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E'K^'l 1^.1:^.^3

^ c ^^7 ^ ^^>-— ^i^

"7^

Works of

riansfield nerriman and
Henry S. Jacoby.

Published by JOHN WILEY & SONS,
43-45 East 19TH Street, New York.

TEtT-BOOK ON ROOFS AND BRIDGES:

F^ut I. Stresses. ?vo, 326 iMges . $3.50

Part II. Graphic Statics. 8vo, 34a pages . . 3.50

Part III. Bridge Design. 8vo, 4^3 pages . . 3.50

Part IV. Higher Structures. 8vo, 385 pages 3.50

By Mansfield Merri.r.an.

TREATISE ON HYDRAULICS. 8vo, 57S.P«ge8 net, 4-00
MECHANICS OF MATERIALS. 8vo, 518 pages . 5.00
PRECISE SURVEYING AND GEODESY. 8vo, 361

pages 2.50

METHOD OF LEAST SQUARES. 8vo, 338 pages . 3.00
ELEMENTS OF SANITARY ENGINEERING. 8vo,

353 pages net, 3.00

ELEMENTS OF MECHANICS. X3mo, 173 pages net, x.oo
STRENGTH OF MATERIALS. x3mo, 156 pages net, x.oo

By Mansfield Merriman and John P. Broolcs.

HANDBOOK FOR SURVEYORS. X3mo, 346 pages 3.00

Edited by Mansfield Merriman and
Robert S Woodward.

SERIES OF MATHEMATICAL MONOGRAPHS:

Eleven Volumes, 8vo each, x.oo

Mansfield Merriman, Editor-in-Chief.

AMERICAN CIVIL ENGINEERS' POCKET BOOK.

x6mo, Morocco, X388 pages .... net, 5.00

By Henry S. Jacoby.

STRUCTURAL DETAILS, or ELEMENTS OF DE-
SIGN IN TIMBER FRAMING. 8vo, 377 pages

net, 3.35

By Henry S. Jacoby.

Published bv McGraw-Hill Book Company
239 West 39th Street, New York
TEXT-BOOK ON PLAIN LETTERING.

Oblong X3mo, 89 pages, 48 plates . . 3*00

A TEXT-BOOK

ROOFS AND BRIDGES

Part II
GRAPHIC STATICS

BY
MANSFIELD MERRIMAN

MBMBBR OF AMERICAN SOCIBTY OF CIVIL EMGINBBR9

HENRY S. JACOBY

FKOFBSSOR OF BRIDGE ENGINEERING IN CORNELL UNtVBRttTir

THIRD EDITION, ENLARGED

TOTAL ISSUE, NINETEENTH THOUSAND

NEW YORK

JOHN WILEY & SONS

London ; CHAPMAN & HALL, Limited

1912

/ ■■. f '.■•.■.:<. '\^ J . MCi ivA . wl '/ >\^

First edition, January, x888.

Second edition, December, 1889.

Third edition, October, 1891 ; reprinted, 1894,

Fourth edition, enlar^d, February, 1896; reprinted, 1897,

Fifth edition, March, 1899; reprinted, 1900, 1901, 1.90a, 1903, 1904.I

Sixth edition, rewritten and enlarged, January, 1905.

Reprinted, 1905, 1906, 1907, 1908, 1909, 19x0, 191 x, X9i2,.

Set up and electrotyped by J. S. Gushing & Co., Norwood, Mass,
Printed and bound by Braunworth & Co., Brooklyn, N. Y,

PREFACE.

The course of instruction in roofs and bridges given to
students of civil engineering in Lehigh University consists of
four parts ; first, the computation of stresses in roof trusses
and in all the common styles of simple bridge trusses ; second,
the analysis of stresses by graphic methods; third, the design
of a bridge, which includes the proportioning of details and the
preparation of working drawings; and fourth, the discussion of
cantilever, suspension, continuous and arched bridges. In the
following pages the second part of this course is presented,
together with additional matter so as to form a tolerably com-
plete treatise on Graphic Statics as applied to the discussion of
common roofs and bridges.

In an elementary text-book of this kind it is not expected
that much will be found that is new except method of arrange-
ment and presentation. Attention is called, however, to the
abbreviated processes employed in some of the diagrams for
wind stresses, to the determination of stresses due to initial ten-
sion, and to portions of the analysis of maximum moments
and shears under locomotive wheel loads as possessing points
of novelty and practical value. These new features are due to
the experienced Instructor whose name appears on the title
page in connection with my own;, the larger portion of the
text has also been written by him, and the cuts and plates
are his work.

IV PREFACE.

The universal approbation expressed concerning the utility
of the blank leaves in Part I leads me to insert them in this
volume also. On these pages students may record in permanent
form the numerical computations which are always requisite
preparatory to graphical analysis, and also make free hand
sketches of some of the stress diagrams required in the problems.
But I regard it as essential that a few well chosen cases shall
be thoroughly and completely worked out as indicated on the
plates, which show the manner in which for many years I have
required students to finish drawings in Graphic Statics. Here,
as in Part I, the minimum as well as the maximum stresses are
determined for most of the examples, and all varieties of load-
ing are treated so that students may be able to work in accord-
ance with all kinds of specifications.

Mansfield Merriman.

Lehigh University, Bethlehem, Pa.,
December i8, 1889.

NOTE TO THE THIRD EDITION.

This edition of Part II is revised on the same plan as the
fourth edition of Part I, the blank leaves being omitted, while
the text is enlarged to nearly double the former length. Por-
tions of Chapters II and III have been rewritten, while Chap-
ters V, VI, and VII are new, and give many improved
methods of graphical analysis.

CONTENTS.

Art.

Chapter I.

PRINCIPLES AND METHODS.

PAGB

The Force Triangle, ....

The Force Polygon.

Conditions of Equilibrium, . ' .

Stresses in a Crane Truss,

Stresses in a Polygonal Frame,

The Equilibrium Polygon,

Properties of the Equilibrium Polygon, .

. 15

Reactions of Beams, . . . .

Simple Beams under Concentrated Loads,

. 20

10.

Simple Beams under Uniform Loads, .

II.

Overhanging Beams, ....

. 24

12.

Center of Gravity of Cross- Sections, .

13.

Moment of Inertia of Cross-Sections,

. 28

14.

Graphical Arithmetic, . . • .

Chapter II.

ROOF TRUSSES.

Art. 15. Definitions and Principles, .
16. Dead and Snow Loads,

33
34

CONTENTS.

PAGE

17. Stresses Due to Dead and Snow Loads, • . 38

18. A Triangular Roof Truss, ... 44

19. Wind Loads. . , . . .4.6

20. A Truss with Fixed Ends, . ... 48

21. A Truss with One End Free, . . .50

22. Abbreviated Methods for Wind Stress, . 54

23. Complete Stresses for a Triangular Truss, . 57

24. Complete Stresses for a Crescent Truss, . 59

25. Ambiguous Cases, . . . . . 6i

26. Unsymmetrical Loads and Trusses, . • 64

Chapter III.

BRIDGE TRUSSES.

Art. 27. Loads on Bridge Trusses,

28. Dead Load Stresses,

29. Live Load Stresses in a Warren Truss,

30. Live Load Stresses in a Pratt Truss, .

31. Snow Load Stresses,

32. Wind Stresses. ....

33. Stresses Due to Initial Tension,

34. Final Maximum and Minimum Stresses,

35. The Bowstring Truss, ...

36. The Parabolic Bowstring Truss,

37. Application of the Equilibrium Polygon,

38. Excess Loads, ....

• 73

n
84
89

• 91

. 102

107

Chapter IV.

LOCOMOTIVE WHEEL LOADS.

39. Standard Typical Loads,

40. Analysis of a Plate Girder,

. 112
114

CONTENTS. Vn

PAGE

Art. 41, Analysis of a Pratt Truss, • • • .119

42. Moments in Plate Girders, • • • 124

43. Shears in Plate Girders, • • • . . 131

44. Simultaneous Moments, • • • • 135

45. Shears in Trusses. ♦ . • • • • 136

46. Floor- BEAM Reactions, • • • • 139

47. Moments in Trusses, • • • • •141

Chapter V.

TRUSSES WITH BROKEN CHORDS.

Art. 48. Points of Division in Panels, • • . 144

49. Position of Wheel Loads, . • • .147

50. Resolution of the Shear, . . , 150

51. Example.— Maximum Chord Stresses, . -153

52. Example. — Maximum Stresses in Diagonals, . 157

53. Example.— Maximum Stresses in Verticals, .. 160

54. Example.— Minimum Stresses in Vertica^-S, . 162

55. Stresses DUE to Wind, • • ^r • 171

Chapter VI.

MISCELLANEOUS TRUSSES.

Art. 56. The Pegram Truss, . • • •

57. The Pennsylvania Truss, • •

58. The Baltimore Truss, . • •

59. Unsymmetrical Trusses, . •

60. Double and Quadruple Systems, •

61. The Greiner Truss,

62. Horizontal Shear in a Beam, . .

63. Roof Truss with Counterbraces, •

64. A Ferris Wheel with Tensile Spokes,

65. A Bicycle Wheel with Tensile Spokes,

175

182

187
187
192
196
198
203
208

via CONTENTS,

Chapter VII.

ELASTIC DEFORMATION OF TRUSSES*

»A6B

Art. 66. The Displacement Diagram, • . • 214

67. Deformation of a Truss, • • • • 216

68. Deflection of a Truss, .... 221

69. Truss Deflection under Locomotive Wheel

Loads, • * • . ... • .223

Appendix.

Answers to Problems* • • • • • • 227

Index, • • • • • • • • • 231

Plates.

TO PACE PACK

L Triangular Roof Truss, • • • • . 58

II. Crescent Roof Truss, . • . . ^ . 60

III. Plate Girder under Locomotive and Train Loads, 118

IV. Pratt Truss under Locomotive and Train Loads, . 122
V. Truss with Curved Upper Chord, . , , 174

VI. Pegram Truss, •..••«, 182

GRAPHIC STATICS.

CHAPTER I.

PRINCIPLES AND METHODS.

Art. I. The Force Triangle.

. A force is determined when its magnitude, direction, and
line of action are known, and accordingly it may be graphically
represented by the length, direction, and position of a straight
line. Forces are given in pounds, tons, or kilograms, while the
lengths of lines are measured in inches or centimeters. If the
scale of force be 4 tons to an inch, a line I.32 inches long v^ill
represent a force of 5.28 tons ; thus the magnitude of forces
will be directly measured by means of the scales used in
draughting.

The resultant of two or more forces is a single force which
produces the same effect* as the forces themselves, and may
therefore replace them. Let two forces P^ and P^ which act
in the same plane upon the point m be represented in magni-
tude and position by the Hnes tnn and mpy and in direction
by the arrows. Let the parallelogram be completed by draw-

ing a line through n parallel n L ^_ ^

to P,, and a line through L

p parallel to P^, and then

let m be joined with their 1^- "Jf

point of intersection. This

line, designated by Ry represents the resultant of the two

2 PRINCIPLES AND METHODS. CHAP. I.

given forces. To find the magnitude of this resultant by the
analytic method, let d be the angle included between P^ and
P^ ; then from either of the triangles composing the figure a
well known theorem of geometry gives

^ = P/ + /\* + 2/^,/>, cos ft

For instance, if P^ = 30 pounds, /\ = 50 pounds, and ^ = 75
degrees, there is found by computation R = 64.6 pounds.

The graphic method of finding the resultant consists of the
following operations : On a sheet of paper, with the help of a
ruler and protractor, from a point m two indefinite lines are
drawn, making an angle of 75 degrees with each other. Using
a suitable scale, say of 40 pounds to the inch, tJie distance mn
is laid off equal to 30 pounds and mp equal to 50 pounds.
With a ruler and triangle the parallelogram is completed, and
m joined with the opposite vertex. This line is now measured
by the scale, and the value of the resultant is found.

It will be seen that it is not necessary to construct the en-
tire parallelogram, since the triangles on the opposite sides of
the diagonal are equal. The triangle above the diagonal can
be constructed by drawing a line through n parallel toP,,
laying off upon it the value of P^ , and then joining its end to
m ; sirhilarly the lower triangle can be independently drawn.
Either of these triangles is called the force triangle.

Usually the lines of action of the given forces form part of a
idiagram upon which it is not desirable to construct the force
/triangle. In this case let any suitable point a be selected, and
ab be drawn parallel and equal to P^ ; then through b let be be
drawn parallel and equal to P^ , and let a be joined with c.
The line ac represents the magnitude of the resultant R, and is
measured by the same scale as that used in laying off ab and
be. The angles bae and bea may be measured by the protractor,
and these are the angles which R makes with P^ and -P, . The

Art. I. THE FORCE TRIANGLE. 3

direction in which the resultant acts is indicated by the arrow
upon^^, and this is seen to be opposed to the directions of
those upon ab and be in following around the triangle. Finally,
the line of action of the resultant R must pass through m, the
point of application of the given forces P, and P^ . Hence,
after constructing the force triangle abc, the resultant R is
found in magnitude, direction, and line of action by drawing
through m a line equal and parallel to ac. *

The above operation is termed composition of forces, P^ and
P^ having been combined into one. The reverse process of
decomposition or resolution of forces may also be effected by
the force triangle. For instance, let R^ in Fig. i be given, and
let it be required to find its components in the directions of
mn and nip. Let ac be drawn equal and parallel to R, and
through its extremities let ab and cb be drawn parallel to the
given directions ; these lines intersect in ^, and when they are
measured by the scale the magnitude of the components will
be known. The directions of ab and be are opposite to that of
ae in following around the triangle. Lastly, through m, the
point of application of R^ let P^ and P^ be laid off upon the
given directions, equal to ab and be^ and the lines of action of
the components are determined.

A number of forces are said to be in equilibrium when no
tendency to motion is produced in the body upon which they
act. In Fig. I suppose a force, P^ , equal and opposite to R, to
be applied at m ; then this force together with P, and P^ will
be in equilibrium, for the last two may be replaced by their re.
sultant Ry which by the conditions specified holds P^ in equilib-
rium. The corresponding force triangle will be abc with the
direction of ac reversed, so that all the forces around the tri-
angle have the same direction ; hence, when three forces are in
equilibrium, they form a closed force triangle.

When three forces whose lines of action lie in a plane and

4 PRINCIPLES AND METHODS. ChAP. I.

intersect in one point are in equilibrium, any one may be de-
termined when two are given. In Fig. 2 let P^ and P^ be given
to find P, . Let ab be laid off equal and
parallel to P^ , and from 6 let 6c be drawn
equal and parallel to P^: then ca, the
closing side of the triangle, represents /\
in magnitude and direction. As its Hne
of action must also pass through m, the
Fiff. «• force P, is drawn equal and parallel to ca,

and in the same direction, thus completing the solution.

Should only one force be given, together with the lines of
action of the other two, their magnitudes and directions may
be found. Let P^ and the lines of action of P^ and P^ be given.
Draw ai parallel to /\ , mark off its length according to scale,
and through its extremities draw lines parallel to P^ and P^ ;
these lines intersect in c, and the length of 6c gives the magni-
tude of /\, its direction being from ^ to ^, while ca represents
P, in the same respects.

The force triangle is the foundation of the science of graphic
statics. By it all problems relating to the composition and
resolution of forces can be solved, when the forces are but
three in number and act in the same plane upon a common
point.

Problem i. Find the magnitude of the resultant of two
forces making an angle of 60 degrees with each other, one
being 25 pounds and the other 40 pounds.

Prob. 2. The lines of action of two forces, of 50 and 30
pounds respectively, make an angle of 120 degrees. What is
the magnitude of the force that holds them in equilibrium
and the angles that it makes with each of them ?

Art. 2. The Force Polygon.

When it is required to find the resultant of a number of
forces acting in the same plane and having a common point of

Art. 2. THE FORCE POLYGOxV. S

application, the resultant of two of the forces may be found by
Art. I, a third force may then be united with it to obtain a sec-
ond resultant, and this
operation continued un- ^"^T"*--

til all the forces are /^ ^ ^ /p ^
combmed. In Fig, 3, x ''•■'- -X\ l^j^
the line R^ is the result- ^

ant of P, and P,, the ^'*f- 3.

line R^ is the resultant of R^ and P^ , and R is the resultant of
R^ and P^, and therefore of the given forces P,, P, , P, ,and
P^ , It is, however, not necessary to construct these resultants
in order to find R, if the dotted lines be drawn parallel and
equal to P^, P^, and P^.

The polygon shown by broken lines is called the force poly-
gon ; the resultant R forms its. closing side, and each of the
other sides represents one of the given forces. The diagram
adcde shows the polygon as it is generally drawn with the diag-
onals omitted. The direction of the resultant is opposed to
the direction of all the given forces in following around the
sides of the polygon ; thus the arrow on ae has the reverse
direction of the other arrows.

The force polygon may therefore be constructed as follows :

Draw in succession lines parallel and equal to the given
forces, each Hne beginning where the preceding one ends,
and extending in the same direction as the force it rep-
resents. The line joining the initial to the final point
represents the resultant in direction and magnitude.

To produce equilibrium with P, , P, , P, , and P^ , a force equal
and opposite to R must be applied at m. This added force in
the force polygon is equal to ea with its former direction re-
versed, and the distance from the initial to the final point in
the construction of the polygon becomes zero.

Hence, if a number of forces lying in the same plane and

6 PRINCIPLES AND METHODS. CHAI\ I.

having a common point of application are in equilibrium, they
will form a closed force polygon, and in passing around it all
the forces will have the same direction.

In either of the above cases it makes no difference in what
order the forces are arranged in the force polygon. Thus in
Fig. 3 the sides of the force polygon are drawn in the order
» Pj y P^j P^yP^jR] but the same value of 7?, both in intensity and
direction, will be obtained if they are drawn in any other order,
as, for example, P^yP^, P^,P^, R. Again, in Fig. 4, let the four

forces which meet at m be in equilibrium ; then taking them in
the order P^y P^y P^y P^ the force polygon abed is drawn, in the
order /\ , P3 , P, , P^ the polygon a'b'c'd! results, and in the order
P^y P, , P3, P^ the polygon a"b''c"d" is found, each of which
graphically represents the given forces. In the last case it is
Seen that two of the lines in the force polygon cross each
other ; this is of frequent occurrence in practical problems.

The force triangle (Art.'l) is but a particular case of the force
polygon, namely, when the forces are but three in number.
The word polygon is hence often used in a general sense as in-
cluding that of the triangle. From three forces in equilibrium
two force triangles may be drawn ; from four forces in equilib-
rium six force polygons can be formed.

Prob. 3. Draw a force polygon for five forces in equilibrium,
and prove that any diagonal of the polygon is the resultant of
the forces on one side and holds in equilibrium those on the
other.

Art. 3. cOxNditions of equilibrium. 7

Prob. 4. Let P^ — icx) pounds, P^ = 175 pounds, and P, = 60
pounds, and let the angles which they make with each other be
P,mP^ = 135°, P^mP^ = 87°, P^mP, = 138°. Draw three force
polygons and determine from each the value of the resultant,
and the angle that it makes with P, .

Art. 3. Conditions of Equilibrium.

When several forces lie in the same plane the necessary and
sufficient conditions of static equilibrium are that there shall
be no tendency to motion, either of translation or rotation.
Analytically this is expressed by saying that the algebraic sum
of the components, both horizontal and vertical, of the forces
must be zero, and that the algebraic sum of the moments of
the forces must also be zero.

When the given forces have a common point of application,
the graphic condition for equilibrium is that the force polygon
must close. For, if it does not close the line joining the initial
with the final point represents the resultant of the given forces
(Art. 2), and this resultant will cause motion ; and if it does
close there exists no resultant. Therefore, if the given forces
which meet at a common point are in equilibrium the force
polygon must close ; and conversely, if the force polygon closes
the given forces must be in equilibrium.

When several forces lying in the same plane have different
points of application, so that their lines of action do not inter-
sect in the same point, and are in equilibrium the force poly-
gon must also close, since no resultant exists. Thus, suppose
the given forces to be four in number, let the directions of two
of these be produced until they intersect and their resultant
found \ this resultant must pass through the point of intersec-
tion of the remaining two forces, since equilibrium obtains.
Hence the rule above established for forces acting at a com-

8 PRINCIPLES AND METHODS. CllAP. I.

mon point applies also to this case, and the force polygon must
cloSe if they are in equilibrium.

If several forces have different points of application, and the
force polygon does not close, the line joining the initial and
final points represents the intensity and direction of the result-
ant. For, as a force can be considered as acting at any point
in the line of its direction two of them may be combined into
a resultant, and this resultant may be combined with one of
the other forces, and so on until the final resultant is obtained
in the same manner as in Art. 2.

If several forces have different points of application, and
the force polygon closes, it is not necessarily true that the
given forces are in equilibrium. For example, let a beam or
stick be acted upon by three forces as shown in Fig. 5, the
>. /) p ^ forces P^, P^, and P^ being equal,

/*, and P^ making an angle of 30
degrees with fhe horizontal and P^
being vertical. It is plain that
^*«^* 5- equilibrium is here impossible, and

yet the force polygon abc closes. Upon reflection it will be
seen that the equilibrium of the beam under the action of the
three given forces can only be maintained by a couple, that is,
by two equal parallel forces acting in opposite directions. It is
because the resultant of the forces of a couple is zero that the
force polygon closes in this case ; and it will be found that in
all instances of non-equilibrium where the force polygon closes
that a couple is necessary to maintain equilibrium.

The above conditions apply to forces lying in one plane. It
IS rare in problems relating to roofs and bridges that forces
acting in different planes need to be considered, and hence in
the following pages it will always be understood, unless
otherwise stated, that the forces under discussion lie in the
same plane.

Art. 4.

STRESSES IN A CRANE TRUSS.

Prob. 5. In Fig. 5, let each of the forces be 100 pounds, and
let the distance between the points of application of P^ and P^
be 4 feet, and between those of P^ and P, be 5 feet. Compute
the magnitude of the forces of a horizontal couple to maintain
equilibrium when the vertical distance between their points of
application is 3 inches. Draw the forces of the couple in both
diagrams of Fig. 5.

Art. 4. Stresses in a Crane Truss.

As an example of the application of the preceding principles
let it be required to graphically determine the stresses in the
members B^y B^, etc., of the crane truss shown in the left-hand
diagram of Fig. 6, due to a load P, acting at the peak. The
member B^ is called the tie, B^ the jib, B^ the post, and B^
the back-stay. The post is vertical and its length is 16 feet,
the length of the jib is 30 feet, of the tie 21.5 feet, and of the
back-stay 20 feet ; from these dimensions the diagram of the
crane truss is constructed. The load P, is taken as 5 tons.

Using a scale of 2 tons to an inch, the construction of the
stress diagram is begun by laying off ab parallel to P, and equal
to 5 tons. Now at the peak the force /\ is resolved into two
forces whose lines of action are in the two members B^ and B^ ;
hence, by Art. i, draw ac parallel to j5, and be parallel to B^ ,
thus obtaining the
force triangle abc \ the
length of ac gives the
stress in B^, and that
of be gives the stress
in -ffj. Next passing

to the apex n the stress Fig. 6.

in B^ is known and those in B^ and B^ are to be found ; hence
from a and c draw parallels to these members and these lines,
intersecting at ^, give ce as the stress in B^^ and ae as that
in B^. This completes the force diagram.

lO PRINXIPLES AND METHODS. ChaP. I.

The next step is to determine the character of the stresses, that
is, whether they are tension or compression. Beginning with
the triangle abc^ which represents the forces acting at the apex
«r, the direction of ab is known to be downward, hence follow-
ing around the triangle (Art. i) the stress 5, acts from b to-
ward Cj and 5, from c toward a ; transferring these directions
to the lines of action at the apex w it is seen that 5, acts to-
ward w, and is therefore compression, while S^ acts away from
«r, and is therefore tension. Passing now to the apex n the
stress S^ in B^ is known to be tension and hence it acts away
from «, accordingly in the force triangle cae it acts from a
toward c\ hence 5^ acts from c to ^, and 5, from e to a; trans-
ferring these directions to the lines of action at n it is found
that 5, is tension and 5^ is compression.

Applying the scale to the lines of the force diagram the
following results are now found, the sign -|- denoting tension
and — denoting compression :

5, = + 6.7, 5, = — 9.35, 5, = — 6.85, 5, = + 10.8 tons.

These are the stresses in the members B^,B^yB^y and B^ due to
the load of 5 tons acting at the peak. If 10 tons were hung at
the peak it is plain that each line of the force triangle would
be twice as long as before, or the stresses in the members would
be double the values above given.

The two parts of Fig. 6 are called the * truss diagram ' and
the * stress diagram ' respectively. Each triangle in the stress
diagram afice corresponds to the forces acting at one of the
apexes in the truss diagram, so that it may be said that the
two figures are reciprocal.

Prob. 6. In Fig. 6 let B^ be vertical and let B^ = 30, B^ = 45,
B^ = 50, and B^ = go feet. Draw the stress diagram and de-
termine the stresses in all the members due to a force of 6 tons
which acts at an angle of 30 degrees to the right of the vertical
drawn through the peak m.

Art. 5

STRESSES IN A POLYGONAL FRAME.

Art. 5. Stresses in a Polygonal Frame.

In Fig. 7 let B^BJB^B^ be a polygonal frame which supportb
the loads P^ and P^ and which is itself suspended by the forces
Pj and P^ acting in two ropes. The frame being in equilibrium
under the action of the exterior forces P^, P^\ P^, and P^, it is
required to find the stresses in the members B^y B^,B^, and B^.

As the exterior forces /\, P,, P,, and P^ are in equilibrium,
the force polygon representing them must close (Art. 3). First,
then, let the force polygon abed be drawn, ab representing P^

Fig. 7.

in magnitude and direction, be representing P^, and so on.
Now at each apex of the polygonal frame there are three forces
which are in equilibrium. Thus at m the force P^ is known,
and if from b and a lines be drawn parallel to B^^ and B^ these
intersect in o, giving bo as the stress 5^ in B^ and ao as the
stress 5^ in B^. Similarly at each of the other apexes the ex-
terior force may be resolved into components in the two given
directions. Thus 5, , 5,, ^3, and S^ are found as the stresses
in B^, B^, B^, and B^ .

To find the character of these stresses it is only necessary to
follow around the sides of each force triangle in the direction
indicated by the given force and then to transfer these directions
to the corresponding apex of the frame. Thus, at the apex H
the direction of P^ is known and the corresponding force Ul-
angle is beo; in this P, acts from b towstrd c^ hence 5. acts from

12 PRINCirLES AND METHODS. CHAP. 1.

c toward o and 5*, acts from o toward b\ transferring these
directions to n it is found that 5, acts toward and S^ away
from n, thus showing S^ to be compression and 5, to be ten-
sion. In like manner it is found that 5, and S^ are also tension.

Pfob. 7. If the frame in Fig. 7 be inverted, draw the force
diagram and determine the character of the stress in each
memb^.

Prob. 8. In Fig. 7 let each of the forces P,, P^, P^, and P, be
vertical and equal to 100 pounds. Let B^ and B^ be horizontal,
the length of the former being 6 feet and that of the latter
being 2 feet. Let the lengths of B^ and B^ be 5 feet. Draw
the force diagram and find the magnitude and character of the
stress in each member.

Art. 6. The Equilibrium Polygon.

When a number of forces acting upon a body do not meet
in a common point the magnitude and direction of their result-
ant is found by the closing line of the force polygon (Art. 3),
but its line of action is not determined. This will now be
found by means of another diagram which is called the equilib-
rium polygon.

Let four forces represented by P^, P^, P^, and P^ be given in
magnitude, direction, and line of action, and let it be required
to fully determine their resultant. Constructing the force
polygon abcde, the length of the closing line ea represents the
magnitude of the resultant, and its direction is from a toward e,
being opposite to those of the other forces in following around
the polygon (Arts. 2 and 3). Now select any point and draw
the lines oa^ ob, oc, od, and oe to the vertices of the force poly-
gon, thus forming five force triangles. In the force triangle
oab the lines oa and ob represent two forces which can hold ab
in equilibrium if their directions be from ^ to ^ and from o to a.
Thus each of the forces in the force polygon can be replaced

r ^'

Art. 6. THE EQUILIBRIUM POLYGON. 1 3

by its components shown by the broken lines ; for example,
/*, has the components 5, and S^ . Now through any point m
on the line of action of /\ draw the lines j5, and B^ parallel to
5, and S^ respectively, and let B^ intersect the line of action of
the force /^ at «. ^r ., j,

parallel to 5«, and -ft/ ^ J^ S,

so on in succession, — &^ — » r' Ar~-5 — y<-' "^

until finally B^ is ^^^^^ ^^'

drawn parallel to 5»
The lines 5^ and B^ Fig. 8.

will intersect at some point r\ through this point draw a line
R parallel to ae and the line of action of the resultant is deter-
mined. For, by the construction the forces 5» and S^ are the
components of the resultant R or ae^ and as their lines of
action are in B^ and B^ the resultant must pass through the
point where they intersect.

If in Fig. 8 there be applied at the point r a. force P^ equal
to R but opposite in direction the forces P^, P^^ P^, P^j and P^
are in equilibrium and the force polygon closes. The polygonal
frame B^B^B^B^B^ thus holds the given forces in equilibrium
by means of the stresses of tension and compression acting in
its members. For the case shown in the figure these stresses
are all tensile, and their- values are given by the lines 5,, 5,,
etc., in the force polygon. The lines of this frame are hence
called an * equilibrium polygon.* The polygonal frame in Fig. 7
is an equilibrium polygon which holds the exterior forces in
balance.

The graphic condition of equilibrium for several forces not
meeting at the same point may now be expressed by saying
that both the force polygon and the equilibrium polygon
must close. If the former closes and the latter does not the
given forces are not a system in equilibrium. For example.

PRINCIPLES AND METHODS.

Chap. I.

Fig. 9.

the three forces P^y P^y and /*, in Fig. 9 are equal in magnitude
and make angles of 120 degrees with each other, but they are

not in equilibrium because their
lines of action do not intersect
in the same point. The force
polygon abc here closes. Select
any point? o and draw the lines
oa, ob, and oc, thus resolving the
force Pj into the components
5i and ^3, the force P^ into S^
and 5,, and the force P, into S^ and 5,. Now select any point
m on the line of action of P^ and draw mr and mn parallel to
Sy and ^3 ; from n, where mn intersects the line of action of P^ ,
draw nr parallel to S^ . Then mr and nr intersect at r which
is not on the line of action of/!,, and hence the three given
forces cannot be held in equilibrium by an equilibrium poly-
gon. In this case it is said that the equilibrium polygon does
not close. If, however, the force P^ be moved parallel to itself
until its line of action passes through r the force polygon
closes and the forces will be in equilibrium.

The point in the plane of the force polygon is called * the
pole,* and the lines oa, oby etc., are sometimes called * rays.*
Since the position of the pole may be selected at pleasure it
follows that for any given system of forces an infinite number
of equilibrium polygons can be constructed. The pole may
be taken either within or without the force polygon as may
be most convenient for the solution of the problem under
consideration.

Prob. 9. Given two forces of 100 and 180 pounds acting at
an angle of 5 degrees with each other, the point of intersec-
tion not being within the limits of the drawing. Find the
magnitude and direction of the resultant by the force polygon,
and its line of action by the equilibrium polygon.

Art. 7. PROPERTIES OF THE EQUILIBRIUM POLYGON.

Art. 7. Properties of the Equilibrium Polygci;:

Let Fig. 10 represent a system of forces P, , P,, . . . P, held
in equilibrium by the jointed frame or equilibrium polygon
whose members are B^, B^, . . . B^. This is constructed by first

Fig. 10.

drawing the force polygon ^^^^^/ which must close, then select-
ing a pole o and drawing the rays oa, ob, . , , of, to which the
members of the equilibrium polygon are made respectively
parallel. The character of the stresses in these members is
determined from the force polygon ; thus in the triangle abo the
directions of bo and oa must be from ^ to ^ and from o to a in
order to maintain equilibrium ; transferring these directions to
the other diagram, it is seen that the stresses in B^ and B^ act
toward the apex m, and hence are compression. Passing nexL'
to the vertex n the stress in B^ is also found to be in compres-
sion, and so on. (Art. 4.)

Let this equilibrium polygon be cut by a section shown by
the broken and dotted line ; the stresses in B^ and B^ , the mem-
bers cut, are given by 5, and 5. in the force diagram, and form
the closing sides of the polygon abcdo and also of the polygon
defao. That is, the stresses in B^ and B^ hold in equilibrium
the external forces P^, P^, and P^ , and also the external forces
P^yP^y and P^ . Therefore the following principle is established:

The internal stresses in any section hold in equihbrium the
external forces on either side of that section. '

l6 PRINCIPLES AND METHODS, CHAP. L

This principle, it will be observed, is the same as that appli-
cable to the internal stresses in a beam (Mechanics of Materials,
Art. is) or to the internal stresses in a truss (Roofs and
Bridges, Part I, Art. 4).

Since the stresses in*^, and B^ hold in equilibrium the exter-
nal forces P^, P^, and P, the resultant of the former mu!st be
equal and opposite to the resultant of the latter. This is also
seen in the force polygon where the line ^?^/ gives the magnitude
of this resultant. To find its line of action let B^ and B^
be produced until they meet in r, and through r draw a line
equal and parallel to ad. Thus results the following important
principle :

The resultant of the external forces on the left of any sec-
tion passes through the intersection of the sides of the
equilibrium polygon cut by that section, its magnitude
and direction being given by the force polygon.

The resultant of the external forces on the right of the sec-
tion is the same in magnitude and line of action as that of those
on the left, but its direction is reversed.

When a system of parallel forces are in equilibrium, the force
polygon becomes a straight line and the equilibrium polygon
has an important special property which will now be deduced.
Let Pj , jP, , and P, in Fig. 1 1 be three downward forces, held in
equilibrium by the two upward forces P^ and P^ ; for example,
the former might be loads acting on a beam and the latter the
reactions of the supports. The force polygon here is abcdea^
the lines aby be, and cd being laid off downward while de and ea
are laid off upward, closing the polygon. Selecting any pole o,
and drawing the rays oa, ob, etc., the equilibrium polygon B^B^
B^B^B^ is formed (Art. 6). Now let any two sides B^ and B^
be cut by a vertical plane, and let the ordinate intercepted be-
tween them be called y. The intersection of these sides pro-
duced gives the point of application of the resultant of the ex-

Art. 7. PROPERTIES OF THE EQUILIBRIUM POLYGON. 1/

ternal forces P, and P, , whose value is given by eb in the force
polygon ; let the horizontal distance from y to this point be
called r, and the resultant be called R. The bending moment

ct^li^-^

Fisr. IX.

M in the given section is then equal to the moment of the re-
sultant of all the forces on the left of that section, or

M=Rr.

Now in the force polygon let the line ok be drawn horizon-
tally through the pole and its value be called H. Then since
the triangle obe is similar to the triangle which has the base y
and the altitude r, and since eb is equal to 7?, we have

r\y\\H\R or Rr = Hy.

Therefore the bending moment of the external forces on the
left of the section is

M=Hy.

The force H is seen to be the horizontal component of each
of the stresses ^<3:, ob^ etc., in the force polygon, that is, of the
stresses in the members B^,B^, etc., in the equilibrium polygon ;
it is called the* pole distance,' and is measured by the same
scale of force as the other lines in the force polygon. The fol-
lowing theorem can hence be stated : '

If a structure be subject to parallel forces, the bending
moment in any section parallel to the forces is equal to
the ordinate^ in the equilibrium polygon multiplied by
the pole distance H in the force polygon.

Hence by adopting suitable scales the values of the bending
moments can easily be found from the diagram. For instance,

1 8 PRINCIPLES AND METHODS. CHAP. I.

if the linear scale used in laying off the positions of the exter-
nal forces be 20 feet to the inch, and if the pole be so selected
that the distance //' is 5 tons, then the moments are measured
by a scale of 5 tons x 20 feet = 100 ton-feet to the inch.

The following is another proof of this important theorem :
Let the line B^ in Fig. 1 1 be produced until it meets the section
in n ; let the vertical distance between n and B^ be called jj/^ , and
that between n and B^ be called j, . Let the lever arms of P^
and /\ with respect to the section be called p^ and p^ . The
bending moment for the section then is

But the triangle whose base is y^ and altitude /^ is similar to
oae, and the triangle whose base is j^ and altitude/, is similar
to oab. Hence P^p^, = Hy^ and /\/, =«-^i ; and accordingly

M^H{y,-y:).
But since y^— y^=. y, this gives M = Hy, which is the same
relation as before deduced.

Prob. 10. Dr»aw an equilibrium polygon for the five vertical
forces given in Fig. 11, taking the pole on the right-hand side
of the force polygon.

Prob. II. Given two parallel forces 12 feet apart and acting
in opposite directions, one being 6 tons and the other 2 tons.
Find by the force and equilibrium polygons the magnitude and
line of action of their resultant.

Art. 8. Reactions of Beams.

By the use of the force and equilibrium polygons the reactions
of the two supports of a beam carrying given loads may be
graphically determined. For example, let the beam in Fig. 12
be subject to two concentrated loads as shown, and be in equi-
librium under the action of these loads and the two reactions.
If the values of the reactions were known an equilibrium poly-

Art. 8.

REACTIONS OF BEAMS.

-By.

~'<B.

gon could then be constructed which should act instead of the
beam to maintain this equilibrium (Art. 7). But since the loads
and reactions constitute a system of forces in equilibrium,
the principle that the equilibrium polygon must close (Art. 6)
furnishes the means of determining the unknown reactions.

Let P^ and P^ be the given loads, and let ab and be be drawn
equal and parallel to them. Since the force polygon must
close the line ca then represents the sum of the two reactions.
Next let any pole
be selected, and
the rays oa, ob, and
oc be drawn, and
parallel to these let
B,, B,, and B^ be
drawn, thus form-
ing part of the equi-
librium polygon.
This polygon can
now be closed by drawing B^ , joining the points where B^ and
B^ intersect the lines of action of the reactions. Finally, in the
force polygon let od be drawn through o parallel to B^ ; thus
determining the point d\ then cd and da are the two reactions,
the former being R^ - and the latter R^ . For, cd is the force
that holds in equilibrium the stresses 5, and Sg in the members
B^ and B^ , and da is the force that holds in equilibrium the
stresses Sj and 5^ in B^ and B^,

As a second example, let it be required to determine the
reactions for the overhanging beam shown in Fig. 13 due to
the two given loads. Laying off ab and be as before, the sum
of the reactions is shown by the Hne ea. Choosing a pole o and
drawing oa, ob, and oe-y the equilibrium polygon is constructed
by taking any point on the line of action of P, and drawing B^
and B^ parallel to oa and ob respectively. Tlien from the point
where B, intersects the Hne of action of P^ the line B^ is drawn

Fig. la.

PRINCIPLES AND METHODS.

Chap. I.

parallel to oc. Now B^ intersects the line of action of R^ at m
and B^ intersects that of ^, at ;^ ; joining mn by the line B^

closes the equilibrium

«4-'

Fig. X3.

polygon, and parallel
to this closing line od
is to be drawn in the
force diagram, thus
determining the two
reactions cd and da.

In both the above figures, any ordinate drawn in the equilib-
rium polygon gives the bending moment in the beam at the
point vertically above it (Art. 7). In Fig. 13 where the sides
of the equilibrium polygon cross there is no ordinate, and this
corresponds to the position of the inflection point in the beam
where the horizontal stresses change from tension to com-
pression.

Prob. 12. Given an overhanging beam as in Fig. 13, its
length being 18 feet, and the distance between the supports 14
feet. Determine the reactions due to three loads, one of
200 pounds at 3 feet from the left end, one of 80 pounds at 6
feet from the left end, and one of 90 pounds at the right end.

Art. 9. Simple Beams under Concentrated Loads.

By applying the principles of the preceding articles the ver-
tical shears and the bending moments may be found for all
Jsections of a beam having only two supports and subject to
any number of concentrated loads. For example, consider a
simple beam 20 feet long, carrying five loads whose positions,
and weights in pounds, are shown in Fig. 14. The reactions
of the supports are found by laying off the loads successively
on the force polygon, or load line a/, the first load being ad,
the second dc, etc. Select the pole <?, draw the rays from 0, and
construct the equilibrium polygon w, n, etc., its closing line

Art. 9. SIMPLE BEAMS UNDER CONCENTRATED LOADS. 21

being ms. Then through o draw og parallel to mSy and the
lines fg and ga will represent the reactions of the right and
left supports.

Between the left support and the first load the vertical shear
equals the reaction ga ; between the first and second loads the
vertical shear is ga ^ ab = gb\ between the second and third

375

loads it is ga — ab — be =^ gc\ and so on. At the fourth load
the shear changes from positive to negative, and at the right
support its value is the reaction gf. The diagram shown in the
figure above the equilibrium polygon gives these shears, and
the manner of its construction is apparent, each step being one
of the loads. This is called the shear diagram.

The ordinates in the equilibrium polygon, or moment dia-
gram, give the bending moments in the corresponding sections
of the beam. It is seen that the maximum ordinate is where
the sides of the equilibrium polygon meet that are parallel to
cd and oe, the rays on opposite sides of and adjacent to og,

22 PRINCirLES AND METIJODS. CllAP. I.

the line parallel to the closing side ms of the equilibrium poly-
gon. As the shear immediately on the left of the correspond-
ing section is positive, being measured from g to d, and that on
the right is negative, being measured from g to e, the important
relation is obtained that the maximum bending moment occurs
at the section where the vertical shear passes through zero.

In making the actual construction for Fig. 14 the linear scale
used in laying off the beam and the positions of the load was
5 feet to an inch, and the force scale used in the force polygon
was 800 pounds to an inch. The pole distance was taken as
I 000 pounds, and hence the moment scale was 5 000 pound-feet
to an inch. Any ordinate in the shear diagram, measured by
the force scale, gives the vertical shear in pounds ; thus,
between the second and third loads the shear is + 300 pounds,
and between the fourth and fifth loads it is — 625 pounds.
Any ordinate in the moment diagram, measured by the
moment scale, gives the bending moment in pound-feet ; thus,
the maximum bending moment is 5 500 pound-feet. Fig. 14,
hov/ever, as here printed, is about one-half the size of the
actual construction.

Prob. 13. Construct the shear diagram and moment dia-
gram for a beam 16 feet long, carrying two loads, each of 4000
pounds, one being at 5 feet from the left end, and the other at
5 feet from the right end.

Art. 10. Simple Beams under Uniform Loads.

Let a simple beam whose span is / be uniformly loaded with
the weight w per linear unit ; then each reaction is equal to
half the total load or ^wL The load may be represented
graphically by the shaded rectangle on the beam whose base is
/ and altitude w»

' For any section at a distance x from the left support the
vertical shear is F= \wl— wx = w{il — x); if V be an ordi-

Art. io. simple beams under uniform loads.

nate corresponding to an abscissa x this is the equation of a
straight line. Thus when ;r = o, F= \wl\ when x = i/,
V = o; and when ;r = /, F = — iw/. The shear diagram is
hence constructed by laying off gi equal to the span, making
^and ik equal to ^ze/Zand joining /with k.

The bending moment in a section distant x from the left
support is M == iw/ . x — wx . ^x = ize;(/ar — jt:'). This is the
equation of a parabola; fori: = oand;r=: /the value of J/ is o;
for X =: i/y M reaches
its maximum value ^w/*.
The moment diagram
may hence be con-
structed by laying off
mn equal to the span,
drawing qr at the middle
equal to the maximum
moment, and then con-
structing the parabola ^»k- 's-
mrn. To do this the lines ms and ns are drawn, rs being
made equal to gr, these are divided into the same number of
equal parts and the points of division joined as shown, thus
determining tangents to the parabola.

If the entire load on the beam were concentrated at the
middle, ada would be the force polygon, and 6/1 and /la the two
reactions. Now let (? be a pole having the pole distance H, and
let the equilibrium polygon msn be constructed. Then from
the similar triangles oaA and msg,

H \ \wl :: \l\ qs.

Hence if H be equal to unity on the scale of force, the ordi-
nate qs has the value \wPy and since qr is \wP the maximum
moment for a single concentrated load at the middle is twice as
great as that due to the same load when uniformly distributed.

Prob. 14. Prove that the sum of all the moments due to a

PRINCIPLES AND METHODS.

Chap. I.

uniform load is two-thirds of the sum of all the moments due
to the same load when concentrated at the middle.

Art. II. Overhanging Beams.

Let a beam be taken with one overhanging end and bearing
a number of concentrated loads as shown in Fig. i6. The
loads are given in pounds and the distances in feet. If in the
force polygon the loads be laid off successively in the order in

Fig. x6.

which they are on the beam, and the equilibrium polygon
m . . n . . stuh^ constructed, the ray og drawn parallel to the
closing line «« will determine the reactions ^ and ^^. The
sides of the equilibrium polygon are found to cross each other
at I, and the ordinates to the right of this point lie on the op-
posite side of the closing line from those on the left. The
ordinates on the left being regarded as positive, those on the
right are negative, and they give the bending moments for all

Art. II. OVERHANGING BEAMS. 25

sections in the beam. The point /, where the bending moment
is zero, is called the inflection point ; on the left of this point
the lower fibers are in tension while on the right they are in
compression.

In order to construct a moment diagram whose ordinates
shall be measured from a horizontal Hne m't' ^ the numerical
values of the reactions should first be computed ; these are
iSj = 60 pounds and i?, = 120 pounds. Now form a force poly-
gon, or load-line, for the reactions and loads by laying off the
reaction R^ from ^ upward to a\ then the first three loads in
succession from a' downward to d\ then the reaction R^ from
d' upward to e\ and finally the remaining loads from e' down-
ward to g'. Take the pole 0' on a perpendicular to the load
line at g'. The equilibrium polygon m\ . n' . , u's't* can then
be constructed, each of whose ordinates is equal to that in the
polygon m . . n . , stu.

The method of constructing the shear diagram on the axis
vuf will be understood without further explanation than that
given in Art. 9. It is seen that the shear passes through zero
at two points, one where the maximum positive moment occurs,
and the other at the right support where the negative moment
is a maximum.

The linear scale used in the actual construction of Fig. 16
was 4 feet to an inch, and the force scale was 60 pounds to an
inch ; the pole distance being icxD pounds, the moment scale
was 400 pound-feet to the inch. In the figure as printed the
scales are one-half these values. By measurement it is found
that the maxinium shear is 60 pounds, the maximum positive
moment 120 pound-feet, and the maximum negative moment
140 pound-feet.

For the case of a uniform load a shear diagram and moment
diagram may be constructed by computing the maximum ordi-
nates and then drawing the straight lines and parabolas (Me-

PRINCIPLES AND METHODS.

Chap. I.

chanics of Materials, Chap. IV). Thus, let Fig. 17 represent a
beam 28 feet long with ends overhanging 4 and 6 feet, and let
the uniform load be 40 pounds per linear foot. The left reac-
tion is found by computation to be 497.8, and the right reaction
to be 622.2 pounds ; these might also be obtained graphically
by the equilibrium polygon, regarding the load on each por-
tion of the beam as concentrated at its center. The shear is
then found to be zero at c^ distant 8.45 feet from the left sup-
port, and at this point the positive moment is a maximum, its
value by computation being i 106 pound-feet. At the left

support the negative moment is
found to be 320 and at the right
support 720 pound-feet. These
moments being laid off by scale
the curves can be constructed
by the method given in Art. 10,
it being known that the end
parabolas have their vertices at
m and q, and that the middle
parabola has its vertex at n.
The inflection points are
equally distant from the place
of maximum positive moment, this distance being 7.45 feet
in Fig. 17. The diagrams thus furnish full information re-
garding the distribution of the shears and moments in the
beam.

Prob. 15. Abeam 20 feet long has two overhanging ends,
each 5 feet long. Draw the shear and moment diagrams due
to a load of 6 tons at one end and a load of 4 tons at the other
end.

Prob. 16. A beam 20 feet long has two overhanging ends,
each 5 feet long. Draw the shear and moment diagrams due
to three loads, one of 6 tons at one end, one of 4 tons at the
other end, and one of 8 tons at the middle.

Art. 12. CENTER OF GRAVITY OF CROSS-SECTIONS.

Art. 12. Center of Gravity of Cross-Sections. ^

In problems relating to the strength of beams it is necessary
to find the position of the neutral axis of the cross-section
(Mechanics of Materials, Chap. III). The neutral axis passes
through the center of gravity of the cross-section, and in find-
ing the position of the center of gravity the equilibrium poly-
gon may be employed. For instance, let the cross-section of a
deck beam shown in Fig. 18 be taken. As this cross-section
has an axis of symmetry AD^ the center of gravity Hes on this
axis. The area is divided into simple geometrical figures or
narrow strips by lines perpendicular to the axis, and at the
centers of gravity of these parts forces proportional to their
areas are applied. The force and equilibrium polygons are
constructed in the usual manner, taking the pole opposite the
center h of the force line and making the pole distance oh
equal to ah. The extreme sides of the equilibrium polygon
m . . n . , q are pro-
duced until they
meet at r, the
special position of
the pole causing
thern to form the
best intersection,
that is, at right an-
gles. According to
Art. 6, the point r
is in the line of action of the resultant of the forces considered,
hence rC drawn parallel to the forces is the line of action of
the resultant. The center of gravity lies on this line and there-
fore at its intersection with the axis AD.

If the surface is very irregular in outline it should be divided
into strips so narrow that the area of each one is equal to the

28 PRINCIPLES AND METHODS. CHAP. L

product of its mean length by its width without appreciable
error. . If there be no axis of symmetry, the process described
above must be repeated for a direction at right angles to the
first, and the center of gravity will lie at the intersection of the
two resultants found.

Prob. 17. Find the center of gravity of the cross-section of
the channel-iron whose dimensions in inches are given in

^ /A Prob. 18. Five parallel forces act

Jh^ ^ j in the same direction. Beginning

at the left their magnitudes are 45,

Fig. 19. 120, 30, 225, and 80 pounds, and

the successive distances between them are 6, 10, 3, and 8 feet.
Determine the magnitude and position of their resultant.

Art. 13. Moment of Inertia of Cross-Sections.

For beams under flexure the bending moment M for any

SI
section equals the resisting moment — with reference to the

neutral axis in that section in which 5 is the unit-stress in the
most remote fiber distant c from the axis and /is the moment
of inertia of the cross-section with reference to the same axis
(Mechanics of Materials, Art. 19).

In Art. 12 the method is given for finding the center of
gravity through which the neutral axis passes, and the moment
of inertia / may be obtained from the same construction by the
^following rule : Let A be the area of the given cross-section,
and A^ the area included between the equilibrium polygon and
the two lines whose intersection determines the center of
gravity; then / is the product of A by A\ or /= AA\ pro-
vided that the pole distance is iA. This rule will now be
demonstrated in connection with a practical example.

Let it be required to find / for the Pencoyd T iron, No. 109,

Art. 13. MOMENT OF INERTIA OF CROSS-SECTIONS.

shown in Fig. 20. The cross-section A .measures 4.04 square
inches, and by the force and equilibrium polygon the neutral
axis EC is found to be 1.32 inches from the base of the T.
Now produce sq until it meets the axis at /. The triangles qtu
and ofe are similar, as their sides are mutually parallel. Let y
be the distance from q to the axis ; then

tu \ y \\ ef \ oh.

But </* equals the area P, laid off to scale, and the oole distance
oh was made equal to \A ; hence,

tu.\A =P,y,

Multiplying this by j/, and remembering that itu.y is the area
of the triangle qtu,

gives,
area qtu

_py

The other triangles
composing the area
between the equilib-
rium polygon and
the lines mr and sr
may be expressed
in a similar manner.
If the area P^ were
of the width dy its moment of inertia would be P^^, and by
adding all the areas together there would be found,

Fig. 30.

A' =

or I=AA\

in which case the broken line m . . . nuqs becomes a curve which
is tangent to the lines mr and sr at the extreme limits of the
given cross-section. This curve may be drawn and the area
A^ determined either by dividing it into strips or by the
planimeter.

30 PRINCIPLES AND METHODS. ClIAP. L

By performing the above operations on a full scale drawing

for the T iron shown in Fig. 20 the area A' was found by the

planimeter to be 1.80 square inches, hence the moment of

inertia is

/= 4.04 X 1.80 = 7.27 inches*.

This agrees very well with the value given in the manufac-
turer's pocket-book, which is 7.26 inches*.

Prob. I9. Determine the moment of inertia of the channel
iron shown in Fig. 19 with respect to an axis through the
center of gravity and normal to the web. Determine also the
moment of inertia for an axis through the center of gravity
and parallel to the web.

Art. 14. Graphical Arithmetic.

In the various operations that are performed on the draw*
ing board cases occur where quantities are to be added, sub-
tracted, multiplied, or divided, and it frequently happens that
this may be done by some simple graphical procedure instead
of by the common methods of arithmetic. Whatever be the
nature of these quantities they are represented by straight lines,
the number of units in the length of a line denoting tne mag-
nitude of the quantity.

To add together several lines it is only necessary to place
them end to end and then measure the total length by the
scale.

To subtract one line from another the first is to be laid off
from the end of the second in the opposite direction and then
their difference can be measured by the scale.

To add algebraically several quantities, some positive and
some negative, the suni of the lines representing the negative
ones is to be found and subtracted from the sum of the lines
representing the positive ones.

Art. 14.

GRAPHICAL ARITHMETIC.

To multiply any line of length a by any quantity repre-
sented by a line of length m, let C^Jfand OF be drawn making
a convenient angle with each other. Lay off Ob equal to unity,
Oa equal to a, and Om equal to m. Then Oy is the required
product a X m. For, by similar triangles,

Oy : Om :: Oa : Ob,
whence Oy =^ OaX Om =• ay^m.

By successive applications of this principle the product of sev«
eral factors may be found.

To divide a line of length m by one of
length bi let Oa be laid off equal to unity, Om
equal to w, and Ob equal to b ; then making
the construction as before, the similar trian-
gles give

Oy^Om-r-Ob^m-^by

and hence Oy is the quotient found by dividing m by b.

By the help of the principle of similar triangles a number
of graphical constructions can be made for adding and multi-
plying fractions. For instance, let it be
required to find the product of the two
fractions -j^ and |. On any line OX lay
off, as in Fig. 22, Oa = 13 and (?^ = 8;
erect the ordinates ab ^= 6 and ^rf = 9 ;
join Ob and Od, Select any abscissa OXj
in this case taken as 10, draw the ordi-
nate xy, and taking Ox' equal to xy, draw the ordinate x'y.
Then the product of the two fractions is one-tenth of ;r'/.
For, by similar triangles,

^-.^ _^ cd _xy
Oa

^ and

Ox ' Oc Ox'^

Multiplying these together, term by term, and inserting the
values, there results

13 8 10

32 PRINCIPLES AND METHODS. CHAP. I.

Measuring x'y by the scale it is found to be 5.2 ; hence the
required product is 0.52.

Prob. 20. Show by Fig. 21 how to graphically transform

2.5
the fraction — to an equivalent fraction having 9 for its de-

5'5
nominator.

Prob. 21. Make the following graphical constructions: {a)
to square a given number, {p) to extract the square root, {c) to
find the reciprocal, the final result in each case to be repre-
sented by the length of a straight line.

Art. li DEFINITIONS AND PRINCIPLES. 33

CHAPTER II.
ROOF TRUSSES.

Art. 15. Definitions and Principles.

A roof truss is a structure whose plane is vertical, and is
supported at its ends by the side walls of the building, being
so arranged that its principal members are subject only to ten-
sile or compressive stresses under the influence of the loads
which it is designed to carry.

The points where the center lines of the adjacent members
meet are the centers of the connections which form the joints.
The joints of the truss are supposed to be perfectly flexible,
and the external forces, consisting of the loads and reactions,
to be applied only at the joints..

For stability the elementary figures composing a truss must
be triangles, since a triangle is the only polygon which cannot
change its shape without altering the lengths of its sides when
loaded at one or more joints.

The * span ' of a truss is the distance between the end joints
or the centers of the supports, and the * rise * is the distance
from the highest point, or peak, to the line on which the span,
is measured.

The ' upper chord ' consists of the upper line of members
extending from one support to the other. Each half of the
upper chord of a triangular truss is sometimes designated the
*main rafter.' The lower line of members is known as the
Mower chord ' or * tie rod/

The *web members* or 'braces' connect the joints of the

34 ROOF TRUSSES. ^ CHAP. II

upper with those of the lower chord, and may be either verti^
cals, diagonals, or radials.

A member which takes compression is called a * strut/
and one that takes tension a ' tie.' The upper chord and
some of the braces are subject to compression while the lower
chord and the rest of the braces are in tension.

The fundamental principles of Graphic Statics as given in
Chapter I apply to the determination of the stresses in trusses
under given conditions of loading, in the manner indicated in
Art. 4. The notation employed in this chapter for designating
truss members and loads differs from
that previously used, the letters being
placed upon spaces instead of upon
lines, and any member is named by the
letters between which it is situated.
Fifir.aa. Thus, in Fig. 23, AC and BC are the

two rafters and CD is the tie rod, while AB designates the
load at the peak.

Prob. 22. The span in the simple triangular roof truss of
Fig. 23 is 26 feet, and the rise is 6 feet 6 inches. Find the
stresses in the members due to a load at the peak of 2 050
pounds.

Art. 16. Dead and Snow Loads.

Four kinds of loads are to be considered in discussing a
truss : the weight of the truss itself, the weight of roof cover-
ing, the snow, and the wind.

The weight of the truss depends upon its span and rise,
the distance between adjacent trusses, the kind of material used
in the construction of the roof covering and the truss, and other
elements of design. This weight is ascertained from the results
of experience. In Ricker*s * Construction of Trussed Roofs,*
page 46, is a table derived from data given by different author!

Art. i6. dead and snow loads. 35

ties which seems to afford the best figures now attainable.
The following formulas give results approximately agreeing
with those found by the use of this table :

For wooden trusses, W = ia/{i -f- 1^/),

For wrought-iron trusses, W = i^/(i -|- -j^/),

*n which / is the span in feet, a the distance in feet between
adjacent trusses, and W the approximate weight of one truss
Vi pounds. The wooden trusses are to have wrought iron ten
«ion members in accordance with the usual practice. It is seen
that they are one-third lighter than the wrought iron trusses.

The roof covering consists of the exterior 'shingling* of
tin, slate, tiles, corrugated iron, or wooden shingles, resting
usually upon timber * sheathing,* which is supported by * pur-
lins,' or beams running longitudinally between the trusses and
fastened to them at the upper joints. In large roofs the sheath-
ing is laid upon * rafters* parallel to the upper chord, the rafters
resting upon the purlins. The actual weight of the roof cover-
ing, rafters, and purlins is to be determined only by computa-
tion for each particular case, but the following values will serve
for preliminary designs and approximate computations. The
weights given are in all cases per square foot of roof surface.

For shingling — tin, I pound ; wooden shingles, 2 or 3
pounds; iron, I to 3 pounds; slates, 10 pounds; tiles, 12
to 25 pounds.

For sheathing — boards I inch thick, 3 to 5 pounds.

For rafters — 1.5 to 3 pounds.

For purlins — wood, I to 3 pounds ; iron, 2 to 4 pounds.

Total roof covering — from 5 to 35 pounds per square foot
of roof surface.

The snow load varies with the latitude, being about 30
pounds per horizontal square foot in Northern New England,
Canada, and Minnesota, about 20 pounds in the latitude of
New York City and Chicago, about 10 pounds in the latitude

36 ROOF TRUSSES. CHAP. II.

of Baltimore and Cincinnati, and rapidly diminishes southward.
On roofs having an inclination to the horizontal of 60 degrees
or more this load may be neglected, as it might be expected
that the snow would slide off.

The weight of a roof truss with that of the roof covering
which it bears is termed the ' dead load ' or ' permanent load.'

For the purpose of securing uniformity in the solution of
the examples and problems given in this book, the following
average values will be used, unless otherwise specified :

For the truss weight — compute from the above formulas.
For the roof covering — 12 pounds per square foot of roof

surface.
For the snow load — 15 pounds per square foot of horizon-
tal area.

The weight of the roof covering, and of the snow which
may be upon it, is brought by the purlins to the joints or
* apexes ' of the upper chords. The weight of the truss itself
is also generally regarded as concentrated at the same points,
the larger part of it being actually applied there. At each
apex of the upper chord there is hence a load called an * apex
load,' and it may be a * dead apex load ' or a * snow apex load.'
In the wooden truss whose outlines are shown in Fig.
24, let the span be 48 feet, the rise 10 feet, and the distance

between the trus-
ses 12 feet. Then
the length of the
rafter is ^24' + 10*
= 26 feet. Each
main rafter is di-
vided into three
equal parts called
'panels.* From the formula the truss weight is found to be
I 670 pounds. The .weight of the roof covering on each rafter

Art. i6. dead and snow 'loads. 37

is26x I2X 12=3 744 pounds. The weight of the snow sup-
ported by the entire truss is 48 X 12 X IS = 8 640 pounds.
The total dead load is I 670 + 2X3 744 = 9 158 pounds,
and the dead apex loads BCy CD, pD\ D'C\ and CB' are
each one-sixth of the total load or i 526 pounds, while AB
and B'A^ are each one-half as much, or 763 pounds. The snow
apex loads are in like manner i 440 pounds and 720 pounds.
The apex load is also called the * panel load.'

If the panels be of unequal lengths, the load at any apex is
found by considering that the weights brought to it by the
purlins are those upon a rectangle extending in each direction
half-way to the adjacent apexes.

When the two halves of the truss are symmetrical the reac-
tions of the supports are equal, each being one-half of the total
load. When the truss is unsymmetrical, the reactions are
found in the same way as for concentrated loads on a beam.
In the above example each dead load reaction is 4 579 pounds
and each snow load reaction is 4 320 pounds.

The reaction and the half apex load acting at each support
produce an effective reaction equal to their difference. This
effective reaction is that due to the other apex loads, and
therefore the "half apex loads at the supports may be omitted
entirely from consideration. The effective reaction for the
above example is 4 320 — 720 = 3 600 pounds, or by disregard-
ing the loads at the supports it is one-half the sum of the full
panel loads, thus, -^(5 X i 440) = 3 600 pounds.

Prob. 23. A wrought iron truss, like Fig. 24, has a span of
60 feet, rise 14 feet, and distance between trusses 16 feet.
Find the apex loads and reactions due to dead and snow loads.

Prob. 24. A wrought iron truss of the above form has a
span of 90 feet 6 inches and a rise of 18 feet 9 inches, the
trusses being 18 feet apart. Compute the apex loads and
reactions.

38 ROOF TRUSSES. CHAP. II.

Art. 17. Stresses due to Dead and Snow Loads.

The wooden truss shown in Fig. 25 has a span of 42 feet,
rise at peak 14 feet, rise at hip 10 feet 3 inches, horizontal
distance from hip to peak 12 feet 6 inches, and a distance
between trusses of 8 feet.

The lengths of the members BG and EL are each
V^ll- 10.25" = 13.32 feet, and of CH dind DK 1^^12.5'+ 3.75*
= 13.05 feet. The weight of the truss is found to be 874
pounds by the formula in Art. 16. The roof covering on
^(? weighs 13.32 X 8 X 12 = I 279 pounds, and on CH is
13.05 X 8 X 12 = I 253 pounds. The snow load on BG is
8.5 X 8 X 15 = I 020 pounds, and on CH is 12.5 X 8 X 15
= I 500 pounds. The apex loads, expressed in pounds, are
then as follows :

AB^EF BC=DE CD Total

Truss, .... 109 218 218 872

Roof covering, . . 640 i 266 i 253 5 065

Dead, 749 1484 1471 5 937

Snow, 510 1260 1500 5040

Each dead load reaction is therefore 2 969 pounds, and each
snow load reaction 2 520 pounds.

The truss diagram, Fig. 25, composed of the center lines
of the truss members, is carefully drawn to as large a scale as
convenient, each joint being marked hy a fine needle point
/and surrounded by a small circle to limit the. lines drawn
toward the point. In the actual construction the diagram was
drawn to a scale of 3 feet to an inch, but the above figure
is reduced to nearly one-seventh of the original size.

The external forces acting upon the truss are in equilibrium,
and hence form a closed force polygon (Art. 2). These forces
being parallel the resulting polygon becomes a straight line,
which is called the * load line.'

Art. 17. STRESSES due to dead and snow loads.

Taking first only the dead load and using a suitable scale
(l 000 pounds to an inch was used on the original drawing), the
apex loads taken in regular order from left to right are laid off
in succession on the vertical load line af in Fig. 26, thus : the
distance ab is laid off equal to the load AB ox 749 pounds,
then be equal to the load
BC, and so on. Next
fm is laid off upward
equal to the reaction
FM or 2 969 pounds,
and ma equal to MAy^f"
thus closing the polygon.

Beginning with the
joint at the left support where the reaction MA and the
load AB are held in equilibrium by the stresses in BG and
MG the polygon representing these forces will also be
closed ; bg- is therefore drawn parallel
to BG and mg^ parallel to MG. The
lengths of bg^ and m£^, measured by
the scale used on the load line, give the
magnitudes of the stresses in these
members. To find the character of the
stresses the direction around the poly- ^l
gon indicated by the upward reaction
is followed, that is, from m to a, a to by
b to g-y and from g to m. Transferring
these directions to the joint considered,
the stress in BG acts toward the joint
and is therefore compression, while that in GM acts away
(rom the joint and is tension.

The stress diagram is continued by passing to the left hip-
joint and constructing the force polygon be kg to represent the
stresses in the members meeting at that joint. Next are con-
structed the polygons for the right support, the right hip-joint.

Fig. 96.

40 ROOF TRUSSES. CHAP. II.

and finally for the joint at the peak. The closing line hk
must be parallel to HKy and the entire diagram be sym-
metrical with respect to a horizontal axis through in. These
closed force polygons are drawn for all the joints taken in suc-
cession because the loads and stresses acting upon each joint
in the truss are in equilibrium.

The points /, ky A, and g, in Fig. 26, may be regarded as
the poles of the various equilibrium polygons which surround
the spaces Z, K^ Hy and C, in Fig. 25. After the application
of the principles is fully understood, it is not necessary even
to consider the entire polygon for any one joint. In deter-
mining the point A, for instance, which corresponds to the
space //, it is found that the points g and c corresponding to
the spaces G and C, that are adjacent to //, are already fixed
in position, whence by drawing ch parallel to CH and gh par-
allel to GH the point h is obtained by their intersection.

The method given above for finding the character of the
stresses requires passing around the perimeter of every force
polygon contained in the stress diagram, unless the truss and
its loading are symmetrical, in which case only one half of
the polygons are so used. The line representing the stress
in each member must in all cases be traced twice, and if no
external force acts at any joint the direction of passing around
the corresponding force polygon must be obtained from the
character of the stress in one of its connected members already
found. Thus, for the middle joint of the lower chord the
stress in GM is known to be tension, and therefore acts away
from this joint; that is, toward the left. The polygon
mghkltn must therefore be followed around in the direction
indicated by the order of the letters here given.

Another method will now be described by which the
character of the stress in any member may be determined
without reference to that in any other member, and by which
considerable time may be saved. For convenient reference

Art. 17. STRESSES due to dead and snow loads.

Figs. 25 and 26 are reprinted on this page, a slight addition
being made to the latter as will be explained, and therefore
it is marked Fig. 26a.

Fig. 95.

The loads AB, BC, CD, DE, and EF, and the reactions
FM 3ind MA, were laid off on the load line in Fig. 26a in the
order just stated; that is, they were taken in regular order
while passing around the truss in Fig. 25 in the direction of
the hands of a watch. As an aid in remembering this fact
the circular arrow is placed aside of the stress diagram. By
passing in the same direction around the left hip joint, for
instance, the letters in the adjacent spaces occur in the or-
der C-H'G-B'C. The corresponding force polygon has the
same lower-case lettets at its vertices, and is to be followed
around so that its letters c-k-g-b-c shall also have the same
order. When the directions thus indicated, c to A, h to g,
etc., are transferred to the joint, ch acts toward it and is
therefore compression, while hgr acting away from it, is ten-
sion, and^^ is compression. Again, let it be required to find
the kind of stress in ML, With reference to the middle joint
of the lower chord these letters occur in the order L-M,
hence the direction of its stress is from / to m in the stress
diagram. As this direction is away from the joint, LM must
be in tension.

42 ROOF TRUSSES. CHAP. II.

As it is evident that trusses supported like that in Fig. 25
have the upper chord in compression and the lower chord in
tension, whatever the arrangement of the web members may
be, it usually remains to find only the kinds of stress in the
web members. In this example they may all be found by
considering the order of the letters in the web spaces with
reference to the middle joint of the lower chord, and in any
other truss having more web members by similarly taking the
letters in their regular succession with reference to the respec-
tive joints of one of the chords. Let the student test this
statement in the next article and observe that the web stresses
will thus form a connected chain which may be rapidly traced.
Let the joints of the lower chord be used first, and those of
the upper chord afterwards, for a com-
parison of their relative convenience.

Considering now the polygon chgmabc
in Fig. 26a, the forces represented by its
k sides are in equilibrium since the poly-
gon is closed. Referring to Fig. 25 it
"i is seen that the stresses in CH^ HGy and
GM are in equilibrium with the reaction
Fig. 37. MA and loads AB and BC on the left of

a section cutting these members. But
chgmfedc is also a closed polygon, hence
the stresses in the same members are in equilibrium with the
reaction and loads on the right of the section. Thus is again
demonstrated the principle that the internal stresses in any
section hold in equilibrium all the external forces on either
side of the section (Art. 7).

The stress diagram for snow load, Fig. 27, is next con-
structed in a similar manner to that for dead load. As the
snow loads are here laid off in the reverse order from the dead
loads, the diagram is situated on the right of the load line
instead of on the left as in the preceding figure.

Art. 17. STRESSES due to dead and snow loads.

The lines in all the diagrams are to be drawn with a well-
sharpened pencil pressed lightly on the paper so as to produce
a very fine line. As soon as an intersection is obtained it is to
be marked with a needle point, enclosed with a small circle,
and designated by the proper letter ; other lines drawn to or
from that point are not to pass within the circumference of this
circle. The triangle and straight edge used in drawing parallel
lines should be so arranged as to require the triangle to be
moved the shortest distance. Special care is needed to hold
the pencil at the same inclination from the beginning to the
end of a line, or the line will not be strictly parallel to the edge
of the triangle.

The following results, expressed in pounds, are now ob-
tained by applying the scale of force to the stress diagrams :

Truss Members.

BG=:EL

CH^DK

CM = LM

GH^KL

Dead load stresses.
Snow load stresses.

-3860
-3500

-3860
- 3 640

+ 25S0
+ 2340

4-1 280

+ 1310

+ 750
+ 600

When but a slight difference is found between the lengths of
any pair of symmetrical lines in the stress diagram the average
of the two is to be used; otherwise the diagram should be
re-drawn.

As a final check the stresses in CH are computed (Roofs and
Bridges, Part I, Art. 5), the lever arm of CH being measured
on the truss diagram and found to be 7.27 feet. The stress
due to dead load thus obtained is 3 861 pounds, and that due
to snow load is 3 640 pounds.

Prob. 25. The upper chord of a wrought iron truss like Fig.
24 is divided into six equal panels, the triangle formed by the
middle panel of each main rafter with the adjacent braces being
isosceles. The span is 78 feet, the rise 19 feet 6 inches, and the
trusses are 16 feet 6 inches apart, center to center. Find the
dead and snow load stresses.

ROOF TRUSSES.

Chap. II.

Art. 1 8. A Triangular Roof Truss.

In the wrought iron truss in Fig. 28 the upper chord is
divided into eight equal panels, verticals are drawn through the
apexes, and the diagonals slope upward toward the center.
Let the span be 80 feet, the rise of peak 15 feet, the lower

chord horizontal, and the distance between adjacent trusses 18
feet, center to center.

The main rafter is found to be 42.72 feet, and each panel
10.68 feet long. The weight of the truss is 9 720 pounds, and
of the roof covering 18 455 pounds, making the total dead load

Art. 1 8. a triangular roof truss. 45

28 175 pounds. The dead apex load is one-eighth of this
amount, or 3 522 pounds, and the half apex loads at the sup-
ports are each i 761 pounds.

Since the inclination of each panel of the upper chord is the

same the snow apex loads are equal, each one being 10 X 18

X 15 = 2 700 pounds. The ratio between any snow apex load

2 700
and a dead apex load being = 0.7666, the stresses caused

by the snow load will bear the same ratio to those due to dead
load, and hence only the dead load stress diagram is required.

The construction is begun by laying off the load line aa!
equal to 28 175 pounds by scale and bisecting it at n. The
half apex loads ab and aV are next laid off equal to i 761
pounds and bb' divided into seven equal parts. The reactions
are a'n and na, closing the polygon of external forces. The
polygons representing the forces acting at each joint are suc-
cessively formed, as explained in Art. 17, by beginning at the
left support, passing to joints alternately on the upper and lower
chords until the peak is reached, then going to the right support
and passing from joint to joint until the peak is reached again.
The last line to be drawn is I' my which must be parallel to L'M
and pass through the intersection tn of the lines Im and nm.
The diagram if accurately drawn will be symmetrical with
respect to nf\ the distances fh^ hky and km will be equal, the
points^,/, and /lying on a line parallel to f'b^ drawn through
a point below b' on the load line at a distance equal to b'c\
Again, eg and c'g^ will pass through the intersection k of nf and
ll\ As a final check the stress in MN may be computed ; thus,

(14088 — I 761)40 — 3 522(30 + 20 + 10) — 5 X 15=0;

from which the value of 5 is -{- 18 784 pounds.

The scale of this stress diagram should be such that the line
bf'wiW not be longer, than the main rafter of the truss diagram.
The dead load stresses in the following table w^ere obtained by

ROOF TRUSSES.

Chap. II.

using scales of 5 feet to ah inch and 4000 pounds to an inch.
The stresses due to snow load are then found by multiplying
the dead load stresses by the ratio 0.7666.

Truss Members.

Dead Load
Stresses.

Snow Load
Stresses.

Upper chord.

BF
CG

Pounds.
-35090

- 35 090

- 30080

- 25070

Pounds.

— 26900

— 23060

— 19220

Lower chord,

FN
HN
KN
MN

+ 32870
+ 28 180
+ 23480
+ 18780

+ 25200
+ 21600
+ 18000
+ 14400

Braces,

FG
GH
HJ

KL
LM

- 3520
+ 5 870

- 5280
+ 7060

- 7040
+ 8450

— 2700
+ 4500
-4050
+ 5410

- 5400
+ 6480

Prob. 26. A wooden truss of the type of Fig. 28 has a span
of 60 feet, rise 15 feet, and distance between trusses 14 feet.
The upper chord is divided into six equal panels. Find the
apex loads, reactions, and the dead and snow load stresses in
all the members.

Art. 19. Wind Loads.

The pressure produced by the wind on a roof surface depends
on the direction and velocity of the wind and on the inclination
of the roof. The wind is supposed to move horizontally, and
a hurricane at 100 miles per hour exerts a pressure of probably
50 pounds per square foot of surface normal to its direction.

Art. 19.

WIND LOADS.

In determining the stresses due to wind it is often specified
that the wind pressure shall be taken at 40 pounds per square
foot of vertical surface.

While the subject is not fully understood, experiments show
that the resultant pressure of a horizontal wind on an inclined
surface may be represented by a normal force varying with the
roof inclination. The following values deduced from experi-
ments give the normal pressure per square foot for a horizontal
wind pressure of 40 pounds per square foot for different inclina-
tions of the roof surface :

Inclin.

Nor. Press.

Inclin.

Nor. Prbss.

Incun.

Nor. Press.

S-i

25°

22.6

45°

36.0

10°

9.6

30°

26.5

50°

38.1

15°

14.2

35°

30.1

55°

39-4

20°

18.4

40'

33.3

60°

40.0

For all inclinations exceeding 60 degrees the normal pressure
is 40 pounds per square foot, and for intermediate inclinations
the pressures are obtained by interpolation. Should the hori-
zontal wind pressure be assumed lower or higher than 40 pounds
the normal pressure is to be changed in the same ratio.

Let it be required to find the wind apex loads for the truss
whose dimensions are given in Fig. 30. The inclination of AB
is found to be 50°
40' and that of
£C 16"* 40', hence
from the above
table the normal
wind pressures are
respectively 38.3
and 15.6 pounds
per square foot. If the trusses be 12 feet apart the total
normal wind pressure on AB is

Fig. 30.

Vg.G" + 11.7' X 12 X 38.3 = 6954 pounds.

ROOF TRUSSES.

Chap. IL

one-half of which is applied at A and one-half at B, as shown.
In the same way the wind upon £C brings at B and C two
normal apex loads, each of i 407 pounds.

The two apex loads at B are then combined by means of the
force triangle, the resultant being 4 711 pounds.

Prob. 27. Find the wind apex loads for the truss in Fig. 25,
using the dimensions given in Art. 17.

Art. 20. A Truss with Fixed Ends.

Roof trusses of short span, especially wooden trusses, gener-
ally have both ends firmly * fixed ' to the supporting walls. The
reactions caused by the wind pressure are inclined and their

horizontal c o m-
ponents tend to
overturn the walls
of the building.

Let the truss in
Fig. 31 have both
ends fixed, the span being 40 feet, the rise of peak 10
feet, the rise of horizontal tie rod 2 feet, and the distance
between trusses 12 feet.

*^^ The length of the main rafter

is found to be 22.36 feet, its in-
clination 26° 34' and the normal
wind pressure (Art. 19) 23.8
pounds per square foot of roof
►^ surface. The apex load BC is

i X 22.36 X 12 X 23.8 =

3 193 pounds = 1.6 tons,

and the half apex loads AB and CD are each 0.8 tons.

As the sum of all the external forces in the direction of the
wind loads must equal zero, the reactions are parallel to them.

Fig. 33*

Art. 20.

A TRUSS WITH FIXED ENDS.

The directions and lines of action of the reactions at the
supports being therefore known it is required to find their
magnitudes, which is done by the method of Art. 8. The
results obtained are 2.2 tons for the left reaction, and i.o
ton for the right reaction. These values may now be checked
by computation. A perpendicular from the right support to
the left reaction measures on the truss diagram 35.78 feet.
Taking moments about the right support,

AL X 35.78 — 0.8 X 35.78 — 1.6 X 24.6 — 0.8 X 13.42 = o,

whence AL = 2.2 tons. In a similar manner, by taking mo-
ments about the left support, the reaction DL is found to be
1.0 ton. The same result is obtained both graphically and
analytically by replacing the apex wind loads by their resultant,
3.2 tons, applied at the middle of the rafter.

The stress diagram is begun by drawing ad normal to the
main rafter and equal to 3.2 tons by the scale of force, while
ad and cd are each made one-fourth of the length of ad. The
diagram is then completed in the manner described in preced-
ing articles. As A and k are found to

coincide it shows that there is no stress in
/^TATwhen the wind blows on the left side
of the truss.

With scales of 3 feet to an inch and i
ton to an inch the results given in the table
were obtained. As a check the stress in GL
is computed, the same value being found
as that given in the table.

If the wind loads be placed on the right-
hand side of the truss the corresponding
stress diagram will be the same as Fig. 32
when revolved about a vertical axis, there-
fore only one stress diagram is required in
this case. Accordingly the stresses in the members of either

Members.

Stresses.

Tons.

-4.68

+ 4.88

-f 1.68

-1.60

+ 3.32

-2.79

+ 2.08

+ 0.52

ROOF TRUSSES.

CiiAr. 11.

half of the truss when the wind blows on the right are the
same as th^ stresses in the corresponding members of the
other half of the truss for wind on the left side. For in-
stance, the stress in GH is -\- 3.32 and in EF is zero for wind
on the right.

Prob. 28. Find the wind stresses in all the members of the
truss in Prob. 26, both ends being firmly bolted to the walls.

Art. 21. A Truss with One End Free.

Changes in temperature cause expansion and contraction in
iron trusses which if both ends are fixed give rise to certain
stresses. To avoid these only one end is fastened to the sup-
porting wall, the
other being mere-
ly supported or
* free/ so that it
may move hori-
zontally in the
direction of the
^*«^-33- length of thetruss.

The free end may rest upon a smooth iron plate upon
which it slides, but this arrangement requires too much
friction to be overcome in the case of heavy roofs, especially if
the plate becomes rusty. Sometimes it is attached to a rocker,
as at A in Fig. 33, and often rollers are employed, as shown at
B. If no friction exists at the free end the reaction there is
vertical.

In determining the stresses due to wind when one end is
fixed and the other free it is necessary to construct two dia-
grams, one for the wind blowing on the fixed side and the
other for the wind load on the free side.

For example, let the truss in Fig. 34 be taken, the span

Art. 21.

A TRUSS WITH ONE END FREE.

being ^6 feet, rise of upper chord 19 feet, rise of lower chore
4 feet, and the distance between trusses 15 feet. The web
members consist of verticals and diagonals as shown, the
chords being divided into eight equal parts.

The inclination of the upper chord is the same as that in
the last article, hence the normal wind pressure per square foot

Fig. 34.

is 23.8 pounds, the apex loads ^(7, CD, and DE are LQtons
each, and the loads AB and £F are 0.95 tons each.

The reaction at the free end being vertical, its value is most
readily found by computation. Taking moments about the
left support and considering the total wind load concentrated
at the middle of the rafter,

7.6 X 21.24 — F[/ X 76 = O, whence FU=: 2.12 tons.

In Fig. 35 let a/ be drawn perpendicular to the loaded
upper chord and made
equal to 7.6 tons; ab
and e/ are then laid
off equal to 0.95 tons
each, and be divided
into three equal parts.
The vertical reaction
/u is next drawn

equal to 2.12 tons, and Fig. 35.

as the other reaction must close the force polygon, ua repre-

52 ROOF TRUSSES. CHAP. H

sents tbe magnitude and direction of the reaction XJA at the
fixed end.

The student should notice particularly that it is not possible
for the reaction at the fixed end to be parallel to the wind
loads when one end rests on rollers.

The stress diagram is constructed in the usual manner by
beginning with the forces acting at the left support, and passing
to joints alternately on the upper and lower chords. After
the joint DEN ML is reached the forces at the peak are taken
instead of passing to the central joint of the lower chord. The
force polygon for the peak is nefon^ the point o being deter-
mined by it. Now passing to the joint below the peak the
corresponding force polygon requires a parallel to UP to be
drawn through Uy and a parallel to OP through o. It is found
however that the former parallel passes through o^ thus closing
the polygon and reducing op to zero. But uo^ of^ and fu form
a closed force triangle which indicates that O belongs to the
entire space between the chords in the right half of the truss ;
therefore in this part of the truss each chord has the same
stress in every panel and no web member is stressed when the
wind blows on the opposite side.

The same thing may be shown in a different manner by be-
ginning at the right support where the reaction FU is held in
equilibrium by the stresses in UT and FT. The force triangle
fuo represents this relation, mo being the stress in UT and fo
that in FT. Passing to the next joint on the upper chord it is
required to draw a triangle two of whose sides shall be parallel
to the straight upper chord and the third side parallel to ST.
This causes the two sides to coincide and the third side to
disappear, hence the stress in FS equals that in FT and the
stress in ST is zero. The same conditions occur at each
joint on the upper and lower chords to the right of the
middle of the truss.

Art. 21.

A TRUSS WITH ONE END FREE.

If the stress diagram be accurately drawn, the point m
marks the intersection of ch, ugy Im, and tint. The points g, A,
ly and n will lie in a straight line and be equidistant.

When the wind blows upon the free side of the roof, as in
Fig. 36, the apex loads are the same as before, and the reaction

Fig. 36.

FU equals 4.68 tons. After fa in Fig. 37 is laid off, fu is

drawn vertically equal to 4.68 tons by scale, its length being

the same as uv in Fig. 35. In this stress diagram which is

completed similarly to

Fig. 35, an represents the

stress in the upper chord,

and un that in the lower

chord, of the left half of

the truss. The braces on

the left of NO are not

affected by the wind on

the right. Fig. 37.

The actual construction of the diagrams for this example
was made to scales of 6 feet to an inch and 2 tons to an inch, <
and the results are shown in Figs. 38 and 39. The stresses are
marked on the skeleton truss diagrams for convenient compari-
son, the members in compression being indicated by heavy
Unes and the tensile members by light 'lines. The checks by
computation give — 6.01 tons for the stress in FO, Fig. 34, and
^ S-OO tons in AN, Fig. 36.

ROOF TRUSSES.

CHAP. II.

It IS seen that greater stresses are produced in the chords ex-
cept EN^ and in the center vertical when the wind blows on the

Fig. 39.

fixed side, while the stresses in the braces to the windw^ard
with the exception of NO^ are the same for the wind blowing
on either side.

The reactions may also be obtained graphically. The direc-
tion of the reaction at the fixed end is unknown, but its line of
action passes through the joint at that end, hence the equilib-
rium polygon should be drawn from that joint to a vertical
through the support at the free end. The ray drawn parallel
to the closing line will then intersect the vertical through f at
the point u.

Prob. 29. Determine the reactions for the example in this
article by the equilibrium polygon.

Prob. 30. Find the wind stresses for the truss in Fig. 28,
using the dimensions given in Art. 18. The right end is to rest
on rollers.

Art. 22. Abbreviated Methods for Wind Stress.

When the lower chord is horizontal the stresses in the main
rafter either to the windward or to the leeward are the sam*?

Art. 22. ABBREVIATED METHODS FOR WIND STRESS.

for the wind blowing either o.z the fixed or on the free side.
The difference between the stresses in the horizontal chord

under the two conditions equals the horizontal component

lav, in Fig. 35) of the wind loads, the tension being less when

the wind blows on the

free side. In such a

case, then, only one

wind stress diagram

is really needed.

Even if the lower
chord is raised, a few
additional lines on the

diagram for wind on Fig. 4a.

the fixed side render the other diagram unnecessary. The full
lines in Fig. 42 are simply a copy of Fig. 35 and the truss
diagram in Fig. 40 is lettered to correspond, the rollers being
at the right support. Now supposing the wind loads to remain
unchanged and the rollers to be transferred to the left support,

5«

ROOF TRUSSES.

Chap. IL

the corresponding^ stress diagram is that shown in broken lines.
This relation of the wind loads, the truss, and the rollers is
illustrated in Fig. 41.

Let a horizontal line be drawn through u until its meets the
vertical through a at u' ^ and join «' with f and a thus giving
the new reactions fu' and u'a. Let u'x be drawn parallel to
bg meeting ug at Xy and u'y parallel \.o fo meeting uo at y. The
line joining x and j/ will be parallel to no.

The stresses in the main rafter are changed when the rollers
are transferred to the left support by the amount gg' = hh' =
//' = nn' = u'x = u'y = 00' ^ those in the lower chord are changed
by ux = uy, and that in NO by xy, while the remaining stresses
are unaltered.

Applying the scale, u'x and u'y are each found to be 1.02
tons, the differences ux and uy are each 4.33 tons and xy meas-
ures 0.90 tons. In the following table the first line contains
the stresses obtained from Fig. 35,* and after subtracting the
changes of stress, the same results are obtained as those derived
from Fig. 37:

STRESSES FOR WIND ON THE LEFT.

Truss Members, . .

Rollers on right, . .
Changes in stresses, .

n.72
1.02

9.66
1.02

7.60
1.02

5-54
1.02

6.01
1. 02

13-52
4-33

10.82
4-33

S.ii
4-33

5.40
4.33

432
0.90

Rollers on left, . >• .

10.70

8.64

6.58

4.52

4-99

9.19

6.49

3.78

1.07

342

Truss Members, . •

BG'

CH'

DL'

EN'

FO'

U'G'

U'K'

U'M'

U'O'

N'O*

For any other type of truss the changes are likewise easily
obtained. When the middle panel ha,s a horizontal tie as in
the example given on Plate I, the form of the auxiliary polygon
u^xzy is somewhat different from the preceding one and the
change for the horizontal tie is measured from k to z. The
following measurements were obtained from the original dia-
gram which was made with a scale of 2 tons to an inch : u'x =r

Art. 23. complete stresses for a triangular truss. 57

u'y = 0.88 tons, kx = ky = 4.29 tons, ^-sr = 4.17 tons, and xz =
yz = 0.40 tons.

Prob. 31. Prepare a table of wind stresses similar to the
above for the example given on Plate I.

Prob. 32. Find the wind stresses for a wooden truss like
Fig. 34 whose span is 74 feet, rise of peak 19 feet, the lower
chord being horizontal and the trusses 16 feet 3 inches apart, j

Art. 23. Complete Stresses for a Triangular Truss.

On Plate I are given the dimensions of a wrought iron roof
truss together with the specified loads. The struts are normal
to the, rafters as shown on the skeleton outline of the truss.
All the diagrams required to determine the stresses due to dead,
snow, and wind loads are shown, and are constructed in the
manner explained in the preceding articles of this Chapter. The
stresses as measured by scale (2 tons to an inch on the original)
are arranged in tabular form.

The preliminary computations give the following results :

Length of rafter, . . . . . 41.15 feet.

Weight of truss, ". . 3.85 tons.

Weight of roof covering, 8.15 tons.

Total dead load, 12.00 tons.

Dead apex load, 1.50 tons.

Dead load reaction, 6.00 tons.

Snow apex load, 1. 14 tons.

Ratio of snow load stresses to dead load stresses, . 0.76.^

Inclination of roof surface, 25^57'.

Normal wind pressure per square foot of roof, 23.3 pounds.

Total wind load, . . ' 7.92 tons.

Wind apex load, . . . ..,.'•... \; . • • ' 1.98 tons.
Horizontal component of total wind load, .• • • 3.47 tons.
Vertical component of total Xvind load, .^.•. • ' 7.12 tons..
Reaction at free end for wind on fixed side, . . 2.20 tons.
Reaction at free end for wind on frec'side, \* . .4.92.'tons.

58 ROOF TRUSSES. ChAP. II.

The reactions are obtained graphically as follows : Supposing
both ends of the truss to be fixed, the reactions due to wind
on the left side are au and ua in the stress diagram marked
* wind on fixed side.' They are obtained by the method of
Art. 8. The pole is at o and the equilibrium polygon is rst, the
wind apex loads being concentrated at apex 2. But since the
right end of the truss rests on rollers the reaction AK of the
right support is vertical and is represented by the line ak or
the vertical component of au. The closing side ka of the
force polygon represents the reaction of the left support. Ap-
plying the scale the value of ak is found to be 2.2 tons, which
is also the value of the vertical component of the reaction of
the left support when the wind blows on the right side.

In order to design a member for the range of stress (Me-
chanics of Materials, Art. 81), it is necessary to know the
minimum stress as well as the maximum stress to which it is
subjected by the combined loads. As the dead load always
acts its effect must be included in finding both the minimum
and the maximum stresses. Snow load always produces stresses
of the same kind as the dead load when the rafters are straight,
and hence is used only in obtaining the maximum. As the
wind cannot blow on more than one side of the roof at the
same time, only one of the wind stresses is to be combined with
the dead, or with the dead and snow load stresses. If in any
member a stress produced by the wind is of a different kind
from that due to dead load, the minimum stress equals
the algebraic sum of the two, but when, as in the present ex-
ample, all the stresses in any member are either tensile or com-
pressive, the minimum equals the dead load stress and the
maximum equals the sum of the dead, snow, and larger
wind load stresses.

It is seen from the table that the maximum chord stresses
are greater on the fixed side than on the free side, while the
maximum stresses in the bracing are the same on both sides.

^IA>

Art. 24. COMPLETE STRESSES FOR A CRESCENT TRUSS. 59

Prob. 33. A wrought iron truss of the type shown in Fig. 28
has a span of 76 feet, rise of peak 18 feet, and rise of tie
MN I feet. The trusses are 16 feet 6 inches apart, their right
ends resting on rollers. Find the maximum and minimum
stresses in all the members.

Art. 24. Complete Stresses for a Crescent Truss.

In the crescent truss whose outline and general dimensions
are given on Plate II the joints on both the upper and lower
chords lie on arcs of circles, and the alternate braces are radials
of the upper circular arc.

The following dimensions and apex loads are obtained either
graphically or by computation, and in some cases by both
methods, one being a check on the other :

Radius of circle containing the joints of the upper

chord, 51*25 feet.

Radius of circle containing the joints of the lower

chord, 99.06 feet.

Length of panels on the upper chord, . •" . . 11.08 feet.

Weight of truss, . . . • 4.12 tons.

Weight of roof covering, 8.52 tons.

Total dead load, 12.64 tons.

Dead apex load, 1.58 tons.

Dead load reaction, 6.32 tons.

Horizontal projection of panels of upper chord,

8.04, 9.44» 10.53, 10.99 feet.

Snow apex loads^ .... 0.48, 1.05, 1.20, 1.29, 1.32 tons.

Snow load reaction, 4.68 tons.

Inclinations of roof surface, . 43** 20', 31'' 05' 18° 30', 6° 10'.

Normal wind pressures per^q. ft., 35.2, 27.3, 17.2, 6.2 pounds.

Total wind loads on the panels, . . 3.12, 2.44, 1.52, 0.54 tons.

The dead load stress diagram is constructed in the same way
as in previous examples, and is symmetrical with respect to a
horizontal axis through k. As the snow apex loads are not

<50 ROOF TRUSSES. CHAP. II.

uniform a separate diagram for stresses due to snow loads is re-
quired, and this one is also symmetrical with respect to a hori-
zontal axis through k, '

The wind apex loads are next obtained by combining half of
the wind loads on the panels adjacent to each apex as illus-
trated in Art. 19. On the truss diagram in Plate II the wind
apex loads are drawn to double the scale of tons in order to
determine their directions with greater precision. The reac-
tions are then obtained by means of the equilibrium polygon
(Arts. 6, 8, and 21). The ray ou\ parallel to the closing side of
the polygon, intersects the resultant of the wind loads au'a at
the point u' giving au' and u'a as the reactions of the right and
left supports if both ends of the truss were fixed, but as the
right end is free its reaction ak is vertical and equal to the
vertical component ol.au\ The load liqe for wind on the free
side is obtained from that for wind on the fixed side by revolv-
ing it about a vertical axis, which operation maybe conveniently
performed by means of a piece of tracing paper. The point u'^
in this diagram is the same as u' in the preceding one, hence
only one equilibrium polygon is required. The reaction ha
of the right support is the vertical component of u"a.

It is observed that when the wind blows on the free side of
the truss, it causes compression in the lower chord from KH'
to the left end. The stresses in the upper chord are also con-
siderably less than for wind on the fixed side, and if the rise
were a little greater it would cause tension in one or more
^panels near the left support.

The truss diagram for this example should be twice as large
as that on Plate Ijfc' so that no line in any stress diagram would
be longer than the truss member to which it is parallel. This
relation was here disregarded as the limited size of the plate
would have reduced the stress diagrams to indistinctness.

" Unless special care is exercised in drawing the wind stress

t:
a

d
t

(
t

Art. 25. AMBIGUOUS cases. 6^

diagrams they will not close. As they lack the check of sym-
metry it is not so easy to locate the error, and therefore it is
best to construct new diagrams until one is obtained that clpses
properly. In all cases the work should proceed from each sup-
port toward the center of the truss.

Upon the completion of the stress diagrams their lines are
measured by the scale of force, the stresses aiTanged in a table
and the maximum and minimum stresses found, as explained in
the preceding article. For instance, the maximum stress in
KB is + 10.8 + 8.5 + 9-7 = + 29.0 tons, and its' minimum
stress is + 10.8 — 3.6 = + 7-2 tons. The maximum stress in
GH'xs — 0.1 -|- 2.3 = + ^'^ ^o"s and the minimum stress in
the same member is — o.i — o.i — 1.7 = — 1.9 tons.

Prob. 34. Find the maximum and minimum stresses in all
the members of Fig. 30, the right end resting on rollers.

Art. 25. Ambiguous Cases.

When, in the determination of the stresses in the Fink
truss shown in Fig. 43, the joint at the middle of the rafter is
reached the load BC and the stresses in BG and GH^x^ known,
leaving three stresses unknown, namely, those in CZ, ZAT, and
KH, As the resultant of the load and the known stresses can-
not be resolved into more than two given directions, another
condition needs to be added.

If the loads AB and CD be equal, as, in this, example, the
symmetrical relation of (7^.andZAr causes them to have equal
stresses and therefore fg and Im lie in the same straight line
parallel to hk. The polygon //^fe/>^A is then readily completed.
If the loads AB and CD be unequal, the panels remaining equal
on the upper chord, the polygon may be. drawn by noting that
the point k must lie midway between thq p,^rallels cl and dm.
This follows from tl^e fact that Zyl/ is, norni^l^to the ^ upper

t>>

ROOF TRUSSES.

Chap: II.

chord and that KL and MN are of equal length and make the
same angle with LM as well as with the upper chord. The tri-
angle lim is hence an isosceles triangle.

If CL and DM be of unequal lengths then both of these
methods fail. A general solution of this problem was given by

WiLLETT in a paper read before the Chicago Chapter of the
'American Institute of Architects, March 22, 1888, which con-
sists in temporarily changing the webbing of the truss.

Let the braces KL and LM be removed and the diagonal
K^^M be substituted as shown in Fig. 44, The load BC and
the stresses in £G and GH being known, those in NK^^ and
K'^C are found from the polygon hgbck'h in Fig. 45. For
the next apex on the upper chord the polygon is k'^cdtnk'^ the

Art. 2$.

AMBIGUOUS CASES.

line mi^' being the unknown stress thus determined Passing
to the joint HK^MNE where three stresses are now known,
the polygon ehk'^mne gives the unknown stresses mn and ne.
As the line mn is now fixed, the original webbing is restored
and the remaining parts of the stress diagram drawn ; hk" is
produced to meet mn at
k^ and kl and ml are
drawn parallel to KL and
ML respectively to meet
ck' produced at /.

The remaining half of
the diagram is not shown
in Fig. 45* If each half of the truss and the loading upon it
be equal to the other, the complete stress diagram will be
symmetrical with respect to a horizontal axis through e.

In the truss whose outline is given in
Fig. 46 it is not possible to begin the
stress diagram by considering the forces
acting at the left support, since three
unknown stresses hold in equilibrium
the known reaction and half apex load.

At the peak the load CC is sup-
ported by two members whose stresses
are de and ec in Fig. 47. The quad-
rilateral bced gives the stresses in the members meeting at
the joint BCED^ and (fVd!e gives those for the corresponding
joint in the right half of the truss. Passing to the joint below
the peak the stress polygon is found to be dedf. For the
left support the reaction, half apex load, and the stresses bd
and df are known, but one stress remains unknown.
As equilibrium exists at this joint the polygon must be
closed by the line fg^ which is also to be parallel to PG^
This completes the diagram, and by following around the

Fiff.4»

ROOF TRUSSES.

Chap. II.

polygons all the stresses are found to be compression except
fgy which is tension.

Suppose the tie FG to be omitted. Each reaction must
then be inclined in order to maintain equilibrium, its horizon-
tal component being equal to fg. The reaction of the right
support is therefore a!f (not drawn) and that of the left
support /^3!.

' Prob. 35. Find -the dead load stresses in the wooden truss
of Fig. 48, the span being 36 feet, the rise 12 feet, and the dis-
tance between trusses 12 feet. Then find the
stresses for a truss like Fig. 46 of the same
Fig. 48. ^ dimensions and compare the results.
Prob. 36. A wooden truss like Fig. 49 has a span of 48 feet
and a' rise of 12 feet, both ends being
fixed. Find the maximuni and mini-
mum stresses, the trusses being 14
feet apart

<^i<N/U>N.

Fig. 49.

Art. 26. Unsymmetrical Loads and Trusses.

In mills^ and shops,' loads are frequently suspended from the
lower chords of the roof trusses, as for instance, lines of shaft-
ing, etc. In such cases it
is convenient to determine
the stresses due td the loads
by means of a separate dia^
gram, as illustrated in Fig
50. The load AB produces
no stress in DE or GH^ while
it causes stresses in the othet
members of the satne kind as
those due to dead, snow, and
pj^ ^^ wind loads, ana hetice their

maximtum stresses. are increased; In specid cases such sua*

Art. 26. unsymmetrical loads and trusses.

pehded loads may even change softie maximam stresses due to
the other loads from compression to tension or from tension
to compression.

When a ceiling is attached to the lower chord it becomes
a part of the dead load and needs ino separate diagram. It is
combined at once! with the^ other loads, in the dead load
diagram, all the loads /and reactions being taken in regular
order around the truss and
laid off on -the' load line,
some portions of which
will be found to overlap.

It was formerly the prac-
tice in England to find the
effect of the wind on roof
trusses by taking vertical
loads of 20 pounds or more
per square foot of horizon-
tal projection, acting upon
one side of the roof. Fig.
51 shows a truss under
such loads and the resulting stress diagram. It is observed
that the stress thus caused in BF is greater than that in CG^
while under normal wind loads they are equal.

In Fig. 52 is given an unsymmetrical truss under the action
of dead load. The stress diagram is hence unsymmetrical and
has also fewer checks upon its construction. The main check
however still remains, which requires that after working from
each support toward the peak the closing line op shall be par-
allel to the member OP. To determine the stresses in an un-
symmetrical truss by the analytic method materially increases
the labor of computation required for a symmetrical truss, but
with the graphic method it makes no difference whatever for
any type of truss.

6fr

ROOF TRUSSES.

Chap. Ti.

Prob. 37. Let the load AB in Fig. 50 be 1.75 tons, and the
dimensions of the truss the same as given in Art. 20. Find
the stresses in all the members.

Fig. 5a.

Prob. 38. Find the wind stresses for the same truss unde^ a
vertical wind pressure of 20 pounds per horizontal square footf
and compare the results with those obtained in Art. 20.

Art. 27. L0AD3 on BRIDGE TRUSSES. &7

CHAPTER III.

BRIDGE TRUSSES.

Art. 27. Loads on Bridge Trusses.

The weight of the floor, lateral bracing, trusses, and all the
pieces that unite and stiffen them, compose the dead load of a
bridge. This weight depends upon the span, width, and style
of the bridge, and upon the live load and unit stresses adopted
in its design, and varies considerably in individual cases ; it is
usually lighter for a highway bridge than for a railroad bridge.

The total weight or dead load of a highway bridge with two
trusses may be expressed approximately by the following em-
pirical formula :

w = 140 + 12^ -{- 0.2^/ — 0.4/,

in which w is the weight in pounds per linear foot, d the width
of the bridge in feet (including sidewalks, if any), and / the
span in feet. The width of a highway bridge in the clear varies
from 16 to 24 feet, which is only exceeded in large cities.

The total dead load of a railroad bridge for a standard gauge
track may be approximately found from the following empiri-
cal formulas :

For single track, w = 560 + $.61,
For double track, w = i 070+ 10.7/.

For spans not exceeding 300 feet these formulas give values
usually a little larger than the actual weights, but sufficiently
exact for the determination of the stresses. For spans greater
than 300 feet they should not be used. The width in the clear
between the trusses of a through railroad bridge is about 13 or

68 BRIDGE TRUSSES. CHAP. Ill

J4 feet for a single track, and 25 feet for a double track.
Wooden and iron bridges of the same strength do not materi-
ally differ in weight.

The live load is that which passes over the bridge, and con-
sists of wagons and foot passengers on highway bridges and
trains on railroad bridges.

The live loads usually assumed for highway bridges, in
pounds per square foot of floor surface, are as follows :

For Country Bridges. For City Bridgss

Spans under 50 feet, 90 100

Spafls 50 to 125 feet, 80 90

Spans 125 to 200 feet, 70 80

Spans over 200 feet, 60 70

This maximum load consists of a dense crowd of people cover-
ing the roadway and sidewalks, and as there is less liability
to crowds on long spans as compared with short spans, and
for country bridges as compared with those in the city, the
load is varied accordingly. Each truss supports one-half of
the wTiole load.

By multiplying the given weight per square foot by the clear
width of roadway and sidewalks the live load per linear foot of
bridge is obtained. This load is to be placed so as to produce
the largest possible stress in any truss member considered.

The live load for a railroad bridge is that of the heaviest
cars and locomotives which pass, or are to pass, over it. When
a bridge is to be designed these loads are generally specified
by the railroad company. Several methods of stating the live
load are in use :

1st. A uniform load per linear foot of single track, the short*
est spans having the heaviest loads, and about as follows:

Span, so, 100, 150, 200, 300, 400 feet.

Load, 4200, 3600, 3200, 3000, 2600, 2400 pounds.

Art. 28. DEAD LOAD STRESSES. 6g

This method was formerly much used, but is now only occa-
sionally employed for computing, the chord stresses.

2d. A uniform train load, varying as above, which is preceded
by one panel of heavy locomotive load. If / be the length of
the panel in feet this preceding load in pounds is for a single
track often taken as 30 000 + 3 Soo/. Sometimes the locomo-
tive load is used for two or three panels in front of the train'
instead of for one.

3d. A uniform train load, varying as above, preceded by one
or two locomotives with their tenders, the weight of these
being taken as concentrated upon, the drivers and other wheels.
This style of loading will be fully explained in Art. 39.

If the bridge have only one track each truss sustains but one-
half the loads above given, but if it have two tracks each truss
sustains the loads as stated, for it might happen that both
tracks would be covered at the same time.

Prob. 39. A highway bridge in the country has a span of 88^
feet, a roadway 16 feet wide, two sidewalks each 4 feet wide,
and each truss has 9 panels. Find the dead and live panel
loads.

Art. 28. Dead Load Stresses.

The principle of the force polygon used in the last chapter
for the determination of stresses in roof trusses may often be
advantageously employed for the analysis of bridge trusses.

As an illustration let a through Pratt truss for a highway
bridge be taken, having 8 panels, a span of 176 feet, and a
depth of 26 feet. Let the bridge have a roadway 2 1 feet wide
and two sidewalks each 6 feet wide. By the formula in Art.
2y the dead load per linear foot of bridge is found to be i 627
pounds, and the dead panel load per truss is 8.95 short tons.
It is required to determine the stresses due to this dead load,
all being supposed to be on the lower chord.

BRIDGE TRUSSES.

Chap. III.

In Fig. 54 let qy be laid off by scale equal to 8.95 X 7 = 62.65
tons, let it be divided carefully into 7 equal parts, lettered in

Fig. 53.

Fig. 54.

the manner indicat-
ed and bisected at
a. The effective
reactions are ya
and aq, the half
panel loads at the
supports being
omitted for conven-
ience. At the left
support the reac-
tion AQ is held in
equilibrium by the
stresses in QB and
BA, and by Art. i
these will form the
closed force trian-

gle aq6. As aq acts in the direction from a to q the other
fgrces must act in the same direction around the triangle, that
Jis, from q to d and from d to a. Transferring these directions
to the joint AQB the stress in QB acts away from the joint and
is therefore tension, while the stress in BA acts toward the
joint and is compression. The construction of the stress
diagram is continued by passing to the joints alternately on
the lower and upper chords until the middle of the truss is
reached, then beginning at the right support and passing to the
joints in the opposite direction until the diagram closes. If

Art. 28.

DEAD LOAD STRESSES.

accurately drawn the diagram will be symmetrical with
respect to a/i. The polygon qrcd being a rectangle shows
the tension in CB to be equal to the load QR, and' the
tension in RC to be equal to that in BQ, The rectan-
gle esad shows the stresses in AD and £S to be equal in
magnitude, the former being compression and the latter
tension as previously determined. Again, a/ equals gl for
a similar reason. It is seen therefore that the stresses in any
two chord members whose adjacent spaces are separated only
by a vertical have the same magnitude.

The compression in the upper chord increases toward the
middle of the truss, and the same is true of the tension in the
lower chord. The diagonals are all in tension but AB, while
the verticals are all in compression but BC. In the web mem-
bers the stresses increase from the middle toward the ends of
the truss, with the exception of BCj which only serves to trans-
fer the load QR to the upper chord.

The following results were obtained from a stress diagram
drawn to a scale of 8 tons to an inch :

Truss Members.

Stresses. I

Truss Members.

Stresses.

Tons.

fAD

- 45.4

rAB

- 4I.I

Upper chord -

AF
^AH

- 568

- 60.6

Diagonals

CD
EF

+ 29.3

+ 17.6

+ 5-9

rBQ

+ 26.5

^BC

+ 9.0

Lower chord -

CR
ES

+ 26.5
+ 45-4

Verticals

DE
FG

- 130
-4.5

Vgt

+ 56.8

.HJ

As a final check the stress in AH is computed thus :

31.325 X 88 -8.95 (66 -1-44 +22) + ^// X 26 = 0,
whence AH =^ — 60.6 tons.

If instead of concentrating all the dead load on the loWer

BRIDGE TRUSSES.

Chap. IIL

I B

C J

LO<fc.rfJ

fd*i

45-4 \

H56.S

454

1^6.5

26 i

chord, it be divided so that panel loads of 2.95 tons be applied
on the upper chord and 6.0 tons on the lower, the stress dia-
gram assumes the form shown in Fig. 56, the broken lines re-
ferring to stresses in the right half of the truss. The loads and

reactions are
taken in regular
order around
the truss and
laid off in suc-
cession on the
load line. The
lower panel
loads are taken
from left to
right, then the
right reaction
followed by the
panel loads on
the upper chord
from right to
left, and finally

the left reaction which closes the polygon. This polygon is
efghh'g'f'e'a'b'c'd'dcbae. The stresses obtained are marked on
the right half of the truss diagram, compression being indicated
by the heavy lines and tension by the light lines.

Comparing Fig. 56 with Fig. 54 it is observed that all the
stresses are the same except those in the verticals whose com-
pression is increased by 2.95 tons — the weight of the upper
panel loads. The tension in KL is accordingly diminished by
the same amount.

A further examination of Fig. 56 shows that the vertical
component of ak is ae^ the reaction ; the vertical component
of Itn is ae — ef—ba\ mn=^ae — ^f—fg — ba\ and so on.
Therefore the following principle is established :

Art. 29. LIVE LOAD STRESSES IN A WARREN TRUSS. 73

For trusses with horizontal chords the vertical component
of the stress in any web member equals the reaction
minus all the loads on the left, that is, equals the verti-
cal shear for that member.

The only exception to this is the vertical KL for reasons
already given in the first part of this article. The above prin-
ciple may be derived from the relation existing between the
stresses in any section of a. truss and the external forces on
either side of that section as demonstrated in Arts. 7 and 17.

The diagram also shows that the difference between the
magnitudes of the stresses in any two chord members equals
the sum of the horizontal components of the stresses in the
web members situated between them. For instance, the dif-
ference between hp and gn is the horizontal component of no^
which also equals the difference between co and gn or between
ph and bm. The horizontal component of any diagonal is
called a chord increment and forms the base of a right triangle
whose height is the vertical shear in that diagonal. (Roofs and
Bridges, Part I, Art. 26.)

Prob. 40. A through Pratt truss of a single track railroad
bridge consists of 7 panels, each 23 feet 2 inches long and 25
feet deep. One-third of the dead load being on the upper
chord, find the stresses in all the members.

Art. 29. Live Load Stresses in a Warren Truss.

'As every load placed upon a bridge truss produces com-,
pression in the upper chord and tension in the lower chord,
the greatest ^hord stresses produced by a live load occur when
every panel point of the chord supporting the floor beams is
loaded. The chord stresses due to a uniform live load are
hence obtained from a diagram exactly similar to that for a
dead load applied only upon one chord. Hence the stress in
any chord member, due to a uniform live load, bears the same

BRIDGE TRUSSES.

Chap. III.

ratio to the dead load stress as that of the corresponding apex
loads, and accordingly either stress may be derived from the
other by using this constant ratio.

In order to investigate the effect of live load on the web
members, let a deck Warren truss of 7 panels be taken, the
span being 126 feet, the depth 12 feet, and the live load i 700
pounds per linear foot per truss. The live panel load is then
15.3 tons.

Placing a panel load at apex i in Fig. 57, the stresses due
to this single load are obtained by drawing Fig. 58 in the usual

A I z 3 4 S 6 A'

Fig. 57.

manner. The reaction a'k is one-seventh of the panel load.

The stresses in the braces are found to be alternately compres-
sion and tension each way from the
load, and on either side the stresses
are the same in magnitude from
the load to the support, their ver-
tical components being equal to
the reaction on that side.

For a panel load at apex 2 the
reaction of the right support will
be twice as great as for the load
>a' at I, and hence the stresses in all
y the braces on the right of apex 2
^^^' 5^* will also be twice as large ; for a

load at apex 3 the stresses on its right will be three times as
great as for the load at i, and so on. Again, a panel load at
apex 6 will produce the same stresses on its left as the load at I
caused on its right, and a load at 3 will produce stresses in the
braces on its left equal to four times those due to the load at

Art. 2g. live load stresses in A warren truss.

6. The stress in each web member due to a single live pari^l
load at any apex may therefore be obtained by taking a simple
multiple of the stress for that member as given by Fig. 58.

In the following table the first and sixth lines are thus filled
out directly with the results scaled off from the diagram (which
was originally drawn to a scale of 5 tons to' an inch), and the,
other hnes by taking multiples of these as indicated above.
The stresses in each column are then combined so as to give
the greatest and least stresses and those due to a uniform live
load throughout.

Web Members.

•
DE

Live panel load at i,
2,
3,
4,

5.
6,

+16.38

+13.65

+10.92

+8.19

+5.46

+2.73

-16.38

-13.65

— 10.92

— 8.19

-546

-2.73

+2.73

-2.73

+2.73
+5.46

-2.73
-5.46
—8.19

+ 13.65

+ 10.92

+8.19

+5.46

+2.73

— 13.65

-.46

— 10.92
—8.19
-5.46
-2.73

+ 10.92
+8.19
+5.46
+2.73

— 10.92
— 8.19
-5.46
-2.73

+8.19
+5.46
+2.73

Live load, greatest,
Live load, least.
Uniform live load.

+57.33

-57.33

+40.95

-2.73

+38.22

-40.95

+2.73

-38.22

+27.30 -27.30

— 8.19' +8.19

+ I9.II — I9.II

+16.35

-16.35

It is found that for any given diagonal all the loads on one
side of it cause one kind of stress, while those on the other side
cause the opposite stress. The maximum stress is hence pro-
duced in a web member when the live load covers the larger
segment of the span, and the minimum stress when the smaller
segment is loaded.

In the construction of stress diagrams for a truss with hori-
zontal chords and equal panels it is not necessary to draw the
skeleton outline of the truss to a large scale. If in this exarr)-
ple ax be laid off by the linear scale equal to some convenient
multiple of the half panel length and ay equal to the same
multiple of the depth of truss, xy will give the direction of half

BRIDGE TRUSSES.

Chap. III.

the web members, and in transferring this direction the triangle
will require very little shifting along a straight edge, thus pro-
moting accuracy. The line ay should be longer than ak. Com-
pleting the rectangle xayz^\\\^ direction of the remaining braces
will be given by az.

The results in the line ' uniform live load ' in the table should
be the same as those derived from a stress diagram made for a
live panel load at every panel point or apex, and may thus be
checked. As such a diagram is required for the chord stresses
it will also answer this purpose.

As the live load cannot act alone, but always in conjunction
with the dead load, the stresses due to the combined loads are
required. These are given in the following table. The dead
apex load on the lower chord is 1.7 1 tons, and on the upper
chord 3.42 tons. The dead load stresses were obtained from
a diagram of 5 tons to an inch.

Web Members.

Live load, greatest,

+ 57.33

-57.33

+ 40.95

- 40.95

+ 27.30

- 27.30

+ 16.35

Live load, least,

-2.73

+ 2.73

— 8.19

+ 8.19

-16.35

Dead load.

+ 20.35

-18.18

+ 13.85

-11.68

+ 7-44

-5.34

+ 1.07

Maximum,

+ 77.68

-75.51

+ 54.80

-52.63

+ 34.74

- 32.64

+ 17.42

Minimum,

+ 20.35

-18.18

+ 11. 12

-8.95

-0.75

+ 2.85

- 15.28

The stresses due to the combined load are obtained by adding
the dead load stresses to each of the corresponding live load
stresses.

By comparing these results with the computations for the
same example in Roofs and Bridges, Part I, Art. 34, it is seen
that they are correct to or within one-tenth of a ton, which is
sufficiently accurate for all purposes of design.

The same method of tabulation might be applied to the

Art. 30. LIVE LOAD STRESSES IN A PRATT TRUSS. JJ

chord stresses, but the diagram for a full live load can be made
in less time.

Prob. 41. Find the maximum and minimum chord stresses
for the above example.

Art. 30. Live Load Stresses in a Pratt Truss.

If a Pratt truss were built having only those diagonals which
are strained under dead load it would be necessary that some
of them resist the compression produced by certain positions
of the live load. As, however, the diagonals are only to be
subjected to tension this is prevented by inserting other diago-
nals inclined in the opposite direction. Panels having two
diagonals are said to be counter-braced and the additional
diagonals are called counter-ties or counter-braces. The main
and counter brace in any panel cannot both be strained at the
same time by any system of loading.

When the counter-ties are called into action by the live load
the stresses in the adjacent verticals are different from what
they would be provided the main braces could withstand com-
pression. This can readily be seen by changing any diagonal
and making the corresponding alteration in the stress diagram.

Let the Pratt truss whose dead load stresses were deter-
mined in Art. 28 be again considered. It consists of 8 panels
each 22 feet long and 26 feet deep. The total width of the
bridge, including sidewalks, is 33 feet. Taking the live load at
80 pounds per square foot of floor surface the panel load per
truss is

J X 22 X 33 X 80 = 29 040 pounds = 14.52 tons.

The truss diagram. Fig. 59, is drawn with the diagonals all
inclined one way, the main ones being on the left of the center
and the counters on the right. Placing one live panel load RR
at apex i the stress diagram. Fig. 60, is constructed which gives

BRIDGE TRUSSES.

Chap. III.

all the stresses due to this load. Fig. 6r gives the stresses due
to a live panel load at apex 7. After rr is laid off equal to the
panel load of 14.52 tons, ru is marked off, by a suitable linear
scale, equal to 26 feet and rt equal to 22 feet, then ut gives the
incHnation of the diagonals. Drawing rs parallel to ut it is found

Fig. 59.

Fig. 60.

to measure 19.02 tons. In both Fig. 60 and Fig. 61 the smaller
vertical lines are one-eighth of the length of rr, or 1.82 tons,
and the smaller diagonals are one-eighth as long as rs, or '2.38
tons. The only part of the stress diagrams actually required
consists of the similar right triangles urt and rrs^ and the Hne
rs is to be carefully determined. Unless these triangles are
very nearly of the same size, as in the above example, the lat-
ter should be made larger than the former.

The live load stresses are then tabulated as explained in
Art. 29. The dead load stresses are obtained from Fig. 62, one-
half of which is like the same part of Fig. 56, while the other
half is changed so as to give the stresses when the counter-

Art. 30. LIVE LOAD STRESSES IN A PRATT TRUSS.

braces alone are inserted in the right half of the truss as
shown in Fig. 59. Two lines in this diagram are marked
hj^ the upper
one measures '
3.0 tons and
represents the
compression in
HJ when the
main ties act on
each side of it
(which occurs
under a full live
load), and the
lower line meas-
ures 1.5 tons being the tension in HJ when the main tie acts
on the left and the counter-tie on the right, as indicated in the
truss diagram.

All of the tabulated results except those in the last two lines
of each table were obtained as if the web members in Fig. 59

Fig. 6a ,

Truss Members.

End Posts.

Main Ties.

Counter-Ties.

AB= PA

Live panel load at

X
8

3
4
S
6

-16.7
-14.3
- 11.9

-9-5
-7X
-4.8
- 2.4

- 2.4
+ 14.3'
+ 11.9
+ 95
+ 7.1
+ 4.8
+ 2.4

- 2.4
-4.8

- 8.4
-4.8

- 7.1

- 24
-4.8

- 71

- 9-5
+ 7.1
+ 4.8
+ 2.4

- 2.4
-4.8

- 7.1
-9.5

- It.Q

- a.4
-4.8

- 7-1
-9-5

- 11.9

+ 11.9
+ 9.5
+ 7.T
+ 4.8
+ 2.4

+ 9.5
+ 71
+ 4.8
+ 2.4

+ 4.8

-14 3

+ 2.4

Uuiform live /oad,

-66.7

+ 47.6

+ 28.5

+ 9.5

- 9.5

-28.5

-476

+ Total,
- Toul,
Dead load,

-66.7
-4t.i

+ 50.0

- 2.4

+ 29.3

+ 35.7

- 7.2

+ 17.6

+ 23.8
- 14.3
+ 5-9

+ 14.3
-23.8

-5-9

+ 7.2

- 35-7
-17.6

+ 2.4
— 50.0
-29.3

Maximum,
Minimum,

— 107.8
- 41.X

+ 79-3
+ 26.9

+ 53.3
+ 10.4

+ 29.7

+ 8.4

BRIDGE TRUSSES.

Chap. III.

Tkuss Mbmbbrs.

Verticals.

Live panel load at

3
4
5
6
7

+ 14.5

+ 1.8
+ 3.6

+ 1.8
+ 3.6
+ 5.5

+ 1.8
+ 3.6
+ 5.5
+ 7.3

+ 1.8
+ 3.6
+ 5-5
+ 7.3
+ q.i

+ 1.8
+ 3.6
+ 5.5
+ 7.3
+ 9.1
+ 10.9

+ 1.8
+ 3.6
+ 5.5
+ 73
+ 9.1
+ 10.9
+ 12.7

. o

o
o

-9-1
-7.3
-55

-3.6
-1.8

-73
-5.5

-3.6
- 1.8

- 5.5
-3.6

- 1.8

-3.6
- 1.8

- 1.8

Uniform live load,

+ 14.5

-21.9

-7.3

+ 7.3

+ 21.9

+ 36.4

+ 509

+ Totol,
- Toul,

Dead load.

+ 14.5

+ 6.0

+ 5.4
-a7.3

-16.5

+ 109
- 18.2

— 7-5

+ 18.2
— 10.9

[- 3.0]
+ 1.5

+ 27.3
- 5.4

+ 10.5

+ 382
- 1.8

+ 19.5

+ S0.9

+ 28.5

Maximum,
Minimum,

+ 20.5
+ 6.0

-438
— 11. 1

- 257
- 30

-9-4
- 30

....

could take either tension or compression. It is now required
to find the actual maximum and minimum stresses due to the
combined loads under the limitation that the diagonals can take
only tension. To avoid repetition only those members are re.
ferred to in the following explanation whose treatment differs
from that of the preceding article.

The minimum stress in GH is zero as the compression due
to the live panel loads at 1,2, and 3 is greater than its dead
load tension. The maximum stress in the counter yAT is + 14.3
— 5.9 = -|- 8.4 tons ; the minimum stress is zero since the dead
load as well as the live panel loads at i, 2, 3, and 4 tend to
^ compress this member. A counter is therefore required in the
fourth and fifth panels of the truss. The counters LM and
NO are not theoretically required because the greatest tensile
stress produced by the live load in each one is not sufficient to
overcome the tendency of the dead load to compress the same.
This is seen also from the fact that the minimum* stresses in
CD and EF are + 26.9 and + 10.4 tons respectively, which

Art. 30. LIVE LOAD STRESSES IN A PRATT TRUSS. 8t

implies that their dead load tension is not reduced to zero un-
der the most unfavorable position of the live load.

In finding the maximum and minimum stresses in any ver-
tical it is necessary to consider whether the adjacent diagonals
shown in the truss diagram really act under the. various con-
ditions of loading. If it is found that they do not act, then the
stresses given by the table for the vertical cannot occur.

The live panel loads at 3, 4, S, 6, and 7 together with the
dead load produce in D£ the maximum compressive stress
equal to — 27.3 — 16.5 = — 43.8 tons, provided the adjacent
diagonals are in tension. Under the influence of these loads
CD and EF are both found to be in tension, and hence the
value just obtained is the required stress. The live loads at
I and 2 acting in addition to the dead load produce the mini-
mum stress of -f 5.4— 16.5 = — ii.i tons for the same reason.

The minimum stress in FG is due to the live panel loads at
I, 2, and 3, which with the dead load give a stress of + I0'9
— 7.5 = -}^ 3.4 tons, provided the adjacent diagonals EF and
GH are in tension. These loads would cause a tension of 22.3
tons in EF and 8.4 tons compression in GH, but as GH cannot
take compression the counter-tie in the same panel is brought
into action. The stress of + 3.4 tons in FG therefore cannot
occur. When the main tie acts on its left and the counter on
its right, the vertical simply supports the dead panel load on
the top chord, hence the minimum stress in FG is — 3.0 tons.
In the same way the minimum stress in H/ is also found to
be — 3.0 tons. Under the live panel loads at 5, 6, and 7 with
the dead load the diagonals GH and /K are strained, hence
the stress of — 10.9 tons in H/ must be added to that pro-
duced by the dead load, + i-S tons, to give the maximum
of — 9.4 tons.

Passing to the verticals on the right of the center it is seen
that the combination of loads which would give either the

82 BRIDGE TRUSSES. CHAP. III.

maximum dt the minimum stress according to the table will
not produce tension in both of the adjacent diagonals, and ac-
cordingly no additional values can be inserted in the table.
The maximum and minimum stresses for these verticals will
be the same as for those on the left of the center. The tabu-
lation for the verticals beyond /// is also shown to be unneces-
sary as KL in Fig. 59 has no diagonal on its right, the counters
LM and NO not being required.

The chord stresses are found in the same way as for dead
load (Art. 28), only the main ties however being inserted in
the truss diagram. The stresses in a Howe truss are deter-
mined in a similar manner to that employed for the Pratt truss,
the diagonals in that case taking only compression.

The following important principles may now be stated, at-
tention to which will materially reduce the work in solving other
problems for cither type of truss mentioned :

The maximum stress in any vertical or main diagonal is
produced when the live load covers the larger segment
of the span.

The maximum stress in any counter diagonal occurs when
the live load covers the smaller segment of the span.

The minimum stresses in both diagonals of a counter-braced
panel are zero.

The minimum stress in the diagonal of a panel not coun-
ter-braced is given when the live load covers the smaller
segment of the span.

The minimum stress in any vertical adjacent to a counter
diagonal equals the dead apex load on the upper chord
of a Pratt truss or the lower chord of a Howe truss.

The minimum stress in a vertical not adjacent to a counter
diagonal is produced when the live load covers the
smaller segment of the span.

Prob. 42. A Howe truss of 12 panels for a through single
track railroad bridge has a span of 123 feet and a depth of 15

Art. 31. sNO.w load stresses. 83

feet. The dead load is 625 pounds per linear foot, one-third
to be taken on the upper chord, and the uniform train load is
I 700 pounds per linear foot per truss. Find the maximunn and
minimum stresses.

Art. 31. Snow Load Stresses.

• In addition to the dead and live load stresses must be con-
sidered those due to the snow and wind. The snow load for
highway bridges is taken lower than for roofs since in the coun-
try it is not probable that the full live load would come on the
bridge while a heavy fall of snow rests upon it, while in towns
the sidewalks are generally cleared of snow. The snow load
may vary from 20 to o pounds per square foot of floor surface
depending upon the climate where the bridge is situated. As
the floor of railroad bridges is open so that but little is re-
tained no snovv load is regarded.

As the snow load is uniforrn the stress diagram is exactly
similar to that for dead load if the latter be taken only on the
chord supporting the floor, or, like the diagram for a full live
load. A separate diagram is hence not required as the stresses
may be obtained from either of those mentioned by graphic
multiplication, the same ratio existing between the stresses as
that of the respective panel loads.

For the example in the preceding article the snow panel
load is

i X 21 X 22 X 15 = 3465 pounds = 1.73 tons,

for a load of 15 pounds per square foot of floor surface for

the roadway only. The ratio of snow to uniform live load

1.73
stresses is therefore = 0.119. This gives a snow load

stress of —8.0 tons in the end post AB, -j- 34 tons in £Fy —
2.6 tons in DEy etc.

BRIDGE TRUSSES.

Chap, in

Prob. 43. A through Pratt truss for a highway bridge in
a village has 12 panels each 11 feet long and 14 feet deep.
The roadway is 18 feet 9 inches wide, and there are two side-
walks each 5 feet wide. Find the stresses due to a snow load
of 10 pounds per square foot.

Art. 32. Wind Stresses.

The greatest stresses due to wind are produced when it blows
horizontally at right angles to the line of the bridge. The sur-
face exposed to wind action is usually taken as double the area
of the side elevation of one truss. If this area be not known
an approximate value may be obtained by taking as many
square feet as there are linear feet in the skeleton outline of
the truss. For railroad bridges the surface of the side of a train,
taken at 10 square feet per linear foot of train, is added to the
above. No similar addition is made for highway bridges as it
is not probable that the live loads would cover them when the
wind is blowing at its maximum rate.

The wind pressure is taken
from 30 to 40 pounds per square
foot and produces its maximum
effect when acting like a live
load. The wind load on the
trusses is divided between the
upper and lower lateral bracing
while the wind load upon the
train is all taken by the lateral
bracing of those chords which
support the floor. The lateral

bracing is generally of the Pratt type, the floor beams acting

as the normal struts in one of the systems.

For an example let the through Pratt truss highway bridge
whose dimensions are given in Art. 28 be again taken. The

Art. 32.

WIND STRESSES.

?=F1

Fig. 66.

side elevation of the outline of the left half of the truss is

shown in Fig. 64, the plan of the upper lateral bracing in

Fig. 63, and that of the lower lateral a^

bracing in Fig. 65. When the wind

blows in the direction indicated

by the arrows and moves from the

right toward the left the diagonals

drawn in full lines are strained and

when it blows in the opposite direction the other set of

diagonals is strained.

The approximate area exposed to wind action is,

176+ 132 + 7 X 26 + 12 X 34.1 = 889 square feet.
Taking the wind pressure at 40 pounds per square foot the total
wind load is

889 X 40 = 35 560 pounds = 17.78 tons,
and the wind panel load is

17.78 -7- (6 + 8) = 1.27 tons.

The chord stresses are determined for uniform wind load
by means of Fig. 66 for the upper lateral bracing and from
Fig. 6y for the lower
system. Only one-
half of each diagram ^
is shown, the other
half being symmet-
rical with it. When
the wind blows in the
opposite direction
the chord stresses on each side of the bridge will exchange values.

For the wind blowing in the direction of the arrows the dia-
grams give the following stresses in tons ; for the upper chord,

A'F>

A'H'

-3.3

- 5.3

-6.0

+ 3.3

+ 5.3

86 BRIDGE TRUSSES.

and for the lower chord,

CHAr. III.

R'C

R'E'

R>G>

-4.7

-7.9

-9.9

— 10.6

+ 4.7

+ 7.9

+ 9.9

In accordance with the simplified construction given in Art.
30 let a horizontal and a vertical line be drawn through s in
Fig. 68. Let st be laid oflF equal to 2 1 feet and su equal to 22
feet, then tus will be the angle which the
diagonals make with the chords. With
a suitable scale of force let sw be made
equal to 1.27 tons and wx drawn parallel
to tu. Applying the scale to wx it is
found to measure 1.84 tons. Dividing
this value by 6 and 8, the number of
panels in the upper and lower chords
respectively, the quotients 0.3 1 and 0.23 tons are obtained which
form the basis of the tabulations for the stresses in the diago-
nals. Dividing the panel load, 1.27 tons, by the same numbers
the corresponding quotients 0.21 and 0.16 tons are found for
the verticals. The following tables are now prepared :

For the upper lateral bracing,

Pig. 08.

Diagonals.

Struts.

DD'

FF'

HH'

DF'

FH'

HJ'

Wind panel load at i
2
3
4
5

+ 1.55
+ 1.24

+ 0.93
+ 0.62
+ 0.31

+ 1.24
+ 0.93
+ 0.62

+ 0.31

+ 0.93
+ 0.62
+ 0.31

- 1.05

- 0.84

- 0.63

- 0.42

- 0.21

- 0.84

- 0.63

- 0.42

- 0.21

- 0.63

- 0.42

- 0.21

Maximum wiiid stresses

+ 4.7

+ 3.1

+ 1.9

- 3.2

- 2.1

- 1.3

Art. 33. STRESSES due to initial tension.
and for the lower lateral bracing,

Diagonals.

Struts.

BB'

EE'

GG'

CE'

GK'

Wind panel load at i

+ 1.61

— 1. 12

....

+ 1.38

— 0.96

....

+ 1.15

+ 1.1S

-0.80

— 0.80

-0.80

4-0.93

4-0.92

— 0.64

- 0.64

— 0.64

+ 0.69

4-0.69

-0.48

— 0.48

+ 0.46

4-0.46

- 0.32

-0.32

— 0.32

-0.32

4-0.23

40-23

40.23

— 0.16

-0.16

— O.I<J

Maximum wind stresses

+ 6.4

4-4.8

4-3.5

+ »-3

-4-5

-3-4

- 2.4.

-1.6

Both diagonals in the same panel have equal stresses due to
wind. The minimum stresses in all the web members are
zero. Since every panel is counter-braced only tensile stresses
in the diagonals and compressive stresses in the struts need to
be tabulated.

Prob. 44. A through single track railroad bridge has a span
of 120 feet. Its trusses are of the Pratt type, have 6 panels,
21 feet deep, and are 16 feet apart between centers. Find the
stresses due to a wind pressure of 40 pounds per square foot,
provided only the wind pressure on the train be considered as
a moving load.

Art. 33. Stresses due to Initial Tension.

In trusses whose diagonals take only tension the counter-
ties are made adjustable in order to be drawn up to a certain
degree of tension when the bridge is unloaded. The stress
thus introduced in these truss members is called initial tension^
and serves to prevent the vibration of the diagonals in the
counter-braced panels under moving loads and to stiflFen the
truss as a whole.

It is required to determine the stresses produced in other
members of the truss when all the counters are subjected to a

BRIDGE TRUSSES.

Chap. III.

given amount of initial tension. The number of counters in
practice is larger than is theoretically required. As the stress
in any counter is equivalent to two external forces, each equal
to the initial tension, applied at the joints united by the
counter-tie and acting toward each other, it may be replaced by
them in this analysis.

In the Pratt truss considered in Arts. 28, 30, and 32 let the
counters in the third, fourth, fifth, and sixth panels be each

subject to an initial
tension of 4 tons.
The truss diagram is
shown in Fig. 69, each
of the external forces
being equal to 4 tons.
In Fig. 70 let these external forces, taken in regular order
around the truss, be laid off, thus forming the closed poly-
gon cdrd'c'g'k'hgc. Since each pair of forces
is in equilibrium the entire system is in equi-
librium and hence there are no reactions at
the supports. No forces being applied at
any of the joints of the first panel the mem-
bers drawn in broken lines may be omitted.
The stress diagram is now completed in the
usuaL way, the characters of the stresses de-
termined, and their magnitudes found by ap-
plying the scale, the results, expressed in tons,
being as follows :

Fig. 70.

Chords. Stresses.

COdinAGN -2.6
DQdiVidHP -2.6

> Verticals. Stresses.

MN — 3.1

CPand QQ' -6.1

Main Ties. Stresses.

NO +4.0
PQ +4.0

The stress diagram shows that the tension in any counter

-affects only the members of the panel to which it belongs.

It produces compression in both chords equal to the horizontal

Art. 34. FINAL MAXIMUM AND MINIMUM STRESSES. 89

component of the initial tension, compression in the verticals
equal to its vertical component, and tension in the main tie
equal to that in the counter. The general effect is therefore to
increase the stresses due to other loads in all the members of
the counter-braced panels except the lower chord, whose
stresses are diminished. In the upper and lower lateral brac-
ing both diagonals in each panel are made adjustable.

It is not necessary to make stress diagrams for these lateral
systems, but simply to draw a right triangle whose base is
parallel to the chords and whose hypothenuse, measuring 4.0
tons, is parallel to one of the diagonals. Applying the scale
the base is found to measure 2.9 tons and the perpendicular
;2.8 tons. The stress in the chords is therefore —2.9 tons
throughout ; in the end struts — 2.8 tons ; in the remaining
struts 2(-- 2.8) = — 5.6 tons ; and in the diagonals -{-4.0 tons.
When the struts are not normal to the chords it is best to con-
struct the complete stress diagrams.

It has not been customary, however, to consider the
stresses caused by initial tension in the counters except those
in the main ties. An examination of the tables in the next
article will show the relation which these stresses bear to the
others and to the final maximum and minimum stresses.

Prob. 45. A through double-track railroad bridge 140 feet
in span has Pratt trusses of 7 panels and 32 feet deep. The
bridge is 28 feet wide between centers of chords. Find the
stresses due to an initial tension of 5 tons in every counter of
the trusses and lateral systems.

Art. 34. Final Maximum and Minimum Stresses.

The final maximum and minimum stresses in any truss
member are the extreme limits of stress to which it is subjected
by all possible combinations of the dead, Hve, snow, and wind
loads, and initial tension. .The larger limit is called the maxi-

BRIPGE TRUSSES.

Chap. IIL

mum and the smaller the minimum stress, and they may have
the same or opposite signs. In finding the maximum and
minimum stresses in the following tables, it is assumed that
the initial tension as well as the dead load is always acting. In
practice the wind stresses in the chords are more frequently
disregarded than taken into account.

In Art. 30 the maximum and minimum stresses due only
to dead and live loads were found for the through Pratt truss
whose stresses due to snow load were found in Art. 31, those
due to wind load in Art. 32, and to initial tension in Art. 33.
The various results are now brought together in the following
table and the final maximum and minimum stresses obtained
by addition. The members are designated as in Figs. 63, 64^
and 65.

Upper Chord.

Lower Chord.

Dead load,

-45.4

-56.8

-60.6

4-26.5

+ 26.5

4-45-4

4-56.8

Live load,

-73.7

— 92.2

-98.3

+ 43.0

+ 43-0

+ 73.7

+ 92.2

Snow load,

-8.8

- II.O

-11.7

+ 5.1

+ S.X

+ 8.8

+ II.O

North wind,

-3.3

-5.3

-6.0

- 4-7

-7.9

-9-9

— 10.6

South wind,

+ 3.3

+ 5.3

+ 4.7

47.9

49.9

Initial tension— Truss,

-2.6

- 2.6

'' —Lateral system,

-2.9

- 2.9

-2.9

Maximum stress.

- 134. 1

— 170.8

— 182.1

+ 71.7

+ 76.4

4- 130-3

4-164.4

Minimum stress,

-48.3

-59.0

-60.8

+ 18.9

+ 15.7

4-30.0

+ 40.7

End
Post.

Main Tibs.

Counter-

TIK.

Verticals.

Dead and live load max.,
Dead and hve load min..
Snow load,
Initial tension.

— 107.8
- 41.1
-8.0

4-79.3

+ 26.9

+ 5.7

+53.3
4-10.4
43.4
+5-0

+29.7

4x.i
45.0

4-8.4

4-5.0

420.5

4 6,0

41.7

-43.8
— II. I
-2.6
-3.1

-25.7
-3.0
-0.9
-6.1

-9.4

- 3.0

- 6.1

Maximum stress.
Minimum stress.

— 115.8
-41.1

485.0
4a6.9

461.7
415.4

435.8
4s.o

4-13.4
+ 5.0

+ 22.2
+ 6.0

- 49-5
-14.2

-32.7
-9.1

-15.5
- 9.1

Art. 35.

THE BOWSTRING TRUSS.

UpfBR Latekal Bracing.

Diagonals.

Struts.

Djy

FF'

HH'

DF'

FH'

HJ'

Wind.
Initial, tension.

+ 4.7
+ 4.0

+ 3.1
+ 4.0

+ 1.9
+ 4.0

-3-a
-5.6

— a.z
-5.6

- «.3
-5.6

Maximum stress,
Minimum stress,

+ 8.7
+ 4.0

+ 7.t
+ 40

+ 5.9
■f4.o

-8.8
-5.6

-7.8
-5.6

-6.9
-5.6

Lower Lateral Bracing.

Diagonals.

Struts.

BB'

EE'

GG'

CE'

EG'

GK'

Wind,

Initial tension,

+ 6.4
+ 4.0

+ 4.8
+ 4.0

+ 3.5
+ 4.0

+ 2.3
+ 4.0

-45
-5.6

- 3.4
-5.6

- 1.6
-5.6

Maximum stress,
Minimum stress,

+ 10.4
+ 40

+ 8.8
+ 4.0

+ 7.5
+ 4.0

+ 6.3
-1-4.0

— 10. 1
-5.6

-9.0
-5.6

-8.0

-5.6

- 7.2
-5.6

Prob. 46. Find the maximum and minimum stresses in the
chords of the above (example, provided the effect of the wind
be disregarded. Also, compute the greatest percentage of re-
duction in the maximum stress of any chord member on this
account.

Art. 35. The Bowstring Truss.

This form of truss is shown in Figs. 71, 72, and 73, and is
frequently used for highway bridges. The apex points of the
upper chord lie upon the arc of a circle. When the bracing is
arranged like that in Fig. 71, the diagonals take only tension,
while the verticals take either tension or compression. In the
truss in Fig. 73, all the web members are made to sustain
either kind of stress. The same is true of the form given in
Fig. 72, with the exception of the middle and end verticals,
which are subject to tension only.

BRIDGE TRUSSES.

Chap. III.

^^p^l><RN^ i

For example, let a truss like Fig. 71 be taken whose upper
panel points lie in the arc of a circle. Let it have 8 panels^

each 14 feet long
on the lower
chord, with a
depth at the cen-
. ter of 16 feet.
^ The bridge has a
*• roadway 22 feet
^ 'K 7^ A""""--^ . wide and two side-

,.<A/ V V V V\>. ; walks each 5 feet

*• wide.
The dead panel load is found to be 4.19 tons, of which 1.40
tons is to be taken on the upper chord and 2.79 tons on the

Fig. 75.
lower. The snow panel load is 1.68 tons. At 90 pounds per
square foot of floor surface the live panel load is 10.08 tons, or
6 times the snow load.

Art. 35.

THE BOWSTRING TRUSS.

Let a truss diagram be drawn as in Fig. 74, containing only
the main diagonals in the left half and the counters in the
right. The depths of the truss at the first, second, and third
panel points are 7.32, 12.23, and 15.07 feet respectively. The
stress diagram
obtained for
dead load is
shown in Fig.
75, that for a
live panel load
at apex i in
Fig. y6, that
for a live panel
load at apex 7
in Fig. ^^y and
that for a uni-
form live load
in Fig. 78. As
a check upon
the construc-
tion of these
diagrams it is
observed that
in Figs. 75 and
78 be and op
are in the same Fig. 77.

vertical line. The same is true of de and mn and of fg
and kl. In general the lines representing stresses in verticals
equally distant from the center of the truss lie in the same
vertical line, or are equally distant from the load line. In Fig.
76 the line de is at the same distance from the load line as
nm in Fig. JT, also nm in Fig. 76 and ed in Fig. ^^ are simi-
larly situated. The same relation exists between the lines
representing stresses in any two verticals occupying symmetri-

BRIDGE TRUSSES.

Chap. IH.

cal positions in the truss. Again, if in Fig. TJ Ik be produced
to meet am and the intersection be called k\ then Ik' will be

f^ equal to fg in Fig. ^6 and

nk' will be equal to fe in

Fig. 76.

In Figs. 75 and 78 hj'
represents the stress in HJ
when the main diagonal is
inserted on its right instead
of the counter shown in
Fig. 74, the point j' being
at the intersection of the
lines ak and hj. Only the
Fig. 78. stresses in web members

are scaled off from Figs. 76 and TT. The snow load stresses are
obtained by dividing those due to uniform live load by six.

The stress diagrams from which the following results were
obtained were drawn to the following scales : The dead load
diagram, 4 tons to an inch ; the diagrams for single live panel
loads, 2 tons to an inch ; and the uniform live load diagram, 10
tons to an inch.

The results expressed in tons are now tabulated as in Art.
30 and the mkximum and minimum stresses obtained, the
effect of wind and initial tension being omitted.

Upper Chord.

Lower Chord.

RB^RC

Dead Load,
Live Load,
Snow Load,

-3X.6
-76.1
- 12.7

- 30.5

- 73-2

- 12.2

— 29.8

- 7^-5
-11.9

— 29,4
-70.6

- 1T.8

+ 28.0

+ 67.4
+ II a

+ 28.8
+ 69.1
+ 11.5

+ 29.2
+ 70.1
+ X1.7

Maximum,
Minimum,

- iao.4
-31.6

- 115-9
- 30.5

- 113.2
— 29.8

- 111.8
- 29.4

+ 106.6
+ 28.0

+ 109.4
+ 28.8

+ III.O

+ 29.2

A'RT. 35.

THE BOWSTRING TRUSS.

Main Diagonals.

Counters.

Live panel load at apex i
7

-9.3

-3.8
- 7.5

— 2.1

- 4.2
-6.4

- X.4

- 2.7

- 4.1
-5.4

-0.9

- 1.8
-2.8

- 37
-4.6

-0.6
-X.3

- 1.9

- 2.5

- 3-2

+ 3.2
+ 2.7

+ 2,1

+ X.6
+ 1.1
+ 0.5

+ 4-2
•4-3.3
4-2.5
+ X.7
+ 0.8

+ 5-2
+ 3.9
+ 2.6
+ X.3

+ 6.6
+ 4.4

+ 2.2

+ 8.3
+ 4.X

-38

+ 10,9

+ Total,

-Toul,
Uniform live load.
Dead load,
Snow load,

+ 11.2
-9.3

4x.9
+ 0.8
+ 0.3

+ 125
-11.3
+ 12
+ 0.6

+ 0.2

+ X3.0
— 12.7

+ 0.3
+ 0.2
+ 0.1

+ 132

-13.6

-04

— o.a

— 0.1

+ 13.4
-T3.8

- 1.4
-0.6

— 0.2

+ 10.9

- X3.3

-2.4

— 1.0

-0.4

Maximum,
Minimum,

+ X2.3

4-X3.3

+ 13.3

+ X3.0

+ 11,8

49.9.

Verticals.

Live panel load at apex

6
7

+ 10.X

+ 4.3
+ 8.6

+ 2.5
+ 4.9
+ 7.3

' + 1.6
+ 3.X
+ 4.6
+ 6.2

+ 1.0
+ 2.0

+ 3.X
4- 4.x
-f 5.x

+ 0.7
+ X.3
+ 2.0
+ 2.7
+ 3.4
+ 4.0

+ 0.4
+ 0.8

+ X.2

+ 1.6

+ 2.1
+ 2.5
+ 29

«-

- 1.2

— x.o
-0.7

- 0.5

— 0.2

— 2.2

— 1.6

— 1.1
-0.5

- 2.9

- X.9

- 1.0

-3-3
- 1.6

-3-0

+ Totol,
- Total,

Uniform live load,

Dead load,

Snow load.

+ 10.1

+ 2.8
+ X.7

+ 12.9
-3.6

4-9.3
+ 2.4

+ 1.5

+ X4.r

+ 9.3
+ 2.4
+ X.5

+ X5 5
-5.8

[+9.3]
+ 97

[+2.5]
+ 2.6

[+ 1.5]

+ T.6

4x53

+ 104
+ 3.0

+ X.7

+ X4.T

-3.0

+ 11.1
43.3
4 x.9

+ XI.5

+ IX.S
4-3.4
+ X.9

Maximum,
Minimum,

+ X4.6
+ 2.8

-f 13^

— 1.2

+ X3.2
-3.0

+ X3 3
- 3.2

- x.9

40.3

+ 14.4
-r 34

The student will have no difficulty in finding the maximum
and minimum stresses in the chords and diagonals. It will be
noticed that counters are required in every panel of this truss.

The application of the principles employed in Art. 30, for
determining the maximum and minimum stresses in the

96 BRIDGE TRUSSES. ChAP. III.

verticals of trusses with counterbraced panels may be further
illustrated by finding those in DE. The greatest compression
in this vertical is shown by the table to be due to the live
panel loads 3 to 7 inclusive, combined with the dead load,
giving a stress of — 3.6 + 2.4 = — 1.2 tons. This stress is
a real one because the adjacent diagonals CD and EF shown
in Fig. 74 are then acting, both of them receiving almost
vtheir maximum stress.

The tension of 12.9 tons given in the line marked * -j* total *
when combined with that due to the dead and snow loads
cannot actually occur because the adjacent diagonals are not
brought into action by the corresponding loads. The greatest
tension in DE will therefore occur under the full live, dead,
and snow loads, unless one or more of the live panel loads on
the right may be removed without causing either CD or EF
to cease acting. If the live load be continuous, as it is cus-
tomary to regard it, the live panel load 7 may be removed
under the above conditions, and the corresponding stress will
be + 9.3 + 2.4 + 1.5 — (— 0.2) = + 13.4 tons. If the live
panel loads 6 and 7 be removed, the diagonal EF ceases to
act since its stress would become + 1.2 +0.64-0.2 --
(+ 1.7 + 0.8) = — 0.5 tons. Were it allowable to consider
the live load as discontinuous, CD and EF would still be
in tension after removing the live panel load 6 only, thus
giving a stress in DE of -\- 9.3 -f- 2.4 +1.5 -— ( — 0.5) =
+ 13.7 tons.

In a similar manner the values of + 12.3 and 4- 0.3 tons
are obtained lovMN, and these are its maximum and minimum
stresses under the condition that the adjacent counter ties LM
and NO are both acting. On account of the symmetry of the
truss the maximum and mininrium stresses in DE have the
same values provided the counter ties are acting in each
adjacent panel.

Two more conditions for DE require attention. The first
is that whea the main diagonal acts on its right and the

Art. 3$. the bowstring truss. 97

counter on its left. The table indicates that this condition
cannot exist under any combination of the given loads. The
second condition occurs when the main tie acts on the left of
DE and the counter on its right. This one is possible, and
requires an additional tabulation.

The live panel load at apex i produces a tension in DE of
1.28 tons, and that at apex 7 of 0.43 ton. The former value
is obtained from Fig. 76 by measuring the distance from d to
the point where the vertical de meets <^/ produced, while the
latter is obtained from Fig. TJ by jneasuring the distance
from d to the intersection of ^/ with the vertical de produced
In the same way, Fig. 75 gives the corresponding dead load
stress of + 2.8, and Fig. 78 (after dividing by six) the snow
load stress of -f- i«7 tons. From the stresses due to the live
panel loads at i and 7 those produced by the loads at apexes
2 to 6 inclusive are found by the method used in the above
tabulation to be -f" 2.6, -|-2.i, +1.7, +1.3, and + 0.9,
when expressed to the nearest tenth of a ton. Since all of
these stresses are tension, it is clear that the maximum will
be caused by the dead and snow loads combined with as
many of the live panel loads as possible without bringing the
main tie on the right of DE into action. The table for the
diagonals indicates that this occurs when the live panel loads
are placed at apexes i to 5 inclusive, and the resulting stress
in DEvB^ + 1.3 + 2.6 + 2.1 + 1.7 + 1.3 + 2.8 + 1.7 = +
13.5 tons. If a similar tabulation were made for MN when
the counter tie acts on its left and the main tie on its right,
the same result of + 13.5 tons would be obtained. !

On comparing the maximum stresses in DE under the three
conditions above described and investigated, the last value
obtained^is seen to be the greatest in magnitude, and hence
the true maximum to be used for both DE and the corre-
sponding vertical MN in the other half of the truss. As the
range of stress from this maximum of + 13.5 to — 1.2, the
minimum in DE^ is greater than to the minimum of + 0.3 in

BRIDGE TRUSSES.

Chap. IIL

MN, the true minimum stress to be used for both DE and
MN is — 1.2 tons.

The stresses in FG when the main tie acts on the left and
the counter on the right are found to be + 0.8, -[-1.6, -j- 2.4,
-j- 1.9, + 1.4, + i.o, and +0.5 for the live panel loads I
to 7 inclusive, + 2.6 for the dead load, and -{- 1,6 tons for the
snow load. The maximum occurs when the live load is at
the apexes i to 6 inclusive, combined with the dead and snow
loads, its value being + I3«3 tons. On comparing this stress
with the values given in the table for FG and KL, the true
maximum for both of these verticals is seen to be + 13.3
tons, and the greatest range of stress to the values of their
minimum stresses shows that the true minimum for both
verticals is -— 3.0 tons.

A similar tabulation might be made for B/^ but this is
unnecessary since the main ties act on each side of it under
the full load. The required stresses were obtained directly
from Figs. 75 and 78 as previously explained, and are inserted
in the table in brackets. The maximum and minimum
stresses are caused by the full load. The largest tension
which can occur in B/ when the main diagonal acts on the
left and the counter on the right is 12.3 tons. The true
maximum and minimum stresses of OP are equal to those of
the vertical BC.

The form of truss shown in Fig. 79 is known as the lenticular
truss and is also used for highway bridges. The paneL-points

of the chords may lie on arcs

of circles, but generally are

placed on parabolas. The

broken lines show the road-

^*'' 'S' way and its connection to the

trusses, the vertical end pieces being heavy posts and the others

tension rods. In determining the stresses the same method is

pursued as for the bowstring truss.

Art. 36, THE PARABOLIC BOWSTRING TRUSS. 99

Prob. 47. A through bowstring truss has six panels each 15
feet long, the depth at the first and fifth panel points being
7.5 feet, at the second and fourth panel points 11.7 feet, and at
the center 13 feet. The dead panel load is 2.5 tons and the
live panel load 7.5 tons. Find the maximum and minimum
stresses due to these loads only.

Art. 36. The Parabolic Bowstring Truss.

When the panel points of the broken chord of a bowstring
truss lie upon a parabola whose vertex is midway between the
supports, the stress diagrams become simpler. Let a parabolic
bowstring truss be taken with the same general dimensions
and loads as given in the preceding article. In the diagrams
like Figs. 75 and 78 the broken lines bed . . . nop become
straight lines, and the points c and rf, e and f, . . .n and 0^ co-
incide. This shows that under a uniform load the stress in the
horizontal chord is the same throughout, the diagonals are not
strained at all, and each vertical carries only the panel load on
the horizontal chord. In the diagrams similar to Figs. 76 and
JT, the points ^, ^,/, h, ky m, and o lie upon a straight line which
intersects the load line produced at the shorter distance ar
from r, thus checking the construction.

In the tabulated stresses for the webbing the sum of the
' + total ' and the * — total ' will give zero for the diagonals
and + 10.08 for the verticals provided the work be done with
the utmost precision. With diagrams like Figs. 76 and ^^ made
to a scale of 3 tons to an inch, the stresses obtained by tabula-
tion for uniform live load averaged 0.05 tons in magnitude for
the diagonals, three being tension and three compression,
and those in the verticals varied on an ayerage o.oi tons
from the true result, some being too large and others too
small.

lOO

BRIDGE TRUSSES.

Chap. III.

The final results in tons are given

in the following table :

Chords.

Maximum

Stkbssbs.

Minimum
Strkssks.

Diagonals.

Maximum
Stkksses.

Verticals.

Minimum
Stresses.

' AD

RB '
RC
RE
RG

- 124.9

- I18.6

- II4.I

- III. 9
+ III. 7

-32.8

- 31.2

- 30-0

- 29.4

+ 29.3

CD
EF
GH

NO '

+ 9-9
+ II.6
+ 12.9

+ 13.4

+ 12.9
+ H.6

BC
DE
FG

^/
KL

MN
OP

+ 2.8
-0.4
- 2.3
-2.9

-2.3

-0.4

-2.3

The minimum stresses in the diagonals are zero, and the
maximum stress in each vertical equals

2.79+ 10.08+ 1-68 = 14.55 tons,
or the sum of the dead, live, and snow panel loads.

If this truss were used as a deck bridge the maximum
stresses in the verticals would be those given in the accompany-
ing table, while the minimum stresses would
equal the dead panel load on the horizon-
tal chord, or — 2.79 tons. It will be ob-
served that the differences between the
minimum stresses in the verticals of the
through truss and the maximum stresses in
the same members of the deck truss equal
twice the dead panel load plus the live
and snow panel loads. The stresses in the remaining mem-
bers' are the same for a deck as for a through bridge except
that the chord stresses change in character.

The properties of the parabola are such as to provide a very
simple and abridged construction for obtaining directly the
maximum and minimum stresses due to dead, live, and snow
loads.

Verticals.

Maximum

Stkesses.

BC and OP
DE-e^n^MN
FG and KL

— 14.5
- 17.7

— 19.6

— 20.2

Art. 36. THE PARABOLIC BOWSTRING TRUSS.

XOI

Let 5 be the stress in the horizontal chord due to the total
uniform load JV (including the half panel loads at the sup-
ports), / the span and d the depth of the truss at the center,
then (Roofs and Bridges, Part I, Art. 39),

S =

Sd'

Substituting for these terms their values for the truss consid-
ered above, the stress due to live load is

^_-8 X 10.08 X 112

70.56 tons.

8X16

Similarly the stress due to dead load is 29.33 tons and that due
to snow load is 11.76 tons.

Now in Fig. 80 on the horizontal line ad with a scale of 4 or 5
tons to an inch, let ad be laid off equal to 29.33 tons, dc equal to

70.56 tons, cd equal to 11.76 tons, and verticals erected at each
point of division. As the depth of the truss is one-seventh of
its span let W and cc^ be made equal to one-seventh of 70.56
or 10.08 tons, and on the span ^V let an outline diagram be
drawn similar to the truss diagram. In the figure one-half is
drawn as a through and the other half as a deck truss. By
measuring the diagonals with the scale of force their maxi-
mum stresses are obtained. Their minimum stresses are zero.

t02 BRIDGE TRUSSES. ChAP. III.

Let the chord members be prolonged until they meet the
verticals through a and d. Each of these lines is divided into
three parts by the four verticals, these parts giving the stresses
due to dead, live, and snow load respectively. For example,
the dead load stress in the horizontal chord is represented by
the part ab^ the live load stress by bc^ and the snow load stress
by cd. The maximum stress in the same member is hence ad^
or 1 1 1.7 tons, and the minimum stress is ab^ or 29.3 tons.

On the through truss draw the horizontal broken and dotted
base line 2.79 tons (the value of the dead panel load) above the
first panel point on the upper chord. By measuring the ver-
ticals extending from this base line to each panel point on the
upper chord, upward being compression and downward ten-
sion,, the minimum stresses in the verticals are found. Their
maximum stresses are each equal to the sum of the dead, live,
and snow panel loads, or 14.55 tons.

On the deck truss let a similar base line be drawn 14.55 tons
above the first panel point from the support on the lower
chord, and the verticals measured from the panel points to the
base line ; thus are found the maximum stresses in the verticals,
all of them being compression.- Their minimum stresses are
■ each equal to — 2.79 tons. The measured stresses arc marked
on the different lines of the diagram.

Prob. 48. A deck parabolic bowstring truss of 10 panels
has a span of 1 20 feet and a depth of 1 5 feet at the center. Find
the maximum and minimum stresses for a dead panel load of 3
tons and a live panel load of 7 tons.

Art. 37. Application of the Equilibrium Polygon.

In the preceding articles of this chapter the method of the
force polygon has been employed exclusively. To illustrate
the application of the equilibrium polygon in the determination

AUT. 37. APPLICATION OF THE EQUILIBRIUM POLYGON. T03

of Stresses let the same example used in Art. 28 be taken, it
being required to find the chord stresses, and afterward the
web stresses due to the dead load. The span is 176 feet, the
depth 26 feet, and the dead panel load 8.95 tons.

Let the truss diagram be drawn to a scale of 10 feet to an
inch and the panel loads laid off successively on the load line
qy in Fig. 81 to a scale of 10 tons to an inch (considerably re-

Pig. 8»

duced as here printed). The effective reactions are ya and aq^
the load line being bisected at a. Let the pole be taken on
a horizontal through a, the pole distance H be made equal to
26 tons, and the equilibrium polygon constructed (Art. 6).
The ordinates at the vertices of this polygon when measured by
the linear scale and multiplied by // give the bending mo-
ments in tons-feet at the corresponding sections of the truss
(Art. 7). The chord stresses are obtained by dividing these
moments by 26 feet, the depth of the truss. For instance, the

104 BRIDGE TRUSSES. ChAP. III.

ordinate nn! measures 45.4 feet, whence the stress in AD or ES is

45.4 X 26

^^-V- = 45.4 tons.

The stresses may therefore be directly obtained by measur-
ing the ordinates with a scale of 10 tons to an inch, the results
being marked on the diagram. Even with a smaller scale than
that indicated above the same stresses, measured to tenths of
a ton, are obtained as in Art. 28.

If the linear scale be 20 feet to an inch, and H be taken as
52 tons, the scale of tons remaining the same, the ordinates

should be measured by a scale of 20 X -^ = 40 tons to an

inch to obtain the chord stresses. Again, if // be laid off by

the Unear scale equal to double the depth of the truss, then the

ordinates are to be measured by double the scale of force or

20 tons to an inch.

As the vertices of the equilibrium polygon lie upon a parab-
ola whose vertex is at k^ the ordinates may be obtained with-
out drawing the equilibrium polygon. The chord stress at the
center of a truss uniformly loaded is

•_ Wl

in which W includes the half panel load at each support. Let
the middle ordinate ik be made equal to

^ 8 X 8.95 X 176 ^^^

5 = cs ^ = 60.0 tons,

8 X 26

let Itn be made equal to ik and divided into the same number
of parts as mk^ in this case four. Drawing radial lines from k
to these points of division their intersections with the corre-
sponding verticals give the required points.

For an odd number of panels in a truss Im should be divided
into as many parts as there are panels in the entire truss, only

Art. 37. APPLICATION OF THE EQUILIBRIUM POLYGON. IDS

the alternate points of division from I to m being used. For
a uniform live load the same method may be employed as
that here given, or the dead and live loads may be combined
in one diagram.

The shear diagram for dead load is shown below the moment
diagram in Fig. 81, the ordinates representing the vertical shear
being limited by the line forming a series of steps from / to y\
If the load were not concentrated at the panel points but uni-
form throughout the ordinates for shear would be measured to
the straight line/V, which intersects the former at the center
of each panel. The inclined line is the most convenient to
use, but usually it is not employed, as the analytic method,
consisting only of a few subtractions or additions, is more
rapidly applied. The lines ab, cd, efy and gh are the stresses in
the diagonals, and de and fg in two of the verticals. To avoid
confusing the diagram cdy ef zxid/g are turned toward the left,
but have the same inclination as the diagonals of the truss.
The stress in BC is one panel load, and that in HJ is zero. If
2.95 tons of the dead panel load be taken on the upper chord,
a compression of that amount is to be added to each of the
above stresses in the verticals.

In order to determine the maximum live load shear in any
panel another shear diagram is necessary. On the horizontal
axis AG in Fig. 82* let the positions of the panel loads be
marked, and their magnitudes laid off on the load line ag. Let
the pole o be placed in a horizontal line through the beginning
of the load line, the pole distance' made equal to the span of
the truss by the linear scale and the equilibrium polygon A'B'C
. ^ IT constructed. Now let the span be so placed that its
light support shall come at F, then every panel point from 3
to 7 inclusive is loaded. By Art. 29 this position of the load
gives the maximum live load shear in EF or DE of Fig. 59.
The ordinate F'F'' is equal to the reaction of the left support

io6

BRIDGE TRUSSES.

Chap. III.

and hence equals the vertical shear in the members named ; for,
the ordinate being contained between the sixth side of the
equilibrium polygon and the first siide produced measures the
sum of the moments of .all the loads between them with
reference to the section through F' (Art. 7). Calling the value

Fig. 83.

of the ordinate y, the sum of these moments equals y X H.
But the section at F' is at the right support of the truss, and
hence the sum of the moments also equals the moment of the
left reaction with reference to this support. Therefore

yXH = Rxl.
and since H was made equal to /,

y = R.
This may also be proved by drawing through the pole a
parallel to the closing line F'l of the equilibrium polygon
forming a triangle which is equal to IF^F" since one side is
equal to its parallel IF*^ and all the sides of both triangles are
mutually parallel. F'F" is hence equal to its parallel R.

The ordinates taken in succession from H' to A' measured
by the scale of force give the maximum live load shears in
each panel of the truss from left to right. The stresses in the
diagonals are obtained from these shears in the manner indi-
cated in Fig. 81. The results by this method are found to be

Art. 38. EXCESS loads. 107

the same as those given in Art. 30 in the line ' -|~ ^otal * for
the diagonals and in the line * — total ' for the verticals. .

. For trusses with inclined chords the moment diagram gives
only the horizontal component of any chord stress, the ordinate
being measured in a section passing through the center of
moments of the chord member. The shear diagram is not
applicable in such cases except for the purpose of finding the
reaction from which the stress in the required web member
may be found by the method of the force polygon.

. It will be observed by the student that the method of th&
equilibrium polygon does' not indicate the character of the
stresses as in the method of the, force polygon. Whether a
web member be in tensiqn or compression is to be determined,
by cutting it by a plane, noting its direction and the sign of
the shear, as w^s done in the analytic method in Part I, Art. 26.

I Prob. 49. Find the maximum and- minimum stresses due to
dead and live loads for the truss in Prob. 42.

Art. 38. Excess Loads.
■ • •

It may be specified that a truss shall be designed to carry a
given load extending over a certain distance in excess of the
uniform live load. In a railroad bridge the excess load would
represent the difference between the locomotive panel load and
the uniform train panel load.

The chord stress due to a single load P distant x from the

left support is a maximum at the load, and for any position of

the load

^_M _ P{1- x)x ^

^ being the bending nioment at the load. If She an ordinate
corresponding to an abscissa x this is the equation of a parab-
ola whose vertex is at the center of the truss. The middle

io8

BRIDGE TRUSSES.

Chap. III.

PI
ordinate has a value of — - and the ordinates are zero at each
Ad

end. The ordinate at each panel point is the maximum stress
in the chord member whose center of moments is at that point.

Although excess loads are not used in determining the
stresses in highway bridge trusses, except for the floor system,

yet as an illus-
tration of the
method let the
Pratt truss in
Art. 30 be
taken and
the maximum
chord stresses
be found for
one excess
''*K-*4. panel load of

5 tons. The ordinates in Fig. 84 are obtained as in the
preceding article, the middle ordinate />t being

^^/Y ^ 5 X 176
4^ 4 X 26

= 8.46 tons.

The other ordinates on either side are 3.7, 6.4, and 7.9 tons.

Let it now be supposed that two excess loads of 5 tons each
and 50 feet apart are specified. Let their distance apart be
taken as 44 feet or two panel lengths. For two equal loads the
maximum stress in a chord naember in the left half of the truss
occurs when one load is at the section passing through its
center of moments and the other load is toward the right. For
the member AF one load is to be placed at apex 3 and the
other at 5. The stress in AF due to the load at 3 is el which
measures 7.9 tons. The stresses for the load at 5 are given by
the- ordipates of the triangle cng^ and since en intersects el at
w, the stress in AF due to this load is em^ or 4.8 tons, and that

Art. 38.

EXCESS LOADS.

109

due to both loads is 7.9 + 4.8 = 12.7 tons. In a similar manner
the other chord stresses are found. The stress in AH may
however be larger when the loads are at 3 and 5, than when at
4 and 6, being in this example 6.4 + 6.4 = 12.8 tons for the
former and 8.5 + 4.2 = 12.7 tons for the latter position,/? being
6.4 tons and fr 4.2 tons.

By inspecting the table in Art. 30 it is seen that the maxi-
mum shear in any web member in the left half of the truss due
to a single live panel load occurs when the load is placed at
the nearest panel point on its right, the shear being equal to
the left reaction. For a single excess load P the vertical shear is

if V be an ordinate corresponding to an abscissa x this is the
equation of a straight line. Thus, when ;ir = o, F = /* and
when AT = /, F= o. From these data the diagram for maxi-
mum shear due to a rolling load may be readily drawn.

For the above example let the span ab be laid off in Fig. 85,
marking the panel points as indicated and erecting verticals at

Fig. 85.

all of these points. Let ca be made equal to 5 tons by scale
and cb joined, then the ordinates to this line give the maximum
shear for each panel. The ordinate at 3 which measures 3.1
tons is the maximum shear in EF and DE and also the stress
in DE. The stress in EF is given by a line drawn through the
upper extremity of this ordinate with the same slope as EF
and is found to be 4.1 tons. The stresses in all the diagonals
may be obtained from cd^- 6.56 tons by the use of a simple
ratio which for EF is \ and for HG is f . These stresses may

BRIDGE TRUSSES.

Chap. IIL

also be obtained directly from the tabulation in Art. 30, by
multiplying the greatest live load stress in any member due to
a single panel load by the ratio of the excess load to the live

panel load. For example, the stress in EF is -j- 1 1-9 X — — =

14.52

+ 4.1 tons and that in DE is -- q.i X — — = — 3.1 tons.

^ 14.52

The same method might be applied to the chords, but would
require more work than that used in this article.

For the two excess loads mentioned above the maximum
stress in EF or DE is produced by placing one load at 3 and
the other at 5. Let ce be drawn equal to | X 5 tons and e
joined withy which is on the line be at a distance of two panels
from b. The ordinates to the line efb give the maximum shears
in each panel due to the two excess loads. The stress in DE
is found to be 5.0 tons and in EF 6.6 tons.

The chord stresses are as follows,

RB and RC

AD 'And RE

AFAndRG

Aff

First excess load,
Addition for second load,

3-7
2.6

6.4
4.2

7.9

4.8

8.5

Two excess loads,

6.3

10.6

12.7

2 X 6.4 = T2.8

The stresses in the verticals are,

First excess load.
Addition for second load,

+ 5.0

— 3.1

- 1.9

- 2.5

- 1.3

- 1.9

- 0.6

- 1.3

-0.6

Two excess loads.

+ 5.0

- 5.0

-3.8

- 2.5

- 1.3

-0.6

and those in the diagonals are.

First excess load,
Addition for second load,

- 5-7

- 4.1

+ 4.9

+ 3.3

+ 4.1
+ 2.5

+ 3.3
+ 1.6

+ 2.5
+ 0.8

+ 1.6

+ 0.8

Two excess loads,

-9.8

4-8.2

+ 6.6

+ 4.9

+ 3.-3

+ ..6

+ 0.8

Art. 38. EXCESS loads. in

These results when combined with the stress given in Art. 34
will increase the maximum stresses in all the members and
reduce some of the minimum stresses in the web members.
The addition of the stresses in the web members due to excess
loads should properly be made in the tables in Art. 30 by-
inserting a line above the dead load stresses and then obtain-
ing the maximum and minimum stresses anew.

If one or more locomotive panel loads be required to precede
the train on a railroad bridge, the maximum chord stresses in
the left half of the truss will be obtained by placing the excess
loads as near as possible to the left support.

' Prob. 50. A Pratt truss for a deck single track railroad bridge
has 6 panels each 13 feet 4 inches long and 13 feet 4 inches
deep. Find the stresses due to one excess load of 6.5 tons
per truss.

112 LOCOMOTIVE WHEEL LOADS. CHAP. IV.

CHAPTER IV.
LOCOMOTIVE WHEEL LOADS.

Art. 39. Standard Typical Loads.

A uniform live load which is carried by the stringers and
floor beams to the panel points of the trusses, giving uniform
live panel loads throughout, is confined mainly to highway
»! 5: 8 ft' 2; o' 0/ ^ 9! of « g' §; §1 oi 8'. ol o;

III S; Si I |i s^ III |; §i I: s §'

» » !b s»; !0| Ij Ij 8j S S S ^ ?! 5: I I

rS rS rS rk ";!. C ^C^C^O % rk A "rk [ff^^

\H\H\4t\4i\ /o± \s\sk\^'. a \?i \4i\4i\H\ cot \s\si\s\3

I! Si I I 11 Si ^ I fi I ii ti Hill

cj) Q U vj ^f^ (^ (?) Q g.^SJ Qv 0000

\si\ 9 \ e \ $± is-isiij-i^a \sir[ j^ \ e \ $i \s\si\j-\3

Fig. 86.

bridges. For railroad bridges it is generally specified that the
live load shall consist of two coupled locomotives, followed by
a uniform train load of a given weight per linear foot of track.
The actual wheel loads thus taken constitute a system of con-
centrated loads whose relations remain unchanged to each
other and to the uniform load following tfrem while passing
over the bridge.

The first of the diagrams in Fig. 86 represents two typical
consolidation locomotives and the second the two typical pas-
senger locomotives and train load specified in 1886 by the Penn-
sylvania Railroad. The numbers above the wheels show their
weights in pounds for both rails of a single track and the num*
bers between them show their distances apart in feet.

Art. 39.

STANDARD TYPICAL LOADS.

A typical locomotive does not really exist, but is used so
that the stresses due to it will be greater than those due to any
locomotives that are likely to be built for some years in the
future. Different railroads specify various typical locomotives
and sometimes actual ones of the heaviest patterns. The
typical locomotives shown in Fig. 86 are about the heaviest
that have been employed for a number of years in designing

Fig. 87.

railroad bridges, while the two coupled consolidation locomo-
tives and train shown in Fig. 87 were first specified in i<56y
by the Lehigh Valley Railroad. The numbers between the
wheels in Fig. 87 indicate inches.

For railroad bridges the ** compromise standard system"
of live loads recommended by Waddell in 1893, after a dis-
cussion of the subject by many engineers, may be here noted
as one which seems likely to be much used in the future.
The typical consolidation locomotives are divided into seven
classes, called T, U, V, etc., and the wheels in each class
have the same spacing. Fig. 88 shows this spacing and the

«»^; •■^1 "-^i (-^j
Vi Vl ^li" ^j^l

^r Z* O! ^'^

§1 ii |i |i

>^ >' > ^'

Oil O, <N|I f\|i

(S OOQQ (!) ^^

^! M^i Vi V| >►!

<i! O! Ci»

^1 Si SI

i» ^»

VI >H ^H >H
^! *^ «Vi o»j

! 4*200 Ibsy
! per lin. ft..
1 1

f - /04'- *->*

Pic:. 88.

loads for class T, while the loads for the other classes are
given in the table. The student should always remember
that the loads in the following table are for both lails of a
single track and correspond in this respect to those given in
Figs. 86 and 87. The distance from the front pilot wheel
to the beginning of the uniform load is in all cases 104 feet.

114

LOCOMOTIVE WHEEL LOADS.

Chap. IV.

and the loads are stated in pounds. Of course in designing a
bridge the exact loads given in the specifications must be
used, but it is hoped that engineers who write specifications
will gradually abandon the awkward typical locomotives with
wheels spaced apart at distances involving odd fractions of a

Class.

Load on Pilot
Wheel.

Load on each
Driving
Whkel.

Load on each
Wheel of
Tenukr.

Total Weight

OF one

Locomotive

and Tender.

Uniform Load
PEK Linear .

Foot.

15 000

25 000

18 000

187 000

3 000

16000

28 000

19 000

204000

3 200

17 000

31 000

20 000

221 000

3 400

18 000

34000

21 000

238 000

3 600

19 000

3700c

22 000

255 000

3 800

. u

20 000

40 000

23 000

272 000

4000

21 000

43000

24000

289000

4 200

foot, for surely this is straining at a precision unwarranted by
actual conditions.

Alternate loads on two axles 7 feet apart are also specified
for each class, the load on both rails of a single track for each
axle being as foUows: Class Z, 40000: Y, 42 000; X, 44000;
W, 46000; V, 48000; U, 50000; and T, 52000 pounds.
These loads represent the heavier drivers of the typical passen-
ger locomotives, and apply only to b^ams of very short spans
and to cross ties in the track on bridges.

Prob. 51. Find the maximum shear in an I beam 15 feet in
span under the load represented in Fig. 87.

Art. 4a Analysis of a Plate Girder.

To illustrate the method of determining the stresses when

the live load consists of concentrated wheel loads, let a deck

, plate girder bridge for a single track railroad be taken, the

span being 55 feet, measured between centers of bed plates,

and the effective depth 5.5 feet. The total weight of both

Art. 40. analysis of a plate girder. 115

girders and of the lateral bracing is 16.45 tons, and the floor
. system is estimated at 420 pounds per linear foot. The live
load is to consist of one Lehigh Valley consolidation locomo-
tive and tender followed by a uniform train load of 4000
pounds per linear foot. It is required to find the maximuni
flange stresses and the maximum shears throughout the girder,
due to the above loads.

The total dead load for each girder is first found to be,

1 f420 X 55 + ,6 \ ^ t„„3^

2 \ 2000 ' -fr^y •»

which is regarded as uniformly distributed. As this is a single
track bridge the live load is divided by two, and for conven-
ience the weights are reduced to tons.

On the axis AN, Plate III, and to a scale of 8 feet to an
inch, are marked the positions of the wheel loads and the
beginning, middle and end of a portion of the train 20 feet in
length. Through these points indefinite verticals are drawn.
On the load line M, at the left of the plate, the wheel loads
are laid off successively to a scale of 10 tons to an inch fol-
lowed by 20 tons — the weight of the 20 linear feet of train.
The pole is chosen at a point above the middle of the load
line, the pole distance being five times the depth of the girder,
or 27.5 feet. It is not necessary to draw the rays from the
pole, as the direction of each ray is determined by the pole and
a point on the load line through which points the edge of the
triangle is passed in the construction of the equilibrium poly-
gon A' B' F L M' N' , As the train load is not concentrated at
its center but is uniformly distributed the required part L'e^N'
of the equilibrium -polygon is a parabola tangent to LM' at U
and to N'M' at N^ (Art. 10). The construction is indicated on
the diagram. The portion A'B' of the polygon is a straight
line parallel to ho and is produced as far as needed.

The left half of the girder is divided into five equal parts

Il6 LOCOMOTIVE WHEEL LOADS. ChAP. IV.

each 5.5 feet in length, and the sections numbered as shown.
After erasing the lines of action of the wheel loads above the
equilibrium polygon, a series of verticals are drawn 5.5 feet
apart. The two verticals each marked cc' are 55 feet apart
and c'c' is the closing side of the equilibrium polygon for the
position cc of the girder. For this position the first driver
stands at section 3 of the girder. The ordinate Pd' represents
the flange stress at section i for this position of the load, the
ordinate Qe' for section 2, and so on. The closing lines a' a'
\.og'g' are drawn, and all points on these lines distant one
space from the left end are united by the curve PP, those dis-
tant two spaces from the left end by the curve QQ^ those dis-
tant three spaces by RR, those distant four spaces by 55, and
those distant five spaces by the curve TT, The ordinates
between the polygon and the curve PP indicate the successive
values of the flange stress at section i as the girder is moved
from left to right with respect to the load, or, in other words,
as the live load passes over the girder from right to left. The
maximum ordinate between, these lines is djrectly over the
first driver, indicating that when the load is placed so that the
first driver stands at section i it will give the maximum flange
stress at that section. The maximum is always located at a
vertex of the polygon, that is, at one of the loads. When the
ordinate is not at 'an intersection through which the upper
curve was drawn its length should be tested by drawing the
closing line for the required position of the load. The pole
distance H being five times the depth of the girder this ordi-
nate must be measured with a scale five times that used on the
load line or 50 tons to an inch. Applying the scale it is found
to be 38.0 tons. The maximum ordinates to QQ and RR are
66.2 and 85.9 tons respectively, both being at the second driver ;
and the maximum ordinates to 55 and TT are 97.0 and 99.6
tons, both being at the third driver.

The center of gravity of the locomotive and tender is 3

Art. 40. ANALYSIS OF A PLATE GIRDER. H/

feet behind the third driver. When the load is so placed that
the center of the girder is midway between these two points,
only these loads being on the girder, the absolute maximum
flange stress will occur in the section at the third driver. The
section is therefore 1.5 feet from the center of the girder and
the flange stress is 100.3 tons. The closing line for this posi-
tion is shown near c'<^.

On an axis equal to the span in length and divided like the
girder into ten parts the flange stresses just found are laid off
as ordinates and a curve drawn through their upper extremi-
ties. Only one half is shown in the lower left side of the plate,
the other half being symmetrical with it.

The diagram for flango stresses due to the dead load is now
constructed below this axis by the method of Art. 35, and by
measuring the ordinates the flange stress for the different sec-
tions of one-half of the girder are found to be 6.3, 11.2, 14.7,
16.8, and 17.5 tons respectively. These are to be added to the
live load stresses in order to obtain the maximum stresses in
the flanges.

To determine the maximum live load shears another equi-
iibrium polygon A"B"D"L"N" is drawn by taking anew pole
distance equal to the span, and as it is convenient to have the
first side A"B" horizontal the pole^' is placed directly opposite
h, the beginning of the load line. From the first driver at C"
the span is laid off toward the right extending to /and then
successive positions ee, dd, etc., of the girder are marked when
it is moved toward the left 5.5 feet at a time. The ordinates
f'z., d"d . . a" a are the corresponding reactions of the left
support (Art. 37) and those portions of the ordinates above
the line uwz are the maximum shears at the sections. The
line uw is parallel to aCy ua equals 4 tons — the load on the pilot
wheel— and wz is a part of B"C'' produced, w being 55 feet
distant from B'\ It is found that the maximum shear h caused

Il8 LOCOMOTIVE WHEEL LOADS. CHAP. IV.

in each section when the leading driver stands at that section
and the load covers the right segment of the girder. For
instance, when the first driver is at section 5 the shear is a"u
or 13.6 tons, when the pilot is placed at 5 the shear, measured
on the ordinate at the left of a"it, is 10.6 tons, and for the sec-
ond driver at 5 the corresponding ordinate (not shown on the
plate) is still smaller. When the first driver is at the left sup-
port the reaction and also the shear at the support are equal to
f'z or 415.0 tons,/''^ lying between the polygon and the second
side B"C" produced, since the pilot is beyond the girder (Arts.
7 and 35). The shears for the sections from o to 5 are 45.0,
38.0, 31.3, 24.9, 18.9, and 13.6 tons respectively. If shears are
desired for intermediate sections they may be measured di-
rectly on the diagram.

The above shears may also be obtained from the moment
diagram used for finding the flange stresses. When the first
driver is at section 5 the closing line of the polygon is a' a' and,
drawing parallel to this a ray through 0^ it is found to cut off
on the load line a reaction of 17.6 tons or a shear of 13.6 tons.
The other rays shown are parallel to the closing lines b'b' to
/y. If one of the series of equidistant verticals did not coin-
cide with the first driver another series of closing lines would
have to be drawn to find the shears in this manner. The pre-
ceding method is usually the most economical in labor, but in
this example the method just given has the advantage.

The shears due to dead load for the left half of the girder
are given by the triangular shear diagram on the right of the
lower end of the load line, and are to be added to the live load
shears. Their values for the sections o to 5 are 7.0, 5.6, 4.2,
2.8, 1.4, and 0.0 tons.

Sometimes an approximate method is employed in which
the pilot wheel is omitted from the system of loads. The dif-
ferences between the true and the approximate shears in this.

Art. 40

Plate

> III.

I Locomotive and Train Loads.

Scale of Fmt.
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. BY CHA

S. HART, 36

VESEY ST N. K.

Art. 41. AN^ALYSIS OF A PRATT TRUSS. HQ

example are indicated on the diagram by the ordinates between
the lines utv and C^'w.

In practice the equidistant verticals corresponding to the
divisions of the girder are drawn upon a separate sheet of trac-
ing paper that may be shifted horizontally over the one on
which the equilibrium polygons are constructed. This facili-
tates some parts of the process very materially. \

If two coupled locomotives be specified instead of one the
maximurn flange stresses will be somewhat greater, as in that
case the drivers of the second locomotive when placed at the
various sections give the positions for maximum moment.
This may be seen from the fact that the straight line B'A' will
be replaced by a broken line starting from B' and bending up-
ward, thus raising the left ends of a number of closing lines and
consequently increasing the ordinates between them and the
polygon.

Prob- 52. A plate girder for a single track bridge has a span
of 31 teet 6 inches and a depth of four feet. Find the flange
stresses and shears due to two coupled Pennsylvania consoli-
dation locomotives.

Art. 41. Analysis of a Pratt Truss.

Let the example whose computations were made in Part I,
Art. 63, be taken in order to compare the accuracy of the
graphic method with the analytic method. The truss is a
through Pratt for a double track railroad, of 140 feet span, hav-,
ing 7 panels, each 20 feet long and 24 feet deep. The dead!
load per linear foot is I 400 pounds or 0.7 tons, and the live load
a Pennsylvania typical passenger locomotive and tender fol-
lowed by a uniform train load of 3 000 pounds or 1.5 tons per
linear foot. Of the dead load 400 pounds per linear foot is to
be taken on the upper chord.

As the construction required by this example (see Plate IV)

1 20 LOCOMOTIVE WHEEL LOADS. CHAP. IV.

is in many respects similar to that in the preceding article, only
those features which are different need explanation. The linear
scale used on the original drawing was lo feet to an inch, and
the scale of force 20 tons to an inch. The pole distance is
taken as twice the depth of the truss or 48 feet. The verticals
above the equilibrium polygon are 10 feet or half a panel length
apart, and the maximum ordinates for the sections through the
panel points are drawn in heavy lines. The ordinates are
measured by double the scale used in laying off the load line
or 40 tons to an inch, and their values marked on the diagram.
The maximum ordinate below the curve 55, drawn for the sec-
tion through the fourth panel point, is smaller than that for
the third, and is the last one required for the chord stresses.

If the Pennsylvania consolidated locomotive be used instead
of the passenger locomotive, the maximum ordinates for the
sections at 3 and 4 will be found at such positions as to place
the engine beyond the bridge, and toward the right of these
ordinates the stress curves will be parallel to the equilibrium
polygon.

In the shear diagram on Plate IV the ordinates at d'\ b'\
etc., represent the reactions of the left support for various
positions of the load since the pole distance o^h is 140 feet, the
length of the span. For instance, the left reaction when the
first driver is at 2 is the ordinate below ^ which measures (by
the scale of 20 tons to an inch) 79.2 tons. The shear in .the
second panel is this reaction minus that portion of the load on
both pilot wheels carried by the stringers to the flo6r beam at
I, or 9.4 tons. Laying this off from b'' downward and drawing
a line parallel to the diagonal CD the stress for that member
is found to be 90.8 tons. If any other load be placed at 2 a
smaller stress is found.

In the fifth panel the positive live load shear Is a maximum
when the second pilot wheel is at the panel point on the

Art. 41. ANALYSIS OF A PRATT TRUSS. 12J

right. The difference between the reaction and the shear is
then 2.2 tons. The construction for two positions of the load
is given for the fourth and fifth panels. The positive live
load shear is a maximum in the sixth panel when the first
pilot wheel is at panel point 6. The shear is then equal to
the reaction, its value being 2.7 tons. The construction is
omitted on the plate. The one shown is that due to the'
second pilot wheel at panel point 6.

The tension in the suspender BC is equal to that portion of
the loads between o and 2 that is carried by the stringers to
the floor beam at i, such loads being brought on as to make it
a maximum. This condition requires the drivers to be near I.
To determine this the small equilibrium polygon is drawn
directly below the locomotive wheels using a pole distance
equal to the panel length. The pole is o". When the first
six wheels are so placed between o and 2 that their center of
gravity is at i, the reactions at o and 2 are each 16.7 tons, as
given by the ordinate above the intersection of the outer sides
produced (Arts. 7 and 37). The reaction at I is therefore
(8 + 8 + 20 — 16.7)2 = 38.6 tons. The first driver is next
placed at i, thus moving the second tender wheel beyond 2,
and the sides of the polygon are produced to meet the vertical
under this load. The reaction at i is then 64 — (9.4+ ^S-o)
= 39.6 tons. When the second driver is placed at i the reac-
tion at I is 64 — (9.6 + 14.8) = 39.6 tons. The greatest live-
load tension in BC is therefore 39.6 tons.

It will be observed by the student that the ordinate of 9.4
tons under the first driver gives the amount that is laid off on
five of the ordinates on the large shear diagram, while the small
ordinate of 2.2 tons under the second pilot is deducted from
three of them,

Another method of finding the stress in BC when the first
driver is at i is shown on the diagram including the load line,

122

XOCOMOTIVE WirEEL LOADS.

CHAP. IV.

in which ^/ is parallel to the closing line/'rf' and ou is parallel
to the closing \mQf'h\

The stress diagram for dead load when the diagonals all
incline one way is shown on the left of the shear diagram.

The stresses thus obtained are for the chords,

AF^nA AH

RB and RC

Dead load

Live load •>••«••••

-58.3
- 130.8

- 70.0
- 154.5

+ 35.0

+ 8r.8

+ 58.3
+ 130.8

+ 189. 1
+ 58.3

+ 70.0

+ 153.5

+ 223.5
+ 70.0

Maximuin

- 189.I
-58.3

- 224.5
— 70.0

+ 116.8
+ 35.0

Minimum ...•

for the diagonals,

Dead load

-54.7

— 127.8

+ 36.5
+ 90.8

-3.5

+ 18.2
+ 59.4
- 15.2

+ 33.5
- 33.5

- 18.2
+ 15.2

- 59-4

- 36.5
+ 3.5

— 90. 8

Live load on right..
Live load on left. . .

IVfaximilm. ••■■•••■

- 182.5
-54-7

+ 127.3
+ 33.0

+ 77.6
+ 3.0

+ 33.5

Minimum.

and for the verticals,

Dead load

Live load

+ lO.O

+ 39.6

— 18.0
-45.6

-4.0

-25.8

+ lO.O
-.11.7

+ 24.0
-2.7

Maximum ...■.■■>.

+ 49.6

+ lO.O

-63.6
-6.3

— 29.8
-4.0

Minimum •■

The live load stress in AH is — 154.5, and that in RG is
4" i53»5 tons, because under the loads which cause both
chord stresses of these magnitudes combined with the dead
load the shear in the middle panel is negative, thus bringing
into action the diagonal which slopes downward toward the

:tri.

h — ■ ^ -- ■ ■

Art. 41. ANALYSIS OF A PRATT TRUSS. I23

left. As, however, the dead load shear is zero, its sign
depends only on the live load and may therefore be most
readily determined by drawing the corresponding closing lines
on the moment diagram on Plate IV and observing the
relative lengths of the ordinates at panel points 3 and 4, as
explained in Art. 9. In both cases the ordinate at panel
point 3 was found to be the greater. In practice the larger
of the two values is often used for both chords of the middle
panel.

The maximum and minimum stresses are found in the
manner explained in Art. 30. The table shows that theoreti-
cally only the middle panel needs to be counterbraced, but
practically counters would be inserted in the third and fifth
panels to provide for some loads which might be allowed to
cross the bridge before it would become necessary to replace
the structure on account of the permanent increase in the
weight of locomotives and trains demanded by the regular
traffic.

As no counter tie is called into action in the third panel by
the specified loads, the minimum stress in the vertical DE is
shown to be — 18.0+ 11.7 = — 6.3 tons, because the live
load stress in DE when the load comes from the left is equal
in magnitude and opposite in sign to that in HJ (which
equals the shear in JK) for the live load coming on the bridge
from the right. Had the counter /AT been required, then the
minimum stress in DE would have been the dead. panel load
on the upper chord, or — 4.0 tons. The stresses in HJ and
KL given in the table are not real, as the adjacent counters'i
JK and LM never act under the given loads.

On comparing these results with those obtained by the
analytic method it is seen that the average difference between
them is about o. i ton, the only difference exceeding this
amount being 0.3 ton for the chord members ^Z> and -^^,
the value obtained by computation being 131. 1 tons. The

124 LOCOMOTIVE WHEEL LOADS. CHAP. IV.

analytic method, however, gives another position of the live
load which also satisfies the condition for the greatest stress
in these chord members, namely, when the second driver
stands at 2. The computed stress for the position is 130.3
tons. Two computations are also required for the stresses in
RB = RC and in GH.

Prob. 53. A through Pratt truss for a single track railroad
bridge has 7 panels each 21 feet 6 inches long and 26 feet
deep. The trusses are 14 feet 8 inches apart center to center
of chords. The dead panel load is 1.93 tons on the upper
and 6.33 tons on the lower chord. The live load is to con-
sist of two coupled Lehigh Valley consolidation locomotives
followed by a uniform load of 4 000 pounds per linear foot.
Find the maximum and minimum stresses.

Art. 42. Moments in Plate Girders.

In Art. 40 the equilibrium polygon was drawn with a pole
distance equal to a multiple of the depth of the girder, mak-
ing it possible to read the values of the flange stresses directly
from the diagram with the scale. This is convenient for a
single example, but if a number of girders are to be analyzed
for the same live load it will economize labor to plat the
equilibrium polygon on a sheet of profile paper by laying off
the moment ordinates under each wheel as taken from a
tabulation diagram similar to the one shown in Part I, Art. 62.
Such an equilibrium polygon, or wheel load moment diagram,
is shown in the line ABC D' in Fig. 89. The first side AB
is horizontal, and coincides with the axis AX. The total
ordinate at any point represents the sum of the moments of
all the wheels on the left of the point with reference to the
point as a center of moments. If the side lying directly
under the pilot (indicated by two sides of a triangle) of the
second locomotive be produced to meet the ordinate at C\

Art. 42.

MOMENTS IN PLATE GIRDERS.

125

the portion of the. ordinate above this line represents the
moment of the entire second locomotive about the head of
the train, while the portion below the side produced repre-
sents the moment of the first locomotive about the same

/ ^J4S6t89 i9 n ta 1314 15 16 tj j8

: o Q Q99 o o 9 .^o Q OOP 9900

FifT. 89.

point, the values of these moments being read off directly
from the sheet with its engraved scale (see Art. 7). As large
a scale of moments should be adopted as possible without
making the sheet too unwieldy to handle conveniently.

In the same article the maximum flange stress for any given
section was obtained by practically finding the stresses for
many different positions of the live load and selecting the
largest one. If the position of the load which produces the
maximum moment, and therefore the maximum flange stress,
could first be determined, it would be necessary to find the
value of only one moment for each section. This will now
be done.

The criterion for the position of the wheel loads which

126 LOCOMOTIVE WHEEL LOADS. CHAP. IV.

produces the maximum moment at any section of a girdei
which supports the load directly, is

in which W is the whole load on the girder, P^ the part of the
load on the left of the section, /' the distance from the section
to the left support, and / the span. This formula niay be
deduced in a similar manner to those in Part I, Art. 6i. It
is really the same formula as that which applies to vertical
sections through the panel points of trusses with vertical
posts, and to the panel points of the loaded chord of those
with inclined web members. In plate girders with stringers
and floor beams the criterion applies only to the sections at
the floor beams. If the cross ties rest directly on the girder,
or if a solid metal floor is adopted which is supported at
points very close together like the cross ties, the load is
regarded as supported directly, and the above formula may
be applied at any vertical section whatever.

To use this criterion let a load line ABCD^ composed of a
series of steps, be constructed on the same sheet as the wheel
load moment diagram, as shown in Fig. 89. The rise in any
step indicates the weight and the position of the correspond-
ing wheel. Any ordinate to this load line gives the sum
of all the loads on its left, and its value can be read off di-
rectly.

On a sheet of tracing paper let the span ab of the girder be
laid off to exactly the same linear scale as that of the profile
paper containing the load line and moment diagram, and let
the half span be divided into the required number of equal
parts and indefinite ordinates erected at these points. Fig.
go shows the completed diagram for a span of 80 feet placed
in position on the load line. The live load used in this
example is given in Fig. 87.

It is very important that when the base lines of the two

Art. 42.

MOMENTS IN PLATE GIRDERS.

127

diagrams coincide, their ordinates shall be truly parallel. The
sections are five feet apart and the 20-foot section is made to
coincide with wheel 4. If this be the proper position for

*Fig. 90.

the maximum moment in the 20-foot section, the equation
P^ = -r-lV must be satisfied. Remembering that the hori-
zontal axis on the left of a is to be considered as a part of
the load line, connect by a straight line the points a and d
where the ordinates at the supports intersect the load line.
Now the ordinate dd equals Wy ab equals /, ae equals /', and

hence W-j is represented by the ordinate ei. If the wheel 4

is just on the right of the section the ordinate eh represents
P\ if just on the left the ordinate ^^ is the load P^ \ and
when it is at the section the load P' may be regarded as
having any value between these limits. The condition is
therefore satisfied, and this position will give the maximum
moment at section 20'. All the possible positions for all the
sections can be determined in 9.. few minutes, by using a small
silk thread, shifting the tracing paper in each case so as to
bring a wheel over the section, stretching the thread as indi-

128 LOCOMOTIVE WHEEL LOADS. CHAP. IV,

cated above and noting whether it intersects that wheel on
the load line. The graphical method of determining position
by means of a load line was first published by Ward BALD-
WIN in Engineering News, Vol. XXII, pages 295 and 345.
See also letter of Dec. 11, 1889, on page 615.

By remembering the condition of loading required for uni-
form and excess loads, it is clear that the wheel loads should
at first be approximately placed so as to cover the entire
girder, and then if most of the drivers are on the left of the
section the load should be moved somewhat toward the right.
If by running off some light wheels on one end heavier ones
may be brought on the girder at the other end without
removing the drivers much farther from the section, the
'moment may probably be .increased. By such considerations
the labor of determining position may be reduced.

The value of the maximum bending moment at section 20^
may now be obtained by drawing the closing line and measur-
ing the ordinate fg, A scale consisting of a separate strip of
profile paper cut from the same sheet is convenient for this
purpose. If the left end a of the closing line were always on
the axis ab, greater precision would be attained by reading the
moment dcy multiplying it by the ratio /' -7- /, which in this
example is one-fourth, and subtracting the moment ef, which
is known and usually marked on the diagram. In plate
girders, however, a is frequently not on the axis. In the
above girder the positions satisfying the criterion for the
various sections were found to be as follows:

Section.

Wheel at Section.

Section.

Wheel at Section.

2, 3

25'

4. 12, 13

10'

2, 3. 4

30'

12, 13

15'

3.4

35'

12, 13

20'

4. 12

40'

5, 8, 9, 10, 12,

It is therefore better to make the scale large enough to insure
the requisite precision when the ordinates are read off directly

Art. 42.

MOMENTS IN PLATE GIRDERS.

129

by a separate scale so that only one value needs to be
recorded. Where more than one position satisfies the con-
dition, the use of the dividers will show which is the largest,
and its value alone needs to be carefully read and recorded.
In order to enable a large vertical scale to be used on the
width of an ordinary sheet of profile paper, the equilibrium
polygon, or moment diagram, was consolidated as shown in
Fig. 91. In this way the vertical divisions on pro-
file paper, plate B, which measure a little more
than 0.8 inch, could be 'taken as 500 thousand
pound-feet instead of 1000 thousand pound-feet

Fig. 91.

as before. The linear scale used was 4 feet to an inch, the
actual working diagrams being therefore about ten times as
large as those here given.

By reference to the above table of positions it is seen that
the ordinates to be measured lie in the left half of the diagram
so that acute intersections of the ordinates with the polygon
are avoided. If this diagram were not also to be used for

130 LOCOMOTIVE WHEEL LOADS. CHAP. IV,

obtaining shears, a further improvement would be made by
inclining the axis downward towards the right, thus bringing
the closing lines more nearly horizontal.

Referring again to Fig. 90 in order to call attention to the
relation between the analytic and graphic methods, it should
be observed that if no load is off the girder at the left, the
ordinate eg is the moment of the reaction while ef is the
moment of the loads on the left of the section, the center of
moments being at some point in the section. If, however,
some loads have passed beyond the left support, then the
corresponding point e will lie on that side of the equilibrium
polygon produced which is intersected by the vertical at the
left support.

In plotting the load line and the equilibrium polygon the
values generally used in practice correspond to the weights on
one rail only, since single track bridges are so much more
numerous than those with a double track. As it is customary
to mark the stresses in pounds on stress sheets, it is preferable
to plat the weights in units of 1000 pounds and the moments
in units of 1000 pound-feet.

If a plate girder has floor beams and stringers, its bending
moment diagram for the live load, corresponding to that
shown above the axis in the lower left corner of Plate III,
wiH be bounded by irregular curves between the moment
ordinates at the floor beams, or panel points. These curves
may be concave as well as convex, and may be replaced by
right lines for all practical purposes. The more precise
determination of intermediate moments will be treated in
Art. 56.

An accurate representation of the dead load bending
moment diagram consists of two parts, one due to the weight
of the floor system, which is concentrated at the panel points
and which may be plotted above the axis; the other due to
the uniformly distributed weight of the girder itself. Fig.

Art. 43.

SHEARS IN PLATE GIRDERS.

131

92 shows the moment diagrams of a through plate girder
whose span is 80 feet, and whose floor system has five panels.

Fig. 93.

The upper diagram gives the moments due to live load, and
the lower one those due to the dead load.

Prob. 54. Construct the diagrams described in this article,
check the positions given in the table, find which position
gives the maximum moment at each section and the values of
these moments.

Art. 43. Shears in Plate Girders.

In Art. 40 it was shown that the maximum shear in any
section occurs when one of the wheels at the head of the loco-
motive is at the section. If M' be the moment of all the
loads on the girder about the right support when wheel i
is at the section and M" when wheel 2 is at the section, the
corresponding values of the vertical shear are

F;=-^, and F;=:-y--P.,

P, being the weight of wheel i. If the load is directly

132

LOCOMOTIVE WHEEL LOADS.

Chap. IV.

supported by the girder there will be some section where the
shear due to both positions is equal. Equating these values
and transposing,

J/" - J/' = P,L

In Fig. 93 let cd be the moment M" and ab the moment
J/'. The distance ac between these moment ordinates is

I I

1^ i [ ,

I I

' 4Q'

Fig. ga.

equal to the distance between wheels i and 2. If F, = F^,
de=i cd— ab=i M" — J/' = Z',/. The position of the sec-
tion where F, = F, may then be found as follows: Place the
girder diagram (constructed on the tracing paper as described
in the preceding Article) on the locomotive moment diagram
(Fig. 89) with its left support at wheel i, and mark on its sec-
tion or ordinate at the right support the distance to the line
Bby which is the side of the equilibrium polygon on the right
of wheel i produced. • This is PJ (Art. 7). Move the former
diagram to the left until the section at the right support is at
wheel 2, and mark the position of wheel i. The two points
marked are shown in Fig. 93, at ^ and b, respectively. Now
move the tracing with the point d remaining on the equi-
librium polygon and with its axis horizontal until a position
is reached where b is also on the polygon. Mark the position
of wheel 2 at/*. It is easily remembered what wheel is to be
marked by noticing that the right-hand ordinate cd is M'\
which by the notation is the moment at the support when
wheel 2 is at the section. At every section, therefore,
between / and e the greatest shear will be produced when
wheel I is at the section, and for all sections between /* and ^
the left support) when wheel 2 is at the section.

Art. 43- shears in plate girders. 133

In the girder under consideration the shears under the first
two wheels are equal at a section a little over 60 feet from the
left support. For the live load here employed no plate girder
has been built of so large a span as to cause wheel 3 at any
section to give a greater shear than wheel 2.

The load is now placed in position for the different sections
successively, and the corresponding moments at the right
support read off. When these are divided by the span the
left reactions are obtained, and from the reactions the shears
are found by subtracting P, in the case of those sections for
whose position /\ lies between the section and the left sup-
port. For the section, at 15' the moment at the right support
is 7 040 thousand pound-feet. This gives a reaction of 7 040
-i- 80 = 88.0, and a shear of 88.0 — 8 = 80.0 thousand
pounds, or 80 000 pounds. For the section o wheel i is off
the girder, and hence the ordinate below the line Bb in Fig.
89 must be deducted from the moment at the right support.
The shear at o is found to be (10 225 — 705) -r- 80 = 1 19.0
thousand pounds. The values of the shears from sections o
to 40' inclusive are 119.0, 105.6, 92.6, 80.0, 68.8, 58.9, 50.1,
41.8, and 33.9, all expressed in units of a thousand pounds.
The form of diagram employed (Fig. 91) indicated at a glance
that a single locomotive followed by a train produces slightly
greater shears at sections 15', 20', 2^5', and 30' than the load
with two locomotives, their values being 80.4, 69.8, 59.8, and
50.4 thousand pounds respectively.

Usually only the maximum shears in one-half of the girder
are required, but in order to show the variation of shear across
the entire girder as well as the minimum values, the shear
diagram due to dead and live load is shown in Fig. 94. The
greatest live load shears (all positive) are laid off above the
axis and the dead load positive she^irs below the axis, so that
the diagram combines their values. The point of zero shear
is at S5«6' or at 15.6 feet from the middle of the girder.

134

LOCOMOTIVii: WllKEL LOADS.

Chap. IV.

The shear may therefore change sign at all points within this
distance on each side of the middle. If the live load were
uniformly distributed, the shear curve would be a
parabola, with its vertex at the right end of the
girder.

When the live load consists of passen-
ger instead of consolidation locomotives
it is possible that for some spans the
section in which V^ = F,"
may be on the , right of
that where F, = F„ in

which event it is necessary to find
where F, = F,. No position for
^^^- ^' maximum shear then requires

wheel 2 to bemt any section, i-^

If the girder is divided intb panels by floor beams the
criterion ;for position for maximum shear is- the same as for
trusses with parallel chords. The necessary formula was'
deduced in Part I, Art. 60, but its graphic application will be
deferred to Art. 45. The live load shear curve would be
transformed into a series of steps like that in Fig. 81, and the
dead- load shear diagram- would consist of two parts, one for
the floor system which is .concentrated rat the panel points,
and the other for ^the. weight of the girder, itself, which is
uniformly distributed.

Prob. 55.^Take the plate girder used in the example in
Art. 49, find where F, = F,, and the values of the maximum
shears due tojone Lehigh Valley consolidation locomotive and
train.

Art. 44.

SIMULTANEOUS MOMENTS.

135

Art. 44. Simultaneous Moments.

In designing the riveting of the flanges to the web of a
plate girder it is necessary to have the horizontal siiear
between them or the increments of flange stress between tlie
sections for which the bending moments and vertical shears
were found. If the sections
are a distance dx apart the
difference of bending moments
dM is a maximum when the
load is so placed as to cause
the vertical shear to be a max-
imum, since from mechanics
dM^ Vdx. When the dis-
tance between the sections is
greater than dx the difference
of bending moments is a max-
imum when the loads are so
placed that the vertical shear
is a maximum in the section Fig. 95.

nearer the middle of the span, and this holds true for uniform
loads until the sections are separated a distance somewhat
less than half the span. As the sections are not taken
farther apart than the depth, which even in short spans is
generally less than one-eighth of the span, the exact value
of the limiting distance referred to need not be determined.

The following table contains the simultaneous bending
moments in each pair of 'adjacent sections of the girder used
in the preceding two Articles when the live load is so placed
as to produce the maximum shear in the section nearer the
middle 6f the girder. The moments are expressed in thousand
pound-feet. ' The differences in this table are to be increased
by the corresponding differences in the dead load bending
moments. The moments themselves are laid off as ordinates

136

LOCOMOTIVE WHEEL LOADS.

Chap. IV.

in Fig. 95, those of each pair being joined by a right line.
The upper curve gives the maximum live load bending

DiSTANCB OF

Sbction from
Support.

Load in Position for Maximum Shbar at Sbction.

DiSTANCB OF

Sbction from
Support.

10'

15'

90'

95'

30'

35'

40'

Feet.

5
10

15
20

530

1315
1610

Feet.
20

480
945

1430
i68o

850
1255

1470
1680

1445
1610

1125
1470

Difference

530

465

405

345

295

250

210

165

Difference

moments. Those due to dead load are laid off below the
axis.

Prob. 56. Find the simultaneous moments in the sections
of the plate girder used in the example in Art. 40 (see Plate
III).

Art. 45. Shears in Trusses.

In a similar manner to that explained in Arts. 43 and 42
for a plate girder the shears and bending moments in the Pratt
truss in Art^ 41 can be found by means of a single diagram
for the locomotive and train load, the positions which give
the maximum stresses being first determined by means of the
load line.

In Part I, Art. 60, the criterion for the position of the live
load producing the greatest vertical shear in any section of a

truss was found to be P'' = — W^, in which W^is the whole

load on the truss, P'' the wheel loads (one or more) on the

panel cut by the section, and m the number of panels in the

truss. Let this equation be transformed into W^= mP'\

When the truss diagram is placed on the load line the value

Art. 45. SHEARS in trusses. 137

of IV can at once be read off on the ordinate at the right
support for any giveh position of the truss with respect to the
loads. If wheel 2 be placed just on the right of a panel point
Wraust be equal to mP^, and if just a little to the left of the
same panel point then W must be m{P^ + -f*,)- The condition
will therefore be satisfied if wheel 2 is at the panel point, and
the value of Wis found to lie between the values mP^ and
m{P^ + -^t)* and similarly for any other wheel.

In this example, which was also treated by the analytic
method in Part I, Art. 63, m= 7, P,= S tons, P, + -P, = 16
tons, and jP, + ^1 + -^i = 3^ tons. The following table is
then arranged :

No. of Wheel at Right Values of /*" Corresponding

End of Panel. * Values of PV.

I 0-8 tons. o- 56 tons

2 8-16 " 56-112 "

3 16-36 " 112-252 "

The moment diagram of the single Pennsylvania passenger
locomotive and train is constructed like that illustrated in
Fig. 89, and the truss diagram (like Fig. 96) is drawn to the
same linear scale on a sheet of tracing paper, but with the
verticals extended as high as the moment diagram. To find

f ^

\
\

N
\

\ \

: I

i i

? J

^ \

r ^

^ J 4 .

Fig. 96. Fig. 97.

the position for maximum shear in Cc and Cd due to this load
shift the truss diagram on that of the load line until wheel 3
is at d. If this is the correct position the load on the truss
must be between 112 and 252 tons. This is found to be the
case, its value being 154 tons. For the greatest positive
shear in /^ (which equals the greatest negative shear in Be)
wheel 2 is placed at g. The load W is then 56 tons, which

138

LOCOMOTIVE WHEEL LOADS.

Chap. IV.

just meets the condition and indicates that wheel I at ^
should also be tried. In the latter case Wis 36 tons. The
positions are all recorded in the following table. In practice
it is preferable to obtain all the positions before reading off
any moments from the equilibrium polygon, that is, to deal
with only the load line at first and with the equilibrium poly-
gon afterwards.

Panel.

Whkel at
Right End
OF Panel.

Moment
AT Right
Support.

Moment at
Right End
OF Panel.

Shear.

Stress.

Web
Member.

15075

188

98.3

— 128.0

moo

188

69.9

4-91.0

7730

188

45.8

- 45.8
4-59.6

Ce
Cd

4940

188

25.9

- 25.9
4-33.7

Dd
De

3885

25.5

19^5

II. 8

4- 15.4

645

2.4

390

2.8

4-3.6

The moments in the third column of the table may now be
read directly from the diagram. Those in the next column
were computed and marked on the diagram when it was con-
structed. They are all expressed in tons-feet. Since the
span is 140 feet and the panel length 20 feet, the shear in Cc
and Cd is (7 730 -^ 140) — (188 -7- 20) = 45.8 tons. As it is
not necessary to know the value of the reaction separately,
and as u.sually the panel length is not such a simple number
as the above, it is better to multiply the moments in the
fourth column by the number of panels in the truss, subtract
the products from the moments in the third column, and
divide the remainder by the span. Thus the shear in Cc and
Cd is (7 730 — 7 X 188) -r- 140 = 45.8 tons. The stresses
are obtained from the shears either graphically or by the aid
of logarithms, as may be more expeditious.

AUT. 46. • FLOOR- BEAM REACTIONS. 1 39

As the method described above h the exact graphic equiva-
lent of the analytic method-given in Part I, Art. 63, the
student should make "a careful comparison between them.
Fig. 97 is placed beside Fig. 96 as an aid in comparing the
results obtained by the analytic and the two graphic methods.

The criterion for position used in this Article applies to all
trusses with horizontal chords and single systems of webbing, ,
like the Howe and the Warren as well as the Pratt truss.
When one or both of the chords are inclined, the maximum
shear does not give the maximum web stresses, since the
chord takes some of the shear. The method of finding the
position of the live load for trusses with inclined chord mem-
bers will be given in Art. 49.

Prob. 57. The Pratt trusses of a single-track through rail-
road bridge have seven panels, each 25 feet 8 mches long and
32 feet deep. Find the shears and stresses in all the web
members except the suspenders due to two Lehigh Valley
consohdation locomotives and train (see Fig. Sy in Art. 39).

Art. 46. Floor-beam Reactions.

In order to find the maximum floor-beam reaction or stress
in the suspender 5^ due to locomotive wheel loads it is neces-
sary to deduce additional formulas. In Fig. 96 let Ra be
the stringer reaction at ^, and R^ the sum of the adjacent
stringer reactions, or the floor-beam reaction, at d. Let P be
the whole load on the two stringers of equal spans ad and 6c,
and^ the distance of the center of gravity from c; let P^ be
the load on ab, and ^' the distance of its center of gravity
from 6. Since the sum of the moments of loads and reactions
about 6 is zero,

Rap-P'g'^o, or R^p^P'g'.
Taking moments about r,

R,2p^R,p^Pg^o.

I40 LOCOMOTIVE WHEEL LOADS. CHAF. IV.

Substituting and reducing,

Pg-2P'g'

If the loads be moved a distance dx to the left both g and g^
will receive an increment dx^ and R^ an increment

,_ Pdx — 2P'dx
dRh = .

Placing the derivative equal to zero gives

P=2P'.

That is, when the live load in both panels is double that in
the panel ab the resulting value of Ri, is a maximum. This
is the same condition as for findmg the maximum bending
moment at the middle of the girder whose span is ac.

The use of the load line gives the position very quickly, and
the moments Pg and P'g' can be read off on the moment
diagram. If Mc be the moment ordinate at r, and Mi, that at
by the value of R may be more conveniently expressed and
remembered as

^ M, - 2M,

P
In the case of bridges where the two panels at the end, /,
and /a, are not equal, the value of Ri, deduceS in a similar
manner is

^^ ~ ^T^^; »

A A

the criterion for loading being that which produces a maxi-
n\um moment at ^ in a girder whose span is ac, /, being the
span of the stringer ab, and p^ the length be.

In applying this condition for loading, one of the heaviest
loads should be placed at b and as large a load brought on
the two panels from ^ to ^ as possible. When wheel 3, in
the example used in the previous Article, is placed at b and

Art. 47.

MOMENTS IN TRUSSES.

141

the thread is stretched to unite the intersections of the
ordinates at a and c with the load line, it crosses the step
representing wheel 3 and hence satisfies the condition. The
moment at ^ is i 170, and that at * is 188 tons-feet, whence
Bd = {i 170 — 2 X 188) -^ 20 = 39.7 tons. When wheel 4
is at If the condition is also satisfied, and the corresponding
stress found to be the same as for the other position.

Prob. 58. Find the stresses in the suspenders in Prob, 57'
in the preceding Article.

Art. 47. Moments in Trusses.

The method used in the preceding Article in obtaining the
maximum shears will now be employed to find the maximum
chord stresses in the same truss and due to the same wheel
loads.

The condition for loading is expressed by a formula
deduced in Part I, Art. 61, which, in slightly modified form,
was given in Art. 42 and its application to a plate girder fully
explained. As there stated, it applies only to the bending
moments in vertical sections through the panel points of the
loaded chords of trusses. The positions are recorded in the
following table for the required sections of the left half of the

Section.

Wheel

Section.

Moment at

Right

Support.

Moment

Section.

Bending
Moment.

Stress.

Chord
Member.

15075

188

1966

81.9

ab = be

16850 — 1 140

476 — 180

1948

14540

1008

3146

131. 1

cd^BC

12610

476

3127

13610

2124

3709

154.5

CD -DE

train

14510

4605

3686

153.6

truss. As for the section Ee the train is at the section the
direction of the thread must coincide with the straight portion

142 LOCOMOTIVE WHEEL LOADS. ChAP. IV.

of the load line representing the train, and hence the left end
of the truss must be placed at the point where the line pro-
duced meets the horizontal axis.

For a truss having equal panels, as in this example, time
may be saved as well as increased precision secured by not
drawing the closing lines and reading the bending moments
directly, but by reading the moments at the right support
and at the given section. The latter, being at a wheel, has
its value marked on the diagram, and is hence quickly
obtained. The subtractive moments for the second position
for section Bb are those due to wheel I, which is off the
bridge.

The computation for the bending moment is very simple

for trusses with equal panels, as it is not necessary to obtain

the value of the reaction separately. For wheel 5 at ^ the

bending moment in the section Cc is

-(14 540) — I 008 = 3 146 tons- feet,

and when divided by the depth the chord stress is 3 146 -5- 24
= 13 1. 1 tons.

Since there is no dead load shear in the middle panel, it is
necessary to find which one of the diagonals is acting for each
of the positions for sections Dd and Ee, The shear equals
the left reaction of the truss minus the loads from a to d^
minus the reaction at d of the stringer de. By producing the
side of the equilibrium polygon immediately on the left of d
when wheel 8 is at ^af and reading the moment intercepted
above this line at e the moment of the stringer reaction at d
about ^ as a center is obtained. Its value is found to be 380
tons-feet. The shear is therefore

^3 ^'^-7x380 _ g^ ^ _ ^ g ^^^^^

140
As the diagonals can take only tension, this shear calls dE
into action, and hence the bending moment for the section

Art. 47- moments in trusses. 143

Dd gives the chord stress in both CD and DE. For the last
position the entire panel is covered by the train load, and
therefore the stringer reaction at ^/ is 15 tons. The shear is

i45io_ 15 =-1.4 tons.

140

Since this also stresses the diagonal dE^ the moment for the
section Ee gives the chord stress in de.

The sign of the shear without its magnitude may be more
quickly determined for each of these positions by drawing the
closing line of the equilibrium polygon and with the aid of
the dividers finding whether the bending moment at d is greater
or less than the simultaneous moment at e. In the former case .
the shear is negative and in the latter positive (see Art. 9).

If the live loading for the greatest moments at the sections
Z?^ and Ee respectively had caused shears of opposite signs in
the middle panel, the required stress in one of the chords of
that p^nel would not have been given by either loading. In
such a case it becomes necessary to shift the load to some
intermediate position for which the bending moment ordinates
at both ends of the middle panel are equal, and the shear is
therefore zero. The required stress may then be obtained from
the moment at either section. In practice, however, the stresses
in both chords of the middle panel are generally assumed to
be equal.

When the center of moments of a chord member is situated
on the unloaded chord of a truss, and does not lie in the
vertical section through a panel point of the loaded chord,
one of the last two formulas of Part I, Art. 61, determines the
condition of loading. An example of their application will be
given in Art. 56. The condition in all cases applies to inclined
as well as to horizontal chords.

Prob. 59. Find the greatest live load stresses in the chords
in Prob. 57 in Art. 45.

146 TRUSSES WITH BROKEN CHORDS. CHAP. V.

through the intersection of B^ and B^ (Art. 7), which by con-
struction coincides with the center of moments. The moment
of the resultant is therefore zero, whence it follows that the
stress in D is also zero.

The position of a concentrated load P causing no stress in
any web member can therefore be found by the following
rule:

Pass a section cutting the web member and a member
of each of the chords. Produce the unloaded chord
member to an intersection with the verticals at the
supports. Join these points with the panel points at
the end of the loaded chord member. The intersec-
tion of these lines gives the required position.

From the manner in which the above investigation was
made it is clear that this rule applies also to a truss in which
both chords are curved, and for webbing whose posts are not
vertical. The rule is therefore stated in its general form and
applies to deck as well as through bridges.

The manner in which the section must be cut to obtain the
stresses in the vertical posts depends upon which diagonals
are acting in the adjacent counter-braced panels (see Part I,
Art. 36, and Part II, Arjts. 30 and 35). The position of P
which produces no stress in the vertical 00 the left oi D in
Fig. 98 when both of the adjacent main diagonals are acting
is somewhat nearer to panel point d.

The results of this lAvestigatiort also show that if the live
load consists of panel loads, all the panel points on the right
of P are to be loaded for the greatest tension in D. If one
excess load is employed- it must be placed at d. For the
greatest tension in the counter dfagorikl in the sarhe panel the
load is similarly placed at c\ and the panel points on the left.
This loading, it will belobserved, does not differ from the
corresponding one for horizdntal chords. •

Art. 49.

POSITION OF WHEEL LOADS.

147

If the live load is uniformly distributed it must extend from
the right support to the position of P for the greatest tension
in Dy and from the left support to the position of P for the
gre?itest tension in the counter diagonal. When the construc-
tion shown in Fig. 98 is applied to trusses with horizontal
chords it gives the positions for true live load shear which
were determined by the anal3^tic metliod in Part I, Art. 56.
The position of locomotive wheel loads which produces the
greatest stress in D will be found in the next article.

Prob. 60. Find the position of P for all the diagonals and
the second, third, and fourth verticals of the bowstring truss
in Fig. 74, Art. 35.

Art. 49. Position of Wheel Loads.

In Fig. 99 the position of P in Fig. 98 which causes a stress
of zero in the diagonal is indicated by the vertical marked z.
Let the stresf>es in the diagonal, lower chord, and upper
chord, cut bya'section through the panel r^/ be. denoted by

Fig. gg.

5, 5i, and S^ respectively, and the depths of the truss at c and
d by h^ and h^. Let the total weight of the wheels (one or
more) on the panel <?^ be P'"^^ and the-distance of its center
of gravity from d be g"\ W being the weight of the entire
load on the truss, and ^ the distance of its center of gravity
from the right support. Let the bending moments at the

148 TRUSSES WITH BROKEN CHORDS. CHAP. V.

upper and lower extremities of D be M^ and M^ respectively.
The remaining terms eniployed in the following discussion are
shown in the figure.

Let the segment of the truss on the left of the section
cutting 5,9 Si and 5, be considered. Resolving horizontally,

5, cos a + 5 sin tf + 5, = o.

.The lever arm of 5, makes the same angle a with the vertical
h^ as 5, makes with the horizontal. Taking moments about d^

M^ + SJi^ cos £x = o» whence 5, cos a = — Jf^ -*- A^;

and taking moments about the panel point at the upper end
of A

M^ — 5,A, =3 o, whence 5, = J/, -5- h^.

After substituting these values above, there follows.

The last equation is an important one, and indicates that the
horizontal component of the stress in any web member equals
the difference of the quotients obtained by dividing the bend«
ing moments at the extremities of the member by the corre-
sponding depths of the truss at those points.

The values of the bending moments are

M,^^ and i/. = -^(/-/0-P'V^

From similar triangles

h^xz^l^xl-^l^ and h^\»^l^\l^\

whence

Art. 49.

POSITION OF WHEEL LOADS.

149

Substituting these values of Af„ J/„ A„ and A„ reducing, and
finally replacing (/, — /,) by /'",

l^ sin Q

If the load advance a distance dx^ both g and g'" receive
an increment of dx^ and the stress 5 receives an increment of

dS^

{Wl'

P"'l,)dx

l^z sin i)

Placing the derivative equal to zero gives the condition
which makes 5 a maximum, which is Wl'^' — P"'l^ = O, or,
when put into more convenient form for use,

P'" =

Wl'

This formula is very convenient to use graphically, and as
it is similar in form to that for ^

maximum moment (Arts. 42 and
47) it is to be treated in like man-
ner. Referring to Fig. 100, which
illustrates the truss diagram
(drawn on tracing paper) placed
in position on the live load mo-
ment diagram, bg represents the

Fig. loa

ISO TRUSSES WITH BROKEN CHORDS. CHAP. V.

total load fF, ob the distance /„ and ^^the distance l"\ The
ordinate di is therefore equal to Wl'" -r- /,, and this must
equal P'^' if the positipn is correct. When the load is so
placed that a wheel is just on the right of the panel point
dy the load P'" is represented by dhy and if just to the
left of it by dk\ hence if / lies anywhere between h and ky or,
in other words, if the thread stretched from o to g cuts the
load placed at the panel point, the criterion for position is
satisfied. A wheel must always be placed at the panel point,
and while usually the first wheel is at the right of ^, it may
sometimes happen* that the condition is met when the first
wheel is a little to the left of o. After the right position is
found the moment ordinates ^/and de are read off as usual.

Prob. 6i. A double track through railroad bridge has
trusses of the type illustrated in Fig. ibo. There are 12
panels, each 30 feet long. The depths at panel points i to 6
inclusive are 29' o", 41' o", 49' 5", ^3' 4", 58' 10", and
60' o" respectively. The live load is Waddell's Com-
promise Standard. Class U (Art. 39). Find the position of
the live load which shall produce the greatest stresses in the
main and counter ties and the posts. Six panels require
counter bracing. Also compare these positions with what
they would be if the truss had parallel chords.

Art. 50. Resolution of the Shear.

In Fig. 100 in the preceding article the stresses S,, 5, and
•S, hold in equilibrium the external forces on the left of the
section cutting these members. These external forces consist
of an upward reaction at a and a downward force at c equal
and opposite to the left reaction of the stringer in the panel
cd. The resultant R of these two forces is an upward force
whose line of action is a little to the left of the support a.
Its position may be readily determined, if desired, by drawing
the closing line ^/and the chord ce of the equilibrium poly-
gon and producing them to their intersection.

Art. 50.

resolIjtion of the shear.

151

Referring now to Fig. lOi, let the resultant R be replaced
by two forces P^ and -P,, the former acting downward at panel

Pig. zox,

Pig. 103.

point d and the latter acting upward at e. The points d and e
are at the extremities of the diagonal cut by the section.
Taking moments about ^, and remembering that the bending
moment at e is M^y

Rr,=P,p', and /> = ^' = ^.
Similarly taking moments about dy

RU^PJ, and 7>=y! = ^.
Taking vertical components,

^ - ^» " ^« - / ■- / •

Since 7? is equal to the vertical shear in the section, the last
member of the preceding equation affords a useful method of
obtaining the vertical shear when the moments are known.

In Fig. 102 the force triangle hfi gives the magnitude and
direction of a force acting in the diagonal which is in equi-
librium with P^ and S„ while the superimposed force triangle
fgk gives the magnitude and direction of a force acting in the
same diagonal which is in equilibrium with P^ and S,. The
polygon lifgkih must therefore express the relation of equi-

152

TRUSSES WITH BROKEN CHORDS. ChaP. V.

librium ' between P„ P,, 5„ 5, and S„ or the polygon hgkih
that between Ry 5,, 5, and 5,. F^ollowing around the trian-
gle in the direction of the known force R as indicated by the
arrows, and transferring these directions to the truss diagram
in Fig. loi, 5, and S are found to be in tension and 5, in
compression. It will be observed that the forces in the poly-
gon hgki follow each other in the same order as they are found
when passing around the segment of the truss. Fig. 103
shows the same construction when the forces are laid off in
the reverse order. As will be illustrated later by an example,
It is sometimes preferable to use the one and sometimes the
other order.

It is evident on inspection that the most convenient and
economical construction of the force polygon in Fig. 102 (or
in Fig. 103) would be to draw it directly on a large scale
truss diagram. In Fig. 104 one such force polygon is drawn

Fig. X04.

in the third panel. The notation shown is well adapted to
promote rapid construction and freedom from confusing the
stresses. The panels are numbered from left to right and the
corresponding numbers are placed at the panel points on their
right. The upper chord members are denoted by [/, the
diagonals by JO, and the lower chord members by Z, the sub-
scripts being those of the panels containing them. The vet"-'
ticals V necessarily have the subscripts of the panel points.
The forces J/, -?-/ and M^ -r-/, equal respectively to the forcds
P^ and P^ in Fig. 10 1, are .laid: off as indicated, and the sides

Art. 51. EXAMPLE— MAXIMUM CHORD STRESSES. 1 53

of the force polygon drawn parallel respectively to the truss
members whose names are placed by their sides. The panel
length/ in this case is equal to the horizontal projection/' of
the diagonal.

In order to obtain the stress in the vertical F,, for example,
the values of Jf, -r-/ and J/, -H/ are found for the proper
position of the live load (which may not be the same as for
D^ and a force polygon drawn as in Fig. 104, This gives C/','
and Z>, for that position of the load, and on constructing the
force polygon for the upper panel point 2 the stress in F, is
determined. The latter polygon may be drawn directly on
the former so as to avoid redrawing the sides U^ and Z>,. To
avoid confusion it is omitted in the figure.

This method may also be applied with advantage to deter-
mine the stresses in the chords when the moments have been
found after placing the live load in its proper position. In
this case it will be desirable to place the force polygon on the
other side of the diagonal, the values of J/~/ being laid off
upward from the lower panel points. This will give a poly-
gon like that in Fig. 103.

Prob. 62. Find the maximum live load stress in the counter
tie of the fourth panel, and the minimum live load stress in
.the second vertical, of the truss in Prob. 61.

Art. 51. Example — Maximum Chord Stresses.

Let the truss in Fig. 104 and on Plate V be that of a single
track through railroad bridge, having seven panels each 27
feet long, and depths at the panel points i, 2, and 3, of 29,
35, and 38 feet respectively. Let the live load consist of
Class U of Waddell's ** Compromise Standard System "
(Art. 39). The dead load will be assumed at i 100 pounds
per linear foot per trussj of which 300 pounds is to be applied
on the upper chord. This ma,kes the panel loads on the lower

154 TRUSSES WITH BROKEN CHORDS. CHAP. V»

chord 21 600 pounds and on the upper chord 8 100 pounds.
Since the dead load stresses in all the members of the truss
except the verticals are the same whether the dead load is all
taken on the lower chord or divided between the chords, and
the stresses in the verticals differ by the amount of the upper
panel loads (Art. 28), the stresses will preferably be obtained
for only lower panel loads of 29 700 pounds, and the maxi-
mum and minimum stresses in the verticals afterward cor-
rected by adding to each of them a compression of 8 100
pounds.

In constructing the load line and equilibrium polygon for
the live load it was found convenient to use the weights on
one rail only, and to adopt scales of 8 feet, 20 thousand
pounds, and 2 000 thousand pound-feet per inch respectively.
Profile paper, Plate A, was used and required a small splice at
the upper right corner in order to extend the curves as far as
necessary. In reading the scales the tenths of the small
divisions of the profile paper were estimated by eye. When
this diagram is used for a double track bridge the stresses
obtained are expressed in tons instead of in units of one
thousand pounds.

The dead load stress diagram for panel loads of 29.7 thou-
sand pounds is shown in Fig. b on Plate V, and the stresses
are marked on the diagram. The character of the web stresses
is also indicated as referred to the small truss diagram. The
computation of the bending moments at the panel points may
ibe arranged as follows, when the panels are all equal : The
product of the panel load and panel length is 29.7 X 27 =
801.9. The half products of the number of panels in each
segment into which the panel points respectively divide the
truss are K^ X 6) = 3, i(2 X S) = 5» i(3 X 4) = 6; and the
bending moments are

Jfj = 3 X 801.9 = 2 406, Jfj = 5 X 801.9 = 4 010,

jjf, = 6 X 801.9 = 4811 thousand pound-feet.

Art. 51. EXAMPLE— MAXIMUM CHORD STRESSES.

iSS

Since/ = 2j feet, the corresponding values of M -^ p are:

-— = 89.1, —- = 148. 5, — =- = 178.2 thousand pounds.
P P P

The following table shows the position of the live load
obtained by means of the load line. The moments in the
third column were read from the moment diagram, and those
in the fourth column were copied from the same diagram,
while the quantities in the fifth and sixth columns were com-
puted from those in preceding columns. The moments are
expressed in units of one thousand pound-feet.

Center

Wheel

Moment

AT Right

Moments.

Section.

Support.

Section.

409S0

480

39400

2622

39010

6331

32 520

8291

35 540

10099

Bending
Moment

5 374

8635

10388

10 292

10 210

199.2
319.8

384.7
381.2

378.1

Remarks.

Z>4 acts
D^ acts
D^ acts

It is necessary to know which diagonal acts in the middle
panel for the last three positions in order to determine which
moments give the stresses in the chords of that panel. As
explained in Art. 47, the vertical shears are found as follows:

(39010 — 7 X I 470) -r- 189 — 166 = — 14.1 ;

(32 520- 7 X 830)- 189- 136 = + 5.2;

(3 554 - 7 X I 240) ~ 189 - 146 = - 3.9.

These results enable the remarks to be inserted in the last
column of the table.

The values oi M -r- p are now laid off on the truss diagram
in Fig. c on Plate V, as there indicated, and the force polygons
completed as explained in the preceding article. The scales
of the original drawing of Fig. c were 6 feet and 80 thousand
pounds per inch respectively. The values of the stresses are
marked on the polygons. The special attention of the

156 TRUSSES WITH BROKEN CHORDS. ChAP. V.

Student is called to the fact that since U^ has its center of
moments at the lower panel point 3 the side of the polygon
ad parallel to U^ must be drawn through d^ the extremity of
M^-x- p laid off on the vertical ordinate passing through the
center of moments. Similarly, as Z, has its center of moments
at the upper panel point 2, the side be must be drawn parallel
to Z, through ^', which lies on the vertical through its center
of moments. Strict attention to this statement is especially
required when the upper panel points are not directly above
the lower ones, in which case the panel points should be
numbered in regular order from left to right, no matter on
which chord they lie. The chords should then have the same
subscripts as their centers of moments.

The side ab of the polygon abed is not the stress in the
diagonal Z>„ because the moments at 2 and 3 used in its con-
struction are not simultaneous.

If it be desired to find by this method whether D^ or Z>/
acts when the moment is a maximum at panel point 3, it can
be done by finding the simultaneous value of M^ -?-/. It is
found to be 2>7^-7' If -^4 be assumed to act, the side L^ will
lie below U^. It is shown as a broken line. By referring
again to Fig. lOi the direction around the polygon is toward
the right of Z4, upward on D^, and toward the left on U^.
Upward on D^ means also toward the right, or away from the
section, and therefore tension, which proves the assumption
to be correct. Again, since {M^ —/) — {M^-r- p) = R, which
equals the vertical shear in the section indicated in Fig. 10 1,
Vthe vertical shear in the middle panel is 370.7 — 384.7 = —
14.0, the difference of o. i from the value given above being
mainly due to the neglect of decimals. Usually the value of
the vertical shear is not desired, but simply its sign, in which
case it may be known as soon as it is seen whether M^ is
greater or less than J/, (see also Art. 9).

As the end post receives its maximum stress under the

Art. 52. MAXIMUM stresses in diagonals.

157

same position of the live load as Z,, its stress may be found
in connection with the chords. Indeed, it may be regarded
as an upper chord member, the polygon of forces becoming a
straight line, as shown on the drawing (Fig. c, Plate V).

The following table gives the maximum stresses in the end
post and the chords due to the dead and live loads only,
expressed in units of one thousand pounds. The minimum
stresses equal the dead load stresses.

Chord
Mbmbbrs.

— 117. 2

— 252.2

— 127.2

— 275.3

1^4

— 126.4
-273.5

L, = L^

Dead load...
Live load...

— 121. 7

— 271.0

T 830
+ 1S5.4

+ 114. 4
+ 247.0

+ 126.4

+ 268.8

+ 395.2

Maximum. . .

— 392.7

-369.4

— 402.5

— 399.9

+ 268.4

+ 361.4

Prob. 63. A truss of the same type as that in the example
given in this article has nine panels, each 24 feet 9 inches
long. The depths at its panel points i, 2, 3, and 4 are
27' o", 35' g'\ 41' o", and 42' 9'' respectively. The dead load
is I 200 pounds per linear foot, of which three-eighths is to be
applied at the upper chord. The live load is Waddell's
Class U. Find the maximum and minimum stresses in tho
chords and end post due to these loads.

Art. 52. Example — Maximum Stresses in Diagonals.

The first step in finding the live load web stresses is to find
the point of division in each panel where a concentrated load
will produce no stress in the diagonal (see Art. 48). In prac-
tice only that portion of each triangle B^B^B^ lying below
cd in Fig. 98 need be drawn after the vertices on the ordi-
nates at a and b are marked off. These points are shown in
Fig. ^on Plate V. In case more than one point is shown in
any panel the left hand one belongs to the diagonal. The
data in the following table are obtained in the same manner as
for the chords after the positions are determined. That relat-

«58

TRUSSES WITH BROKEN CHORDS, CHAP. V.

Bending

BINDING

Whkbl at

Moment

Right

Moment

AT Right

(Af) AT

(A/) AT

Panel.

End

AT klGHT

End

Left

Right

Left ^.

Right — .

SUFPORT.

End of

Panel.
4

40 980

Panel.
480

Panel.

O — I

5374

J99.2

I — 2

28 830

230

4117

8004

152.5

296.4

2-3

19 840

230

5669

8273

210.0

306.4

3—4

12 2SO

230

5263

6787

194.9

251.4

4-5

6331

230

36 J 8

4292

134.0

159.0

5 501

3143

3849

116. 4

142.5

5-6

2 5»«

230

1793

1921

66.4

71. 1

I 960

1400

1 600

51.8

59-3

ing to the first panel is inserted here, although it was also
included in the table in the preceding article, the end post
being treated as a chord member. The moments are expressed
in units of one thousand pound-feet. The panels are indi-
cated by the panel points in this table as a guide to the sub-
scripts which properly belong to the corresponding values
of M, The moments at the right support for the panel 4—5
are given with greater precision than the rest because it hap-
pened that these had been computed and marked on the
diagram.

Attention is again called to the fact that the vertical shear
in any panel may be found by taking the difference between
the corresponding quantities in the last two columns.

In testing for position in panel 5-6 it was noticed that the
thread just touched the edge of the step when wheel 3 was
placed at panel point 6. When wheel 3 is placed at panel
point 5, wheel I is a little on the left of the point of division,
but the condition of loading is satisfied. If the chords were
both horizontal the positions would be 4, 4, 3, 3. 2 and 2 in
the successive panels, no panel having two positions of the
live load.

The values oi M -^ p are next laid off on the verticals
through the panel points in Fig. d of Plate V. The values
belonging to each panel are marked inside of the panel to

Art. 52. MAXIMUM stresses in diagonals.

1 59

guard against confusion. This danger is not great, however,
as it will be noticed that at each vertical the ordinate referring
to the panel on the right is considerably less than that for the
panel on the left. After completing the force polygons the
stresses in the diagonals are scaled off and marked on the
diagram. As the portion of the trusS on the left of the sec-
tion through any diagonal is considered, and the lower chord
is always in tension, the direction of passing around the poly-
gon is toward the right on L and toward the left on CT, and
therefore if the direction along D is toward the right it indi-
cates tension. This is seen to be the case for all but one of
the polygons shown on the plate.

In the fifth panel two polygons are drawn, the left one for
wheel 3 at panel point 5, and the right one for wheel 2 at 5.
The latter is placed on the other side of the diagonal to avoid
interference. This position is not convenient for diagonals,
as will be shown in the next article. In the sixth panel both
polygons refer to the position of wheel 2 at panel point 6, the
left hand one being drawn in order to show the influence on
the construction by changing diagonals. The stress in D^
when P^ is at 6 is — 21.8.

The maximum and minimum stresses expressed in units of
one thousand pounds are given in the following table, the end
■post being omitted, as its stresses were given in the preceding
article.

Diagonals.

Dead load

Live Inad from the rijjht
Live load Irom the left. .

Maximum.

Minimum

A '

A(=A')

A'(= A')

+ 46.2
+ 127.5
— 22.2

+ 19.6
+ 91.0

+ 69.6

— 20.7
+ 47. r

+173.7
+ 24.0

-I-II0.6

-f 69.6

+ 26.4

Prob. 64. Find the maximum and minimum stresses in the
diagonals in Prob. 63 in Art/ 51.

i6o

TRUSSES WITH BROKEN CHORDS. CHAP. V.

Art. 53. Example — Maximum Stresses in Verticals.

The position for the maximum stress in F, is either wheel
4, wheel 12, or wheel 13 at panel point i (Art. 46). When
wheel 13 is at I, the same wheels of the second locomotive
are placed on the first two panels and in the same position as
those of the first locomotive when wheel 4 is at panel point i,
together with one additional wheel; hence it is not necessary
to find the stress due to the latter position. The greatest
stress occurs when wheel 13 is at i, and equals

[3 730 — (2 X 780)] -f- 27 = 80.4 thousand pounds tension.

If it is desired to employ a similar method for F, as that
described in the preceding article, let the first two panels be
regarded as a truss, and the load placed in proper position for
maximum moment at panel point i. The bending moments
are then M. = o, J/, = ^(3 730) — 780 = i 085, and M^ = o.
M^-T- p -=2 1085-^27 = 40.2. The vertical shear in each
panel (disregarding signs) is therefore 40.2, and the floor beam
reaction, or the stress in F„ equals their sum, or 40.2 + 40.2
= 80.4 thousand pounds.

The respective points of division in the third, fourth, and
fifth panels, where a concentrated load produces no stress in
the vertical at the left of the panel, are the right hand ones
shown in Fig. d of Plate V. The position and other necessary
data given in the following table are found in exactly the
same way as for the diagonals.

Wheel AT

Moment

Bending

Brnding

Right

Moment

Left^.

Right ^.

Panel.

End

AT Right

Right

{M)KT

(A/) AT

OF •

Support.

End of

Left End

Right End

Panel.

Panrl.

OF Pankl.

of Panel.

2—3

18 340

5 240

7780

194. 1

288.1

3—4

II 050

4736

6234

175.4

230.9

4—5

5501

3 143

384c,

I16.4

142.5

Art. 53. MAXIMUM stresses in verticals.

161

If the greatest live load stress in V^ were due to the same
position of the load as for Z>„ it would only remain to draw
(on the diagram in the third panel of the truss in Fig. d of
Plate V) the line marked W^ parallel to that member of the
truss in order to complete the force polygon for the upper
panel point 2. The magnitude and character of the simul-
taneous stress in V^ is marked on the diagram. If a force
polygon like that one be drawn for the values oi M -^ p in the
first line of the above table, the stress in V^ is found to be
— 61.2 thousand pounds. The construction is omitted on
the plate to avoid confusion, as it would partly cover the dia-
gram already drawn. In the same way the greatest live load
compression in F, is obtained. Its value is — 41,5. As the
stress in V^ (— 26.2) is less than that in F,, and since V^ and
V^ are symmetrically located in the truss, the compression to
be used for V^ is the same as for F, and will occur when the
live load comes on the bridge from the left.

A reference to the lower force polygon in panel 5 of the
same figure will now explain why it is not desirable to place
all the force polygons for the web members on the right of
the diagonals. It is seen that the side of the polygon giving
the stress in V^ lies on the vertical V^, which is not a con-
venient arrangement. In the next panel the polygon is
placed on the right of D^ so as to preserve uniformity in lay-
ing off the values oi M -r- p downward, and as no construction
is needed for the stress in the vertical adjacent to that panel.
The compression in the verticals is usually not required on
the right of the middle of the truss.

Verticals.

Dead load

+ 29.7
+ 80.4
- 8.1

- 4.1

- 61.2

- 8.1

+ 14

- 41.5

- 8.1

Live load ••

Correction for division of dead panel loads.

Maxi murn

+ 102.0

- 73.4

-35.6

l62

TRUSSES WITH BROKEN CHORDS. CHAP. V.

The maximum stresses are given in the preceding table, the
correction being applied on account of having taken the dead
panel loads, as explained in Art. 51. The stresses are
expressed in units of one thousand pounds.

Prob. 65. Find the maximum stresses in the verticals in
Prob. 63 in Art. 51.

Art. 54. Example — Minimum Stresses in Verticals.

In Arts. 30 and 35, as well as in other places, attention
has been called to the fact that the stresses in the verticals of
a truss with counter-braced panels depend upon the diagonals
which are acting simultaneously in the adjacent panels. The
influence of the diagonals not only affects the magnitude and
character of the stress for any given position of the live load,
but also the rate of change in the stress as the live load passes
across the bridge. In order to determine what position of the
live load will produce the minimum stresses in the verticals of
the truss employed in the three preceding articles, let the

s#omplete cycle of changes in the stress in V^ (Fig. 105) be
traced as the locomotives and train pass across the bridge from
right to left.

When wheel i is at the right support the stress in V^ is
dimply that due to the dead load. As the live load advances
the combined dead and live load stress in V^ gradually dimin-
ishes at an increasing rate, until a position is reached when
the stresses in both diagonals -O, and 2?/ are zero. Mean-

Art. 54. MINIMUM STRESSES IN VERTICALS. 163

while the stress in the vertical has passed through zero from
compression into tension. The tension increases at a reduced
uniform rate until the stresses in D^ and Z>/ are both zero. If
this is not possible, then the tension increases until Z?, becomes
a minimum. As the load advances the tension in V^ at first
diminishes and afterwards increases until the stresses in D^ and
Z>/ again become zero (if possible). During this interval the
stress in V^ has passed through zero twice, so that it is again
tension, but larger in magnitude than before. The rate of
change at the beginning and end of the period are also more
rapid than in either the preceding or the succeeding one.
The tension now increases at a reduced uniform rate until the
stresses in both D^ and D^ are again zero. As the load
advances until it covers the entire bridge, the stress in V^
diminishes and passes through zero the fourth time and into
compression. The rate of decrease was itself a decreasing
one being greater at the beginning of this period, and nearly
if not quite zero at the end. It will now be very slightly
reduced until the head of the train arrives at the left support,
when it will remain constant until the rear of the train begins
to pass over the bridge. As the train continues to pass off
the bridge the compression in F, increases at a variable rate
until it reaches the maximum value, and then gradually di-
minishes again to the value of the dead, load stress. The abso-
lute maximum compression in F, was not reached in this
passage of the live load, but will occur when it crosses the
bridge from left to right.

During this cycle there were several periods during which
the diagonals whose upper extremities are at panel point 5
were not acting, and when the stress in V^ was therefore to
be obtained by drawing the force polygon for that upper panel
point. It is evident then that the tension in V^ is the greatest
when the compression in £/, and U^ is the largest possible
without calling D^ into action. As the maximum stresses in
£/, and U^ occur when the entire bridge is covered with the

164 TRUSSES WITH BROKEN CHORDS. CHAP. V.

live load, the required position may be obtained from this one
by moving the train backwards until the main diagonal in the
panel which is adjacent to the vertical, and on the side toward
the middle of the bridge, shall just cease to act. For deck
bridges this statement would, of course, need modification.
In the present example D^ does not act under any position
of the live load, but the statement in the preceding paragraph
was so framed as to apply also to V^ by making the corre-
sponding changes in the subscripts of D and D\

The required position for the greatest tension in V^ , or its
minimum stress, was found by trial to be that when wheel i
is 3 feet to the left of panel point 4. The moment at the
right support is 9860, and those at panel points 4 and 5 are
30 and 1460 thousand pound-feet respectively. The live load
bending moments at these points are therefore 5604, and
5583. Adding those due to dead load (Art. 51), J/^ =
10 415, and M^ = 9S93. When divided by the panel length
of 27 feet, the quotients are 385.7 and 355.3 thousand pounds.
When these are laid off on Fig. r, Plate V, the resulting force
polygon is reduced to two straight lines, indicating that there
is no stress in D^. The corresponding polygon for D^ is
drawn in broken lines. On drawing a parallel to U^y as
shown, and thus completing the force polygon for the upper
panel point 5, the stress in V^ may be measured by scale.
The direction of passing around the polygon is evident, since
U^ and U^ are both known to be in compression. The com-
bined stress is -f- 30.5 thousand pounds. If M^ is divided by
38 feet and M^ by 35 feet, the quotients are both 274. 1
thousand pounds, which being the horizontal component of
U^ and U^ shows also that there is no stress in the diagonals
and checks the graphic construction.

After some experience this position can be found with but
few trials, and will not require much time if all the operations
are performed by graphics. In Fig. 106 let the depths of

Art. 54. minimum stresses in verticals. 165

the truss at panel points 4 and 5 (38 and 35 feet respectively)
be laid off on one side of an angle, and some convenient
number, as 50, on the other side to the same scale. Join a
and b with /. With the same scale which was used in draw-
ing the equilibrium polygon for 10^
the wheel loads, lay off the bend- v'^^^vxV
ing moments M^ and 'M^ due to i^''^^^^^?N\ ^\
dead load. Now assume a posi- ^''\^^'^'' n^ n^

tion of the live load, draw the ^ ^f.-' \g ^^

closing line of the equilibrium ^ fI ioT **

polygon, and with the dividers

transfer the bending moments due to the live load and lay
them off above the others. If the position is correct, the
lines ce and de^ parallel respectively to af and bf^ will intersect •
each other on the line of. If the head of the locomotive is
too far to the left, they will intersect below the line. If oe
be measured by the scale of moments and divided by 50
feet (assumed as above), the quotient will be the horizontal
component of the stresses in U^ and U^.

If portions of two trains cover certain panels at each end
of the bridge, a stress will be caused in V^ which is a little
larger than the value given above, and which can be found as
follows: Let the required positions of two trains approaching
each other be that illustrated in Fig. jo/. The diagonals in

.....;:::-.^rf :-?::::.•.•.....%

»< r- — ^

Fig. 107.

the fifth panel are omitted, since there must be no stress in
the diagonals of that panel for a maximum tension in V^y as
proved in the preceding portion of this article. Let P^ be the
resultant of the loads transferred to the truss at panel points
5 and 6 by the floor system, together with the dead panel

l66 TRUSSES WITH BROKEN CHORDS. CHAP. V.

loads at those points, and g its distance from the right sup-
port. Let c be the intersection of the chord members U^ and
Z,, and d the intersection of U^ and Z,. If a section be passed
through £/, and Z^, the stresses in those members hold in
equilibrium the forces P^ and the reaction B^ and therefore
their resultant. The resultant of the stresses in U^ and Z
passes through c\ and therefore the resultant R of P, and B
must be equal and opposite to it, and applied at the same
point. The value of R is readily found by taking moments
about 7, whence R = P^g ~ r.

If a section be now passed cutting C/",, V^, and Z, the
stresses in these members hold in equilibrium the same forces
P^ and B as the stresses in U^ and Z^, since the dead load at
the upper panel point 5 is zero in this case. Substituting R
for P^ and B^ denoting the stress in V^ by 5, and taking
moments about dy there results — Rr' -|- 5>fe = o, whence
5 = Rr' -T- ife = P^S'^' -T- kr. But k, r, and r' are constant;
therefore ttie position of the live load to give the tension in
Fi must be such as to render P^g 2l maximum. This shows
that the stress in V^ is independent of the distribution of the
train load on the left, and it may therefore consist of the rear
portion of a preceding train.

Referring now to Fig. 108, which shows the truss diagram
in position on the load line and moment diagram, the ordinate
bf'is the moment at the right support of the truss due to the
locomotives of the. right-hand train. If the chord 41 be pro-
duced to ey be will represent the moment of the live panel load
at 4 about ^ as a center. The ordinate ef will therefore
represent P^g, less the moment due to the dead panel loads at
5 and 6. As this last moment is constant,^ ^/ must be made
a maximum. It is also evident that heavy loads should be
placed at $, and usually the head of the locomotive will not
pass beyond the panel. The possible positions are therefore
quite limited, and on applying the test it is found that when

Art. 54. MINIMUM STRESSES IN VERTICALS.

167

wheel 3 is at panel point 5, </"= 6330 — (3 X 230) = 5640.
In this equation 6330 equals bf as read from the diagram,
and 230 equals the ordinate 5/. Similarly, for wheel 4 at 5,

Fijf. 109.

efz=, 7260 — 3(480) = 5820, and for wheel 5 at 5, </"= 8290
— (3 X 830) = 5800. In the required position, therefore,
wheel 4 must be placed at panel point 5.

Assuming that the train on the left is also in its right posi-
tion, the closing line of the equilibrium polygon is hfy while
if that train is off the bridge the closing line is af. The
equilibrium polygon for the truss is shown in Fig. IO0, the
ordinates at the panel points being the same as in Fig. 108,
and all the sides straight lines. By adding the moments due
to dead load below those which are due to live load the
polygon becomes kmnofh. By Art. 7 the intersection of the
sides no and A/ produced (Fig. 109) is on the line of action of
the resultant R of the forces on the right of the section, and

lC8 TRUSSES WITH BROKEN CHORDS. CHAP. V.

therefore this intersection lies in the same vertical as the
intersection of U^ and Z.. The position of R was shown in
Fig. 107. Let the intercept bg at the right support, between
the chords U^ and Z, produced (Fig. 108), be denoted by z'
and that in the same vertical between no and A/ produced
(Fig. 109) by y, the depths of the truss at 4 and 5 by h^ and
A,. There follows, il// : h^ = M^^ \h^'=, y \s^^ whence

m: _/

A, " z"

If the stress in Z7j be denoted by S\ and the angle between
l/^ and a horizontal by «, and the length of l/^ by /,, / being
the panel length,

S' = ^^^ and -^=/..
A, cos a cos a •

Substituting,

y _ py h y

5' =

COS a pz' cos a z'* p*

An inspection of Fig. 109 shows that in order to determine
y it is not necessary to know M* and J//, and therefore not
necessary to consider either the weight or the position of the
train on the left. When that train is not on the bridge the
closing line is af^ and therefore the corresponding bending
moments M^ and M^ will also determine y. Remembering
that the moment ^/(Fig. 108) for wheel 4 at panel point 5
was found to be 7260, that the moment 5/ is 480, and that
the bending moments at 4 and 5 for dead load are 48 11 and
4010 thousand pound-feet respectively,

J/, = 4^ X 7260 + 481 1 = 8960,
and

M^^\y. 7260 — 480 + 4010 = 8716.

Art. 54. MINIMUM STRESSES IN VERTICALS. 169

As M^ and M^ are respectively 3 and 2 panel lengths from the
right support,

y ( 3 X 8716) - (2 X 8960) .u ^ A

-r=^ ' -^r" — = 304.7 thousand pounds.

The stress in U^ can now be found by graphics in the fol-
lowing manner: On Fig. c of Plate V draw the line Jg' parallel
to U^ intersecting the vertical V^ at g^\ join g^ with the upper
panel point 4; lay ofiy -f-/ = 304.7 on F» as indicated, and
draw if parallel to ^4. The line/S represents the stress in U^.
The force polygon for the upper panel point 5 can next be
obtained by dxdiVJingjk parallel to U^. On measuring 5^ by
scale the stress in F, is found to be + 3i»5 thousand pounds.

This result may be checked as follows: Let a stress diagram
be drawn giving the stress in D^ (Fig. 105) when the reaction
at the right support is i.o. It is found to be 1.017. By the
method described in the preceding article, let the stress in
D^ be found for the above values of M^ and M^, Its value is
+ 22,8. To reduce this stress to zero the reaction at b (Fig.
108) must be increased 22.8 -^ 1.017 = 22.4 thousand pounds.
This requires a moment at the left support a of 22.4 X 189
= 4234 thousand pound-feet. If this is to be produced by a
train apprpaching from the left, its wheel I must be about
half a foot on the left of panel point 2, as shown in Fig. 108.
If, however, it be produced by the rear end of a preceding
train, the train must cover a distance of 65 feet from the left
support. The bending moment J// = 8960 -j- 3/7 X 4234
= 1077s, and MJ = 8716 + 2/7 X 4234 = 9926. If these
values are respectively divided by the depths h^ = 38 and
A^ = 35 feet, each quotient gives the same horizontal compo-
nent of 283.6 thousand pounds iot U^ and £/.. See Fig. r,
Plate V.

The greatest stress in i?/ was produced by wheel 3, being
placed at 5. The construction for this position is also given

I70

TRUSSES WITH BROKEN CHORDS.

Chap. V.

on the Plate, the resulting stress in F. being -f- 30.8 thousand
pounds.

The stresses in the diagonals in the center panel become
zero the second time when the live load covers the entire
bridge, and therefore the greatest tension in V^ or in F, occurs
when [/^ has its maximum stress. By laying off M^ -5-/ =
178.2, which is" due to dead load, above d, and constructing
the triangle ^^ the stress in F, is found to be + 44*3 thou-
sand pounds. If it were attempted to apply the method out-
lined above, y -r-/ would be 609.8, which would give a stress
in [/^ greater than that under full load, which is not possible;
and if the position of the train approaching from the left, which
reduces the stress in D^ and Z>/ to zero, were determined, it
would be found to conflict with that of the other train. The
maximum tension occurs under full load only for the vertical
adjacent to a center panel, or for the middle vertical of a
truss with an even number of panels.

The minimum stress in f^, occurs under dead load only.
The accompanying table gives the final minimum stresses
after applying the correction on account of dividing the dead
panel loads.

Verticals.

From diagram •

Correction for division of dead panel loads.

Minimum stress.

^, = ^.

r^ = K,

F, = r.

+ 29.7

- 8.1

+ 315
- 8.1

+ 44.3

- 8.1

+ 21.6

+ 23.4

+ 36.2

If the number of panels in the truss were 9 or more, the
second vertical from the right support would be adjacent to
two panels requiring no counter bracing. In such cases the
minimum stress in the vertical is obtained in exactly the same
way as the maximum, except that the load covers only the
smaller segment of the span.

It is apparent that to secure precise results the methods

Art. 55.

STRESSES DUE TO WIND.

171

outlined in the example of the four preceding articles, and
illustrated on Plate V, must be drawn to a large scale.
Results which shall answer all the purposes of design may,
however, be readily secured with reasonable care on drawings
which are not unwieldy in size.

Prob. 66. Find the minimum stresses in the verticals in
Prob. 63 in Art. 51.

Art. 55. Stresses due to Wind.

The method of finding the stresses in the upper lateral
system of a bridge whose trusses have broken upper chords is
exactly the same as for horizontal chords after the panel loads
due to the pressure of the wind on the trusses are computed.

In order to illustrate the method of obtaining the stresses
in the trusses due to the wind, let the same bridge be em-
ployed whose dimensions were given in Art. 51, the distance
between the centers of trusses being 16 feet. It is usually
specified that for bridges of this span the upper lateral system
shall be designed for a stationary wind force of 150 pounds
per linear foot. For this example the wind load per panel is
therefore 1.50 X 27 = 4050 pounds.

Now let the middle three panels of the upper lateral system
be considered as separated from the end panels and supported

d e

Pig. zio.

at C and F of the windward truss (see Fig. 1 10), and the
corresponding points C and F' of the leeward truss.

The panel loads at D and E tend to overturn this portion

172 TRUSSES VTITH BROKEN CHORDS. ClIAP. V.

of the upper lateral system, the overturning moment for each
half being 4050(38 — 35) = 12 150 pound-feet. This must
equal the moment of the equal and opposite vertical reactions
at C and C or at F and F'y depending upon which half is
considered. The lever arm of the couple formed by these
reactions is 16 feet, the distance between the centers of
trusses. Each reaction equals 12 150-=- 16= 760 pounds,
those on the windward side being downward, and on the
leeward side upward. The same vertical reactions would be
caused by vertical loads of 760 pounds acting downward at
D^ and E' and upward at D and E. In addition to the
vertical reactions referred to above there are also horizontal
reactions of 4050 pounds at C and F. Hence the same
stresses will be produced in the vertical trusses if the horizon-
tal panel loads of 4050 pounds at D and E are replaced by
vertical loads of 760 pounds, acting upward at D and £, and
downward at D' and E\ together with horizontal wind panel
loads of 4050 pounds applied at C and F respectively.

Next consider the entire upper lateral system, with the
changes made as just indicated and supported at the points
B, Gy G'y and B\ Horizontal wind loads of 4050 -|- 4050 =
8100 pounds are acting at C and /% and these produce an
overturning moment in each half of the system of 8100(35 —
29) = 48 600 pound-feet. These loads may therefore be
replaced by vertical loads of 48 600 -j- 16 = 3040 pounds, act-
ing upward at B and G and downward at B' and G'y together
with horizontal wind loads of 8100 pounds at B and G
'respectively.

The end posts and portal bracing connect the upper lateral
system to the supports of the bridge, and may be regarded as
apart of it. The horizontal wind loads of 8100 + 4050 =
12 150 pounds acting at B and G cause an overturning
moment for each half of the bridge of 12 150 X 29 = 352 350
pound-feet, and may therefore be replaced by vertical loads-

Art. 55. . STRESSES due to wind. 173

of 352 350 H- 16 = 22 020 pounds, acting, upward at B and G
and downward at B' and G\ together with horizontal wind
forces of 12 150 pounds applied at the support of each end of
the bridge. It is generally assumed that the feet of the end
posts react equally.

The above analysis shows that the upper lateral system
transfers the horizontal wind loads directly to the supports,
and that the stresses in the leeward truss due to the overturn-
ing moment of the wind pressure which is applied at the panel
points of the same system may be obtained by applying
vertical loads of 22 020 pounds at B' and G\ 3040 pounds at
C and F', and 760 pounds at D' and E' respectively. Since
the resulting stresses always act in conjunction with and are
less than the dead load stresses, their values in the windward
truss are the same as those in the leeward truss with the signs
changed. Practically, the wind transfers a part of the dead
load from the windward to the leeward truss.

If the upper chord were horizontal the stresses in the truss
would be obtained by applying vertical loads only at B' and
G' (or at B and G) equal to 3 X 4050 X 29 ~ 16 = 22 020
pounds. For the sake of simplicity the analysis in this
article was made with the implied assumption that the speci-
fied wind pressure for the upper lateral system was concen-
trated at the windward panel points, and that the diagonals
take only tension. The wind pressure should, however, be
equally divided between the windward and leeward panel
points, but the computed vertical panel loads will not be
modified by it nor by the fact that the upper lateral system
may be designed with stiff diagonals.

If the lower chord be some distance above the level of the
bridge supports the wind loads of the lower lateral system
will likewise produce an overturning moment. The wind
pressure on the train also tends to overturn the bridge about
an axis through its leeward supports. In obtaining the trans-

174 TRUSSES WITH BROKEN CHORDS. CHAP. V.

fer of load from the windward to the leeward supports of the
bridge the lever arm of the wind pressure equals the difference
in elevation between the center of pressure on the train and
the axis of rotation. The elevation of the center of pressure
is variously specified to be from 7i to 8i feet above the base
of the rail. The vertical panel loads which may replace the
wind pressure on the train in finding the stresses in the
trusses must be applied at the panel points of the loaded
chord and regarded as a moving load. Each of these panel
loads equals the product of the wind pressure on one panel
length of the train by the distance of the center of pressure
above the plane of the lateral system divided by the distance
between the trusses.

The stresses in the trusses, the floor, and lateral systems ol
a bridge due to the curvature of the track are treated in a
paper by WARD BALDWIN in Transactions of the American
Society of Civil Engineers, Vol. XXV, page 459, Nov. 1 891,
entitled ** Stresses in Railway Bridges on Curves." The
paper contains a practical example in which the stresses are
computed.

Prob. 67. Find the stresses in the truss in Fig. 1 10 due to
the wind panel loads as computed in this article. The left end
of the truss is toward the north.

F\g^

. \

v.^

Art. s6. the fegram truss. 175

CHAPTER VI.
MISCELLANEOUS TRUSSES.

Art. 56. The Pegram Truss.

The Pegram truss has a curved or broken upper chord, and
posts whose angles with the vertical increase from the middle
of the truss to its ends. The horizontal projection of the
upper chord is about one and one-half panel lengths shorter
than the lower chord, but both chords are divided into the
same number of panels. The panel points of the upper
chord lie upon the arc of a circle. The form, proportion,
and relative economy of this type of truss is discussed by the
inventor in Engineering News, Vol. XVIII, pages 414 and
432, December 10 and 17, 1887.

Fig. 1 1 1 shows the skeleton diagram of a Pegram truss of
seven panels. On account of the inclined posts the notation
is necessarily modified from that used in Chap. V. The

^ rUsorUig

Fig. XXI.

panel points are numbered in regular order from left to right.
The subscript for any chord member is the number of the
panel point which is the center of moments. The subscript
for any post or tie is the panel point at the right extremity of

176 MISCELLANEOUS TRUSSES. CHAP. VI.

the member. The middle panel of the lower chord will be
designated by Z, or Z,, according as its center of moments for
a given position of the load is at 7 or 8. In order to distin-
guish the middle panel of the upper chord from the adjacent
members which have their centers of moment at the points 6
or 9, the designating letters are primed.

When the center of moments for any chord member of a
truss is not in the same vertical as a floor beam, the method
of determining the position of the locomotive wheel loads and
the corresponding maximum moment described in Art. 47
does not apply. In Part I, Art. 61, the required criterion
for position,

was deduced, in which Q is the load in the panel cut by the
vertical through the center of moments, P^ the load on the
left of this panel, W the whole load on the truss, q the hori-
zontal distance from the center of moments to the left -end of
the panel containing Q, p the panel length, /' the distance of
the center of moments from the left support, and / the span
of the truss.

In order to satisfy this criterion a wheel has in most cases
to be placed at the left end of the panel containing the aggre-
gate load Qy although it will often be satisfied when a wheel
is placed at the right end of this panel.

Fig. 112 shows the left end of a Pegram truss in position
on the load line and moment diagram of the wheel loads for
the maximum moment at the panel point 3. The line au
produced passes through the point where the vertical at the
right support cuts the load line. Wheel 4 is at the floor
beam at panel point 2. The ordinate 4r represents the load
P' '\' Q. If wheel 4 be just to the rigKt of the floor beam
the ordinate 21 or the equal ordinate mj represents P', while

Art. 56.

THE PEGRAM TRUSS.

177

md represents P + ^Q. If, however, wheel 4 be just on the

left of the floor beam, the ordinate 2h equals F and mf equals

P' + 7 G* The ordinate me equals y W and the position is

therefore correct when wheel 4 is at the floor beam and the
point e falls between the
points d and / where the
lines ri and rh respectively

cut the vertical through the ? [ / M

■■P\ V

Fig. xxa.

center of moments 3. The point r is at the intersection of
the load line with the vertical through panel point 4.

With the load in this position the equilibrium polygon for
the truss is composed of the straight sides ak, koy os, st, etc.,
and the closing line av. The bending moment is therefore
given by the ordinate c^. In a similar manner the positions
of the live load and the corresponding moments are obtained
for all the panel points of the upper chord of the given truss.

The position of the live load and the bending moments for
the sections through the panel points of the lower chord are
determined in the manner described in Art. 47. For those
points the second term of the left-hand member of the above
criterion for position becomes zero.

As an example a truss of seven panels will be used whose
span is 207 ft. The horizontal projections of the posts P,,

178

MISCELLANEOUS TRUSSES,

Chap. VI.

/i, P„ and P, are 22.500, 15.595, 9.336, and 3.467 feet
respectively, while their vertical components measure 25.000,
32.349, 37.273, and 39.717 feet. The live load consists of
Waddell'S Class U. The dead load is estimated at 1050
pounds per linear foot per truss, 300 pounds being taken on
the upper chord. As the average horizontal projection of the
upper chords is 23.14 feet, the upper panel loads are 23.14 X
300 = 6942, or say 6950 pounds. The panel loads on the
lower chord are 750 X 207 -7- 7 = 22 180, or say 22 200
pounds.

The following table gives the data for all the live load
chord stresses, the moments being expressed in units of a
thousand pound-feet, the stresses in thousands of pounds, and
the lever arms in feet :

Center

Moments.

Wheel

Section.

Moment
AT Right
Support.

Moment

Section.

Bending
Moment.

Lever
Arm.

Stress.

Chord
Mhmber.

2
4
6

9
12

13
t4
15
16

47220
46570
43070

4«^090
35 420
38 500
40470

480
3 245
6331

7 261

8 291
10099
II 286

6266

10 061
12 128

12 063

11 949
II 901
II 840

25.96
34.53
39-14
39.717

39.717

- 241.4

- 291.3

- 309.9

- 305.4

I
3
5
7

Wheel 4 at point 2
Wheel 4 at point 2
Wheel 13 at point 6
Wheel 12 at point 6
Wheel 13 at point 6
Wheel 16 at point 9

4 790
8 270

10 710
12 090
12 000

11 SSo

25.00
32.349

37.273

39.717

+ 191.6
+ 255.6
+ 287.3

-f- 299 I

The upper part of this table is arranged like that in Art. 47.
For wheel 14 at the section through panel point 9, the shear
is positive in the middle panel, while for the two preceding
and the two succeeding positions the shear is negative. The
moments at 9 for the last two positions cannot be used, since
the diagonal in the middle panel which extends to 9 is not

Art. 56. The pegram truss. 179

then brought into action. The stresses in the fifth and sixth
lines of the table are not computed, since the corresponding
bending moments are less than 12 128 thousand pound-feet.

The bending moments in the lower part of the table were
read directly from the diagram after drawing for each position
the closing line of the equilibrium polygon, and the side of
the polygon which lay below the center of moments. For
the last three positions the shear in the middle panel is nega-
tive, and hence only the last moment can be used to obtain
the stress in the lower chord of that panel.

Since the ties of this truss do not have equal horizontal
projections as in the case treated in Chap. V, it is found to
be more convenient to obtain the chord stresses directly by
dividing the bending moments by the corresponding lever
arms. As the lengths of all the members of the truss must
be computed, the lever arms may be obtained by very little
additional computation, or they may be measured on a large
scale truss diagram with sufficient precision.

The stresses due to the dead load in the upper chord are
— 105.8, —132.7, —140.8, and — 139.4 thousand pounds, and
in the lower chord, +85-3> +116.4, +132. i, and 139.4
thousand pounds.

In determining the positions of the live load for the web
stresses, the points of division in each panel are found by the
method explained in Art. 48, and the results are marked on
the truss diagram on Plate VI. It will be observed that the
points for the posts are on the right of those for the ties
except in the case of Ti, and P,„ where the counter tie is not
required. The positions of the locomotive wheel loads and
the remaining data required to construct the force polygons
for the web stresses are given in the following table. The
.same units are employed as in the preceding table.

Let the student make a careful comparison between this
tible and those in Arts. 52 and 53. Attention is also called

i8d

MISCELLANEOUS TRUSSES.

Chap. VI.

to the fact that only for one member was more than one posi-
tion of the load found to satisfy the criterion. The bending

Bending Moments

Pakbl

{M) AT EXTKBMITIBS

Web
Mbmbbrs.

Points

AT THE

Bxtrbm-

Whbbl

Panbl

OF Web Member.

Length of
Horizontal
Projection

Right ^.

ITIBS.

Point.

At Left
End.

At Right
End.

K^U

O— I

4 at 2

4790

22.500

212.9

7-.

1—2

4 at 2

4790

6 270

7.071

677.3

886.8

2—3

3 at 4

4 800

7 200

15.595

307.8

461.7

7-4

3—4

3 at 4

7 200

9370

13.976

515. 1

670.3

4-5

2 at 6

6 150

7090

9.336

658.7

759.3

3 at 6

6600

7580

9.336

706.8

811. 9

5-6

3 at 6

7580

9675

20.235

374-6

478.1

6-7

2 at 9

5560

5750

3.467

1608

1659

7—9

3 at 9

6340

7960

26.104

242.9

304.9

r,!

8— II

3 at II

4040

4980

30.038

122.3

150.8

7-.,

II — 12

3 at 13

2 010

2090

13.976

143.8

149.5

Pit

12 — 13

3 at 13

2090

2 180

15.595

134.0

139.8

moments in the fourth and fifth columns belong to the sec-
tions through the panel points given in the second column,
and were read directly from the diagram like those in the
lower part of the preceding table for chord stresses.

The ordinates whose values are given in the last two
columns are laid off on the truss diagram on Plate VI, and
the force polygons drawn as explained in the preceding
chapter. On account of the great range of these values, and
in order that the force polygons may be drawn to as large a
scale as possible, different scales are employed. The original
truss diagram was drawn to a scale of 6 feet to an inch, and
the following scales were used for the force polygons in
regular order from left to right: 60, 200, 100, 200, 200, 80,
30c, 50, 30, 30, and 40 thousand pounds to an inch. The
stress in P^ for the position of wheel 2 at panel point 6 is
— 66.2 thousand pounds, the corresponding force polygon
not being shown- on Plate VI.

Art. 56. THE PEGRAM TRUSS. 181

The greatest live load tension in P^ (or in P^ occurs either
when one or both of the adjacent upper chord members are
subject to their maximum stresses. In this example their
maximum stresses occur simultaneously when wheel 12 is at
panel point 6, and while the tie T^ is acting. The force tri-
angle at panel point 7 therefore gives the corresponding live
load stress in P^ of -|- 32.0 thousand pounds.

The greatest tension in P„ occurs when wheel 5 is at panel
point II. The moment ordinate at the right support is 9475
and at II is 830, whence P^g ^ 9475 — (830 X 3) = 6985
(see Art. 54). The bending moments in the sections through
8 and 11 due to the dead load are 5526 and 4575 thousand
pound-feet respectively. The bending moments due to the
dead load plus the above live load on the right end of the
truss are

^ . 114.81Q
Jf,= 5526+^^.9475= 10782;

Jf„ = 4575 +^.9475 - 830= 10 5 14.

The value of/' is found graphically to be 10030 thousand
pound-feet. The horizontal projection p' of U\^ is 23.702
feet, and hencey -=-/' = 423.2 thousand pounds. On laying
this off on the vertical from panel point 10 to /, and drawing
^'parallel to ^8 as explained in Art. 54,7—10 gives a stress
of — 333.7 thousand pounds in U\^ at the' time when P„ is
subject to its greatest tension. If the force polygon for panel
point 10 be next constructed the stress in P„ will be found.
At this point the stresses in U\^, U,,, and P,„ together with
the dead panel load of 6.95 thousand pounds, are in equi-
librium. The polygon is drawn on the truss diagram, and
gives a stress of -|- 31.4 thousand pounds in P^,.

The maximum and minimum stresses in the web members
du« to the dead and live loads may now be obtained in the
usual manner:

I82 MISCELLANEOUS TRUSSES.

For the ties.

Chap. VL

7-.

7-ii

Dead load '•

-f 56.9
+ 141. 2

4- 33.7
4- 10&.0

- 19.3

4- 17.0
+ 89.4

+ 74.5

— 21.0
4-56.4

Live load from the right

Live load from the left

Maximum stress

4- 198. 1
4- 56.9

4- 141.7
4- 14.4

4- 106.4

4- 74-5

4- 35.4

Minimum stress. ... .•••

For the posts,

Dead load

Live load from the right. .
Live load frorri the left. . .
Full live load

Maximum stress

Miiiiiiium stress

127.2

286.7

— 36.2

— 108.6
4- 35.5

- 9.1

- 67.4

4- 7.4

- 39-0
4-32.0

- 4139

— 127.2

144.8
0.7

-76.5
[4- 31 4j

— 31.6
4- 39-4

It will be observed that if the live load were just a little
greater that all the posts except the ones at the end oi the
truss would be subject to reversal of stress,

Prob. 68. A Pegram truss for a through railroad bridge has
five panels, and a span of 150 feet The horizontal projec-
tioHL of the posts are 20.000, 11.995, and 4.335 feet, while
their vertical components are 22.000, 26.505, and 28.740 feet.
The live load is Waddell's Class U. The dead load is 950
pounds per linear foot per truss, 275 pounds being taken on
the upper chord. Find the maximum and minimum stresses
in all the members.

Art. 57. The Pennsylvania Truss.

This type of truss is illustrated in the skeleton diagram of
Fig. 113. ,It is derived from the Pratt truss with a curved
upper chord by subdividing its panels by means of subverticals

u;.

u..

I83

De
U%
U\

Mi
Mi

Li
Li

M.
M

gre
tru
1
five
tioi
the
Th
poi
the
in ;

Art. 57.

THE PENNSYLVANIA TRUSS.

183

and short diagonals. The vertical broken lines indicate struts
which support the upper chord members at their middle
points, and the corresponding horizontal lines serve to give
lateral support in the plane of the truss to the long vertical
posts. These are no real truss members, and are omitted in
the diagrams employed in finding the stresses in £he truss.
In this case the counter diagonal eG does not coincide with,
the short diagonal FG, although in a number of trusses
whicJ have been erected the panel eEGg is counterbraced by

connecting the points e and /^ with a tie. The detail drawing
of such a truss may be found in Engineering News, Vol.
XXIII, page 249, March 15, 1890.

The stress in ^/due to locomotive wheel loads is found in
the same manner as for the Pratt truss, the center of moments
being at E, -The stress in ^equals that in ef^ as may be seen
from the force polygon for the panel point /.

By the method employed in Part I, Art. 61, a criterion for
the position of the live load may be obtained which will give
the stress in EG^ and which indicates that the wheel loads
from a to e, plus twice those from e to/, shall equal Wl' -j- /,
W^ being the whole load on the truss, /the span, and /' the
distance from the left support to the center of moments g.
To satisfy this criterion a wheel load must usually be placed
at e^ although sometimes it may be satisfied when a wheel is
zX f. In view of the examples given in Chap. V and in the
preceding article, the student should have no difficulty in
making the graphic construction required by this criterion.

l34 MISCELLANEOUS TRUSSES. . CHAP. VI.

As the section cutting EG and only two other members
must pass on the left of/, the bending moment for EG must
equal the moment of the left reaction of the truss, minus the
moments of the loads transmitted by the floor system to the
truss at the panel points ^ to ^ inclusive; or, in other words,
the required bending moment exceeds that at the vertical
section through^ by the moment of the panel load at/. In
the graphic determination, if the line joining the points where
the verticals through e and / meet the moment curve of the
live-load diagram, be produced to the vertical at^, and the
ordinate from this point of intersection to the closing line be
read off, the required moment will be obtained.

For the main tie EF^ the position is found by the method
given in Arts. 48 and 49, the auxiliary lines ^, and B^ of Fig.
98 being drawn in this case through the points e and / of
Fig. 113. .The force polygon is then constructed as in Art.
50 by using the moments at E and g^ the points where the
diagonal EF meets the upper and lower chords respectively.

The maximum stress in Ee occurs also when the head of
the locomotive is in the panel </*, and hence there will be no
simultaneous live-load stress in DE, The section through
Ee must therefore cut CE and ef, and in finding the point of
division in </*the chord member CE must be produced as in
Fig. 98. If in any case the position for Ee should be the
same as for EF^ the stress in the former may be obtained
from the force polygon already constructed for the latter by
completing the polygon for the stresses meeting at the panel
point E. If not, then a new force polygon for the simul-
taneous stress in EF must be drawn.

The panel load at / is suspended from F by the subvertical
Ffy while the members EG^ EF, and FG form a secondary
truss which serves to transfer this panel load to the panel
points -£ and G. As the panels are equal one half of the load
at /is transferred to E and G respectively. Since the posts

Art. 57. THE PETIT TRUSS. 183

Ee and Gg are both vertical, the stress in Fg is exactly the
same as if the wheel loads in the panels ef and fg were trans-
ferred to the panel points e and ^ by a stringer whose span
is eg. The stress in Fg is therefore found by the method
given in Chap. V, after considering the members Ff and FG
removed. The greatest stress in the counter tie eG (or Gi) is
obtained in a similar manner to that in Fg,

The preceding statement also shows that the maximum
stresses in i^and FG occur when the floor beam reaction is a
maximum. For consolidation locomotives this usually re-
quires the second or third driver to be placed at the floor
beam. If the panels are long, the reaction will be greater
under the corresponding position of the second locomotive,
for one or two of the tender wheels of the first locomotive
may then be brought on the panel at the left. The tension
in FG equals one half of the floor beam reaction multiplied by
the secant of the angle which FG makes with the vertical.

The stress in the vertical Cc depends not only upon, the
floor beam reaction at r, but also upon that at ^, one-half of
the latter being transferred to c by the secondary truss abcB.
By employing the same method as in Art. 46, the following
formula may be deduced for the stress in Cc due to the loco-
motive loads:

P '

in which M^ is the moment of all the loads on the first three
panels about d sls a, center, and M^ the moment of the loads
on the first two panels about r as a center, and / the panel
length of the three equal panels ad, 6c, and cd» The values
of Mc and M^ can be read off directly from the live-load
moment diagram. The corresponding position of the load
requires the wheel loads in the first two panels to be equal to
two-thirds of the load on the three panels. It will be
observed that this is the same position as that for the maxi-

i86 MISCELLANEOUS TRUSSES. ChAP. VI.

mum moment at ^ of a beam or truss whose span is ad. As
the Kve-load diagrams are always constructed with the head
of the locomotive toward the left, the maximum stress should
also be found in Kk and compared with that in Cc, the larger
value being used for botn members. The tension in Bb
equals the floor-beam reaction at b.

If instead of the short tie FG a short strut eF be inserted,
the auxiliary truss will then be efgF^ which will transfer the
panel load at /.to the points e and g. The moment of the
stress in fg about the center E will then be the bending
moment in the vertical section through E plus the moment of
the panel load at /. The corresponding criterion for position
will require the wheel loads from a to f minus the wheels
from/ to ^ to equal Wl' — /, in which VV, /, and /' have the
same significance as before, whence /' equals the horizontal
distance from the left support to Ey which is the center of
moments. The stress in EG is the same as if the secondary
truss were omitted, and the stringer extended from e to g.
The methods described above for finding the stresses in EF
and Fg will now have to be exchanged.

The construction of the stress diagram for the dead load
offers no difficulty in either case, and will therefore not be
illustrated.

The form of this truss as shown in Fig. 113 is sometimes
modified by reducing the two panels at each end to one,
thereby omitting the subverticals at b and /. This arrange-
ment was used in the Susquehanna River bridge at Havre
de Grace, Md. Another modification was adopted in the
Merchants' bridge, at St. Louis, whereby the point B was
raised so as to bring it into the curve of the upper chord.
See Engineering News, Vol. XXII, page 578, Dec. 21, 1889.

Prob. .69. The truss in Fig. 113 has a span of 283 feet, the
depths at C, E, and G being 42, 47, and 48^ feet respec-
tively. Find the stresses in all the members of the truss due

Art. 59. UNSYMMETRICAL TRUSSES. 187

to Waddell's Compromise standard, Class U, the bridge
having a single track.

Art. 58. The Baltimore Truss.

The Baltimore truss is a special case of the Petit truss when
the upper chord is horizontal. The chord stresses for both
trusses are found in exactly the same manner. The method
used for the web stresses of the Petit truss also apply to the
Baltimore truss, but for most of the web members it is prefer-
able to make the comparison with the methods employed for
the Pratt truss.

If the upper chord in Fig. 1 13 were horizontal the stresses
in EF and in Ee would be the same as if the truss were of the
Pratt type with 14 panels, the load on the truss being 14
times the load in the panel ef. For the stress in Fg the load
on the truss must equal 7 times the load in the panel, and the
stress would be the same as if the truss were of the Pratt type
with only 7 panels. The stresses in i^and Bb are equal to
those in the suspender of a Pratt truss having the same panel
lengths, while those in Cc and FG are the same as for the
Petit truss.

Prob. 70. A Baltimore truss of a single-track through
bridge has the same span, number of panels, and live load as
the truss in Prob. 69, its depth being 47 feet 2 inches. The
counter-ties Ge and Gi are, however, omitted, and the mem-
bers Eg and Ig counterbraced. Find the live load stresses in
all the members.

Art. 59. UNSYMMETRICAL TRUSSES.

Fig. 114 represents the side elevation of the two unsym-
metrical Pratt trusses of a through railroad bridge, together
with the plans of the upper and lower lateral systems. The
floor beams are perpendicular to the center line of the bridge,
and they are placed at equal distances apart from each other,

i88

MISCELLANEOUS TRUSSES.

Chap. VI.

and from the points midway between the bearings of the end
stringers. The end posts are inclined so that their horizontal
projections are equal to the distance between the floor beams,
and this necessitates shortening the end panel of the upper
chord at one end of each truss, and lengthening that at the

Fig. 114.

other end by an amount equal to one-half the longitudinal
component of the skew. The end panels of both chords are
therefore equal at the same end of each truss, and the sus-
penders are inclined. Sometimes the trusses of skew spans
have been^ built whose end posts are not equally inclined, but
this is very unusual.

As the two trusses are equal, but with their ends reversed,
it is necessary to find the stresses in all the members of one
truss for all the loads. Since the load line and equilibrium
polygon for the locomotive wheel loads are usually arranged
for the load advancing from the right, the live load stresses

Art. 59. UNSYM METRICAL TRUSSES. 1 89

are found in all the members of the left half of each truss,
except the counters whose stresses are found in the right half.

The dead load stresses in the trusses and the stresses in the
lateral systems due to that portion of the wind load which is
regarded as stationary are as readily determined by the
graphic method as for symmetrical trusses, although in the
analytic method the numerical work of computation is ma-
terially increased.

The stresses in the lateral systems due to the pressure of
the wind on the trusses are found by means of a stress diagram
like Fig. 67 in Art. 32, but extended to all the members of
each system. For the wind panel loads which are treated as
a moving load (representing the pressure of the wind on the
train), the stresses in the chords may also be obtained by a
stress diagram, while that in the lateral struts and ties may
be found by means of tables similar to those in Art. 32. The
relation between the values in the different lines of the table
are, however, not as simple as for trusses whose panels are all
equal. For instance, if the nearer truss at is on the windward
side, the braces of the lower system brought into action are
ab'bccd'de'e . . . hh^iy and a wind panel load at g will cause
stresses in members on its left equal to those due to a panel
load at h multiplied by the ratio of ^/ -7- hi. These products
are most conveniently obtained by graphical arithmetic (Art.
14).

Another method, which is preferable in most cases, is to
draw a diagram giving the stresses in all the members for a
reaction of i.o at the left support a, and then to multiply the
stress in each brace by the corresponding reaction produced
by the panel loads which make its stress a maximum. These
reactions are most quickly determined by means of an equi-
librium polygon whose closing line will shift as one panel load
after another is taken away from the full load for which it is
at first constructed.

190

MISCELLANEOUS TRUSSES.

Chap. VI.

The position of the wheel loads is obtained in the same way
as for symmetrical-trusses, it being assumed, however, that
the loads are distributed along the center line 00' of the track
as shown in Fig. 114, and that the trusses have their supports
at and o\ Although this method is approximate, it generally
gives the correct position. In case a very slight shifting of
the loads would dissatisfy the criterion, it may be well also
to find the stress when the next wheel is placed at the corre-
sponding panel point and compare the results.

In the bridge represented in Fig. 114 let the span be 146
feet, /J^= cd=de= ef=fg = gh = 18' 3'',^^ = BC = 13'
7i'', ki=^ GH = 22' loj", the width 16 feet between centres
of chords, and the depth 24 feet. For the greatest stress in
the chord members ab and be due to Waddell*s Class U
(Art. 39) the required position is that with wheel 3 at panel
point b. Fig. 115 shows the left portion of the truss diagram
placed in this position on the
load line and moment diagram
of the live load. The inter-
section of the line ov with the

Fi«. X15.

vertical through the center of moments B is seen to lie
between the points where mw and nw cross the same vertical.

Art. 59. UNSYMMETRICAL TRUSSES. I9I

As the floor system distributes the load to the panel points
of the truss the lower sides of the equilibrium polygon for the
truss under this position of the load are the right lines, or
chords, op, pr, rs, etc., the points 0, p, r, and s being on the
live load polygon. The point is the regular panel length
of 18' i" on the left of b. The closing line of the polygon
must end in a vertical through the left support of the truss at
a, and therefore at the intersection k of the vertical ka with
the line op. In a similar manner the right end of the closing
line (not shown in Fig. 115) is located at the intersection of
the chord or side of the equilibrium polygon whose extremi-
ties lie on the verticals through h and 0' (see Fig. 1 14) with
the vertical through i. The bending moment in the section
through B is measured by the full line ordinate with arrows
at its extremities (Fig. 115). This moment divided by the
depth of the truss gives the stress in be. As the stresses in
ab and aB, however, hold in equilibrium only the reaction at
a, the moment of the stress in ab about B is equal to the
moment of the reaction, and hence i3 measured by the ordi-
nate when produced to y, its intersection with the side op (or
kp^ produced.

The stress in the end post aB is preferably obtained by
dividing the stress in ab by the sine of the angle which aB
makes with the vertical. For the remaining web members
except the suspenders, the only modification required of the
method described in Art. 45 for trusses with equal panels is
that the moment at the right support must be read to the
point corresponding to that described in the preceding para-
graph as the right end of the closing line of the equilibrium
polygon for the truss.

The floor-beam reaction is the same as if the panel ab of
the truss were equal to be, for the deduction of the formula in
Art. 46 indicates that the panel lengths introduced are really
the spans of the corresponding stringers, and while usually

ig2 MISCELLANEOUS TRUSSES. CHAP. VI.

they are equal to the panel lengths of the truss, this is not
always the case. The tension in £b equals the product of the
floor-beam reaction by the secant of the angle which £b makes
with the vertical.

Fig. 1 1 5 also shows the left-hand portion of the diagram of
truss a'B' H*i' superimposed upon the other. The closing
line is k'uy k^ being at the intersection oi po produced with
the vertical a'k\ The moment of the chord stress in a'b'
ip' coincides with b) is measured by the ordinate below B' and
is indicated by arrows at its extremities. By producing this
ordinate to /, on the chord rp produced, it gives the moment
of the stress in b'cf .

It will be observed that the same position of the live load
was used for the end chord members of both trusses. If the
point had been moved to a^ the criterion would not have
been satisfied by placing wheel 3 at ^, but only by putting
wheel 4 at b. The moment for the latter position, however,
is less than that for the former.

Prob. 71. Find the maximum and minimum stresses in the
above example due to the given live load and a dead load of
900 pounds per linear foot, one-fourth to be taken on the
upper chord.

Art. 60. Double and Quadruple Systems.

For trusses having more than one system of webbing it is
assumed that each system is affected only by the loads which
it carries.

In the double system Warren truss in Fig. 1 16, the loads

A^ B c D E r

P^, P„ and P^ are carried by the full line diagonals, and the
loads, Q,y !2a» !2»> and Q^ by the diagonals drawn in broken

Art. 6o. double and quadruple systems.

193

lines. The truss is therefore regarded as composed of two
separate trusses having common chords. The stresses in each
system may then be determined and the results combined.
In this case, however, either the dead load stresses in all the
members or the live load stresses in the chords may be found
by means of one diagram. The reaction at the left support
is in equilibrium with the stresses in Aa, Ba^ and ab^ but the
compression in Aa is known, since it equals the reaction due
only to the loads j2» thus leaving but two unknown stresses.
The stress diagrams may therefore be readily constructed.
The maximum live load stresses in the diagonals are obtained
by considering each system separately.

For the Whipple truss in Fig. 1 17 (which is a double inter-
section truss of the Pratt type) both dead and live load

B C D E F H L__TfC

/ ^

Fig 117.

stresses must be found for each system, the division into
systems being somewhat different, however, for the required
stresses in the chords and web members. For the chord
stresses the division may be made into the two symmetrical
systems shown in Fig. 118, provided the live load is uniform

B D F H 7/C

ace ft J ""l ^ ^ ^ ^

Fig. X18. Fig. xxg.

throughout. If the live load is not uniform, or if it consists
of excess loads combined with a uniform train load, the division

194 MISCELLANEOUS TRUSSES. CHAP. VL

may be made similar to that shown in Fig. 119, care being
taken to insert those diagonals near the middle, which are in
tension under the combined dead and live loads. If the
stresses are obtained only for the left half of the truss, the
excess loads may require an additional diagonal to slope
downward toward the left in the lower diagram of Fig. 119.
It is clear that only those dead and live load stresses in any
, chord member may be added together, which were obtained
under the same conditions; that is, with the same diagonal of
the given panel acting in each case.

When the division is made into systems, or component
trusses, which are unsymmetrical, there is some ambiguity in
the stresses due to the fact that the suspenders are attached
to the panel points B and K which are common to both sys-
tems. As reasonable an assumption as any is to regard the
panel loads at b and k to be equally divided between the two
systems. The same stresses will, however, be obtained if
both suspenders be considered as a part of only one and the
same system, and this arrangement is also more convenient
in finding the stresses.

If these two methods of division be compared for a uniform
load throughout, the greatest difference in chord stresses is
found to be not quite four per cent, most of them being much
less. The difference may be reduced one-half by considering
the suspender Bb as belonging only to that component truss
which contains the adjacent diagonal Bc^ and the suspender
Kk as being a part of the system containing the diagonal Kj.
This arrangement reduces the shear at the middle to the
minimum value possible in each case, and requires the con-
struction of only one stress diagram for the chord members,
since one system equals the other with its ends reversed.

For the stresses in the web members the systems are divided
as in Fig. 120, all the ties sloping one way except those near
the right end, where it is certain that no counters are needed.

Art. 6o. double and quadruple systems.

19s

Only the loads supported by one of the systems are considered
in finding the stresses due to both dead and live loads in any
web member of that system. The method employed is that

Pig. xao.

s>i Art. 30, the labor of tabulation being materially lessened
by noticing the general statements made in that article in
regard to the maximum and minimum stresses in web mem-
bers. On account of the ambiguity in stresses the panel loads
at b and k may be placed on either system, so as to produce
the maximum and minimum stresses in any given web mem-
ber.

When one excess load is used in connection with uniform
panel live loads, it must be placed on that system which gives
the greatest chord stresses; and for two excess loads the first
may be on one system and the second on the other, depending
upon their distance apart and the panel length. For maxi-
mum stresses in the webbing the excess loads are always
placed at the head of the train.

If concentrated wheel loads are to be employed it will be
best to always place the first driver at the panel point. Each
system is regarded as acting independently, and as being
strained only by the loads transferred to it by the stringers
and floor beams. As part of the weight of the pilot is carried
by the stringer to the other system, that part is disregarded
in obtaining the stresses in the bracing. For the chord
stresses the locomotives are so placed as to produce the
greatest moment at the middle of the truss, and the weights

196

MISCELLANEOUS TRUSSES.

Chap. VL

transferred to each system are used only in determining the
chord stresses for that system.

A quadruple Warren truss or lattice girder is treated in

[xyxyxxi

Pig. laz.

a similar manner to the double system Warren, and also
requires but one diagram for dead load.

Prob. 72. A double system deck Warren truss of 100 feet
span has 10 panels, and is 10 feet deep. The dead load per
linear foot per truss is 560 pounds, and the train load 1800
pounds, which is preceded by two heavy locomotive panel
loads of 65 000 pounds each. Find the maximum and mini-
mum stresses in all the members.

Art. 61. The Greiner Truss.

The truss shown in Fig. 122 was designed in 1894 by J. E.
Greiner, Engineer of Bridges of the Baltimore and Ohio
Railroad, for overhead highway purposes so as to be adapted

B CD E F

Fig. 199.

to the use of material taken out of old bridges combined with
rails worn out in track service. It consists essentially of a
bowstring truss with a horizontal upper chord combined with
a Pratt truss whose diagonal ties are omitted, so that the
supports of the former are at the upper ends of the end posts
of the latter.

The stresses in the web members above the bowstring

Art. 6i. the greiner truss. 197

Be' de^ F 2s^ first cletermined by regarding them as members of
the bowstring truss only, the supports being at B and F.
The stresses in the end posts are the same as if the truss were
of the usual Pratt type not in combination with the bowstring,
while the stresses in the verticals Bb^ c'cy e'e^ and Ff 2X^ the
corresponding floor-beam reactions.

The stress in any member of the upper chord BF is the sum
of its stresses when considered successively as a member of
the bowstring and Pratt systems, the load in the second case
being the panel loads at b and /, together with loads at B and
F equal to the reactions in the first case. This stress is a
maximum when the load covers the entire truss and its posi-
tion satisfies the same criterion, which would be used if the
truss were of the pure Pratt type. The corresponding
stresses in the bowstring Bc'de' F^ and in the lower chord are
easily found, as each of these chords belongs only to one of
the component trusses.

Since the diagonal bracing of the Pratt truss is replaced by
the bowstring truss, it is necessary not only to determine the
stresses in each system separately, but also to consider the
effect on the latter when the load on the former is not
symmetrically disposed. As the panel points B^ dy and F
are common to both systems, the Pratt truss may first be
regarded as having counterbraced long diagonals Bd and dFy
and after the stresses in Bd and ^//^are found for the unequal
loads transferred to B and F^ or the panel loads at b and /,
or both of these conditions combined, the stress in each of
these members may be replaced by two equal and opposite
external forces applied at its extremities, as was done in Art.
33, Fig. 69, in replacing the initial tension in the counters.
As the stress in any case will be compression in one and ten-
sion in the other long diagonal, care must be taken to give
the equivalent external forces the proper direction. The
stresses in the bowstring truss due to these forces must then

198 MISCELLANEOUS TRUSSES. CHAP. VI.

be combined with those obtained by treating each system
independently. Attention is again called to the important
fact stated in the preceding article, that only those stresses
must be combined which occur simultaneously under the same
conditions.

If the dead and live loads are uniform and the bowstring is
parabolic, the maximum stresses in the two horizontal chords
are uniform throughout. This fact indicates why this type
of truss is adapted to the use of track rails in these chords.
The details of a truss of this type, in which, however, the
parabola is in compression, were published in the Railroad
Gazette, Vol. XXVII, page 602, Sept. 13, 1895.

Prob. 73. A highway bridge with a clear roadway of 15
feet has a span of 57 feet. The truss is of the same form
and number of panels as Fig. 122, and its depth is 7 feet.
Find the stresses due to a live load of 80 pounds per square
foot.

Art. 62. Horizontal Shear in a Beam.

Let it be required to construct a diagram showing the dis-
tribution of the hocizontal shear in a beam. This may be
conveniently illustrated by an example, such as occurs in the
.design of a deepened beam. A deepened beam consists of
two timbers of rectangular cross-section placed above each
other and united by keys or brace blocks so as to make the
timbers act like a single stick. By this means the combined
strength of the timbers is double that secured when they act
separately.

Let the loads to be supported exclusive of the weight of
the beam consist of three concentrated loads of 4000, 8000,
and 6000 pounds respectively, and a uniformly distributed
load of 2000 pounds per linear foot, extending over a portion
of the span as indicated in Fig. 123. A single bending
moment diagram for both concentrated and uniform loads

Art. 62.

HORIZONTAL SHEAR IN A BEAM.

199

J_iJL

80Q0

iiiiijliiiiliililll

lbs.|per lin. ft.

iitiiiiiTiiiiniiTi

2'\

Pig. X33.

200 MISCELLANEOUS TRUSSES. CHAP. VI.

may be constructed by treating separately the portions of the
uniform load which lie between the concentrated loads. The
loads are taken in succession from left to right and laid off on
the load line (not shown). The left reaction A is found by
computation to be 22 900 pounds, and by laying this off on
the load line the closing line of the equilibrium polygon may
be made horizontal, if so desired, by taking the pole directly
opposite the point of division of the reactions (see Arts. 9
and II). The equilibrium polygon acefgiba is first drawn
by regarding the portions of the uniform load as concentrated
at their centres of gravity. The final form is then obtained
from this by constructing a parabola (Art. 10), tangent to the
right lines ce and efzX, the points rf and / respectively, these
points being directly below the extremities of the 4-foot por-
tion, and a second parabola tangent to fg and gi at the
points /and h.

The vertical shear diagram a!k . . . Im . . nb^ \s drawn next.
The shear passes through zero, where Im crosses a'b'y and on
measuring the moment ordinate directly above this it is found
to be 126400 pound-feet. If b be the breadth and d the
depth of the rectangular section of the beam, and 875 pounds
per square inch the working unit stress in the outer fibres,

bd^ = (126400 X 12 X 6) -7- 875 = 10400 inches'.

If b be assumed as about ^d, the beam will require two
timbers 14 X 14 inches in section.

The weight of these timbers for a length equal to the span,
at 3 pounds per foot board measure, equals i960, or say 2000
pounds. The bending moment at the middle of the beam
due to this weight is 5000 pound-feet. The corresponding
moment diagram is drawn as explained in Art. 10 by making
the parabola arb tangent to as and sb, the ordinate at s being
2 X 5000 = 10 000 pound-feet. The shear diagram a'pqb^
for the weight of the beam is added to the other by laying off

Art. 62. HORIZONTAL SHEAR IN A BEAM, 20I

the reactions in the opposite direction from the axis a'b' .
The shear due to all the loads passes through zero at where
Im crosses /y, which is a little to the right of the point found
before, and is 8.52 feet from the left support. The maximum
bending moment directly above this point is now measured,
and its value of 131 300 pound-feet obtained. As the resist-
ing moment of the beam slightly exceeds this amount no
correction is necessary.

If Sk is the unit horizontal shear, and V the vertical shear
in any section of the beam, b and d the breadth and depth of
the rectangular cross-section, the following relation is given
by mechanics (Mechanics of Materials, Art. 78).

•^*~2 bd'

In order to obtain the horizontal shear for a distance dx along
the beam, Sk must be multiplied by bdxy giving

SJbdx-^.Vdx.
2d

If the total horizontal shear be required between any two sec-
tions of the beam, it is necessary to integrate this expression
between the given limits of x. F is a function of ;r, and the
integral of Vdx is the area of the vertical shear diagram
between the given sections.

The total horizontal shear between a' (the left support) and

<? is -^ . Mr^,, sincey Vdx = J dM = M. Substituting the

values found above, the total horizontal shear for either the
portion c^o, or ob' equals (3 X 131 300 X 12) -r- (2 X 28) =
84 420 pounds. If four keys or brace blocks of equal strength
are to be employed to resist this shear, each block must be
designed to take a pressure of 21 105 pounds. In order to
determine their location, it is necessary to divide the vertical
shear diagram on each side of the zero shear into four equal

202 MISCELLANEOUS TRUSSES. CHAP. VI.

parts. This is most readily done by dividing it into narrow
strips, say a foot wide by scale, finding the area of each one,
and, beginning at the point o, adding each area to the sum of
the preceding ones. The total area should equal the maxi-
mum moment. These areas are laid off as ordinates on the
axis a''o' in such a way that the length of the ordinate at any
section of the beam represents the area of the vertical shear
diagram from that section to the point of zero shear. By
dividing the last ordinate a"t into four equal parts and draw-
ing parallels to a'^o through these points as indicated by
broken lines, their intersection with the curve passing through
the extremities of the ordinates gives the positions required.

The corresponding diagram for the right-hand portion is
drawn above the axis o'^b" . All the positions of the brace
blocks are marked on the bottom line of Fig. 123. Numbers
I, 2, and 3 are respectively 4' 9", 2' \\'\ and i' 4V from 4,
which is at the left support; while numbers i', 2', and 3' are
5' ioi"» 3' S"» and i' ^^' from 4' at the right support. It
will be observed that in the middle portion of the beam no
brace blocks are required in this case for almost one-half of
the 3pan.

When moving loads are substituted for stationary loads the
length of the middle space is materially reduced, for in that
case the maximum horizontal shear in any section does not
occur when the load covers the entire beam except for the
sections at the supports. If the cross-section of the beam is
not uniform, it is necessary to construct the diagram so that
the ordinates shall represent the corresponding sums of the
horizontal shears directly. * By using the general form of the
equation the distribution of the horizontal shear in a beam of
any cross-section may be similarly shown by a diagram.

Prob. 74. Two deepened beams having an effective span
of 22 feet carry a single track railway across a culvert. The
weight of the* track is to be, assumed at 400 pounds per linear

Art. 63. ROOF truss with counterbraces, 203

foot, and the live load at 3600 pounds per linear foot. Thef
beams are to be of timber weighing 45 pounds per cubic foot,
and with an allowable unit stress in the outer fibers of 1500
pounds per square inch. Find the positions of the brace
blocks or keys, provided eight are used in each beam.

Art. 63. Roof Truss with Counterbraces.

Fig. 124 shows the skeleton diagram of a roof truss with
straight upper chords and a curved lower chord. The upper
chord on each side is divided into six equal panels, and at the
panel point marked 2 the strut is normal to the upper chord.
The end panels of the lower chord are parallel to the adjacent
upper chord members and the panel points i, 2, . . . 5 . . .
2', i' lie on the arc of a circle. With the exception of the
end ones, the lower chord members are also equal. The
ventilator covers a panel on each side of the peak.

The dead panel loads i to 7 on the upper chord are 2500,
33SO» 3350. 3350» 2750, 2150, and 3600 pounds, and the
corresponding loads 6 to 5 on the lower chord are 1400, 600,
600, 600, 600, and 600 pounds, somewhat less than half the
weight of the truss being regarded as concentrated at the
panel points of the lower chord. The snow panel loads are
2700 pounds, excepting those at I, 5, and 6 which are 1350
pounds. The horizontal wind pressure against the vertical
side o — I is 5600 pounds, and against 5 — 6 is 4800 pounds^
The normal wind panel load is 6300 pounds, distributed to
the panel points in the usual manner.

The stress diagrams are drawn at first for the truss with its
counterbraces omitted as described in Art. 17, the additions
required on account of the counters (in this example, counter-
ties) being made afterwards. Only one of the stress diagrams
is shown, that due to wind on the fixed side of the truss being
given in Fig. 125. As the direction of the reaction at the
fixed end is not known, the equilibrium polygon used to de-

204

MISCELLANEOUS TRUSSES.

Chap. VI.

termine the reactions must have its left-hand vertex at that
support and its right-hand vertex on the vertical line of action
of the reaction of the right support. To reduce the number
of sides of the equilibrium polygon, and thereby to secure
greater accuracy, it is desirable to concentrate the wind pres-
sures on the two vertical and the two inclined surfaces at
their respective centers. Should this arrangement cause a
vertex of the equilibrium polygon to fall beyond the limits
of the drawing even with the most favorable position of the

Fig. 134.

pole, it may be remedied by replacing the horizontal and
normal pressures on the ventilator by their resultant. To-
avoid ambiguity in stress, the member ^ J/' is attached to the
panel point 7 so as not to receive or transmit any direct stress
due to the wind.

To determine the stresses in the members adjacent to the
second panel of the truss when the counter-tie DE acts,
imagine the main diagonal removed and complete the corre-
sponding force polygon for the panel points at its extremities,
the additional lines required being made with short dashes
instead of full lines. In a similar manner the additions are
made due to the counters FG, HI^ and those in the other half
of the truss. It will be observed, for instance, that the
polygon hhii in the stress diagram has all of its sides respeg-

Art. 63. ROOF TRUSS WITH COUNTERBRACES.

205

tively parallel to the members of the panel containing the
diagonal ties HI, while the two points marked i are on the
same parallel to NI, and the two points h are on the same
parallel to AH.

The stress in AH when the counter acts is represented by
the shorter ah, while its stress when the main tie acts is the
longer ah, as is evident
at a glance on observing
with which point // the
broken line hi or the full
line hi is connected. As
the lines in the stress dia-

gram, which give the
stresses in the strut con-
necting the upper panel point 4
PiK. X35 -vvrith the lower panel point 3,
have the same small or lower-case letters at their extremi-
ties as the capitals which designate the spaces adjacent to
the strut, an inspection of the truss diagram will indicate
which letters are to be used in any given case. When the
counter acts on the left and the main tie on the right the
strut is designated as FH and the corresponding stress is
fh. If the counters act on both sides the strut is FI 2ind its
stress//. The character of the lines representing the stresses
in the main and counter ties afford a check against errors,
for the stress diagram above shows that £^i is the stress

206

MISCELLANEOUS TRUSSES.

Chap. VI.

in the strut when the acting adjacent ties are the main tie
FG^ and the counter tie HL As an additional example, the
stress in the next strut to the left is eg when the acting
adjacent ties are the main DE and the counter FG.

Before measuring the stresses in the chords and struts the
stresses in the main and counter ties are tabulated so as to
determine how many counters are actually required. The
following tables give all the stresses for the ties in three panels
and the maximum and minimum stresses in the remaining
ties, the stresses being expressed in units of one thousand
pounds.

Ties.

Main.

Counter.

Main,

Counter.

Main.

Counter.

Dead load .tt •«• • •

+ 17.8
4- 10.8
\- 14.8
- 2.4

- 19.8

— 12.0
-16.4
+ 2.7

4- 16.8
- 8.1

— 1.0

— 12.6
4- 6.1

+ 14.5
+ 9.4
+ 33-7
- 12.7

- 10.3

- 6.7
7 23-9
+ 90

Snow load

Wind on fixed side

Wind on free side

Maximum

Minimum.... -. -

-f 43.4
4- 15.4

+ '9-3

+ S.I

t57S
4- 1.8

Ties.

B'C»

lyE'

F'G'
(m)

FfQ'
(c)

H»r

J'K»

Maximum

4-98.1
+ 24.4

+ 78.4
+ 34.1

+ 44.3
+ 14.3

+ 16.4

+ 3-1

+ 40.8

+ M5

4-18.9

+ 68.3
+ 5-9

Minimum ..«•

These tables show that only one panel in each half of the
truss needs counterbracing.

The following tables give all the stresses in four chord

Chord Members.

Main.

Counter.

Main.

Counter.

Dead load

+ 30- 1
- .33-5

+ 305
-f 18.9
+ 67.8
- 42.2

+ 46.0
+ 27.8
+ 68.3
- 37.7

+ 46.9
+ 28.7
+ 80.2
- 43.5

- 52.4

- 3»-9

+ X3.6

- 51.5

- 31-2

Snow load .•••..

Wind on fixed side

Wind on free side

Maximum

7 33.5
+ 30.x

4-117.3
- II. 7

4-142. X

-154.7

- 43.?

+ 3.4

Art. 63. ROOF truss with counterbraces.

207

Chord Mkmbbrs.

A£'

AH'

J'U

Maximum

Minimum :..

-123.7
- 29.2

- 93.8

- 40.8

-154-3
- 25-4

-125.7
- 53.3

-M4-4
- 42.9

-"5.4
- 5'-2

-116. a
- 41.7

— 109.2

- 42-7

Chord Members.

Maximum
Minimum..

AF'

-126.0

• S3. 4

NB'

- 3.3

Niy

NG'

-f 69.4 4-ITO.O

+ 30.S 4- 380

+106.3
-h 7

NI'

-f 93 7
+ 19.6

NK=. NK'

-f- 751
4- 4-5

members, and the maximum and minimum stresses in the
others. These tables indicate that two members of the lower
chord are subject to both tension and compression, and must
be designed accordingly.

Before measuring the stresses in the struts or posts it is
desirable to prepare a table showing which diagonals are act-
ing under the six possible combinations of load. Such a
table for the ties FG and F'G is shown below, the acting tie
being indicated by an asterisk. It being remembered that no
counters act in any other panels, it is apparent that the
stresses to be obtained from the diagrams for the posts of the
counterbraced panels are those given in the first one of the

Dead load

Snow load

Wind on fixed side..
Wind on free side. . .

G'H'

E'F'

- 3-0

-3.8

- 38

- 30

— 2.2

- 3-»

- 3>

— 2.2

- 2.4

- 3-1

- 2.4

- 1.6

- 3.5

- 2.5

- 1.6

- 6.7

-15.6

- 3-5

- 5-a

-14.2

-6.7

F'G'

G'H'

E'F'

- 3.0

- 5-4

- 5-7

- 7-3

- 9-7

- 12.1

•

- 30

- 5-4

- 9.7

- 12. 1

- 7.4

- 9.0

- 3.8
-6.9
-19.4
-22.5

-11

--11

—18.0
— ai.i

Dead + snow loads

Dead 4-wind on fixed side
Dead 4- snow-}- wind fixed
Dead 4- wind on free side
Dead + snow 4- wind free

Max
Mini

mum

— la.x
- 30

-22. s
- 3'

— ai.i
- 31

— 12.1
- 3«c>

mum

2o8

MISCELLANEOUS TRUSSES.

Chap. VI.

following tables. The ties which act on each side of the
posts, are indicated by m and c for the main and counter
diagonals respectively. In some trusses three or four stresses
may be required in some of the posts for the dead and snow
loads and two for the wind loads. From this data the second
portion of the last table may now be filled out, and the maxi-
mum and minimum stresses selected by inspection. It will
.be observed that the final values for £F a,re the same as for
£^F\ and that for GH and G^H' the minimum stresses are
equal, while the maximum ones diSer but slightly.

If the counter FG were omitted, and the main tie FG were
then replaced by a member which could take both tension and
compression, it would change the minimum stress in GH to
+ 0.4, while if the main tie were omitted and the counter
replaced by a counterbraced member it would change the
maximum in GH to — 12.7, the maximum in £Fto — 9.0,
and the minimum in £F to + i«8 thousand pounds. It would
seem therefore that it would be preferable to counterbrace
the panels in this case instead of the members that would
otherwise require it.

The following table gives the stresses in the remaining ^posts
and in the middle suspender.

AB'

CD'

KK'

Maximum — ....
Minimum

-65.4
- 16.5

- 48.6

— 22.0

- 37.2

— 12.0

- 3».5

- 13.4

- 44.6

- 3-7

- 35.2

— 10.9

+ 1.3

Prob. 75. Find the maximum and minimum stresses in the
crescent roof-truss treated in Art. 24, provided the panels be
counterbraced with diagonals which take compression only.

Art. 64. A Ferris Wheel with Tensile Spokes.

The skeleton diagram of a small Ferris wheel with eight
apexes and supported at the hub is given in Fig; 126. The
broken circular line indicates the rack where the power is

Art. 64. A FERRIS WHEEL WITH TENSILE SPOKES.

209

applied to rotate the wheel. The spokes are designed to take
only tension, and equal loads are applied at all the apexes.
The crown of the wheel tends to sag under the influence, of
the load at apex i ; but as this would produce compression in
the spoke IB it will not act, and therefore the load is sup-

A '^^

Pig. X37.

ported by the segments AI and AB of the rim. These
stresses are obtained by constructing the force polygon for
the apex i as shown in the stress diagram at the right. The
force polygons for the remaining apexes may now be con-
structed in regular order. The long vertical ib is the reaction
of the support and the diagram shows that it is in equilibrium

2IO MISCELLANEOUS TRUSSES. CHAP. VI.

with the stresses in all the spokes, which truly expresses the
relation of the forces at the hub. In this article the resist-
ance due to friction in the bearings is not taken into account.

" An examination of this diagram shows that under uniform
load the stresses in the spokes and in the segments of the rim
gradually increase from the top to the bottom of the wheel,
those in the rim being compression throughout. When a
segment of the rim is in a horizontal position at the top of
the wheel its stress is — i.oP, P being the apex load. The
stress diagram for this position is not given. As it revolves
its stress gradually increases to the maximum value of
— 3.92/^ when the segment reaches the position o{ A£, the
stress now remains unchanged until the position of AF is
reached,, and then gradually diminishes until the segment is
again horizontal at the top. If the wheel had 16 spokes the
compression in any segment of the rim would vary between
the limits 2.4i4Pand 7.689/^.

The stress in aiiy spoke remains zero during the interval
between its two upper positions which make an angle with
the vertical equal to one-half the angle between the spokes,
and then gradually increases until it reaches its maximum
value of -f- 4^» when the spoke is below the hub in a vertical
position. It is interesting to observe that the maximum
stress in any spoke is independent of the number of spokes
when that number is not less than four. The construction of
the diagram also indicates that the stresses in both rim and
spokes are independent of the size of the wheel, except so
far as the apex loads may depend upon it.

Fig- 1^7 gives the wheel and stress diagrams for the case
when only three of the observation cars are occupied. The
horizontal reaction applied on the circular rack which is used
to rotate the wheel is found by equating to zero the sum of
the moments of all the external forces with reference to the
centre of rotation of the wheel. In the stress diagram the

Art. 65. A BICYCLE WHEEL WITH TENSILE SPOKES. 211

points i and b which coincide indicate no stress in the vertical
spoke IB, while the inclined line ib is the reaction IB of the
supports, for, since the spoke has no stress the letter / really
applies to the entire space on the left of the inclined arrow
designating the reaction. It will be noticed that most of the
stresses on the loaded side are considerably greater than those
on the other.

The Ferris wheel at the World's Columbian Exposition in
1893 was 250 feet in diameter and had 36 spokes. An article
on this subject applying the graphic method to water wheels
of a given type as well as to other structures, and including a
description of the main features of a wheel of the same mag-
nitude as that at the Exposition, but designed so that its
stresses should be statically determinate, may be found in
Zeitschrift fiir Bauwesen, Vol. XLIV, page 586 (1894).

Prob. 76. Determine tha stresses in all the members of a
Ferris wheel with 8 tensile spokes for a load Pat each apex,
when one of the spokes is vertical, and also when one of the
rim segments is horizontal.

Art. 65. A Bicycle Wheel with Tensile Spokes.

Bicycle wheels are usually constructed with spokes of very
light steel rods and stiff rims of metal or wood. The number
of spokes is generally 32. A wheel with one-half that number
of spokes is represented in Fig. 128. It carries a load W^at
the hub, and is subject to an equal reaction at the point where
it rests upon the ground.

The reaction BC tends to produce compression in the spoke
attached to the same point of the rim, but as it can take only
tension it will not act, and hence may be considered as removed
for the time being. The force triangle abc accordingly ex-
presses the condition of equilibrium at apex i between the
reaction BC and the direct stresses in AB and -4C The
sides ab and ac are parallel to the chords of the arcs AB and

212

MISCELLANEOUS TRUSSES.

Chap, VI.

AC respectively. These stresses are equal, and their value is
— 2.563 W. The complete stress diagram is found to have a
form similar to that of the wheel, except that the rim is com-
posed of straight segments, and shows that the tension in all
the spokes except BC is W^ while the compression in all the
segments of the rim is 2.563 W. The perimeter of the stress

o n

{'A

t. o ?

Piff. X99.

diagram expresses the condition of equilibrium at the hub
between the load and the stresses in all the spokes.

The direct compression in any segment causes flexure, the
fibers on its inner side being in compression, and those on the
outside in tension. When the wheel rests on the ground at
any intermediate points of a segment of the rim, the segment
serves as a beam to transfer the reaction to the adjacent
apexes or panel points of the wheel truss, the resulting
flexure causing tension in the inner fibers and compression in
the outer ones. Fig. 129 gives the position when the support
is midway between the apexes i and 16. The direct stresses
in the wheel members may therefore be found by replacing
the reaction by the two upward forces at i and 16, each equal

Art, 65. A BICYCLE WHEEL WITH TENSILE SPOKES, 313

to i W. The stress diagram is readily constructed, as both
Bpokes QB and BC are not acting. The tension in the
remaining spokes is 0.5 10 W^, and the compression in all the
segments of the rim except AB is 1.307 W^ that in AB being
1.207 IV.

It is seen, therefore, that in passing from the position in
Fig. 128 to that in Fig. 129 the direct stresses in the entire
rim and in all the spokes except two, are reduced nearly one-
half, and during the next thirty-second of a revolution the
stresses increase again to their former value. This cycle of
changes occurs 16 times in every revolution for each segment
of the rim and 14 times for the spokes. In addition to this
the stress in each spoke changes from IVto o and back again
to Win passing from the position of QB to that of CD in Fig.
128. Further, the bending moment due to the reaction of
the ground, as well as that due to the direct stress in each
segment of the rim, passes through a similar cycle of changes
16 times in each revolution. The form of the stress diagram
shows that if the number of spokes is doubled the direct stress
in the rim segments is almost doubled, while the bending
moment is reduced nearly 75 per cent. On the other hand,
the magnitude of the stress in the spokes is independent of"
their number.

The changes in stress caused by attaching the spokes, in
the customary manner, in series tangent to opposite skies of
a hub of given diameter instead of radiating to its center, can
readily be found by applying principles heretofore given.

Prob. yy. Find the stresses in a bicycle wheel with 32
spokes for both positions .given in Figs. laS and 129 when
the load I^is 150 pounds.

2^14: : JEJ-ASTIC PEFOKMATIOM OF TRUSSES. CHAP. Vlli

CHAPTER VII.
* ELASTIC DEFORMATION OF TRUSSES.

: Art. 66. The Displacement Diagram.

TKe change in length A. of any member of a truss which is
subject to given loads and reactions may be computed by the
well-known'formula (Mechanics of Materials, Art. 5)

in- which /is the length of the given member, A the area of
its cross-section, E the coefficient of elasticity of the material
of which it is composed, and P the total stress in the member.
In case S'tr esses due to temperature are to be taken into
account the above value of \ must be combined with the quan-
tity atlt in which a is the coefficient of linear expansion for a
change of one degree, and / the rise or fall of temperature
expressed in degrees.

As a'truss is composed of triangles, the method of finding
the displacement of its panel points due to any given loads
may be illustrated by showing how to determine the displace-
ment of one panel point when two others with which it is
connected byltruss members are known. Let the panel point
^'in Fi^. 136 be connected with a arid b by members whose
lengths are /^ and /j respectively. Let the stress in ac be a
compression which produces a shortening of A, in its length,
while the stress in 6c is tension and A., is the corresponding
elongation. The magnitudes and directions of the displace-
ments of a and i are represented by the lines aa' and 6b\

Art. 66,

THE DISPLACEMENT DIAGRAM.

2ii

Let a'c^ be drawn parallel and equal to ac. Since the. stress
in ac is compression, A, must be laid off from c^ towards the
point a\ which for the moment is to be regarded as fixed.
This shortening, indicated by a heavy full line, is very much
exaggerated in the figure, for if it were laid off to the same
scale as a!c^ it would not be visible. With a! as a center and
the reduced length /, — A, as a radius let an arc be described.

Fig. 130.

Fig. 131.

The panel point c must lie somewhere on this arc. Because
the elastic deformations of the truss members are, however,
very small, the tangent to the arc may be substituted for the
arc itself. A perpendicular to aV, is therefore drawn at the
end of the line marked A,. Similarly, b' c^ is drawn parallel to
bc^ and its length increased by its elongation A^ and a perpen-
dicular erected at its extremity. The point t' is therefore at
the intersection of these two perpendiculars, and the line ctf
(not drawn) represents the displacement of c in magnitude and
direction.

In view of the exceedingly small values of A as compared
with / it is desirable to exclude from the diagram that portion
which contains the lines representing the lengths of the mem-
bers themselves. This can readily be done, as it is seen that
the lines cc^ and cc^ representing the displacements of a and b^
the lines A, and A,, and the perpendiculars at the extremities
of A, and A, form a closed polygon. In Fig. 131 it is drawn
separately to thrice the scale of that in Fig. 130, the pole O
in the former replacing the point c in the latter. The dis-

ELASTIC DEFORMATION OF TRUSSES. CHAP. VII.

placements of the panel points a^ 6, and c all radiate from the
pole 0.

Such a diagram is called a displacement diagram. In its
construction especial care must be exercised in observing the
directions in which the values of X are to be laid off, constantly
referring to the panel points of the truss diagram, which for
the time being are considered as fixed. The lengths of the
perpendiculars whose intersections locate the successive panel
points need not be measured.

Prob. 78. A weight of 15 000 pounds is suspended from a
ceiling at points 10 feet apart by means of two wrought-iron
bars, one being one inch square and 9 feet long, and the otler
3/4 inch square and lO feet long. Find the displacement of
the point where the weight is attached to the bars.

Art. 67. Deformation of a Truss,

It is required to find the displacements of the panel points
of a wooden king-post truss whose span is 16 feet and depth
8 feet, which carries a load of 12 000 pounds at panel point
'6 (Fig. 132). The data required for the construction of the
displacement diagram is given in the following table. In
computing \ the value of E was assumed i 500 000 pounds.

Mbmbrr.

Strkss.

Lbngtk.

Cross-sbction.

Mbmbbr.

ab :s be

aB^ Be

Pounds.
+ 6000
- 8490
+ 12000

Inches.
96

135.75
96

Square Inches.

36
64
36

Inches.

-h 0.0107
— 0.0120
+ 0.0213

Number.

1 and 5

2 and 4

For the sake of illustration let the point a be fixed, and the
point c be regarded as perfectly free to move horizontally,
although in practice such a short span is fixed at both sup-
ports, since the horizontal movement of c due to the load is
too small to require a movable support;

Art. 6y.

DEFORMATION OF A TRUSS.

2\J

The displacement diagram may be constructed by beginning
at any panel point and regarding as fixed its own position as
well as the direction of one member attached to it. Let the
point a (which is actually fixed) and the direction of ab be so

Fig. 135.

regarded. In Fig. 134 the point af will therefore coincide
with the pole O and A, will be laid off toward the right, that
is, in the direction of a toward b on the truss diagram. For
convenience the values of \ are marked on the truss diagram,
and when they are laid off in Fig. 134 they are marked by
the same numbers as the corresponding members in Fig. 132.
After V is thus determined the displacement of B is next

•2 1 8 ELASTIC DEFORMATION OF TRUSSES. CHAP. VI L

found by regarding a and b in the triangle aBb as fixed. The
elongation A, is laid off upward from b\ and the shortening A,
downward from a\ the intersection of the perpendiculars
giving B\ In the same way d is located. The lines OB'
and Oc' are the displacements of B and c.

In order to show the deformed truss under the conditions
assumed (that a and the direction of ab are fixed), the dis-
placements are laid off on Fig. 132 to one-tenth of the scale
employed in Fig. 134, and the corresponding points joined
by broken' lines. The student will observe that the deforma-
tion shown is greatly exaggerated, and hence the members
seem to have unduly altered their lengths.

The primary conditions of the problem, however, require
that c shall inov^e only in a horizontal line, and therefore the
entire truss rriust be revolved about ^ as a center until c^ falls
into the iiorizontal through c, As the arc thus described is
very small, compared with the radius ac^^ which in turn differs
but very little from aCy a perpendicular to ac from r, may be
substituted for the arc without appreciable error.

In Fig. 135, to which were transferred the displacements
Ob'y 0B\ and Oc without the construction lines, the corre-
sponding path of rotation of c is represented by c'c" which is
drawn perpendicular to ac in Fig. 132, and continued until it
meets the line Oc"\ which is drawn parallel to the direction
in which the panel point c is free to move. In this example
that direction is horizontal, and happens to coincide with the
line aCy but the statement here given is so framed as to apply
equally to inclined lines of motion of panel points supported
by expansion rollers or rockers. When the successive dis-
placements Oc' and c'c'" are combined, the resultant displace-
ment is Oc'" , The displacement of B due to the rotation of
the truss is B'B'" , which is perpendicular to aB (in Fig. 132),
and whose length is proportional to its distance from the
center a. That is, B'B'" : c'c" = aB : ac, from which the

Art. 6/. deformation of a truss. 219

length of B^B"' may be conveniently found by similar tri-
angles. Similarly, as <3:<J equals one-half of aCy Vb'" equals
one-half of ^V". The resultant displacefnehts at^e then- repre-
sented in direction and magnitude by 0B"\ Ob"! ^zxiA Oc"\
and as^ is the center of rotation, a'" coincides with ^' and
Oa'" is zero. ^

In Fig. 133 the final position of the deformed truss is shown
in broken lines, the resultant'displacements BB^, bb^, arjd cc^
being laid off parallel to and equal to one-tenth of the lengthy
of OB'"y Ob"\ and Oc'" in Fig. 135. If the deformation
were not exaggerated the truss dia.gr Sim aB^c^b^ in Fig. 133
would be equal to aB^cJb^ in Fig. 132 in both form and
dimensions.

For the purpose of simplifying the- construction let the,
three parallelograms in Fig. 135 be -completed, and the points-
a" , B'[, c"j and b" joined by lines as indicated. Jhe- lines
B"a",b"a'\ and c"a" (parallel and equal. to B'B"\b'b"\ and
c'c'") represent the displacements of panel points \5, ^,- and c
due to rotation about a, and are respectively perpendicular
and proportional to aBy aby and ac of the truss diagram, and
therefore it follows that- the diagram a"B"c"b" is similar to
aBcb^ and all of their lines are mutually perpendicular. This
important fact furnishes a means of determining the final dis-
placements on Fig. 134 in a very simple manner, as follows:
Through c' draw c'c" parallel to the constrained line of motion
of the panel point Cy and draw a"c" perpendicular to ac (in
Fig. 132), and intersecting c'c" at c" . On a'c" draw a dia-
gram similar to the truss diagram. The required displace-
ments are then given by the directions and distances of the
points B' y b' y and c' from B" y b" y and^" respectively. It is
thus seen that the points B" y b'\ and c" in Fig. 134 corre-
spond to what may be regarded as successive positions of the
shifted pole O l\\ Fig. 135, which conception aids the memory
in reading the directions correctly.

220 ELASTIC DEFORMATION OF TRUSSES. ClIAP, VII.

It is very desirable in practice to reduce the displacement
diagrams to their most compact form. It will both diminish
the errors due to slight inaccuracies in the directions of the
intersecting perpendiculars as well as allow increased pVecision
by the use of a larger scale. This result may be secured by
beginning the construction with the line which suffers the
minimum change in direction under the influence of the given
>loads. In simple trusses some line may be found at the
middle which does not change its direction at all, provided
the truss and the loading are both symmetrical with reference
to a vertical section at the middle, or which changes but little
in unsymmetrical trusses. For a bridge truss with an equal
number of panels, the middle vertical is such a member, or
the chords of the middle panel when the number of panels is
odd. As an illustration of the effect thus produced, the dis-
placement diagram for the above truss when the direction of
the middle vertical and the position of either of its extremities
is assumed to be fixed, is given in Fig. 136 to the same scale
as that in Fig. 134. Since a perpendicular to ac through a!
meets a horizontal through c' at a' , the diagram a!'B"c"b" is
thereby reduced to zero. If the scale were doubled the
vertical dimension of this diagram would not be quite equal
to that of the one previously drawn. With a larger number
of panels the difference is still greater. In this case the
diagram makes a direct comparison between the displacements
of all the panel points.

On applying the scale and protractor to the original draw-
ing the displacement of B and b were found to be 0.0296 and
0.0501 inches, their directions being inclined I2i° and 21^^
to the vertical. The angles were read only to the nearest
quarter degree. As the lower chord is horizontal, the dis-
placement of c is the sum of the elongations A, and A, which
equals 0.0214 inch.

In order to avoid the excessive labor of making corrections

Art. 68. deflection of a truss. 221

in the cross-sections of tension members to allow for the effect
of rivet holes in riveted shapes or of mortices or other cuts in
timber, it is customary to use the gross sectional areas and to
reduce somewhat the coefficient of elasticity.

Prob. 79. A single-track through Pratt truss railroad bridge
has 5 panels, each 23.1 feet long and 25 feet deep. Using
the same notation as in Fig. 96 in Art. 45, the members have
the following areas of cross-section in square inches : BC =
CD, 27.99; ^^ = *^» 16.44; ^^» 17.62; aB, 2g.7Sl Bb, 11.58;
Bcy 10.5; Cc, 11.58; Cd=cD, 8.04. The tie ^^ and the
chord cd are composed of eye-bars of medium steel, and the
rest of the members are built up with soft steel shapes.
Find the displacements of the panel points of the lower chord
due to a live load of 4000 pounds per linear foot combined
with two excess loads of 26 000 pounds two panel lengths
apart. (Use values of E of 29 000 000 and 26 000 000 pounds
for the medium and soft steel.)

Art. 68. Deflection of a Truss.

While the displacement diagram gives the actual displace-
ments of the panel points in the plane of the truss, their
vertical components only are generally required. When
bridge trusses are erected they are cambered so that under
their maximum load the panel points of the loaded chord shall
not fall below a horizontal line joining the panel points at the
supports. This camber is secured by shortening the tension
members by an amount equal to the elastic elongation due to
the sum of the live and dead load stresses, when the live load
is so placed as to produce the maximum moment at the mid-
dle panel point, plus an allowance for clearance in the case of
pin-connected joints. The compression members are length-
ened in a similar way. (See Part III, Art. 62.) The maxi-
mum stresses must not be employed throughout because they
are not simultaneous.

If to these changes of length for the members there be
added the values of X caused by the dead load only, due

222 ELASTIC DEFORMATION OF TRUSSES. ClIAP. VII.

regard being paid to their respective signs, and the correspond-
ing displacement diagram be drawn, the vertical components
of the displacements will give the deflections of the several
panel points when the bridge supports only the dead load,
and their values may be used for comparison with the
observed deflections. Roof trusses supporting horizontal
ceilings are cambered in a similar manner.

In the example used in the preceding article, the deflection
of b is found to be 0.0489 and of B is 0.0276 inch. The
diagram shows that when two points are directly above each
other their deflections should differ by the change in length
of the member connecting them. This may serve as a useful
test of the accuracy of the drawing.

Prob. 80. Fmd the deflection due to the live load, of the
bridge whose data is given in Prob. 79.

Art. 69. Truss Deflection under Locomotivb
Wheel Loads.

The most convenient way to find the stresses in the truss
members whether the chords are both horizontal, or either
one or both of them are arched, is the following: Let the
position of the live load be found which produces the maxi-
mum moment at the panel point at, or nearest to, the middle
of the truss (Art. 47 or 5 i) and with the truss diagram in this
position on the equilibrium polygon let the closing line be
drawn as well as the chords of the pol3'gon whose horizontal
projections equal the successive panel lengths in magnitude
and position. The extremities of these chords connect the
points two and two where the verticals of the truss diagram
intersect the equilibrium polygon. By drawing ra^s parallel
to these chords through the pole they will cut off on the ver-
tical load line the panel loads for this position of the wheel
loads, and a ray parallel to the closing line will divide the

Art. 69. UNDER LOCOMOTIVE wh£el loads.

223

reactions. The stress diagram, which is similar to that for
dead load, may now be completed in the usual manner.

The following table gives the required data for determining"
the deflection of the double-track through bridge No. 77 of
the second division of the Baltimore and Ohio Railroad, due

Mbmbbrs.

Stress.

Lbngth.

Cross-sectioh.

Member.

xooo lbs.

Inches.

Square Inches.

Inches.

Number.

ab = be

+ 287.0

321

36.0

-p O.0S82
4- 0.0866

11,8

+ 360.0

321

46.0

+ 356.0

321

46.0

+ 0.0856

ef^fg

+ 266.5

321

36.0

-f- 0.0819

18, 21

- 369.0

329

65.12

- 0.0717

CD^DE

- 414.0

321

71.52

- 0.0715

2, 13

— 365.0

329

65.12

— 0.0709

- 413.0

464.7

76.44

— 0.0966

+ 140.0

336

17.52

-0.1033

+ 106.0

464.7

23.56

- 0.0804

+ 12.5

408

30.6

- 0.0064

+ 87.5

5 19. 1

34.0

- 0.0514

408

20.6

+ 94.5

519,1

34.0

+ 0.0555

4- 5.0

408

30.6

. + 0.0024

+ 130.0

464.7

23.56

4- 0.0986

4- 106.0

336

17.52

4- 0,0782

-4

- 386.5

464.7

76.44

— 0.0903

to the specified live load of two B. & O. typical consolidation
locomotives and train. The form of the truss is shown in
Fig* 1 37* The lower chord consists of eye-bars of medium
open-hearth steel, while the upper chord and web members are
built up of shapes of soft steel. There are no counters. The
stresses were found in the manner described above, the stress
diagram being drawn to a scale of 50 000 pounds to the inch.
The value of the coefficient of elasticity was assumed as
26000000 for soft and 29 000 000 pounds for the medium
steel.

Tlie displacement diagram, shown in reduced size in Fig.
138, was constructed by assuming the position of </and the
direction of dD as fixed. As the elongation of this member
is zero, it is the same condition as to assume that the position

224

ELASTIC DEFORMATION OF TRUSSES. ChAP. VII

of D and the direction of Dd are fixed. The truss mem^
bers are numbered in the order in which their values of
X were used in constructing the diagram. As the left end of
the truss is fixed and the right end rests on expansion rollers,
a" coincides with a\ and g'' lies in a horizontal through ^'

^ r^ .r

directly above a\ The distance a"g" is too small a span to
permit a diagram similar to the truss diagram to be drawn
without confusion of points. As, however, only the deflec-
tions are desired, the necessity for this diagram may be
obviated by the following construction of a deflection polygon.
Let a^ be obtained by projecting a' across on the vertical
through a, and similarly for^^. The intersection of the line
joining a^ and g^ with any vertical, as for example that
through r, gives a point c^ whose height is the same as ^'
would have been if the diagram ^" . . . g'^ had been drawn.
By projecting b'c'd'e' and/' on the corresponding verticals and

Art. 69. UNDER LOCOMOTIVE WHEEL LOADS. 22$

joining them as indicated, a polygon will be obtained whose
ordinates at the panel points represent the corresponding
deflections of the panel points of the lower chord. The values
of the deflections in inches are marked on the diagram. The
scale of the original displacement diagram was 0.060 inch to
an inch. A deflection polygon for the panel points of the
upper chord might also be drawn, if desired; but as these
points are united to the lower chord by verticals, their deflec-
tions may be obtained by subtracting the elongations of the
verticals from the corresponding deflections marked on the
diagram.

Prob. 81. What change in the deflection would be caused
in the truss used in this article by substituting medium steel
eye-bars for Be and Cd with sectional areas of 24.0 and 20.25
square inches respectively?

APPENDIX.

ANSWERS TO PROBLEMS. "

Prob. I. 56.8 pounds, making an angle of 37° 35' with the
smaller force.

Prob. 2. 43.6 pounds, 36° 35' and 83° 25'.

Prob. 4. 107.8 pounds and 129° 20'.

Prob. 5. 2 800 pounds.

Prob. 6. 5, = + 5.56, 5, = - 3.27, 5, = - 5.86, and 5, = +
5.95 tons.

Prob. 8. 5, = — 43.6, 5, = 5^ = -j- 109. i, and 5, = + 43-6
pounds.

Prob. 9. Resultant = 279.7 pounds, and angle with greater
force = 1° 45'.

Prob. II. Resultant == 4 tons, is parallel to forces und 6 feet
from greater force.

Prob. 12. 1 77. 1 and 192.9 pounds.

Prob. 13. Maximum shear = ± 4000 pounds, maximum
moment = -\- 20000 pounds-feet.

Prob. 15. Maximum shear = — 6 tons, maximum moment
= — 30 tons-feet.

Prob. 16. Maximum shear = — 6 tons, maximum moment
= — 30 tons-feet.

Prob. 17. 0.70 inches from the back of channel iron.

Prob. 18. 500 pounds, 15.27 feet from the first force.

Prob. 19. 1= A X A' = y.io X 10.81 = 76.75 inches*, and
r = A X A^' = 7.10 X 0.51 = 3.62 inches*.

227

228 APPENDIX.

Prob. 22. Stress in AC = BC = — 2290 pounds, and in
CD = + 2 050 pounds.

Prob. 23. Apex loads = 1.48, 1.20 tons; reactions = 444^
3.60 tons.

Prob. 24. 2.79, 8.37, 2.04, and 6.12 tons.

Prob. 25. Apex loads = 2.15 and 1.61 tons. Dead load
stresses are: in upper chord, — 12.03, —9.78, — 10.10; in
lower chord, -j- I0'75> +6.45; and in braces, — 2.30, —2.30,
-j- 4.29 tons. The corresponding snow load stresses are,
-9.00, -7-32, -7-56; +8.05, +4-83; -1.72, -1.72,
-j- 3.21 tons.

Prob. 27. AB = 2 035, BC = 2 750, and CD = 814 pounds,
the normal wind pressures being 38.2 and i $.6 pounds per square
foot.

Prob. 28. Apex load = 1.86 tons ; reactions = 3.84 and 1.74
tons. Stresses in upper chord, — 5.82, — 6.75, —• 5.36, — 3.49,

— 3-49» — 349; »n lower chord, +6.51, + 443> + 2.35»
+ 2.35, + 2.35 ; and in braces, - 2.08, + 2.94, - 3.12, + 3.75,
o, o, o, and o tons.

Prob. 32. Apex load = 2.05 tons ; reactions at free end, 2.30
and 4.99 tons. Stresses for wind on fixed side are : in upper
chord, — 8.93, — 7.45, — 5.98, — 4.51, — 5.04; in lower chord,
+ 11.22, + 8.98, + 6.73, + 4.48; and in braces, — 2.52, + 1.15,

— 3.21, -|- 2.30, —4.1 1, +3.45, and o tons. Lower chord
stresses are diminished 3.75 tons for wind on free side.

Prob. 34. Dead apex loads = 0.70, 1.40, 1.40, etc. ; snow apex
loads = 0.43, 1.08, 1.30, etc. ; and wind apex loads = 1.74, 2.35,
and 0.70 tons. Maximum stresses in upper chord, — ii.o,

— 10.3, — 8.7, — 9.4; in lower chord, + 9.4, + 8.4, + 6.4; and
in braces, + i-9» + 2.7, + 1.7, + 3-3 tons. Minimum stresses,

— 3-9» - 3-3, - 34» - 4-0 ; - 0.7, + 0.8, +2.3 ; + i.o, - 1.2,

— 0.2, and — 0.5 tons.

Prob. 36. Dead, snow, and wind apex loads are 0.91, 0.84,
and 1.49 tons. Maximum stresses in upper chord, — 14.45,

APPENDIX. 229

— 11.38, --9.56; in lower chord, +13-96, +9.41; and in
braces, — 3.82, + 2.06, — 3.21, + 3.44 tons. Minimum stresses,
-5.09, -4.07, -3-39; +4.S5» +3-34; -1.02, +0.55.

— 0.85, and + 1.21 tons.

Prob. 37. cd^ce^ — 4.31, eg = ch =z ^ 1.73, ^^= -f 3.91,
bf= + 1.26, *A = + I.S7, de = o, e/= + 3-S5» ^= +0.39.
and £-/i = o tons.

Prob, 38. Apex loads = 0.6, 1.2, and 0.6 tons. Designating
the members as in Fig. 5 1 , the stress ^y = —4.15,^^= —3.61,
dk = dl = — 2.06, ef = -\- 3.76, e/t = + 1.50, ^/ = + 1.88,
/g= — 1.08, g/i = + 2.36, Ak = + 0.47. and kl = o tons.

Prob. 39. Dead panel load per truss = 1.9S, live = 4.7a tons.

Prob. 40. Panel loads = 2.83 on upper, and 5.67 tons on
lower chord. Stresses in upper chord, — 39.4, — 47.3, — 47.3 ;
in lower chord, +23.6, +23.6, +39.4, +47.3; and in web
members, — 34.7, + 5.7, + 23.2, — 11.3, + ii-6, o, and o tons.

Prob. 41. Maximum stresses in upper chord, — 46.6, — 123.3,

— 169.2, — 184.5; in lower chord, +92.0, + i53-3> + 183.9
tons. Minimum stresses, —12.2, —31.5, —43.0, —46.8;
+ 23.1, +38.5, +46-2.

Prob. 43. Panel load = 0.79 tons. Stresses in the chords,
3.4,6.2, 8.4, 9.9, 10.9, 1 1.2; in diagonals, —5.5, + 4-5» + 3-S>
+ 2.5, +1.5, +0.5; and in verticals, +0.8, —2.8, —2.6,

— 1.2, — 0.4, and o tons.

Prob. 44. Panel load due to truss is 1. 074, and that due to
train is 4.0 tons. Stresses in upper chord for south wind, O,
+ 2.0 ; for north wind, — 2.0, — 2.7 tons. In the lower chord,
o, +15.9, +25.4; —15.9, —25.4,-28.5 tons. Maximum
wind stresses in diagonals of upper lateral system, + 2.6, -j- 0.9;
in struts, — 1.6, — i.i tons. In lower lateral system, -|- 20.3,
+ i3-3» + 7-3 ; — 12.7, -- 8.3, and — 5.1 tons.

Prob. 46. The greatest reduction of stress is in RC, and
equals 4.7 -r- 76.4 = 6.2 per cent.

Prob. 47. Maximum stresses in upper chord, — 56.0, — 53.2,

250 APPENDIX*

— 52.1 ; in lower chord, + 50.0, + 50LO, + 51.3 ; in main ties»
+ ^-3* + ^-7 • i^ counter ties, + 7-3» + 8-0 ; and in verticals,
-j- lO.o, 4- 9-4» + 9-0 tons. Minimum stresses in chord, — 14.0,

— 13.3. — 13.0, -f 12.5, -f 12.5, -r 12.8 ; in diagonals, o; and
in verticals, + 2.5, + 0.2, — 0.3 tons.

Prob. 48. Maximum stresses in the chords, — 100.0, + '09.7,
+ ^05.9,4- 103.0, -j- loi.i, -p 100. 1 ; in main ties, 4- 9.0, + 10.2,
-j- 10.9 ; in counters, -j- j.7, -f- 9.0, -*- 10.2, and in verticals,

— lo.o, — 12.4, — 14.2, — 15.2 tons. Minimum stresses in the
chords, —30.0, -f32.9» -f 3i-8, + 30.9» + 30-3. +30-O; »n
diagonals, o ; and in verticals, — 3.0 tons.

Prob. 51. 42.5 tons.

Prob. 52. Dividing the span into eight equal parts, the
flange stresses at the sections are 0.0, 16.0, 26^, 32.6, and 35.1
tons. The absolute maximum is 35.2 tons at 4 inches from
center of girder. The shears are 20.1, 16.3, 12.8, 9.4, and 6.1
tons.

Prob. 53. Maximum stresses in chords, 86.2, 138. i, 167.1 ;
in end post, — 135.2; in main tics, -f 95-4' +^•^ -r 28.0; in
counter ties, + 3o» +28.0; and in verticals, -f 40-8' —48.3,
and — 23.5 tons. Minimum stresses in the chords, 20.5, 34.2,
41.0; in end posts, —32.2; in main ties, -{- 17.9, o, o; in
counter ties, o, o ; and in verticals, -|- 6.3, — 1.9, and — 1.9 tons.

INDEX.

Abbreviated method for wind stress,

54
Absolute maximum flange stress, 117
Ambiguous stresses, 61, 194
Analysis of a plate girder, 114
Pratt truss, 119
Apex loads, 36
Arithmetic, graphical, 30
Arrow, circular, 41

Baldwin, W., 128, 174
Baltimore truss, 187
Bending moments, 17, 20

scale of, 17
Bicycle wheel, 211
Bowstring truss, 91, 99
Braces, 33
Bridge trusses, 67-111

Camber, 221
Cantilever beams, 24
Center of gravity, 27
Character of stresses, lO, II, 39
Chord increment, 73
stresses, 153
Chords, 33
Circular arrow, 41
Composition of forces, 3, 4
Concentrated loads, 20
Counter braces, 77, 82, 93, 162, 203

ties, 77, 87, 96, 123, 204
Crane truss, 9
Crescent truss, 59
Cycle of stresses, 162

Dead loads, 34, 67
Dead load stresses, 38, 69
Deepened beam, 198
Definitions, 33

Deflection of a truss, 221, 222
polygon, 224
under locomotive loads,
222
Deformation of trusses, 214-225, 216
Diagonals, maximum stresses, 80,
82. 95, 157
minimum stresses, 80,

82,95
Displacement diagram, 214, 217, 223
Double intersection truss, 193
systems, 192

Elastic deformation of trusses, 214-

225
Ends of trusses, 48, 50
Equilibrium, 3

at each apex or joint,

10, II
between external

forces, 38, 42
conditions of, 7, 13
forces in, 3, 5
polygon, 12, 40, 102,
115,117,120,124,129,
145, 154. 167,177,190
Equilibrium polygon, propert'^s of,

15
Excess loads, 107
External forces, 38, 42

23?

232

INDEX.

Ferris wheel, 208
Fink truss, 61
Floor beam, 120, 130

reactions, 139
Force polygon, 4, 152

representation of a, I

triangle, i

Forces, composition of, 3, 4

in equilibrium, 3, 5

resolution of, 3, 4

resultant of, i, 2, 4

Graphical arithmetic, 30
Graphic method, advantages, 189
Greiner, J. E., 196
Greiner truss, 196

Horizontal shear, 198
Howe truss, 82, 139

Inflection point, 25, 26
Initial tension, 87
Internal stresses, 38, 42

Joints, 33

equilibrium at, lo, II

Keys, 198

Kingpost truss, 216, 221

Lateral bracing, 84, 171, 189
Lattice girder, 196
Live loads, 68

Live load stresses in Pratt truss, 77
Warren truss,

73
Load line, 20, 38, 115, 126
Loads on bridge trusses, 67
Locomotive loads, 11 2-143

standard typical,

112
Waddell's com-
promise stand-
ard, 113

Maximnm chord stresses, t$$
moment, 21, 25
shears, 105, 109, 131
stresses, 58, 61, 75, 80, 89^
89,94,122,182,
206
in diagonals, 157
in verticals, 161
Methods, i-3», 54

Minimam stresses, 58, 61, 75, 80, 82,
89, 94, 122, 182,
206
in verticals, 163
Miscellaneous trusses, 175-213
Moment diagram, 21, 23, 25, 103, 130,
135. 198
wheel load, 117,
124, 129
Moment of inertia, 28
Moments in plate girders, 124

in trusses, 141, 155, 177,

183, 187, 190
simultaneous, 135

Normal wind pressure, 47
Notation, 34, 152, 175

Overhanging beams, 24

Panel, 36

loads, 37, 69

effect of divided, 73
point, 76
Parabola, construction of, 23
Parabolic bowstring truss, 99
Pegram truss, 175
Pennsylvania truss, 182
Plate girder, analysis, 114
Points of division in panels, 144
Pole, 14

distance, 17
Polygonal frame, il
Position of wheel loads, 147

for maximum floor beaa
reactions, 139

INDEX.

233

Positiou for maximum moments in

plate girders, 125

for maximum moments in

trusses, 141, 143, 176, 183,

187, 190. 195

for maximum shears in

plate girders, 132, 134
for maximum shears in

trusses, 137, 139
for maximum web stresses
in trusses, 137, 139, 147,
164, 184, 187, 190, 195
PfSts, 146. 156, 173
I ratt truss, 84, 88. 139

analysis, 119
live load stresses, 77
Principles, 1-32, 33
Purlins, 35

Quadruple system, 193

Rafters, 35, 44
Range of stress, 58
Rays, 14

Reactions, determination of, 4S, 54^
58. 189

effective, 37, 70

floor beam, 139

of beams, 18
Reciprocal figures, 10
Representation of forces, I
Resolution of forces, 3, 4

of the shear, X51
Resultant of forces, I, 2, 4, 5, 13, 16,

150, 166, 167
Rise of truss, 33
Roof trusses, 33-66
Roof truss with counter braces, 203

Shear diagram, 21, 23, 25, 105. 118,
120, 134, 200
division of area, 201
Shears in plate girders, 131

in trusses, 13(3
Sheathing, 35

Shingling, 35

Simple beams, 20, 23

Simultaneous moments, X35

Snow loads, 34, 35

load stresses, 83

Span, 33

Stress diagram, construction of, 9,
39, 43, 45. 49» 50,
52, 60, 61, 70, 75t
79. 93. 204, 209,
211
defined, 10

Stresses, determination of, 10, ix, 39
due to track curvature, 174

Stringers, 120, 130

Strut, 34

Subverticals, 182

Suspender, 121

Tabulation of web stresses, 75, 78,
86, 95, 206, 207

Tie, 34

Train loads, 69

Triangular roof truss, 44, 57

Truss, defined, 33
diagram, 10
with fixed ends, 48
with one end free, 50

Trusses with broken chords, 144-174

Typical locomotive, 112

Uniform loads, 22
Unsymmetrical loads, 64

trusses, 64, 187

Vertical shear, 21, 22, 25, 73, 142,
151. 158
sign of, 21, X43
Verticals, maximum stresses, 81, 82,
95. 161
minimum stresses, 8x, 82,

95, 163
relation of stresses to diag*
onals, 81, 95-98, 146, 162

Warren truss, 73, 139

234

INDEX.

Warren truss, doable system. ig2

quadruple system, 196

Web members, 33

Weight of highway bridges, 67
railroad bridges, 67
roof covering, 35
foof trasses, 3s

Wheel loads, locomotive, 1x9-143,
147, 222
(See also Position of
wheel loads.)
Whipple truss, 193
Wind loads, 46

stresses, 48, 50, 54, 84, 17X1 204

A TEXT-BOOK ON ROOFS AND BRIDGES

MANSFIELD MERRIMAN and HENRY S. JACOBY.

Part I, Stresses in Simple Trusses*

Sixth Edition, Octavo, Cloth, Price $2.50.

Chap. I, Stresses in Roof Trusses. Chap. II, Bridge Trusses under Dead Loads.
Chap. Ill, Bridge Trusses under Live Loads. Chap. IV, Final Stresses for
Bridge Trusses. Chap. V, American Bridge Trusses. Chap. VI, Bridge
Bracing, Members, and Floors. Chap. VII, Deflection and Least Work.
Chap, VIII, Miscellaneous Structures.

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with Broken Chords. Chap. VI, Miscellaneous Trusses. Chap. VII, Elastic
Deformation of Trusses.

Part in. Bridge Design*

Fifth Edition, Octavo, Cloth, Price $2.50.

Chap. I, History and Literature. Chap. II, Principles of Economic Design. Chap.
Ill, Bridge Contracts and Office Work. Chap. IV, Fabrication and Erection.
Chap. V, Tables and Standards. . Chap. VI, Details of Plate-girder Bridges.
Chap. VII, Design of a Plate-girder Bridge. Chap. VIII, Details of Rail-
road Pin Bridges. Chap. IX, Design of a Pin Truss Bridge. Chap. X,
Highway Bridges. Chap. XI, Railroad Riveted Bridges.

Part IV, Higher Structures.

Third Edition, Octavo, Cloth, Price $2.50.

Chap. I, Continuous Bridges. Chap. II, Draw Bridges. Chap. Ill, Cantilever
Bridges. Chap. IV, Suspension Bridges. Chap. V, Three-hinged Arches.
Chap. VI, Twd-hinged Arches. Chap. VII, Arches without Hinges. Chap.
VIII, References to Literature.

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CHAPMAN & HALL, Limited, London.

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Black's United States Public Works Oblong 4to,

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Burr's Elasticity and Resistance of the Materials of Engineering 8vo,

Byrne's Highway Construction 8vo.

Inspection of the Materials and Workmanship Employed in Construction.

16mo,

Church's Mechanics of Engineering 8vo,

Mechanics of Solids (Being Parts I. II. Ill of Mechanics of Engineer-
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Du Bois's Mechanics of Engineering.

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Fowler's Ordinary Foundations 8vo, 3 50

* Greene's Structural Mechanics 8vo, 2 50

Holley's Analysis of Paint and Varnish Products. (In Press.)

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2 00

6 00

5 00

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* Martin's Text Book on Mechanics. Vol. I. Statics 12mo, 1 25

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Maurer's Technical Mechanics 8vo, 4 00

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Part III. A Treatise on Brasses, Bronzes, and Other Alloys and their

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Tillson's Street Pavements and Paving Materials 8vo, 4 00

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Waterbury's Cement Laboratory Manual 12mo, 1 00

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Wood's (De V.) Treatise on the Resistance of Materials, and an Appendix on

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Steel 8vo, 4 00

RAILWAY ENGINEERING.

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Roberts' Track Formube and Tables 16mo. mor. 3 00

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DRAWING.

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♦ Bartlett's Mechanical Drawing 8vo, 3 00

• ** '* " Abridged Ed 8vo, 150

♦ Bartlett and Johnson's Engineering Descriptive Geometry 8vo, 1 50

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Elements of Drawing. (In Press.)

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Durley's Kinematics of Machines 8 vo, 4 00

Emch's Introduction to Projective Geometry and its Application 8vo, 2 60

Hill's Text-book on Shades and Shadows, and Perspective 8vo, 2 00

Jamison's Advanced Mechanical Drawing 8vo, 2 00

Elements of Mechanical Drawing 8vo, 2 50

Jones's Machine Design :

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• Kimball and Barr's Machine Design 8vo, 3 00

MacCord's Elements of Descriptive Geometry 8vo, 3 00

Kinematics; or. Practical Mechanism 8vo, 5 00

Mechanical Drawing 4to, 4 00

Velocity Diagrams 8vo, 1 50

McLeod's Descriptive Geometry Large 12mo, 1 60

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Moyer's Descriptive Geometry 8vo, 2 00

Reed's Topographical Drawing and Sketching 4to, 6 00

* Reid's Mechanical Drawing. (Elementary and Advanced.) 8vo, 2 00

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Smith (A. W.) and Marx's Machine Design 8vo, 3 00

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Tracy and North's Descriptive Geometry. (In Press.)

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Elements of Machine Construction and Drawing 8vo, 7 50

Elements of Plane and Solid Free-hand Geometrical Drawing .... 12mo, 1 00

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Weisbach's Kinematics and Power of 'Transmission. (Hermann and

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Wilson's (H. M.) Topographic Surveying 8vo, 3 50

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Andrews's Hand-book for Street Railway Engineers 3X5 inches mor. 1 25

Anthony and Ball's Lecture-notes on the Theory of Electrical Measure-
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Anthony and Brackett's Text-book of Physics. (Magie.) Large 12mo, 3 00

Benjamin's History of Electricity 8vo, 3 00

Betts's Lead Refining and Electrolysis 8vo, 4 00

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Crehore and Squier's Polarizing Photo-chronograph 8vo, 3 00

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Dawson's "Engineering" and Electric Traction Pocket-book. . . . 16mo, mor. 5 00
Dolezalek's Theory of the Lead Accumulator (Storage Battery), (von Ende.)

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Duhem's Thermodynamics and Chemistry. (Burgess.) 8vo, 4 00

Flather's Dynamometers, and the Measurement of Power 12mo, 3 00

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Gilbert's De Magnete. (Mottelay ) 8vo, 2 50

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Landauer's Spectrum Analysis. (Tingle.) Svo, 3 00

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Norris and Dennison's Course of Problems on the Electrical Characteristics of

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Reagan's Locomotives: Simple, Compound, and Electric. New Edition.

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LAW.

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Law of Contracts , 8vo, 3 00

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Emch's Introduction to Projective Geometry and its Application 8vo, 2 50

Fiske's Functions of a Complex Variable Svo, 1 00

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Mafhematical Monographs. Edited by Mansfield Merriman and Robert

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Bacon's Forge Practice 12mo.

Baldwin's Steam Heating for Buildings 12mo,

Ban* and Wood's Kinematics of Machinery 8vo,

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* " " " Abridged Ed 8vo,

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* Burr's Ancient and Modem Engineering and the Isthmian Canal 8vo,

Carpenter's Heating and Ventilating Buildings 8vo,

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Treatise on Toothed Gearing 12mo,

Dingey's Machinery Pattern Making 12mo,

Durley's Kinematics of Machines 8vo,

Flanders's Gear-cutting Machinery Large 12mo,

Flather's Dynamometers and the Measurement of Power 12mo,

Rope Driving 12mo,

Gill's Gas and Fuel Analysis for Engineers 12mo,

Goss's Locomotive Sparks 8vo,

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Hering's Ready Reference Tables (Conversion Factors') 16mo, mor.

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Hutton's Gas Engine 8vo,

Jamison's Advanced Mechanical Drawing 8vo,

Elements of Mechanical Drawing 8vo,

Jones's Gas Engine 8vo.

Machine Design :

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1 50

2 50

3 00

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Leonard's Machine Shop Tools and Methods 8vo,

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Mahan's Industrial Drawing. (Thompson.) 8vo,

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Richards's Compressed Air 12mo,

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Church's Mechanics of Engineering 8vo,

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Keep's Cast Iron Svo,

* King's Elements of the Mechanics of Materials and of Power of Trans-

mission Svo,

Lanza's Applied Mechanics Svo,

Lowe's Paints for Steel Structures 12mo.

Maire's Modem Pigments and their Vehicles 12mo,

$1 25

5 00

2 00

3 00

2 50

4 00

5 00

4 00

1 50

3 50

2 50

12 50

3 50

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2 00

3 00

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3 00

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1 00

3 00

1 50

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1 25

8 00

2 00

3 60

2 50

1 50

3 00

Maurer's Technical Mechanics 8vo,

Merriman's Mechanics of Materials 8vo,

* Strength of Materials 12mo.

Metcalf 's Steel. A Manual for Steel-users 12mo,

* Murdock's Strength of Materials 12mo.

Sabin's Industrial and Artistic Technology of Paint and Varnish 8vo.

Smith's (A. W.) Materials of Machines ] 2mo,

* Smith's (H. E.) Strength of Material 12mo.

Thurston's Materials of Engineering 3 vols., 8vo.

Part I. Non-metallic Materials of Engineering 8vo,

Part II. Iron and Steel 8vo.

Part III. A Treatise on Brasses, Bronzes, and Other Alloys and their
Constituents 8vo,

* Waterbury's Laboratory Manual for Testing Materials of Construction.

12mo,

Wood's (De V.) Elements of Analytical Mechanics 8vo.

Treatise on the Resistance of Materials and an Appendix on the

Preservation of Timber 8v<^ 2 00

Wood's (M. P.) Rustless Coatings* Corrosion and Electrolysis of Iron and

Steel 8vo. 4 00

STEAM-ENGINES AND BOILERS.

Berry's Temperiature-entrppy Diagram. Third Edition Revised and En-
larged 12mo.

Camot's Reflections on the Motive Power of Heat. (Thurston.) 12mo,

Chase's Art of Pattern Making 12mo,

Creighton's Steam-engine and other Heat Motors Svo.

Dawson's "Engineering" and Electric Traction Pocket-book. .. . lOmo. mor.

* Gebhardt's Steam Power Plant Engineering Svo,

Goss's Locomotive Performance Svo,

Hemenway's Indicator Practice and Steam-engine Economy 12mo,

Hirshfeld and Barnard's Heat Power Engineering. (In Press.)

Hutton's Heat and Heat-engines Svo,

Mechanical Engineering of Power Plants Svo,

Kent's Steam Boiler Economy Svo,

Kneass's Practice and Theory of the Injector Svo,

MacCord's Slide-valves Svo,

Meyer's Modem Locomotive Construction 4to.

Miller, Berry, and Riley's Problems in Thermodynamics Svo, paper.

Moyer's Steam Turbine Svo,

Peabody's Manual of the Steam-engine Indicator 12mo,

Tables of the Properties of Steam and Other Vapors and Temperature-
Entropy Table Svo.

Thermodynamics of the Steam-engine and Other Heat-engines. . . . Svo.

* Thermodynamics of the Steam Turbine Svo,

Valve-gears for Steam-engines , Svo,

Peabody and Miller's Steam-boilers Svo,

* Perkins's Introduction to General Thermodynamics 12mo.

Pupin's Thermodynamics of Reversible Cycles in Gases and Saturated Vapors.

(Osterberg.) ; 12mo,

Reagan's Locomotives: Simple, Compound, and Electric. New Edition.

Large 12mo,

Sinclair's Locomotive Engine Running and Management 12mo.

Smart's Handbook of Engineering Laboratory Practice 12mo,

Snow's Steam-boiler Practice Svo.

Spangler's Notes on Thermodynamics 12mo,

Valve-gears Svo,

Spangler, Greene, and Marshall's Elements of Steam-engineering Svo,

Thomas's Steam-turbines Svo,

Thurston's Handbook of Engine and Boiler Trials, and the Use of the Indi-
cator and the Prony Brake Svo,

Manual of Steam-boilers, their Designs Construction, and Operation Svo,

Manual of the Steam-engine 2 vols., Svo,

Part I. History, Struccure, and Theory Svo,

Part II. Deagn, Construction, and Operation Svo,

2 50

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10 00

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10 00

6 00

Wehrenfennig's Analsrsis and Softening of Boiler Peed-water. (Patterson.)

8vo, $4 00

Weisbach's Heat, Steam, and Steam-engines. (Du Bois.) 8vo, 5 00

Whitham's Steam-engine Design 8vo. 6 00

Wood's Thermodynamics, Heat Motors, and Refrigerating Machines. . .8vo, 4 00

MECHANICS PURE AND APPLIED.

Church's Mechanics of Engineering 8vo, 6 00

Mechanics of Pluids (Being Part IV of Mechanics of Engineering). . 8vo, 3 00

* Mechanics of Internal Work 8vo, 1 50

Mechanics of Solids (Being Parts I, II, III of Mechanics of Engineering).

8vo, 4 50

Notes and Examples in Mechanics 8 vo, 2 00

Dana's Text-book of Elementary Mechanics for Colleges and Schools .12mo, 1 50
Du Bois's Elementary Principles of Mechanics:

Vol. I. Kinematics 8vo, 3 50

Vol. II. Statics Svo. 4 00

Mechanics of Engineering. Vol. I Small 4to, 7 50

Vol. II Small 4to, 10 00

* Greene's Structural Mechanics Svo, 2 50

* Hartmann's Elementary Mechanics for Engineering Students 12mo, 1 25

James's Kinematics of a Point and the Rational Mechanics of a Particle.

Large 12mo. 2 00

* Johnson's (W. W.) Theoretical Mechanics 12mo, 3 00

* King's Elements of the Mechanics of Materials and of Power of Trans-

mission Svo, 2 50

Lanza's Applied Mechanics Svo, 7 50

* Martin's Text Book on Mechanics. Vol. I, Statics 12mo, 1 25

* Vol. II. Kinematics and Kinetics 12mo, 1 50

* Vol. III. Mechanics of Materials 12mo, 1 50

Maurer's Technical Mechanics Svo. 4 00

* Merriman's Elements of Mechanics 12mo, 1 00

Mechanics of Materials Svo, 5 00

* Michie's Elements of Analytical Mechanics Svo, 4 00

Robinson's Principles of Mechanism Svo, 3 DO

Sanborn's Mechanics Problems Large 12mo, 1 50

Schwamb and Merrill's Elements of Mechanism Svo, 3 00

Wood's Elements of Analytical Mechanics Svo, 3 00

Principles of Elementary Mechanics 12mo, 1 25

MEDICAL.

* Abderhalden's Physiological Chemistry in Thirty Lectures. (Hall and

Defren.) Svo, 5 00

von Behring's Suppression of Tuberculosis. (Bolduan.) 12mo, 1 00

* Bolduan's Immune Sera 12mo, 1 50

Bordet's Studies in Immunity. (Gay.) Svo, 6 00

* Chapin's The Sources and Modes of Infection Large 12mo, 3 00

Davenport's Statistical Methods with Special Reference to Biological Varia-
tions 16mo. mor. 1 50

Ehrlich's Collected Studies on Immunity. (Bolduan.) Svo, 6 00

* Fischer's Nephritis Large 12mo, 2 50

* Oedema Svo, 2 00

* Physiology of Alimentation Large 12mo, 2 00

* de Fursac's Manual of Psychiatry. (Rosanoff and Collins.) . . . Large 12mo, 2 50

* Hammarsten's Text-book on Physiological Chemistry. (Mandel.) Svo, 4 00

Jackson's Directions for Laboratory Work in Physiological Chemistry. .Svo. 1 26

Lassar-Cohn's Praxis of Urinary Analysis. (Lorenz.) 12mo, 1 00

* Lauffer's Electrical Injuries 16mo, 50

Mandel's Hand-book for the Bio-Chemical Laboratory 12mo. 1 50

* Nelson's Analysis of Drugs and Medicines l2mo, 3 00

* Pauli's Physical Chemistry in the Service of Medicine. (Fischer. )..12mo, 1 25

* Pozzi-Escot's Toxins and Venoms and their Antibodies. (Cohn.). . 12mo, 1 00

Rostoski's Serum Diagnosis. (Bolduan.) 12mo, 1 00

Ruddiman's Incompatibilities in Prescriptions Svo. 2 00

Whys in Pharmacy 12mo, I 00

Salkowski's Physiological and Pathological Chemistry. (Omdorff.) .. ..Svo, 2 50

* Satterlee's Outlines of Human Embryology 12mo. $1 25

Smith's Lecture Notes on Chemistry for Dental Students 8vo, 2 50

* Whipple's Tyhpoid Fever Large 12mo, 3 00

* Woodhull's Military Hygiene for Officers of the Line Large 12mo. 1 50

* Personal Hy^ene 12mo, 1 00

Worcester and Atkinson's Small Hospitals Establishment and Maintenance,
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Hospital 12mo, 1 26

METALLURGY.

Betts's Lead Refining by Electrolysis 8vo, 4 00

Bolland's Encyclopedia of Founding and Dictionary of Foundry Terms used

in the Practice of Moulding 12mo. 3 00

Iron Foimder 12mo, 2 60

Supplement 12mo, 2 50

* Borchers's Metallurgy. (Hall and Hayward.) 8vo, 3 00

* Burgess and Le Chatelier's Measurement of High Temperatures. Third

Edition 8vo, 4 00

Douglas's Untechnical Addresses on Technical Subjects 12mo, 1 00

Goesel's Minerals and Metals: A Reference Book 16mo, mor. 3 00

* Iles's Lead-smelting 12mo, 2 50

Johnson's Rapid Methods for the Chemical Analysis of Special Steels,

Steel-making Alloys and Graphite Large 12mo, 3 00

Keep's Cast Iron 8vo, 2 60

Mstcalf 's Steel. A Manual for Steel-users 12mo, 2 00

Minet's Production of Aluminum and its Industrial Use. (Waldo.). . 12mo, 2 50

* Palmer's Foundry Practice Large 12mo, 2 00

* Price and Meade's Technical Analysis of Brass 12mo, 2 00

* Ruer's Elements of Metallography. (Mathewson.) 8vo, 3 00

Smith's Materials of Machines 12mo, 1 00

Tate and Stone's Foundry Practice 12mo, 2 00

Thurston's Materials of Engineering. In Three Parts 8vo, 8 00

Part I. Non-metallic Materials of Engineering, see Civil Engineering,
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Part II. Iron and Steel 8vo, 3 60

Part III. A Treatise on Brasses, Bronzes, and Other Alloys and their

Constituents 8vo, 2 50

U Ike's Modem Electrolytic Copper Refining 8vo, 3 00

West's American Foundry Practice 12mo, 2 50

Moulders' Text Book 12mo. 2 50

MINERALOGY.

* Browning's Introduction to the Rarer Elements 8vo,

Brush's Manual of Determinative Mineralogy. (Penfield.) 8vo,

Butler's Pocket Hand-book of Minerals 16rao, mor.

Chester's Catalogue of Minerals 8vo, paper,

Cloth,

* Crane's Gold and Silver 8vo,

Dana's First Appendix to Dana's New "System of Mineralogy". .Large 8vo,
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Large 8vo,

Manual of Mineralogy and Petrography. 12mo,

Minerals and How to Study Them 12mo,

System of Mineralogy Large 8vo, half leather.

Text-book of Mineralogy 8vo,

Douglas's Untechnical Addresses on Technical Subjects 12mo,

Eakle's Mineral Tables 8vo,

* Eckel's Building Stones and Clays 8vo,

Goesel's Minerals and Metals: A Reference Book 16mo, mor.

* Groth's The Optical Properties of Crystals. (Jackson.) 8vo,

Groth's Introduction to Chemical Crystallography (Marshall) 12mo,

* Hayes's Handbook for Field Geologists , 16mo, mor.

Iddings's Igneous Rocks 8vo,

R6ck Minerals 8vo,

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Johftnnsen's Determination of Rock-forming Minerals in Thin Sections. 8vo.

With Thumb Index $5 00

* Martin's Liaboratory Guide to Qualitative Analysis with the Blow-

pipe , 12mo, 60

Merrill's Non-meUllic Minerals: Their Occurrence and Uses 8vo. 4 00

Stones for Building and Decoration 8vo, 5 00

* Penfield's Notes on Determinative Mineralogy and Record of Miners* Tests.

8vo. paper, 50
Tables of Minerals, Including the Use of Minerals and Statistics of

Domestic Production 8vo, 1 00

* Pirsson's Rocks and Rock Minerals 12mo, 2 50

* Richards's Synopsis of Mineral Characters 12mo, mor. 1 25

* Ries's Clays: Their Occurrence, Properties and Uses 8vo, 5 00

* Ries and Leighton's History of the Clay-working industry of the United

States 8vo. 2 50

* Rowe's Practical Mineralogy Simplified 12mo, 1 25

* Tillman's Text-book of Important Minerals and Rocks 8vo, 2 00

Washington's Manual of the Chemical Analysis of Rocks 8vo, 2 00

MINING.

* Beard's Mine Gases and Explosions Large 12mo, 3 00

* Crane's Gold and Silver 8vo, 5 00

* Index of Mining Engineering Literature 8vo, 4 00

* 8vo, mor. 5 00

* Ore Mining Methods 8vo, 3 00

* Dana and Saunders's Rock Drilling 8vo, 4 00

Douglas's Un technical Addresses on Technical Subjects 12mo, 1 00

Eissler's Modem High Explosives 8vo. 4 00

* Gilbert Wightman and Saunders's Subways and Tunnels of New York. 8vo, 4 00

Goesel's Minerals and Metals: A Reference Book 16mo, mor. 3 00

Ihlseng's Manual of Mining. 8vo, 5 00

* Iles's Lead Smelting 12mo. 2 50

* Peele's Compressed Air Plant Svo. 3 50

Riemer's Shaft Sinking Under Difficult Conditions. (Coming and PeeIe.)8vo, 3 00

* Weaver's Military Explosives Svo, 3 00

Wilson's Hydraulic and Placer Mining. 2d edition, rewritten 12mo, 2 50

Treatise on Practical and Theoretical Mine Ventilation 12mo. 1 25

SANITARY SCIENCE.

Association of State and National: Food and Dairy Departments, Hartford

Meeting, 1906 8vo,

Jamestown Meeting. 1907 8vo,

* Bashore's Outlines of Practical Sanitation 12mo,

Sanitation of a Country House » 12mo,

Sanitation of Recreation Camps and Parks 12mo.

* Chapin's The Sources and Modes of Infection Large 12mo,

Polwell's Sewerage. (Designing, Construction, and Maintenance.) Svo,

Water-supply Engineering Svo,

Fowler's Sewage Works Analyses 12mo,

Fuertes's Water-filtration Works 12mo,

Water and Public Health 12mo,

Gerhard's Guide to Sanitary Inspections 12mo,

* Modem Baths and Bath Houses Svo,

Sanitation of Public Buildings 12mo,

* The Water Supply, Sewerage, and Plumbing of Modem City Buildings.

Svo,

Hazen's Clean Water and How to Get It Large 12mo,

Filtration of Public Water-supplies Svo,

* Kinnicutt, Winslow and Pratt's Sewage Disposal Svo,

Leach's Inspection and Analysis of Food with Special Reference to State

Control Svo,

Mason's Examination of Water. (Chemical and Bacteriological) 12mo,

Water-supply. (Considered principally from a Sanitary Standpoint).

Svo,

* Mast's Light and the Behavior of Organisms Large 12mo,

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* Merriman's Elements of Sanitary Engineering 8vo,

Ogden's Sewer Construction 8vo,

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* Ogden and Cleveland's Practical Methods of Sewage Disposal for Res-

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Parsons's Disposal of Municipal Refuse 8vo,

Prescott and Winslow's Elements of Water Bacteriology, with Special Refer-
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* Price's Handbook on Sanitation 12mo,

Richards's Conservation by Sanitation .8vo,

Cost of Cleanness 12mo,

Cost of Food. A Study in Dietaries 12mo,

Cost of Living as Modified by Sanitary Science 12mo,

Cost of Shelter 12mo,

Richards and Woodman's Air, Water, and Food from a Sanitary Stand-
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* Richey's Plumbers', Steam-fitters', and Tinners' Edition (Building

Mechanics' Ready Reference Series) 16mo, mor.

Rideal's Disinfection and the Preservation of Food 8vo,

Soper's Air and Ventilation of Subways 12mo,

Tumeaure and Russell's Public Water-supplies 8vo,

Venable's Garbage Crematories in America 8vo,

Method and Devices for Bacterial Treatment of Sewage 8vo,

Ward and Whipple's Freshwater Biology. (In Press.)

Whipple's Microscopy of Drinking-water 8vo,

* Typhoid Fever Large 12mo,

Value of Pure Water Large 12mo,

Winslow's Systematic Relationship of the Coccaceae Large 12mo,

MISCELLANEOUS.

* Burt's Railway Station Service 12mo,

* Chapin's How to Enamel 12mo,

Emmons's Geological Guide-book of the Rocky Mountain Excursion of the

International Congress of Geologists Large 8vo,

Ferrel's Popular Treatise on the Winds .' 8vo,

Fitzgerald's Boston Machinist 18mo,

* Fritz, Autobiography of John 8vo,

Gannett's Statistical Abstract of the World 24mo,

Haines's American Railway Management 12mo,

Hanausek's The Microscopy of Technical Products. (Winton) 8vo,

Jacobs's Betterment Briefs. A Collection of Published Papers on Or-
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Metcalfe's Cost of Manufactures, and the Administration of Workshops,. 8vo,

* Parkhurst's Applied Methods of Scientific Management 8vo,

Putnam's Nautical Charts 8vo,

Ricketts's History of Rensselaer Polytechnic Institute 1824-1894.

Large 12mo,

* Rotch and Palmer's Charts of the Atmosphere for Aeronauts and Aviators.

Oblong 4to,

Rotherham's Emphasised New Testament * . . Large 8vo,

Rust's Ex-Meridian Altitude, Azimuth and Star-finding Tables. ...•••• 8vo,

Standage's Decoration of Wood, Glass, Metal, etc r . 12mo,

Westermaier's Compendium of General Botany. (Schneider) 8vo,

Winslow's Elements of Applied Microscopy 12mo,

HEBREW AND CHALDEE TEXT-BOOKS.

Gesenius's Hebrew and Chaldee Lexicon to the Old Testament Scriptures.

^ (Tregelles.) Small 4to. half mor. 5 00

Green's Elementary Hebrew Grammar , 12mo, 1 25

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