority of pins is to be
found in the prevention of eccentric transmission of the direct stresses, a condition which
is difficult to accomplish in riveted connections.

The excessive bending stresses carried from floor-beam connections to the vertical
trusses can hardly.be estimated and should be avoided as far as possible by the introduc-
tion of flexible connections in preference to the rigid type so much used in the past.

The secondary stresses, which will now be considered, are those caused by the elastic
deformations taking place in the plane of the frame itself and due to the three primary
causes above enumerated.

ART. 61. SECONDARY STRESSES IN THE PLANE OF A TRUSS DUE TO

RIVETED CONNECTIONS

Every elastic structure must undergo certain deformation when subjected to loads.
If the structural members are rivet-connected at the ends, then this deformation is
taken up by the members themselves in the production of certain bending moments which
are resisted by the rigid panel connections.

Figs. 6lA and 61 B show the several possible combinations of single and double flexure
with compression and tension as they might occur in any structure.

The straight chord AB, in each case, represents the position of the member on the
supposition of frictionless pin connections, while the curved line indicates the member
as it would be distorted by the bending moments and axial forces resulting from the fixed
riveted connections.

Assuming such a member to be released at the ends by sections passed close to the
panel points, then the internal stresses may be replaced by external forces P a and P b .
Each of these may be resolved into two forces and a moment acting at the original fixed
ends. Thus P a is equivalent to S a , A and a moment M a while P b is replaced by S b , 11
and a moment M b . The conditions of equilibrium require that P a =P b and hence that,

228

KINETIC THEORY OF ENGINEERING STRUCTURES CHAP. XIII

S a =P a cosd=S b =P b cos6; also A=B=P sin 6 and M a +M b -Al=0. Since is
always very small S a S b =P a -=P b , which assumption is entirely permissible.

In the case of compression members the concave side of the elastic curve is always

FIG. 6lA Tension Members.

toward the resultant P, and hence a maximum moment may occur at some point inter-
mediate between A and B. For tension members the convex side is toward the force,
and the maximum moment must occur at one end of the member.

FIG. 61s Compression Members.

In any case then, the stresses produced in a member by the bending moments M
and M b , are readily determined provided these moments are known. Of the several

ART. 61

SECONDARY STRESSES

229

methods proposed for the evaluation of the end moments, the analytic method of Professor
Fr. Engesser is probably the most direct and is the one given here.

For any structure of w members there will be 2m unknown end moments to be
evaluated. These are derived from the angular deflections r, which the respective fixed
ends of the members undergo as a result of certain distortions Ja, J/?, Jf, etc., produced
in the angles of the frame by the work of the external forces and reactions, as per Eqs. (37n).

The first step in the development of the method is then to derive the relation between
the moments M a and MI, and the deflection angles r a and r b for a column on two fixed
supports.

The following assumptions are made in the interest of simplifying the theory for
practical applications.

Since in well designed members the shear, direct thrust and temperature have a very
slight effect on flexure, these factors are neglected. However, for poorly designed members
with small moments of inertia and slender dimensions, the deflections due to normal
thrust etc., may become very considerable and these assumptions might not hold.

It is further assumed that the axes of all the members are situated in the same
plane, and that these axes are centric for all members meeting in a point. Also, that the
external loads are all applied at the panel points, neglecting for the present the flexure
produced in the members by their own weight.

Since the fixed ends, Fig. 61c, do not of themselves produce any direct thrust or stress
in the member ~AB (not being absolutely fixed in space), therefore, the column involves
only two redundant conditions which are the moments M a and M b . The principal system
is then a beam on two supports.

FIG. 61c.

The effect of the direct stress S is not considered here because for well designed
members with comparatively large moments of inertia this was found by Manderla to
be entirely negligible. Otherwise a moment Sij would enter the general moment as
given by Eq. (6lA).

Eq. (7 A) now gives for A =0, A n = l/l and A b = -l/l,

A=A -A a X a -A b X b =-^+^,

and the moment M m , about any point ra, becomes

M m =

a = (X b - X a ) + X a ,

(61A)

230

KINETIC THEORY OF ENGINEERING STRUCTURES CHAP. XIII

giving

_ x , 'dM m _x

- ~ I -dX b ~T

Then according to Eq. (15L), the virtual work of the moment M a is

~j7rr I "nXl ~-^X a + lX a -^X b + -^X a X a -rjj(2X a + X b ) .

EiL\_Zi Zi o o ^ 'QtiiL

A similar process will furnish r b and calling X a =M a and X b =M b , then for any
member of single curvature

and

(61n)

When the moments M a and M b act in the same direction then the elastic curve
will have a point of counterflexure and a similar derivation to the one just given will
furnish the values

and T b =-^=-ft2M b M a ) (61c)

6EI

Considering a triangle as the fundamental element of a frame, the distortions Act,
Aft and Af in thethree angles resulting from the changes in the lengths of the members,
expressed in terms of unit stresses, were determined in Art. 37. It is now necessary to
show the relation between these angle distortions and the- deflection angles T at each end
of each member forming the triangle.

FIG. 61o.

Referring to Figs. 6lD and 6lE, it is easily seen that

(6lD)

Fig. 6 ID represents an impossible condition because at least one member of the
three composing the triangle must undergo double curvature as shown by Fig. 61 E and
as required by Eq. (3?E) , whereby

0. . (61s)

ART. 61

SECONDARY STRESSES

23 1

The elastic distortions Aa, J/3 and Jj- are given by Eqs. (37o), and may be considered
known for all angles of a frame and for one simultaneous case of loading and stress.

While Eqs. (37D) were derived on the assumption of frictionless pin connections
between all members, they are now used to determine the amount of distortion in each
angle which is prevented from taking place by virtue of the rigid riveted connections.

Two equations of the form of Eqs. (61c), may be written for each side of a triangle
or for each member of a frame. Using the designations indicated by the double sub-
scripts in Fig. 6lD, where the first refers to the apex or panel point and the second to
the far end of the adjacent member, then by inserting the deflection angles r into Eqs.
(6 ID), the following formulae are obtained for any triangle ABC'.

QEJa =

. . (6lF)

cb +r ca ) = (2M A +M bc ) +

*a lb

+M ac )

In the Eqs. (61r), the six moments are unknown, and hence a solution is impossible
unless at least three of them may be independently derived from other conditions to be
developed later.

According to Eqs. (Glo) the sum of the three angle distortions J for any triangle,
must equal the sum of the six deflection angles r and by the condition of Eq. (6lE), both
sums must then be equal to zero, hence

(61c)

Also, for any apex angles meeting in the same panel point, the sum of the angle
distortions J must equal the sum of the deflection angles r, hence for any panel point

2J = 2r, ........ .... (6lH)

and therefore, if one of the deflection angles r is known, the others around that panel
point may be successively found by applying equations of the form of Eqs. (6lD).

For any determinate frame, there will thus be as many unknown deflection angles
- as there are panel points, and for p panel points there will be p equations of the form
of Eqs. (61n), each involving one unknown deflection angle T.

The following values

M ab l c

r J

6/ c

"* ca'b

(61 1)

232 KINETIC THEORY OF ENGINEERING STRUCTURES CHAP. XIII

are now substituted into Eqs. (61r), as a matter of convenience, to obtain the following
formulae for any triangle ABC, thus

EAa=K ba +2(K ac +K ab )+K ea , . . . / ,~ : . . (61j)
EAp=K ab +2(K bc +K ba )+K cb , . . , .'._. . (61 K)
EA r =K bc +2(K cb +K ca )+K ac . . t -. \ . [. . ".. (61 L )

Adding these three equations and imposing the condition of Eq. (6 IE), then for
any triangle ABC,

=Q. ....,'-., (6lM)

Assuming now that the three quantities K^, K^ and K ba are known, then the remain-
ing three may be obtained from the above equations as follows:

from Eq. (61J), K ca =EJa-2(K ac +K ab ) -K^, . .>. . .. (Glx)

from Eqs. (6 IK) and (<ol^) , K^^E^+K^-K^+K^, . . . ; . .' (61o)
from Eqs. (6lL) and (6lM), K cb =EAr+k ab +K ba -K ca . . t . . . . (61r)

By the conditions of static equilibrium, the sum of all the moments about any panel
point must be zero, from which p equations of the following form are obtained, using
the values from Eqs. (61i). Thus, for any panel point A

~ ''

including all the members meeting in this point. See Fig. 61n and Eq. (61u) for the
manner of forming these equations.

The following procedure will furnish a solution for all the bending moments of a frame,
using Eqs. (6 IN) to (61q) : In the first triangle, evaluate three of the quantities K in
terms of the other three, using Eqs. (61 N, o, P). Two of the values K so found, also
belong to the adjacent or second triangle and a third value K in this second triangle
may be found from Eq. (6lQ), making three values known to find the other three by
applying Eqs. (6lN, o, P) to the second triangle.

This process is continued throughout the series of triangles up to the last one, where
two extra moment condition Eqs. (61q) become available and these suffice to deter-
mine the three first assumed values K. This is merely a process of successive elimination,
eminently suited to the present problem.

When the values K are thus found then the several moments M are obtained by sub-
stitution into Eqs. (61i). See the example below for symmetric structures with symmetric
loading.

The solution of the whole problem is possible, because for any frame with m members
and p panel points, there will be (ra 1) triangles furnishing f(w 1) equations of the
form of Eqs. (6lN, o, P) and p equations of the form of Eq. (61q) making in all |(rw 1) +p

ART. Gl SECONDARY STRESSES 233

available equations from which to find 2m unknown moments M. Hence, if the prob-
lem is determinate, there must be as many equations as there are unknowns and

2m = |(m-l)+p, giving 2p=m+3 (61n)

which by Eq. (3x) is actually true for any determinate frame composed of triangles
and, therefore, the solution is possible.

It was previously pointed out that the assumptions regarding the signs of the
deflection angles r as illustrated in Fig. 6lD, are impossible, and that not all these angles,
nor the moments M in the members of a triangle can have the same algebraic sign on
account of the condition imposed by Eq. (6lE). Therefore, the computed results must
furnish deflection angles r and moments M of both signs.

However, to avoid complicated rules, it is necessary to adopt some conventional
or standard figure and then reverse the assumed deflections wherever the computed angles
T are negative. For this reason the conventional figure of distortion is assumed as indicated
in Fig. 61r, wherein the members in alternate triangles I, III, "V, etc., are made to
present a concave line outward, indicating positive moments M.

FIG. 61r.

The even triangles II, IV, VI, etc., must then represent negative moments, but
in order that Fig. 61r may represent the conventional assumption of all positive moments,
the Eqs. (3?D) are divided through by minus one when applied to the even triangles with
convex lines outward.

Hence, for the odd triangles with positive moments M, Eqs. (3?D) are applied in

the form

E Ad = (f a ~fb) COt r + (f a ~f c ) COt /? 1

EJp = (f b -fc) cota+(/ 6 -/a) cot r \, (61s)

E A Y - (f c -fa) cot p + (f c -f b ) cot a 1

and for the even triangles the negative moments are also made positive by using Eqs.
(61s) in the form

= (/6-/ a ) cot r + (fc-fa) cot/? I
(f c -fb) cota + (/-/ fc ) cot r , (6lT)

= (fa -fc) cot p + (f b -fc) cot a

234

KINETIC THEORY OF ENGINEERING STRUCTURES CHAP. XI! I

and then all moments, represented by Fig. 61r, may be considered positive. The sub-
scripts a, b, and c in Eqs. (61s) and (Gli), refer to the sides a, b, and c opposite the
respective angles a, {3, and 7-.

Since the above assumptions do not represent real or possible conditions, some of
the moments M, resulting from the final computation, must develop negative signs.
Thus, if M ac and M^ in Fig. 61o, should turn out with negative signs, then the real
deformations of the triangle would be as represented in Fig. 6 IG.

FIG. 61n.

In evaluating Eqs. (61cj) for the several panel points, it is necessary to observe the
relative signs of the moments in the several members according to the following direc-
tions. Thus, for any panel point, C, Fig. 61n, the clockwise moments should be treated
as positive and the counter clockwise moments as negative to obtain the following
equation :

= +M oa -M ob +M oc -M od =Q,

or in terms of the values K

6/0

ion

Qlob

lob

'On

(61u)

The arithmetic operations required in the solution of problems are somewhat lengthy
but not difficult especially when the values 7 and I of the members are well rounded off,
which is always permissible in these computations.

The secondary stress resulting from the bending moments M =QKI/l, produced in any
member by the riveted connections, may now be found from Navier's law, Eq. (49M), as

My_ QKy

(61v)

where y is the distance from the gravity axis of the member to the extreme fiber and
f s is the unit secondary stress caused by the bending of the member due to the moment
Mg, or Mb applied at one end of the member.

The total maximum unit stress in the member is then ff s , where f=S/F, which
is the primary unit stress due to the direct loading.

ART. 01 SECONDARY STRESSES 235

Thus for a compression member, the maximum total stress occurs on the side where
f s is negative.

The eccentricities e, Fig. 6lA or 6lB, produced at either end of any member ~AB by
the secondary moments, may be found from

and *!_. ...... (61W)

The secondary stresses here considered may vary in percentage of the primary stresses
from 5 to 100 per cent, depending on the type of truss and the details of design employed
for the members.

The subject is, therefore, one of considerable importance and cannot be dismissed
in the usual manner as taken care of by the safety factor.

The question of maximum secondary stress in any member has not yet received
consideration here, though it is a matter of vital importance.

It does not follow that the maximum secondary stress occurs" simultaneously with
the maximum primary stress for any particular member. The results of computations
show this to be quite true for the chords, but not for the web members, though there
seems to be no criterion available to indicate the position of a train of loads to produce
maximum secondary stress in a member.

The method of influence lines, while not impossible, becomes practically prohibitive
on account of the labors involved. The only reasonable treatment which suggests itself
is to analyze the structure first for total maximum load and then for a few typical posi-
tions for the moving load and from these results select the probable maximum values
for each member.

Since the entire truss must be completely solved for each simultaneous position
of the live load, this appears to be the only practicable advice to give in view of the
immense labors involved in such analyses.

The above method is now illustrated by the solution of a problem.

Example. A single-track, through Pratt truss, designed by Messrs. Waddell &
Hedrick, in 1899, for the Vera Cruz and Pacific Ry., Mexico, is selected. The truss has
six panels with a span of 145 feet, a depth of 30 feet, and the distance between trusses
is 17 feet. The equivalent uniform live load for WaddelTs class W was used. The case
here investigated is for maximum total load over the entire span.

The cross-sections F and the moments of inertia I are computed and tabulated
together with the lengths I, widths b and distances y for all the members of the truss.
These together with the stresses S, due to the total maximum load Fig. 61i, are given
in Table 61 A, where the unit stresses f=S/F and the functions 61/1 and 6y/l are com-
puted. Impact is not considered.

A live load of 54000 Ibs. and a dead load 20800 Ibs. per panel were assumed for

tke computation of the stresses.

While it is not necessary to know how the members were designed as long as F and
7 are given for each, yet it may be of some interest to the reader to have this
Figs. 61 J thus represent the sections of the several members.

236

KINETIC THEORY OF ENGINEERING STRUCTURES CHAP. XIII

FIG. 61i.

B-D.

D-F.

A-C-E.

E-G.

B-C.

T*8

D-EANOF-G.

B-EAnoD-G.

II 1

FIGS. 61J.

C E

FIG. 6lK Unit Stresses and Cotangents

ART. 61

SECONDARY STRESSES

237

TABLE 6lA
DATA FOR THE TRUSS FIG. 61i.

Mem.

F
Gross,

Sq.in.

Gross,

in.<

/
in.

In.

Lbs.

f- S
f- F

Lbs. sq.in.

6v
1

15.9

139

290

12.37

6.19

+ 158,400

+ 9960

2.88

0.128

15.9

139

290

12.37

6.19

+ 158,400

+ 9960

2.88

0.128

26.9

282

290

12.5

6.25

+ 252,060

+ 9370

5.84

0.129

26.5

750

290

15.0

7.5

-252,060

-9510

15.52

0.155

29.4

805

290

15.0

7.5

-283,500

-9640

16.66

0.155

33.5

1464

462

16.25

8.34

-245,000

-7310

19.01

0.108

17.6

323

462

12.0

6.0

+ 156,200

+ 8875

4.19

0.077

17.6

323

462

12.0

6.0

+ 48,400

+ 2750

4.19

0.077

13.7

119

360

12.37

6.19

+ 68,000

+ 4960

1.98

0.103

14.7

288

360

12.0

6.0

- 44,200

-3010

4.80

0.100

14.7

288

360

12.0

6.0

6,800

- 460

4.80

0.100

6 = width of member in the plane of the truss.

y = distance from neutral axis to extreme fiber of section.

I is taken about the gravity axis perpendicular to the plane of the truss.

The distortions in all the angles of the several truss triangles are now computed
from Eqs. (61s) and Eqs. (6lT), using the unit stresses/ from Table 6lA and the cotangents
of the angles, all as shown in Fig. 6lK.

Applying Eqs. (61s) to the odd triangles I, III and V, and Eqs. (6lT) to the
even triangles II and IV the following values Ed are obtained.

Triangle I. Eqs. (61s) :

EAai=( 4960-9960)0.0 +( 4960+7310)1.24 ==+15,215
i=( 9960 +7310)0.806 + ( 9960-4960)0.0 ==+13,920
i = ( -7310 -4960) 1.24 +(-73 10 -"9960) 0.806= -29,135

Triangle II. Eqs. (6 IT):

= ( 9960 -8875)0.806 + ( 4960- -8875)1.24 =- 3,980

= ( 4960-9960)0.0 +( 8875-9960)0.806=- 875

2 = ( 8875-4960)1.24 +( 9960-4960)0.0 =+4,855

238 KINETIC THEORY OF ENGINEERING STRUCTURES CHAP. XIII

Triangle III. Eqs. (61s):

= ( -3010 +9510)0.0 +(-3010-8875)1.24 =-14,737
= ( -9510-8875)0.806 + ( -9510 +3010)0.0 = - 14,818
= ( 8875+3010)1.24 +( 8875+9510)0.806= +29,555

00
Triangle IV. Eqs. (6 IT) :

= ( 9370 -2750)0.806 + (-3010 -2750) 1.24 =- 1,806
= ( -3010 -9370)0.0 + ( 2750 -9370)0.806 = - .5,336
= ( 2750+3010)1.24 +( 9370+3010)0.0 =+ 7,142

00
Triangle V. Eqs. (61s) :

= (- 460+9640)0.0 +(- 460-2750)1.24 -- 3,980
= ( -9640 -2750) 0.806 + (-9640+ 460)0.0 = - 9,986
-( 2750+ 460)1.24 +( 2750+9640)0.806= +13,966

The condition Eq. (6lE) must be satisfied for each triangle as indicated above, offer-
ing a complete check on the numerical results.

FIG. 6lL Values EA.

The values EA are now entered on the diagram Fig. 6lL, where the conventional
assumptions for the moments M are indicated according to Fig. 6 IF. The moment
Eq. (61q), is then written out by following the directions given under Eq. (61u).

Using the values for 67/7 from Table 6lA in Eq. (61q) the following equations are
obtained :

ART. (51 SECONDARY STRESSES 239

For panel point A,

"ob 'oc

and after inserting the values from Table 61 A, this gives

19.01^=2.88^ or K ac =Q.QQK ab . ...... (1)

For panel point B,

2 M b = - M te + M bc - M be + M bd = 0,
giving

- ig.Oltffc, + 1.98# 6c -4. l9K be + 15.52K bd =0,
or

K bd = 1.23^-0.12^ +0.27 K be ......... (2)

For panel point C,

2M c = M Ctt -M c6 +M ce =0,
giving

2.8&K M - 1.98^ +2.8K C9 =0,
or

^ ca . . . ........ (3)

For panel point D,

2M d = -M db +M de -M dg +M df =0,
giving

- 15.52K db +4.SOK de -4. l9K dg + 16.66 K df =0,

(4)

For panel point E,

SM e = -M ec +M eb -M ed +M en =0,

giving

-2.88K ec +4.19# e6 -4.8QK ed +5.84K eg =0,

... (5)
For panel point F, noting that for symmetrical loading Mf d =M fh , then

panel point G, where for symmetric loading AT,, =--M gi and M gd = M ghi then

-r /A^

,11? 71 r Hf . , ... 1 71/f ?IT . ffon

For

giving , , v n
-5MK ge +4.19X ffd -4.80^/ +4.19A,, -o.84^ fft - =0,

#^--2.433^+1.746^. (7)

240

KINETIC THEORY OF ENGINEERING STRUCTURES CHAP.XIJI

When there is no middle vertical FG, then, for symmetric loading, an equation of
the form Eq. (6lM) is written out for the center triangle DGH to supply the last condition.

For an unsymmetric truss or unsymmetric loading, the several equations must be
extended over all pin points of the structure.

Assuming now that K ab , K^ and K ac , Fig. 6lL, are known, then Eqs. (6lN, o, P)
will furnish the values K^, K bc and K cb for the first triangle ABC, and by applying this
process successively to all the triangles it becomes possible to express all the values K
in terms of K ab and K^ by including the moment equations (1) to (5) just evaluated.
The two last Eqs. (6) and (7) will then serve to find K ab and K^, as will be shown later.,
and these values substituted back will furnish all the unknowns K.

In applying Eqs. (6lN, o, P) to the several triangles, it is decessary to use the standard
lettering employed in the derivation of these formulae and hence each triangle is sketched
in Figs. 6lM, to exemplify the process. The angle a must be so located in a triangle
that it is included between the two adjacent known K's, and the angle /? must be adjacent
to the third known K, while f is between the adjacent unknown K's. This then deter-
mines the vertices (A), (B) and (<?), shown in parenthesis, for the standard lettering.

FIGS. 6lM.

The designations of the angles in Fig. 6lK, for the solution of the values EA were
selected so as to comply with the conditions just described. The arrows in the Figs.
6lM indicate the values K found from a previous triangle and from one new moment
Eq. (6lQ) above evaluated as (1) to (5).

Eqs. (6lN, o, P) and Eqs. (1) to (5) thus give:

At panel point A, by Eq. (1) : K ac =Q.QK ab .

By Eq. (6lN), K M

By Eq. (61o), K be

By Eq. (61p), K A =EA n +K ta> +K l>a -K ea ^ -44,350 + 16.2^

At panel point C, by Eq. (3), K ee =O.Q9K A -K M = -45,816+26.4^+2.38^.

I By Eq. (6lN), K ec =EJa 2 -2K ee -2K A -# 6c = 147,2 17 -76.6^-6.76^

II \ By Eq. (61o), K bc =E4p 2 +K cc -K hc +K ec -=71,3Ql -41.6^-2.38^

By Eq. (61p), K eb =EA r2 +K cb +K bc -K ec = - 157,577 +84.2# a6 +6.767^.

ART -6l SECONDARY STRESSES 241

At panel point B, by Eq. (2) : ^ M = l-23X te -0.12X 6c +0.27X 6e = 15,780-10.2X o6 +0.83X 6([ .

[ ByEq. (61 N ), K db =EJa 3 -2K bd -2K be -K eb = -31,502 + 19.4X a6 -3.66X 6o
AIII j ByEq. (61o), K ed ~EJ i 8 3 +K M -K eb +K db = l 27,037 -75.0X a6 -9.59Xj >a

I By Eq. (61p), K de =E4 r3 +K be +K eb -K db = -25,129 +23.2X a6 +8.04X 6a .

At panel point E, by Eq. (5) :

K eg =0.493X ec -0.72X e6 +0.82X ed =290,203 - 159.9X n6 - 16.07X 6a .

I" By Eq. (6lN), K ge =EJa^-2K eg -2K (id -K de = -811,157+446.6X a6 +43.28X 6a
AIV I ByEq. (61o), K dg =EJ l 3 4 +K eg -K de +K ge = -501,161 +263.5X^ + 19.17X6,,

I By Eq. (61p), K gd =E4 r4 +K ed +K de -K ge =920,207 -498.4X tt6 -44.83X 6a .

An panel point D, by Eq. (4) :

K d1 =0.932X d6 -0.288X de +0.252X dff = -148,414 +77.80 K ab -O.gOX^.
ByEq. (6lN), K fd =EAa5-2K df -2K dg -K gd =374,963 -184.2X a6 +8.29X 6a
AV ByEq. (61o), K g f=E^s + K df -K ad + K fd = -703,644 +392.0X ab +52.22X fca

By Eq. (61?) , K fa =EJ r5 +K dg +K gd -K fd =58,049 -SOJX^ -33.95^.

The moment Eqs. (6) and (7) for panel points F and G now furnish independent
values for X/ ff and K g f and by combining these with the last two equations of AV, the
values K ab and X^ may be found as follows:

By Eq. (6), X /ff =6.942X /d =2,602,993 -1278.7X a6 +57.55X 6a .

By Eq. (7), ^--2.433^+1.746^-3,580,226-1056.81^-183.57^.

Substituting these values of Kf g and K g j into the last two equations of AV, then,

2, 544,954 -1228.0X aft + 91.50X^=0,
4,283,870 -2348.8X a6 -235.79X fea =0,

from which K ab = 1947.5 and K ba = -1676.8.

These values of K ab and X^ are now substituted into the above twenty equations,
furnishing all the values X as follows:

(1) K ab = 1947.5

(2) K ba =- 1676.8

(3) X ac =6.6X ab = 12,854

(4) # = 15,215- 29,602+ 1,677= -12,710

242 KINETIC THEORY OF ENGINEERING STRUCTURES CHAP. XIII

(5) K bc = 29,135- 16,748+ 3,354= 15,741

(6) K cb =- 44,350+ 31,549- 3,354 = -16,155

(7) K cc =- 45,816+ 51,414- 3,991= 1,607

(8) K ec = 147,217-149,178 + 11,335= 9,374

(9) K be = 71,391- 81,016+ 3,991= - 5,634

(10) K eb =- 157, 577 + 163 ,979 -11, 335=- 4,933

(11) K bd = 15,780-19,864- 1,392- - 5,476

(12) Kdb=-- 31,502+ 37,781+ 6,137= 12,416

(13) K ed = 127,037-146,062 + 16,081 = - 2,944

(14) K de =- 25,129+ 45,182-13,481= 6,572

(15) K eg = 290,203-311,405+26,946= 5,744

(16) K ge = -811,157 +869,753 -72,572 = -13,976

(17) X d ,= -501,161 +513,166 -32,144 = -20,139

(18) K gd = 920,207-970,634+75,171= 24,744

(19) K df =- 148,414 + 151,515+ 1,509= 4,610

(20) K id = 374,963-358,730-13,901= 2,332

(21) K gf = -703,644 +763,420 -87,562 = -27,786

(22) K ta = 58,049- 98,738+56,927= 16,238

Eq. (6lM) now furnishes a convenient check on the above numerical results, by tak-
ing the sum of all the six values K for each triangle.

Thus for AI, the values (1) to (6) give

S#i =30,542 -30,542 =
and for All, the values (5) to (10) give

2K 2 = 26,722 -26,722=
and for AIII, the values (9) to (14) give

2# 3 = 18,988 -18,987= ?.
and for AIV, the values (13) to (18) give

2 A' 4 =37,060 -37,059 = 1
and for AV, the values (17) to (22) give

2^5=47,924-47,925= -1

The secondary stresses due to the bending at each end of each member are now found
from Eq. (61v) as given in the following Table 6lB.

ART. (51

SECONDARY STRESSES

243

TABLE 6lB
SECONDARY STRESSES DUE TO BENDING

Mem.

End.

6y
I
in.

/.- 5 ?

Eq. (61v).
Lbs. sq.in.

Lbs.
sq.in.

67 K

" SI
in.

Mem.

End.

E
D
E
G
D
G
D
F
G
F

6.V
I
in.

'.= 6J P
Eq. (61y),
Lbs. sq.in.

/-I

Lbs.
sq. in.

61 K

~ SI
in.

AC
EC
CE
BE
BD

A
B
A

C
B
C
C
E
B
E
B
D

0.103

+ 1,948
- 1,677
+ 12,854
-12,710
+ 15,741
-16,155
+ 1,607
+ 9,374
- 5,634
- 4,933
- 5.476
+ 12,416

200
172
1645
1627
1621
1664
205
1200
434
380
849
1924

-7310
-7310
+ 9960
+ 9960
+ 4960
+ 4960
+ 9960
+ 9960
+ 8875
+ 8875
-9510
-9510

0.15
0.13
0.23
0.23
0.46
0.47
0.03
0.17
0.15
0.13
0.34
0.72

ED
EG
DG
DF
GF

0.100

- 2,944
+ 6,572
+ 5,744
-13,976
-20,139
+ 24,744
+ 4,610
+ 2,332
-27,786
+ 16,238

294
657
741
1803
1551
1905
715
361
2779
1624

-3010
-3010
+ 9370
+ 9370
+ 2750
+ 2750
-9640
-9640
- 460
- 460

0.32
0.72
0.13
0.32
1.76
2.16
0.28
0.14
19.62
11.46

0.128

0.129

0.103

0.077

0.128

0. 155

0.077

0.100

0.155

The total stress on the extreme fiber is f s +/, noting that no increase was made in / for buckling
effect in compression members.

The actual signs of the values K, and hence also of the moments M^&KI/l, now being
determined, the real character of the distortions may be represented diagrammatically
in the following Fig. 6 IN.

For compression members the most severely stressed fibers will occur on the side
where /. is negative and for tension members on the side where / is positive. Thus in
the compression member AB, the critical stressesoccur on the upper side at A and on the
lower side at B, while for the tension member AC, these occur on the upper
and on the lower side at C.

244 KINETIC THEORY OF ENGINEERING STRUCTURES CHAP. XIII

ART. 62. SECONDARY STRESSES IN RIVETED CROSS FRAMES OF TRUSSES

The analysis of cross frames, so far as this is possible, presents many obstacles, some
of which cannot be overcome owing to the variable character of the external loads.

The bending effect due to all forces acting on a frame, even when direct stresses are
neglected, leads to very complicated formulae of more or less questionable value, and
no attempt is made here to discuss the general problem.

A cross frame is usually subjected to the dead load of the bridge floor concentrated
on the floor beam; the live load, impact and centrifugal force applied to the floor; wind
pressure against the moving load and the vertical trusses; and unequal deflection of the
main trusses due to a variety of causes, but particularly to unequal temperature.

The magnitudes of the secondary stresses depend of course on the details of construc-
tion and bracing employed in any special frame, so that many different forms would
require investigation. However, only two principal types will be discussed and the
same formulae are applicable to both through and deck bridges.

In the following, the unbraced and one type of braced cross frames are considered
first for direct loading and then for wind effect.

(a) Dead and live load effects. Unbraced cross frame with rigid post connections. The
construction, shown in Fig. 62A with lettered dimensions, is analyzed by assuming
equality between the unknown bending moments induced in the posts by the symmetric
loads P, as follows: M = M 2 and M] =M S , positive as indicated when they produce
compression on the outer fibers of the posts.

Call M the bending moment on the floor beam due to symmetrically placed loads
P acting on a simple beam which rests on two supports C and D.

Also call mi the moment at any point of the post AC distant x from A ; m?. the
moment at any point of the strut AB; and m^ the moment at any point of the beam
CD. Then from Fig. 62A.

m l =M 1 +( M ~ Ml \x, m 2 =M 1 and m 3 =M +M. . . . (62A)
\ h /

Considering only the effect due to bending, by neglecting shear and direct thrust,
the virtual work of deformation for the entire frame would be by Eq. (15n)

~ Cmfdz. .- . : . . . . . . . . (62n)

-&/3./0

-&/3

Substituting the values from Eqs. (62A) into Eq. (62p,) then

which integrated by considering everything constant except x, gives

?^.+ * \M *b+2M f b Mdx+ P^WI, (62c)

/2 bl3\_ J Jo

ART. 62

SECONDARY STRESSES

245

which is the general expression for any case of loading where M must be evaluated for
each case.

Now since the unknown moments M and MI must have such values as will make
the first differential derivative of W with respect to each, equal to zero, then after some
reduction

'dW

and

3M 3/i

_ ^ =f)

- " "

(62n)

?*" m * II

-j

,-""

.-I, I,-

50 ITIj Ij -

FIG. 62A.

FIG. 62B.

Solving Eqs. (62n) gives

,. 6l\o . \ oil

= -Mn(-r^-+2 -TJ-

and

from which

(62s)

Mi =

^i- 1 Mdx

3/i6

(62p)

-1

1/2 '-/\ W 3

For the symmetric loading shown in Fig. 62A, the integral Mdx becomes

and for a uniform load p per foot of floor beam,

fb f)b 3

(62G)

246 KINETIC THEORY OF ENGINEERING STRUCTURES CHAP. XIII

MQ and MI may thus be found from Eqs. (62F) and (62r), and from these the moments
TOI, m-2 and ra 3 , for any point of the frame, are given by Eqs. (62 A).

The maximum fiber stress in the posts must be combined with the direct stress sus-
tained as a truss member.

The upper strut receives a compression of (MoMJ/h and a bending moment MI
over its entire length.

The floor beam will receive a bending moment of Mo+M and a direct stress of
(Mi-Mo) /h.

Example. Given the cross frame at DE, Fig. 61i, with the following dimensions:
h =28 ft., b -=17 ft., a = 7 ft., P -59000 Ibs., h =226 in. 4 = 0.0109 ft. 4 , 7 2 = 1134 in. 4 =0.0547
ft., 4 and 7 3 = 15262 in. 4 =0.736 ft. 4 The style of the frame is as shown in Fig. 62A.

Using the value from Eq. (62c) in Eq. (6.2r), and substituting the above data, then

3X00109 59000
__ 28X0.736* 4 (U n __

/3X0.0109X17 W3X0.0109X17 \ - >

V 28X0.0549 )\ 28X0.736 /

and from Eq. (62s)

For t/=5.5 inches, Eq. (61v) then gives the secondary stress in the post DE.

t ,, ,Miy 1499X12X5.5

for the upper end D, f d = -^~ = =438 Ibs. per sq.in.;

t ., . ,M v 3541X12X5.5

for the lower end E, f e = p- = = 1034 Ibs. per sq.m.

1\ 22u

The actual deformation of this frame is indicated in Fig. 62s.

A braced cross frame with rigid post connections Fig. 62c, will now be treated as in
the previous case.

Neglecting shear and direct stress as before, and dealing only with the effect due
to bending, then Eq. (62B) becomes

Integrating this expression as above, and then differentiating first with respect
to MQ and then with respect to MI and placing these derivatives equal to zero, the follow-
ing equations are obtained:

~~ = A! M + 2hi M l + 2eMi - 0.

ART. 62 SECONDARY STRESSES

Solving these two zero equations and noting that h=h\+e, then

247

i M

and

Mdx

(62n)

/*6

wherein I Mcfo is given by one of Eqs. (62a) and the moments MQ and M\ may

thus be found.

The stresses in the upper struts then become

M I M,

= -- and /S e / = -- 1
e e

1L.

-m,
-I.

> .. _

FIG. 62c. FIG. 620.

and the stresses in the post will be

and

FIG. 62E.

Example. Taking dimensions as in the previous example and making hi =19.5 ft.
then M =4770 ft.-lbs., and ^=1661 ft.-lbs. giving /, = 1393 Ibs. per sq.in., and f c =

Ibs. per sq.in. .

(b) Wind effects. Unbraced cross frame with rigid post connections.

wind loads on the trusses of a bridge may be carried to the abutments by means of com-
plete horizontal trusses in the planes of the top and bottom chords respectively, pro
vided the end reactions of the top chord wind system can be carried to the abutments
by suitable end postal bracing. In this case the intermediate cross frames suff<
little or no distortion and hence carry no bending effects.

However when the wind pressure along the top chord is carried down to the I
torn chord locally at each cross frame, then the latter must be distorted and thus resi
bending In this case the total wind pressure is carried to the abutments through the
bottom chord lateral system. The external forces on a cross frame are then as repn

248

KINETIC THEORY OF ENGINEERING STRUCTURES CHAP. XIII

sented in Fig. 62D. The wind loads w are transmitted through the posts to the lower
lateral system, producing bending effects as shown in Fig. 62E.

Assuming that the wind loads w are equal and calling T the tangential stress at the
point of counterflexure and S the normal stress at the same point, then

H=w =

(62i)

The general work equation for the frame, considering bending effects only, is the
same as Eq. (62fi) in which the moments mi, m 2 and ra 3 must be evaluated for the wind
forces now acting.

The moments of the external forces about A, B, C, D and 0, from Fig. 62o (count-
ing clockwise moments positive) are respectively

M 3 =- T(h -ho) -Sb =H(h ~ho) = -

(62J)

M 2 = Th Q Sb 2wh = Hh because
-Sb-2w(h-ho)=0
The moments mi, m 2 and m^ thus become

mi=Mi + ( ^- - J x=H(x +hoh) with origin at A
\ h I

( ^-r

-\x=H(hho)(-rl} with origin at A

~with origin at C

(62K)

Substituting these values into Eq. (62s) then
2# 2 C h ,

W =ETJ O (x+h ~v

which gives by integration,

- dx

(62L)

Since &o must be so chosen that the internal work will be a minimum, this particular
value may be obtained by equating to zero the differential derivative of W with respect
to ho, thus obtaining

^ W h SRI QM _i_ b (i> i \ j_^ & n
5r- =7-(6rto 3ft) +- r (h -hi) +-j- =0,

from which

b_ Oh b'

/3 /I /2

(62M)

ART. 62

249

Substituting this value of ho into Eqs. (62X), will give the moments at all points
of the frame and from Eqs. (62j) the stress S and the moments at the ends of the post
may be found.

Example. For the cross frame of the first example and a horizontal wind load of
Ibs. the following values are obtained:

oo
2S

3X28
(XOl09

0.0547 /

17 6X28 17_
0.736 0.0109 0.0547

- 14.25 ft.

Substituting values into Eqs. (62j), then,

Mi = -H(h-ho) = -3100(28-14.25) = -42,625 ft.-lbs.
MO =Hho =3100 X 14.25 -44,175 ft.-lbs.
_ 2w(h -ho) _ 6200(28 - 14.25)

b 17

This gives for the stress at the bottom of the post AC,

= -5197 Ibs.

M y_S
' /i F

44175X12X5.5 5197 , 100ftn .

~T4~7 = 12,899 3o3 Ibs. per sq.m.

226

For a braced cross frame with rigid post connections Fig. 62r, the analysis is con-
ducted in precisely the same manner as in the previous case, making as before, H =w = T.

Counting clockwise moments positive, then the moments of the external forces about
the points E, C and are respectively

M l = -T(h l -ho) = -H(h l -ho) '

M = Tho=Hho .... , -. .. . . (62N)

-Sb-2w(h-ho)=0

250 KINETIC THEORY OF ENGINEERING STRUCTURES CHAP. XIII

The moments at any point x from C then become:

for Cl, Mo + ( Ml ^ M -}x = H &o -*) ,

i,\

(62o)

for EA, Mi (x -h,) =H(h l -h ) (-},

e \ e /

for CD, Mo ~x =HhJl ~ .

These values inserted into the general work Eq. (62B) give for the frame

i .
--- dx+--

from which is obtained after integrating

i *e^^H
*o) q- + Q

o J o

Taking the first differential derivative with respect to /? and equating the same
to zero then

3fto ^i
from which

MM
0= - .............. ^ ^

Example. Given a cross frame as per Fig. 62r, for which ft =28 ft., 7^ =19. 5 ft.,
6 = 17 ft., e=8.5 ft., /i =226 in. 4 , 7 3 = 15262 in. 4 and w>=3100 Ibs. to find /% M and M l .
From Eq. (62p)

19.5(56 + 19.5)

ho= - iT><226 = 10 - 96ft -'
2(28+39)+^

and from Eqs. (62N)

MI = -3100(19.5-10.96) = -26,474 ft.-lbs.
M =3100X19.96 =33,976 ft.-lbs.

^=-^^0) = -6215 Ibs.
6

The maximum stress in the post ~AC is then

My _S^ _ 88976X12 XBJ _^18 _ 9928 _ 423 Ibs. per sq.in.
/i /'ac 226 14.7

The stresses in the top struts and diagonal are not so easily determined when the
double latticed type of bracing is used. In the present case with one diagonal AF, the

ART. 03 SECONDARY STRESSES 251

stresses are determined by passing a section tt through the three members and the point
of counterflexure, and taking moments about A, then,

P-^T 3100(28-10.96)

_T(hhv) -eEF=Q, or EF=- ^ - = -62i5 Ibs.

o.o

The stress in AJ3 is found by taking moments about F, obtaining,

- T(hi -h ) -Sb -we +eAB =0,
whence

T ( hl _^) + Sb +we 3100(19.5 -10.95) -6215X17 +3100X8.5 =
AB= - T 8.5

Also from the vertical shear on the section

ji a =0, or TF = --^- = +^a = !3,842 Ibs.

In concluding the subject of cross frames it might be added that good designs should
aim at deep floor beams and slender vertical posts, as is clearly indicated by Eq. (62n) ,
which shows that M is diminished when 7 3 is increased and increased when I r is increased.

It has been stated by Mr. Grimm, " Secondary Stresses in Bridge Trusses," p. 80,
that the assumption of fixed connections between floor beams and posts is not verified
by investigations. The author suggests an explanation for this by calling attention
to the fact that recent tests indicate that compression members as formerly built are
not nearly as stiff against buckling as was supposed and furthermore, a slight initial
deformation approaching the elastic curve which the stressed member might attain,
would almost obviate secondary stresses in the cross frames. That such deformations
actually exist, or might be produced in overloaded posts by a permanent set, there can
be little doubt, and hence it is quite easy to understand why some of the high theoretical
stresses do not appear to exist when the actual stresses are measured with mstrum

ART. 63. SECONDARY STRESSES DUE TO VARIOUS CAUSES

(a) Bending stresses in the members due to their own weight. The maximum bending
moment, in any member, resulting from its own weight, when supported on fnctionless
pin bearings at the ends, will occur at the center of the member and i

(63A)

where I is the length of the member, p its weight per unit of length and 6 is the angle
which the member makes with the horizontal.

WhenThe member is fixed at the ends and there is no direct stress the bendmg moments
M at the ends of the member, and M c at the center, become respe,

and

252 KINETIC THEORY OF ENGINEERING STRUCTURES CHAP. XIII

In any case these moments produce a unit stress / on the extreme fibers of the mem-
bers which is given by Navier's law, Eq. (49M) as

(630

where y is the normal distance from the neutral axis to the extreme fiber.

. It is thus seen that M increases as the square of I while / is inversely proportional
to 7. Hence, long members of small moment of inertia may receive severe secondary
stress due to their own weight.

When the member is in compression or tension, the combined bending stress on the
extreme fiber, due to the direct stress and the uniform load, must be investigated. This
cannot be accomplished by algebraic summation of the separate bending effects, because
axial compression increases the deflection due to cross binding, while axial tension dimin-
ishes this deflection. The bending stress, resulting from the simultaneous loading must,
therefore, be found.

The analysis for riveted end connections, while very complicated, is not usually
necessary. For pin-connected members where the cross bending effect is always more
severe, the following approximate solution is given.

A horizontal compression member, with centric pin connection at each end, is shown
in Fig. 63A. Let M c designate the maximum bending moment which, in this case, occurs
at the center of the member. Calling p the weight of the member per unit of length,
S the axial stress, I the length of the member and 3 the deflection at the center, then

I ,'.'.-.. . . (63D)

_ = _ 8Sd
~48ET~3MEI P
from which

Assuming the effect of the direct stress on the deflection d to be the same in character
as that of the uniform load p, which is not quite true, then

(63E)

(63F)

48EI -5S1 2 '
This value of d substituted into Eq. (63D) gives

6pm

ART. 63 SECONDARY STRESSES 253

The combined direct and bending unit stress on the extreme fiber of the member
then becomes by Eq. (49M)

(030)

where F is the cross-section and y is the normal distance from the neutral axis to the
extreme fiber of the member.

Horizontal tension members with pin-connected ends, when similarly treated, lead
to the following formulae:

, 6pPE7
" "* M <=48W+ (63H)

The combined direct and bending stress on the extreme fiber is again found by

Eq. (63o).

Example- Given an eye bar 2 X 15 in., making F =30 sq.in.; 5=600,000 Ibs.; Z=660
in.; 7=562.5 in. 4 ; p=8.5 Ibs. per in.; and E= 29, 000 ,000 Ibs. per sq.in.

Then from Eqs. (63n), 3 =0.4824 in.; and M e --= 173 ,450 in. Ibs. The stress on the
extreme fiber then becomes by Eq. (63c)

,600,000 173,450X7.5 =2313 lbg .

J 30 562.5

showing that the bending stress alone increases the tension in the lower fiber by 11.5

per cent.

(b) Secondary stresses in pin-connected structures. In all usual stress computations
the question of how equilibrium is established between the stresses in the several mem-
bers meeting at a pin or panel point is not considered. It is thus assumed that the
individual members extend to the pin centers with full effective sections where the stres
are suddenly balanced in a point.

In reality the case is very different and no such balance in the stresses can be accom-
plished The nearest approach to a realization of the ideal condition is in the case of two
abutting compression members and then only when there are no bending stres

be transmitted. ,, , ,

When all the members are connected by pins according to the usual methods of
construction, the stresses are transmitted past the panel points in very much the same
manner as for riveted connections on account of the factional resistance on the pins
created when the structure is distorted by superimposed loads^ The advantages uu^
claimed for pin connections are not all realized in pract.ce and wh, e cent connections
are best made by this stvle of panel joint there may be suffic.ent fnctional resistance on
the pin to produce secondary bending stresses qnite equal to those occurring m rive.

is thus an erroneous idea to regard a pin-connected structure as free from secondary

254 KINETIC THEORY OF ENGINEERING STRUCTURES CHAP. XIII

stresses. This is never so and the rnalysis may be even more complicated, though the
results are probably more reliable than for riveted structures.

The exact behavior of a member with pin bearings depends on many circumstances,
especially the diameters of the pins. When these are sufficiently large the member can-
not turn under load, and secondary stresses must result. These may be partially relieved
by the vibratory action of the live loads which would probably allow the structure to
adjust itself for some mean condition of loading and reduce the secondary stresses to a
minimum. However, temperature changes would enter as a disturbing element to pre-
vent such favorable action from establishing permanent relief.

The frictional resistance offered by a pin connection is now analyzed.
Let S =the stress in any member.

R =the total frictional resistance on the surface of contact between the pin

and the member.
r=the radius of the pin.
^o the angle of repose, making tan ^o = ( a=the coefficient of friction between

the two metals.

c =the eccentricity of S required to resist the moment of the frictional force R.
e=the actual eccentricity of S required to resist the bending moment M a due

to secondary stress.

The line of stress S, Fig. 63s, is supposed to suffer a displacement c from the pin
center A of an amount which will make the moment Sc exactly equal to the opposing
moment of the frictional resistance R.

FIG. 63B.

For a frictionless pin the eccentricity c would be zero and the stress would pass through
the pin center.

Hence from the figure

Sc=Rr and c=r sw.p, ......... (63i)

showing that c is independent of the stress S.

When c>e no rotation will take place as a result of deformation of the truss due to
load effect, and when c < e the member will turn on the pin and the full amount of bend-
ing cannot be developed by the frictional resistance.

The secondary stresses in a pint-connected structure are found in precisely the same
manner as shown in Art. 61, for riveted connections, except that for all members where

ART. 63

SECONDARY STRESSES

255

c<e the limiting value c will govern and the secondary stress becomes accordingly less.
Eq. (61w) gives the value of e=QKI/Sl, furnishing the limiting value

^ Sic Sir sin p
= = ~

and by Eq. (61v) the limiting secondary stress becomes

f _M a y_Syrsmp
Js j- j

(DDK)

Regarding the values of p, which are purely empiric, no good experimental data
seem to be available, though it is preferable to choose rather high values from 12 to 14,
making sin p =0.2 to 0.25.

(c) The effect of eccentric connections. It frequently happens that the line of stress
through a member is not coincident with its neutral axis, thus giving rise to eccentric
connections.

It might be said that such connections should never be tolerated, and hence their
effect in producing secondary stresses would not require investigation. However, there
may be rare cases where eccentricity in unavoidable and proper provision must then be
made for taking care of the bending stresses thereby produced.

These bending stresses are not merely additional stresses, but the secondary stresses
previously found on the assumption of centric connections will be incorect on account
of the changes which must take place in the deflection angles r and the angles a, (3, j-,
etc., between the truss members, as a consequence of the eccentricity.

Calling c a and c b the eccentricities of the stress at the two ends of a member AB, Fig.
63c, then the moments M a =S ab c a and M^S^ will produce deflection angles r by
Eqs. (61c), which become

I t^tr n if \ ^db/'

(63L)

256 KINETIC THEORY OF ENGINEERING STRUCTURES CHAP, xill

The values Er a and Er b , resulting from eccentricity, are now incorporated with their
proper signs into the values EJa, and EA$ of Eqs. (6lN, o, p) and the secondary stresses
are' then computed in the ordinary way as described in Art. 61.

Finally, the stresses f 8 on the extreme fibers must be combined with the direct stresses
/ resulting from the bending moments SabC a and SbaC b . It should be observed that the
stresses f s and / frequently have opposite signs, in which case the eccentricity effect may
partially neutralize the secondary stress due to riveted connections.

It is sometimes important to investigate the effect due to the eccentricity of the
end supports where the friction of the expansion bearings may produce considerable
bending stress in the end members. Good designs will, however, obviate any serious
stresses from this source.

(d) Effect of temperature and erroneous shop lengths. The manner of dealing with
these effects in the case of indeterminate structures has received attention in previous
chapters. For determinate structures having end supports and pin. connections absolutely
frictionless, a small change in geometric shape, due to temperature or erroneous shop
lengths, would have no appreciable effect on the stresses in the members.

However, when the panel connections are not frictionless then any effect, as unequal
temperature, etc., which is productive of a deformation of the structure, will also create
secondary stresses.

The unit stresses / which result from any deformation of the structure due to any
changes Al in the lengths of the members, may be found from the proportion

Al f d

whence f=jE=Eet.-'. . . ._ . .' . . . (63M)

Knowing the stresses / for all the members of the structure due to any simultaneous
condition of temperature or loading, the values EAa and K, and the consequent secondary
stresses, may all be found in the manner described in Art. 61.

ART. 64. ADDITIONAL STRESSES DUE TO DYNAMIC INFLUENCES

(a) Forces acting longitudinally on a structure. The moving loads in passing over a
structure exert a certain horizontal impact which is transmitted through the floor system
to the main trusses. When the brakes are set the amount of this impact may become
considerable arid increases with the number of braked wheels on the span and .the mag-
nitude of the moving loads. The forces thus applied to the structure are called tractive
forces and produce certain additional stresses in the truss members.

The effect which the tractive forces exert on the trusses depends on the relative posi-
tion of the roadway to the points of support. For a bridge, supported on the level of the
loaded chord, all the additional stresses are confined to this chord and the floor system.
When the supports are on some other level then all the truss members will be stressed,

ART. 64

SECONDARY STRESSES

2:,7

and this gives rise to a great variety of cases depending on the type of truss considered.
Two cases will be presented for illustration.

Let 7"= the tractive force on one rail under any wheel load P.

/=the coefficient of friction between the braked wheel and the rail, usually
taken between 0.15 and 0.2, and varying greatly according to weather
conditions.

Then the total tractive force, for n wheels on the bridge, becomes

(64A)

For a through bridge, the compression in the bottom chord members will then be
increased by the tractive force producing the additional stress 2 X 1 T, where x is the dis-
tance from the fixed end of the span to the head of the train.

All top chord and web members remain unaffected. When the train approaches
from the roller end the bottom chord receives compression and the stress increases uni-
formly from the roller end toward the fixed end.

For a deck bridge, as shown in Fig. 64A, the reactions due to a tractive force 2T 1
becomes:

A=jST; B=-j2T; H = VT (64 B )

FIG. 64A.
The additional stresses in the members of panel 1-2 are found as follows:

For top chord,

17=*-

fi 2

h 2

I T

V" '*2/ ^X-*-

I or bottom chord E*-

M, Ax, -g(fto-e)-eSg!T

cog

For diagonal
For vertical

D = (A-Lsin X) sec 6
F=A-Lsin A

(64c)

The angles Jl and are regarded positive as shown in the figure. ^ The tractive
force is taken positive in the direction from right to left, supposing the tram to appn
the span from the roller end at B.

The maximum value for the additional stress is found for the same position o
train loads as is used to produce maximum primary stress in any particular member.

258 KINETIC THEORY OF ENGINEERING STRUCTURES CHAP. XIII

These additional stresses become less important for long span, massive, structures
where the dead load is large in proportion to the live load. For smaller structures, the
chord members, adjacent to the hinged end of the span, may receive excessive additional
stresses requiring increased area or provision for reversal of stress.

The effect of tractive forces on the floor system is not taken up here.

When the loaded chord is on a grade, the total weight G of the combined dead and
live loads on the span exerts a certain component G sin a longitudinally along the struc-
ture and in a down grade direction, where a is the grade angle. This component is trans-
mitted to the hinged support and should be considered in designing this bearing, especially
if the support happens to be a steel tower.

(b) Forces acting transversely to a structure. They are wind pressure, lateral
vibrations and centrifugal force on a curved track. These will be separately discussed.

Wind pressure is the force due to wind against the total area of all structural elements
of both trusses, the floor system and live loads, as presented in elevation. While the
wind may have any direction, it is presumed that at times this direction will become
normal to the exposed area and thus attain maximum intensity.

Experiments have shown that the normal wind pressure may reach a value of 50
Ibs. per sq.ft., which amount is assumed in figuring the wind stresses in the unloaded struc-
ture.

Usually it would be unwise to cross a bridge when the wind has sufficient velocity
to produce such pressure, and hence the maximum pressure to be assumed for the loaded
structure need not exceed about 30 Ibs. per sq.ft., applied to the total area of the struc-
ture, the moving load, and the portion of the leeward truss which is not covered
by the moving load. While both trusses may not receive the full intensity of pressure
it is customary to make no deduction on this account.

The wind loads falling on the truss members are considered as concentrated loads
acting on the several pin points of the structure, and in estimating the stresses in
the web members of the horizontal wind trusses or lateral trusses, these loads are treated
as moving loads. The wind pressure on the live load is always considered as a moving
load applied to the wind truss of the loaded chord. The total maximum wind load will
govern for the chord stresses.

The end reactions of the wind trusses must eventually be carried to the main truss
supports through the end portals or sway bracing. Sometimes only one wind truss is
used, and then the wind loads of the other chord are carried to this wind truss by
sway bracing at each panel point, throwing the entire wind load on the one lateral system.

Since the center of gravity of the wind load on the moving load is always above the
plane of the lateral system of the loaded chord, it is necessary to consider the overturn-
ing effect of these loads in producing unequal vertical loading of the two main trusses.
Figs. 64B show five cases which are met with. In Figs, a, b and c one wind truss is pro-
vided and a sway brace at each panel, while in Figs, d and e there are two wind trusses.

Let Wi =the wind pressure per panel on the live load.

hi =the lever arm of w\ to the plane of the horizontal wind truss.

Wz =the wind pressure per panel on the bridge.

Ji2=the lever arm of w 2 to the plane of the horizontal wind truss.

ART. 04

SECONDARY STRESSES

2o9

Then the overturning moment is

+w 2 h 2 in Figs, a and b }

fc TW i' with one

w 2 h 2 in l H ig. c

in Figs, d and e with two wind trusses

truss

. 64n)

and the vertical load V, acting down on the right-hand truss and up on the left-hand truss
becomes V=M/b. Depending on the direction of the wind, this leads to positive and
negative stresses of equal intensity in each vertical truss. Hence, V is treated as a
live load for the vertical truss to obtain the additional stresses due to overturning effect

rr
h,

t f

(a)

(c)

(d)

FIG. 64s.

of the wind on the moving live load, though it might be better to determine the stresses
separately for the wind loads Wi and w 2 . Calling V\ and V 2 the respective vertical wind
loads, pd the live load per panel and gd the dead load per panel, then for parallel chords
the additional wind stresses become SdVi/pd+SiV 2 /gd, where Sd is the dead load and
Si the live load stress in any member.

It should be noted that the case in Fig. c is most favorable, while the one in Fig. b
is most unfavorable in the production of loads V, and may give rise to stresses amounting
to as high as 30 per cent of the primary stresses in certain members.

The stresses in the wind trusses may be found by applying the methods given in
Chapter XII, and those occurring in the portal bracing and sway system were analyzed
in Art. 62s.

All of these additional stresses resulting from wind pressure in the primary members
must finally be combined with the primary stresses to obtain total stresses.

260 KINETIC THEORY OF ENGINEERING STRUCTURES CHAP. XIII

The additional wind stresses increase in importance with the length of the span,
and may assume enormous proportions in large structures. It should be mentioned,
however, that the frequency in the occurrence of maximum wind stresses is extremely
rare, and their simultaneous combination with maximum primary stresses is very
improbable.

In view of these considerations it would seem proper to emplo}^ higher allowable
unit stresses, say such as are permissible for quiescent loading, possibly restricting the
total sections of the primary members more nearly to those required only for maximum
dead and live loads. For single-track bridges exceeding a span of 200 ft., and for double-
track structures of over 350 ft. .span, the wind stresses must be fully considered, while
for smaller structures the end posts and end chord sections should be appropriately strength-
ened, making perhaps little allowance in the other members.

Lateral vibrations, while they exist, do not appear to produce any serious increase in
the primary stresses of a bridge. This is certainly true for highway structures, and in the
case of railroad bridges, this effect may be reduced to insignificance by making appropriate
provisions to keep the moving loads steady while passing over the span.

The magnitude of the vibratory forces here considered depends on the velocity of
the moving load, the character and type of locomotive, the condition of the track and
the relative stiffness and rigidity of the floor system and main trusses. Short spans
and narrow structures, also skew bridges might receive considerable vibratory stress,
especially in the end panels. Coned wheels and loose or spread rails are very objectionable.

It is needless to say that factors of such uncertain constitution and variable nature
are not susceptible to analysis, and a direct measurement (even if it could be made) of
such additional stresses would merely indicate what might occur in certain cases, but
the results from different experiments could not be accurately correlated with the
surrounding circumstances.

The remedy thus lies in obviating, so far as possible, the conditions producing dele-
terious effects rather than in any attempt to estimate their magnitude and weigh down
the structure by allowing extra material in the required cross-sections. If this can be
done, then the slight additional stresses due to lateral vibrations may safely be neglected.
The vertical effect of impact is considered elsewhere in this article.

Centrifugal force due to curved track. This problem necessarily deals only with rail-
road bridges and includes the centrifugal force acting transversely to the structure and
producing stresses in the stringers and in the lateral system of the loaded chord; the
overturning effect of the centrifugal force producing stresses in the stringers, floor beams
and in the main trusses; and the unequal distribution of the vertical loads on the floor
and between the main trusses due to the eccentricity of the curved track axis with the
straight bridge axis. This latter problem was solved by influence lines in Art. 29, for
the case of a skew plate girder on a curve, and will receive no further attention
here.

The amount of the centrifugal force acting transversely to the bridge is given
by the formula

C=^-P=cP, (64B)

ART. (54

SECONDARY STRESSES

261

where v is the velocity of the moving train; P is any moving load on the bridge;
#=32.2 ft. per second, being the acceleration due to gravity; and R is the radius of
curvature of the track. The units are foot, pound, second. For velocities in miles
per hour, and curvature in degrees D, c =0.00001 \7v 2 D.

The following table gives values of c=v 2 /gR for curves of one to ten degrees and
velocities which may be regarded as maximum for such curves.

Degree of Curve,
D

v Miles per
Hour.

Degree of Curve,
D

r Miles per
Hour.

0.042

0.176

0.078

0.189

0.110

0.198

0.136

0.204

0.158

0.206

In computing the additional stresses S c in the main truss members resulting from
centrifugal force, the maximum total stress might be obtained from the train giving
the maximum primary stresses rather than from the train moving with the highest
velocity. For short spans the maximum combination is likely to occur from passenger
trains, while for long spans this may be true for freight trains moving at lower
speed.

The methods of finding the stresses in the main trusses and in the lateral system
of the loaded chord are precisely the same as given for wind loads.

Since the centrifugal force C is a linear function cP of the moving load, therefore
the horizontal forces coming on the lateral truss of the loaded chord will, at all points,
be c times the moving load. The same is true of the overturning effect separately
considered, where the line of action of C is above the plane of the lateral truss and
passes through the center of gravity of the load P, which
may be assumed at the same level as the center of pressure
of the wind load.

The stresses in the lateral truss of the loaded chord will
then be found in the ordinary way for loads cP considered
as moving live loads. When these loads P are uniformly
distributed then the stresses found for loads cp will ^be
cp/w times those found for the uniform moving wind
load iv.

The stresses in the main trusses, due to overturning
effect of the centrifugal force C, are analyzed in the manner
previously explained for wind loads. Fig. 64c represents the
condition at a panel point assuming the load P and the
centrifugal force C to be loads per panel and applied at the FIG. 64c.

center of gravitv of the moving load.

The track is eccentric with the truss and the floor beam is superelevated to suit

262 KIXETIC THEORY OF ENGINEERING STRUCTURES CHAP. XIII

the curved track. The resultant K, of P and C, intersects the plane of the lower
laterals at a distance e from the truss axis and may there be resolved into forces P
and C =cP as they would act on the main and lateral systems.

The reactions Vi and V 2 , representing the vertical loads which are transmitted
to the main trusses, then become

P
and Vi=

With these loads Fj and V^, which vary for each panel point on account of the
variable eccentricity e, the method given in Art. 29 may be advantageously used to
find the combined stresses in the main trusses by influence lines and that is the best
solution for this problem. It should be noted that the load Vi becomes maximum
for quiescent loading, while V% attains maximum for the highest velocity of the
train.

(c) Forces acting vertically on a structure. This embraces several causes which
together exert a more or less severe dynamic influence on the stringers, floor beams
and vertical trusses of a bridge, all included under the general term impact.

So far as these can be separately enumerated, they may be classified as follows:
(1) The true impact due to the instantaneous application of a moving load to a structure
at rest; (2) the impact resulting from the unbalanced condition of the engine drivers;
irregularities in the surfaces of the rails and imperfections in the rolling stock, etc.;
(3) elastic vibrations of the structure which tend to increase the effect of dynamic
loads otherwise applied to a rigid body; and (4) vertical centrifugal forces occasioned
by the vertical curvature (deflection and camber) of the track.

Owing to the extremely variable conditions and combinations of circumstances r
it is practically impossible to analyze separately the magnitudes of these several causes
and their resulting effects, though a brief discussion may serve to point out precau-
tionary measures tending to reduce the impact stresses in bridges under traffic. Some
very valuable tests have been conducted during late years, both in Europe and America,
which give a very good idea of the sum total effect produced on railroad bridges by
the moving loads, and from these, formulas have been deduced which make it possible
to provide suitable strength in designing bridges.

The additional stresses, resulting from this combination of causes, are a function
of the velocity and magnitude of the moving load, and hence assume minor importance
when dealing with highway structures.

The true impact due to the instantaneous application of live loads is really not a
serious source of additional stresses, because forces are transmitted through a structure
with a rapidity approaching the velocity of sound, while the moving train could not
exert its effect in so short a space of time.

Theoretically, a load instantaneously applied would produce a dynamic effect
twice as great as the static load, but in reality this cannot take place and there will

ART. 64 SECONDARY STRESSES 263

always be a sufficient lapse of time between the application of the loads :md tin-
necessary elastic deformation of the structure to Obviate to a great extent this extreme.
dynamic impulse. The vibrations accompanying the moving loads really constitute
the element of danger.

The most unfavorable conditions prevail in short span structures and in the wd>
members subject to a rapid change of stress.

The causes enumerated under (2) and (3) are really of the most serious character,
and while much may be accomplished by a careful maintenance of the track and
rolling stock, a certain average condition will usually prevail beyond which no remedial
measures will be possible or feasible.

In point of design and construction of a railroad bridge the following suggestions
may be offered: The bridge and its approaches back for some distance should be
straight and when curves are absolutely unavoidable speed restrictions would seem
proper. The rails should be long and the joints and tie connections of the best and
most rigid construction, carefully maintained. The connections at the abutments
require the most attention, avoiding uneven settlements and loose rails. The intro-
duction of continuous roadbed over bridges is very desirable as affording some elasticity
to absorb the impact rather than to transmit the same to the structural portions of
the floor and trusses. A similar effect is produced by long ties over widely spaced
stringers. Very rigid floors and rails directly on stringers may be classed as
objectionable.

The greater the proportion of dead to live load and the longer the span, the less
will be the effect due to impact.

The avoidance of synchronous vibrations between the moving load and the
structure is of the utmost importance and may be practically accomplished by properly
stiffening all the members and by providing thorough bracing in the lateral and sway
systems.

The effect of vertical centrifugal force might be separately estimated, but the
amount is small and the attending conditions are too uncertain to warrant
this.

As early as 1859, Gerber proposed a general formula providing a certain reduction
in the allowable unit stresses to cover these several sources of dynamic impact. This
formula is

wherein /' is about the elastic limit of the material and S t and S d are the respective
live and dead load stresses in any member.

The following formula are at present in use and give the factor by whic
live load stress should be multiplied, so as to include the dynamic effects considered
under the present, heading. Calling S t the live load stress, S d the dead load stress
and I the strength of span for chord members, or the loaded distance producing
maximum stress in a web member, then <p is. given as follows:

264

KINETIC THEORY OF ENGINEERING STRUCTURES CHAP, xill

(1) C. C. Schneider, 1887, railroad bridges,

, _ 1 300

1+300 1+300

(2) C. C. Schneider, electric roads,

,_ t 150 _l+4oO
1+300 ~r+300

(3) J. A. L. Waddell, railroad bridges,

, 400 _/+900

* Z+500 "Z+500

(4) J. A. L. Waddell, highway bridges,

,_ 1 100 _Z+250
^ Z + 150 ~r+150

(5) H. S. Pritchard, 1899, railroad bridges,

, Si 2S t +S d t

Y = + o . o = o , o ; for counters 0=2

(6) J. Melan, railroad bridges,

_
+30 1+30

(7) J. Mdan, secondary roads,

(8) F. Engesser, railroad bridges,

(9) F. Engesser, highway bridges,

fgg _ A 2

^ = l-50+ y 3300 ; for Z>33, = 1.50

(10)* American Railway Eng'r and Maintenance of Way Ass'n, 1910,

, _l 2 +40,000
* 1 2 + 20,000

Table 64A gives values of as obtained from each of the above ten formula for
a few even lengths I.

Only those values of are comparable which are intended for the .ame cla of
structures as (1), (3), (5), (6), (8) and (10). It is clearly seen that formula (lO),ives
the highest values for short spans I and the lowest values for long spans, noting that in

'. . . (64a)

* Where I is the span length and all members receive the same impact.

ART. 64

SECONDARY STRESSES

265

this formula I represents the length of span and not the loaded distance as in formula?
(1) to (9).

VALUES OF

TABLE 64 A.
AS FOUND FROM THE VARIOUS EQS. (64c)

I Feet.

(l)

(2)

(3)

(4)

(5)

(6)

(7)

(8)

(9)

(10)

1.97

1.48

1.78

1.63

Cannot

1.73

1 .55

1.97

1.66

1.99

1.86

1.43

1.73

1.50

1.44

1.33

1.69

1.50

1.89

100

1.75

1.38

1.67

1.40

solved

1.32

1.24

1.67

1.50

1.67

200

1.60

1.30

1.57

1.28

without

1.24

1.18

1.67

1.50

1.33

300

1.50

1.25

1.50

1.22

knowing

1.21

1.16

1.67

1.50

1.18

400

1.43

1.21

1.44

1.18

the

1.20

1.14

1.67

1.50

1.11

500

1.38

1.19

1.40

1.15

stresses

1.18

1.13

1.67

1.50

1.07

600

1.33

1.17

1.36

1.13

1.18

1.13

1.67

1.50

1.05

The American Railway Engineer and Maintenance of Way Association formulae
is undoubtedly entitled to the greatest confidence, being based on very extensive
experiments which were carried out by the committee on impact tests under actual
conditions of traffic. The formula is not as yet officially adopted by the association.

However, no allowance was made for secondary stresses in the members and hence
the formula (10) may be said to include more than actual impact effect which is
probably true of all the Eqs. (64o).

(d) Fatigue of the material. Based on the classic experiments of Wohler (1859-1870)
which were continued by Professor J. Bauschinger, a formula was proposed by Launhardt
(1873) and later modified by Weyrauch, aiming to apply the principles of the fatigue
of material to the design of bridge members.

Wohler's law states that for a large number of repeated load applications, rupture
of a material is produced under stress which is below the ordinary (static) breaking
strength of that material. The conditions under which Wohler's experiments were
made differ so radically from those attending the actual operating conditions of bridges,
that it is questionable whether anyone is justified in applying the Launhardt- Weyrauch
formula? to bridge designs.

Wohler's load repetitions followed in quick succession and were continued without
interruption (several million times) until failure was produced. Bridge members are
subjected to a repetition of stress which is always followed by a rather long period of
rest, and few structures, even under the heaviest traffic, are retained in service long
enough to receive several million applications of the moving load. Also, a well-designed
bridge is never calculated for stresses approaching e ven the elastic limit, while Wohler
bases all his observations on stresses exceeding this limit.

In addition to these facts, modern experiments made under conditions which
respond quite closely with bridge practice, though limited in extent, point uniformly

con

against the existence of fatigue in bridges.

The above mentioned Launhardt-Weyrauch formulae have been extensively used

266 KINETIC THEORY OF ENGINEERING STRUCTURES CHAP. XI I]

in designing bridge members and are still retained in many specifications. However,
this practice does not seem justifiable, especially when allowance is made for impact and
secondary stresses generally.

The fatigue formula undoubtedly served a good purpose in the days when so many
important factors were neglected, but at the present time the aim should be to make
proper allowance for all the known stress elements and thereby reduce the " factor of
ignorance " to a minimum.

(e> Unusual load effects may be produced by loads of unusual magnitude applied
at rare intervals or by loads which occur under unusual circumstances.

It is possible that the maximum combination of moving loads, wind pressure and
temperature effects may take place, under which the structure might be stressed to its
utmost capacity, while under ordinary conditions of traffic the stresses would be far
below those for which the design was made. The maximum load basis for a design is
then an unusual load effect for the average use of a structure.

At times it may be necessary to transport some abnormal piece of freight or a train
of locomotives, which would exceed the loads assumed in the computations. This may
be done without danger even if the members are stressed quite near the elastic limit,
provided the design is made with some forethought. For most ordinary structures such
a practice might prove very disastrous.

It should be observed that, while the stress in any member for a particular position
of a moving load is exactly proportional to the intensity of this load, the combination of
maximum and minimum stresses for such member, which fixes the required cross-
section, might not be tlms proportional. In the case of chord members, all governing
stresses are proportional to the loads because the critical positions of the loads are
alike for all chords, that is, for maximum chord stress the whole span is fully loaded
while the minimum chord stress occurs for dead load only.

Hence, the chord sections are directly proportional to the total moving load, while
most of the web sections are not so proportional and increase more rapidly than the
loads. This is particularly the case with all counter web members wherein the stress
is the difference between that produced by the moving load and that due to the dead
load. It also applies to the sections of such web members as are subject to stress reversal.

This is an important item in making provision for a future increase in train loads,
where 25 to 50 per cent additional carrying capacity might easily be secured by providing
a slight increase in the counters and web members having stress reversals. Many old
structures would still be usable for considerable overload had such provision been made
in their design.

The best way to provide for such overload in view of future increase in train loads,
is to design the structure for the given case of loading and allowable unit stresses and
then to increase the sections of the counters, and web members with stress reversals,
for the desired overload, which of course could not exceed a reasonable amount according
to the limitations imposed by the chord sections.

A structure so designed might then carry say 30 per cent overload without imposing
unduly on certain few web members, while the ordinary structure would fail by the
buckling of these members long before the chords had received any serious stresses.

iART. 05 SECONDARY STRESSES 267

In the same way a highway bridge may be designed to carry an occasional overload
W a certain heavy road roller, etc., without increasing the sections of the chords and
[main web members, but by providing for the extra counter stresses due to the excess load.

To the second class of unusual load effects belong derailments, collisions with drift-
wood during times of high water and possible settlements of piers or abutments.

The absolute prevention of such occurrences may be classified with the impossible.
However, the ultimate destruction of a structure when they do occur may be safeguarded
by applying certain precautionary measures which should never be overlooked.

Thus the bridge floor should be made sufficiently strong to allow a derailed train
to pass over without breaking through, and the wheels should be guided between guard
rails or by other means to prevent collision with the main truss members, necessitating
certain clearances for through bridges to accomplish this. In the case of deck bridges
t may be advisable to utilize the top chord as a barrier or protection to prevent the train
'rom leaving the structure.

The design of the floor should, therefore, aim at the use of solid web, riveted
girders and braces in preference to built-up frames, and the soft, ductile varieties of
steel are more desirable than the harder, brittle varieties.

In cases where high water may reach the bottom chord, slender eye-bar members
ire decidedly objectionable.

ART. 65. CONCLUDING REMARKS

(a) Features in design intended to diminish secondary and additional stresses. The

'ollowing suggestions should be given careful consideration:

Skew structures and bridges on curves should never be built except in very rare
cases where no other disposition is possible. These types should be regarded as measures
)f last resort.

The axes of all members of a truss should be in the same plane and should intersect
.n a point at all connections.

Special attention should be directed toward a careful design of all connections with
i view of reducing the secondary stresses. Thick gusset plates and large diameter rivets
materially reduce the sizes of these plates and are thus desirable from this aspect.

The widths of members, in the plane of the truss, should not be chosen larger than
s absolutely required to secure proper connections and stiffness against buckling -in
compression members. It may thus be desirable to taper compression members from
ihe center toward both ends.

The cross-sections of members should be so chosen that the material is concentrated
as far from the neutral axis as possible, thus securing the largest moments of inertia
for the smallest over all dimensions. The cross form is thus the least advantageous,
while a square box form is most desirable. In rare instances the cross form may be
acceptable, when the width of a member is determined by other conditions. Secondary
stresses occurring simultaneously in the plane of the truss and in a cross frame are additive
in members of the box type while in the cross form they are not additive, a consideratK
which may become important at times.

268 KINETIC THEORY OF ENGINEERING STRUCTURES CHAP. XIII

The secondary stresses may be reduced to a certain degree by shortening all tension
members and lengthening all compression members by amounts Al, determined for the
case of maximum total load. This practice is generally followed and provides a camber
in the unloaded structure which exactly disappears under the full load when the structure
assumes its true geometric shape.

The plane of the lateral system should coincide with that of the chords and the
plane of the floor should be as close as possible to that of the lateral system.

The web members of the lateral systems should present considerable stiffness against
buckling, making l/r not greater than 140 for all these members, whether in compression
or tension.

The end portals, or cross frames, should be heavy to carry the wind reactions and
dynamic effects to the supports.

Floor beams should be made deep to reduce secondary stresses in the cross frames-
When this cannot be done then flexible connections should be provided between the
truss members and the beams, a type which is best suited to large bridges, while riveted
connections are desirable for small spans.

The stringers should be made heavy and continuous and should be designed tot
transmit the tractive forces to the panel points of the loaded chord instead of to the floor
beams by inserting proper tie members between the stringers. Long ties over widely
spaced stringers tend to relieve impact vibrations.

The use of pin connections is not admissible for bridges of less than 200 ft. span,
and even in larger structures the advantages to be gained may not be so easily demon-
strated. Some benefits may be derived from pins for the web system but certainly not
for the compression chord.

According to prevalent practice in designing pins, the diameters are usually so large
that the friction produces secondary stresses quite equal to those resulting from riveted;
connections. Perhaps this condition might be remedied.

Also pin-connected columns are not as stiff as those with riveted ends, though the
great convenience offered by pin connections during erection and the saving of material
and other advantages in point of design, speak greatly in their favor.

The advisability of using the higher classes of steel for the main truss members of
large structures and employing soft steel for the floor system was previously mentioned. '

In choosing between different styles of trusses, those of the statically determinate
class should always receive preference, other things being equal. The primary stresses
will usually be less than in similar indeterminate systems, especially when temperature
stresses are included. Yet the deformations and secondary stresses may be less and
frequently the connections may be simpler for the indeterminate types.

The use of light-colored paints is advisable to reduce temperature effects, especially
in structures involving redundant conditions.

(b) Final combination of stresses as a basis for designing. If the several stresses
resulting in a bridge member from all causes could be absolutely known, then there is nqj
good reason why a design should not be based on a unit stress of say four-fifths the elastic
limit of the material and still allow ample leeway for some deterioration and unusual effects.

The low unit stresses of one-third to one-half the elastic limit, generally employed

ART. 65 SECONDARY STRESSES 269

are intended to cover, more or less blindly, the unknown stresses, on the presumption
that these are in a way proportional to the known primary stresses. In reality, few,
if any, of the secondary and additional stresses are directly related to the primary
stresses, but are produced by totally different causes.

Our knowledge of the properties of material, while of a purely empiric nature,
undoubtedly approaches the truth as closely as the knowable accuracy of the moving
loads would require. The behavior of full-size bridge members has also been investigated
to an extent which should make it possible to design such members with a reasonable
expectancy of developing a strength commensurate with that observed on test specimens.
This was perhaps not possible in the past, but should be expected when the results of the
elaborate tests of the American Society for Testing Materials become available.

If the stresses in the members of a structure are determined with similar exactness
and combined into totals representing the maximum and minimum stresses for each
member as a result of all known material causes, and on a basis of equivalent quiescent
loads, then there is no valid argument why the safe unit working stress / should not be
taken at least equal to two-thirds the elastic limit of the material.

If also the main members and counters, subjected to reversals of stress, are designed
for a 30 per cent overload, then such a structure should still be usable even for a 30
per cent increase in the assumed moving loads.

Furthermore, since the combined effect of the maximum values of all the known
primary, secondary and additional stresses is one of very rare occurrence, this provides
still greater safety and subjects the structure to a less severe average usage than that
intended by the maximum combination of all loads.

Assuming then that the following stresses have been computed, or approximately
estimated when no other means is afforded, then the required area for any member may
be determined from the formulae given below.

Let Sd =the stress in any member due to dead load.
Sz=max. stress in this member due to moving load.
S'i=mm. stress in this member due to moving load.
5,0= the max. wind stress due to Wi and w 2 , Art. 64b.
S r =the stress due to tractive forces, Art. 64a.
S t =the temperature stress when redundant conditions exist.
S c =the stress due to centrifugal force from curved track. Art. 64b.
^=the impact coefficient or moving load factor Eq. (64c).
Z=the length of a member in inches.
r=the least radius of gyration in inches.

/=the allowable unit stress = two-thirds the elastic limit of the material.
M 8 =the bending moment produced by rigidity of joints, weight of the member,
eccentricity, etc., representing the total secondary stress, Arts. 61, 62, 63.

Then the required area F for any tension member becomes:

t-8m4-8*+&l+-rar. ( 65A ^

270 KINETIC THEORY OF ENGINEERING STRUCTURES CHAP. XIII

Whenever the area My/fr 2 can be added to the area given by the first term of Eq.
(65A) to obtain F without materially altering y and r, then it may be done, otherwise
a new section must be chosen to satisfy the equation. Also, that combination of the
stresses in the parenthesis which gives maximum must be used.

Eq. (65A) will also apply to any compression member when / is reduced for buckling
effect according to a modified form of Mr. T. H. Johnson's formula, for values of Z/r< 120:

For hinged ends and soft steel of 30,000 Ibs. per sq.in. elastic limit,

/= 20,000 -88^- 1

I ..... ..... (65B)
for flat ends, / = 20,000 -7 1 1 - \

For hinged ends and medium steel of 35,000 Ibs. per sq.in. elastic limit,

/=24,000-105-

(65c)

for flat ends, /=24,000- 85-

When S'i is opposite in sign from Sd and Si, then the member must be capable of
carrying an additional stress l.3</>'S'i in the place of <l>Si, using the maximum combina-
tion of the same sign as S'i in Eq. (65A). In this case, which is one of stress reversal,
the factor 1.3 should also be applied to the stress </>Si. The impact factors and <//
being determined by one of the formula? in Eqs. (64c) for the load lengths producing
the respective stresses Si and S'i. However, this should not be construed to mean that
the area must be designed to carry the total loads LSflS'i+US^Si as if simultaneously
applied.

Aside from the additional 30 per cent in the live load stresses, members with stress
reversals may not require any increase in area, though the design should always be made
so as to provide for the two kinds of sitress.

CHAPTER XIV
MITERING LOCK GATES

ART. 66. THE NATURE OF THE PROBLEM

A mitering lock gate is in principle a three-hinged arch, so placed that the hinges
have vertical axes, and the arch elements are acted upon normally by horizontal water
loads. Each half arch is called a gate leaf and the hinges are replaced by compression
joints adjacent to the quoin ends and miter post, see Figs. 6?A and 68A.

Each leaf is designed to swing about a vertical axis at the quoin end, thus permitting
the gate to be opened when the pressures on up- and downstream sides are equalized,
a condition which prevails when the gate is swung in air or when the water level is equal
on both sides.

The fastening on the top of the quoin post is called anchorage, and is usually in
tension while the gate is open. At -the same time the entire weight of a leaf is supported
on a pintle located at the foot of the quoin post. See Figs. 68A.

The maximum stresses on the anchorage and the pintle are encountered when the
gate is swung in air, while the maximum stresses in the structural elements of the gate
occur when the gate is closed against a full head of water on the upstream side with no
water on the downstream side, which is the case when the lock chamber is empty.

The ordinary working condition of a pair of gate leaves is less severe, being due to
a full head of water on the upstream side and a counter pressure on the downstream
side produced by the water inside the lock arid sufficient to float the largest vessels for
which the lock is intended. These two water levels are usually called the upper and
lower pool levels.

In order that a closed gate may act as a barrier or structural element against a
water head, the bottom horizontal girder is made to bear against a horizontal miter sill
extending over the full length of each leaf. The amount of pressure exerted by each
leaf against this miter sill is somewhat problematic, depending on the adjustment of
the sill to the gate. Usually this adjustment is so made that the pressure is just sufficient
to make a water-tight joint, though it may approach zero or may be made to take the
entire reaction due to the full head of water.

Hence, a variety of conditions may occur, ranging from zero sill pressure to the full
hydrostatic' head. Usually these two extreme assumptions are made and the resulting
stresses in the gate are found for each case. The actual condition during operation
may then be assumed at some intermediate state of sill contact which can only 1
surmised but never absolutely known.

272 KINETIC THEORY OF ENGINEERING STRUCTURES CHAP. XIV

The structural elements of a single gate leaf consist of a series of horizontal girders
connected vertically by a sheathing or skin plate and a number of vertical intermediate
braces or stiffening girders. The horizontal girders of both leaves thus constitute a
series of three-hinged arches transmitting the water pressure through the quoin hinges
to the lock walls. The vertical system of stiffeners may be regarded as a set of trusses
connected with the horizontal arches and thus constituting a series of vertical continuous
girders, the supports of which are the elastic horizontal girders.

If the horizontal girders were designed to carry exactly their proportionate water
loads and at the same time be allowed to deflect in proportion to these loads, then for
no contact between the miter sill and the vertical stiffeners, the entire work would fall
on the horizontal girders and the vertical stiffeners would do no work and would thus
be superfluous.

However, for practical reasons, the top horizontal girders are always designed much
stronger than is actually required for the water load and also the other girders are not
all different, owing to commercial sizes of rolled shapes and the saving in manufacture
resulting from duplication of like parts. Also, some sill pressure must be provided to
accomplish water tightness of the gate. Therefore, the vertical system of stiffeners
must necessarily do a certain amount of work in distributing these inequalities between
the horizontal girders and the vertical stiffening system. The reactions, which the
horizontal girders thus present to each vertical girder, may be considered as redundant
reactions of a vertical continuous girder on n flexible horizontal supports, where n is
the number of horizontal arch girders.

The vertical girders are not very definite structural elements, being made up of a
rather disconnected system of web bracing between the horizontal arches. The total
vertical stiffness of a gate is thus produced by the combined effect of the sheathing
plates, the quoin and miter posts and the intermediate vertical girders. If these were
all combined in one vertical system so as to represent one vertical girder for a < gate leaf,
then the average stiffener per foot of gate could be approximated by dealing with an
average vertical section of the gate leaf. Such a hypothetical vertical girder could then
be treated as a continuous girder, on n horizontal supports consisting of n three-hinged
arches.

The problem would thus be to find the magnitudes of these n redundant supports
Xi to X n of the vertical girder for a certain water load (usually the maximum) together
with the elastic displacements d\ to d n of the n supports.

Since the contact at the miter sill cannot be absolutely adjusted, it is best to inves-
tigate two cases; one with full contact against the miter sill, giving the maximum sill
pressure; and the other where the miter sill pressure is just zero and there is theoretically
no contact.

It is certain that all possible cases must be included between these two extremes.
However, in practice it is customary to design the bottom horizontal girder to carry
the total water load without sill contact and to design the sill amply strong to carry the
same full load, but to adjust the gate so that the sill pressure is only sufficient to make
a water-tight joint.

The author has proposed details of construction obviating such duplication of

ART. 67 MITERING LOCK GATES

273

strength in the gate and miter sill by the introduction of a flexible sill connection whereby
the water pressure against the sill can never exceed a certain allowable quantity.

ART. 67. THE THEORY OF THE ANALYSIS

The principal system. The problem is analyzed according to the solution of the
general case of redundancy presented in Art. 55, using the graphic method for all deflection
polygons. Since there are only two cases of loading requiring investigation it is best to
employ the general Eqs. (55A) and (55B) necessitating two solutions, each involving n
equations of n unknown quantities X.

The effect of temperature will be neglected in the present, though it could be
determined in accordance with Art. 56.

To make the problem perfectly definite, it is first necessary to decide on the principal
system with its determinate reactions, and then to apply the redundant forces along with
the external loads following the general scheme illustrated in Figs. 7 A to ?E, Art. 7.

The case of full contact at miter sill is illustrated in Figs. 67A and B.

Fig. 67A shows a lock gate in plan, with the water loads p, per linear foot, applied
to the upstream surface. The resultants R, of the loads p on each leaf, give rise to
arch thrusts R' and R" at the quoin ends and a horizontal thrust H acting between the
miter posts. The gate is supposed to undergo elastic deformations, as indicated by the
dotted deflection curves. The lock walls take up the arch thrusts.

Each horizontal girder receives a total load P which is proportional to the depth
of water below the surface. These loads are easily computed for all the horizontal girders
and are assumed to be known.

Fig. 67fl shows a projected elevation of the gate with the several loads P applied
to the respective horizontal girders. The bottom girder rests against the miter sill over
its entire length.

The vertical girder AB represents an element one foot thick lengthwise of the gate
leaf and may be regarded as a plate girder whose chords are portions of the sheathing
plate and all material making up the average vertical stiffness of the leaf, while its web
may be neglected.

This vertical girder is now treated as the principal system and the problem consists
of determining the stresses in, and the deflection of, this girder.

The determinate supports for this vertical girder (or principal system) are the sill
reaction B, which is immovable in the present case, and the reaction A offered by the
top horizontal arch girder. All the other horizontal girders 2 to n (shown by dotted
lines) are now considered removed and the redundant reactions X which they present
to the vertical girder are shown as external forces Xi to X n .

The principal system is thus the vertical giider on two determinate supports A
and B, with A elastic and B immovable. The external loads on the principal system
are the water loads P and the redundant reactions X.

The yielding condition of the support A may be obviated by introducing another
redundant reaction Xi and including the top horizontal arch as a part of the principal
system with immovable supports at the quoin ends D and E. The reaction A is then

274

KINETIC THEORY OF ENGINEERING STRUCTURES CHAP. XIV

-^^- :: '^'

" ]

'I J

ii L CI

1 |

FIGS. 67A, B Principal System, Full Loading, for Case of Full Contact at Miter Sill.

<-'_"

&**

FIG. 67c Principal System, Full Loading, for Case without Contact at Miter Sill.

AET. 67

MITERING LOCK GATES

275

regarded as the resultant of the immovable reactions A' and A", and the deflection di of
the point of application of X lf can then be evaluated in terms of X it making 3 l =di ^X^
The deflection curve of the vertical girder is indicated by a heavy dotted line.

FIG. 67o Principal System, Condition X=0. For Case of Full Contact at Miter Sill.

The case without contact at the miter sill is illustrated in Fig. 67c, where the principal
system consists of the vertical girder A B supported on two yielding supports. Herwe,
by taking in the top and bottom horizontal arches as part of the principal system the
fixed reactions become the four points D, E, F and G, while the elastic supports A and

FIG. 67E-Principal System, Condition X,= l. For Case of Full Contact at Miter Sill.

B undergo certain deflections ^ and 9 , which are evaluated in terms of the redundants
A'i and X 6 The reaction B is the resultant of the arch thrusts B' and B .

This case presents no new difficulty except to add one redundant condition X n

276

KINETIC THEORY OF ENGINEERING STRUCTURES CHAP. XIV

and another water load P n to those involved in the previous case. No further con-
sideration is, therefore, given to the case without sill contact.

The conventional loadings of the principal system are illustrated in Figs. 67o and E,
for the case of full contact at the miter sill.

In Fig. 6?D the condition X =0 is represented. This embraces the total water
loads on the principal system and the determinate reactions A and B resulting there-
from, when all the redundants X are removed.

Fig. 6?E shows, for example, the condition X 3 = l, where all loads are removed from
the principal system except a unit load applied in the place of ^3 and acting in a direction
opposite to the direction assumed for the redundant reaction X 3 . Similarly all other
conventional loadings from Xi =1 to X n = \ are applied one at a time. It was not deemed
necessary to show them all by figures.

The external forces. Eqs. (55A) may now be applied to the principal system as
above illustrated, choosing the case of n horizontal arches with, or without, miter sill
contact. The only distinction between the two cases is that X n =0 for full contact,
otherwise the general formula apply to both cases.

' 5

^rv

n
X

V, .

FIG. 67F.

Let Fig. 67r represent a general case of n redundants, showing the principal system
on two determinate supports A and B. Also, let m be any panel point about w r hich
the moment M m may be desired to determine the stresses in the chords of the vertical
girder.

Eqs. (55A) for n redundants and moments about m become:

A=A -A l X l -A 2 X 2 -A 3 X 3
B=B B\X\ B 2 X 2 B^Xs
m = M mo -M ml X! -M m2 X 2 -M m3 X 3

A m X m . . A n X n

B m X m . . B n X n

(67A)

The lettered constants in Eqs. (67A) are evaluated as follows:

ART. 67

MITERING LOCK GATES

277

Condition.

X m =l

A --

_!'A*

AS= /*7

Am = ~h~

1-hm+i

Am+i = ^

A =

/ ^2\

'. = hm I 1 7" )

V hS

--^( l -h.)

The tabular values substituted into Eqs. (67 A) give after considerable reduction:

. . . . (67B)

The last of Eqs. (67s) for m = l and h m =h lf gives M A =0; and for m=n and
h m =h n =0, gives M fi =0, which results must follow from the conditions of equilibrium.

The same expression for M m may also be obtained by taking the moments about
m of all the external forces to either side of a section through m.

The redundant reactions X. The general work Eqs. (55n), when the effect of tempera-
ture is neglected, will furnish n equations for finding n redundant reactions X, as follows:

etc.,

etc., etc.,

+X 2 d n? . +X 3 3 n3

2_4+etC.,+^"n^2n

etc., etc., etc., to

(67c)

wherein the subscript m has all values successively from 1 to In all ^es of double
subscripts the first one always indicates the point of application of the conventional

278 KINETIC THEORY OF ENGINEERING STRUCTURES CHAP. XIV

load while the second one refers to the location of the deflection. It should also be
remembered that all deflections d, with like double subscripts, are always equal; thus
^2-6=^6-2 and d m5 = 5m , etc.

The double subscript bearing d's are found from deflection polygons drawn for the
principal system, and for each case of conventional loading, requiring n such deflection
polygons for n redundants. Thus, the several double-subscript-bearing deflection ordinates
3 of the first of Eqs. (67c) are all obtained from a deflection polygon drawn for condition
Xi =1 acting on the principal system.

The deflections 8 l} d 2 , <5 3 to 3 n are the actual displacements of the points 1 to n, of the
principal system. Since these points are the points of support offered by the horizontal
arch girders to the vertical girder, they represent the deflections really produced in the
arch girders by the redundant reactions ,X\ to X n respectively. Hence di, 3 2 , 3 to d n
cannot be deternined until the redundants Xi to X n are known, though they may be
expressed in terms of the latter without increasing the number of unknowns.

Let di =the deflection of horizontal girder I at the point of support of A"i and due
to the conventional load X\=l.

d-2. =the deflection of horizontal girder II at the point of support of X% and due to the
conventional load ^2 = 1. Similarly for ^3 to d n . When several horizontal girders are
of like sections then their deflections d become equal, thus obviating the necessity for
drawing n such deflection polygons.

When the conventional deflections d\, d%, d% to d n are known, then the actual deflec-
tions di, 82, 3 to d n , due to the redundant reactions Xi to X n , become (by the law of pro-
portionality between cause and effect) :

fli-di^i; 8 2 =d 2 X 2 ; 8 3 =d 3 X 3 ; etc., to 8 n =d n X n . . . .' (6?D)

It should be noted that in the particular case here considered, the deflection d\ =d\-\
because the vertical girder is not deflected by the load X\=\ while the total displacement
of the principal system at point 1 is that due to the deflection of the horizontal arch, and
the vertical girder merely moves with the .arch. Similarly when there is no sill contact
then d n =8 nn . These conditions make it possible to draw the deflection polygons for
X\ = 1 and X n = 1 for the vertical girder.

The conventional loading for the principal system (being the vertical girder of 1
ft. thickness measured lengthwise of the horizontal girder) is taken as a unit force of say
one kip = 1000 Ibs. The corresponding conventional loading for the horizontal arch
girders should then be one kip per foot of wetted length of the arch. Whatever loads are
used, this ratio between the conventional loads of the horizontal and vertical systems
must be maintained.

The values from Eqs. (6?D) are substituted into Eqs. (67c), which latter are then
solved by any process of elimination to obtain the n redundants X.

When these redundants X are thus evaluated for any particular case of water loads
PI to P n , then the functions A, B, M m and di to d n may all be obtained by substitutions
in Eqs. (67B) and (67o) , and the problem is considered solved.

ART. 08

MITERING LOCK GATES

279

The moment M m , will in turn furnish the chord stress in the vertical girder for the
center of moments m. The reaction B will represent the pressure against the miter sill
which must be considered in designing this sill.

ART 68. EXAMPLE-UPPER GATE, ERIE CANAL LOCK NO.

The example originally chosen for analysis was one of the huge Panama Canal lock
gates with sixteen horizontal arches. After completing the drawings and computations
it was deemed advisable to use a smaller gate for illustrative purposes as the theory would
be less obscured by the rather extensive tabulated computations. Accordingly one of
the small 1909 steel gates of the Erie Canal was used.

Figs. 68A show one leaf of this gate in elevation and two sections. There are six
horizontal girders and the upstream side is covered by a f-inch sheathing plate, while the
downstream side is open.

The material is soft steel with a modulus of elasticity #=29,000 kips per sq.in. The
quoin and miter-post cushions are made of white oak which when saturated with water
becomes quite soft. According to the best available data, the modulus EI, for such
material, may be taken at about 30 kips per sq.in.

The cross-sections of the horizontal girders are shown in Figs. 68B, from which it is
seen that girders II to V are exactly alike, so that there are only two different types of
horizontal girders requiring analysis.

Since the sheathing plate is continuous between the girders, a certain portion of this
plate will act with the flange material of each horizontal girder. It is difficult to say
just how much of the sheathing will actually act in that manner, but in the present analysis
a width of sixteen inches is thus included with the upstream flanges of the horizontal
girders except for girder I, where the sheathing stops at the girder web, and only eight
inches are included here.

The cross-sections F and moments of inertia I, of these horizontal girder sections,
are computed from Figs. 68B and given in Table 68A.

TABLE 68A
VALUES OF F AND I FOR THE HORIZONTAL GIRDERS.

At Points

Girder I.

Girders II to V.

F
sq.in.

7
in. 1

sq.in.

in.<

2 and 10

3 to 9

26.45
26.91

3740.6
4120.1

30.96
31.30

4054.6
4466.0

Fig. 68c shows the average vertical girder and the water loads for the gate.
This girder is made up of the material which gives vertical stiffness to the gate and
includes the sheathing plate on the upstream surface and such plates and angles as

280

KINETIC THEORY OF ENGINEERING STRUCTURES CHAP. XIV

GENERAL DRAWING OF MITERING LOCK GATE

DOWNSTREAM ELEVATION.

wood

Cushio

DOWNSTREAM SIDE OPEN.

SECTION ON GIRDER IV.

FIG. 68A.

SECTIONS OF HORIZONTAL GIRDERS.
GIRDER I. , GIRDERS IITO,V.

Z9"

POINTS ZfclO.

AT POINTS 3T09.

FIGS. 68s.

AT POINTS 2 It io.

AT POINT* 3 TO 9.

ART

MITERING LOCK GATES

281

extend vertically over the downstream surface. The total web mav be considered

^ f the leaf

The total volume of this material is then divided by the length of the gate leaf
(24.3 ft.) giving the average section of the vertical girder per horizontal foot of the gate

r-roo.5

i -30.3 *
ATCRSURTAC _

MITER SILL.

The loads p are th water pressures in kips per foot- of horizontal arch:
The loads P= 27 parrthe total pressures for the horirontal girders.

FIG. 68c.

is shown in Fig. 68c. The vertical girder is thus considered as an average strip of the
;ate, one foot thick lengthwise of the leaf.

The water loads p are the actual pressures (in kips) per horizontal foot of gate and
xtend vertically from center to center of the panels between the arches. The loads
*=-27p are the total pressures for the horizontal girders of wetted length 27 ft. Hence
ili-P would be the total water pressure on one gate leaf, while ?p would be the total
3ad on the vertical girder with an exposed surface of 1 ho.

282

KINETIC THEORY OF ENGINEERING STRUCTURES CHAP. XIV

The deflections d of the horizontal arches are now found for the conventional loading
according to the previous article. This loading should be one kip per horizontal foot
of wetted length of the arch, corresponding to a conventional loading of one kip for the
vertical girder one foot wide horizontally. However, in the present problem a conven-
tional load of q = l/l -=1/27 =0.037 kip will be used for the horizontal girders, giving
deflections l/l times too small. That is, the conventional load should have been I kips
on each leaf, uniformly distributed, but instead of this one kip was distributed over each
leaf of 27 feet wetted length.

The deflection of a three-hinged arch under a uniformly distributed load is made
up of a direct cross bending of the beam or half arch, and a deflection of the arch crown
due to axial compression, as shown in Fig. 68D.

Let N av be the average axial thrust in the half arch of length I and cross section F.
Also, let d c be the deflection at the crown normally to DC due to a shortening Jl in the
length / as a result of the compression N av .

Then

i * V fl ft j

= EF and

tan a EF tan a '

(68A)

FIG. 680.

The effect of d c on the arch is merely to lower the crown to C' , leaving the beams
DC and CE otherwise unchanged and producing the dotted deflection line DC'E. With
this must be combined the deflection due to bending and shear in the beam, finally
producing the deflection curve DC'E, with ordinates d measured normally to the girder
DC. The bending deflection will be determined from the bending moments, by the
method given in Art. 39 for solid web beams.

Before proceeding to this it may be well to say a few words regarding the particular
deflection d which is to be used in the solution of the problem, since this point may
involve a considerable element of doubt.

In the first place the deflections must be determined in the direction normally to
the girder DC, because that is the direction of action of the water loads P and of the
redundant reactions X. Ordinarily the deflections for any three-hinged arch would be
taken normally to the span DE.

The principal system was chosen as an element of average vertical cross-section
of the gate and the deflection properly belonging to such a hypothetical vertical girder

ART. 08

MITERING LOCK GATES

283

would necessarily be an average deflection. Hence, the value of d which is actually
used is obtained by dividing the area DCC', of the deflection curve DC*, by the length
/ of the leaf. This average deflection d av , is shown in Fig. 68D under the assumed vertical
girder. See also Figs. 68r.

The bending moments for the various points of a horizontal girder, due to the load
9 -=0.037 kip per foot of wetted length, are now determined by drawing a resultant
polygon through the girder and then taking the moments of the resultant normal thrust
about the center of gravity of the section, as shown in Fig. 68E.

The loads </i to q<n are first combined into a force polygon furnishing the resultant
R and the reactions R' and H, which latter can be determined when the points D and C
through which the resultant polygon shall pass, are fixed. The point C is taken in the
center of the joint and the point D is taken so that the resultant R' is normal to the
hollow quoin. The resultant polygon is 'then easily drawn and the offsets v, between
it and the gravity axis of the girder, are thus found. The axial thrusts N are then scaled
from the force polygon and from these and the eccentricities v the moments M =Nv are
computed in Table 68s .

The elastic loads w=MAx/l are also determined and from these the required
deflection polygon is drawn.

TABLE 68s
ELASTIC LOADS w FOR HORIZONTAL GIRDERS.

Pt.

TV
Kips.

inches.

M = Nv
Kip in.

Horizontal Girder I.

Horizontal Girders II to V.

/
in. 4

Ax
Inches

M4x

""7"

/
in.<

M
I

4x
Inches.

Mix
__

D, 1

1.38

0.0

0.00

0.0

40.00

0.0126

40.00

0.0116

1.396

1.7

2.373

3740.6

0.00063

24.00

0.0377

4054.6

0.00058

24.00

0.0348

1.40

7.4

10.360

4120.1

0.00251

32.64

0.1142

4466.0

0.00232

32.64

0.1054

1.40

13.2

18.480

4120.1

0.00449

32.64

0.1643

4466.0

0.00414

32.64

0.1514

1.40

16.4

22.960

4120.1

0.00558

32.64

0.1877

4466.0

0.00514

32.64

0.1728

1.40

17.4

24.360

4120.1

0.00591

32.64

0.1845

4466.0

0.00545

32.64

0.1697

1.40

15.8

22 . 120

4120.1

0.00537

32.64

0.1542

4466.0

0.00495

32.64

0.1421

1.40

12.0

16.800

4120.1

0.00408

32.64

1000

4466.0

0.00376

32.64

0.0920

1.40

6.0

8.394

4120.1

0.00204

24.00

0.0266

4466.0

0.00188

24.00

0.0246

1.39o

0.5

0.698

3740.6

0.00018

33.00

0.0030

4054.6

0.00017

33.00

0.0028

C, 11

1.37o

0.0

316.84

0.9848

316.84

0.9072

284

KINETIC THEORY OF ENGINEERING STRUCTURES CHAP. XIV

0.5

I KIR

FIG. 68E.

ART. 68 MITERING LOCK GATES 285

There are two different kinds of horizontal girders, thus making two such deflection
polygons decessary, one for Girder I and one for Girders II to V. The sections are given
in Figs. 68s and Table 68A. The increments Ax are measured along the span Z)C such
that 2Ja;=L All dimensions are now taken in inches and loads in kips.

According to Art. 39, an equilibrium polygon drawn for the elastic loads w with a
pole distance equal to "=29,000 kips, will represent the deflection polygon for the
moments M, with ordinates measured to the scale of lengths of the drawing.

In Figs. 68r, the pole distances were made equal to H= E/20 ,000 = 1.4 5, giving
deflections 20,000 times actual to the scale of lengths.

The moment deflection polygon is constructed by plotting the values M/I, from
Table 68s, as ordinates at the points 1 to 11, using any convenient scale, then finding
the centers of gravity g of the several trapezoidal areas of this M/I polygon and applying
the loads Wi to WIQ at these centers. The equilibrium polygon drawn for the w forces
is then the moment deflection polygon, with deflections measured parallel to the direction
taken for the w loads, in this case perpendicular to the chord of the girder.

The scale for the loads and the force polygon is immaterial so long as it is convenient
and the pole distance H must be laid off to the scale of loads. The pole may be
anywhere to suit a convenient closing line D'C'.

The actual moment deflection of girder I at point 6 is thus

-0.00188 ind,

The actual deflection due to shear at the same point 6, which is the point of maximum
moment, is by Eq. (39D) ,

5X24 ' 36

=
6

_
2EFi 2X29000X30X0.370

=0.00019 inch,

where Fj is the area of the web plate and M 6 is taken from Table 68s.

The deflection for point 6, due to shear alone, is thus about 10 per cent of the
deflection for the same point and due to moments. Hence, all the deflection ordinates
are increased 10 per cent to obtain the combined moment and shear deflection polygon.

The crown deflection d c , Eq. (68A), due to rib shortening by axial thrust, is now

found. . .

Since a portion of the gate, at the quoin and miter posts, is made of oak, which i:
probably saturated when the gate is in operation, it is necessary to make allowance for
this condition in computing Al.

For an average thrust JV- 1.4 kips, and a net Z-= 292 inches, F -26.91 sq.m., and
=29 000 kips per sq.in. for the steel part of the gate and Z'=24.8 inches, F'=24
=864 sq.m-, and E l =30 kips per sq.in. for saturated oak, then for tan a =0.367.
Eq. (68 A) gives for the steel and oak portions

1.4 X292 1.4X24.8 =0.00499 inch,

de = 29000 X26.91 X0.36V ^30 X 864 X0.367

286 KINETIC THEORY OF ENGINEERING STRUCTURES CHAP. XIV

DEFLECTION POLYGONS OF HORIZONTAL GIRDERS
For Conventional Loading of One Kip Uniformly Distributed Over Wetted Length

_* 5. <JL T .AN T. . POAV G ON .

--- r .W>TTD* LO.CTH '- 17 FT. "T ' ~ ' ---

j
DEFLECTION ^OLYGON FOR GIRDER I.

M/I POLYGON FOR GIRDERS II TO V.

DEFLECTION "POLYftOIH FOR GIRDERS II TOY

h^a**C'

All deflections are 20,000 times actual
measured to the scale of lengths.

FIGS. 68F.

ART. 68

MITERING LOCK GATES

and 20,000 times actual in feet becomes

^=0.00499(^5^) =8.32 feet.
\ iz /

This crown deflection d c =8.32 ft. is now incorporated in the deflection polygon
for Girder I,J^ig. 68r, by laying off the ordniate C 7 C 77 =d c and then drawing the "final
closing line D'C" to complete the deflection polygon. The ordinates written in this
diagram represent deflections 20,000 times actual in feet.

In similar manner the deflection polygon for girders II to V is drawn.

9/ &) v 1 2

The actual moment deflection at point 6 is now "' =0.00171 inch, and the

20,000

actual deflection due to shear at the same point is

&MR 5x2436

d * =2EF\ = 2 X29000 X 30 X0.375 =0 - 00019 inch >

being about 11 per cent of that due to moments. Hence, the deflection ordinates are
increased 11 per cent to obtain the combined moment and shear deflection polygon.

The crown deflection d c is found as before, and all numerical values are the same
except the cross-section F of the girder, which is now 31.3 sq.in. Hence

1.4X292 1.4X24.8 . '

c 29000 X 31.3 X 0.367 + 30X864X0.367

or 20,000 times actual in feet gives d c =S.l5 feet.

This value of d c is now used to complete the deflection polygon for girders II to V.

The average deflection d av for each girder is now computed from the ordinates taken
from the deflection polygons Figs. 68F, and the distances Az between these ordinates. The
values Az were taken in inches and the areas finally divided by the length Z in inches,
which eliminates the unit of length. The computations are given in the following table :

TABLE 68c
DEFLECTIONS d av FOR THE HORIZONTAL GIRDERS.

Horizontal Girder I.

Horizontal Girders II to V.

Areas

Inches.

Feet.

dJz.

Feet.

dJj.

26.5

+ 1 . 58

20.94

+ 1 . 50

19.88

27.5

1.58+3.12

64.63

1.50+2.96

61.33

27.5

3.12+4.64

106.70

2.96+4.38

100.93

31.2

4.64 + 6.10

167.54

4.38+5.80

158.81

32.5

6.10+7.30

217.75

5.80+6.92

206.70

32.4

7.30+8.14

250.13

6.92+7.76

237.82

32.3

8.14 + 8.52

269.06

7.76+8.20

257.75

31.8

8.52 + 8.68

273.48

3.20+8.35

263.15

27.0

8.68+8.62

233.55

8.35+8.30

224.78

26.3

8.62 + 8.48

224.86

8.30+8.22

217.24

21.8

8.48+8.32

183.12

8.22 + 8.15

178.43

316.8

Tntn.1

2011.76

Total

1926.82

The deflection ordinates d are 20.000 times actual in feet.

288 KINETIC THEORY OF ENGINEERING STRUCTURES CHAP. XIV

This gives for horizontal girder I,

2011.76 .. 6.35X12

d av = 3168 =6.35 feet, or actually 2000Q -- 0.00381 inch,

and for girders II to V,

89 fi

- =6-082 feet, or actually - =0.00365 inch,

which values will be used in Eqs. (67o) to obtain the -deflections d^ to 5 , observing
that d'a v for arches II, III, IV and V is the same quantity.

Attention is called to a slight oversight on Figs. 68F, where the deflection ordinates
should have been measured at the numbered points 2 to 10, instead of on the lines of
the w loads. The effect on the results is scarcely appreciable and therefore, the error was
not corrected.

The conventional deflections d of the principal system are determined graphically in Figs.
68c and 68n, for the case of full contact at the miter sill. This requires drawing a deflection
polygon for the principal system for each of the five cases of conventional loading.

The conventional loading for the horizontal arches was made equal to 1 kip, uni-
formly distributed over the wetted length of a gate leaf, being 1/1 = 1/27 =0.037 kip
per foot of arch. Hence the same load \/l =0.037 kip must be employed as the con-
ventional load for the principal system.

The deflection of the principal system is made up of the deflection of the vertical
girder and the elastic displacements of the two determinate supports A and B. For
full contact at the miter sill the B support is assumed as being immovable and the A
support is subject to the elastic displacement of the top horizontal arch caused by a con-
ventional load X = 1/1 acting on the principal system. The elastic displacement of the
point A is determined from the deflection d av just found for the top horizontal arch.

For the condition X\ l/l, there are no moments produced in any part of the vertical
girder and hence the deflection polygon for this case of loading would show no bending
effect of the vertical girder, but merely an elastic displacement of the A support which
was previously found to be d\-\=d av for the top horizontal girder. Hence, this con-
ventional deflection polygon is easily drawn and becomes a triangle aa'b, Fig. 680, wherein
the ordinate aa'=d av -= di_i for the top horizontal girder. This deflection line also
furnishes the values d l _ 2 , ^1-3, ^1-4 and <?j_s, which are respectively equal to d 2 -i, <V-i,
^4_i and d$_i. The latter are the end ordinates at A of the several deflection polygons
to be drawn for the conditions X 2 = l/l to Xr, = \/l.

Hence, the deflections of the principal system are easily found when the deflection
lines of the vertical girder are once drawn for each of the several conventional loadings.

The deflection lines for the vertical girder are drawn, again using the method for
plate girders, Art. 39. However, no allowance is made for shear because the web is too
insignificant.

A set of w loads is now computed for each of the loadings X z --=l/l to X 5 = l/l, and
observing that the moment of inertia 7=971.6 in. 4 , for the vertical girder, is constant,
it is preferable to make these loads equal to MAx for a pole distance H =EI, which would
give actual deflections to the scale of lengths.

AUT. 68

MITERING LOCK GATES

289

For simplicity in determining the moments M and the w loads, the conventional
loads X = l are used, giving moments 27 times too large. The deflections for the loads
A 7 " = 1/7 are then obtained from

MAx

=^Tfr, using a pole distance

Zt Ji/l

H =27EI =27 X29000 X971.6 =760,762,800,

where the dimensions are kips and inches, giving actual deflections in inches to the scale
of lengths. Instead, however, the pole was made #=27#//30,000 =25,360 for deflec-
tions 30,000 times actual to the scale of lengths, and w loads 27 times actual.

w loads for X 2 = 1 Kip.

w loads for X 3 1 Kip.

M for -X" 2 = l Kip.

w-MAx

M for X 3 = l Kip.

w = MAx

inches.

Kips and inches.

A 2 = 0.2341 B. = 0.7659

4 3 = 0.4305 B 3 = 0.5695

M t = 0.7659 X 0.00= 0.0

M! = O.O

48.00

w,= 882.0

u>!= 655.2

M 2 = 0.2341X157.0 =36.75

M 2 = 0.5695 X 48.00= 27.3

40.25

w 2 = 1289.6

w 2 = 1561.7

M 3 = 0.2341 X 1 16.75 = 27.33

M 3 = 0.5695 X 88.25= 50.3

40.25

w> 3 = 910.5

w 3 = 1674.4

MI = 0.2341 X 76.5 =17.91

vl/ 4 = 0.4305= 76.50= 32.9

40.25

w 4 = 539.4

w t = 976.1

M 5 = 0.2341 X 36.25= 8.49

M, = 0.4305 X 36.25= 15.6

36.25

u> 5 = 153.9

w> s = 282.8

M 6 = 0.2341X 0.0 = 0.0

SM> =3775.4

M 8 = 0.0

~Lw = 5150.2

w loads for X 4 = 1 Kip.

w loads for X b = 1 Kip.

Ax
inches.

M for X 4 = l Kip.
Kips and inches.

w = MAx

MforX 5 =l Kip.
Kips and inches.

-**

A 4 = 0.627 4 = 0.373

/1 5 = 0.823 B & = 0.177

48 00

M, = 0.0

Wl = 429.6

u>,= 204.0

M 2 = 0.373 X 48.0 =17.9

w 2 = 1022.4

M 2 = 0.177 X 48.00= 8.5

w; 2 = 485.0

M 3 = 0.373 X 88.25X32.9

w 3 = 1626.1

M 3 = 0.177 X 88.25 = 15.6

M> 3 = 770.8

40.25
36.25

M 4 = 0.373 X 128.5 =47.9
M 5 = 0.627 X 36.25 = 22.7

w> 4 = 1420.8
M> 5 = 411.4

M 4 = 0.177 X 128.5 =22.7
M 5 = 0.177 X 168.75 = 29.9

w t = 1058.6

tw s = 541.9
Su> = 3060.3

290 KINETIC THEORY OF ENGINEERING STRUCTURES CHAP. XIV

The several conventional deflection polygons, for loadings X^l/l =0.037 kip, are now
drawn as illustrated in Figs. 68c and 68n. The factor 30,000 was introduced to obtain dia-
grams of convenient size to the scale of the drawing, and it is worthy of comment that the
highest desirable accuracy may be obtained from small drawings by an appropriate choice
of the scale and this factor. The scale employed for the force polygon is entirely immate-
rial so long as it is convenient, noting that the w loads and the pole must be measured
to the same scale. The pole distance remains the same for all the diagrams and is 25,360
w units.

The moment diagrams were drawn for the conventional loadings X = 1 to correspond
with the moments in Table 68D, and serve merely to locate the centers of gravity gi, <7 2 ,
etc., which are the points of application for the w loads.

The value d av =0.00381 inch, as previously found for the top horizontal arch, now
furnishes <5i_ 1 =30,000 d av = 114.3 inches, from which the deflection line for X^=l/l is
drawn. This diagram then furnishes the deflections #1 -2=^2-1, di_ 3 -=d s _i, ^ 1 _ 4 =o 4 _ 1
and ^1-5=^5-1, which are necessary in fixing the closing lines of the several other deflec-
tion polygons.

With this explanation the drawings should be readily understandable and will
furnish all the deflections d required in solving Eqs. (67c) .

The problem is solved completely for the case of full contact at the miter sill, and
the method of deriving the deflections for the case of no sill contact is shown in Figs. 68H.

When there is no sill contact then the bottom support of the vertical girder becomes
the bottom horizontal arch in the same manner as the principal system was previously
f ormed for the top support at A , requiring now the determination of d' av for the bottom
arch (for a loading q =0.037 k. per foot of arch) as was done for the other arches. This
d'^ then gives <5e -6=30,000 d' av for the principal system and furnishes the means of draw-
ing the deflection polygon for condition X 6 = l/l from which ^0-5=^5-6, ^6-4=^4-6,
e~tc., are obtained. These deflections then serve to complete the conventional deflection
polygons for the case of no contact at the miter sill in the manner shown for the two
conditions X 4 = l/l and X 5 = l/l in Figs. 68n. Hence, all the work previously accom-
plished in solving the case for full contact is also usable for the case of no contact, though
the final deflections d are different in the two problems.

The formation of the final equations for the solution of the redundants X is now pos-
sible by introducing the numerical values for all the deflections d into Eqs. (67c) , remem-
bering that these were all taken 30,000 times actual, which in no wise interferes with their
use in the equations.

The numerical terms 27> TO l7n , Zp m d 2m , etc., must first be computed for the actual
water loads p per foot of the horizontal arches. These loads are given on the diagram
Fig. 68c and represent the actual water pressures on the girders for vertical depths extending
from center to center of the respective spaces between arches. Thus the load pi is the
water pressure per horizontal foot of arch and over a depth extending from the water
surface down to a line 24 inches below the top arch. The load p% is the pressure per
horizontal foot of the arch included between the depths 38 inches and 82.12 inches below
the surface, and so on for each girder.

The double-subscript-bearing deflections are taken from the deflection polygons in

ART. 08 MITERING LOCK GATES

DEFLECTION POLYGONS FOR THE PRINCIPAL SYSTEM

FOR FULL CONTACT AT MITER .
THE CONVENTIONAL LOAD *1/1= 1/27O.O37 KIP CON.

291

* 2 MOMENT DjIAGRAM FORjX^=IKIR

All deflections are 30.000 times actual measured to the scale of lengths.
FlGS. 68G.

292

KINETIC THEORY OF ENGINEERING STRUCTURES CHAP. XIV

DEFLECTION POLYGONS FOR THE PRINCIPAL SYSTEM

FOR FULL CONTACT AT MITER.
THE CONVENTIONAL LOAD = 1/1= 1/27= O.037 KIP CONCENTRATED.

. . ._!. .._,

upstream chord.

Gravity axis.

tn~

4>"

4 r

L AX=4ft^

I W-'iS J

4o'.'z5 >

' 4o'/25 >

< 36.25

MOMENT DIAGRAM FOR|X = I

I 21

__jPi
oi'o,

nit

MOMENT DIAGRAM ^0^X5= I KIP

I DEFLECTION POLYGON FOR X ft 1/l- 0.037 K- NO MITER SILL CONTACT.

SCALES

BO
LOADS.

1000 O 5000 10,000 \

All deflections ire 30,000 times actual measured to the scale of lengths.

FlGS. 68H.

ART. 68

MITERING LOCK GATES

293

Figs. 68a to H, using the values 30,000 times actual, in inches. The deflections di to S 5
are obtained according to Eqs. (67o), using 30,000 times the deflection d av , found for the
horizontal arches.

All these numerical values are now entered in Table 68E, and the sum products
SpwAm are computed, giving all required data for writing out the general equations
for the redundants.

TABLE 68E
COMPUTATION OF THE TERMS K = Zp m dam FOR EQS. (67c)

3 = davX

Pt.

s im

Pm "im

Pm 3 tm

S 3m

Pm 3 vn

3 4 m

Pm 3 <m

s sm

Pm*sm

Eq.(67 D )

Kips.

0.314

114.3

35.890

87.6

27.506

65.2

20.473

42.8

13.439

20.2

6.343

114.3Xi

1.150

87.6

99.590

69.6

80.040

53.8

61.870

37.0

42.550

18.4

21.160

109. 5Xi

1.783

65.2

116.252

53.8

95.925

43.6

77 . 739

30.7

54.738

15.4

27.458

109. 5Xj

2.475

42.8

105.930

37.0

91.575

30.7

75.983

22.0

54.450

11.0

27.225

109.5X4

3.015

20.2

60.903

18.4

55.476

15.4

46.431

11.0

33 . 165

5.6

16.884

109. 5X.

2Pm im =

418.565

*PA =

350.522

*WW-

282.496

*Pm*m =

198.342

***,-

99.070

The summations cover all the redundants from 1 to 5 and the subscript m has all successive values from 1 to 5. All
the values d are 30,000 times actual in inches.

FINAL EQUATIONS FOR THE REDUNDANTS

Eq. (1) Xi^.j
Eq. (2) X^,,

Eq. (3) X^s-i
etc.

<? 2 _ 2 + X 3 <5 2 -

(68B)

etc.

After substituting the numerical values from Table 68E, the following equations
are obtained :

(1) 114.3Xi +87.6X2+&5.2X 3 +42.8X4+20.2X 5 + H4.aX 1 =418.56

(2) 87.6X! +69.6X 2 +53.8X 3 +37.0X 4 + 18.4X + 109.5X 2 =350.52

(3) 65.2X! +53.8X2 +43.6X3 +30.7X 4 + 15.4X 5 + 109.5X 3 =282.50 [ . (68c)

(4) 42.8Xi +37.0X 2 +30.7X 3 +22.0X4 + 1 l.OXs + 109.5X4 = 198.34

(5) 20.2X! +18.4X2 + 15.4X3 + H.OX4+ 5.6X5 + 109.5X5= 99.07]

The terms are now collected in the above equations and the solution is carried out
in the following tabular forms:

29t

KINETIC THEORY OF ENGINEERING STRUCTURES, CHAP. XIV
TABLE 68F SOLUTION OF THE EQUATIONS

Operations.

Coefficients of the X'a.

Numerical
Term
K.

A'.i

Eq. (1)..

228.6

87.6

65.2

42.8

20.2

418.56

Eq. (2)

87.6

179.1
33.55

53.8
24.97

37.0
16.39

18.4
7.74

350.52
160.31

Jg-frEq- (l)] = 0.383[Eq. (1)]

Eq. (2),=Eq. (2)-0.383[Eq. (1)]

145.55

28.83

20.61

10.66

190.21

Eq. (3).. 65.2

53.8

153.1

30.7

15.4

282.50

65 2
[Eq. (l)] = 0.285[Eq. (1)]

24.97

18.58

12.20

5.76

119.29

Eq. (3)! = Eq. (3)-0.285[Eq. (1)]

28.83

134.52

18.50 '

9.64

163.21

9C OQ

^ [Eq. (2) 1 ] = 0.198[Eq. (2),]

5.71

4.08

2.11

37.66

Eq. (3) 2 = Eq. (3), -0.198[Eq. (2),]

128.81

14.42

7.53

125.55

Eq. (4).. 42.8

37.0

30.7

131.5

11.0

198.34

42 8
[Eq. (l)] = 0.187[Eq. (1)]

16.39

12.20

8.00

3.78

78.27

Eq. (4), = Eq. (4)-0.187[Eq(l)J

20.61

18.50

123.50

7.22

120.07

f)f\ f*~t

^- 5 [Eq. (2)J = 0.142[Eq. (2),]

4.08

2.92

1.51

27.01

Eq. (4) 2 = Eq. (4) 1 -0.142[Eq. (2)J
14 42

14.42

120.58

5.71

93.06

12881 " q ' ( 3 )2-l- - 112 L Ec l- ( 3 )2-l

.62

.o4

14 .Do

Eq. (4), = Eq. (4) 2 -0.112[Eq. (3)J

118.96

4.87

79.00

Eq. (5) 20.2

18.4

15.4

11.0

115.1

99.07

20.2
228^ [Eq. (l)] = 0.0884[Eq. (1)]

7.74

5.76

3.78

- 1.78

37.00

Eq. (5) 1 = Eq. (5)-0.088[Eq. (1)]

10.66

9.64

7.22

113.32

62.07

1 f\ f\f\

145.55 ^ q ' (^iJ- - 0732 !^ !- ( 2 )J

. 11

.51

.7o

id. yz

Eq. (5) 2 = Eq. (5),-0.073[Eq. (2)J

7.53

5.71

112.54

48.15

12881^ q ' ^J- - 0585 ^- ( 3 ):J

.o4'

.o4

Eq. (5) 3 = Eq. (5) 2 -0.058[Eq. (3)J

4.87

112.10

40.81

118%^ q ' ^ 3 " q ' ^ ^

Eq. (5) 4 = Eq. (5) 3 -0.04irEq. (4)3]

111.90

37.58

ART. 68

MITERING LOCK GATES

295

The solution of the equations is completely given in Table 68r, which also illustrates
the method. The successive operations are expressed in the first column in unmis-
takable language, and the order of performing the computations is to work down the
table first entering the given equations, then obtaining the products / [Eq. (1)] down
through each sub-table, and finally making the subtractions. At this point apply the
following checks : Note that the symmetric values must be the same and that the terms
included between the double-ruled lines and those in the columns headed K are not
checked. Therefore, check the latter terms by repeating the numerical work and check
the symmetric terms by inspection. Then proceed with the work of obtaining the
products/" [Eq. (2)i] and repeat the same programme above outlined.

The final Eq. (5) 4 gives the value of ^5 = 0.3358 kip per horizontal foot of horizontal
girder V. The other redundants are then determined from the equations in the last
sub-table of Table 68F by successive substitutions as indicated in the following Table 680.

Terms.

.sr,

K-

99.07

62.07

48.15

40.81

37.58

x t c>

-38.65

-38.02

-37.79

-37.64

37.58

x t c t

- 7.16

- 4.70

- 3.72

-37.64

x 3 c s

-13.58

- 8.50

-41.51

3.17

= 0.3358

x 3 ' 17

.AjCj

18.73

51.22

.64

4 ~4 87

6 64

-78.12

10.85

= 0.6509

7.53

10 85

20.95

= 0.8818

2 10.66

20.95

= 1.0178

1 20.2

= 1.0371ki

ps per horizontal

"oot of gate.

C = the coefficients from the last of Tables 68F.

The redundants X are thus expressed in kips per horizontal foot of gate, representing
the average intensity of these forces for the average vertical stiffness of a gate leaf.

The reactions A and B and the moments M for this average vertical girder, may now
be found from Eqs. (67e) employing the values in the following table.

296

KINETIC THEORY OF ENGINEERING STRUCTURES CHAP. XIV

TABLE 68n
REACTIONS AND MOMENTS, VERTICAL GIRDER

Point.

p
Kips per ft.

X
Kips per ft.

p-X
Kips per ft.

h
Inches.

hi -h
Inches.

(p-X)h
Kip-inches.

(p-X)(hi-h)
Kip-inches.

0.314

1.0371

-0.7231

205.0

0.0

-148.236

0.0

1.150

1.0178

+ 0.1322

157.0

48.0

+ 20.724

6.336

1.783

0.8818

0.9012

116.75

88.25

105.192

79.513

2.475

0.6509

1.8241

76.5

128.5

139.536

234 . 380

3.015

0.3358

2.6792

36.25

168.75

97.120

452.115

Totals

8.737

3 . 9234

+ 214.336

772.344

Eqs. (67fi) then give for A, B and M 3

779
-h) =i_'

=3.767 kips.

341.848 = 196.057 kip-inches,

t , My 196.057X23.4 __,_ .
giving for f=j= -- =- ~ - = 4.727 kips per sq.m. on the most extreme fiber of

J- \j i J. .O

the vertical girder at the panel 3.

In like manner the stress may be found for any other point of the vertical girder,
using the results given in Table 68n.

B+PQ represents the pressure which the miter sill must sustain on the assumption
of full contact.

Finally the static condition

././ -. . . . . . . (680)

must be fulfilled by the above numerical results.

The deflections of the vertical girder are now readily found from Eqs. (67c) and the
redundants X which are now known.

The deflections d av of the horizontal girders were found for a load of 1 kip uniformly
distributed over the wetted length of 27 ft. Hence, for a load of 27.X" uniformly dis-
tributed over the same length, the deflection would be 27d av X. Accordingly the
deflections d to d$ become

<?! =27d av Xi =27 X0.003S1 X 1.037 =0.107 in.

d 2 =27d'a V X 2 =27 X0.00365 X 1.018 =0.100 in.

3 3 =27d v X 3 =27 X0.00365 X0.882 =0.087 in.

<? 4 =27d'a V X 4 =27 X 0.00365 X 0.651 =0.064 in.

d 5 =27d' m ,X 5 =27 X0.00365 X0.336 =0.033 in.

ART. 68 MITERING LOCK GATES 297

An approximate idea of the deflection curve at any vertical ra other than the
average, may be obtained by assuming that the same proportionately exists between
the deflections d av and d m as between di and d m , etc., where the point ra designates any
vertical section of the gate. Thus in Fig. 68r, for a vertical girder at point 8, the
deflection d 8 =8.68 ft. for the top horizontal arch while d av for that arch was 6.35 ft.,
giving a ratio d 8 /d av = 8.68/6.35 = 1.367. For the arches II to V the ratio between
similar ordinates becomes dg/d ap = 8.35/6. 082 = 1.373.

Hence, in considering a vertical girder at point 8, where the deflections in the
horizontal arches are about 37 per cent greater than the average, the effect on all the
conventional deflections would be to increase them approximately by the same constant
percentage. This would be equivalent to multiplying Eqs. (68c) by a constant 1.37
and since the water loads p remain unchanged the resulting redundants X would not
be materially different than for the average vertical girder.

However, the deflections di to 5 would be increased throughout by the same 37 percent.

Since the ratio d m /d av is not necessarily constant, though nearly so in the present
example, this proposition cannot be accepted for more than it is worth, and approximate
results only can be expected.

Where a rigid determination is desirable, it would be necessary to repeat the whole
computation for the actual deflections of another principal system with vertical girder
at point 8. This would not involve very much labor, however, beyond solving Eqs.
(68c) for a new set of numerical coefficients.

A summary of the steps employed in the solution of the above problem -is given in
closing this subject.

Horizontal arches. Draw a moment deflection polygon for each of the different
arches, using a conventional load of one kip uniformly distributed over the wetted length
of each leaf. These deflections are then combined with the arch deformation for axial
thrust or rib shortening for the same loading, to obtain the total displacements of the
several points of the arches with respect to the fixed quoin post supports.

These total displacements are then averaged for one gate leaf by dividing the area
of the displacement polygon by the straight length of the girder, giving the deflection
d av for each arch. Then d=d av X, and for the top arch d av =d l . li while for the bottom
arch d' av =d nn when there is no sill contact and d' av =0 when full sill contact is assumed.

The principal system. Draw conventional deflection polygons for the principal
system, using conventional loads of 1 /I kips for each condition X = l/l where I is the
wetted length of one gate leaf as above. This gives the conventional deflections of all
panel points of the vertical girder for each condition of loading, remembering that the
vertical girder represents a strip of gate one foot thick and of average cross-section.

The water loads p per horizontal foot of gate for each arch, are then used as the
actual loads in computing the numerical terms Sp m 8 am for each condition X-l/l. _
resulting values X are then the redundant reactions of the average vertical gir
against the horizontal arches per linear foot of the latter.

When the vertical girder is a truss frame instead of a plate girder, then the method
of deflections given in Art. 36 must be used, neglecting the effect of the web member*
as the nature of the problem does not warrant the extreme accuracy impl*
web system is considered.

CHAPTER XV
FIXED MASONRY ARCHES

ART. 69. GENERAL CONSIDERATIONS

The fixed arch is statically indeterminate in the third degree, or, as commonly
expressed, it involves three redundant conditions, provided the arch ring may be treated
as an elastic body.

According to the theory of elastic deformation, these redundant conditions are
determined from the elastic properties of the material, while for the older " line of
thrust " method the redundants are merely approximated by assigning certain assumed
conditions to be fulfilled by the line of thrust.

A line of thrust is defined as an equilibrium polygon for any case of simultaneous
loads acting on the arch ring. Hence, for any section, the adjacent elements of the
line of thrust will represent the resultant of all the external forces on either side of the
section, including the reactions.. From this it follows that the sum of the moments of
all external forces about any point on the line of thrust, or resultant polygon, must always
be zero.

Therefore, if the external redundant reactions were known, then the real resultant
polygon could be drawn, but since the redundants can be found only from the theory
of elasticity, the line of thrust method fails to give a direct solution.

A resultant polygon may always be drawn through three given points, as will
be shown later. Hence, by successive trials, a resultant polygon may be approximated
so as to fulfill certain requirements which are now discussed. The line of thrust is
henceforth called the resultant polygon, this being a more appropriate designation.

The theory of elasticity fixes the position of the resultant polygon in terms of the modu-
lus E of the material and the elastic deformation of the arch ring. According to the
older method an infinite number of resultant polygons may be constructed for the same
arch and same loading, and it becomes a question to decide which of all possible lines
is the most probable.

Hagen (1844 and 1862), according to his " theory of the most favorable distribution
of stress," defines the most probable resultant polygon as the one for which the vertical
projections of the minimum distances, between the sides of this polygon and the bound-
ing lines of the arch ring, become equal.

Culmann (1866) advanced the theory that of all the possible resultant polygons,
the most probable one must approach the arch center line in such manner as to reduce
the stresses in the critical sections to a minimum. Carvallo (1853) and Durand-Claye
(1867) adopted this theory with slight modifications.

ART. 69 FIXED MASONRY ARCHES 299

The theory of elastic deformation was introduced by Navier in 1826, by his analysis
of the stresses on an arch section, in which he assumes a combined thrust and bending
moment distributed over the section. According to Navier, the stress on the extreme
fiber of any arch section becomes

f _N My_N M

J ~^ .........

where N is the normal thrust on the section, M the bending moment, F the area, y the
distance from the neutral axis to the extreme fiber and / and W the moments of inertia
and of resistance, respectively, of the cross-section. The formula when applied to curved
beams, or arches, is not strictly accurate, but when the radius of curvature is large in com-
parison with the depth of the arch ring, a condition which always obtains in ordinary
arches, then the error due to omitting curvature is extremely small and is never considered.

The experiments of Bauschinger (1834-93), Koepcke (1877), and the Austrian Society
of Engineers and Architects (1895), have substantiated the correctness of the theory of
elasticity within knowable limits of the elastic properties of building materials.

Nearly all modern writers on this subject adopt Navier's law as the basis of investiga-
tion. The various arch theories by Winkler (1867), Mohr (1870), Belpaire (1877), J.
Weyrauch (1878), F. Engesser (1880), Mueller-Breslau (1880), and those of American
writers, differ principally in manner of presenting the subject. Each author has con-
tributed something in the direction of rendering the theory of elasticity more usable, and
this is especially desirable even at the present time.

Professor Winkler was the first to prove that the most favorable resultant polygon in an
unsymmetric arch must intersect the arch center line in at least three points, while -for
symmetric arches four such intersections are necessary. He also, proved that according
to the theory of elasticity the most probable resultant polygon is the one for which the
I'esidual departures from the arch center line become a minimum, according to the method
of least squares. This then furnishes the real criterion with which to judge an arch design
in connection with certain other limitations to be discussed later.

The common graphic solution is nothing more than the application of the principles
of the three-hinged arch to the fixed arch by assuming the location of the hinged points,
at the crown and springing. The location of these hinged points may be altered until
such a resultant polygon is found as will approach the arch center line in the manner
required by the theory of elasticity.

The most probable resultant polygon, when found, must remain within the middle
third of the ring, when tensile stress is not permissible. This follows from Eq. (69A).
Hence it is clear that the most economic masonry arch is one for which the center line
coincides most closely with the resultant polygon drawn for the case of average or half
total live load, provided the arch ring has sufficient thickness at all points to prevent
excessive unit stresses due to unsymmetric loading. This idea will be followec
treating of methods for preliminary designs.

Formerlv an arch problem was considered solved, when, for a given span, rise and
loading a resultant polygon could be found which, for the critical position of the live
load, remained entirely within the middle third of the ring. However, this is far from

300 KINETIC THEORY OF ENGINEERING STRUCTURES CHAP. XV

constituting an acceptable solution, as will be shown in the following. The fact that most
arches have stood so well is no defense of this method of designing, especially when it
is understood that the usual factor of safety employed in such designs ranges between
ten and sixty, rarely going below twenty.

Furthermore, solid spandrel arches are usually designed with very high safety
without considering the additional carrying capacity of the masonry in the spandrel,
which is often sufficient to sustain the loads without regard to the ring.

The author adds the criticism that few fixed arches exist which are not disfigured by
unsightly cracks and this, in conjunction with the very wasteful expenditure of mate-
rial necessitated by low unit stresses, constitutes a fertile field for doubt regarding the
expediency and justification of building fixed arches in this progressive age.

While it is freely admitted that these masonry structures, many of which are monu-
mental in character, have stood well, and will undoubtedly continue to stand, every
engineer must agree that a cracked arch is neither an achievement nor 'the ultimate aim
of his ambition, to say nothing of the dissatisfaction which such a blemish constitutes
to both engineer and owner.

Practically, the objections to fixed masonry arches may seem trivial and mostly
a matter of sentiment, while theoretically they are important. The remedy for all these
objections is to introduce hinged joints during construction or preferably for all time.
This has been practiced in Germany since 1880, and has given excellent satisfaction,
though many fixed arches are still being built. This will continue to be the case so long
as we do not come to recognize that " new truths are better than old errors." *

The theory ot elasticity is most applicable to arches of small depth of ring and high
rise, and the analysis of stresses will be more accurate the nearer the shape of the arch
center line coincides with the resultant polygon for average loading.

This is true when dealing with the arch ring alone, as the ring is the only portion of
an arch which could be expected to follow the laws of elastic deformation. Hence, in order
that the theory shall be at all applicable, the relation between the arch ring and its span-
drel filling must be such as to allow the ring perfect freedom to undergo elastic
deformations.

The style of structure which best satisfies this requirement is one where the arch
ring supports the roadway on a succession of piers, producing what is called an open
spandrel filling. Even here the roadway should be provided with expansion at one end.

Any other style of spandrel will introduce further redundants which are wholly
beyond analysis, because the spandrel is then subjected to bending and direct stress
the samfe as the ring 1 .

Therefore, to apply the theory of elasticity or in fact any theory, to a solid spandrel
fixed arch is, in the author's opinion, a mere waste of time which should be expended in
a more profitable pursuit. In the following, nothing but open spandrel arch bridges
are dealt with.

While the theory of elasticity is Undoubtedly the only trustworthy basis for the

* See paper by the author on Three-Hinged Masonry Arches, Trans. Am. Soc. C. E., Vol. XL, 1898,
p. 31.

ART - 70 FIXED MASONRY ARCHES 301

analysis of fixed arches, and should always be employed in designing structures of any
importance, it does not follow that the older methods of investigation, and many of the
empiric rules in common use, are all worthless. On the contrary, the preliminary design
can be carried out most efficiently by the aid of these simpler methods, and the final
design should then be tested by the application of the more exact method of the
theory of elasticity.

The theory of fixed arches has reached a status of perfection quite in keeping with
the nature of the problem, the still existing uncertainty being a function of the material
and other circumstances, depending on the rigidity of the abutment foundations, condi-
tions of erection, etc., all of which can never be definitely known nor be entirely eliminated
in the best designs.

When the loading is not known with considerable accuracy, as for arches, sustain-
ing high earth banks, the design must necessarily remain more or less indefinite, though
the elastic property of the superimposed earth assists greatly in distributing pressures
and rendering conditions more favorable. Also in dealing with small arches, such as cul-
verts and road crossings, the resultant polygon theory will probably continue to be the
only method of approximate analysis.

ART. 70. MODERN METHODS OF CONSTRUCTION

Many difficulties are encountered in the construction of fixed masonry arches, owing
particularly to insufficient elasticity in masonry. The natural deformations in the arch,
caused by shrinkage of the masonry or concrete, due to the setting process, stress and
temperature, usually cause cracks, which, while rarely of a serious character, are reasons
for discouragement to the engineer, who has probably applied eveTy known precaution
to prevent their occurrence.

So long as there are no abutment displacements after completion of an arch, the
above difficulties may be fairly well controlled. However, when an arch was designed
for certain allowable unit stress on the extreme fiber at the critical points, then, if the
structure is to be regarded safe for all time, the original stress must not be exceeded even
when the material increases in strength (as by setting of the cemjent) or when the arch
or its abutments undergo slight elastic or permanent displacements. For this there
is absolutely no assurance, though it is an essential necessity in the general assumption.

If an arch ring could be built in such manner that its resultant polygon would pass
through the center of the ring at the crown and springing points at the time of releasing
the falsework, a large proportion of the redundant stress could be prevented. In other
words, the bending moments at the critical points would then be almost zero for sym-
metric loading, reserving the strength for the unsymmetric live load and other contingencies
affecting the shape of the arch ring.

By the ordinary process of construction, the arch is commenced at the abutments
and the load is gradually applied to the falsework, thus distorting the latter and causing
bending stresses in the completed portion of the ring. The final settling of the ring,
produced by the contraction of the mortar or concrete, as a result of compression and
shrinkage during setting, even when the abutments remain immovable, are sufficient

302 KINETIC THEORY OF ENGINEERING STRUCTURES CHAP. XV

to create serious initial stresses, thus destroying to a large extent the usefulness of the
structure and frequently cracking the arch.

Several methods have come into use by which these initial arch stresses may be more
or less completely prevented. One of the oldest of these consisted in setting the arch
stones on the entire falsework, spacing the joints by small strips of wood, and lastly
filling all the joints simultaneously with mortar. This is still a very good programme for
construction of small arches, but not so well adapted to large structures, owing to the
excessive stresses produced in the falsework, frequently causing considerable settlement
before the ring can be closed. However, the falsework could be wedged up just previous
to filling the joints.

Another method, frequently employed in the construction of brick and concrete
arches, is to close a complete ring adjacent to the falsework, and after allowing this
course to set firmly, the remainder of the arch is completed, thus carrying the load by
the first ring rather than by the falsework. By this means the settlement during
construction becomes very slight, but, as is readily seen, the intrados is excessively
stressed, a condition which cannot be averted and which is highly objectionable.

A programme of construction frequently followed on large arches is to commence work
simultaneously at two, four or six points of the ring and closing at three, five or seven
points, respectively. The larger the span the greater the number of points of commence-
ment.

The most modern method consists of the introduction of temporary flexible or
hinged joints at the crown and springing and following the last-mentioned programme
of construction. These flexible joints are made of stone with curved roller-like surfaces;
or iron blocks may be used. By this means the resultant polygon becomes determinate
during the period of construction. After the falsework is removed and the structure
has assumed a normal condition of stress, these open joints are grouted with cement
mortar.

Sheet lead or lead blocks have also been used for this purpose, allowing sufficient
surface of contact on the arch center line to carry the pressure without causing the lead
to flow. A safe allowable pressure for lead is from 1000 to 1500 Ibs. per sq. in.

The author would recommend an alloy of lead with from 5 to 10 per cent of copper
added, thus permitting an allowable unit stress of about 2500 Ibs. per sq. in. The
last mentioned method gives very good results in preventing cracks and excessive
stresses, during construction, but subsequent abutment displacements or other changes
cannot be compensated and still remain as a serious objection to this class of structure.
Eventual settlements in foundation masonry, contraction of mortar or concrete due
to setting and drying out; compression due to loading, elastic deformations caused by
temperature and load effects all these influences are ever present to create stresses
which, in spite of all precautions during construction, may attain dangerous proportions
and make it utterly impossible to estimate the ultimate strength of a fixed masonry
arch.

However, it should be remembered that, for masonry arches, the live load is generally
a small fraction of the dead load, and for this reason an arch which is sufficiently strong
to sustain its own weight permanently will carry temporary live loads with perfect safety.

ART. 70 FIXED MASONRY ARCHES

Masonry arches possess the redeeming feature that when stressed to the breaking
limit the masonry generally chips near the surface, thus relieving the stress and allowing
the resultant polygon to return to a more favorable position. In this manner an arch
which has become distorted may readjust itself to a new condition of stress.

Hence fixed masonry arches, in view of their past history, cannot be condemned
on account of insufficient safety which they afford, but because masonry is not a suitable
material from the criterion of inability to withstand elastic deformations well. A fixed
plate girder arch would be a far more rational structure than a fixed masonry arch, though
the former would require frequent painting and even then might deteriorate and thus
outlive its usefulness earlier than the masonry structure. But from an engineering
standpoint the plate girder would probably be more satisfactory because it would not
open up cracks and thus bring discredit to the designer and builder.

In this connection it may be of interest to recall the recommendations proposed by
the Austrian Society of Engineers and Architects, which, if strictly carried out, would
limit the application of fixed masonry arches to comparatively short spans with bed
rock foundations. These recommendations require the fulfillment of the following
conditions :

1. The abutments must be perfectly rigid.

2. The falsework must retain its form during the period of construction of the
arch ring.

3. The masonry must be of the best quality.

4. The construction of the arch ring must be most carefully conducted.

5. The falsework must not be released until the cement has thoroughly set.

6. When the falsework is finally released, it must be done gradually and uniformly.

7. To this the author adds that the arch ring should be closed at the lowest possible
temperature.

The use of sand jacks or sand pots, which was introduced in 1854 during the construc-
tion of the Austerlitz Bridge, in Paris, offers a very novel and efficient means of releasing
falsework.

The necessity of these recommendations is clearly understood after what has been
said regarding the theory of fixed masonry arches and its practical applications and
limitations.

The two first conditions can be realized only when rock foundations are available.
The other requirements can generally be fulfilled by exercising proper care and by
permitting nothing other than first-class materials and workmanship.

From the above review of the subject and actuated by many years of practical
experience, the author ventures the following opinion as to the general advisability of
adopting the fixed type of masonry arch for any particular case in hand: If the span
i is moderate and the rise h comparatively high, such that l/h will be between two and
four, then for bed rock foundation and open spandrels the fixed masonry arch 'may be
chosen as an appropriate type of structure. The design for such a bridge should be
based on a factor of safety of at least ten, allowing no tension unless carried by steel
reinforcement, and after the preliminary design is completed it should receive a final
analysis according to the theory of elasticity.

304 KINETIC THEORY OF ENGINEERING STRUCTURES CHAP. XV

When all this is followed by a most careful and approved system of construction,
it may frequently happen that the results will not prove entirely satisfactory to the
engineer. However, such a structure will be certain to last well, to cost little or nothing
for maintenance, and to be safe practically for all time, even though small cracks may
appear at any time after removing the falsework.

On the other hand, when the spandrel filling must for some reason be made solid.
and bed rock foundations are not available, also when the span is great and the rise
small, a fixed masonry arch should never be built. A hinged arch is the only rational
solution for such a case in the light of modern engineering experience, and that type
can be employed only when the spandrels can be so arranged as to allow of proper
expansion or rotation at the hinged points.

All this applies strictly to bridges and riot to small arched culverts and floors where
the hinged type is in many ways impracticable and the fixed type, preferably of reinforced
concrete, has given good satisfaction.

It is thus seen that the successful construction of a large masonry arch may justly
be considered a masterpiece of engineering achievement and requires far more intimate
knowledge and experience than is required for a steel structure of the same span.

Masonry and concrete are materials which are solely adapted to carrying compressive
stress. Hence, it is rational to reinforce this material to better resist tensile stress by
the introduction of steel, but this should be confined within proper limits and with the
sole aim of developing thereby the otherwise latent capacity of the material to do work
in compression.

When, therefore, steel is used to reinforce concrete in compression, the result is
bound to be wasteful, unprecedented, and a monument to engineering misconception.
Designs of this kind should be discountenanced, both from an aesthetic as well as from
an engineering standpoint, because they imply a misapplication of the true purpose of
both masonry and steel.

ART. 71. PRELIMINARY DESIGNS

On the supposition that a fixed masonry arch is feasible only when the foundations
are rigid, it is fair to assume that the fixed conditions will remain unchanged even
though the transmission of stress may call forth very slight elastic changes in the abut-
ments.

In any problem the given data are usually the live load to be carried, the clear span
and the clear rise of the intrados and the allowable unit stress. The elevation of the
roadway is also known.

The preliminary design would then include a reasonably accurate determination of
the thickness and shape of the arch ring at all points over the span.

To accomplish this, it is necessary first to decide on the general outlines of the
bridge, particularly on the design of the spandrel filling and the roadway, which should be
completely detailed before proceeding to the design of the arch ring. By this means
a large portion of the dead load can be accurately estimated, leaving the bare arch ring
as the only variable portion of the dead load.

ART - 71 FIXED MASONRY ARCHES

305

Since the dead load for masonry arches is usually large compared with the live
load this procedure is clearly indicated.

The next step should be to determine the axis of the arch ring so that it will
coincide with the resultant polygon for an average load, which latter is taken as the
dead load plus half the uniformly distributed live load. In preliminary designs the live
load, whatever it may be, is always taken as uniformly distributed.

The shape of the arch center line thus obtained will be the one yielding the most
economic sections for the unsymmetric maximum and minimum live loads.

However, this center line cannot be directly found because the resultant polygon
upon which it depends must be determined before the dimensions of the arch ring and
its weight are definitely known.

Hence, this part of the problem must be solved by successive approximations and
trials. We must seek to find such an arch ring for which the resultant polygon will
coincide approximately with the center line.

With the aid of approximate formulae, the thicknesses of the arch ring at the crown
and abutments are determined, and from these the radius of curvature at the crown is
approximated to find the first arch center line. The weight of this arch ring is now
estimated and combined with the weight of roadway arid half live load, and a resultant
polygon is then drawn through the centers of the crown and springing sections. This
then gives a criterion as to the modifications required to make a second approximation
for a correct center line. This process is continued until a center line is found which
coincides closely with the resultant polygon.

The maximum stress for unsymmetric loading is now estimated from another
resultant polygon drawn for the critical quarter points and this then furnishes the
criterion for the thickness of the arch ring. If this stress is sufficiently close to the
allowable stress, the preliminary design may be accepted and the final rigid analysis
may then be undertaken. Otherwise further alterations in the dimensions must be
made until the preliminary design is sufficiently close to the assigned requirements.

The approximate formulae and method of conducting this investigation will now
be given, and this will constitute a complete solution as far as is possible by the use of
the older, resultant polygon method.

Fig. 7lA represents a symmetric arch ring of unit thickness. If the arch were
slightly unsymmetric, the present solution might answer for the symmetric portion and
the resultant polygon could afterward be extended into the unsymmetric portion which
was omitted in the first analysis. However, this could not be done in the final rigid
analysis by the theory of elasticity and should never be considered for badly unsymmetric
structures.

All pressures and loads will be expressed in terms of cubic feet of masonry per square
foot of surface, so that all areas on the drawing icpresent true relations of loads from
which actual weights are obtained by multiplying the areas by the weight of one cubic
foot of masonry. This is the most convenient unit for graphic solutions, as it does away
with estimating the weight of the arch ring and spandrel filling.

The following data are supposed to be given : The system of loading; width of roadway ;
span and rise of intrados; and the allowable unit stresses of the masonry or concrete. ,

306

KINETIC THEORY OF ENGINEERING STRUCTURES CHAP. XV

The first step should be to sketch a general outline of the structure in sufficient
detail to design the roadway and the spandrel filling. This will be considered completed
and will not be discussed here.

If the bridge is to be for a single track railway for Cooper's E 60 loading, span 150
ft., rise 50 ft., and width of arch ring 13 ft., then the uniformly distributed load will be
1070 pounds per square foot, and taking the weight of one cubic foot of concrete equal
to ?- = 140 Ibs., then the load in terms of cubic feet of masonry would be 1070-:- 140=7.64
cubic feet = p. This includes 67 per cent for impact.

The floor, ballast, ties, rails, etc., complete, will weigh 710 Ibs. per sq. ft. according
to a design previously made, and this gives the dead load of floor q =710/140 =5.07
cu.ft. of masonry. Since the floor contains considerable steel, its actual thickness is
not q, but in the example chosen it was only 4 ft.

The preliminary design of the arch ring is now undertaken, referring to Fig. 7lA,
for the lettered dimensions.

LIVE LOAD

r~- ^ r

^T
^^ d

-~r. v > .r-

^, \ 1 _,-"

~~*T

FIG. 7lA.

The thickness of the arch ring at the crown is approximated from the formula? of
A. Tolkmitt,* which are rational but approximate, and, therefore, furnish much better
results than purely empiric formula?..

Thus for a uniform live load p/2 over the whole span, and an allowable unit stress
0.4/, which is supposed to be 0.4 as great as for the unsymmetric critical loading, the
crown thickness should be

4+T5)'

t for feet or meters, (71 A)

where the allowable unit stress / is likewise expressed in cu. ft. of masonry, as was done
for the loads.

* Leitfaden fuer das Entwerfen Gewoelbter Bruecken, 1895.

ART - 71 FIXED MASONRY ARCHES 307

For the unsymmetric live load, extending over half the span to the center, the
following value of Z) is obtained :

where G = $(q + ?-+*} f for feet or meters, (71s)

\ 2i Au/ j

When the value from Eq. (7 IB) exceeds that given by Eq. (71 A) then the larger
dimension should be adopted.

It is clearly seen that the full allowable unit stress / could not be taken in the
above formulae, because this average case of loading will stress the arch ring only about
half as much as when the resultant polygon passes through the middle third points. In
all arch designs where tensile stresses are prohibitive, this condition will govern.

The next step is to choose a preliminary shape for the intrados such that, for the
case of dead load plus half live load over the whole span, the resultant polygon will
exactly coincide with the arch center line.

At this point all theory fails and the judgment and experience of the engineer must
guide in making a suitable first approximation. However, the following practical
suggestions will serve a valuable purpose.

Theoretically the intrados can never be a true circle nor a true parabola for any arch
which is made to follow the resultant polygon, as above required, for economic reasons.

The resultant polygon becomes a parabola when the total load is uniform- per foot
of arch. This can never happen in a real arch, even if the spandrel filling were neglected,
because the stresses in the arch ring increase from the crown toward the abutments,
thus necessitating a variable thickness of ring, increasing with the stress.

On the other hand the resultant polygon becomes a circle when the superimposed
load at the springing points becomes infinite, a condition which is never attainable.

Hence an arch of economic design and shape to fit the resultant polygon for average
loading will have an intrados which ^neither a circle nor a parabola, but a curve
lying between these two.

Having thus established the limits between which the true intrados must be situated
and knowing from experience that the true line falls nearer to the mean of the two curves
than to either one of them, for all open spandrel arches, it becomes an easy matter to
approximate the intrados.

Hence, for open spandrels with roadway supported on piers or columns, the intrados
may safely be taken half way between the circle and the parabola, both drawn through
the three given points of the intrados.

The radius of the circle is given by the formula

and the parabola passing through the same three points is

which gives the coordinates y =ho/l and x =/ /4 for the quarter points, see Fig. 7U.

308 KINETIC THEORY OF ENGINEERING STRUCTURES CHAP. XV

From this a mean point at each quarter span is readily determined and a three-
center curve is then made to pass between these two and the given crown and abutment
points.

When the spandrel filling is solid, then the intrados approaches nearer to the circle
and for a plain arch ring without any spandrel loading, the true intrados approaches
the parabola. Between these limits the designer will soon be able to make very close
approximations even for the first trial.

To draw the extrados, no good rules can be given, though a fair approximation can
be made by applying the formula for the thickness of any point of an arch ring in terms
of the thickness DO at the crown, as follows:

= V / .- . . . /;.;. (71B)

cos 9

For flat arches this gives good approximations, but for semicircular arches the
formula should not be used for angles greater than 30 from the crown. Even for
<=30 the values are a little large and for (/> =90, D = GO , which is clearly impossible.

However, the 30 from the crown include the most important portion of the arch
ring, so that the remainder of the extrados can be approximated with sufficient accuracy
for the first trial.

Another method is to sweep an arc with radius r'=r+Z)o through the center of the
crown joint, which will be a good approximation for a center line between crown and
quarter points.

It is not prudent to spend more time in preliminary guessing, but with the dimensions
above given proceed at once to the graphic analysis.

For this purpose the arch should be drawn on a scale of at least 1 : 100, using a scale
1 ft. =3000 cu. ft. of masonry for the scale of forces. For small structures larger scales
may be employed.

The arch ring is now divided up into suitable sections corresponding with the panel
lengths chosen for the spandrel piers and the areas of these sections and of the piers are
then computed from the drawing.

Having thus found all the areas, including the uniform floor load q per ft. of bridge,
we may proceed to draw the resultant polygon for a live load p/2 extending over the
whole span. This polygon should be passed through the center points of the crown
and abutment joints and if it coincides quite closely with the assumed arch center line,
then the investigation may be continued for the unsymmetric loading, otherwise the
shape of the arch ring must be corrected and the first resultant polygon must be recon-
structed. A second trial nearly always gives acceptable results, and sometimes the first
trial is sufficiently close. For symmetric loading the resultant polygon is drawn for
the half span only.

With the dead loads previously found and a live load covering half the span and
extending past the crown to the load divide i for the critical point m, see Fig. 7lA, now
pass a resultant polygon through the points a, b, and n. The point a is the outer third
point of the arch section at A, the point b is the inner third point at B and the point n
is about Do/8 above the center line at the crown. This is the first approximation.

ART. 72 FIXED MASONRY ARCHES 309

If this resultant polygon remains within the middle third at every point and crosses
the axial line in at least three points, then the design is acceptable. If the polygon
does not touch the middle third point at the two critical sections m and s, Fig. 7lA,
then the ring is too thick. If the polygon goes outside the middle third at one of these
sections and remains inside at the other, then the point n should be shifted and a new
polygon should be passed through a, b, and n. However, if the resultant polygon
passes outside the middle third at both critical sections, then the arch ring must be
made thicker and the investigation is then repeated.

Lastly the stresses at the four sections a, m, n, and b must not exceed the allowable
unit stress as given by Eq. (69A).

When all these requirements are fulfilled, the crown section should be tested for
full live load over the entire span, which is easily done from the loads already found for
the half span, since this is a case of symmetric loading.

This then constitutes a complete design according to the ordinary graphic method,
and the arch ring so found should now be subjected to a rigid analysis according to the
theory of elasticity which follows.

The above outline will be exemplified more fully when presenting a problem at the
close of this chapter.

The criterion for position of moving loads to produce maximum and minimum stress
is discussed in another <place.

ART. 72.

Introductory. A fixed, solid web arch must be treated as a structure with three
external redundant conditions according to Eq. (3c) and Fig. SF. Hence there is very
little difference between a fixed framed arch and a fixed solid web or masonry arch,
since the complications arising from external redundancy are alike in both cases while
the internal stresses must be found by the methods peculiar to frames and isotropic
bodies, respectively.

The method of treating the external redundant conditions involves the use of
deflection polygons for the determinate principal system, and in this connection a
further difference exists in the manner of determining the elastic loads for the two types
of fixed arches.

The elastic loads w for a framed structure are easily expressed in terms of the
geometric relations existing between the members and the angles composing the frame.
For isotropic bodies the elastic loads are functions of the cross-sections obtained by
integration and expressed in terms of moments.

Hence, with due regard to these differences, \vhich are easily distinguished, both
framed and solid web, fixed arches may be analyzed in precisely the same manner.
However, the general subject of solid web and masonry arches involves many consider-
ations not met with in framed arches. For this reason the solid web arch will be treated
in full detail in the following articles.

Regarding the reliability of the method, which is based on the theory of elasticity,

310

KINETIC THEORY OF ENGINEERING STRUCTURES

CHAP. XV

the author would add here that for steel arches, either framed or solid web, the method
is entitled to full confidence, while for stone and concrete it is reliable when the design
is based on the condition that tensile stress does not occur. In attempting to design
in concrete for equal tensile and compressive stress by reinforcing with steel, the method
must be regarded as more or less approximate, since it involves the modulus E of the
material, which is quite uncertain owing to the heterogenous character of masonry and
concrete generally, and especially when the latter is combined with steel. The intro-
ductory remarks of Art. 52 are especially applicable here.

General relations of the external forces to the principal system. The considerations
under this heading, in Art. 52, are again applied to the analysis of the solid web arch,
and the principal system is chosen as a determinate beam on two supports.

FIG. 72A.

Fig. 72A shows the principal system for an unsymmetric arch with the redundant
conditions acting as external forces on the left-hand abutment. Similar redundants
act on the right-hand abutment, and these are equal and opposite to the set shown for
the left-hand end. See Figs. 52n and 52c.

The arch ring aa'bb', loaded with a single load P, is referred to coordinate axes
(x, y) choosing the y axis vertical and the x axis making some angle /9 with the horizontal.
The origin is supposed to be known and is fixed by certain geometric relations to be
established later.

The ordinate y m , of any point ra of the gravity axis or axial line of the arch ring,
is measured vertically Irom the x axis, while the abscissa x m> of this point, is measured
horizontally from the y axis instead of parallel to the x axis. This is a mere matter of
convenience.

The span AB=l is taken as the line joining the intersections between the axial line

ART. 72 FIXED MASONRY ARCHES 311

and the verticals through a'\ and b'\. The support at a\ is assumed as hinged while
the one at b\ is made movable for the principal system.

The single load P then produces reactions RI and RZ, intersecting in the point C
and passing respectively through the' unknown points a^ and 61 on the verticals through
the supports a\ and b\. The triangle aiC&i thus becomes a resultant polygon with
the closing line a\b\. The reaction RI may be resolved into the vertical component A
and the haunch thrust H' along a\b\. The reaction R% may be similarly resolved into
the vertical reaction B and the thrust H', which latter is equal and opposite to H'
acting at a\.

The vertical reactions A and B are the same as for a simple beam of span I on
determinate supports. Hence,

A =j(l-e) and ' B =*. . ....... (72 A )

Also, the moment for any point m of the simple beam, equals

-\f i\f

osa or tf--i--, .... (72 B )

where K is the ordinate ig through m, and H is the horizontal component of H' .

Therefore, the resultant polygon a\Cb\ becomes fixed whenever H' and the closing
line a\b\ are found.

The origin is connected to the arch along aa' by the rigid disc aa'O and to this
origin are applied two equal and opposite forces H f , which are equal and parallel to the
original hanuch thrust acting at a\. The equilibrium of the principal system and of
the fixed arch thus remains undisturbed.

Suppose now that all the external forces to the left of a section tt' act on the
principal system only, and that the three forces H' and the vertical reaction A are
applied to the rigid disc and are thence transmitted to the principal system. Then the
force H' at a] and the opposing force H' acting at 0, form a couple with lever arm
z cosa, producing a moment X a =H'z cosa. The other force H', acting at and to
the right, may be resolved into two components X b and X c , where X b is vertical along
the y axis, and X c acts along the x axis. The external forces to the left of the section
and applied to the principal system are then P, A , X b , X c , and a moment
Xa=H'z cos a = Hz . Of these the two forces X b and X c and the moment Xa constitute
the redundant conditions while the forces P and A are known, and all are applied to
the principal system to the left of the section tt'. A similar set of external forces (not
shown) acts on the principal system to the right of this section.

The moment of all the external forces abou'c any point m of any structure, involving
three external redundant conditions, is given by Eq\ (7 A) as

M m ^M om -M a X a -M b X b -M c X c , ....... (72c)

wherein

M om =-A (li x m )Pd=the moment about m due to the load P acting on the
principal system, This is condition X=0.

312 KINETIC THEORY OF ENGINEERING STRUCTURES CHAP. XV

M a = l is the moment about m due to the moment X u = l applied to the principal

system. Condition X a 1 .
Mb = 1 x m is the moment about m due to the force X b = 1 acting on the principal

system. Condition X b = 1 .
M c = l-y m cos /? is the moment about m due to the force X c = l acting on the

principal system. Condition 'X c = 1 .

Substituting these values into Eq. (72c), gives the following fundamental moment
equation for fixed arches:

M m =M om -l-X a -x m X b -y m cospX e , . (72D)

Before M m can be determined for any point m of the arch, the three redundants
X a , Xb, and X c must be evaluated from three simultaneous work equations of the form
of Eqs. (44s) which may be made to apply to the present problem by so locating the
(x, y) axes that d ab =d ba =Q, ^=^=0 and d bc =d cb =0.

The Redundant conditions for a single load P 1 then become

^w>f " "wZ> " "me

for fixed abutments and omitting temperature effects.

Neglecting the term involving the axial thrust N in Eq. (15N) the deflections in
Eqs. (72E) may be expressed as virtual work in terms of the moments M om , M a = l,
M b =x and M e =y cos/?, as follows:

~M om du C Mldu Cdu

_ fM om M a du _ (
dma ~J~~~ET ~J

rM om M b du fM om xdu C

8* =) - - Er -J -^7; fa - J

j~. r ~\/r1j,. r-.vj,.

(72r)

CM om M c du CM om y cos 8du C M 2 c du A/ 2 cos 2 3du

O mc =J ^J E j -I ^ = J E J~ = J gf-

Introducing the elastic loads

du du du

v7 =w a, x~T^T =w b, and y-j^f =w c (72c)

Hit til ' hi

into Eqs. (72r), then substituting these values into Eqs. (72E) and replacing the inte-
grations by summations, then

V _m _ - om b -^ _ mc _ j omc . .

~~~ ; Ac ~~

The modulus E, being involved in both numerator and denominator, of Eqs. (72n),
has no effect on the redundants except when temperature and reaction displacements
are included.

If the axial thrust is to be considered, then the deflections due to this effect alone

ART. 72

FIXED MASONRY ARCHES

313

and for N a =Q, A\ = 1 sin (f> and A' c = 1 cos </>, according to Eq. (15N), become for
du=dx/cos (f):

/A T ~ / fl ' 2 rk A

-fij- = J E p CO^A

fN c 2 du _ fcos 2 fi du _ C l cos < Ax
CC= J ~Ej r= J EY Jo ~EF~

(72j)

These values added for total effect includin axial thrust in the determination of

the redundants will change Eqs. (72n) to the following:

X a =-

I,M om w c

(72K)

EF cos (f>

Eqs. (72n) and (72x) being written for a single load P = l represent the equations
of the influence lines of the redundants.

If axial thrust is to be considered in Eqs. (72K) it will be well to compute the two
functions in the denominators and thus include them in the pole distances when drawing
the Xb and the X c influence lines.

It should be pointed out that in general, axial thrust becomes important for very
flat arches, in which case the function S sin 2 <f)Ax/EF cos $ becomes very small, while
cos <f>Jx/EF becomes large. Hence it is never necessary to include the thrust function
in determining X^, while for X c it may take on considerable proportions.

Regarding the curvature of an arch, which is involved in the integral J du and which

could be fully considered should the necessity arise, it must be remembered that the
depth of the arch ring is. almost without exception, very small compared with the
radius of curvature of the axial line, and no appreciable error is committed when this
effect is entirely neglected. Some comparative analyses have been worked out by
various investigators, which show quite conclusively that this error is smaller than the
knowable accuracy of the strength of materials, to say nothing about temperature
stresses and abutment displacements, which may be estimated but never become known.

For these reasons, which are considered ample, the question of axial curvature will
not be treated here.

The location of the coordinate axes was to be so chosen that the displacements
o<j6 = ^ar"=^c ;== 0, thus permitting the use of the simplified Eqs. (44s) instead of the
involved form of Eqs. (8b) .

According to Eq. (15is T ) and noting that N a =0, these displacements become

fM a M b du Cxdu v

ab== J TH~ = J ~w = ^ x '

CM a M c du_ (ydu
dac ~J El "JET ^ y

CM b M c du fxydu v
bc = J ~KL~ = J FT = ^ x

, o = 2w c =0
w=0

314

KINETIC THEORY OF ENGINEERING STRUCTURES

CHAP. XV

If the coordinate axes are located so as to comply with the conditions of Eqs. (72L)
then the origin of these axes must be the center of gravity of the elastic loads w a because
%xw a =Q and 'i/w a =Q. Also the axes must be conjugate in order that the centrifugal
moment ^lxyw a 0.

The manner of computing the elastic loads and then of finding the positions of the
axes with respect to any unsymmetric arch ring follows here.

ART. 73. COORDINATE AXES AND ELASTIC LOADS

Unsymmetric arches. The elastic loads w a depend on the geometric shape of the
arch sections while the loads Wkxw a and w c =yw a are seen to be functions of the w a
loads and of the coordinates of the axial points of the arch. Hence the w a loads may
be found for any arch ring of given sections while the w^ and w c loads require for their
determination a careful location of the coordinate axes in compliance with the conditions
given by Eqs. (72b) .

The first step is then to evaluate the w a loads for finite distances Ju, which is done

by dividing up the arch ring into an
even number of lengths and
puting El for each section,

W a = I w a = I -=-7 are found by summa-
J Jo Jiil

tion according to Simpson's rule. This will
answer all practical purposes and will be
the more accurate the shorter the lengths
Au or the greater the number of sections
taken.

The manner in which this is done
will now be illustrated, as the accuracy
of the analysis depends almost entirely on the values W a . See Fig. 73 A.

Instead of dividing the axial line into small lengths Ju, it is preferable to make
these spaces equal horizontally, so that Ju = Ax/cos (f>, where < is the angle which the
normal section makes with the vertical. The Eqs. (72c) then become

Au Ax xAx , yAx

after com-
the values

Ix'^ix*

T
I

X , ^

T N *

i ' i -

I't-M

B
+\

***

x^x^x-*.x4

*-"

i -

FIG. 73A.

Wa El El

w b =xw a =

El cos

and

El cos

> [ (73 A)

Each half of the arch is now spaced off into an even number of spaces Ax, beginning
at the crown. When the spaces Au appear to get too large they may be shortened
toward the haunches of the arch, but it is best to keep as many as possible of the Ax alike.

Now tabulate the depths D of the arch ring and angles < for the axial points from
A to B and figure the reciprocal values of El cos <j> for each. If the section is one of
a steel girder the real moment of inertia is to be used while for a masonry arch it is best
to take a section of width unity and depth D making I=--D 3 /12.

Simpson's rule is now employed to sum the values w a in order to obtain finite

r/* I x /

w=== ~~~~ for the

ART. 73

FIXED MASONRY ARCHES

315

The quantities 1/EIcoscf) are treated as parallel ordinates to some irregular curve
and the volume of length I, which is divided into an even number n of equal spaces Ax-,
is then iven from

W a =-^-[w +4wi+2w 2 +4 : W3+2w 4 to 4u' n _i

(73s)

One such compression can be written for any even number of equal spaces Ax, Ax',
or Ax".

To find W a =\w a for each of the sections, Simpson's rule for three quantities is
then applied. This is the well-known prismoidal formula for volume and is used here
to figure volumes between successive mean areas taking the originally computed values
w for the middle terms.

Since the loads W a are wanted for the points m themselves and not between these,
we interpolate mean values between consecutive pairs of computed values and then
figure the sums W a , using the computed W a as the middle term in the formula.

Thus the formula,

Ax
w a3 =-^-[w 2 +4:W 3 +w 4 ], (73c)

as here applied, becomes

Ax \w 2 +w s , w 3 +WA~\

w ^=-^- -V--+4w 3 +- M, (730)

6 [ 2 2 J

when the interpolated means are used. This gives a much better check on Eq. (73s)
than could be obtained by the use of Eq. (73c).

Beginning at A, and making - I .=w _i, - -=wi_ 2 , etc., the several values

of W a are as follows:

Ax'
i'Wo = ^-\2wo +WQ _il = half load for an end section

D
W -

W a2 = - _

a3 = [MJ 2 _ 3

etc.,

w ^x
W a s=- Q -

etc.,

etc.

Ax"

_ 8 +2lV 8 ] +[2Wg +W> 8 _9

etc.

=- -Wn -12

(73E)

316 KINETIC THEORY OF ENGINEERING STRUCTURES

and as a final check the sum of all W a loads must be by Eq. (73s)

CHAP. XV

Xi A X ' j x

w a =-Hi0 +wi +w 2 ] + -x-
o o

4 +4w 5 +2w 6

Ax"
+4w 7 +w 8 ] + [w s +4w g +2w w +4wn

/dx C Ax

w a . these are now used
El cos Jo

to determine the center of gravity and the angle /? which the x axis makes with the
horizontal. The y axis was taken vertically and is thus fixed as soon as the origin is
located.

FIG. 73s.

The center of gravity is located with respect to any assumed rectangular axes
(z, v), Fig. 73B, with origin at the axial point A the same as shown in Fig. 73A. It is
most convenient to assume the z axis vertical and the v axis horizontal. The coordinates
(z, v) are found for each of the several axial points m previously used for the computation
of the W a loads, and all are tabulated together. Finally the moments zW a and vW a are
found and from these the coordinates l\ and z, 7 of the origin are obtained from

/ v rr a , . *-*a

Z I=^TI^; and -^

(73F)

This also fixes the y axis which is parallel to the vertical z axis through the center of
gravity 0. The x axis, while passing through 0, makes some angle /? with the horizontal
such that I,xyW a =Q, according to the last of Eqs. (72L).

The (x, y) coordinates may be derived from the (z, v) coordinates when the angle
,5 is determined. Taking /? positive when measured to the right of the origin and above
the horizontal, as shown in Fig. 73s, then

x =li v and y =z z ' +x tan /?

(73o)

ART. 74

FIXED MASONRY ARCHES

317

The angle /? is found by substituting the value of y from Eq. (73c) into the condition
equation %xyW a =Q, giving, as in Eq. (52j),

(73H)

The abscissae x being known from Eqs. (73c) the values HxzW a and 2x 2 W a are
readily found and tan /? is then obtained from Eq. (73n) and the new axes and (x, y)
coordinates are thus determined.

For symmetric arches the above calculations are greatly simplified and the coordinate
axes are readily located as follows: The y axis, being the axis of symmetry, is practically
given when the shape of the arch is given, and the angle 0=0 because for symmetric
forces the centrifugal moment ( xyW a =0 for any pair of rectangular gravity axes.

Hence the y axis and the direction of the x axis are given and the origin is located
by computing the ordinate z ' from any assumed horizontal v axis, giving for symmetric

arches,

SzW
' ; 0=0; and y=z-z r ....... (73j)

In any case the W a , W b , and W c loads are now readily figured and from these the
equations for the influence lines for the redundant conditions X a , Xb, and X c are evaulated.

Solid web circular arches which are only very slightly unsymmetric may be approx-
imately analyzed by extending the short half to a point of symmetry and treating the
structure as a symmetric arch. However, when the difference in the elevations of the
springing points is appreciable, such approximations should not be made.

The center of gravity may be found graphically by combining the W a loads into
two equilibrium polygons, for the loads acting first vertically and then horizontally
through the centers of gravity of the arch sections. The two resultants thus obtained
will intersect in the required point 0. The moments of inertia 5ix 2 W a and Hiy 2 W a may
be obtained from the same polygons by Professor Mohr's inethod.

ART. 74. INFLUENCE LINES FOR X a , X 6 , X c , AND M m

Having located the coordinate axes (x, y) to fulfill the requirements d a b^ae ^bc =
the Eqs. (72E) and (72H) now become applicable to the present problem, giving

X a =

~"2W ~JT~ riar

j vv & J-^a

1<\ <N

U mo (-* mo

~ ~^rW, = ~tf7 = ^ 6 " ;

cos

. f\

-Jj '/cm

(74A)

These Eqs. (74A) which are written for a load P m = l, thus represent the equations
for the influence lines of the redundants X a , X b , and X c .

318 KINETIC THEORY OF ENGINEERING STRUCTURES CHAP. XV

For X a = l, the first of Eqs. (74A) gives l-^W a = l'd ma = ^M om W a . Therefore,
the special moment M om equals unity and this equals the moment X a = 1 acting on the
principal system, which would follow from Maxwell's law. Hence, an equilibrium
polygon drawn for the loads W a with pole unity, would give the deflection polygon
for d ma due to a moment X a = I applied to the principal system. Also, the ordinates
of a similar equilibrium polygon drawn with a pole H a = ^W a would represent the
ordinates rj aw , of the X a influence line, measured to the scale of lengths of the drawing.

The influence lines for Xj, and X c are similarly found from equilibrium polygons
drawn for the elastic loads Wb and W& with pole distances respectively equal to 2xW b =H b
and cos pEyW c =H c . When axial thrust is to be considered then the pole distance
H c =cos p^yW c + S cos 2 <j>du/EF =cos ^yW c + 2 cos <j>Ax/EF from Eq. (72K) .

The M m influen.ce line for moments, about any point m- of the axial line and a
moving load P m =l, is now derived from Eq. (72D). Accordingly the ordinates jj m ,
of the M m influen.ce line, must represent the algebraic summation of the influences due
to Mom, X a , Xb, and X c expressed by the equation

y m =M m =M om -X a -x m X b -y m cos t 8X c ...... (74s)

The influence line for M om is the ordinary moment influence line drawn for the
point m and for a simple beam on two determinate supports at a'\ and b'i, Fig. 72A.
It is a different line for each axial point m.

The influence lines for X n , X b , and X c remain the same for the same structure, but
the ordinates of the Xb and X c lines, according to Eq. (74s), require multiplication by
the variable coordinates x and y cos {3, which differ for each axial point m.

Hence the M m influence line is best constructed by computing the ordinates
(X a +x m X b +y m cos 3X C ) for the- particular values of x m and y m cos/? and plotting these
ordinates negatively from the M om influence line.

The M m line, so found, will serve to determine the bending moment for the axial
point m due to any position of a system of concentrated loads.

The normal thrust N m , and stresses on any normal arch section through m, are
then computed as described in Art. 49 and further discussed below by involving the
moments about the kernel points e and i.

ART. 75. TEMPERATURE STRESSES

The effect ,of a uniform rise in temperature is to expand the principal system
uniformly in all directions. But since a uniform vertical expansion is not resisted by
the fixed abutments, it may be assumed that the whole temperature stress is produced
by horizontal expansion. This manifests itself by a bending moment extending over
the entire arch ring, and produced by resisting abutments.

Thus for a rise in temperature, the extreme elements of the extrados at the crown
are in tension while at the haunches they are in compression, when the arch is not
otherwise loaded. The simultaneous condition of the intrados elements is exactly the
reverse.

ART. 75 FIXED MASONRY ARCHES 319

In Eqs. (72E) the temperature term was neglected. It is now separately considered
without regard to the effect of the loads P.

According to Eqs. (44c), which apply to the present problem, the redundant
temperature stresses, for rigid abutments, become:

X at =^=0; A" M =A'=0; and X ct = d ^ ....... (75A)

Oaa Obb O cc

The first two are placed equal to zero b.ecause uniform temperature changes cannot
produce rotation nor -vertical deflection of the principal system, which latter is here
regarded as a simple beam on two supports.

The distance over which the change d ct is cumulative must be the projection of the
span / on the x axis, hence d ct =etl/cos^ and from Eqs. (72F) d cc = 2?/ cos 2 (3W C , giving
for the last of Eqs. (75A) :

' where w --* w '- (75B)

The moment about any axial point m due to temperature alone, may now be found
from Eq. (72o), wherein M ow =0, X at ^0, X bt =Q, and X^ is given from Eq. (75s).
Then

-M mt =y m cos ?X et =^ 2 *. . . . , .. . (75c)

According t'o Eq. (75c) the only variable affecting the moment M mt is the ordinate
y m . Hence for y m =0, M mt =0 and for y m maximum, M mt becomes maximum. The
points where the x axis intersects the axial line are thus points of zero moment for
temperature effects, while the crown and haunch points receive maximum effects.

The temperature effects must finally be combined with the dead and live load
stresses to obtain absolute maxima and minima.

It is a well-established fact that when masonry arches open up cracks this generally
occurs during the coldest winter weather. This is because masonry or concrete is
scarcely ever placed during freezing temperatures, but generally in the warmest summer
months. Hence, the variations in temperature from that existing at the time of closing
the arch ring are nearly always greater in the negative direction. Thus, if the ring was
closed at 70 F. the completed structure might at some future period attain a temperature
of perhaps 90, but it is quite certain to fall to about 10 if the climate is severe.
This then would subject the arch to temperature stresses resulting from +20 and 80
from the closing temperature.

When it is considered that a lowering in temperature will induce tensile stresses
in the spandrels and thus cause rupture at the points of least strength, it is clear why
this phenomenon is of such frequent occurrence. Rising temperature would tend to
increase the compressive stresses, but probably not to any alarming extent.

320 KINETIC THEORY OF ENGINEERING STRUCTURES CHAP. XV

ART. 76. STRESSES ON ANY NORMAL ARCH SECTION

This subject is fully discussed in Art. 49 and illustrated by Fig. 49s and will not
be repeated here except in so far as to show the application of Eqs. (49s) to the fixed
masonry arch.

For any rectangular arch section the kernel points e and i are located on opposite
sides of the axial line at equal distances k=D/Q from the axis. The kernel point i for
the intrados is situated above the gravity axis, while the kernel point c for the extrados
is below this axis as shown in Fig. 49s.

The moments M e and M t - are found from Eq. (74s) in which the coordinates of the
axial point m are replaced by the coordinates (x e , y e ) for an extrados kernel point and by
the coordinates (xi, yj) for an intrados kernel point. This is permissible because this
equation is perfectly general and is true for any point in the plane of the structure.

The moment equations for the kernel points e and i, of any normal section, are
thus

M e =M oe -[X a +x e X b +y e c.oa t 3X e ] j (

Mi = M oi -[X a + XiX b + yi cos pX e ] I

Designating all influence line ordinates by y and using proper subscripts as a
distinctive feature, then the ordinates of the kernel point moment influence lines may
be expressed as

wherein )j , x, and y are special for each normal arch section it' while the ordinates jj a ,
jjfe, and rj c are the same for all influence lines but differ for each point of the span.

When the kernel moments are known for any rectangular section and due to any
position of a moving train of loads, the stresses on the extreme fibers of the arch section
and the normal thrust N and its distance v from the axial line, may be determined
from Eqs. (49s) thus:

J)2 '

(76c)

i=5 ; F^l-D; and /=~

where D is the depth and F is the area of a normal arch section of unit thickness.

ART. 76

FIXED MASONRY ARCHES

321

In Eqs. (76c) M and Mi have opposite signs when TV acts between the two kernel
points e and i. The stresses f e and /; take their signs from M e and Mi respectively,
and compression is regarded as a negative stress. The offset v is positive when measured
from the gravity axis toward the extrados and when so applied determines a point on
the resultant polygon for that section.

It is not necessary to draw the M, n influence lines because the M- e and Mi Tines
cannot be derived from the M m line, nor is it possible to determine N and v when M m
alone is known except by constructing a resultant polygon.

Hence the stresses on any arch section are best found from the kernel point moments
M e and Mi of that section. If the redundants X a , X^, and X c are evaluated from their
respective influence lines for a certain position of a train of loads, then the kernel
moments, for any section, may be computed from Eqs. (76A). However, it is preferable
to conduct the entire solution by means of influence lines, as will be illustrated
later.

Since different portions of the arch ring are situated in the four quadrants of the
coordinate axes, it may be well to show how to pass from the coordinates of any axial
point m to those of the kernel points of the normal section through ra.

For a rectangular arch section k e =ki D/Q, so that the following relations may
be written out, from Fig. 76A, in terms of the lettered variable dimensions D and <f>
for each point, thus:

For point A, x e = x-s'

For point m,

For point n j

x e = x -~- sin

Xi = x+-^-sm
6

x e =0

and
and
and
and
and
and

sr COS

For point m', x e = x -f-^- sin and

x -^ sin
6

and

For point B, -x e = -x + sin (j> and

yi = y ~i~ ~TT cos <^>
D

7/ e = y COS

COS (f>

D ,
= y cos (p

;= 2/ + " COS^>

x ~-si

and iji = y + cos 96

. (76D)

322

KINETIC THEORY OF ENGINEERING STRUCTURES CHAP. XV

The general equations for the moment influence lines of the two kernel points may
then be expressed in terms of the ordinates rj and coordinates (x, y) of the axial point.
Thus

D , A J D

r /

-Tioe |^a ^

COS

, (76E)

where the coordinates with their proper signs are to be taken from Eqs. (76D).

ART. 77. MAXIMUM STRESSES AND CRITICAL SECTIONS

The maximum stress for any given section is determined from the moment influence
line for that section by placing the maximum train of live loads over the positive portion
of the influence line. Similarly the minimum stress in found by loading the negative
portion of the moment influence line.

Any point for which the TJ ordinate of a moment influence line- is zero, must be a
load divide, and there is no satisfactory way of locating the load divides other than by
drawing the influence lines. This will be illustrated in connection with the problem
which follows.

By the method of influence lines we are thus enabled to find the maximum and
minimum stresses for any particular section.

However, this does not afford sufficient knowledge regarding the safety of a structure
unless those sections are examined which receive the greatest maximum stresses for the
most unfavorable positions, of the live load considering the structure as a whole. These
sections of greatest maxima are called critical sections, and for any given structure all of

ART. 77 FIXED MASONRY ARCHES 323

the critical sections must be investigated while all other sections are of minor importance
and usually receive no consideration.

Unfortunately the theory of elasticity does not afford a direct solution for finding
the critical sections. The critical sections might be located after a number of sections
all along the arch ring had been examined, and the points of greatest maxima would
correspond to the critical sections. However, this would involve much labor and from
the nature of the case is not warranted, since the stresses in an arch ring always change
gradually and not abruptly. An approximate knowledge of the location of the critical
sections is, therefore, all that is required and this may be had from a discussion of the
moment equation.

Thus the general equation for moments about any point m by Eq. (74s) is

M m =Mom -[X a +x m X b +y m cos pX c ]

wherein the parenthetic quantity may be positive or negative, depending on the location
of the point m.

Since the stresses are direct functions of M e and M{ for a given section, then these
moments and stresses attain maximum values simultaneously:

1. For X a negative and Xj, and X c both positive, when y m =0 or when x m =0.
This would locate three critical sections, one at the crown for x m =0, and two on the x
axis for y m =Q.

2. For X a , X b , and X c all positive, M m may become maximum when y m has its;
greatest negative value or when both x m and y m ha-ve their greatest negative values.
This locates two other critical points, one in each abutment.

Hence, there are five critical sections in every arch which must be carefully examined
for maximum and minimum stresses, and the positions of the moving loads for these
limiting stresses are given from the influences lines drawn for the five critical sections.
In the case of symmetric arches, only three critical sections require investigation.

The two critical sections for ?/=0 fall very close to the quarter points of the arch,
a fact which is important in making preliminary designs by the method of resultant
polygons. The load divide may then be approximately located in the manner given
for three-hinged arches by treating the crown as a hinged point.

The maximum stresses are not appreciably altered by taking the quarter point
critical sections a little to one side or the other of the real theoretical point, while the
stresses are greatly affected by the shape of the axial line of the arch ring.

Thus for maximum economy, the axial line should be a line of zero moments for the
case of average loading. That is M m should be zero for every point of the axis of the ring
when the arch is carrying a dead load and half the maximum live load over the whole svan.

324

KINETIC THEORY OF ENGINEERING STRUCTURES

CHAP. XV

ART. 78. RESULTANT POLYGONS

The resultant polygon for any case of applied loads P may be located by finding
the corresponding values of X a , X^, and X c from the three influence lines for these
redundants, having previously located the (x, y) axes.

The redundants X c and X b , Fig. 78A, were taken as the components of H' , respectively
coincident with the x and y axes. The angle /?, which the x axis makes with the hori-
zontal, is given by Eq. (73n) and the y axis was taken vertically. Hence the horizontal
component H , of X c , is

H=X c cos{3 and, from Fig. 72A,

tan a=

H' =

X c cos /?
cos a

cos a

X a _X a
H' cos a ~ H

Co =^! ^ (tan a -tan /?)
ti

X a =X c z cos fi=X b z -
c

(78A)

These dimensions fix the closing line a\b\ on the two end verticals of the abutments,
also the haunch thrust H' , all in terms of X a , Xb, and X c .

The reactions R\ and #2 may be found from the vertical reactions A and B and
the thrust H'. The vertical reactions are those due to loads P acting on a simple beam

= ^(le) P/l and B =^Pe/l. The

and may be computed from Eqs. (72A) as
reactions R\ and R% must then intersect on the line of the resultant R of all the applied
loads P.

A force polygon, Fig. 78A, is drawn by laying off all the loads P in proper succession,
dividing this load line into the parts A and B and at the dividing point drawing a line
parallel to aib\ of length equal to H'. This determines the pole of the force polygon
from which the reactions RI and R 2 and the resultant polygon through i and bi are
easily drawn. See also the force polygon in Fig. 52A, showing how the pole is located
when H', A , and B are given.

The resultant polygon thus found must intersect the arch center line at least three
times for unsymmetric loading on an unsymmetric arch, and four times for a symmetric
arch, provided the arch center line was so chosen as to be coincident with the resultant
polygon drawn for the case of average loading. This average loading consists of the
total dead load, plus half the live load uniformly distributed over the entire span. It

ART. 78

FIXED MASONRY ARCHES

325

will be so understood whenever average loading is referred to in connection with arch
designs.

The most favorable resultant polygon is the one which coincides with the axial line
of the arch ring. Hence the most economic shape for an arch is the one for which the
axial line is chosen to be the resultant polygon drawn for the case of average loading
as just defined. The live load, in its critical positions, will then produce minimum
values for the moment M m =N m v by making the offsets v, between the axial line and
the resultant polygon, all minimum. The absolute minimum for M m =N m v can occur

FIG. 78A.

omy for v=Q at all arch sections, and this condition can be produced only for one case
of loading, which is taken as the average loading, when maximum economy is to be
achieved in the design.

For all cases of live loads other than the average, the moments M m --=N m v could
never become zero for all points of the axial line, though these moments would be
minimum for an arch properly designed for the average loading.

326

KINETIC THEORY OF ENGINEERING STRUCTURES

CHAP. XV

ART. 79. EXAMPLE 150 FT. CONCRETE ARCH

(a) Given data and preliminary design. The data. Required to design a single
track railway arch of 150 ft. clear span and 50 ft. clear rise with profile of surface and
foundations as shown in Fig. 79fi. The loading to be Cooper's EGO, allowing full
impact according to formula (1), Eqs. (64c), 300 -r-[l +300], where I is the portion of
span covered by loads, for any case of loading.

The material shall be concrete, mixed in the proportion of one part American
Portland cement to two parts sand to four parts crushed granite or limestone. Cubes
of 12 inches on each side, at the age of 6 months, shall sustain a compressive stress
of not less than 3600 Ibs. per sq. in. and briquettes 90 days old, mixed one part cement
to two parts sand, shall not break below 300 Ibs. per sq. in. in tension.

Accordingly the weight of one cu. ft. of concrete is assumed at 140 Ibs., and for
stresses between 100 and 600 Ibs. per sq. in. "=3,670,000 Ibs. per sq. in., and

13'

FIG. 79A Pier of Roadway 3'X 13'.

e =0.0000054 per 1 F. The allowable unit compressive stress will be 500 Ibs. per sq.
in. or 515 cu. ft. of concrete per sq. ft. of surface. All pressures will be expressed in
cu. ft. per sq. ft., a unit which is very convenient and nearly the same as Ibs. per sq. in.
Thus if 7--=144 Ibs., then Ibs. per sq. in. =cu. ft. per sq. ft.

Design of roadway It is .proposed to place the track on a floor built of concrete
and reinforced by rolled I-beams of sufficient strength to carry the whole wheel loads
over spans 10 ft. in length. This floor system is supported by concrete piers 3 ft. thick
and 13 ft. wide and these in turn transmit the loads to the arch ring. This floor is shown
in section Fig. 79A.

Each I-beam must carry two 30,000 Ib. wheel loads spaced five feet apart and
producing a maximum bending moment of

M =30,000X60-30,000X30=900,000 in. Ibs.

Allowing 97 per cent impact for a 10 ft. span, this moment becomes 1,733,000 in.
Ibs., and for an allowable unit stress of 16,000 Ibs., the required section modulus is

ART. 79 FIXED MASONRY ARCHES 327

A// 16,000 = 111 in inches. One 20-inch I, 65 Ibs., has a section modulus of 117, which
is ample.

Weight of floor. 2 ft. of ballast at 120 Ibs. per cu. ft. . .3850 Ibs.

Concrete '. .5000 "

2 I-beams and cor. bars 150 "

2 rails 200 "

Total per foot, of roadway 9200 Ibs.

Total per sq. ft. of arch ring q=7lO Ibs. =5.07 cu. ft.

Live loads. E60 loading for 150 ft. span, all locomotives, gives 8400 Ibs. per ft. of
one track, and this, distributed over a 13 ft. wide arch, gives 640 Ibs. per sq. ft. Adding
67 per cent, impact for 150 ft. span, then the total live load p becomes 1.67x640 = 1070
Ibs. per sq. ft. -=7.64 cu. ft. concrete per sq. ft.

Preliminary design by resultant polygons. The methods outlined in Art. 71 are
applied here and the design is shown in Fig. 79s.

Taking dimensions above given and assuming that the stress for dead load plus half
uniform live load will be about 0.4 of the allowable, then Eq. (71 A) gives

and Eqs. (71s) give D =6.00 ft. which value is accepted for the crown thickness of the
arch.

Eq. (71c) gives 81.25 ft. for the radius of the intrados at the crown, and the quarter
point for the parabola passing through the same three points at crown and springing,
is from Eq. (7 ID)

y J* = 12.5 feet and x -? =37.5 feet.
4 4

The point on the parabola is plotted at s, Fig. 79s, and the point s' is on the circle
drawn with radius 81.25 ft. The point s", midway between s and s', is chosen and a
three-center curve is drawn through s" and the given crown and springing points.

The thickness of the arch at the point s" is found to be 6.9 ft., using Eq. (7lE).

The remainder of the ring is drawn according to good judgment and the thickness
at the springing was taken at 10 ft.

The arch ring is now divided up into an even number of sections as described in
Art. 73, so that the same subdivisions may be used throughout the final analysis. The
dead loads are then computed and tabulated together with the live loads as required
for the construction of the resultant polygons. All this is given in Table 79*.

328

KINETIC THEORY OF ENGINEERING STRUCTURES CHAP. XV

h-l
fe

ART. 79

FIXED MASONRY ARCHES

329

TABLE 79A
DEAD LOADS AND EQUIVALENT UNIFORM LIVE LOADS

Ordinates
V
from
.4

ft.

Length
on
Axis,

ft.

DEAD LOADS IN CUBIC FEET.

Live
Load,
P

cu.ft.

Q+P

cu.ft.

Arch
Sections,
sq.ft.

Piers.

sq.ft.

Road-
way.

sq.ft.

Total Q.
cu.ft.

0- 0.0
1-5.9
2-12 A
3-18.9
4-25.4
5-31.9
6-38.4
7^4.9
8-51.4
9-57.9
10-64.4
11-71.9
12 79 4

10.00
9.30
8.60
8.00
7.60
7.20
6.90
6.64
6.46
6.30
6.20
6.10
6.00

193.44
145.40
115.47
95.82
87.33
91.62

108.00
64.80
42.00
21.00
14.80
6.00

65.91
65.91
65.91
65.91
65.91
76.05
Totals

<3 1 = 367.35
#3 = 276.11
Q 5 = 223.38
Q 7 = 182.73
#, = 168.04
Q u = 173.67

99.32
99.32
99.32
99.32
99.32
114.60

^ = 417.01
P 3 = 325.77
P 5 = 273.04
P 7 =232.39
P 9 = 217.70
P u = 230.97

P' 23 = 466.67
P' n = 375.43
P' 1B = 322.70
P' 17 = 282.05
P'is = 267.36
P' 13 = 288.27

20.80

18.10

16.00

14.40

13.84

15.02

1391.28

611.20

1696.88

2002.48

Above loads are all for an arch ring one foot thick.

The resultant polygon for the symmetric loading Q+^P. given in Table 79A as loads PI
to PI i, may now be drawn so as to pass through the center of the arch ring at the joint
a and at the crown n. This polygon is shown in Fig. 79B, as a fine line almost coincident
with the center line, and the shape of the arch ring is thus found to be acceptable. Had
this coincidence not been so close then the shape of the arch would require modification.

For the above case of symmetric shape and loading, the resultant polygon for the
half span only is required, which is best constructed by computing the horizontal thrust
instead of finding it graphically. This is done by computing the moments of the forces
PI to PII about n and dividing the sum of these moments by # = 2P to obtain the lever
arm x of R. The vertical component of the reaction at A is equal to R and hence H is
found by taking moments about n of all the external forces to the left of n. This gives

H=Q

1696.88(79.4-45.83)
50.4

= 1130.2 cu.ft.

The force polygon is thus easily constructed and the resultant polygon is then drawn
through the two section centers and the crown.

The resultant polygon for maximum one-sided loading, is now constructed. To do this
the load divide for the quarter point m is found as indicated in Fig. 79s by drawing lines
am and bn which intersect in the required point i.

The full load p =7.64 cu.ft. per foot is allowed to cover the span from the right up to
the point i and the left portion is acted on by dead loads only. The loads Q\ to QQ and
P'n to P'zz, from Table 79A, are now united into a force polygon with assumed pole

330

KINETIC THEORY OF ENGINEERING STRUCTURES

CHAP. XV

distance E'\ and an equilibrium polygon a'cb' is drawn for the purpose of finding the
reactions R\ and R'-z and the correct pole 3 .

It is then required to pass a resultant polygon through the three points a, n and
6, where a, and b are middle third points and n is D/8 above the axial line. This is done
as follows: project the points a, n and b down vertically on the equilibrium polygon,
giving the points a', c and b' respectively. Draw a'c and b'c. Then in the force polygon
make 0> 2 c' \\ a'c and O^cf' \\cb' thus establishing the points c' and c". Now draw the lines
c'0 3 || an, and c"0 s \\bn giving the intersection 3 as the required pole and the line U^T [| ~ah
will be the true pole distance. An equilibrium polygon drawn through the point a, using
the force polygon OsMN, will then pass through the other two points n and b. The loads are
divided at T into the reactions .4 = If rand B =NT&s per Eqs. (72A) &ndO s M=R'i and
3 N*R>3.

By the construction given in Fig. 79s, the correct pole H' =1220 cu.ft., and the
required resultant polygon anb are found. The polygon intersects the arch center line in
three points and remains just inside the middle third of the arch ring at all critical sections.

It is seen by inspection that a more favorable resultant polygon could not be drawn,
hence the solution is considered completed provided the stresses resulting from this load-
ing do not exceed the allowable limits.

The stresses are now found from Eq. (69A) which, for a rectangular section of depth
D and width unity, becomes

(79B)

where v is the lever arm of the normal N about the axial point of the section.

TABLE 79B
STRESSES FOUND FROM PRELIMINARY DESIGN

Section.

V
ft.

ft.

UNSYMMETKIC LOAD Q + P.

LOADS Q + P OVER WHOLE SPAN.

cu.ft.

ft.

fe
cu.ft.

cu.ft.

N
cu.ft.

ft.

fe
cu.ft.

cu..t.

0.0

10.00

1980

1.67

396.0

0.0

2372

237.2

41.3

6.77

1390

1.13

0.0

410.7

79.4

6.00

1220

0.75

406.6

0.0

1266

422.

422.0

94.4

6.20

1234

1.03

398.0

0.0

158.8

10.00

2290

1.67

0.0

458.

2372

237.2

It is thus seen that the stresses are all well on the safe side and the preliminary
design is, therefore, acceptable. Had it been impossible to construct a resultant polygon
for the unsymmetric loading, which would remain within the middle third of the arch
ring, then the thickness of the ring would have to be altered accordingly. Also, if the
unit stresses had exceeded the allowable limits the thickness of the arch would have to
be increased.

ART. 79 FIXED MASONRY ARCHES 331

The structure thus designed and found adequate to carry the required loads safely
is now subjected to a rigid analysis by applying the foregoing formulae, derived from the
theory of elasticity.

(b) Analysis of the preliminary design by the theory of elasticity treating the
structure as symmetric. The computations are all given in tabular form, thus
illustrating more clearly the method of conducting such analyses and furnishing a record
of the numerical work which is well suited for reference and checking.

The computations for the elastic loads W, and the location of the (x, y) axes are carried
out in Table 79c, following closely the method described in Art. 73.

The original subdivisions of the arch ring are still retained and each space is again
divided into two, making 24 sections in the whole ring between a and b.

The programme carried out in the table is as follows: Compute the loads W a
from Eq. (73A) then determine z ' and transform the original (v, z) axes to the (x, y) axes
according to Eqs. (73j). Finally compute the Wb and W c loads by Eqs. (73A), and the
pole distances H a , H b and H c from Eqs. (74A). This then furnishes all the data for con-
structing the X a , X b and X c influence lines.

The y axis is known to be the axis of symmetry and must, therefore, pass through
the crown joint n or 12. The location of the x axis is found by computing the ordinate
2,,' of the origin from some assumed v axis which was taken 26.4 ft. below the springing
points and 24, Fig. 79u.

In Table 79c, the dimensions D, z and sec <j> are tabulated from the drawing and from
these the values I/I, w a and zw a are derived. The abscissa; x and increments Ax are also
taken from the drawing, and the other values result from performing the operations
indicated in the headings. It should be noted that the constant modulus E was not
considered in computing the elastic loads W so that all these loads are E times actual.

The secants are computed from

R 115.8 ,

sec = 7^-^ = ^r^~ f or P om ts between a and ra,

2+OO.O 2 +00.;)

sec & 7= '^-^ for points between m and n.
2+0.9

The sums 'w a = W a , 2zw a =zTF , etc., are computed by Simpson's rule for distance
increments Ax, according to Eqs. (73 E).

The ordinate z ' , of the center of gravity or origin of coordinates, is then found to
be 22TF a /2PF a =64.72 ft. Also, the sum 2zzTF a =0 when extended over the whole
arch, hence tan /3=0, making y=zz '.

The functions xw a , x 2 w a , yw a and y 2 w a are now figured and from these the sums
2.ru? =Wb, H>x 2 w n =xWb, Hyw n = W c and ^y 2 w a =yW c are again derived by applying
Simpsons's rule for the distance increments Ax. It should be noted that for the whole
arch ring 2 JF& =0 and 2JF =0, serving as a check on the computations. If these sums d6
not become zero, then errors have occurred which must be rectified. Small discrepancies
will always be found owing to insufficient accuracy in the data and not carrying sufficient
decimals. These may be adjusted by proportionate distribution. However, this must

332

KINETIC THEORY OF ENGINEERING STRUCTURES

CHAP. XV

w
o

PH
C

05 *<

*~ w o

s tf

3 H

^ H
N|

^ 02

w
i

fe
O

I I

, s >

"si >i

C5COOOr(H>-H

IOOC5OOO1 I i-l i l

ooooooooooooo

I I I I I I +

1 +

?O(MOC5t-HCDGOTtl'Ttl' IT-HT^T- 1

i I O5 iO CC CO LO Cl i i CO

(N(N(N(MIM(MGOGOOOOOOOXC

U5 t>- O i i

Oi-Hi-HlMIMfNINIMCOCOCOeOIN
OOOOOOOOOOOOO

co co co co co co

pppooooooopop
ooooooooooooo

ooppppppppppp
odo'do'o'ooooo'oo

I *

N O

II -S

55 -I

was omil

o
H

ART. 79

FIXED MASONRY ARCHES

333

be confined to small errors not exceeding ^ per cent. Mistakes must be corrected and not
distributed.

Lisa's

FIG. 79c. X Influence Lines for "Symmetric Arch.

The influence lines for X a , X 6 , and X c were drawn in Fig. 79c, using the results from
Table 79c, and neglecting the effect Scos (j>Jx/EF due to the axial thrust N on the pole
distance H c of the X c influence line. If this effect is to be considered then the value
* cos (f>4x/EF would be determined in the manner later indicated in Table 79M, omitting

334

KINETIC THEORY OF ENGINEERING STRUCTURES

CHAP. XV

the constant E, since it was also omitted in computing the values w a . This is permissible
because E being involved as a reciprocal factor in all elastic loads, as per Eqs. (72o),
may be considered as canceled in Eqs. (72ii). It must be remembered, however, that the
elastic loads are then actually E times too large, a point which must be considered in
computing temperature effects by Eqs. (7oB).

The X influence lines are the equilibrium polygons for the W loads when certain
pole distances are employed in the respective force polygons. It is generally convenient
to draw these diagrams so that the influence line ordinates may appear a certain number
of times actual when scaled to the scale of lengths of the drawing.

The X a influence line is the equilibrium polygon for the W a loads when the
pole distance is made equal to H a = HW a =6.564, as given by Table 79c. In the
present case the pole was made ^TF a so that the ordinates >j a of the X a influence

FOR INTRADOS KERNEL POINTS.
37/0 m n

FIG. 79o. Mo Influence Lines for Symmetric Arch.

line become twice actual to the scale of lengths. The scale of the force polygon is
chosen so as to give a reasonably large figure and to permit of accurate construction
of the equilibrium polygon, otherwise the scale of loads need not bear any specified
relation to the scale of ordinates. As a check on the drawing, the outer rays of this
influence line must intersect on the y axis, in fact this is the graphic method of locating
this axis.

The Xb influence line is the equilibrium polygon for the Wb loads when the pole dis-
tance is made equal to H b = 2xW b =2X5482.9 = 10965.8, from Table 79c. As this pole
is very large in comparison with the 2^ = 107.4, it is advisable to take Hb=Q.QlHxWb
= 109.66, making the ordinates r jb one hundred times actual to the scale of lengths. The
scale of the Wb loads was again chosen as a matter of convenience. The closing line
AB divides the influence line into a positive and a negative area.

The X c influence line is the equilibrium polygon for the W c loads when the pole
is made equal to # c =cos 8^yW c =2X560.9 = 1121.8. Here also, the pole is too large

ART. 79

FIXED MASONRY ARCHES

335

to furnish a well shaped force polygon and hence the pole was made # C =0.02?/TF C ,
giving influence line ordinates >j c fifty times actual to the scale of lengths. The scale
of W c loads is also chosen as a matter of convenience.

The M influence lines are ordinary moment influence lines drawn for the simple span
AB and the several kernel points of. the critical sections. Fig. 79o. The ordinates may
be made actual to the scale of lengths or, as in the case of the M ot line, the ordinates were
made five times actual.

The M; and M e influence lines for the kernel points of the critical sections are drawn
by plotting the coordinates computed according to Eqs. (76E). See Fig. 79E and
Tables 79E and 79F.

The coordinates of the kernel points i and e of the three critical sections at t, m and n
of the arch ring, are computed from the coordinates (x, y) of the axial points, using Eqs.
(76o) as given in Table 79D.

TABLE 79D
DATA RELATING TO THE CRITICAL SECTIONS

AXIAL POIXT.

KERNEL, POINTS.

D .

Point.

sin c

COS <

x i

x e

ft.

ft,

ft.

9.75

56 30'

78.05

-36.19

1.36

0.89

79.4

-35.3

76.7

-37.1

6.90

30 30'

41.8

0.0

0.62

0.99

42.4

+ 0.99

41.2

- 0.99

6.00

00'

0.0

+ 12.08

0.00

1.00

0.0

+ 13.08

0.0

+ 11.08

From the coordinates of the kernel points Table 79o, and the ordinates in Tables
79E and 79r, taken from the X a , X b , X c and M influence lines, t) e and TH are computed
for each section of the span and for the critical sections t, m and n.

Thus, for the section at m, Eqs. (76B) may be written

(79c)

The computations are all indicated in the headings of the tables and no further
comment is necessary here.

With the use of M f and M e influence lines for the three critical sections of a symmetric
arch, the stresses in these sections may be found for any possible case of loading.

The resultant polygon for any particular case of simultaneous loading, is located
when the redundants A" a , X b and X c are known. Eqs. (78A) then give the necessary
dimensions from which the resultant polygon may be drawn.

Table 79o gives the computations for the resultant polygon for the case of average
loading Q +%P, over the entire span. This is the case for which the resultant polygon
should coincide with the axis and the results are comparable with those on Fig. 79s.

336

KINETIC THEORY OF ENGINEERING STRUCTURES

CHAP. XV

The lettered dimensions in the table are all shown in Fig. 7SA, but the resultant polygon
is not drawn for this case.

The influence line ordinates y a , TJ& and TJ C are taken from Fig. 79c and the sums of the
products of the loads and ordinates give the respective redundants X from which z ,c
and c 0} also A and B are readily found by Eqs. (78A).

TABLE 79s
MOMENT INFLUENCE LINES FOR EXTRADOS KERNEL POINTS. SYMMETRIC ARCH

Pt.

r]a
ft.

r/b
ft.

f)c

ft.

SECTION t
J2=>?o [i}a+76.7i}6 37.l7j c ]

SECTION m.

T)m = ijo - [>?a+ 4 1 .2-fjb - 0.99r; c ]

SECTION n.

T)n=T)o [ija+ll.l'jc]

76.7 T) b

37.lTjc

fit

41.27JC

0.99i? c

Tim

H.lTJc

0.0

2.9

0.031

0.014

2.61

2.38

- 0.52

-2.15

4.5

1.28

-0.01

0.33

2.9

0.14

-0.14

6.0

0.062

0.056

9.4

2.55

-0.05

0.90

6.3

0.62

-0.32

8.8

0.086

0.120

2.39

6.59

- 4.45

-8.55

14.2

3.54

-0.12

1.98

9.5

1.33

-0.63

11.4

0.102

0.200

19.1

4.20

-0.20

3.70

12.8

2.22

-0.82

13.9

0.110

0.296

2 17

8.44

-10.98

-9.19

24.0

4.53

-0.29

5.86

16.0

3.29

-1.19

16.0

0.111

0.396

29.0

4.57

-0.39

8.82

19.3

4.40

-1.10

17.8

0.105

0.496

1.95

8.05

-18.40

-5.50

27.5

4.33

-0.49

5.86

22.6

5.51

-0.71

19.4

0.095

0.588

25.9

3.91

-0.58

3.17

25.9

6.53

-0.03

20.7

0.079

0.672

1.73

6.06

-24.93

-0.10

24.2

3.25

-0.67

0.92

29.1

7.46

0.94

21.7

0.058

0.734

22.7

2.39

-0.73

-0.66

32.3

8.15

2.45

22.4

0.030

0.782

1.48

2.30

-29.01

+ 5.69

21.0

1.24

-0.77

-1.87

36.0

8.68

4.92

22.6

0.000

0.800

19.2

0.00

-0.79

-2.70

40.0

8.88

8.52

22.4

-0.030

782

1 23

-2.30

-29.01

10.14

17.4

-1.24

-0.77

-2.99

36.0

8.68

4.92

21.7

-0.058

734

15.6

-2.39

-0.73

-2.98

32.3

8.15

2.45

20.7

-0.079

0.672

0.98

-6.06

-24.93

11.27

14.0

-3.25

-0.67

-2.78

29.1

7.46

0.94

19.4

-0.095

0.588

12.4

-3.91

-0.58

-2.51

25.9

6.53

-0.03

17.8

-0.105

0.496

0.77

-8.05

-18.40

9.42

10.8

-4.33

-0.49

-2.18

22.6

5.51

-0.71

16.0

-0.111

0.396

9.3

-4.57

-0.39

-1.74

19.3

4.40

-1.10

13.9

-0.110

0.296

0.55

-8.44

-10.98

6.07

7.7

-4.53

-0.29

-1.38

16.0

3.29

-1.19

11.4

-0.102

0.200

6.1

-4.20

-0.20

-0.90

12.8

2.22

-0.82

8.8

-0.086

0.120

0.32

-6.59

- 4.45

2.56

4.6

-3.54

-0.12

-0.54

9.5

1.33

-0.63

6.0

-0.062

0.056

3.1

-2.55

-0.05

-0.30

6.3

0.62

-0.32

2.9

-0.031

0.014

0.10

-2.38

- 0.52

0.10

1.5

-1.28

-0.01

-0.11

2.9

0.14

-0.14

0.0

All ordinates ij in above table are expressed in feet.

Table 79n presents the case of unsymmetric loading shown in Fig. 79s. The results
ere plotted in Fig. 79r, and the resultant polygon is finally drawn in accordance with the
method given in Art. 78.

The arch being symmetric /3=0 and Zi =2 hence with the values of X a , Xb and X c
from Table 79n, the resulting data in that table were found.

In Fig. 79r the arch ring and the (x, y] axes are given to locate the points a and b
on the verticals through the end supports and then to draw the resultant polygon.

ART. 79

FIXED MASONRY ARCHES

337

' I 2

AT SECTION t.
1-158'.8

8 \> 10 II U l 14 IS 16 17 18 >? 30 21 22

LENGTHS.

Ijllll.M.I I I __l___L_ 1|

10 IO 20 30 40 SOFT.
OROI NATES.

I//

itifrados kernel points.
extrados kernel points.

FIG. 79E. M Influence Lines for the Kernel Points. Symmetric Arch.

338

KINETIC THEORY OF ENGINEERING STRUCTURES. CHAP. XV

TABLE 79F
MOMENT INFLUENCE LINES FOR INTRADOS KERNEL POINTS. SYMMETRIC ARCH

Pt.

Tja
ft.

T,b
ft,

1)c

ft.

SECTION t.
rj = ^,0 [rja + 79.4,6 35. 3, c ]

SECTION m.
,m = ,o- [,3+42.4,6+0.99,0]

1
SECTION n

T)n=T)o [,<z + 13. Irjc]

79.4,6

35.3,6

rjo

42.4,6

0.99, c

13.1,0

0.0

2.9

0.031

0.014

0.0

2.46

- 0.49

- 4.87

4.5

1.31

0.01

0.28

2.9

0.18

-0.18

6.0

0.062

0.056

9.4

2.63

0.05

0.72

6.3

0.73

-0.43

8.8

0.086

0.120

0.0

6.83

- 4.24

-11.39

14.3

3.65

0.12

1.73

9.5

1.57

-0.87

11.4

0.102

0.200

19.3

4.32

0.20

3.38

12.8

2.62

-1.22

13.9

0.110

296

0.0

8.73

-10.45

-12.18

24.2

4.66

0.29

O . oC

16.0

3.88

-1.78

16.0

0.111

0.396

28.1

4.71

0.39

7.00

19.3

5.19

-1.89

17.8

0.105

0.496

o.c

8.34

-17.51

- 8.63

26.6

4.45

0.49

3.86

22.6

6.50

-1.70

19.4

0.095

0.588

25.0

4.03

0.58

0.99

25.9

7.70

-1.20

20.7

0.079

0.672

0.0

6.27

-23.72

- 3.25

23.5

3.35

0.67

-1.22

29.1

8.80

-0.21

21.7

0.058

734

22.0

2.46

0.73

-2.89

32.3

9.61

0.99

22.4

0.030

0.782

0.0

2.38

-27.60

+ 2.82

20.3

1.27

0.77

-4.14

36.0

10.24

3.36

22.6

0.000

0.800

18.6

0.00

0.79

-4.79

40.0

10.48

6.92

22.4

-0.030

0.782

0.0

-2.38

-27.60

7.58

16.8

-1.27

0.77

-5.10

36.0

10.24'

3.36

21.7

-0.058

0.734

15.1

-2.46

0.73

-4.87

32.3

9.61

0.99

20.7

-0.079

0.672

0.0

-6.27

-23.72

9.29

13.6

-3.35

0.67

-4.42

29.1

8.80

-0.21

19.4

-0.095

0.588

12.0

-4.03

0.58

-3.95

25.9

7.70

-1.20

17.8

-0.105

0.496

0.0

-8.34

-17.51

8.05

10.6

-4.45

0.49

-3.24

22.6

6.50

-1.70

16.0

-0.111

0.396

9.1

-4.71

0.39

-2.58

19.3

5.19

-1.89

13.9

-0.110

0.296

0.0

-8.73

-10.45

5.28

7.5

-4.66

0.29

-2.03

16.0

3.88

-1.78

11.4

-0.102

0.200

6.0

-4.32

0.20

-1.28

12.8

2.62

-1.22

8.8

-0.086

0.120

0.0

-6.83

- 4.24

2.27

4.4

-3.65

0.12

-0.87

9.5

1.57

-0.87

6.0

-0.062

056

3.0

-2.63

0.05

-0.42

6.3

0.73

-0.43

2.9

-0.031

0.014

0.0

-2.46

- 0.49

0.05

1.4

-1.31

0.01

-0.20

2.9

0.18

-0.18

0.0

All ordinates in above table are expressed in feet.

The construction is made by laying off z down from the origin on the y axis to
obtain the point s. Then lay off c up from s to determine s' on the y axis. Through
s' draw a line parallel to the x axis and where this line intersects the vertical at A gives
the point a on the resultant R'i. The line as prolonged to intersect the vertical at B
gives the point 6 on the resultant R' 2 . Knowing A , B and H' , the resultants R'\ and
R*2 can now be drawn, and their intersection must fall on the resultant R, distant r from
the 2 axis.

The force polygon is constructed by laying off the load line MN and_then finding T,
which is the point of division between A and B . Through T draw 0\T\\ob and lay
off the force H' , thus locating the pole Oi, and giving the reactions R'i and R' 2 . The
equilibrium polygon drawn through a and with pole 0\ will then pass through b as a final
check. The resultant polygon for unsymmetric loading intersects the axial line three
times, as it should.

ART. 79

FIXED MASONRY ARCHES

339

HM
O"

CO
C3 Q

p ^
o

H >-} !

H g

3
3

fe
O

RESULTING DATA.

II
^2.

O 4*

o ->

. 3

*- O5

^ ic

2 4
II ^ c
I- II -

1 ^ %

~ "
Jfcj **?

w II
06 133 *
2 a
II ^
- H
u
O
h

05
-1&5 -S

11 P

<? *;

o3 **"!

- 1 ."' 3

s a i 1

1 -U 1 1

5 o S II
5 II ci OSIN

.0 t^ ||

^-N-! ^

II II II
? K ^

a
^
+

II
ftf

* O i-H

(X) rH 00

1^- OS Tt< Tfi O5 O
(N (N CO CO (M (M

-1 O Tt*

XI ^H 00

lO O5 O
CO 00

1C CO O O CO ^C
' 1 ^f CO 00 TJI i I

3 C5 >C

* co

1C M

2 II

ft.|

CO <M CO
O5 O O

O O CO CO O O (
* <M O5 O (M * <

O (N CO

n o O5

1C 1C ^

+ H

^2^

<N 00 O
i-H (N CO

* r^ co CD i^ Tti c

(N rH r-l <N t

1 III

D GO (N
O (M rH

1 1

rH rH >/?
+ 1 *

co o r^

t~- TjH Tfl Ttl * t^ 1

^ O CO

O5 t- Tfl
O CD OS
<M 00 t-~
^H (N CO

co co * -*i co co -

CO O h- I> O CO .
i 1 1C i 1 rH 1C r- 1 1

rfi TJI tc >C -<^ * C

* r-- o

S CO O
^ GO (N

O (N rH

f:^ e

3 ""On IN*;

O 00 O

* O O O l> * <

D 00 O

^ I *

H| ^ 3

I> i CO
1-1 (N t^
TJH CO C^

(N t^ I-H T-H t^ (N C
CO 1 CO CO -4 CO t
(M (N (N (N (N (N C

O >C t^

>. C^ rH

^ CO -*

>^o

1C iO 1C

CO cO CO

1C 1C O O >C 1C '

co co co co co o c

C 1C 1C

O CO CD

a 1 |CI

O C5 O5
^* ^^ ^

C5 C5 t^ t^ O O5 C
TjH TfH >C 1C Tt< Tjl

35 C5
* Tj( TfH

73 -S" * i '

TjH 1 Tt<

t> O t^ 1^ O t^ -

* rH T(H

go> "3
0,3 3

i^ co eo

CO t^ <M

(M 00 CO CO 00 (M C

oo co i-~ t^ cc oo c

CC t^
^ t>- CO

R-

* O CO

r-H C^ 05

O ^H (M

CO (M (N (M C^J CO C

o> t^ oo oo t^ O5 c

rfn CO t^ t-- CO Th C

o o **

R (M rH

^ rH O

o o o

o o o o o o c

-f^>

rH CO O

O O i i

>C C5 O O C5 1C C
O t^ CO CO t^ O i-
T-I O O O O I-H '

D CO rH

H x co

H O O

o o o

o o o o o o c:

1 1 I

1 1 1

O 00 O5

00 t- rH TfH t>- 00 C

s oo as

C-

(M QO CO

t^ O (M iM O b- C

r-4 <N (N C^l <N i-H r

5 00 (M

i i co ic

t>. os i i co >c t>- c

i 1 I 1 rH T 1 T-

5 rH CO
H IM N

340

KINETIC THEORY OF ENGINEERING STRUCTURES

CHAP. XV

gO 1

O ^J

-^ co
CO

2 8

co cs

00 ^H

n . <H

oo *

CO I-H
(N

co o

a i

ft!

05 -3

CO"3<N(NOOCOCOOOcOiCTt< OS

OOCOCON.OOCOOO3OOOOO CM

COi-HCMOtNCNiOr i (iCi lie
I-H C^ I-H O1 I s - t* CO OS I-H OS 1C CO

00 OS CO 1C CD C^ CO C^l -H C^
<M <N (M CO * >C> !> O
CO

Oi O^ Oi 0i O5 O^ O5 O^ Ol O5 O^ 03
iOGO^-*Tfl^-^-tCDOCOcDO5C<l

^* CO CO C'l ^^ *-O ^O CO t^* C^ ^O CO

rH rH r-H CO O^ ^^ ^^ CO 00 *O O *O

COcOOi-HC s 1C^I^-COO5^ (M

O CO

o^ ^^ t^ oo r^ ^o ^o c*j ^H co oo t^ oo co

O I-H

(M i-H iH I (N (M CO CO

' ' I I I I + I

iCC5iC(Nt-t--t--COC5iOCOCO
CD<NOiCr>.iCiCCOi IOOO1C

i-H C^ CO CO CO CO CD 1C 1C "^ CO i-H 1C

CD CO C^l 00 OO OO I s * C3 C^l 1C CD 00 ^^

cD(^-CMOOCOOOOOCDOO<Mt^CO O

CO<N<Ni-Hi-H<N(M(M(MCOCOT}<

OOOOOCOCOCOCOCOCOCO
CO CO CO CO CO Tt^ ^^ OS OS OS OS OS

T"H I-H os os os os os

i-H Jt

t^ co co' (N oo co co oo' CM' co' CD f^

CD t^- C^ OO CO t^- l> CO 00 CN t^ CO

CO<M<Ni ii ii Ii I i I r- i<N<MCO

TfOCOCDtNCMC^CJCDCDOTti
i-H(NOsOsr^OOOOt^O3OS(Mi-H

Oi itNi^cor tco^cMi-HO
O O O O O O O O O O O O

i (CDOiCOSOOOSiCOcOi-H

coooi-HOt^cocot^-Oi-Hooco

OOl-Hl-HOOOOl-Hr-HOO

oooooocdooooo

I I I I I I

OS OO OS OO t^- ^ "^ l>- 00 OS OO OS
<No6cOt^O<N<NOf~COOO<N

1-Hl-H(N<N(N(Ml-Hl-H
rl CO 1C I> OS i I CO 1C f>- OS T-l CO

ART. 79

FIXED MASONRY ARCHES

341

342

KINETIC THEORY OF ENGINEERING STRUCTURES

CHAP. XV

The moments M e and M, and the resulting stresses on any critical section, and for any
case of loading, may be found from the moment influence lines Fig. 79E, drawn for the
kernel points of the critical sections.

The stresses for the two cases of loading, Q+^P and the one-sided load Q+P, are
computed in Tables 79i and 79J, using the influence line ordinates from Tables 79E and
79r, or from the diagrams Fig. 79E. At the bottom of the tables the values R, N, v, i g
and fi are computed for each section, using Eqs. (76c) .

The sections t and t', at the abutments, are so located that the intrados kernel points
i fall on the verticals through the end axial points, thus avoiding negative TJ O values in
Eqs. (79c), and other difficulties affecting the X influence lines.

It is seen from Table 79i, that the actual resultant polygon for the- case of average
loading really coincides very closely with the arch center line as designed. The offsets
v being + and show that the polygon intersects the center line four times, as it should.

MOMENTS M e AND Mi FOR THE CRITICAL SECTIONS, LOADS Q+-

SYMMETRIC ARCH

Loads

SECTION I.

SECTION m.

SECTION" n.

Pt.

M e

cu.ft.

ft.

417.0

-2.15

- 4.87

- 896.6

-2030.8

0.33

0.28

137.6

116.8

-0.14

'-0.18

- 58.4

75.1

325.8

-8.55

-11.39

-2785.6

.-3710.9

1.98

1.73

645.1

563.6

-0.63

-0.87

- 205.3

- 283.4

273.0

-9.19

-12.18

-2508.9

-3325.1

5.86

5.35

1599.8

1460.6

-1.19

-1.78

- 324.9

- 485.9

232.4

-5.50

- .63

-1278.2

-2005.6

5.86

3.86

1366.9

897.1

-0.71

-1.70

- 165.0

- 395.1

217.7

-0.10

- 3.25

21.8

- 707.5

0.92

-1.22

"200.3

- 265.6

4-0.94

-0.21

4- 204.6

- 45.7

231.0

4-5.69

+ 2.82

4-1314.4

4- 651.4

-1.87

-4.14

- 432.0

- 956.3

4.92

4-3.36

1136.5

+ 776.2

231.0

10.14

7.58

2342.3

1751.0

-2.99

-5.10

-690.7

-1178.1

4.92

4-3.36

1136.5

4- 776.2

217.7

11.27

9.29

2453.5

2022.4

-2.78

-4.42

- 605.2

- 962.2

4-0.94

^0.21

+ 204.6

45.7

232.4

9.42

8.05

2189.2

1870.8

-2.18

-3.24

- 506.6

- 753.0

-0.71

-1.70

- 165.0

- 395.1

273.0

6.07

5.28

1657 . 1

1441.4

-1.38

-2.03

- 376.7

- 554.2

-1.19

-1.78

- 324.9

- 485.9

325.8

2.56

2.27

834.0

739.6

-0.54

-0.87

- 175.9

- 283.4

-0.63

-0.87

- 205.3

- 283.4

417.0

0.10

0.05

41.7

20.9

-0.11

-0.20

- 45.9

83.4

-0.14

-0.18

- 58.4

75.1

R =

3393.5

9'. 75

4-3341.1

-3282.4

6'. 90

4-1116.7

-1998.1

Z> =

6'. 00

4-1175.0

-1018.0

2037.9

1354.3

JV =

1096.5

4-0'. 014

-0'.325

v =

4-0'. 071

- 210.9

- 207.2

- 140.8

- 251.8

- 195.9

- 169.7

Eqs. (76c),

f O'M e

/e= -^ =

,
*

v 1S positive when measured from the axis

2N D* ' * W '

toward the extrados.

All loads in above table are expressed in cubic feet of masonry and the stresses are cubic feet per square foot. Cubic
feet per square foot X 0.972 = pounds per square inch.

The stresses in Table 79j are those due to the case of loading shown in Fig. 79r, and
hence the offsets v are those which the resultant polygon in that figure actually present.
The smallness of the scale, however, prevents a very close agreement. The resultant
polygon for the unsymmetric loading is seen to intersect the center line three times as
it should, since two values v are negative and three positive.

ART. 79

FIXED MASONRY ARCHES

343

TABLE 79j

MOMENTS Me AND M FOR CRITICAL SECTIONS. TOTAL DEAD AND ONE-SIDED LIVE

SYMMETRIC ARCH

SECTION I.

SECTION m.

SECTION n.

Ft.

Loads

Q and P',

rje

fji

M e

Tje

T)i

M e

fje

rji

M e \ Mi

cu.ft.

ft.

Qi = 367.4

- 2.15

- 4.87

- 789.9

-1789.2

0.33

0.28

121.2

102.9

-0.14

-0.18

51.4

- 66.1

Q 3 = 276.1

- 8.55

-11.39

-2360.7

-3144.8

1.98

1.73

546.7

477.6

-0.63

-0.87

- 173.9

- 240 . 2

Ob = 223.4

- 9.19

-12.18

-2053.0

-2721.0

5.86

5.35

1309 . 1

1195.2

-1.19

-1.78

- 265.8

-397.6

Qi= 182.7

- 5.50

- 8.63

-1004.9

-1576.7

5.86

3.86

1070.6

705.2

-0.71

-1.70

- 129.7

-310.6

Qo= 168.0

- 0.10

- 3.25

- 16.8

- 546.0

0.92

-1.22

154.6

- 205.0

+ 0.94

-0.21

+ 157.9

- 35.3

Pn' = 288.3

+ 5.69

+ 2.82

+ 1640.4

+ 813.0

-1.87

-4.14

-539.1

-1193.6

4.92

+ 3.36

1418.4

+ 968.7

Pis' = 288.3

10.14

7.58

2923.4

2185.3

-2.99

-5.10

-862.0

-1470.3

4.92

+ 3.36

1418.4

+ 968.7

Pis' = 267.3

11.27

9.29

3012.5

2483 . 2

-2.78

-4.42

- 743 . 1

-1181.5

+ 0.94

-0.21

+ 252.3

- 56.1

PIT' = 282.0

9.42

8.05

2656.4

2270.1

-2.18

3.24

-614.8

- 913.7

-0.71

-1.70

- 200.2

-479.4

P 19 ' = 322.7

6.07

5.28

1958.8

1703.8

-1.38

-2.03

-445.3

- 653.0

-1.19

-1.78

- 384.0

-574.4

P 2 i'= 375.4

2.56

2.27

961.0

852.2

-0.54

-0.87

-202.7

- 326.6

-0.63

-0.87

- 236.5

-326.6

Pi/ = 466.7

0.10

0.05

46.0

23.3

-0.11

-0.20

- 51.3

- 93.3

-0.14

-0.18

- 65.3

- 84.0

3508.3

9'.75

+ 6973.9

+ 453.2

6'.90

-256.1

-3556.1

D =

e'.oo

+ 1740.2

-632.9

2006.4

1434.8

1186.6

+ 1'.851

v =

-1'.328

v =

+0'.467

- 440.2

t28.5

+ 32.3

- 448.2

- 290.0

-105.5

Ft.

Loads
Q and P',

cu.ft.

SECTION m'.

SECTION V.

ft.

M e

ft.

M e

Qi = 367.4

-0.11

-0.20

40.4

- 73.5

0.10

0.05

36.7

18.3

Os = 276.1

-0.54

-0.87

- 149.1

- 240.2

2.56

2.27

706.8

626.7

#6 = 223.4

-1.38

-2.03

- 308.3

- 453.5

6.07

5.28

1356.0

1179.6

Q 7 = 182.7

-2.18

-3.24

- 398.3

- 591.9

9.42

8.05

1721.0

1470.7

Q 9 = 168.0

-2.78

-4.42

- 467.0

- 742.6

11.27

9.29

1893.4

1560.7

Pii' = 288.3

-2.99

-5.10

- 862.0

-1470.3

10.14

7.58

2923.4

2185.3

Pis' -288.3

-1.87

-4.14

- 539.1

-1193.6

5.69

2.82

1640.4

813.0

Pis' = 267. 3

+0.92

-1.22

+ 245.9

- 326.0

-0.10

- 3.25

26.7

- 868.7

Pn' = 282.0

5.86

+ 3.86

1652.5

+ 1088.5

-5.50

- 8.63

-1551.0

-2433.7

Pi 9 ' = 322.7

5.86

5.35

1891.0

1726.4

-9.19

-12.18

-2965.6

-3930.5

P 2 i' = 375.4

1.98

1.73

743.3

649.4

-8.55

-11.39

-3210.5

-4275.8

P 24 ' = 466.7

0.33

0.28

154.0

130.7

-2.15

- 4.87

-1003.4

-2272.8

3508.3

D-

6'. 90

+ 1922.5

-1496.6

Z>=

9'.75

+ 1520.5

-5927.2

if-

1485 . 3

2291.4

v =

+ OM44

-0'.962

- 242.3

- 188.7

- 96.0

- 374.1

The influence line ordinates ij for points TO' and tf are symmetric with those of points m and t. See note, foot of
Table 79i.

344

KINETIC THEORY OF ENGINEERING STRUCTURES CHAP. XV

By referring to Fig. 79E, it is seen that this position of loading gives maximum stresses
for the section I and almost maximum for section m, but not for the other sections. In
the present instance the effects at the five critical sections were found for one position
of the moving load, giving a simultaneous condition of stress, but not maximum stresses
at all sections.

The maximum stresses for each of the critical sections must be computed for special
positions of the moving train as illustrated in Table 79K, for the crown section.

In the general investigation of stresses it is best to obtain the dead load moments
separately for each section and combine these with the live load moments which are found
without regard to impact and then increased for impact prior to adding the dead load
moments, as shown in Table 79K.

TABLE 79K

INVESTIGATION FOR MAXIMUM STRESSES AT THE CROWN SECTION "n."

SYMMETRIC ARCH

Dead

Total Dead Load Moments.

No.

l-I'L 1

Max. + Live Load Moments.

Point.

Loads

vv neel

Live

from

Loads,

ije

1,1

M e

Head of
Train.

f)e

'Ji

M e

cu.ft.

ft.

cu.ft.

ft.

367.4

-0.14

-0.18

- 51.4

- 66.1

16.5

0.00

-1.19

0.0

- 19.6

276.1

-0.63

-6.87

-173.9

-240.2

33.0

1.30

+ 0.10

42.9

+ 3.3

223.4

-1.19

-1.78

-265.8

-397.6

33.0

2.55

1.10

84.2

36.3

182.7

-0.71

-1.70

-129.7

-310.6

33.0

4.10

2.55

135.3

84.2

168.0

+ 0.94

-0.21

+ 157.9

- .35.3

33.0

6.10

4.45

201.3

146.9

173.7

4.92

+ 3.36

854.6

+ 583.6

21.4

6.40

4.80

137.0

102.7

173.7

4.92

+ 3.36

854.6

+ 583.6

21.4

4.40

2.80

94.2

59.9

168.0

0.94

-0.21

157.9

- 35.3

21.4

2.55

1.00

54.6

21.4

182.7

-0.71

-1.70

-129.7

-310.6

21.4

1.30

0.00

27.8

0.0

223.4

-1.19

-1.78

-265.8

-397.6

0.0

0.00

-1.19

0.0

27fi 1

fi^

087

74O 9

ft A

fff \f A

367.4

\J . UO

-0.14

. 01

-0.18

. U

- 51.4

HI . ,

- 66.1

Max.+ L.L.=

+ 777.3

+ 435.1

84%

652.9

+ 365.5

R =

2782.6

Z> =

6'. 00

+ 783.4

-932.4

N =

857.9

TotaFD.L.=

783.4

-932.4

v =

-0'.087

Total max. + A/ =

2213.6

-131.8

-130.6

-155.4

max.A' r =

1190.8

=-<y.88

max./=

-369.0

- 22.9

All stresses / are in cubic feet per square fook Multiply by 0.972 to reduce to pounds per square

300

inch. Impact for 56 feet is =0.84.
oUO+56

The live load moments are found by placing the train of wheel loads for Cooper's
E 60 loading over the positive portions of the M e and M., influence lines for the section
in question to obtain the simultaneous maximum positive moments for the kernel points.
Similarly the negative portions of these influence lines will yield the maximum negative
moments for the same section.

ART. 79 FIXED MASONRY ARCHES 345

Since the loads are distributed over a roadbed 13 feet in width the train of wheel
loads, for the present purpose, is obtained by dividing the standard axle loads by 13, and
also converting the loads into cubic feet of masonry, all as shown in Fig. 79o. The load
diagram must of course be drawn to the same horizontal scale as the influence lines, which
is not the case in the figures as here reproduced.

The wheel loads, in kips, are converted into cubic feet of masonry by multiplying
kips by 1000/140-7.14.

The maximum stresses at the crown section n are computed in Table 79ic. The dead
load moments are first found and these are combined with the live load moments which
latter are increased for impact. The maximum stresses are found from the total maximum
moments.

The live load diagram Fig. 79c, was placed over the influence line for section n, Fig.
70E, so that the first load came at the point 8, which is the zero point for positive ordinates.
One locomotive then covered the distance from point 8 to 16, and the remainder of the
span was considered empty. The same position was used for both M e and Mi moments,
because Eqs. (76c) apply only to simultaneous kernel moments.

UNIFORM J.OAO OP 0.461 kiM p.tT.

V $' 5' S'

FIG. 79c. Cooper's E 60 Loading for an Arch Ring One Foot Thick.

Comparing the result for max. f e = 369 cu.ft., obtained here with the value given
in Table 79j as 290 cu.ft., the relative effect of the two cases of loading is clearly seen.

Each of the critical sections should be examined in the manner shown in Table
79K, and the positive as well as negative influence line areas should be considered as sepa-
rate cases.

The temperature stresses are easily determined according to the method given in
Art. 75.

Referring back to the given data, E =3,670,000 Ibs. per sq.in. =3,755,720 cu.ft.
per sq.ft.; =0.000,005,4 per 1 F; =-80; Z = 158.8 ft; and cos /? = !, then etf =
-0.0686 ft., and by Eq. (75c)

dlE 0.0686 X3,775,720

where the value cos 2 /?t/TF c is taken from Table 79c for W c loads which are E times
actual.

The stresses on any section are then found by evaluating the kernel moments for
temperature and combining these with the kernel moments previously found for the case
of maximum loading.

346 KINETIC THEORY OF ENGINEERING STRUCTURES CHAP. XV

Thus, for the haunch section t, Table 79o gives D =9'. 75, y e = -37'. 1 and t/ t -= -35'.3
and maximum M e and Mi are taken from Table 79j as follows:

M et = -37.1 X 230.9= -8566.4
M e from Table 79 j- +6973.9

Total M e =-1592.5

.... , 6X1592.5 , inn ,.

giving f e = + , = + lOO.o cu.ft.

9.7o~

Also,

Mil= -35.3X230.9= -8150.8
Mi from Table 79 j = + 453.2

Total M t =-7697. 6

. . , 6X7697.6 ,. e ,,

giving/, = ., = 458.8 cu.ft.

9.75"

Similarly for the crown section n where Z>=6'.0, y e = ll'.OS and i/; = 13'.08, then for
the maximum moments from Table 79K

M et = 10.08 X 230.9 = + 2558.4
M e from Table 79K - + 2213.6

Total M e = +4772.0 or/ e = -795.3 cu.ft.,

and M it = 13.08 x 230.9 = + 3020.2

Mi from Table 79K = - 131.8

Total M;= +2888.4 or /= +481.4 cu.ft.

It is readily seen that these stresses are excessive and would undoubtedly cause
cracks on the tension elements occurring in the intrados at the crown and in the extrados
at the haunches.

(c) Analysis as under (b), treating the structure as unsymmetric. It is presumed
that for good and sufficient reasons the B abutment had to be founded on a lower stratum
of rock, a condition which developed after the original design for the symmetric arch
was accepted. The question to be answered here is whether such a change in the location
of a foundation would seriously alter the resulting stresses and necessitate a change in
the design.

The location of the (x, y) axes and computation of the elastic loads is given in Tables
79L and 79.M, following the method outlined in Art. 73, and using the same data as in the
previous case of the symmetric arch with the addition of two sections at the right-hand
end, producing the unsymmetric condition illustrated in Fig. 79u. .

The (z, v) axes are located through the axial points of the haunches, which is the
same position previously chosen for these axes in dealing with the symmetric condition.

The dimensions D, v and z are tabulated in Table 79L, and from these the values
1/7, sec (j>, w a , vw a and zw a are computed, using the formulae for sec < above given, p. 331.

ART. 79

FIXED MASONRY ARCHES

347

O O O C5

co TP co

+ I I I I I I I I I I I I I

B
tf

i + r I I I I J M I I I I

J5 N 'T* * .p "5

h-*. W I! *-H iO 00 C^ CM ^O 00 C^ CM CO t^ ^2 *"i CO I^ CO CM CO 00 *O C^ ^D GO ^O 01 i~H Cw CO **"* PH

*" f-H T i I-H I-H C^4 CM C^ C^ CO CO CO CM CM C^J C^l i~~i *~H I-H i^ CM M

t>. .^

!>. CO OXOOCOCOCOQOOOOO' i^QOO' iLOOOOOCOT-HOOr^OOO * "5

CC - C^^CCOCN' iiOiOC v J^ ( O l OQO'^t | OOC^lT-HCM <^ fOOC^QO'^O'~'Ot^- CO

O5 Ti* S

r 1 P

K fe^ OOOOOOOOOOOOOOOOOOOOOOOOOOO CD i "-

^ ^ PN

fa ^

^ ~ T ^ ^ ^ ^ ^ ~ ~ il, " ~ ~ ^-, ^. ^- .- ^ ^ ~ ^ N a

?) OOi-Hi-H'-t(N(MCOCOCOCO^Tt < TtiCOCOCOCOCMCM'-li-H' lOOOO

O g Or-HCOCOGO(M_COO^OOCOOOTt<COOOp^HCMr-(pl> p-HOOt--^H M C5

H 2. __^ SS S

S PI 1 c^^i^^^^ooot-Cicoooococnt-ooo^^LCic^^as-H T T ^

-j i"^ OOOOOOOOOOOOOOOOOOOOOOOOOOO

do'dddooodoooooooooooooooooo

CO I -

"- coooooOOTc^t^^^^c^doocM^^^t^cM'osboQGOcocD'-o

f^ II *j p p p p p c5 p p p p p p p p p p p p p p p p p p p ~

^ r-i i >~H o' o' o o o' o d d o d o o' o' o o' o o' d d o o o' d o o o o

o
p

2 CMCO^f^O^COCOcOt-t^-t^-t^-t t t- t^-t 5>cOCD'O l O^COCM> I

p ^

t-H

1 1

^~* OOOOOOO-^cO. OOOOOOOcOrPOOpOOOOOO

___^ co^f

O I-H (N CO -^ 1C CO t^'OO
PH

348

KINETIC THEORY OF ENGINEERING STRUCTURES

TABLE
COMPUTATION OF W FORCES AND POLE

Pt.

W a

XZW a

.V.

= xz W a

Eq. (73c),

?/ 2 2 ' ~\~ X T ' ill 3

XW a

x tan 3

2 2 '

ft.

0.0232

49.8

153.2

155.1

446.6

-4.22

-37.17

-41.39

1.885

0.0251

66.6

142.6

414.5

885.0

-3.91

-28.37

-32.28

1.892

0.0285

85.0

135.2

551.2

877.8

-3.57

-20.27

-23.84

1.963

0.0325

101.4

126.4

656.0

817.5

-3.24

-13.57

-16.81

2.027

0.0354

110.4

110.5

716.8

718.8

-2.90

- 7.77

-10.67

1.978

0.0394

118.3

96.0

765.2

622.3

-2.56

- 2.77

- 5.33

1.945

0.0429

119.7

78.8

774.2

511.5

-2.22

+ 1.53

- 0.69

1.839

0.0458

114.4

60.6

739.5

393 . 9

-1.89

5.13

+ 3.24

1.666

0.0477

102.0

42.6

661.0

278.3

-1.55

8.03

6.48

1.425

0.0499

86.1

27.3

556.8

178.8

-1.21

10.23

9.02

1.166

0.0513

65.2

14.6

450.6

102.8

-0.88

11.73

10.85

0.865

0.0530

38.0

4.6

283.2

38.0

-0.49

12.93

12.44

0.497

0.0555

8.0

0.2

51.2

5.2

-0.10

13.23

13.13

0.104

0.0530

- 22.8

1.7

-168.6

16.3

+ 0.29

12.93

13 . 22

-0.298

0.0513

- 50.7

8.8

-349.1

62.8

0.68

11.73

12.41

-0.674

0.0499

- 72.3

19.2

-467.3

126.4

1.02

10.23

11.25

-0.980

0.0477

- 89.2

32.6

-577.8

213.4

1.36

8.03

9.39

- 1 . 246

0.0458

-102.7

48.8

-663.6

317.5

1.69

5.13

6.82

-1.494

0.0429

-109.3

65.7

-706.6

426.6

2.03

1.53

3.56

- 1 . 679

0.0394

-109.3

82.0

-706.8

531.7

2.37

- 2.77

- 0.40

- 1 . 798

0.0354

-103.0

96.2

-668.5

625.8

2.71

- 7.77

- 5.06

-1.845

0.0325

- 95.3

111.7

-615.4

722.4

3.04

-13.57

-10.53

-1.906

0.0285

- 80.4

120.9

-521.2

784.9

3.38

-20.27

-16.89

- 1 . 856

0.0251

- 63.3

128.8

-388.0

782.1

3.72

-28.37

-24.65

- 1 . 798

0.0174

- 35.6

104.6

-214.3

621.6

4.02

-37.17

-33.15

-1.349

0.0129

- 16.7

89.8

- 99.6

544.1

4.33

-48.07

-44.74

-1.076

0.0131

0.0

104.5

- 8.2

301.0

4.64

-63.57

-58.93

-1.170

+ 620.3

11953.1

All elastic loads W

The sums ^w a = W a , ^vw a =vW a and Hzw a =zW a are now computed by Simpson's
rule according to Eqs. (73E), using the lengths Jv for the horizontal distances between
the arch sections.

Eqs. (73r) then give the coordinates l\, z ' of the new origin with respect to the
(v, z) axes. Thus

2vW a 545.7264

6.7149

=81.272 feet,
=63.574 feet,

and x=li v.

ART. 79

FIXED MASONRY ARCHES

349

79M

DISTANCES. UNSYMMETRTC ARCH

ft.

z^-r,

yw a

2 yWa We

y 2

cos <f>

2 cos <j>Jx

Point.

5.56

-0.960

- 2.76

1713.1

39.74

110.5

052

0.158

5.9

11.77

-0.810

- 5.04

1042.0

26.15

163 . 4

0.064

0.394

6.5

12.77

-0.679

- 4.43

568.3

16.20

106 8

0.077

0.500

6.5

13.13

-0.546

- 3.54

282 6

9.18

60.6

090

0.583

6.5

12.88

-0.378

- 2.47

113.8

4.03

27.4

0.101

0.657

6.5

12.61

-0.210

- 1.36

28.4

1.12

8.3

0.113

0.732

6.5

11.92

-0.030

- 0.13

0.5

0.02

1.0

0.123

0.799

6.5

10.80

+ 0.148

+ 0.95

10.5

0.48

3.7

0.135

0.872

6.5

9.26

0.309

1.99

42.0

2 00

13.3

0.144

0.930

6.5

7.56

0.450

2.90

81.4

4.06

26.3

0.153

0.987

6.5

5.99

0.557

3.89

117.7

6.04

42.6

0.158

1.110

7.5

3.71

0.659

4.91

154.8

8.20

61.0

0.163

1.225

7 .5

0.73

0.729

5.39

172.4

9.57

70.7

0.167

1.250

7.5

- 2.22

0.701

5.22

174.8

9.26

68.8

0.163

1.225

7.5

4.65

0.637

4.45

154.0

7.90

55.3

0.158

1.110

6.5

6.35

0.561

3.62

126.6

6.32

40.8

0.153

0.987

6.5

- 8.08

0.448

2.89

88.2

4.21

27.4

0.144

0.930

6.5

- 9.66

0.312

2.01

46.5

2.13

- 14.1

0.135

0.872

6.5

- 10.86

0.153

0.99

12.7

0.54

4.1

0.123

0.799

6.5

- 11.63

-0.016

- 0.08

0.2

0.01

0.8

0.113

0.732

6.5

- 11.98

-0.179

- 1.16

25.6

0.91

6.9

0.101

0.657

6.5

- 12.31

-0.342

- 2.21

110.9

3.60

24.4

0.090

0.583

6.5

- 12.05

-0.481

- 3.14

285.3

8.13

54.2

0.077

0.500

6.5

- 10.94

-0.619

- 3.75

607.6

15.25

92.6

0.064

0.393

5.9

8.03

-0.577

- 3.44

1098.9

19.12

114.2

0.047

0.275

5.9

6.53

-0.577

-3.52

2001 . 7

25.82

158.7

0.032

0.187

5.9

3.40

-0.772

- 2.18

3472.7

45.49

124.5

0.020

0.064

+ 118.69

+ 39.21

1482.4

19.511

-118.69

-39.21

.#:,*.

= 1480.6

are E times actual.

When the ordinates x are found then the quantities xzw a and x 2 w a and their sums
xzW a and xWi, are determined in Table 79M, using Simpson's rule in summing for the
horizontal increments Ax.

Eq. (73n) then gives

620.3

tan 8 =

11953.1

= -0.0519.

Eq. (73o) now furnishes the ordinates y and from these the quantities yw a and
y 2 w a are computed. As a check STF&=0 and SW C =0, and any small discrepancies
must be adjusted.

350

KINETIC THEORY OF ENGINEERING STRUCTURES CHAP. XV

ART. 79

FIXED MASONRY ARCHES

351

The W loads were again derived without including E so that these loads are all E
times actual, a fact which should not be overlooked if temperature stresses are to be
found from Eq. (75c).

In Table 79M the quantities cos (f>dx/D are given, showing that the pole distance
H c =cos 52?/TF c + 2cos (f>Ax/D from Eq. (72ic), would be increased by 1.3 per cent if axial
thrust is included in the determination of X c . This would tend to diminish X c , and the
ordinates of the X c influence line.

The influence lines for X a , X b and X c are shown in Fig. 79i, noting that the length of
span /, is now taken as the horizontal distance between the intrados kernel points of the
haunch sections. This was done to obviate the negative M moments otherwise encoun-
tered here and leading to complications which are rather far reaching. In the previous
case of symmetric arch, the real haunch sections could not be examined becuase the span
was chosen as the distance between the verticals through the axial points and the sec-
tion t was accordingly taken as the nearest approach to the haunch section. It is thus
interesting to note the two methods of treating this matter. The length does not
enter into the computations, but must be considered in drawing all the influence
lines.

The detailed description of the method of drawing these influence lines need riot
be repeated here, and the reader is referred back to the treatment of the symmetric arch
in the first part of this article.

Axial thrust was not included in Fig. 79i, thus making all the results of both investiga-
tions comparable. However, the effect is so small that it may be regarded as a negligible
quantity in the present case.

Having located the coordinate axes, this determines the critical sections, and the data
required for the computation of the ordinates for the M e and Mi influence lines are given
in Table 79N.

TABLE 79N. DATA RELATING TO THE CRITICAL SECTIONS UNSYMMETRIC ARCH

Coordinates,

Coordinates Extrados Kernel

Coordinates Intrados Kernel

Axial Point.

Points.

T> '

D .

D ,

g sm <j>

g COS $

x e

cos @y e

COS &/j

ft.

10.0

58 51'

+ 81.27

-41.39

1.43

0.86

+ 79 . 85

-42.25

-42.20

+ 82.70

-40.53

-40.48

6.9

31 04

+ 41.20

0.00

0.59

1.00

+ 40.61

- 1.00

+ 41.79

+ 1.00

6.0

0.00

+ 13.25

0.00

1.00

0.00

+ 12.25

+ 12.23

0.00

+ 14.25

+ 14.23

7.2

34 58

-43.90

0.00

0.69

0.98

-43.21

- 0.98

-44.59

+ 0.98

14.7

73 11

-89.33

-58.93

2.35

0.71

-86.98

-59.64

-59.56

-91.68

-58.22

-58.14

/?=2 58' and cos = 0.9987 from Table 79L

The M influence lines for the kernel points of the several critical sections are shown
in Fig. 79J, the data for the construction of these being taken from Table 79x. It should
be noted that for the haunch sections A and B, M ot -=0.

352

KINETIC THEORY OF ENGINEERING STRUCTURES CHAP. XV

1=174138

FIG. 79i. X Influence Lines for Unsymmetric Arch

ART. 79

FIXED MASONRY ARCHES

353

From the coordinates of the kernel points in Table 79N, and the ordinates in Tables
79o and 79p, taken from the X a , X b , X c and M influence lines, Figs. 79i and 79J, the T) e
and j)i are computed for the odd points and for the five critical sections according to Eqs.
(76B). The operations are all indicated in the headings of Tables 79o and 79p, thus
requiring no further description here.

These ordinates, plotted in Fig. 79K, furnish the M e and M{ influence lines from
which the moments and stresses for any position of the live loads may be found for each
of the five critical sections. In practice it is well to compute all the ordinates TJ
instead of merely the odd ones as given here.

FOR INTRADOS KERNEL POINTS.

FORCXTRAOOS KERNEL POINTS,

FIG. 79j. M Influence Lines for Unsymmetric Arch.

Table 79o, gives the computations for the resultant polygon for the same case of
loading previously used in Table 79n, adding one additional load P^ 5 at a point on the
right abutment. The results of this computation are shown in Fig. 79n, where the resultant
polygon is drawn through the points a and b which are located on the verticals through
the intrados kernel points of the A and B sections.

It should be noted that c and c are both negative, indicating that they must be meas-
ured up from the line ah when found. Hence z is measured down on the y axis to
locate the point s and c =ss' is then laid off upward from s. The line as'||z
axis, locates the point a at the left abutment and the line as prolonged fixes the point 6
at the right abutment. The force polygon is therl easily constructed by laying off
the reactions A and B and finding the pole 0\ by drawing the line 0\T \\ ab and equal in
length to H',

354

KINETIC THEORY OF ENGINEERING STRUCTURES CHAP. XV

TABLE 79o. MOMENT INFLUENCE LINES FOR

Point.

ft.

fib
ft.

rye
ft.

SECTION A.

r> A = r> - [ij a + 79.857/6 - 42.27j c ]

SECTION TO.
rjm = ijo [i)a + 40.6j?& l.Oqc]

r lo

79.85 t)b

-42.2r,c

T>0

+ 40.67,6

T)m

3.5

0.036

0.012

2.74

2.87

- 0.51

- 3.12

5.2

1.46

0.2t>

9.8

0.082

0.100

2.52

6.55

- 4.22

- 9.61

15.2

3.33

2.17

15.2

0.102

0.252

2.31

8.14

-10.63

-10.40

25.1

4.14

6.01

19.7

0.092

0.424

2.10

7.35

-17.89

- 7.06

31.0

3.74

7.98

22.9

0.064

0.580

1.88

5.11

-24.48

- 1.65

28.0

2.60

3.08

25.0

0.014

0.692

1.64

1.12

-29.20

+ 4.72

24.5

0.57

-0.38

25.5

-0.044

0.720

1.40

- 3.51

-30.38

9.79

21.0

-1.79

-1.99

24.5

-0.100

0.652

1.18

- 7.99

-27.51

12.18

17.4

-4.06

-2.39

22.1

-0.132

0.520

0.97

-10.54

-21.94

11.35

14.2

-5.36

-2.02

18.7

-0.148

0.360

0.75

-11.62

-15.19

8.86

11.1

-6.01

-1.23

14.3

-0.136

0.196

0.53

-10.86

- 8.27

5.36

8.0

-5.52

-0.58

9.1

-0.100

0.070

0.32

- 7.99

- 2.95

2.16

4.8

-4.06

-0.17

3.9

-0.046

0.016

0.13

- 3.67

- 0.68

0.58

2.0

-1.87

-0.01

All ordinates in above

TABLE 79p. MOMENT INFLUENCE LINES FOR

Point.

T/a
ft.

>?&
ft.

T)c
ft.

SECTION A.
^ = 0-[7j a +82.77j6-40.5rjc]

SECTION m.

fjm = rjo [f)a + 41 .87J6 + 1.0)jf]

S2.7rj b

-40.57y c

r >A

1)0

41.87/6

l.OlJc

TJm

3.5

0.036

0.012

2.98

- 0.49

- 5.99

5.4

1.50

0.01

0.39

9.8

0.082

0.100

6.78

- 4.05

-12.53

15.4

3.43

0.10

2.07

15.2

0.102

0.252

8.44

-10.21

-13.43

25.2

4.26

0.25

5.49

19.7

0.092

0.424

7.61

-17.17

-10.14

30.0

3.85

0.42

6.03

22.9

0.064

0.580

5.29

-23.49

- 4.70

27.0

2.68

0.58

0.84

25.0

0.014

0.692

1.16

-28.03

+ 1.87

23.8

0.59

0.69

-2.48

25.5

-0.044

0.720

- 3.64

-29.16

7.30

20.2

-1.84

0.72

-4.18

24.5

-0.100

0.652

- 8.27

-26.41

10.18

17.0

-4.18

0.65

-3.97

22.1

-0.132

0.520

-10.92

-21.06

9.88

14.0

-5.52

0.52

-3.10

18.7

-0.148

0.360

-12.24

-14.58

8.12

11.0

-6.18

0.36

-1.88

14.3

-0.136

0.196

-11.25

- 7.94

4.89

7.8

-5.68

0.20

-1.02

9.1

-0.100

0.070

- 8.27

- 2.84

2.01

4.8

-4.18

0.07

-0.19

3.9

-0.046

0.016

- 3.80

- 0.65

0.55

2.0

-1.92

0.02

0.00

All ordinates in above

It is seen that the resultant polygon intersects the axial line four times, showing con-
sistency with Professor Winkler's theorem.

As a final check, the normal thrusts TV, offsets v and stresses /are computed in Table
79n for the same case of loading; used in Table 79Q, and the close agreement of the results
shows both solutions to be satisfactory. The only stresses, however, which are maximum,
are those on the section A, the others are simply for the simultaneous case of loading
and have no special interest.

The stresses on the haunch section A are thus found to be/ e = 549 and/ t -= +152.9

ART. 79

FIXED MASONRY ARCHES

355

EXTRADOS KERNEL POINTS UNSYMMETRIC ARCH

SECTION n.

, = ,- [, a +12.23,c]

SECTION m'.

SECTION B.

Point.

r/o

12.23 7j c

r/n

-43.2,6

-.098, c

T)m

fjo

-87.0,6

-59.56, c

r ,B

3.7

0.14

+ 0.05

1.9

-1.56

-0.01

-0.03

0.18

- 3.13

- 0.71

0.52

10.6

1.22

-0.42

5.5

-3.54

-0.10

-0.66

0.54

- 7.13

- 5.96

3.83

17.2

3.08

-1.08

9.1

-4.41

-0.25

-1.44

0.89

- 8.88

-15.01

9.58

24.2

5.19

-0.69

13.0

-3.97

-0.41

-2.32

1.24

- 8.00

-25.25

14.79

31.1

7.10

+ 1.10

16.4

-2.76

-0.57

-3.17

1.59

- 5.57

-34.54

18.80

38.5

8.46

5.04

20.3

-0.60

-0.68

-3.42

1.99

- 1.22

-41.22

19.43

40.8

8.80

6.50

24.6

+ 1.90

-0.71

-2.09

2.38

+ 3.83

-42.88

15.93

34.2

7.97

+ 1.73

28.4

4.32

-0.64

+ 0.22

2.76

8.70

-38.83

8.39

28.1

6.36

-0.36

32.0

5.70

-0.51

4.71

3.10

11.48

-30.97

0.49

22.0

4.40

-1.10

33.4

6.39

-0.35

8.66

3.47

12.88

-21.44

- 6.67

15.6

2.40

-1.10

23.7

5.88

-0.20

3.72

3.82

11.83

-11.67

-10.64

9.5

0.85

-0.45

14.4

4.32

-0.07

+ 1.05

4.17

8.70

- 4.17

- 9.46

4.0

0.20

-0.10

5.8

1.99

-0.02

-0.07

4.50

4.00

- 0.95

- 2.45

table are expressed in feet.

INTRADOS KERNEL POINTS UNSYMMETRIC ARCH

SECTION n.
'Jn= >?o [)Ja+ 14.23^ c ]

SECTION m'.
5}m /==) ?o -[Jjo 44.6i;6 + 0.987j c ]

SECTION B.
?B = 0-[?-9l.7ij6-58.14ij c ]

Point.

rio

14.23ij c

Tjn

-44.67?6

0.98)? c

Tim

-91.7>j6

-58.14/jc

3.7

0.17

+ 0.03

1.8

-1.61

0.01

-0.09

-3.30

- 0.70

+ 0.50

10.6

1.42

-0.62

5.4

-3.66

0.10

-0.84

-7.52

- 5.81

3.53

17.2

3.58

-1.58

8.9

-4.55

0.25

-2.00

-9.35

-14.65

8.80

24.2

6.03

- 1 . 53

12.4

-4.10

0.41

-3.61

-8.44

-24.65

13.39

31.1

8.25

-0.05

16.0

-2.85

0.57

-4.62

-5.87

-33.72

16.69

38.5

9.85

+ 3.65

19.6

-0.62

0.68

-5.46

-1.28

-40.23

16.51

40.8

10.24

5.06

23.8

+ 1.96

0.71

-4.37

+ 4.03

-41.86

12.33

34.2

9.28

0.42

27.8

4.46

0.64

-1.80

9.17

-37.91

4.24

28.1

7.40

-1.40

31.2

5.88

0.51

+ 2.71

12.10

-30.23

- 3.97

22.0

5.12

-1.82

33.4

6.60

0.35

7.75

13.57

-20.93

-11.34

15.6

2.84

-1.54

24.2

6.06

0.20

3.64

12.47

-11.40

-15.37

9.5

0.99

-0.59

14.6

4.46

0.07

0.97

9.17

- 4.07

-14.20

4.0

0.23

-0.13

6.0

2.05

0.02

0.03

4.22

- 0.93

- 7.19

table are expressed in feet.

cu.ft., which are quite excessive as compared to those previously obtained for section
t of the symmetric arch, where f e = 440 and J\ = +28.5 cu.ft.

If the investigation of stresses were extended to all the critical sections it would
be found that the dimensions found safe for the symmetric arch are now quite insufficient
when the arch is treated as an unsymmetric structure using the same axial line as before.
The shape of the ring might be so adjusted as to reduce the stresses within the required
limits by computing a resultant polygon for the average loading Q+%P, and shifting
the axial line by the amounts v obtained for the several critical sections.

356

KINETIC THEORY OF ENGINEERING STRUCTURES CHAP. XV

^^rrr^r^ SECTION n. ^r^rrri'

rf rl'"-T T^jNa ip n la is K i5/fcJ"'T T

m 5*7 8S X ] T / T>I7 18 19 ;

intrados kernel points. \

extrados ke

rnel points. \ //uml
\ / '~~^

/ LENGTHS.

/ l...iliml I I I I l|

/ 10 O |O ZO 3O 4O SO PT.

OROINATCS.

III!

SFT.

v'
FIG. 79K. M Influence Lines for the Kernel Points. Unsymmetric Arch.

ART. 79

FIXED MASONRY ARCHES

357

TABLE 79q.
THE RESULTANT POLYGON FOR LOADS Q + P UNSYMMETRIC ARCH

Dead

Live

Total

r,(Q + P) for

Moments about a

Point.

r ia

r, c

Loads.

x a

x b

X c

r(Q+P)'

ft.

cu.ft.

ft.

3.5

0.036

0.012

367.4

0.0

367.4

1285.9

13.23

4.40

5.9

2167.7

9.8

0.082

0.100

276.1

0.0

276.1

2705.8

22.64

27.61

18.9

5218.3

15.2

0.102

0.252

223 .4

0.0

223.4

3395.7

22.82

56.30

31.9

7126.5

19.7

0.092

0.424

182.7

0.0

182.7

3599.2

16.81

77.46

44.9

8203.2

22.9

0.064

0.580

168.0

0.0

168.0

3847.2

10.75

97.44

57.9

9727.2

25.0

0.014

0.692

173.7

114.6

288.3

7207 . 5

4.03

199 . 50

71.9

20728.8

25.5

-0.044

0.720

173.7

114.6

288.3

7344.8

-12.68

207.59

86.9

25053.3

24.5

-0.100

0.652

168.0

99.3

267.3

6548.9

-26.73

174.28

100.9

26970.6

22.1

-0.132

0.520

182.7

99.3

282.0

6232.2

-37.22

146.64

113.9

32119.8

18.7

-0.148

0.360

223.4

99.3

322.7

6034.5

-47.76

116.17

126.9

40950.6

14.3

-0.136

0.196

276.1

99.3

375.4

5368.2

-51.05

73.58

139.9

52518.5

9.1

-0.100

0.070

367.4

99.3

466.7

4247.0

-46.67

32.67

152.9

71358.4

3.5

-0.040

0.012

680.0

84.0

764.0

2674.0

-30.56

9.17

165.9

126747.6

4272.3

60490.9

-162.39

1222.81

428890.5

= #

= X a

=x b

=X C

= M

Z = 174'.38, r, = 82'.7, /
Then H = X c cos /?= 1221.22 cu.ft,

H' = = 1242 cu.ft.

cos a

c = -X&=-23.19ft.
n

r = 7r= 100.388 ft.
ti

tan 13= -0.0519, cos/3 = 0.9987

tan a = tan

-77

= -0.1849

= = 49.53 ft.

-^Xi-- 11.00 ft.

B =

=2494.6 cu.ft.

A = R B = 1777 7 cu.ft.

The investigation for temperature is not repeated here except to evaluate the moment
Eq. (75c) for this case using the same data just given for the symmetric arch.

For cos,/? =0.9987 span 1 = 174.4 ft., and 2t/ W c = 1482.6 from Table 79M, then

etl = -0.0752 ft. and
_0.0752X3,775,720

from which the temperature stresses for any section are readily obtained by computing
the moments about the two kernel points of the section.

A complete investigation of the stresses in aiiy arch would then consist in finding
the kernel moments due to the dead loads, the kernel moments due to the train of con-
centrated live loads, Fig. 79o, making proper allowance for impact; and finally the kernel
moments due to temperature effects; all for each of the five critical sections. These
moments for the same section are algebraically combined into the total M e and M t - moments
and from them and Eqs. (76c) the maximum stresses are found.

358

KINETIC THEORY OF ENGINEERING STRUCTURES

CHAT. XV

TABLE 79E

COMPUTATION OF MOMENTS AND STRESSES ON THE CRITICAL SECTIONS
FOR LOADS Q+P UNSYMMETRIC ARCH

SECTION A.

SECTION m.

SECTION n.

Point.

Loads

Q + P

fie

i?e

M e

cu.ft.

ft.

367.4

- 3.12

- 5.99

-1146.3

-2200.7

0.25

0.39

91.9

143.3

+ 05

+ 0.03

+ 18.4

+ 11.0

276.1

- 9.61

-12.53

-2653.3

-3459.5

2.17

2.07

599.1

671.5

-0.42

-0.62

- 116.0

- 171.2

223.4

-10.40

-13.43

-2323.4

-3000.3

6.01

5.49

1342.6

1226.5

-1.08

-1.52

- 241.3

- 339.6

182.7

- 7.06

-10.14

-1289.9

-1852.6

7.98

6.03

1457.9

1101.7

^0.69

-1.53

- 125.1

- 279.5

168.0

- 1.65

-4.70

- 277.2

- 789.6

3.08

0.84

517.4

141.1

4-1.10

-0.05

+ 184.8

- 8.4

288.3

+ 4.72

+ 1.87

+ 1360.8

+ 539.1

-0.38

-2.48

- 109.5

- 715.0

5.04

+ 3.65

1453.0

+ 1052.3

288.3

9.79

7.30

2822 . 4

2104.6

-1.99

-4.18

- 573.7

-1205.1

6.50

5.06

1874 . 1

1458. S

267.3

12.18

10.18

3255.7

2721 . 1

-2.39

-3.97

- 638.8

-1051.2

-1-1.73

-1-0.42

-1- 462.4

+ 112.3

282.0

11.35

9.88

3200 . 7

2786.2

-2.02

3.10

- 569.6

- 874.2

-0.36

-1.40

- 101.5

394.8

322.7

8.86

8.12

2859 . 1

2620 . 3

-1.23

-1.88

- 396.9

- 606.7

-1.10

-1.82

355.0

587.3

375.4

5.36

4.89

2012.1

1835.7

-0.58

-1.02

- 217.7

- 382.9

-1.10

-1.54

- 412.9

- 578.1

466.7

2.16

2.01

1008 . 1

938.1

-0.17

-0.19

- 79.4

- 88.7

-0.45

-0.59

- 210.0

- 275.4

25*

764.0

0.42

0.40

320.9

305.6

-0.01

0.00

7.6

- 0.0

-0.08

-0.11

- 61.1

- 84.0

Totals

4272.3

10'.

+ 9149.7

+ 2548.0

6'.90

+ 1415.7

-1739.7

6'. 00

+ 2369.8

- S3.S

1980.5

1372.0

A" =

1226.9

v =

+ 2'.953

v =

-OM19

v =

-f 0'.932

- 549.0

+ 152.9

- 178.4

- 219.7

- 395.0

- 14.0

SECTION in'.

SECTION B.

Point

Loads.

Q + P

T ie

M e

tfi

cu ft.

ft.

367.4

-0.03

-0.09

11.0

- 33.0

0.52

0.50

191.0

183.7

276.1

-0.66

-0.84

- 182.2

- 231.9

3.83

3.53

1057 . 5

974.6

223.4

-1.44

-2.00

- 321.7

- 446.8

9.58

8.80

2140.2

1965.9

182.7

-2.32

-3.61

- 323.9

- 659.5

14.79

13.39

2702.0

2446.4

168.0

-3.17

-4.62

- 532.6

- 776.2

18.80

16.69

3158.4

2803.9

288.3

-3.42

-5.46

- 986.0

- 1574 . 1

19.43

16.51

5601.7

4759.8

288.3

-2.09

-4.37

- 602.5

-1259.9

15.93

12.33

4592.6

3554.7

267.3

+ 0.22

-1.80

+ 58.7

- 481.1

8:39

4.24

2242 . 6

1133.4

282.0

4.71

+ 2.71

1328.2

+ 764.2

0.49

- 3.97

138.2

-1119.5

322.7

8.66

7.75

2794.6

2500.9

- 6.67

-11.34

-2152.4

-3659.4

375.4

3.72

3.64

1396.5

1365.4

-10.64

-15.37

-3994.3

-5769.9

466.7

+ 1.05

0.97

+ 490.1

452.7

- 9.46

-14.20

-4415.0

- 6627 . 1

25i

764.0

-0.05

0.02

- 38. 2

15.3

- 1.76

- 6.80

-1344.6

-5195.2

Totals

4272.3

D =

7'. 20

+ 3070.0

- 374.0

D =

14'. 7

+ 9917.9

-4548.7

1435.0

;V =

2952.4

+ 940

+ 0' . 909

- 355.3

43.3

- 274.7

- 126. C

Eqs. (76c,

-Jj {

D 2

v is positive when measured

from the axis toward the extrados. All loads in above table are expressed in cubic feet of masonry
and the stresses are cubic feet per square foot. Cubic feet per square foot X 0.972 = pounds per square
inch.

BIBLIOGRAPHY

THEORY OF ELASTICITY

HOOKE. Philosophical tracts and collections, 1678. Hooke's law of proportionality between

stress and strain.

BERNOULLI, 1667-1748, was the first to give an analytic solution for the elastic curve.
EULER. Methodus invemendi hneas curvas, 1741-4.
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LAGRANGE. Mecamque analytique, Paris, 1788.
YOUNG. A course of lectures on natural philosophy and mechanical arts, 1807 Gives Young 3

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POISSON. Memoire de 1'Academie des Sciences, VIII, 1829.
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LAME. Lemons sur la theorie mathematique de 1'elasticite, 1852.
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CASTIGLIANO. Theorie de 1'equilibre des systemes elastiques, 1879.
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GRAPHOSTATICS

PONCELET. Cours de Mecanique industrielle, 1826-9.
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CLERK MAXWELL. On reciprocal figures and diagrams of forces. Phil. Mag., 1864.
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Boehmen, 1868. Introduces influence lines.

360 BIBLIOGRAPHY

MOHR. Beitrag zur Theorie der Holz und Eisenkonstruktionen. Zeitschr. des Arch.- und Ing.

Vereins, Hannover, 1868. Introduces influence lines.
CREMONA. Le figure reciproche nella Statica Grafica, 1872.
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MOHR. Ueber Geschwindigkeitsplane und Beschleunigungsplane. Civiling., 1887.
W. HITTER. Anwendungen der graphischen Statik, 1890-1906.

GENERAL TREATISES

SCHWEDLER. Theorie der Brueckenbalkensysteme. Zeitschr. fuer Bauwesen, 1851.
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1870, 1874, 1875, 1877.

WEYRAUCH. Theorie und Berechnung der Kontinuierlichen und einfachen Traeger, 1873.
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Formaenderung von Fachwerken. Civiling., 1875.
FRAENKEL. Ueber die unguenstigste Einstellung eines Systemes von Einzellasten auf Fachwerk-

traeger mit Hilfe von Influenzkurven. Civiling., 1876.
FOEPPL. Theorie des Fachwerks, 1880.

MUELLER-BRESLAU. Graphische Statik der Baukonstruktionen. 1881-1908.
SWAIN. On the application of the principle of virtual velocities to the determination of the deflec-
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MELAN. Beitrag zur Berechung statischunbestimmter Stabsysteme. Zeitschr., des Oestr. Arch.

und Ing. Vereins, 1884.
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1887.
LAND. Beitrag zur Ermittelung der Biegungslinien ebener und elastischer Stabwerke. Civiling.,

1889.

L. v. TETMAJER. Angewandte Elastizitaets und Festigkeitslehre. 1888, 1902.
MUELLER-BRESLAU. Festigkeitslehre, 1893.
LANDSBERG. Beitrag zur Theorie des raemulichen Fachwerks. Zentr. der Bauverwaltung. 1898

and 1903.

MEHRTENS. Statik der Baukonstruktionen. 3 vol., 1903-5.
MOHR. Abhandlungen aus dem Gebiete der technischen Mechanik. 1905.
KECK-HOTOPP. Elastizitaetslehre, 1905.
Handbuch der Ingenieurwissenschaften. Der Brueckenbau.

SECONDARY AND ADDITIONAL STRESSES

ENGESSER. Ueber die Druchbiegung von Fachwerktraegern und die hierbei auftretenden zusaetz-

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ASIMONT. Hauptspannung und Sekundaerspannung. Zeitschr. fuer Baukunde, 1880.
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BIBLIOGRAPHY 361

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LANUSBERG. Beitrag zur Theorie des Fachwerks. Zeitschr. des Arch.- und Ing.-Vereins, Han-
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HIROI. Statically Indeterminate Bridge Stresses, 1905.

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GRIMM. Secondary Stresses in Bridge Trusses, 1908.

SPECIAL TREATISES ON ARCHES

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FRAENKEL. Berechnung eiserner Bogenbruecken. Civiling., 1867, 1875.

MOHR. Beitrag zur Theorie der elastischen Bogentraeger. Zeitschr. des Arch.- und Ing.-Ver..
Hannover, 1870, 1874, 1881.

WINKLER. Beitrag zur Theorie der elastischen Bogentraeger. Zeitschr. des Arch.- und Ing.-Ver.,
Hannover, 1879.

WEYRAUCH. Theorie der elastischen Bogentraeger. Zeitschr. fuer Baukunde, 1878.

ENGESSER. Theorie und Berechnung der Bogenfachwerktraeger, 1880.

MUELLER-BRESLAU. Theorie und Berechnung der eiserner Bogenbruecken, 1880.

W. RITTER. Der elastische Bogen berechnet mit Hilfe der graphischen Statik, 1886.

MUELLER-BRESLAU. Graphische Statik der Baukonstruktionen, 1892, 1907.

WEYRAUCH. Die elastischen Bogentraeger, 1897.

TOLKMITT. Leitfaden fuer das Entwerfen gewoelbter Bruecken, 1895.

MEHRTENS. Statik der Baukonstruktionen, 3 vols., 1903-5.

Handbuch der Ingenieurwissenschaften, Vol. I, 1904.

INDEX

Abutment displacements, fixed framed arches . 188
for general case of re-
dundancy 208

in two-hinged arches,

156, 165, 168

stresses due to 32

Additional stresses due to dynamic influences . 256

Arches, bibliography of special treatises on 361

fixed framed 178, 197

fixed solid web or masonry 298

three-hinged 72,. 78

two-hinged 153, 166, 176

without hinges 178, 298

Beams, deflection of 117, 118, 123, 125, 126

work of deformation for 41

Betti's law 28

Bibliography of treatises, etc 359

Cantilever bridges by influence lines 67

Castigliano's law, derivative of work equation . 30, 31

Centrifugal force due to curved track 260

Changes in the angles of a triangle due to

stress 107

Choice of the redundant conditions 202

Clapeyron's law 2, 14

Column formulae 270

Combination of stresses as a basis for design-
ing 268

Continuous girder over four supports 130

three supports. 129, 143, 148
Critical loading for max. moments, direct

loading 59

Critical loading for max. moments, indirect

loading 60

Critical loading for max. shear, direct loading. . 62

, indirect loading 62

PAGE

Dead-load stresses in determinate structures . . 210
Definitions of terms used throughout this work . vii

Deflection influence lines 49

of a solid w r eb beam 117

polygons according to Prof. Laud . . 107
polygons according to Prof. Mohr,

100, 104
polygons for determinate structures. 99

polygons for the loaded chord 104

Deformations 11

Deformation, work due to 1, 29

Derivative of the work equation; Castigliano's

law... .\ 30, 31

Designing, combination of stresses as a basis

for 268

Determinate structures, dead load stresses in .. 210
displacements of

points 18

displacement influence

lines for 122

influence lines for. 46, 65

stresses in 210, 218

Direct and indirect loading 47, 48, 50

Displacement influence lines, determinate

structures 122

for a cantilever. . 124
for a simple beam
ortruss.123,125,126

Displacements, horizontal 114

of points for determinate struc-
tures 18

Distortion of a statically determinate frame by

graphics.

Distortions due to changes in the lengths of

members 87

Double intersection trusses by influence lines . . 65

Dynamic impact 262

work of deformation due to. ... 43

influences ' 256

363

364

INDEX

PAGE

Eccentric connections, stresses produced by. . . 255

Effects due to unusual loads 266

Effect of shop lengths on determinate structures 256
indeterminate struc-
tures 208

Elastic deformations 11

Elastic loads w in terms of angle changes. 109, 111

Equation of an influence line 47

Equations, solution of simultaneous 295

Externally indeterminate, definition of 6

External redundant, influence line for one . 128, 140

Fatigue of material 265

Features in design intended to diminish secon-
dary stresses 267

Fixed framed arches 178

effect of abutment dis-
placements 188

example 189

influence lines for the re-

dundants 184

location of the coordinate

axes 182

resultant polygon for .... 187
stress influence lines for.. 185
temperature stresses in . . 188

Fixed masonry arches 298

coordinate axes and
elastic loads, 314, 331, 346

critical sections 322

determination of the

redundants 309

example of symmetric. . 326
example of unsymmet-

ric 346

influence lines for the

redundants. 317, 333, 351
maximum stresses,

322, 344, 357
modern methods of con-
struction 301

preliminary design for,

304, 326
resultant polygons,

324, 335, 357
stresses on any normal

section 320, 344

temperature stresses in,

318, 345, 357

Framed structures, theorems, laws and formulae
for 11

General case of redundancy 203

Girder continuous over four supports 130

three supports. 129, 143, 148
fixed at one end, supported at the other . 140

Glossary of terms vii

Graphostatics, bibliography 359

H
Horizontal displacements 114

Impact due to dynamic effects ...............

formulae ...........................

Indeterminate, externally ...................

internally ....................

Indeterminate straight beams, work of defor-
mation ....................

structures by Maxwell's law. ...

by Mohr's work equa-
tion ..............

effect of abutment
displacements. 32,
effect of shop lengths
effect of temperature
influence lines for,

127,

influence line for one
redundant .......

stress influence lines
for ...... 137, 139,

Influence line, equation for ..................

defined .......................

for cantilever bridges ..........

deflections ................

determinate structures,

46, 65,
direct loading ..............

displacements of determinate
structures ...............

double intersection trusses ...
indeterminate structures. 127,
indirect loading ............

moments ................ 49

one external redundant. 128,
one internal redundant .....

one redundant condition. . . .

shear ................... 43

262

264

41
24

208
208
206

140
127

140

47
46
67
49

218

122
65
140
50
, 50
140
128
127
, 50

INDEX

365

PAGE I

Influence line for skew plate girder on curved

double track 80

stresses in truss members ... 52

reactions 48, 50

redundants in fixed masonry

arches 317, 333

three-hinged arches 72

three-hinged masonry arches 78
three-hinged solid web

arches 78

trusses with subdivided

panels 83

two redundant conditions. . . 130
Internal redundant, influence lines for one .... 128

Internally indeterminate 6, 7

Introduction 1

Isotropic bodies, theorems, laws and formulae
for 34

Kernel moment influence areas for three-
hinged solid web arch 79

Kernel moment influence areas for two-hinged
solid web arches 159

Lateral trusses, stresses due to wind, etc . . 258, 261

vibrations 260

Least work, theorem of 29

Live load stresses 218, 219

author's method 219

Load divide for a truss 51

Loading, direct and indirect . 47, 48, 50

Masonry arches, see under Arches.

Maxwell's theorem of reciprocal displacements . 27

Menabrea's law of least work 29, 31

Mitering lock gates 271

example 279

theory of the analysis .... 273

Mohr's rotation diagram 89

work equations 16, 17

Moment influence lines 49, 50

Moving load stresses 218, 219

Multiple redundancy 132, 137

simplification of influence
lines for 132, 135

N
Nature of secondary stresses 226

Plate girder continuous over four supports .... 130
over three supports,

129, 143, 148

deflection of 117

fixed at one end and supported at

the other 140

Positions of a moving load for max. moments. . . 57

for max. shears 62

Preface v

Principle of virtual velocities 2, 16

work 1, 15

Reaction conditions 5

influence lines 48, 50-

summation influence lines 54

Redundancy, simplification of influence lines

for 132, 135

solution of the general case of ... 203-
stress influence lines for struc-
tures involving same . 137,139,140

Redundant conditions. 5

influence lines for one ... 128
influence lines for two . . . 130

on the choice of 202

Ritter's method of moments 211

Rotation diagram, Mohr's 89

of a rigid frame about a fixed point. . 89

Secondary and additional stresses, bibliog-
raphy 360

Secondary stresses 226

as effected by certain de-
signs 267

concluding remarks on 267

due to riveted connections,

227, 235

due to various causes 251

in members due to eccentric

connections 255

in members due to fheir own

weight 251

in pin-connected structures. 253
in riveted cross-frames, load

effects 244

in riveted cross-frames, wind

effects 247

the nature of 226

Shear influence lines 49, 50

Shearing stress, work of deformation due to. . 38

366

INDEX

Simplification of influence lines for multiple

redundancy 132, 135

Skew plate girder on curved double track by

influence lines 80

Solid web beam, deflection of 117

Solution of simultaneous equations 295

Special applications of influence lines to inde-
terminate structures 140

Statically determinate frame, distortion of by

graphics 87

structures, deflection

polygons for 99

indeterminate structures 140, 194

influence lines for. 140
methods of pre-
liminary designing 194

Stresses by Ritter's method of moments 211

the method of influence lines 218

stress diagrams 213

Stresses, combination of as a basis for designing 268

due to abutment displacements 32

in statically determinate structures,

210, 218

on any arch section 157, 320

Stress influence lines for indeterminate struc-
tures 137, 139, 140

for truss members 52

Structural redundancy, tests for 6,7

Summation influence lines for reactions . . 54

Temperature effects on determinate structures . 256
indeterminate struc-
tures 206

stresses follow Menebrea's and

Castigliano's laws 31

fixed framed arches 188

fixed masonry arches,

318, 345

girder on three supports . 146
truss on three supports .. 152
two-hinged arches,

156, 165, 167

Tests for structural redundancy 7

Theorem of least work 29

Theorems relating to work of deformation .... 29

Theory of elasticity, bibliography 359

Three-hinged framed arches by influence lines . 72
solid web and masonry arches by
influence lines. . 78

Tractive forces, effect of j

Trusses with subdivided panels ;

Truss on three supports, problem 1

Two-hinged arch with cantilever side spans. . . 170

tension member 1

framed arch, abutment displace-
ments 168, 1

deflection for any

point 168, 1

example 166, 1 9

stress* influence areas

for 168, 1

temperature stresses

in 167, 17"

with column supports I.'/
X a influence line for,

166, 1

solid web 'arch, abutment displace-
ments 156, ]

example 153, 1

kernel moment in-
fluence areas . . . . ]
temperature stresses,
156, ]

X a influence line for,
154,

Unusual load effects.

Virtual velocities , 2,

work, principle of 1 ,

Williot displacement diagrams

Williot-Mohr diagrams

Wind pressure 2

stresses in cross-frames

in main and lateral trusses 2

Work equations for any frame 14

for isotropic solids

of deformation 1

due to dynamic impact.. .
due to shearing stress ....

due to moments

for any indeterminat i

straight beam

theorems relating to

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