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A TEXT BOOK 



ON 



GRAPHIC STATICS 



BY 



CHARLES W. MALCOLM, C. E., 

Assistant Professor of Structural Engineering, University of 

Illinois; Associate Member American Society of Civil 

Engineers ; Member Society for Promotion 

of Engineering Education. 



FIRST EDITION. SECOND THOUSAND 



\. 



NEW YORK AND CHICAGO 
THE MYRON C. CLARK PUBLISHING CO. 



LONDON 

£. & F. N. Spon. Ltd., S7 Haymarket 
1911 






' V J « « 






THE NEW YORK 

PUBLIC UBRARYl 

536085 

A8T0A, Lenox ANO 
TIU>CM fCtUNDAflOlltL 

H 1912 L 



COPYRIGHT. 1909, 

BY 

CHARLES W. MALCOLM. 















*•• 



• • • • • 






# • • • • 

• •• ••• • 

* • • • # • 









PREFACE. 



This text was prepared for the author's students in elemen- 
tary Graphic Statics and Stresses. It has not been the object of 
the writer to discover new principles, but rather to present the 
subject clearly and logically. Many texts on Graphic Statics are 
open to the criticism that they have a tendency to state principles 
and give constructions without proofs, and the student is there- 
fore compelled to memorize propositions and constructions without 
being taught the underlying principles. It has been the aim of 
the writer to give the proofs, together with full explanations of 
the constructions. No attempt has been made to give elaborate 
solutions which have little or no practical applications. Particular 
attention has been given to the order of presentation; and the 
text has been divided into chapters, articles, and numbered 
sections to facilitate easy reference. 

Most of the material in Part I and Part II has been used in 
printed form by the author's students for several years. It is 
hoped that the material in Part II, Part III, and Part IV will be 
found of assistance to the practicing engineer. 

The author gratefully acknowledges his indebtedness to 
Ira O. Baker, Professor of Civil Engineering, University of 
Illinois, for many valuable suggestions and criticisms. 

Urbana, Illinois, 

August, IS, 1909. 



CONTENTS. 



PART I. GENERAL PRINCIPLES. 

CHAPTER I. DEFINITIONS. 

PAOX 

Dynamics: Kinetics, Statics. Graphic Statics. Bigid Body. Best and 
Motion: Translation, Rotation. Force. Elements of a Force: 
Magnitude, Direction, Line of Action, Point of Application. Con- 
current and Non-concurrent Forces. Coplanar and Non-coplanar 
Forces. Equilibrium. Equivalence. Resultant. Equilibrant. Com- 
ponents. Composition of Forces. Resolution of Forces. Couple. 
Force and Space Diagrams. Notation 3 

CHAPTER II. CONCURRENT FORCES. 

Art. 1. Composition of Concurrent Forces 8 

Resultant of Two Concurrent Forces. Force Triangle and Force 
Polygon. Resultant of Any Number of Concurrent Forces. 

Art. 2. Resolution op Concurrent Forces 11 

To Resolve a Force into Two Components. 

Art. 3. Equilibrium of Concurrent Forces 12 

Solution of Problems in Equilibrium. 

CHAPTER III. NON-CONCURRENT FORCES. 

Art. 1. Composition of Non-concurrent Forces 16 

Non-concurrent Forces. Resultant of Two Non-parallel, Non-con- 
current Forces: Forces Intersecting Within the Limits of the 
Drawing ; Forces Intersecting Outside of the Limits of the Drawing. 
Resultant of Any Number of Non-concurrent Forces: Forces 
Having Intersections Within the Limits of the Drawing; Forces 
Having Intersections Outside of the Limits of the Drawing. Funic- 
ular Polygon: Pole, Pole Distance, Rays, Strings. Resultant of 
Any Number of Non-parallel, Non-concurrent Forces — Resultant a 
Couple. Resultant of Any Number of Parallel Forces. Closing of 
the Funicular Polygon. 

V 



VI CONTENTS. 

PAOB 

Art. 2. Resolution op Non-concurrent I'orces 23 

Resolution of a Force Into Three Non-parallel, Non-concurrent Com- 
ponents Having Known Lines of Action. Resolution of a Force 
Into Two Parallel Components Having Known Lines of Action: 
Line of Action of Known Forces Between the Lines of Action of 
the Two Components; Line of Action of Known Forces Outside of 
the Lines of Action of the Two Components. 

Art. 3. Equilibrium op Non-concurrent Forces 26 

Conditions for Equilibrium of Non-concurrent Forces. Use of 
Force and Funicular Polygons in Solving Problems in Equilibrium. 
Problem 1, Parallel Forces. Problem 2, Non-parallel Forces. Prob- 
lem 3, Non-parallel Forces. Inaccessible Points of Intersection. 
Problems. Special Method. Problems. 

Art. 4. SPECLA.L Constructions for Funicular Polygons 34 

Relation Between Different Funicular Polygons for the Same 
Forces. To Draw a Funicular Polygon Through Two Given Points. 
To Draw a Funicular Polygon Through Three Given Points. 

CHAPTER IV. MOMENTS. 

Art. 1. Moments op Forces and op Couples 38 

Moment of a Force. Positive and Negative Moments. Moment 
Area. Transformation of Moment Area. Moment of the Resultant 
of Two Concurrent Forces. Moment of the Resultant of Any Num- 
ber of Concurrent Forces. Moment of a Couple. Moment of the 
Resultant of Any System of Forces. Moment of a System of 
Forces. Condition of Equilibrium. 

Art. 2. Graphic Moments 43 

Graphic Representation of the Moment of a Force. Moment of 
Any System of Forces. Moment of Parallel Forces. Problems. 

CHAPTER V. CENTER OF GRAVITY OF AREAS. 

Geometrical Areas: Parallelogram, Triangle, Quadrilateral, Circular 
Sector, Circular Segment. Irregular Areas. Determination of the 
Centroid of Parallel Forces. Graphic Determination of the Cen- 
troid of a System of Parallel Forces. Center of Gravity of an 
Irregular Area 47 

CHAPTER VI. MOMENT OF INERTIA. 

Art. 1. Moment op Inertia op Parallel Forces 52 

Definition. Moment of Inertia of a System of Parallel Forces: 
Culmann's Method; Mohr's Method. Relation Between Moments 
of Inertia About Parallel Axes. Radius of Gyration. Graphic 
Determination of the Radius of Gyration. 



CONTENTS. VU 

PAGE 

Abt. 2. Moment of Inertia of Abeas 58 

Moment of Inertia of an Area. Approximate Method for Finding 
the Moment of Inertia of an Area. Radius of Gyration of an Area. 
Accurate Method for Finding the Moment of Inertia of an Area. 
Relation Between Moments of Inertia of an Area About Parallel 
Axes. Moment of Inertia of an Area Determined from the Area of 
the Funicular Polygon (Mohr's Method). 



PART IL FRAMED STRUCTURES- 
ROOF TRUSSES. 

CHAPTER VII. DEFINITIONS. 

Framed Structure. Types of Framed Structures: Complete Framed 
Structure, Incomplete Framed Structure, Redundant Framed Struc- 
ture. Roof Truss: Span, Rise, Pitch, Upper and Lower Chords, 
Web Members, Pin-connected and Riveted Trusses. Types of Roof 
Trusses: Fink Truss, Quadrangular Truss, Howe Truss, Pratt 
Truss, Cantilever Truss 65 

CHAPTER VIII. LOADS. 

Art. 1. Dead Load 69 

Construction of a Roof. Dead Load: Roof Covering; Purlins, 
Rafters, and Bracing; Roof Trusses; Permanent Loads Supported 
by the Trusses. 

Art. 2. Snow Load 72 

Art. 3. Wind Load 73 

Wind Pressure. Wind Pressure on Inclined Surfaces: Duchemin's. 
Hutton's, and The Straight Line Formulae. 

CHAPTER IX. REACTIONS. 

Art. 1. Reactions for Dead and Snow Loads 75 

Problem : Joint Loads, Reactions. Snow Load Reactions. Effective 
Reactions. 

Art. 2. Reactions for Wind Loads 78 

Wind Load Reactions: Truss Fixed at Both Supports, Reactions 
Parallel, Horizontal Components of Reactions Equal; One End of 
Truss Supported on Rollers, Rollers Under Leeward End of Truss, 
Rollers Under Windward End of Truss. 



• • • 



Vlll CONTENTS. 

CHAPTER X. STRESSES IN ROOF TRUSSES. 

PAGE 

Art. 1. Definitions and General Methods for Determining 

Stresses 84 

Definitions: Tension, Compression, Shear. General Methods for 
Determining Stresses: Algebraic Moments, Graphic Moments, Alge- 
braic Resolution, Graphic Resolution. Notation. 

Art. 2. Stresses by Algebraic Moments 86 

Method of Computing Stresses by Algebraic Moments. Problem. 

Art. 3. Stresses by Graphic Moments 91 

Method of Computing Stresses by Graphic Moments. Problem. 

Art. 4. Stresses by Algebraic Resolution 95 

Method of Computing Stresses by Algebraic Resolution. Problems : 
Forces at a Joint; Forces on One Side of a Section. 

Art. 5. Stresses by Graphic Resolution 99 

Method of Computing Stresses by Graphic Resolution. Problems: 
Loads on Upper Chord; Loads on Lower Chord. 

CHAPTER XL WIND LOAD STRESSES. 

Art. 1. Both Ends of Truss Fixed— Reactions Parallel 105 

Problem. 
Art. 2. Both Ends op Truss Fixed — ^Horizontal Components of 

Reactions Equal 108 

Problem. 
Art. 3. Leeward End of Truss on Rollers 110 

Problem. 
Art. 4. Windward End op Truss on Rollers Ill 

Problem. 

CHAPTER Xn. STRESSES IN CANTILEVER AND UNSYMMETRI- 

CAL TRUSSES— MAXIMUM STRESSES. 

Art. 1. Stresses in Cantilever and Unsymmetrical Trusses 114 

Stresses in a Cantilever Truss. Problem. Unsymmetrical Truss — 
Combined Stress Diagram. Problem. 

Art. 2. Maximum Stresses 118 

Problem 1. Problem 2. Maximum and Minimum Stresses. 

CHAPTER XIII. COUNTERBRACING. 

AiiT. 1. Definitions AND Notation 125 

Definitions. Counterbracing. Notation. 



CONTENTS. IX 

PAGE 

Abt. 2. Stresses in Trusses with Couxterbracinci — Separate 

Stress Diagrams 129 

Problem 1, Truss with Parallel Chords. Problem 2, Truss with 
Non-parallel Chords. 

Art. 3. Stresses in Trusses with Counterbracing — Combined 

Stress Diagram 139 

Truss with Parallel Chords. Truss with Non-parallel Chords. 

CHAPTER XIV. THREE-HINGED ARCH. 

Definition. Reactions Due to a Single Load. Reactions. Reactions and 
Stresses for Dead Load. Wind Load Stresses for Windward Seg- 
ment of Truss. Wind Load Stresses for Leeward Segment of Truss. 143 

CHAPTER XV. STRESSES IN A TRANSVERSE BENT OP A 

BUILDING. 

Construction of a Transverse Bent. Condition of Ends of Columns: 
Columns Hinged at Base and Top; Columns Hinged at Top and 
Fixed at Base; Columns Fixed at Top and Base. Dead and Snow 
Load Stresses. Graphic Method for Determining Wind Load Reac- 
tions. Wind Load Stresses — Columns Hinged at Base. Wind Load 
Stresses — Columns Fixed at Base 150 

CHAPTER XVI. MISCELLANEOUS PROBLEMS. 

Stresses in a Grand Stand Truss. Stresses in a Trestle Bent. Eccen- 
tric Riveted Connection 160 



PART III. BEAMS. 

CHAPTER XVII. BENDING MOMENTS, SHEARS, AND DEFLEC- 
TIONS IN BEAMS FOR FIXED LOADS. 

Art. 1. Bending Moments and Shears in Cantilever, Simple and 

Overhanging Beams 167 

Definitions. Bending Moment and Shear Diagrams for a Cantilever 
Beam: Cantilever Beam with Concentrated Loads; Cantilever 
Beam with Uniform Load. Bending Moment and Shear Diagrams 
for a Simple Beam: Simple Beam with Concentrated Loads; 
Simple Beam with Uniform Load. Bending Moment and Shear 
Diagrams for an Overhanging Beam : Overhanging Beam with Con- 
centrated Loads; Overhanging Beam with Uniform Load. 



X CONTENTS. 

PAGE 

Art. 2. Graphic Method for Determining Deflections in Beams. . 178 
Explanation of Graphic Method — Constant Moment of Inertia. 
Practical Application. Deflection Diagram — Variable Moment of 
Inertia. 

Art. 3. Bending Moments, Shears, and Deflections in Restrained 

Beams 187 

Definitions. Bending Moment, Shear, and Deflection Diagram for a 
Cantilever Beam. Bending Moment, Shear, and Deflection Diagram 
for a Beam Fixed at One End and Supported at the Other. Sim- 
plifled Construction. Bending Moment, Shear, and Deflection Dia- 
grams for a Beam Fixed at Both Ends. Algebraic Formulae. 

CHAPTER XVIII. MAXIMUM BENDING MOMENTS AND SHEARS 

IN BEAMS FOR MOVING LOADS. 

Beam Loaded with a Uniform Load: Maximum Bending Moment; 
Maximum Shear. Beam Loaded with a Single Concentrated Load: 
Maximum Bending Moment; Maximum Shear. Beam Loaded with 
Concentrated Moving Loads: Position for Maximum Moment; 
Position for Maximum Shear 199 



PART IV. BRIDOES. 

CHAPTER XIX. TYPES OF BRIDGE TRUSSES. 

Through and Deck Bridges. Types of Bridge Trusses: Warren, Howe, 
Pratt, Baltimore, Whipple, Camels-Back, Parabolic Bowstring, 
Petit. Members of a Truss: Main Trusses, Lateral Bracing, 
Portals, Knee-braces and Sway Bracing, Floor System, Pedestals, 
Connections 209 

CHAPTER XX. LOADS. 

Art, 1. Dead Load 214 

Weights of Highway Bridges. Weights of Railroad Bridges. 

Art 2. Live Load 215 

Live Load for Light Highway Bridges. Live Load for Interurban 
Bridges. Live Load for Railroad Bridges: Uniform Load; Con- 
centrated Wheel Loads; Equivalent Uniform Load. 

Art. 3. Wind Load 219 

Per Linear Foot of Span. Per Square Foot of Surface. 



CONTENTS. XI 

CHAPTER XXL STRESSES IN TRUSSES DUE TO UNIFORM 

LOADS. 

PAGE 

Art. 1. Stresses in a Warren Truss by Graphic Resolution 221 

Problem. Dead Load Stresses. Live Load Stresses : Chord Stresses ; 
Web Stresses; Maximum and Minimum Live Load Web Stresses. 
Maximum and Minimum Dead and Live Load Stresses: Chord 
Stresses; Web Stresses. Loadings for Maximum and Minimum 
Stresses. Simplified Construction for Live Load Web Stresses. 

Art. 2. Stresses in a Pratt Truss by Graphic Resolution 228 

Problem. Dead Load Stresses. Live Load Stresses : Chord Stresses ; 
Web Stresses; Maximum and Minimum Live Load Web Stresses. 
Maximum and Minimum Dead and Live Load Stresses: Chord 
Stresses; Web Stresses. Loadings for Maximum and Minimum 
Stresses. Howe Truss. 

Art. 3. Stresses by Graphic Moments and Shears 238 

Problem. Dead Load Chord Stresses. Live Load Chord Stresses. 
Dead Load Web Stresses. Live Load Web Stresses. Maximum and 
Minimum Dead and Live Load Stresses: Chord Stresses; Web 
Stresses. 

Art. 4. Stresses in a Bowstring Truss — Triangular Web Bracing. 243 
Problem. Chord Stresses: Dead Load Chord Stresses; Live Load 
Chord Stresses. Web Stresses: Dead Load Web Stresses; Live 
Load Web Stresses. Maximum and Minimum Dead and Live Load 
Stresses: Chord Stresses; Web Stresses. 

Art. 5. Stresses in a Parabolic Bowstring Truss 247 

Problem. Chord Stresses: Dead Load Chord Stresses; Live Load 
Chord Stresses. Web Stresses: Dead Load Web Stresses; Live 
Load Web Stresses. Maximum and Minimum Dead and Live Load 
Stresses: Chord Stresses; Web Stresses. Proof of Construction 
Shown in Fig. 129: Stresses in Diagonals; Stresses in Verticals. 

Art. 6. Wind Load Stresses in Lateral Systems 255 

Upper Laterals: Chord Stresses; Web Stresses. Lower Laterals: 
Chord Stresses, Fixed Load, Moving Load; Web Stresses, Fixed 
Load, Moving Load. Maximum and Minimum Stresses in Upper 
Laterals. Maximum and Minimum Stresses in Lower Laterals. 

Art. 7. Stresses in Trusses with Parallel Chords by the Method 

OP Coefficients 259 

Algebraic Resolution — Method of Coefficients. Loading for Maxi- 
mum and Minimum Live Load Stresses. Conclusions. Simplified 
Method. Coefficients and Stresses in a Warren Truss. Coefficients 
for a Pratt Truss. Coefficients for a Baltimore Truss. 



Xll CONTENTS. 

CHAPTER XXII. INFLUENCE DIAGRAMS, AND POSITIONS OF 
ENGINE AND TRAIN LOADS FOR MAXIMUM MOMENTS, 

SHEARS, AND STRESSES. 

PAGE 

Influence Diagrams. Position of Loads for a Maximum Moment at 
Any Point in a Beam, or at Any Joint of the Loaded Chord of a 
Truss with Parallel or Inclined Chords. Position of Loads for a 
Maximum Moment at Any Joint of the Unloaded Chord of a Truss 
with Parallel or Inclined Chords. Position of Loads for a Maxi- 
mum Moment at a Panel Point of a Truss with Subordinate 
Bracing. Position of Loads for a Maximum Shear at Any Point 
in a Beam. Position of Loads for a Maximum Shear in Any Panel 
of a Truss with Parallel or Inclined Chords. Position of Loads for 
a Maximum Stress in Any Web Member of a Truss with Inclined 
Chords. Position of Loads for a Maximum Floorbeam Reaction. 267 



CHAPTER XXIII. MAXIMUM MOMENTS, SHEARS, AND STRESSES 

DUE TO ENGINE AND TRAIN LOADS. 

Art. 1. Maximum Moments, Sheaes, and Stresses in Any Partic- 
ular Girder or Truss 285 

Maximum Flange Stresses and Shears in a Plate Girder: Flange 
Stresses; Shears. Maximum Chord and Web Stresses in a Pratt 
Truss: Chord Stresses; Web Stresses. 

Art. 2. Maximum Moments, Shears, and Stresses in Girders and 

Trusses op Various Types and Spans 295 

Load Line and Moment Diagram. Application of Diagrams in Fig. 
149 to Determining Maximum Moments in Plate Girders, or at 
Joints of the Loaded Chord of a Truss with Parallel or Inclined 
Chords. Application of Diagrams in Fig. 149 to Determining 
Maximum Moments at Panel Points in the Unloaded Chord of a 
Truss with Parallel or Inclined Chords. Application of Diagrams 
in Fig. 149 to Determining Maximum Shears: Maximum Shears 
in Beams and Girders; Maximum Shears in Trusses. Application 
of Diagrams in Fig. 149 to Determining Maximum Web Stresses in 
Trusses with Inclined Chords. Determination of Maximum Stresses 
in a Trass with Subordinate Bracings— Petit Truss. 



GRAPHIC STATICS. 



INTRODUCTION. 

This text will treat oif the general principles of Graphic 
Statics, and of the application of these principles to the solution 
of some of the problems of especial interest to the Civil and 
Structural Engineer. For convenience, the subject will be divided 
into four parts as follows: 

Part I. General Principles. 

Part II. Framed Structures — Roof Trusses. 

Part III. Beams. 

Part IV. Bridges. 



PART I. 



GENERAL PRINCIPLES. 



CHAPTER I. 



DEFINITIONS. 



1. Dynamics is the science that treats of the action of 
forces upon a body at rest or in motion. Its two main 
branches are kinetics and statics. 

Kinetics treats of the motion of bodies and of the laws gov- 
erning the production of motion by forces. 

Statics treats of the action of forces under such conditions 
that no change of motion is produced in the bodies acted 
upon. It is therefore the science of equilibrium — the science 
by the aid of which are determined the forces necessary to 
maintain a body in its state of rest or motion, notwithstand- 
ing disturbing tendencies. 

2. Graphic Statics has for its object the deduction of the 
principles of statics and the solution of statical problems by 
means of geometrical constructions. 

3. Rigid Body. A rigid body is a body which is incapable 
of change in shape or size when acted upon by forces. Such 
a body is only imaginary, in that all solids possess elasticity 
to a greater or lesser degree. Many solids, however, closely 
approximate a condition of rigidity and for practical purposes 
may be considered rigid if the forces acting upon them are not 
great enough to cause rupture. A well designed and properly 

3 



DEFINITIONS. 



Chap, I. 



constructed truss approximates a rigid body, in that it acts as 
a whole to resist external forces. 

Particle or Point, A particle or point is the smallest con- 
ceivable rigid body. 

Every problem in statics presupposes the existence of a 
rigid body or particle upon which the forces act. 

4. Rest and Motion. Rest is the relation existing between 
two particles when a line joining them does not change either 
in length or in direction. Motion is the relation existing between 
two particles when the line connecting them changes either in 
length or in direction. 

There are two kinds of motion, viz.: translation and 
rotation. 

Translation. Translation is the motion corresponding to the 
change in length of the line connecting two particles. The 
motion of a body is translation when every point in the body 
travels in a straight line. 

Rotation. Rotation is the motion of a body when all points 
in the body, that change their positions at all, describe con- 
centric circles in parallel planes. The common normal to 
these planes, which contains the centers of all the circles, is 
called the axis of rotation. If the axis of rotation is fixed in 
position, the motion is pure rotation. 

There may be compound translation, as that of a body 
sliding across a moving car ; or compound rotation, as that of 
the earth revolving around its own axis and also about the 
sun; or combined translation and rotation, as that of a ball 
rolling along a straight path. 

Two bodies may also be at rest with respect to each other 
and in motion with respect to a third body. Rest, then, means 
motionlcssness with respect to a definite body of reference, 
which in this work is understood to be the earth. 

5. Force. Force is an action exerted upon a body tending 
to change its state of rest or motion. 

6. Elements of a Force. In order that a force may be 
completely known, four characteristics of it, called elements, 
are essential, viz. : 



f 

I 



DEFINITIONS. 5 

1. Magnitude, 

2. Direction. 

3. Line of Action, 

4. Point of Application, 

7. Magnitude. The magnitude of a force is given by 
stating, numerically, the ratio of its effectiveness in producing 
motion to that of the unit force. The magnitude of a force 
may be expressed, graphically, by the length of a line, the 
magnitude of the force being the ratio of the given length to 
the unit length. 

8. Direction. Direction is the specification as to which 
of the two ways along the line the force tends to produce 
motion. Direction is expressed, graphically, by an arrow 
placed on the line representing the force, indicating which 
way the force tends to produce motion. 

9. Line of Action. The line of action of a force is the 
path along which the force tends to produce motion. The 
line of action is expressed, graphically, by the position of the 
line representing the force. 

10. Point of Application. The point of application is the 
place (considered as a point) where the force is brought to 
bear upon the body. The point of application is on the line 
of action of the force, and together with its position locates 
the line. 

11. Concurrent and Non-concurrent Forces. Concurrent 
forces are those whose lines of action meet in a point. Non-con- 
current forces are those whose lines of action do not meet in a 
point. Whether or not a given system of forces is concurrent 
or non-concurrent affects rotation only, since translation is 
entirely independent of the position of the point of application 
of the forces. 

12. Coplanar and Non-coplanar Forces. -Co planar forces 
are those whose lines of action lie in the same plane. Non-coplanar 
forces are those whose lines of action do not lie in the same 
plane. Except where statements are of a general character, 
coplanar forces only will be treated in this work. 

13. Equilibrium. A system of two or more forces is in 



6 DEFINITIONS. Cha 

equilibrium when the combined effect of the forces prodi 
no change in the body with respect to its state of rest 
motion. 

14. Equivalence. Two forces or systems of forces are 
equivalent when they have identical effects upon the body 
acted upon, with respect to its state of rest or motion. 

15. Resultant. The resultant of a system of forces is the 
simplest system which is equivalent to the given system. 
Usually the given system is equivalent to a single force, 
although this is not always the case. 

16. Equilibrant. A single force, or the simplest system, 
which will exactly neutralize the effect of a system of forces, 
is called the equilibrant of the system. The equilibrant is 
numerically equal to the resultant, but acts in an opposite 
direction. 

17. Components. Any one of a system of forces having 
a given force for its resultant is called a component of that 
force. It is evident that a force may have any number of 
components. 

18. Composition of Forces. Composition of forces is the 
process of finding, for a given system of forces, an equivalent 
system having a smaller number of forces than the given 
system. The process of finding a single force to replace the 
given system is the most important case of composition. 

19. Resolution of Forces. Resolution of forces is the pro- 
cess of finding, for a given system of forces, an equivalent 
system having a greater number of forces than the given 
system. The process of finding two or more forces which are 
equivalent to a given force is the most important case of reso- 
lution. 

20. Couple. A couple consists of two equal, parallel 
forces, opposite in direction, having different lines of action. 
The arm of the couple is the perpendicular distance between 
the lines of action of the two forces. 

21. Force and Space Diagrams. In the solution of prob- 
lems in graphic statics, it is usually most convenient to draw- 
two separate figures, one of which shows the forces in magni- 






NOTATION. 



tude and direction and the other in line of action. The former 
is called the force diagram, and the latter the space diagram, 

22. Notation. In solving problems, graphically, it is 
often convenient to draw both the force and the space dia- 
grams, and these diagrams are so related that for every line 
in one there is a corresponding line in the other. The solution 
is greatly facilitated if a convenient system of notation is 
adopted for the diagrams. 

In the force diagram, each line represents a force in mag- 
nitude and direction, and this line will be designated by plac- 
ing a capital letter at each ex- ^ 
tremity of the line. In the 
space diagram, the correspond- 
ing line represents the line of 
action of the force, and this 
line of action will be marked 
by the corresponding small 
letters, one on each side of the 
line. An arrow placed on the line indicates the direction of 
the force. The sequence of the letters representing the force 
also indicates the direction of the force ; thus, the force repre- 
sented by AB acts in a direction from A towards B. This 
system oiF notation is illustrated in Fig. i, AB representing the 
force in magnitude and direction, while its line of action is 
marked by the letters a b placed as shown. 



Fig. 1. 



CHAPTER 11. 



CONCURRENT FORCES. 



The subject matter given in this chapter will be divided 
into three articles, as follows: Art. i, Composition of Concur- 
rent Forces; Art. 2, Resolution of Concurrent Forces; Art. 3, 
Equilibrium of Concurrent Forces. 



Art. I. Composition of Concurrent Forces. 

23. Resultant of Two Concurrent Forces. Given the two 
concurrent forces represented in magnitude and direction by 
the lines BA and BC. It is required to find their resultant. 



Fig. 2. 

Let BA and BC (Fig. 2, a) represent two concurrent forces 
in magnitude and direction, acting at B. It is required to find 
their resultant BD. Complete the parallelogram of forces 
(Fig. 2, a) by drawing the line CD parallel to BA, and AD 
parallel to BC; then the line BD, connecting the points B and 
D, will represent the resultant of. the two forces in magnitude. 
The direction of the resultant will be as indicated by the arrow 
placed on the line BD. (The proof of the above will not be 

8 



Art. 1. COMPOSITION OF CONCUERENT FOECES. 9 

given here as it may be found in any elementary treatise on 
mechanics.) 

It is unnecessary to construct the entire force parallelo- 
gram, for, draw AB (Fig. 2, b) equal and parallel to BA (Fig. 
2, a), and BC equal and parallel to BC. Then AC will be equal 
and parallel to BD ; for, by construction, the triangle ABC is 
equal to the triangle ABD and has its corresponding sides 
parallel;. hence, AC is equal and parallel to BD the required 
resultant. Likewise, CA (Fig. 2, c) is the resultant of the two 
concurrent forces BA and BC. It should be noted that the two 
forces act around the triangle in the same direction, and that 
the resultant acts in a direction opposite to them. 

It is thus seen that any two concurrent forces may be com- 
bined into a single resultant force, and further that it is imma- 
terial in which order the forces are taken so long as they act 
in the same direction around the triangle. 

24. Force Triangle and Force Polygon. The triangle 
shown in either Fig. 2, b, or Fig. 2, c, is called a force triangle, 
and its principles form the basis of the science of graphic statics. 
If the figure has more than three sides it is called a force polygon, 

25. Resultant of Any Number of Concurrent Forces. The 
resultant of any number of concurrent forces may be found 
(i) by the method of the force triangle, or (2) by the method 
of the force polygon. 

( I ) Solution by Force Triangle. Let AB, AC, AD, AE, and 
AF (Fig. 3) represent in magnitude and 
direction a system of forces meeting at A. 
It is required to find their resultant R. 

The method explained in § 23 may 
be applied to any number of forces. 
Commencing at B (Fig. 3), the extrem- 
ity of the force AB, draw the line BG 
equal and parallel to the force AC; 
then AG, acting in the direction 
shown, is the resultant of the forces 
AB and AC. In like manner, from the 
point G, draw the line GH equal and ^^^ 3 




10 



CONCURRENT FORCES. 



Chap. II. 



parallel to the force AD ; then AH, acting in the direction shown, 
is the resultant of the forces AG and AD, or, since AG is the 
resultant of AB and AC, then AH is also the resultant of the 
forces AB, AC, and AD. In like manner, AI is the resultant of 
the forces AB, AC, AD, and AE ; and AJ, acting in the direction 
shown, is the required resultant R of the given system of forces 
AB, AC, AD, AE, and AF. 

(2) Solution by Force Polygon, Let AB, BC, CD, DE, and 
EF (Fig. 4) represent in magnitude and direction a system of 
forces meeting at O. It is required to find their resultant R. 




FiQ. 4. 

The given system of forces is represented in magnitude 
and direction in Fig. 4, b, and in line of action in Fig. 4, a. 
Commencing at any point A (Fig. 4, b), draw in succession the 
lines AB, BC, CD, DE, and EF, parallel respectively to the 
lines of action ab, be, cd, de, and ef, representing the given 
forces in magnitude and direction, each' force beginning at the 
end of the preceding one. Then AF, the line connecting the 
starting point of the force polygon with the end of the last 
force, represents in magnitude and direction the required 
resultant R. For, inserting the dotted lines AC, AD, and AE, 
which divide the polygon into triangles, it is evident that AC, 
acting in the direction shown, represents in magnitude and 
direction the resultant of the two forces AB and BC. Like- 
wise, AD represents in magnitude and direction the resultant 
of the forces AC and CD, or in other words, AD is the result- 



', 



-^rt. S. EESOLUTION OF CONCURBENT FORCES. 11 

ant of the forces AB, BC, jand CD. In like manner, AE is the 
resultant of the forces AB, BC, CD, and DE ; and AF, acting 
in the direction shown, represents in magnitude and direction 
the required resultant R of the given system of forces. The 
line of action of R passes through O, and its direction is as 
indicated by the arrow. 

It will be seen by referring to the force polygon that all 
the forces except the resultant act around the force polygon 
in the same direction, and that the resultant acts in a direction 
opposite to them. By taking the forces in a different order 
from that shown in F'ig. 4, it will be found that the order in 
which the forces are taken is immaterial, so long as the given 
forces act around the force polygon in the same direction ; 
since the positions of the initial and final points remain the 
same. 

If the points A and F coincide, the force polygon is said to 
be closed. 



Art. 2, Resolution of Concurrent Forces. 

26. It is readily seen from what has been given in the 
preceding sections that, to resolve a given force into any num- 
ber of components, it is only necessary to draw a closed 
polygon, one side of which represents in magnitude the given 
force and is parallel to its line of action ; then the other sides 
of the polygon will be parallel to, and will represent in mag- 
nitude, the components into which the given force ^is resolved. 
The given force will act around the force polygon in a direc- 
tion opposite to that of the components. 

27. To Resolve a Given Force Into Two Components. It 
is evident that this problem is indeterminate unless other con- 
ditions are imposed; as an infinite number of triangles may 
be drawn with one side of the given length. 

There are four cases of the resolution of a force into two 
components, corresponding to the four cases of the solution of 
a plane triangle, which may be stated as follows : 



12 



CONCURRENT FORCES. 



Chap. II. 



(i) It is required to resolve a given force into two com- 
ponents which are known in line of action only. 

(2) It is required to resolve a given force into two com- 
ponents which are known in magnitude only. 

(3) It is required to resolve a given force into two com- 
ponents, one of which is known in line of action only, and the 
other in magnitude only (two solutions). 

(4) It is required to resolve a given force into two com- 
ponents, one of which is completely known, while the other is 
completely unknown. 

The solution of the first case is given below, the other 
three cases being left for the student to solve. 

Case I, Let AB (Fig. 5) represent in magnitude and direc- 
tion the known force, and let ac and cb represent the lines of 
action of the two components meeting at O. It is required to 
find the unknown elements of the two components. 





. Fig 6. 

From the extremities of the known force AB (Fig. 5), draw 
the lines AC and CB, parallel respectively to ac and cb, inter- 
secting at the point C ; then AC and CB represent in magni- 
tude the two components. The directions of these components 
are shown by the arrows. 



Art. 3. Equilibrium of Concurrent Forces. 



28. Equilibrium of Concurrent Forces. A system of con- 
current forces is in equilibrium if the resultant of the system 
is equal to zero; since this is the condition which must exist if 
no motion takes place. Referring to Fig. 4, it is seen that if 



Art, 3. EQUILIBRIUM OF CONCURRENT FORCES. 13 

the resultant is zero, the force polygon is closed; hence the 
following proposition : If any system of concurrent forces is 
in equilibrium, the force polygon must close; and conversely, 
if the force polygon closes and the forces act in the same 
direction around the polygon, the system is in equilibrium. 
This is equivalent to the algebraic statements that the summa- 
tion of the horizontal components of the forces is equal to 
zero, and that the summation of the vertical components is 
equal to zero. 

If the system of forces is not in equilibrium, the resultant 
is represented in magnitude and direction by the closing line 
of the force polygon, this resultant acting around the force 
polygon in a direction opposite to the other forces. If a force 
having the same magnitude but acting in an opposite direction 
is substituted for the resultant, the system is then in equili- 
brium. It has been shown that the resultant acts around the 
force polygon in a direction opposite to the given forces; and 
therefore the force that will hold the given system in equi- 
librium, called the equilibrant, acts around the force polygon 
in the same direction as the given forces. 

29. Solution of Problems in Equilibrium. The fact that 
the force polygon for a system of concurrent forces is closed if 
that system is in equilibrium, furnishes a method for solving 
problems in equilibrium when all the elements of the forces 
are not known. 

The following problems illustrate the general principles 
employed in the solution of concurrent forces in equilibrium : 

Problem i. Given a system of five concurrent forces in equi- 
librium, three of which are completely known ; of the remain- 
ing two, one is known in magnitude only, and the other in line 
of action only. It is required to find the unknown elements of 
the two forces. Give two solutions. 

Problem 2. Given a system of five concurrent forces in equi- 
librium, three of which are completely known, the other two 
being known in line of action only. It is required to find the 
unknown elements of the two forces. 

Problem j. Given a system of five concurrent forces in equi- 



14 



CONCURRENT FORCES. 



Chap. II. 



librium, one of which is completely unknown. It is required 
to fully determine the unknown force. 

Problem 4. Given a system of five concurrent forces in equi- 
librium, three of which are completely known, the other two 
being known in magnitude only. It is required to find the 
unknown elements of the two forces. Give two solutions. 

The solution of the first problem is given below, the others 
being left for the students to solve. 

Problem i. Let the forces AB, BC, and CD (Fig. 6) be com- 
pletely known, let DE be known in magnitude only, and EA 
in line of action only. It is required to find the unknown ele- 
ments of the two forces. 





FiQ. 6. 

First construct the portion of the force polygon ABCD 
(Fig. 6), using as sides the known forces AB, BC, and CD, 
taking care that the forces act progressively around the force 
polygon. Complete the polygon by drawing the line EA 
through A, parallel to the known line of action of EA. Then, 
using D as a center and the known magnitude of DE as a 
radius, draw an arc of a circle intersecting EA at the point E. 
The lines DE and EA will then represent in magnitude the 
forces DE and EA, these forces acting around the force 
polygon in the same direction as the known forces AB, BC, 
and CD. The entire force polygon is ABCDE, a closed 
polygon; and since all the forces act around the force polygon 
in the same direction, they are in equilibrium. ABCDE' is 
also a true form of the force polygon, and gives correct values 



\ 



Art. S, EQUILIBRIUM OF CONCUREENT FORCES. 15 

for the unknown forces ; since all the conditions of the problem 
are fulfilled, i.e., the force polygon is closed, and the forces 
act continuously around the polygon. Another solution, which 
gives different values for the unknown forces, is indicated by 
the force polygon ABCDEi. 



CHAPTER III. 



NON-COXrURRENT FORCES. 



The subject matter given in this chapter will be divided 
into four articles, as follows: Art. i, Composition of Non- 
concurrent Forces; Art. 2, Resolution of Non-concurrent 
Forces; Art. 3, Equilibrium of Non-concurrent Forces; and 
Art. 4, Special Constructions for Funicular Polygons. 



Art. I. Composition of Non-concurrent Forces. 

30. Non- concurrent Forces. In many engineering prob- 
lems, the forces acting upon the body or structure do not meet 
at a point, but are applied at different points along the struc- 
ture. Such forces are non-concurrent, and their lines of action 
may, or may not, be parallel. 

31. Resultant of Two Non-parallel, Non-concurrent 
Forces. The determination of the resultant of two non- 
parallel, non-concurrent forces requires one of two somewhat 
different methods of solution, depending upon whether the 
given forces intersect inside or outside the limits of the drawing. 

(i) Forces Intersecting Within the Limits of the Drawing, 
If the two forces are not parallel, their lines of action must 
intersect at a point, which may be taken as the point of appli- 
cation of each force. The two forces may therefore be treated 
as concurrent forces, and their resultant may be determined 
as in § 23. 

(2) Forces Intersecting Outside of the Limits of the Draw- 
ing, Let AB and BC (Fig. 7) represent in magnitude and direc- 

16 



'Art. 1. 



COMPOSITION OF NON-CONCURRENT FORCES. 



17 



tion two non-parallel forces whose lines of action ab and be 
intersect outside of the limits of the drawing. It is required 
to find their resultant. 





Fig T. 

Draw the force polygon ABC (Fig. 7), and connect the 
points A and C by the closing line AC; then AC, acting as 
shown by the arrow, represents in magnitude and direction 
the resultant of the two forces AB and BC, its line of action 
being as yet unknown. To determine this line of action, an 
auxiliary construction is necessary. From any point O, draw 
the lines OA, OB, and OC connecting the point O with the 
points A, B, and C of the force polygon. These lines represent 
in magnitude and direction components into which the forces 
AB and BC may be resolved. Thus, AB is equivalent to the 
two forces represented in magnitude by AO and OB, acting 
in the directions shown by the arrows. Likewise, BC is 
equivalent to the two forces represented in magnitude and 
direction by BO and OC. To determine the lines of action of 
these components, start at any point on the line of action ab, 
and draw ao and ob parallel respectively to AO and OB. 
Prolong ob until it intersects the line of action be; and from 
the intersection of ob and be, draw the line oc parallel to OC 
to intersect the line ao. The lines ao, ob, bo, and oc are then 
the lines of action of the components AO, OB, BO, and OC, 
into which the original forces have been resolved. Now the 
forces represented by OB and BO are equal, have the same 
line of action, and act in opposite directions; hence they 
neutralize each other and may be omitted from the system, 



18 NON-CONCURRENT FORCES. ^^"P- ^^^^ 

thus leaving the two forces represented in magnitude and 
direction by AO and OC and in line of action by ao and oc, 
respectively. The resultant of these two forces, which have 
been shown equivalent to the given forces AB and BC, is the 
force represented in magnitude and direction by AC and in 
line of action by ac, drawn through the intersection of ao and 
oc, parallel to AC. 

The method explained above is of great importance and 
should be thoroughly understood by the student; as the prin- 
ciples employed in it will be of great use in solving succeeding 
problems. 

32. Resultant of Any Number of Non-concurrent, Non- 
parallel Forces. There are two methods in general use for 
finding the resultant of any number of non-concurrent forces. 
These methods embody the principles explained in § 31, and 
are merely applications of the method explained in that sec- 
tion. They depend upon whether the given forces have inter- 
sections (i) inside of the limits of the drawing, or {2) outside 
of the limits of the drawing. The first method to be described 
is limited in its application, but permits of a simpler construc- 
tion for some problems. The second method, however, is of 
more general use; as it may be employed for finding the 
resultant of parallel, as well as non-parallel, forces. 

(i) Forces Having Intersections Within the Limits of the 
Drawing, Let AB, BC, CD, and DE (Fig. 8) represent in mag- 
nitude and direction a system of four non-parallel, non-con- 
current forces ; and let ab, be, cd, and de, respectively, repre- 
sent their lines of action. It is required to find the resultant 
of the system of forces. 

Applying the method explained in (i) § 31, the resultant of 
the forces represented by the lines AB and BC (Fig. 8) is repre- 
sented in magnitude and direction by AC, acting as shown by 
the arrow, and in line of action by ac, drawn through the 
intersection of ab and be, parallel to AC. This resultant may 
then be combined with the force represented by CD, giving 
as their resultant the force represented in magnitude and direc- 
tion by AD and in line of action by ad, drawn through the 



Art. 1. 



COMPOSITION OF XOX-CONCUKRENT FORCES. 



19 



intersection of ac and cd, parallel to AD. In like manner, AD 
may be combined with the force DE, giving as their resultant 
the force represented in magnitude and direction by AE and 
in line of action by ae, drawn through the intersection of ad 
and de, parallel to AE. This last force AE represents in mag- 





Fia. 8. 

nitude and direction the resultant of the given system of forces 
AB, BC, CD, and DE, its line of action being ae. 

It should be borne in mind that AE does not represent the 
magnitude and direction of any actual force. By the resultant 
AE is meant a force which, if applied, would produce the same 
effect upon the body as the given forces. 

(2) Forces Having Intersections Outside of the Limits of 
the Drawing, Let AB, BC, CD, DE, and EF (Fig. 9) represent 




Fia. 0. 



in magnitude and direction a system of non-concurrent forces, 
and let ab, be, cd, de, and ef. respectively, represent their lines 



20 NON-CONCURRENT FORCES. Chap. Ill, 

of action. It is required to find the resultant of the given 
system of forces. 

The resultant of the given system of forces may be found 
by applying the method explained in ('2) § 31. Construct the 
force polygon ABCDEF, and draw the closing line AF. Then 
AF, acting in the direction shown by the arrow, represents in 
magnitude and direction the resultant of the given system of 
forces. To find its line of action, assume any point O, and 
draw the lines OA, OB, OC, OD, OE, and OF. Then con- 
struct the polygon whose sides are oa, ob, oc, od, oe, and of 
(see (2) § 31). Note that the two lines, which represent the 
components into which each force is resolved by the lines 
drawn from the point O to the extremities of that force, are 
respectively parallel to the two lines which .meet on the line 
of action on that force. To find the line of action of the 
resultant, prolong the extreme lines oa and of until they inter- 
sect, and through the point of intersection draw af parallel to 
AF. This is the line of action of the required resultant; for, 
all the components, into which the given forces are resolved 
by the lines drawn from the point O to the extremities of the 
forces, are neutralized (as shown by the arrows) except the 
two forces represented in magnitude and direction by AO and 
OF and in line of action by ao and of, respectively. The 
resultant of these two forces, which are equivalent to the g^ven 
system, is found in magnitude by completing the force triangle 
OAF and in direction by making the resultant act around the 
force triangle in a direction opposite to the other two forces. 
Since the line of action of the resultant must be on the line 
of action of each force, it must act through the common point 
of each line, viz : their point of intersection. 

33. Funicular Polygon. The polygon whose sides are oa, 
ob, oc, od, oe, and of (Fig. 9) is called the funicular polygon 
(also called the equilibrium polygon). 

Pole, The point O (Fig. 9) is called the pole of the force 
polygon. 

Pole Distance, The perpendicular distance from the pole to 



Art. 1. COAirOSlTlON OF NON-CONCUUKENT FORCES. 21 

the line representing the force in the force polygon is called the 
pole distance. 

Rays. The lines OA, OB, OC, OD, OE, and OF (Fig 9), 
drawn from the pole O to the extremities of the lines repre- 
senting the forces in the force polygon, are called rays. The 
rays terminating at the extremities of any side of the force 
polygon, represent in magnitude the two components which 
may replace the force represented by that side. 

Strings, The sides oa, ob, oc, od, oe, and of of the funicular 
polygon are called strings. The strings are parallel to the 
corresponding rays of the force polygon, and are the lines of 
action of the forces represented by the rays. 

Referring to the diagram (Fig. 9), it is seen that for each 
ray in the force polygon there is a string parallel to it in the 
funicular polygon; and further, that the two rays drawn to 
the extremities of any force in the force polygon are respect- 
ively parallel to the two strings which intersect on the line of 
action of that force. Keeping in mind these facts will greatly 
facilitate the construction of the funicular polygon. 

34. In both cases given in § 32, the resultant has been 
found to be a single force. This may not always be true, and 
the simplest system that will replace the given system may 
be a couple ; as will be shown in the following section. 

35. Resultant of Any Number of Non-parallel, Non-con- 
current Forces. — Resultant a Couple. Upon determining the 
resultant of a given system of forces, it may be found that the 
first and the last sides of the funicular polygon are parallel. If 
this is the case, the constructions shown in § 32 do not deter- 
mine the line of action of the resultant. Referring to Fig. 9, 
suppose the pole is taken on the line AF; then the first and 
the last strings of the funicular polygon are respectively 
parallel .to AO and OF, and are therefore parallel to each other. 
In this case the difficulty is avoided by taking the pole at some 
point not on the closing line AF. There is, however, one par- 
ticular case in which AO and OF will be parallel no matter 
where the pole is taken. Again referring to Fig. 9, it is seen 
that AO and OF will be parallel if the points A and F coincide. 



22 



NOX-COXCURRENT FORCES. 



Chap. Ill, 



These two rays will then represent equal and opposite forces, 
which cannot be combined into a simpler system unless their 
lines of action coincide. If their lines of action ao and of are 
coincident, the two forces, being equal and opposite in direc- 
tion, neutralize each other, and their resultant is equal to zero. 
If their lines of action are not coincident, the system reduces 
to a couple. Even if the lines of action of the forces repre- 
sented by AO and OF are coincident,' the forces may still be 
considered as a couple with an arm equal to zero; hence the 
following proposition : If the force polygon for any system 
of forces closes, the resultant is a couple. 

By changing the starting point of the funicular polygon, 
the lines of action of the forces represented by AO and OF 
will be changed ; and by taking a new pole, either their magni- 
tudes, or directions, or both their magnitudes and directions 
may be changed. Hence, it is seen that any number of couples 
may be found which are equivalent to each other; since they 
are equivalent to the same system of forces. 

36. Resultant of Any Number of Parallel Forces. Let 
AB, BC, CD, DE, and EF (Fig. 10) represent in magnitude 
and direction a system of parallel forces ; and let ab, be, cd, de, 
and ef represent their lines of action, respectively. It is 
required to find the resultant of the given system. 



b b 






a ' 
a 



d 
o* 



f 






-f -tAn 



J 



1 TT 'D 
I I I > 

I I I «L 
III' 

R« I A • 

• , 1 1 

Y • ' 

' ' I V 

ITi.iC 
• I 






V 









Pig. 10. 



Construct the force polygon ABCDEF, which in this case 
is a straight line; since the forces are parallel. Then AF, 
acting in the direction shown by the arrow, represents in 



/irt, ^. UESOLUTION OF NON-CONCURRENT FORCES. 23 

magnitude and direction the resultant of the given system of 
forces. To determine its line of action, assume any pole O, 
and draw the rays OA, OB, OC, OD, OE, and OF. Then 
construct the funicular polygon whose sides are oa, ob, oc, od, 
oe, and of, and prolong the extreme strings oa and of until 
they intersect. Through this point of intersection, draw af 
parallel to AF, which gives the line of action of the resultant. 
For, the given system of forces may be considered to be 
replaced by the forces represented in magnitude and direction 
by AO and OF and in line of action by ao and of; since the 
other forces represented by the rays neutralize each other. 
The resultant of these two forces is given in magnitude and 
direction by the closing line AF of the force triangle OAF, 
and in line of action by af, acting through the intersection of 
ao and of. 

37. Closing of the Funicular Polygon. The given system 
of forces represented in Fig. 9 has been shown to be equivalent 
to the two forces represented in magnitude and direction by 
AO and OF and in line of action by ao and of, the first and 
last strings of the funicular polygon. In general these lines of 
action are not parallel, but it may happen that they are parallel 
or that they coincide. In case they comcide, the funicular 
polygon is said to be closed. 



Art. 2. Resolution of Non-concurrent Forces. 

38. The problem of resolving a given force into two or 
more non-concurrent components is indeterminate unless 
additional data are given concerning the magnitudes and the 
lines of action of the required components. A given force 
may be resolved into three non-parallel, non-concurrent com- 
ponents, or into two parallel components, provided the lines 
of action of these two components are given. For a greater 
number of components the problem is indeterminate. 

39. Resolution of a Force Into Three Non-parallel, Non- 
concurrent Components Having Known Lines of Action. Let 



24 NOX-CONCUIIRENT FORCES. Chap. III. 

AB (Fig. ii) represent in magnitude and direction a given 
force, and let ab be its line of action. It is required to resolve 
this force into three components acting along the lines cb, do, 
and ad. 



Fio. 11. 

Since the given force AB may be assumed to act at any 
point in its line of action, let its point of application be taken at 
the intersection of ab and cb. Prolong the lines of action ad 
and dc until they intersect, and connect this point with the 
point of intersection of ab and cb. Resolve the given force 
AB into two components acting along ac and cb (§ 2y) ; then 
AC and CB, acting as shown by the arrows, will represent in 
magnitude and direction two components of AB. In like man- 
ner, resolve the force represented by AC, whose line of action 
is ac, into two components acting along the lines ad and dc. 
These two components are given in magnitude and direction 
by the lines AD and DC, drawn parallel respectively to ad 
and dc. Hence, since the given force AB is equivalent to the 
two forces AC and CB, and AC is equivalent to the two forces 
AD and DC; therefore, the force AB is equivalent to the 
three forces represented in magnitude and direction by AD, 
DC, and CB. 

If the line of action of the given force does not intersect 
any of the given lines of action within the limits of the draw- 
ing, the given force may be replaced by two components, each 
component then resolved by the above method, and the result- 
ing forces combined. 

40. Resolution of a Force Into Two Parallel Components 
Having Known Lines of Action. There are two special cases 



Art, 2, 



RESOLUTION OF NON-CONCURRENT FORCES. 



25 



of this problem depending upon whether the Hne of action of 
the given force is (i) between the lines of action of the two 
components, or (2) is outside of the lines of action of the two 
components. 

(i) Line of Action of Known Force Between the Lines of 
Action of the Two Components, Let AB (Fig. 12) represent in 
magnitude and direction the given force, and let ab be its line 
of action ; also let ac and cb be the lines of action of the two 
parallehcomponents. It is required to find the magnitudes and 
directions of the two components. 

The method explained in (2) § 31 may be used to find the 
resultant of two parallel forces; and conversely, it may be 
used to resolve a force into its two components. 



a 






^o 
a" 



^b 



tc 



I -. '-^ 



(a) 



B 



' ^ — 






(b) 



Fig. 12. 



Assume any pole O (Fig. 12), and draw the rays OA and 
OB. At any point on ab, the line of action of the force AB, 
draw the string ao parallel to AO, and ob parallel to OB. Pro- 
long the two strings until they intersect ac and cb; join these 
points of intersection by the string oc; and from the pole O, 
draw the ray OC parallel to the string oc, cutting the line AB 
at the point C. Then AC and CB, acting in the directions 
shown by the arrows, represent in magnitude and direction 
the two components whose lines of action are ac and cb, 
respectively. For, the diagram shown in Fig. 12, b is the 
force polygon for the two components AC and CB and their 
resultant AB ; conversely, AB is resolved into its two com- 
ponents AC and CB. 

A practical application of this case is the determination of 



26 NON-CONCURRENT FORCES. ^^^P- ^^^« 

the magnitudes of the two reactions of a simple beam loaded 
at any point with a single concentrated load. 

The line oc is called the closing string of the funicular poly- 
gon, and the ray OC is called the dividing ray of the force poly- 
gon ; since it divides the given force into its two components. 

(2) Line of Action of the Known Force Outside of the Lines 
of Action of the Two Components. The solution of this case will 
be left to the student. 



Art. 3. Equilibrium of Non-concurrent Forces. 

41. Conditions for Equilibrium of Non-Concurrent Forces. 
It has been shown that a system of concurrent forces is in 
equilibrium if the force polygon closes; as the resultant is then 
equal to zero. In order that a system of non-concurrent forces 
be in equilibrium, it is necessary that the force polygon close, 
but this single condition is not sufficient to insure equilibrium. 
For, referring to Fig. 9, it is seen that the given system of 
forces may be reduced to two forces represented in magnitude 
and direction by AO and OF and in lines of action by ao and 
of, respectively. For equilibrium, these two forces must be 
equal, must have the same line of action, and must act in 
opixosite directions. It is readily seen that the two forces are 
equal and act in opposite directions if the points A and F 
coincide; or in other words, if the force polygon closes. In 
order that they have the same line of action, the first and last 
strings, ao and of, of the funicular polygon must coincide. 
From the above, it is seen that for equilibrium of a system of 
non-concurrent forces : 

(i) The force polygon must close, 
(2) The funicular polygon must close. 
Therefore, a system of non-concurrent forces is in equilib- 
rium if both the force and funicular polygons close ; and con- 
versely, if both the force and funicular polygons close, the sys- 
tem is in equilibrium. For, if the force polygon closes, the 
two forces AO and OF (Fig. 9) are equal and opposite in 



Art. S, 



EQUILIBRIUM OF NON-CONCURRENT FORCES. 



27 



direction, and if the funicular polygon closes, they have the 
same line of action and neutralize each other. 

42. Use of Force and Funicular Polygons in Solving Prob- 
lems in Equilibrium. It has been shown in Art. 2 that the 
force and the funicular polygon construction is especially 
adapted to the solution of problems involving non-concurrent 
forces. The two conditions necessary for equilibrium of non- 
concurrent forces, viz. : that the force polygon must close and 
that the funicular polygon must close, furnish a convenient 
graphical method for the solution of problems in equilibrium. 
In order that such problems may be solved, it is necessary that 
a sufficient number of forces be completely or partly known to 
permit of the construction of the complete force and funicular 
polygons ; otherwise the problem is indeterminate. 

The general method of procedure is as follows: First con- 
struct as much of the force and funicular polygons as is pos- 
sible from the given data ; then complete these polygons, keep- 
ing in mind the facts that both polygons must close, and that 
the forces must act continuously around the force polygon. 

The application of the above principles to the solution of 
some of the most important cases arising in practice will now 
be given. 

43. Problem i. Parallel Forces. Given a system of par- 
allel forces in equilibrium, all being completely known except 
two, these two being known in line of action only. It is 
required to find the unknown elements of the two forces. 









t-.--*^ C 



o^ 



:^ 



O 

'e 



/ 
/ 



y 



e 



Q4^ €^ 



^>^ 



At t 

i I 



% 

♦ I 

I I 



Fig. 13. 



28 NON-CONCURRENT FORCES. Chap, III, 

Let the known forces be represented by the three loads AB, 
BC, and CD (Fig. 13), applied to the beam along the lines ab, 
be, and cd, and let the unknown forces be the supporting forces 
of the beam, EA and DE. 

Draw the portion of the force polygon ABCD containing 
the three known forces laid off consecutively. From the con- 
dition that the force polygon must close, the combined magni- 
tudes of the two supporting forces must be equal to DA and 
must act in the direction from D towards A. To determine the 
magnitude of each supporting force, take any pole O, and 
draw the rays OA, OB, OC, and OD. Then, commencing at 
any point on ea, the line of action of the left supporting force, 
construct the portion of the funicular polygon whose sides are 
oa, ob, OC, and od, drawn parallel respectively to the rays OA, 
OB, OC, and OD. Prolong od until it intersects de, and close 
the funicular polygon by drawing the line oe. Also, from the 
pole O, draw the ray OE parallel to the closing line oe, cutting 
DA at E. Then EA and DE, acting in an upward direc- 
tion, represent in magnitude and direction the two supporting 
forces. 

To thoroughly understand this solution, the student should 
follow through the constructions from the principle of the tri- 
angle of forces. Thus the force AB is resolved into two com- 
ponents represented in magnitude and direction by AO and 
OB and in line of action by ao and ob, respectively. In like 
manner, the force BC is resolved into the two components 
BO and OC, acting along the lines bo and oc ; and the force 
CD is resolved into the two components CO and OD, acting 
along the lines co and od, respectively. The two forces BO 
and OB neutralize each other; since they are equal, act in 
opposite directions, and have the same line of action. For the 
same reasons, EO neutralizes OE, and CO neutralizes OC. 
The three given forces are therefore equivalent to the two 
forces AO and OD, acting along the lines ao and od, respect- 
ively. Now in order to have equilibrium, the resultant of AO 
and the left supporting force EA must neutralize the resultant 
of OD and the right supporting force DE. The lines of action 



Art. S. 



EQUILIBRIUM OF NON-CONCURRENT FORCES. 



29 



of these two resultants must coincide in eo in order that they 
may neutralize each other. Since the resultant EO and the 
two forces AO and EA all intersect at a point, they must 
form a force triangle AOE (§ 23), of which AO is the known 
side. Then EA, acting in a direction from E towards A, rep- 
resents the left supporting force in magnitude and direction. 
In like manner, it may be shown that DE represents the right 
supporting force in magnitude and direction. 

44. Problem 2. Non-parallel Forces. Given a system of 
non-parallel forces in equilibrium, all completely known except 
two. Of these two, the line of action of one and the point of 
application of the other are known. It is required to find the 
unknown elements of the two forces. 

Let the known forces be the three wind loads AB, BC, and 
CD (Fig. 14) acting on the roof truss, and let the unknown 
forces be the two supporting forces of the truss. The line of 
action de of the right supporting force is given as vertical, and 
the point of application of the left supporting force ea is given 
at the left end of the truss. 





Fig. 14. 



Construct the portion of the force polygon ABCD, contain- 
ing the three known wind forces AB, BC, and CD laid off 
consecutively; and from any pole O, draw the rays OA, OB, 
OC, and OD. The side DE of the force polygon must be par- 
allel to the known direction de ; but as its magnitude is 
unknown, as is also the magnitude of the supporting force 
EA, the polygon cannot be closed. Since the only known 



30 



NON-COXCURREXT FORCES. 



Chap. III. 



point in the line of action of the left supporting force is its 
point of application at the left end of the truss, let the funicu- 
lar polygon be started at this point. Draw the strings oa, ob, 
oc, and od parallel respectively to the rays OA, OB, OC, and 
OD, keeping in mind the fact that the lines of action of the 
two components, into which each force is resolved by the rays, 
must intersect on the line of action of that force. Produce the 
string od until it intersects de, the line of action of the right 
supporting force, and close the funicular polygon by drawing 
the line oe. From the pole O, draw the ray OE parallel to 
oe, and from D draw the line DE parallel to* the known direc- 
tion de ; their intersection then determines the point E of the 
force polygon. Close the force polygon by drawing the line 
EA ; then DE and EA, acting in the directions shown by the 
arrows, represent in magnitude and direction the two support- 
ing forces. The line of action of the left supporting force is 
found by drawing the line ea parallel to EA, through its 
known point of application at the left end of the truss. 

45. Problem 3. Non-parallel Forces. Given a system of 
non-parallel forces in equilibrium, all completely known except 
three, these three being known in line of action only. It is 
required to determine the unknown elements of the three 
forces. 

Let AB, BC, and CD (Fig. 15) represent the forces com- 
pletely known, and let DE, EF, and FA represent the three 
forces which are known in line of action only. 





Pig. 15. 



Since the resultant of any two forces must pass through 



Art. 3. 



EQUILIBRIUM OF NON-CONCURRENT FORCES. 



31 



their point of intersection, let two of the unknown forces, 
whose lines of action are represented by ef and fa, be replaced 
by their resultant acting through their point of intersection. 
The problem then becomes identical with Problem 2, de being 
the line of action of one unknown force; and the intersection 
of ef and fa, the point of application of the other unknown 
force. The construction of Problem 2 having been made, and 
the resultant EA of the two forces found, the magnitudes of 
these forces, whose lines of action are represented by ef and fa, 
may be found by resolving this resultant into two components 
parallel to these known lines of action. In Fig. 15, EA repre- 
sents this resultant in magnitude and direction, and EF and 
FA represent the two components into which it is resolved. 
Since the force DE is given by the construction of Problem 2, 
the three unknown forces are therefore represented in magni- 
tude and direction by DE, EF, and FA. 

In these problems and in those given in the preceding sec- 
tions, it should particularly be noted that the forces should be 
used in such an order that those which are completely known 
are consecutive. 

46. Inaccessible Points of Intersection. In Problem 3 and 
in other problems, it may happen that the point of intersection 
of the two forces falls outside of the limits of the drawing. 
When this is the case, and when it is required to draw a line 
from a given point through the point of intersection of two 
forces whose lines of action intersect outside of the limits of 
the drawing, the following geometrical construction may be 
employed : 

Let XX and YY (Fig. 16) be two 
lines intersecting outside of the 
limits of the drawing. It is re- 
(j^uired to draw a line through their 
point of intersection from a given 
point P. 

Draw any line PA through the 
point P to intersect the lines XX 
and YY at the points A and B, ^jq jq 




32 



NON-CONCURRENT FORCES. 



Chap. Ill, 



respectively. Also, draw the lines AC and PC intersecting 
on YY to form the triangle ACP. From any point A' on 
XX, draw the lines A'P' and A'C, parallel respectively to 
AP and AC, and from C draw the line C'P' parallel to 
CP to intersect A'P' at P'. Then PP', prolonged, will 
pass through the point of intersection of XX and YY. For, 
since the triangles ABC and BCP are similar to the triangles 
A'B'C and B'CF, respectively, therefore AB : A'B' : : 
CB : C'B', and BP : B'P' : : CB : C'B', which gives AB : A'B' : : 
BP : BT', thus proving that the three lines XX, YY, and PP' 
meet in a point. 

47. Problems, (i). A rigid beam 16 feet in length rests 
horizontally upon supports at its ends and carries the follow- 
ing loads: its own weight of 200 lbs., acting at its center, 
and three loads of 150 lbs., 80 lbs., and 120 lbs., acting at 
points which are distant 4 ft., 10 ft., and 13 ft., respectively, 
from the left support. Find the upward pressures, or reac- 
tions, at the supports. 

(2). Given the roof truss and wind loads as shown (Fig. 

17). The truss is fixed to the 
wall at the right end, and is 
supported on rollers at the left. 
It is required to find the mag- 
nitude and line of action of thp 
right reaction and the magni- 
tude of the left. (The left re- 
action must be vertical; since 
the left end of the truss is on rollers and can have no hori- 
zontal component.) 

(3). A uniform bar 24 inches long, weighing 20 lbs., has 
one end resting against a smooth wall, and is supported on 
a smooth peg, which is 13 inches from the wall. The bar is 
held in equilibrium at an angle of 60 degrees with the ver- 
tical by a weight W suspended from the free end of the bar. 
It is required to find the weight W, and the pressures against 
the bar by the wall and the peg. 




Fig. 17. 



Art S. 



EQUILIBKIUM OF XOX-COXCUaBEXT FORCES. 



33 



48. Special Method. If any system of forces in equilib- 
rium is divided into two groups, the resultants of the two 
groups must be equal, must act in the opposite directions, and 
must have the same line of action. These facts suggest a 
special method for solving certain problems, which in some 
cases is simpler than the general method of constructing the 
force and funicular polygons. 

Problem. Let AB (Fig. 18) represent a force which is com- 
pletely known; and let BC, CD, and DA be known in line 
of action only, the lines of action of the four forces being ab, 
be, cd, and da. 





Fig. 18. 

The resultant of the forces whose lines of action are ab 
and be must pass through their point of intersection ; also, the 
resultant of the forces whose lines of action are cd and da 
must pass through the intersection of cd and da. For equi- 
librium, these two resultants must be equal, must have oppo- 
site directions, and must have the same line of action; hence 
each must act through the line ac. Draw AB representing the 
magnitude and direction of the known force, and from A and 
B, the extremities of the line AB, draw lines parallel respect- 
ively to ac and be. Then BC represents the magnitude and 
direction of the force whose line of action is be; and AC, act- 
ing in the direction from A towards C, represents the magni- 
tude and direction of the resultant of AB and BC. But for 
equilibrium, CA, acting in the opposite direction, i. e., from C 
towards A, must represent the resultant of the forces acting 
along the lines cd and da ; hence these forces are represented 
in magnitude and direction by CD and DA, drawn from the 
points C and A, parallel respectively to cd and da. 



A 



34 NOX-COXrURREXT FORCES. Chap. III. 

Problem. A beam with rounded ends has one end resting 
against a smooth wall and the other end on a smooth floor. 

The weight of the beam is 60 lbs. acting through its center. 
Neglecting the friction of the beam against the wall and floor, 
what force applied horizontally at the lower end of the beam 
would support it at an angle of 60 degrees with the hori- 
zontal ? 



Art. 4. Special Constructions for Funicular Polygons. 

49. Relation Between Different Funicular Polygons for 
the Same Forces. It will now be shown that if two funicular 
polygons are drawn, using the same forces and force polygons 
but different poles, the intersections of corresponding strings 
of these funicular polygons will meet on a straight line which 
is parallel to the line joining the two poles. 

Let AB, BC, and CD (Fig. 19) represent in magnitude and 
direction a system of forces, and let ab, be, and cd, respect- 
ively, be their lines of action. 

X 






^ P /b 



.^0 <..4 " / 









Fig. 19. 



Draw the force polygon ABCD for the given system of 
forces, and with the pole O, draw the rays OA, OB, OC, and 
OD ; also, with any other pole O', draw the rays O'A, O'B, 
O'C, and O'D. Construct the funicular polygons whose sides 
are oa, ob, oc, and od, having the strings respectively parallel 
to the rays OA, OB, OC, and OD ; also, construct the funicu- 
lar polygon whose sides are o'a, o'b, o'c, and o'd, having the 



Art. 4, 



FUNICULAR POLYGON TUKOUGH TWO POINTS. 



35 



strings respectively parallel to the rays O'A, O'B, O'C, and 
O'D. Produce the strings oa and o'a until they intersect. In 
like manner, produce the other corresponding strings until 
each pair intersects. Draw the line XX through these points 
of intersection; then XX will be parallel to OO', the line 
joining the two poles. For, the quadrilaterals whose sides are 
AB, OA, OO', O'B ; and ab, oa, XX, o'b have by construc- 
tion three sides and two diagonals respectively parallel each 
to each ; AB parallel to ab, OA parallel to oa, O'B parallel to 
o'b, O'A parallel to o'a, and OB parallel to ob: hence the 
fourth sides OO' and XX are also parallel. The same relation 
may be proved for the quadrilaterals whose sides are OB, BC, 
CO', OO' ; and ob, be, co', XX, and so on. 

This relation between the two funicular polygons is em- 
ployed for drawing the line of pressure of an arch. 

50. To Draw a Funicular Polygon Through Two Given 
Points. Let AB, BC, and CD (Fig. 20) represent in magni- 
tude and direction a system of forces acting on the beam ; and 
let ab, be, cd, respectively, be their lines of action. It is 
required to draw a funicular polygon passing through the two 
points M and N, which points are on the lines of action of 
the two reactions. 

A 





Fig. 20. funiculab Polygon Through Two Points. 

Assume any pole O', and draw the rays O'A, O'B, O'C, 
and O'D. Construct the funicular polygon whose sides are 
o'a, o'b, o'c, and o'd, and close the polygon by drawing the 



36 



NON-COXCURREKT FORCES. 



Chap. III. 



closing line o'e. From the pole O', draw the dividing ray 
O'E, cutting AD at the point E. Then DE and EA represent 
the two reactions whose lines of action are de and ea, respect- 
ively. Now the point E will remain fixed, no matter where 
the pole is taken; since the two reactions must be the same 
for the given system of forces. From E draw the line EO 
parallel to the line connecting the two given points M and N. 
Then the new pole O, for the funicular polygon passing 
through the two given points, must be on this line ; since OE 
is the dividing ray of the force polygon, drawn parallel to the 
closing string MN. With the new pole O (at any point on 
OE), commence at M, and draw the funicular polygon whose 
sides are oa, ob, oc, od, and oe, which must pass through the 
given points M and N. 

51. To Draw a Funicular Polygon Through Three Given 
Points. Let AB, BC, CD, and DE (Fig. 21) represent in 
magnitude and direction a system of forces acting on the 
beam; and let ab, be, cd, and de, respectively, be their lines 
of action. It is required to draw a funicular polygon passing 
through the three given points a, b, and c. 





Fig. 21. Funicular Polygon Through Three Points. 



Assume a system of forces to be acting upon a beam of 
such a length that the two reactions will pass through the 
two outside points. This length is determined by drawing 
lines through the two outside points parallel to the equilib-^ 
rant of the given forces. 



Art. 4, FUNICULAR POLYGON THROUGH THREE POINTS. 37 

Take any pole O', and draw the rays O'A, O'B, O'C, 
O'D, and O'E. Commencing at a, construct the funicular 
polygon a b' c', and close the polygon by drawing the closing 
line ac'. From the pole O', draw the dividing ray O'C, cut- 
ting EA, the equilibrant of the given forces, at C. Then 
EC and CA represent the two reactions whose lines of action 
are ec' and c'a, and which pass through c and a, respectively. 
Now draw the line DA, which represents the equilibrant of 
the three forces to the left of b. Through b, draw the line 
bb' parallel to DA. Connect the points a and b', and the 
points a and b by the lines ab' and ab, which are the closing 
strings of the funicular polygons for the forces to the left of b. 
Through O', draw the dividing ray O'B' parallel to ab', cut- 
ting DA at B'. Then DB' and B'A represent the reactions of 
the forces to the left of b, acting through the points b and a, 
respectively. Point C is common to all force polygons for 
the given system of forces, and point B' is common to all force 
polygons for the forces to the left of b. From the points C 
and B', draw lines parallel respectively to ac (the line con- 
necting two of the given points a and c) and ab, intersecting 
at O. Then these lines are the dividing rays corresponding 
to the closing strings of the required funicular polygon, and 
their intersection O will determine the pole for this polygon. 
With this pole O, commence at a, and draw the required funic- 
ular polygon, which must pass through the three given points 
a, b, and c. For, if the pole O is on the line CO, the funicular 
polygon must pass through the points a and c; and if it is 
on th^ line B'O, the funicular polygon must pass through the 
points a and b. 

This construction and that shown in § 50 may be used to 
determine whether or not the line of pressure in a masonry 
arch remains within the middle third at all joints. 



CHAPTER IV. 



MOMENTS. 



This chapter is divided into two articles, viz.: Art. i, 
Moments of Forces and of Couples, and Art. 2, Graphic 
Moments. The first article treats of the general principles of 
the moments of forces and of couples ; and the second treats 
of the graphic determination of moments. 



Art. I. Moments of Forces and of Couples. 



52. Moment of a Force. The moment of a force about 
any point is the product of the magnitude of the force into 
the perpendicular distance of its line of action from the given 
point. The point about which the moment is taken is called 
the origin (or center) of moments; and the perpendicular 
distance from the origin to the line of action of force is called 
the arm of the force. 



P 



P 



A 



B 



10 

Moment-+ Pa 
(a) 



Moment—Pb' 
(b) 

Fig. 22. 



V / 

^ / 

w ^d^ V 1 

O C 

Fig. 23. 



53. Positive and Negative Moments. Rotation may be in 
either of two opposite directions, viz. : in the direction of the 
hands of a clock, or opposite to the direction of the hands of a 

38 



^'^' ^- MOMENT AREA. 39 

clock. The first will be called positive ; and the second, nega- 
tive. The sign of the moment of the force is taken the same 
as that of the direction of the rotation it tends to produce. 
Thus the moment of the force P (Fig. 22, a) about the point 
O is equal to -\-P2Ly and the moment of P' (Fig. 22, b) about 
the point O' is equal to — P'a'. 

54. Moment Area. The moment of a force may be repre- 
sented by double the area of a triangle, the vertex of which is 
at the origin of moments and the base a length in the line of 
action of the force equal to the magnitude of the force. Thus 
the moment of the force AB (Fig. 23) about the point O is 
numerically represented by twice the area of the triangle 
OAB, which is equivalent to the area of the rectangle ABCD. 

55. Transformation of Moment Area. In comparing the 

moments of forces, it is often conven- 
A B' ient to transform their moment areas 

1***.^ "I o into equivalent areas having a common 

base. The moments are then to each 
other as the altitudes of their respect- 
ive moment areas. Thus in Fig. 24, let 
^ Fig 24* ^ ^^^ moment area represented by ABCD 

be transformed into an equivalent area 
having its base equal to DC. 

Connect A and C, and from C draw CA' parallel to C'A, 
and prolong it until it intersects DA, prolonged, at the point 
A'. Complete the. rectangle A'B'C'D, which is the required 
moment area equivalent to ABCD. For, since the triangles 
ADC and A'DC are similar, AD:A'D::DC:DC; hence, 
ADXDC=A'DXDC, or in other words, the area ABCD= 

the area A'B'CD. 

^^ . 

56. Moment of the Resultant of Two Concurrent Forces. 

Proposition. The moment of the resultant of two concurrent 
forces about any point in their plane is equal to the algebraic 
sum of their separate moments about the same point. 

Let DA (Fig. 25) represent the resultant of the two con- 
current forces BA and CA whose lines of action intersect 
at A, and let O be the origin of moments. It is required to 



j 



40 MOMENTS. Chap. IV. 

prove that the moment of DA about O is equal to the alge- 
braic sum of the separate moments of BA and CA about the 
same point. 

Connect the points O and A by the line OA, and from 0, 
draw OE perpendicular to OA. Draw the lines DE, CF, and 

f BG parallel to OA, and join the points 
7? O and B, O and C, and O and D 
/'/Vn^ by the lines OB, OC. and OD, re- 

pL ^i-/.-Y.i*c spcctivcly. Then (§ 54), the moment 

® I" V*" V-jSsr^ V / of BA about O is equal to twice the \ 

\ /y/-"' ^s\/ ^^^^ ^^ ^'^^ triangle OBA = +OAX 
Q^ A "" ^^^* Likewise, the moment of CA 

Fig. 25. about O is equal to twice the area of 

the triangle OCA = +OAXOF; and the moment of DA 
about O is equal to twice the area of the triangle ODA = 
H-OA X OE. But OE = OF + FE, or since FE = OG by 
construction, then OE = OF f OG. Therefore, OA X OE = 
OA (OF + OG), or the moment of DA about the point O is 
equal to the algebraic sum of the moments of BA and CA 
about the same point. 

57. Moment of the Resultant of Any Number of Concur- 
rent Forces. Proposition, Tlic moment of the resultant of 
any number of concurrent forces about any point in their 
plane is equal to the algebraic sum of their separate moments J 
about the same point. 

The proof of this proposition follows directly from that 
given in § 56, and is simply an extension of that proof. 

58. Moment of a Couple. The moment of a couple about 
any point in the plane of the couple is equal to the algebraic 
sum of the moments of the two forces composing the couple 
about the same point. 

It will be shown that the moment of a couple is a O 

equal to the product of one of the forces into the . 

perpendicular distance between the lines of action of p 
the forces. 



Let O (Fig. 26) be the origin of moments, and p' 
let P and P' be the two equal forces of the couple. p^^ ^o" 



Art. 1. 



MOMENT OF THE RESULTANT OF FORCES. 



41 



Then the moment of the couple is equal to — P' (a + b) -|- Pb 
= — P'a (since P is equal to P'), or the moment of the couple 
is equal to the product of one of the forces into the perpendicular 
distance between the two forces. Since O is any point in the 
plane of the couple, it is evident that the moment of the couple is 
independent of the origin of moments. 

59. Moment of the Resultant of Any System of Forces. 
Proposition. The moment of the resultant force, or resultant 
couple, of any system of coplanar forces about any point in 
their plane is equal to the algebraic sum of the separate 
moments of the forces composing the system about the same 
point. 

Let AB, BC, CD, and DE (Fig. 27) represent in magnitude 
and direction any system of forces, and let ab, be, cd, and de 
represent their lines of action, respectively. 





Pia. 27. 



Assume any pole O, and draw the rays OA, OB, OC, OD, 
and OE. Also, draw the funicular polygon whose sides are 
oa, ob, oc, od, and oe. Now, AB may be replaced by AO and 
OB, 'acting along ao and ob ; BC may be replaced by BO and 
OC, acting along bo and oc ; CD may be replaced by CO and 
OD, acting along co and od ; and DE may be replaced by DO 
and OE, acting along do and oe. (By AB, BC, etc., are meant 
the forces represented in magnitude and direction by AB, BC, 
etc.). Now (§ 56), no matter where the origin of moments is 
taken : 



42 MOMENTS. Chap- ^^• 

Moment of AB = moment of AO + moment of OB ; 
" BC = " " BO + " " OC; 
" CD = " " CO + " " OD; 
" DE = " " DO + " " OE. 

Since the forces represented by OB and BO are equal in 
magnitude, opposite in direction, and have the same line of 
action, their moments are equal but have opposite signs. In 
like manner, the moments of OC and CO, and of OD and DO 
are equal but have opposite signs. The addition of the above 
four equations shows that the sum of the moments of AB, 
BC, CD, and DE is equal to the sum of the moments of AO 
and OE. Now the resultant of the given system of forces I 
may be either a resultant force or a resultant couple (§32 and * 

§ 35). In the former case, the resultant of the system is the 
resultant of AO and OE, which is AE; and its moment is 
equal to the algebraic sum of their moments, or to the moment 
of AE (§ 56). In the latter case, which occurs only when E 
coincides with A, the resultant is composed of the forces AO 
and OE. These forces are equal, act in opposite directions, 
and form a couple whose moment is equal to the algebraic 
sum of the moments of AO and OE. The proposition is there- 
fore true in either case. 

It should be noticed that the proof here given applies to 
any system of forces, whether parallel or non-parallel. 

60. Moment of a System of Forces. Definition. The 
moment of a system of forces is equal to the algebraic sum of 
the moments of the forces composing the system. 

61. Condition of Equilibrium. Proposition. If a given 
system of forces is in equilibrium, the algebraic sum of their 
moments about every origin must be equal to zero. For, in 
order that the given system of forces shown in Fig. 27 be in 
equilibrium, AO and OE must be equal, must act in opposite 
directions,' and must have the same line of action. The sum 
of their moments, which is equal to the sum of the moments 
of the given system of forces, is therefore equal to zero. The 
converse of this proposition, if the algebraic sum of the 
moments about every origin is equal to zero, the system of 



^rt, g. GRAPHIC MOMENTS. 43 

forces is in equilibrium, is also true. For, if the sum of their 
moments is not equal to zero, the system must have either a 
resultant force or a resultant couple. If the resultant is a 
force, then its moment is not equal to zero unless the origin 
is taken on its line of action ; and if the resultant is a couple, 
then its moment is not equal to zero for any origin, no matter 
where taken. Therefore, if the sum of the moments about 
every origin is equal to zero, the system has neither a resultant 
force nor a resultant couple, and must be in equilibrium. 



Art. 2. Graphic Moments. 

62. Graphic Representation of the Moment of a Force. 
Proposition. If through any point in the space diagram a line 

is drawn parallel to a given 









On 1 . 



X 

V. 

l_J > 



Fio. 28. 



force, the distance intercepted 
upon this line by the two 
strings corresponding to the 
z^O components into which the 
given force is resolved by the 
rays multiplied by the pole dis- 
tance of the force is equal to 



the moment of the force about the given point. 

Let AB (Fig. 28) represent in magnitude and direction a 
force whose line of action is ab. Also, let P be the origin of 
moments, and H the pole distance of the given force. From 
the pole O, draw the rays OA and OB, and from any point on 
ab, draw the strings oa and ob parallel respectively to OA 
and OB. Also, from the origin P, draw a line parallel to the 
line of action of the given force AB, intersecting these strings 
and cutting off the intercept y. It is required to prove that 
the moment of the given force is equal to H multiplied by y. 

Let h be the perpendicular distance from P to ab. Then 
the moment of AB about P is equal to — AB X h. From sim- 
ilar triangles, H : h : : AB : y ; therefore H X y = AB X h, 
which proves the proposition. 



44 



MOMENTS. 



Chap, ir. 



It should be noticed that the intercept always represents 
a distance and is measured to the same scale as the distances 
in the space diagram, and that the pole distance always repre- 
sents a force magnitude and is measured to the same scale 
as the forces in the force diagram. 

63. Moment of Any System of Forces. Proposition. The 
moment of any system of coplanar forces about any origin in 
the plane is equal to the distance intercepted, on a line drawn 
through the origin of moments parallel to fhe resultant of all 
the forces, by the strings which meet upon the line of action 
of the resultant, multiplied by the pole distance of the 
resultant. 

Let ABCDE (Fig. 29) be the force polygon, and let the 
polygon whose sides are oa, ob, oc, od, and oe be the funicular 
polygon for the given system of forces AB, EC, CD, and DE. 
Also, let R represent the resultant of the system, let H be the 
pole distance of this resultant, and y the distance intercepted, 
on a line drawn through the origin of moments P parallel to 
the resultant, by the strings meeting on the line of action of 
the resultant. It is required to prove that the moment M of 
the given system of forces is equal to H X y« 









The moment of the resultant is equal to the moment of the 
given system of forces (§ 59). Then the moment M of the 
given system is equal to R X h, where h is the perpendicular 
distance from the origin of moments to the line of action of 



Art, tS. 



MOMENT OF PABALLEL FORCES. 



45 



tlie resultant. From similar triangles, H : h : : R : y ; therefore, 
H X y = R X h, which proves the proposition. 

Referring to the space diagram (Fig. 29), it is seen that 
the resultant R tends to produce rotation in the direction of 
the hands of a clock, and is positive. The magnitude of this 
moment decreases as the origin approaches the line of action 
of the resultant, until it becomes zero when on the line of 
action of R. If the origin passes through this line of action, 
the moment is negative. 

64, Moment of Parallel Forces. The method explained 
in § 62 and § 63 is especially useful when it is desired to find 
the moment of any, or all, of a system of parallel forces. For 
example, it is required to find the moment at any point in a 
beam caused by several loads on the beam. 

Let M-N (Fig. 30) be a simple beam loaded with the loads 
AB, BC, CD, and DE. It is required to find the moment at 
any point P due to the forces to the left of P. 

A 



M 



R. 



b b 






d 
PI 






Nm 



N 



Rz 



i.w: \ 



C 



F^ 









^L'^' 



Fio. 30. 



Construct the force and funicular polygons for the given 
system of forces, as shown in Fig. 30, and determine the magni- 
tudes of the reactions Rj and Rg. Produce each string of the 
funicular polygon until it intersects a line drawn through P 
parallel to the resultant of all the forces. Let H be the pole 
distance for the given forces, which in this case is the same 
for all the forces; since the force polygon is a straight line. 
Then (§ 62), considering the forces to the left of P: 



« " BC " 



46 MOMENTS. ^^«P- ^''• 

Moment of R^ about P = +H X nir; 

AB " P ^ H X mn; 

BC " P = — H X no; 

CD " P ^ H X op. 

The addition of these equations gives the sum of the moments 
of Ri, AB, BC, and CD = H (+ mr — mn — no — op) = + H 
X y, which is equal to the moment in the beam at the point P. 
In like manner, it may be shown that the moment of the forces 
to the right of P is equal to — H X y. Since P is any point 
along the beam, it is seen that the moment at any point is 
equal to the ordinate of the funicular polygon, cut off by a line 
through the given point parallel to the resultant of all the 
forces, multiplied by the pole distance. 

The summation of the moments of all the forces to the left 
of any point in the beam about that point is called the bending 
moment. 

The ordinate represents a distance and is measured to the 
same scale as the beam, while the pole distance represents a 
force and is measured to the same scale as the forces in the 
force polygon. 

65. Problems. Problem i. Assume a system of five non- 
parallel, non-concurrent forces, and choose any origin in their 
plane. Determine their separate moments, and also the 
moment of their resultant by the method of § 62 and § 63. 

Problem 2, Given the simple beam loaded as shown in Prob- 
lem I, § 47. Find the moment at a point 12 feet from the left 
end by the method of § 64. 

Problem j. Given the simple beam loaded at the same points 
as in Problem i, § 47. The loads have the same numerical 
values as in Problem i, but make the following angles with 
the axis of the beam (all angles being measured counter-clock- 
wise) : load at 4-ft. point, 315° ; load at center, 270° ; load at 
lo-ft. point, 240° ; load at 13-ft. point, 225°. It is required to 
find the bending moments in the beam at points which are 
distant 6 ft., 11 ft., and 14 ft., respectively, from the left end of 
the beam. Both reactions are parallel to the resultant of all 
the loads. 



CHAPTER V. 

CENTER OF GRAVITY OF AREAS. 

In designing static structures, it is frequently necessary to 
deal with the center of gravity of areas. The center of grav- 
ity of some areas may be found by simple geometrical con- 
structions; while for others, it is convenient to divide the 
figure into elementary areas, and, treating these areas as a 
system of parallel forces, to locate the point of application of 
the resultant of the system. The point of application of the 
resultant of a system of parallel forces acting at fixed points 
is frequently referred to as the center of parallel forces, but is 
more appropriately called the centroid; and is the point 
through which the line of action of the resultant passes in 
whatever direction the parallel forces are assumed to act. 

Methods will be given in this chapter for finding the center 
of gravity of some common geometrical areas, also of irregu- 
lar areas. 

66. Geometrical Areas. The center of gravity of some of 
the most common geometrical areas will now be located. The 
proofs of the constructions will not be given, as they may be 
found in any treatise on plane geometry ; but the student should 
be able to supply the demonstrations. 

(i) Parallelogram, The center of gravity of a parallelo- 
gram is at the point of intersection of the two bisectors of 
the opposite sides of the figure. 
Thus let ABCD (Fig. 31) be a paral- 
lelogram, and let ac and bd be the 
bisectors of the opposite sides ; then 
the center of gravity of the figure is 
at G, the point of intersection of ac 
and bd. f^«- ^l 

47 




48 



CENTER OF GRAVITY OF AREAS. 



Chap. V, 





Fio. 33. 



( 2 ) Triangle. The center of gravity of a triangle lies on 

a line drawn from any vertex to the 
middle of the opposite side, and is 
therefore at the intersection of anv 
two such lines. The center of gravity 
is at a distance from the vertex equal 
to two-thirds of the length of any 
bisector. Thus let Ba (Fig. 32) be 

one of the bisectors; then the center of gravity G divides the 
line Ba so that BG is equal to two-thirds Ba. 

(3) Quadrilateral. Let ABCD (Fig. 33) be a quadrilateral 
whose center of gravity is required. A 
Draw BD, and let E be its mid- 
dle point. Join E with the points 
A and C. Make EM equal to one- 
third EA, and EN equal to one- 
third EC. Join the points M and 
N, and make MG == NH ; then G 
is the center of gravity of the quadrilateral. 

(4) Circular Sector. Let ABCO (Fig. 34) be a circular 

sector whose center of gravity is re- 
quired. On AO make Aa equal to one- 
third AO, and describe the arc abc. 
Bisect the sector by the line OB. Draw 

^bd tangent to the arc abc at b, and 
make bd equal to the length of one- 
half of the rectified arc abc. Join the 
points O and d, and from c draw 

ce parallel and from e draw eG perpendicular to OB. Then G 

is the center of gravity of the sector. 

(5) Circular Segment. Let ABC (Fig. 35) be a circular 
segment whose center of gravity is required. Draw the 
extreme radii AO and CO, thus completing the sector ABCO. 
Draw the middle radius BO. Locate the center of gravity of the 
triangle AOC, and that of the sector ABCO by the methods pre- 
viously explained. Draw ab perpendicular to CO, and from g 
and g', the centers of gravity of the triangle ACO and the sector 




CENTROID OF PARALLEL FORCES. 



49 



Gk^ 



.i.;:vQ 



A4r 



FiQ 35. 



ABCO, respectively, draw gd and 

g'c parallel, in any direction. Make 

g'c equal to ab and gd equal to the 

q.w,--^^^ \ ^ length of the rectified arc BC. Join 

^c "l'!-:^« ^^^ points d and c, and prolong the 

d line dc to cut BO at G, which is the 
center of gravity of the sector. 

67. Irregular Areas. The cen- 
ter of gravity of an irregular area 
may be found by dividing the figure into elementary areas, re- 
placing these areas by parallel forces, and then locating the 
centroid of the resultant of the system of forces. The centroid 
of a system of parallel forces having fixed points of applica- 
tion has been defined as the point through which the line of 
action of the resultant of the system passes, in whatever direction 
the forces are assumed to act ; and a method will now be given 
for finding this centroid. 

68. Determination of the Centroid of Parallel Forces. 
Let ab and be (Fig. 36) be the lines of action of the two paral- 
lel forces AB and BC, and let ac be 
the line of action of their resultant AC. 
Draw any line MN perpendicular to 
the lines of action of the given forces, 
intersecting them at the points M, O, 
and N. Since AC is the resultant of 
AB and BC, the moment of AC about 
any point in their plane is equal to the 
sum of the moments of AB and BC 
about the same point. If the origin of moments is taken on 
the line of action ac, then the moment of AC is equal to zero ; 
and therefore — AB X MO -f BC X ON =0, or ABXMO= 
BC X ON. The latter equation shows that MN is divided into- 
segments which are inversely proportional to AB and BC. 
Any other straight line cutting ab and be will also be divided 
into segments which are inversely proportional to AB and 
BC. If the two forces act in opposite directions, then ac will 
lie outside of ab and be; but the above statement will hold 



M 



N 



1 



Fig. 36. 



50 CENTER OF QBAVITY OF AREAS. ^'*«P- ^• 

true. Now suppose the lines of action ab and be tc be turned 
through any angle about the points M and N, respectively; 
then the line of action ac will always pass through O. For, if 
the points M and N remain fixed, the line of action ac of the 
resultant will always divide the line MN into segments which 
are inversely proportional to the two forces AB and BC. 
Therefore, the point O will always remain fixed. The point 
O is called the centroid of the parallel forces AB and BG for 
the fixed points M and N. 

The centroid of any number of parallel forces having fixed 
points of application may be located by first finding the 
centroid of any two of the forces, and then, taking the resultant 
of these two forces as acting at the centroid thus determined, 
by finding the centroid of this resultant and one of the remain- 
ing forces, etc. 

69. Graphic Determination of the Centroid of a System of 
Parallel Forces. Since the centroid must be on the line of 
action of the resultant of the given system of forces, to locate 
this centroid it is only necessary to find the line of action of 
the resultant for each of two assumed directions of the forces; 
then the intersection of these two lines of action is the centroid 
of the given system. The forces may be assumed to be turned 
through any angle ; but the greatest accuracy is secured if the 
two resultants are made to act at right angles to each other. 

70. Center of Gravity of an Irregular Area. The method 
explained in § 69 may be used to find the center of gravity of 
an irregular area. The figure may be divided into areas whose 
centers of gravity are known, and these areas may then be 
taken as parallel forces acting through their respective centers 
of gravity. The problem then resolves itself into finding the 
centroid of a system of parallel forces having fixed points of 
application. 

Let PQRSTUVW (Fig. 37) be the area whose center of 
gravity is required. Divide the figure into three rectangles, as 
shown, whose areas are represented by the forces AB, BC, 
and CD. Then take the point of application of these forces 
at the centers of gravity of their respective areas, and draw 



CENTER OF GRAVITY OF IRREGULAR AREAS. 



51 



their lines of action parallel to the side PQ. (The lines might 
have been drawn parallel in any direction, but are drawn as 
indicated for convenience). Construct the force and the 
funicular polygons for these forces, and locate the line of action 
of their resultant Rj. Then take the lines of action of the 
forces AB, BC, and CD parallel to the side PW, construct 
their force and funicular polygons, and locate the line of 
action of their resultant Rj. The point of intersection of the 

ABC D 



a 















\ 



/ 1 \©i - 



:w^ 



^5l 












T 



W 



■H: 



/ 






V 



6 :::>o 



Fig. 37. Center op Gbavity op an Area. 



lines of action of the two resultants is the center of gravity of 
the area PQRSTUVW. For, the center of gravity of the forces 
which represent the respective areas must be on the line of 
action of every resultant, and is therefore at the point of inter- 
section of any two of them. 

If the given area has an axis of symmetry, then its center 
of gravity must be on this axis; and if it has two such axes, 
its center of gravity must be at their point of intersection. 



CHAPTER VI. 



MOMENT OF INERTIA. 



The moments of inertia of most of the standard steel sections 
used in engineering have been computed, and may be found 
in tables published by the manufacturers of such sections. The 
moment of inertia of an irregular shaped section is often 
required, and the moment of inertia of such a section may be 
readily found by graphic methods. The elementary areas com- 
posing the section may be treated as parallel forces, and the 
moment of inertia of these forces may then be found by means 
of the force and the funicular polygons. This chapter will 
therefore be divided into two articles, as follows: Art. i. 
Moment of Inertia of Parallel Forces, and Art. 2, Moment of 
Inertia of Areas. 



Art. I. Moment of Inertia of Parallel Forces. 

71. Definition. The moment of inertia of a force with 
respect to any axis is the product of the magnitude of the force 
into the square of the distance of its point of application from 
the given axis. The moment of inertia of a system of parallel 
forces with respect to any axis is the sum of the moments of 
inertia of the separate forces composing the system about the 
same axis. 

The student should notice that the moment of inertia as 
defined above is simply a mathematical product which fre- 
quently occurs in the solution of engineering problems, and in 
no sense involves the inertia of a body or mass. Since the 

52 



-4rt. i. culmann's method. 53 

moment of inertia is so frequently required in the solution of 
problems, it is desirable to determine this product for all sec- 
tions likely to occur in practice. 

72. Moment of Inertia of a System of Parallel Forces. 
There are two methods in common use for finding the moment 
of inertia of forces, viz. : Culmann's, and Mohr's. In Culmann's 
method, the moment of inertia is determined by first finding 
the moment of the given forces, and then the moment of this 
moment, by means of the force and the funicular polygons. In 
Mohr's method, the moment of inertia is determined from the 
area of the funicular polygon. 

73. Culmann's Method. Let the given system of parallel 
forces and the axis with respect to which their moment of 
inertia is to be determined lie in the same plane. The moment 
of inertia of any one of the given forces may be found as fol- 
lows: First find the moment of the force about the given 
axis ; then assume a force, equal in magnitude to this moment 
and acting in a direction corresponding to the sign of the 
moment, to act at the point of application of the original force, 
and find the moment of this new force about the given axis, 
which is the moment of inertia of the given force. The algebraic 
sum of the moments of inertia of all of the forces is the moment 
of inertia of the given system. The application of the above 
principles to a problem will now be shown. 

Let ab, be, cd, and de (Fig. 38) be the lines of action of 
the given system of forces AB, BC, CD, and'DE whose magni- 
tudes and directions are as shown in the force polygon. Also, 
let the line XX, which is drawn parallel to the lines of action 
of the given forces, be the axis with respect to which their 
moment of inertia I is required. 

Construct the force polygon ABCDE, and the funicular 
polygon whose sides are oa, ob, oc, od, and oe, for the given 
forces. Then the moment of the force AB with respect to the 
axis XX is equal to A'B' X H (§62). Also, the moments of 
the forces BC, CD, and DE are respectively equal to B'C X H, 
CD' X H, and D'E' X H. It is seen from Fig. 38 that the 
distinction between positive and negative moments is given if 



54 



MOMENT OF INERTIA. 



Chap. VI. 



the sc(|iicncc of ihc letters is observed in reading* the intercepts. 
Now take these moments as forces, applied along the lines of 
action of the original forces. Thus, let the force represented 
by A'B' X H be applied along ab, let the force represented by 
li'C X H be applied along be, etc. The line A'B'C'D'E' may 
be taken as the force polygon for the second system of forces ; 
but it must be borne in mind that each force in this polygon 
must be multiplied by H to give its true magnitude. Assume 
any pole O', and construct a new funicular polygon whose 
sides are o'a', o'b', o'c'. o'd', and o'e'. Since the moment of 
inertia of the system of forces is required, and since the sum 
of the separate intercepts is equal to the intercept between the 
extreme strings, it is only necessary to prolong the extreme 
strings o'a' and o'e' until they intersect the axis XX at the 

X 










. r-^ b: -y 



a*^-' 
-?>• 



a 



I 



A 

B 



D 



X 



— -^>o 



Fig. 38. Moment of Inertia — Cclmann's Method. 



points M and N, respectively. The moment of this second 
system of forces is then equal to MN X H X H', which is the 
moment of inertia of the given system. For, A'E' X H is the 
moment of the original system of forces AB, BC, CD, and DE 
(§ 63), and is equal to the summation of the products of each 
force into its distance from the axis XX ; also MN X H X H' 



Art, 1. 



MOIIUS METHOD. 



55 



is the moment of this moment, and is equal to the summation 
of the products of each force into the square of its distance 
from the axis XX, which by definition is the moment of 
inertia I of the given system of forces. 

It should be noted that H is a line in the force diagram, 
therefore H represents a force and is measured to the same 
scale as the given forces; while H' and MN are lines in the 
space diagram, represent distances, and are measured to the 
same scale as the distances in the space diagram. In choosing 
H and H', they should be taken of such units of length that the 
numerical force which H represents and the distance which 
H' represents may be easily multiplied together. 

74. Mohr's Method. Let ab, be, cd, and de (Fig. 39) be 
the lines of action of the given forces AB, BC, CD, and DE, 
whose magnitudes and directions are as shown in the force 
polygon. It is required to find the moment of inertia of the 
system about the axis XX, which axis is parallel to the lines of 
action of the given forces. 




B 

C 
D 



-^:h-^ 



- — - ' --^^v ' I 






E^ 



Fia. 39. MoMiENT of Inertia — Mohr's Method. 



Construct the force polygon ABCDE, and with the pole O 
and the pole distance H, draw the funicular polygon whose 
strings are oa, ob, oc, od, and oe. Prolong these strings until 
they intersect the axis XX at the points A', B', C, D', and 
E', respectively. Then the moment of AB about XX is equal 



56 MOMENT OF INEKTIA. Chap. VI, 

to the intercept A'B' multiplied by the pole distance H ; and 
the moment of inertia of AB about the same axis is equal to 
the moment of this moment, which is equal to A'B' X H X m. 
But A'B' X ni equals twice the area of the triangle whose 
base is A'B' and whose vertex is on ab; and therefore the 
moment of inertia of AB about the axis XX is equal to the 
area of this triangle multiplied by 2H. In like manner, it may 
be shown that the moment of inertia of BC about the axis XX 
is equal to the area of the triangle whose base is B'C and 
whose vertex is on be, multiplied by 2H ; also that the moment 
of inertia of CD is equal to the area of the triangle whose base 
is CD' and whose vertex is on cd, multiplied by 2H ; and 
further that the moment of inertia of DE is equal to the area 
of the triangle whose base is D'E' and whose vertex is on 
de, multiplied by 2H. Adding the moments of inertia of these 
forces, it is seen that the moment of inertia of the given system 
is equal to the area of the funicular polygon multiplied by twice 
the pole distance, i.e., I is equal to the area of the polygon whose 
sides are oa, ob, oc, od, oe, and E'A', multiplied by 2H. 

75. Relation Between Moments of Inertia About Parallel 
Axes. A graphic proof will now be given for the proposition 
that the moment of inertia I of a system of parallel forces 
about any axis parallel to the forces is equal to the moment of 
inertia I ^.^ about an axis through their centroid plus the moment 
of inertia I of the resultant of the system about the given axis. 

The moment of the resultant R of the system of forces 
shown in Fig. 39 about the axis XX is equal to A'E' X H ; 
and the moment of inertia I, of R about XX is equal to the 
moment of this moment, which is equal to A'E' X H X h 
(where h is the perpendicular distance between XX and the 
line of action of R). But A'E' X h is equal to twice the area 
of the triangle whose base is A'E' and whose vertex is on the 
line of action of the resultant R. Therefore, the moment of 
inertia I of R about XX is equal to the area of the funicular 
triangle whose base is A'E' and whose vertex is on the line 
of action of R, multiplied by 2H. In like manner, if the axis 
is taken to coincide with the line of action of R, i.e., through 



-4rt. i. RADIUS OF GYRATION. 57 

the centroid of the system, it may be shown that the moment 
of inertia leg. is equal to the area of the funicular polygon 
whose sides are oa, ob, oc, od, and oe, multiplied by 2H. 
Now the area of the triangle plus the area of the polygon is 
equal to the total area of the figure whose sides are oa, ob, oc, 
od, oe, and E'A'. But the total area of this figure multiplied 
by 2H is equal to the moment of inertia I of the given system 
of forces about the axis XX ; therefore, I = leg. + Ir . 

It will be shown that the moment of inertia of a system 
of forces about any axis is equal to the moment of inertia 
of the given system about a parallel axis through its centroid 
plus the product of the magnitude of the resultant of 
the system into the square of the distance between the two 
axes. For, by definition, the moment of inertia Ir of the 
resultant of the system about the axis XX is equal to Rh* 
(where h is the perpendicular distance from the line of action 
of R to the axis XX). Substituting this value of Ir in the 
equation I = Ic.g. +Ir gives I = Ic.£. + Rh^ which proves 
the proposition. 

By taking the axis through the centroid of the system and 
then moving it, first to one side, then to the other side of this 
centroid, it is readily seen (Fig. 39) that any movement of the 
axis out of the centroid increases the total area included 
between the extreme strings meeting on the resultant. There- 
fore, the moment of inertia of a given system of parallel forces 
is a minimum about an axis through the centroid of the 
system. 

76. Radius of Gyration. The radius of gyration of a sys- 
tem of parallel forces about any axis is the distance from the 
axis to a point through which the resultant of the system 
must act in order that the moment of inertia may be the same 
as that of the given system. An equation will now be derived 
for the radius of gyration in terms of the moment of inertia 
and the resultant of the given system. Let R, the resultant 
of the system, be substituted for the given forces, and let this 
resultant act at such a distance r (which by definition is the 
radius of gyration of the system) from the axis that its 



58 MOMENT OF INERTIA. Chap. VL 

moment of inertia Ij remains the same as that of the original 
system of forces, I. Then I^ = I = Rr^, 



or r* = - ~. which gives r = %/ -p- 




77. Graphic Determination of the Radius of Gyration. 

The moment of inertia of the system of forces shown in Fig. 38 
is equal to H X H' X MN. If r is the radius of gyration of 

the given system, then (§ 76), r2= J.= IL>liL^lM^. 

R R 

Now if H (Fig. 38) is taken equal to AE = R, then the preced- 
ing equation becomes r2 = H'XMN, which gives r== 
VH' X MN. The length of this radius of gyration may be found, 
graphically, as follows : Draw AB = MN, and BC = H'. 

With AC as a diameter, construct the 
semi-circle ADC; and from B, draw 
the line BD perpendicular to AC, in- 
tersecting the semi-circle at D. Then 
from geometry , AB : BD :: BD : BC, 
or AB X BC = BD\ Substituting the 
^^^' ^^' values of AB and BC, we have 

MN X H' ="BD», or BD=VH' X MN, which has been shown 
equal to the radius of gyration r. 

In the above construction, H has been taken equal to 
AE = R. If H had been taken equal to nR, then AB (Fig. 40) 
should have been taken equal to nMN, or BC equal to nH'. 

Art. 2. Moment of Inertia of Areas. 

78. Moment of Inertia of an Area. The moment of inertia 
of an area about any axis is equal to the summation of the 
products of the differential areas composing the area into the 
squares of the distances of these differential areas from the 
axis. The moment of inertia of an area may be found by 
dividing the given area into elementary areas, treating these 
elements as parallel forces, and finding the moment of inertia 



Art. £. APPROXIMATE METHOD — MOMENT OF INERTIA. 



59 



of the forces. There are two somewhat similar graphic 
methods for finding the moment of inertia of an area, one of 
which gives an approximate value, and the other an accurate 
value. 

79. Approximate Method for Finding the Moment of 
Inertia of an Area. An approximate value for the moment of 
inertia of an area about any axis may be obtained as follows : 
Divide the given area into small elements by lines drawn 
parallel to the axis, and let these elementary areas be replaced 
by forces numerically equal to them, acting at their respective 
centers of gravity. Then find the moment of inertia of this 
system of forces, which will be approximately equal to that of 
the given area. The smaller these elementary areas are taken, 
the more nearly will the moment of inertia of the parallel 
forces which represent them approximate that of the given 
area; and if they could be taken as infinitesimal in size, their 
moment of inertia would be the true value for the moment of 
inertia of the area. 

The application of the above principles to the determina- 
tion of the moment of inertia of an area will now be shown. 



\ ' 

V 

\ 
\ 

\ 



X 

I 



N /?"». ^ 



V V 



\ / 



^ / 1 \ 



/die \ 




D 
C 
D 






y-y3r^;>o 



H 



■•1 




Fia. 41. 



60 MOMENT OF INERTIA. Chap. VI, 

Let the area shown in Fig. 41 be the area whose moment of 
inertia about the axis XX is required. 

Divide the section into four rectangular areas as shown, 
and assume the forces AB, BC, CD, and DE, numerically 
equal to these areas and parallel to the axis XX, to act at the 
centers of gravity of the respective areas. Construct the force 
polygon ABCDE for these forces, and with the pole distance 
H (in this case H is taken equal to AE) and the pole O, draw 
the funicular polygon corresponding to this pole. Prolong 
the strings of the funicular polygon until they intersect the 
axis at the points A', B', C, D', and E', and with the pole 
distance H' and the pole O', draw the second funicular polygon, 
using the same lines of action as before. Prolong the first and 
last strings of this funicular polygon until they intersect the 
axis XX at the points M and N. Then H X H' X MN is equal 
to the moment of inertia of the forces AB, BC, CD, and DE, 
(§ 73) > ^^d is approximately equal to the moment of inertia of 
the given area. 

A more nearly correct value for the moment of inertia of 
the area might have been obtained if the area had been divided 
into smaller strips by lines drawn parallel to XX; as the 
distance of each elementary area from the axis would then 
have been more nearly equal to the distance of its center of 
gravity from the axis. 

80. Radius of Gyration of an Area. The radius of gyra- 
tion of an area about any axis is the distance from the axis to 
a point at which if all the area was concentrated the moment 
of inertia would be the same as that of the given area. 

An approximate value for the radius of gyration of the area 
shown in Fig. 41 may be obtained by applying the method 
explained in § yy. Thus, lay off BC (Fig. 41, a) equal to MN, 
and AB equal to H'. Then on AC as a diameter, construct a 
semi-circle, and from B draw the line BD perpendicular to AC, 
cutting the semi-circle at the point D. Then BD is approxi- 
mately equal to the radius of gyration of the given area. It 
should be remembered that the pole distance H (Fig. 41) was 
taken equal to AE. If H had been taken equal to n X AE, 



,Art. S. ACCURATE METHOD — MOMENT OF INERTIA. 61 

then AB should have been taken equal to nH', or BC equal 
to nMN. 

8i. Accurate Method for Finding the Moment of Inertia 
of an Area. An accurate value for the moment of inertia of an 
area about any axis may be obtained if it is possible to divide 
the area into elementary areas whose radii of gyration about 
axes through their centers of gravity and parallel to the given 
axis are known. For, if the radius of gyration of each elemen- 
tary area about an axis through its center of gravity and also 
the distance of its center of gravity from the given axis are 
known, then its radius of gyration about the given axis may 
be determined. Now, if forces numerically equal to these areas 
and acting parallel to the axis at distances equal to their 
respective radii of gyration about the given axis are sub- 
stituted for these areas, and the moment of inertia of these 
forces is determined, the result will be the moment of inertia 
of the given area. For, by definition, the radius of gyration of 
an area is the distance from the axis to a point at which the 
entire area must be concentrated in order that the moment of 
inertia niay remain the same as about the given axis. 

The application of the above method to the determination 
of the moment of inertia of an area will now be shown by a 
problem. Let the area shown in Fig. 42 be the area whose 
moment of inertia about XX is required, and let AB, BC, CD, 
and DE represent in magnitude the elementary areas into 
which the section is divided. 

The lines of action of these forces may be found as follows : 
Draw PR (Fig. 42) perpendicular to XX and equal to the 
distance from XX to the center of gravity of the elementary 
area shown; also, at the point R, draw QR perpendicular to 
PR and equal to the radius of gyration of this area about an 
axis through its center of gravity and parallel to XX. Connect 
the points P and Q, and with the line PQ as a radius, draw 
the arc QS. Then PS = PQ is the radius of gyration of the 
elementary area about XX, and is the distance from the axis 
to the point of application of the force AB. For, the moment 
of inertia of the elementary area about an axis through its 



62 



MOMENT OF INERTIA, 



Chap. FI. 



center of gravity and parallel to XX is equal to AB X QR' 
{from the definition of the radius of gyration), and the moment 
of inertia of this area about the axis XX is equal to 
ABX"QR* + ABxTR'=AB(QR' + PR*)=AB X~P^ = 
AB X PS' (§75)' In I'l^e manner, the lines of action of the 
forces BC, CD, and DE, which represent the other elementary 




areas, may be found. Now apply the method explained in § 73 
and § 79, and find the moment of inertia of these forces (see 
Fig. 42). Then the moment of inertia of these forces, which 
is equal to the moment of inertia of the given area, is equal to 
H X H' X MN. 

The correct value for the radius of gyration of the area 
shown in Fig. 42 is given by the line BD (Fig. 42, a). The 
construction for finding this radius of gyration is similar to 
that used for Fig. 41, a, which is explained in § 80, except that 
the values for H' and MN, which have been found by the con- 
struction of Fig. 42, are used in Fig. 42, a. 



Art. S. 



MOMENTS OF INERTIA ABOUT PARALLEL AXES. 



63 



82. The table shown in Fig. 42, b gives the algebraic for- 
mulae for finding the moments of inertia and radii of gyration 
of several common sections. 



Section 



^d— 



i.-dL.! 



.1 



' d ' 







L.d-J 




Moment of Inertia 



\z 
\z 



fed' 

12 



bdibd? 
12 



0.049 d^ 



0.049(didr) 



Raoiv/s or Gyration 



d 
V12 



^^ 



dSdf 

12 



A. 

Viz 



V 



I2(bd-bidi) 



4 



^ 



d^tif 



Fig. 42, b. 

83. Relation Between the Moments of Inertia of an Area 
About Parallel Axes. If the areas are used instead of the 
forces which represent them, then the relation shown in § 75 
for the moment of inertia of a force about any axis, in terms of 
its moment of inertia about a parallel axis through its center 
of gravity, becomes, the moment of inertia of an area about 
any axis is equal to its moment of inertia about a parallel axis 
through its center of gravity plus the product of the area into 
the square of the distance between the two axes. The applica- 
tion of this relation is of great importance in designing sec- 
tions. 

84. Moment of Inertia of an Area Determined from the 
Area of the Funicular Polygon (Mohr's Method). In the 



64 MOMENT OF IXERTIA. Chap, VL 

examples that have been given for finding the moment of 
inertia of an area, only Culmann's method has been used, but 
Mohr's method might have been employed instead. If the 
lines of action of the forces had been taken as acting at the 
centers of gravity of the elementary areas, an approximate 
value for the moment of inertia would have been found ; and if 
each force had been taken as acting at a distance from the 
given axis equal to the radius of gyration of the elementary 
area about the given axis, then an accurate value would have 
been found from the area of the funicular polygon. 



PART II. 



FRAMED STRUCTURES-ROOF TRUSSES. 



CHAPTER VII-. 

DEFINITIONS. 

85. Framed Structure. A framed structure is a structure 
composed of a series of straight members fastened together at 
their ends in such a manner as to make the entire frame act as 
a rigid body. 

The only geometrical figure which is incapable of any 
change of shape without a change in the length of its sides is 
the triangle; and it therefore follows that the triangle is the 
basis of the arrangement of the members in a framed structure. 

86. Types of Framed Structures. Framed structures may 
be divided into the three following classes: (a) complete 
framed structures, (b) incomplete framed structures, and (c) 
redundant framed structures. 

(a) Complete Framed Structure, A complete framed 
structure is one which is made up of the minimum number of 
members required to form the structure wholly of triangles. 
This is the type which is usu- 
ally employed, and which will 
receive the most attention in 
this work. A simple form of a 
complete framed structure is 

shown in Fig. 43. ' Fiq. 43. Complete. 

65 




66 



DEFINITIONS. 



Chap, Vll. 




Fig. 44. Incomi'Lete. 



fb) liicomplete Framed Structure, An incomplete framed 
structure is one which is not wholly composed of triangles. 

A simple form of an 
incomplete framed 
structure is shown 
in Fig. 44. Such a 
structure is stable 
only under symmet- 
rical or specially ar- 
ranged loads, 
(c) Redundant Framed Structure. A redundant framed 
structure is one which contains a greater number of members 
than is required to form the structure wholly of triangles. If 
the second diagonal is added to a quadrilateral, then the added 
diagonal is a redund- 
ant member; but if 
the member is capa- 
ble of resisting only 
one kind of stress, 
then the redundancy 
is only apparent. Fig. F'«- 4*' ». redundant. 

44, a shows a redundant structure; but if the diagonals are 
made of rods and can therefore take tension only, then the 
redundancy is only apparent; as only one diagonal will act 
at a time. 

87. Roof Truss. A simple roof truss is a framed struc- 
ture whose plane is vertical and which is supported at its 




Upper C hord 



Members 




Fia. 45. Roof Truss. 



TYPES OF EOOF TRUSSES. 



67 



ends. The ends of the truss may be supported upon side 
walls, or upon columns. A common form of a simple roof 
truss is shown in Fig. 45. 

Span, The span of a truss is the distance between the end 
joints, or the centers of the supports, of the truss. 



8 F=bnels 

(Q) 



16 Panels 
(d) 



With Ventilator 



Circular Chord 
(h) 



10 Panels 
(b) 



12 Panels 

(C) 





20 Panels 





Cannbered 
Fink Trusses *^^ 





Quadrangular 
(1) 





Pratt 
(k) 





Saw Tooth Cantilever 

<l) (m) 

FiQ. 46. Types of Roof Trusses. 



68 DEFINITIONS. C^^^P- ^I^- 

Rise, The rise is the distance from the apex, or highest 
point of the truss, to the line joining the points of support of the 
truss. 

Pitch, The pitch is the ratio of the rise of the truss to its 
span. 

Upper and Lower Chords. The upper chord consists of the 
upper Hne of members of the truss, and extends from one support 
to the other, through the apex. The lower chord consists of the 
lower line of members, and extends from one support to the other. 

Web Members. The web members connect the joints of the 
upper chord with those of the lower chord. A web member 
which is subject to compression is called a strut, and one which 
is subject to tension is called a tie. 

Pin-connected and Riveted Trusses. A pin-connected truss is 
one in which the members are connected with each other by means 
of pins. A riveted truss is one in which the members are fastened 
together by means of plates and rivets. The latter type is the 
more common, while the former is used for long spans. 

88. Types of Roof Trusses. Several types of roof 
trusses are shown in Fig. 46. 

When a building is to be designed, the conditions govern- 
ing the design should be carefully studied before deciding 
upon the type of roof truss to be used; as a truss which is 
economical for one building may not prove so for another. 

The Fink truss is very commonly used, and is well 
adapted to different spans ; as the number of panels may be easily 
increased. 

The Quadrangular truss shown in Fig. 46, 1 is well adapted 
to the use of counterbracing. 

The Howe and Pratt trusses shown in Fig. 46, j and Fig. 
46, k, respectively, are common type, and may be used with 
a small rise. 

As a rule, trusses with long compression members are not 
economical. 



CHAPTER VIII. 



LOADS. 



The loads that must be considered in the discussion of a 
roof truss may be classed as follows: (i) dead loads; (2) 
snow loads ; and (3) wind loads. A discussion of these loads 
will be given in the three following articles. 

Instead of considering separately the dead, snow, and 
wind loads, an equivalent vertical load is sometimes taken. 
This method is efficient in some cases if intelligently used, 
but should not be used by the beginner. 

Art. I. Dead Load. 

A short description of the common type of roof construc- 
tion, together with the terms employed, will now be given, 
preliminary to the discussion of dead loads. 

89. Construction of a Roof. A roof includes the covering 
and the framework. There are a number of materials used 
for roof coverings, among the more common being slate, tiles, 
tar and gravel, tin, and corrugated steel. Sheathing is com- 
monly used in connection with roof coverings, but it is often 
omitted when corrugated steel is used. 

The covering is usually supported by members called 
jack-rafters, which in turn are supported by other members 
called purlins. The purlins run longitudinally with the build- 
ing and are connected to the trusses, generally at panel points. 
The jack-rafters are usually made of wood, while the purlins 
may be either of wood, or of steel. If the distance between 
the trusses is great, the purlins may be trussed. 

69 



70 lOADS. Chap, VI 11. 

A system of szvay bracing is generally used to give rigidity 
to the structure. This bracing may be in the plane of the 
upper chord, in the plane of the lower chord, or in both planes. 
Sometimes the sway bracing is made continuous throughout 
the entire length of the building, and at other times the 
trusses are connected in pairs, depending upon the rigidity 
required. 

The trusses may be supported upon masonry walls, upon 
masonry piers, or upon columns. When the trusses rest upon 
masonry and the span is short, the expansion and contraction 
of the trusses are provided for by a planed base plate ; but for 
long spans, rollers are usually employed. 

go. Dead Lioad. The dead load includes the weights of 
the following items: (i) roof covering; (2) purlins, rafters, 
and bracing; (3) roof trusses; and (4) permanent loads sup- 
ported by the trusses. 

(i) Roof Covering. The weight of the roof covering varies 
greatly, depending up<n the materials employed; but it may 
be closely estimated if the materials used in its construction 
are known. The approximate weights of some of the common 
roof coverings are given in the following table : 

Slate, without sheathing 8 to 10 lbs. per sq. ft. 

Tiles, flat 15 to 20 lbs. per sq. ft. 

Tiles, corrugated 8 to 10 lbs. per sq. ft. 

Tar and gravel, without sheathing. . 8 to 10 lbs. per sq. ft. 

Tin, without sheathing i to 1.5 lbs. per sq. ft. 

Wooden shingles, without sheathing 2 to 3 lbs. per sq. ft. 

Corrugated steel, without sheathing i to 3 lbs. per sq. ft. 

Sheathing, i inch thick 3 to 4 lbs. per sq. ft. 

(2) Purlins, Rafters, and Bracing. Wooden purlins will 
weigh from 1.5 to 3 pounds, and steel purlins from 1.5 to 4 pounds 
per square foot of roof surface. 

Rafters will weigh from 1.5 to 3 pounds per square foot of 
roof surface. 

The weight of the bracing is a variable quantity, depending* 
upon the rigidity required. If bracing is used in the planes of 



^rt. 1, DEAD LOAD. 71 

both the upper and lower chords, its weight will be from 0.5 to i 
pound per square foot of roof surface. 

(Jg) Roof Trusses. The roof trusses may be either of wood 
or of steel. Steel trusses are now commonly used for spans of 
considerable length, although wooden trusses are still used for 
short spans; but the rapidly increasing cost of wood, and the 
difficulty of framing the joints so as to develop the entire 
strength of the members, has led to a general use of steel for 
trusses. 

The weights of wooden trusses are given by the following 
formula, taken from Merriman's "Roofs and Bridges," which 
is based upon data given in Ricker's "Construction of Trussed 
Roofs." 

W=4aL(i+^^jL), (i) 

where W = the total weight of one truss in pounds, 

A = the distance between adjacent trusses in feet, 
L = the span of the truss in feet. 
The weights of steel roof trusses vary with the span, the 
pitch, the distance between trusses, and the capacity or load 
carried by the trusses. The following formula is given in 
Ketchum*s "Steel Mill Buildings," and is based upon the actual 
shipping weights of trusses : 

W=— AL(i + — ^), (2) 

45 5\'A" 

where W = weight of the steel roof truss in pounds, 

P *= capacity of the truss in pounds per square foot 

of horizontal projection of the roof, 
A = distance, center to center of trusses, in feet, 
L = span of truss in feet. 
These trusses were designed for a tensile stress of 15000 lbs. 

per sq. in., and a compressive stress of 15000 — 55 — lbs. per 

sq. in. ; where 1 = length, and r = radius of gyration of the 
member, both in inches. The minimum sized angle used was 



72 LOADS Chap, VIIL 

2" X 2" X J4"> 2i"d the minimum thickness of plate was one- 
fourth inch. 

Dividing equation (2) by AL, we have 

where w is the weight in lbs. per sq. ft. of the horizontal pro- 
jection of the roof. 

Short span simple roof trusses may weigh somewhat less 
than the values given by the above formulae, especially if the 
minimum thickness and minimum size of angles are used, but 
such trusses are too light to give good service. 

(4) Permanent Loads Supported by the Trusses. It is im- 
possible to give figures which will be of much value for the 
weights of permanent loads supported by the trusses. If the 
roof truss supports a plastered ceiling, the weight of the ceil- 
ing may be taken at about 10 lbs. per sq. ft. If other loads 
are supported by the truss, they should be carefully consid- 
ered in estimating the weight of the truss. 



Art. 2. Snow Load. 

91. Snow Load. The amount of the snow load to be con- 
sidered varies greatly for diflferent localities. The weight of 
the snow which must be taken into account is a function of 
both the latitude and the humidity, and those factors should 
be carefully considered in determining the weight to be used 
for any particular locality. The maximum wind load and the 
maximum snow load will probably never occur at the same 
time ; and if the maximum wind is taken, then a smaller snow 
load should be used. For a latitude of about 35 to 40 degrees, 
the writer recommends a minimum snow load of 10 pounds, 
and a maximum snow load of 20 pounds per square foot of the 
horizontal projection of the roof; the former to be used in 
connection with a maximum wind load, and the latter when 
the wind load is not considered. 



-4*'*- ^' wind load. 73 

Art. 3. Wind Load. 

92. Wind Pressure. The pressure of the wind against a 
roof surface depends upon the velocity and direction of the 
wind, and upon the inclination of the roof. The wind is 
assumed to move horizontally, and the pressure against a flat 
surface, normal to the direction of the wind, may be found by 
the formula 

P = 0.004 V^, (4) 

where P is the pressure in pounds per square foot on a flat 
normal surface, and V is the velocity of the wind in miles 
per hour. 

The pressure on a flat vertical surface is usually taken at 
about 30 pounds per square foot, which is equivalent to a wind 
velocity of 87 miles per hour. This assumed pressure is suffi- 
cient for all except the most exposed positions. 

93. Wind^ Pressure on Inclined Surfaces. The normal 
wind pressure upon an inclined surface varies with the inclina- 
tion of the roof; and several formulae have been derived for 
finding the normal component. Th^ best known of these are 
Duchemin's, Hutton's, and The Straight Line formulae. 

Duchemin's formula is 

p»=PTr^. (5) 

I +sm A 

where P^ is the normal component of the wind pressure, 
P is the pressure per sq. ft. on a vertical surface, 

A is the angle which the roof surface makes with 

the horizontal, in degrees. 

Mutton's formula is 

P^ = Psin A,i-««co»A.i (6) 

where Pn, P, and A are the same as in Duchemin's formula. 
The Straight Line formula is 

Pn = -^A. (7) 

45 
where P., P, and A are the same as in Duchemin's formula. 



74 



Chap. VIII. 



Duchemiti's formula is based upon carefully conducted 
experiments, gives larger values, and is considered more reli- 
able than Hutton's. The Straight Line formula is preferred 
by many on account of its simplicity, and gives results which 
agree quite closely with experiments. 





1 Mill! 


1 






,---■ 


r^- 






Duchemin 
Hut ton 








-^ 


^■. 









1, 


^■' 


'■■•< 


^ 






■^V 


A^ 


S="' 












1 


1 


V 


n'Y 


/■ 































-^-' 












V 




•' 1 


>'> 




--' 


















v^>, 


py. 


0*' 
















1 


4^ 


Ay 


^- 














,^ 




.f 


■i 


















^/ 






y 


y}r 




r^^ 


-::r' 










QV 


'X 




r-^. 




















.■■A 


V 


■<- 


0>!r- 


fiji 














/, 


1 


r 


^> 


" 1 
















■■/ 


.^i> 


\^ 




















'.y 


X 


■A 




1 




















P 


b 


r- 




r 










V. 


'-' 






u 


"1 




-jj 










^^ 






Iff 


It 


r^ 




IT 














-«i 


r 


j-^ 


H 




-« 









40 



g>0 



5 10 15 eo 25 30 35 40 45 50 55 SO " 
Angle,A.Roof Makes with Horizontol in Degrees. 

Fig. 4T. NOBMtL n'iKD Load on Roof bt DirFEREKT Pormulab, 

Fig- 47 gives values for the normal wind pressure P^ in 
pounds per square foot, in terms of the pressure on a vertical 
surface and of the angle which the roof surface makes with 
the horizontal. The use of this diagram will greatly, lessen 
the work required to find the normal wind pressure when the 
pressure on a vertical surface and the inclination of the roof 
are known. 



CHAPTER IX. 

BEACTIONS. 

The reactions are the forces which if applied at the center 
of the bearings of a truss would hold in equilibrium the weight 
of the truss and the loads supported by it. The reactions are 
numerically equal to the pressures exerted by the truss against 
its supports. As the method of finding the reactions for vertical 
loads differs somewhat from that for inclined loads, the determi- 
nation of these reactions will be considered in separate articles, 
the first article treating of the reactions for dead and snow loads, 
and the second of the reactions for wind loads. 

Art. I. Reactions for Dead and Snow Loads. 

Before the reactions can be determined, it is necessary to 
find the loads that are supported by the truss. The purlins are 
usually placed at the panel points of the upper chord, and the 
loads are considered to act at these panel points. The method 
of finding the joint loads and the dead load reactions will now 
be shown by the solution of the following problem. 

.94. Problem. It is required to find the dead load reac- 
tions for the truss shown in Fig. 48. The truss has a span of 
48 feet, a rise of 12 feet, and the distance apart, center to center 
of trusses, is 14 feet. The roof is of slate, laid on sheathing, 
which is supported by wooden rafters. The purlins and the 
truss are of steel. 

The solution will be divided into two parts ; the determina- 
tion of (a) the joint loads, and (b) the reactions. 

75 



76 



BEAOTIONS. 



Chap, IX. 




(a) Joint Loads. Referring to the table given in § 90 (i), 

it is seen that the weight 

of the slate covering may 
be taken at 10 pounds, and 
the weight of the sheath- 
ing at 3 pounds per square 
Fio. 48. foot of roof surface. From 

§90 (2), it is seen that the weight of the purlins and bracing 
may be taken at 3 pounds, and the weight of the rafters at 2.5 
pounds per square foot of roof surface. This gives a total 
weight of 18.5 pounds per square foot of roof surface, exclusive 
of the weight of the truss itself. 

The length of one-half of the upper chord =V 24^ -|- 12^ feet 
= 26.83 feet. This length is divided into three equal parts by 
the web bracing, making the length of each panel of the upper 
chord equal to 26.83 -f- 3 = 8.94 feet. Since the joint loads arje 
taken at panel points, and the trusses are spaced 14 feet apart, 
the joint load supported by the truss, exclusive of the weight 
of the truss, = 8.94 X 14 X 18.5 = 2315 pounds. 

The weight of the roof truss per square foot of horizojital 
projection for a capacity P of 40 pounds (which is about right 
for the given roof truss) is given by formula (3), § 90, and is 
about 3 pounds. This is equivalent to a joint load of 8 X 14 X 
3 = 336 pounds. The total joint load is therefore equal to 2315 
+ 336 = 2651 pounds, or say 2650 pounds. Referring to Fig. 
48, it is seen that the loads acting at the joints B, C, D, E, 
and F are full panel loads; while those acting at A and G 
support but half the area, and are half panel loads. The loads 
acting at B, C, D, E, and F are each equal to 2650 pounds; 
while those acting at A and G are each equal to 1325 pounds. 

(b) Reactions, The problem now is to find the reactions at 
the ends of the truss loaded as shown in Fig. 49, or in other 
words, it is required to find the two forces acting at the ends 
of the truss, which will hold in equilibrium the given loads. 

Construct the force polygon ABCDEFGH for the given 
loads on the truss. In this case, the force polygon is a straight 
line, and is called the load line. Assume any pole O, construct 



Art. 1. 



DEAD LOAD REACTIONS. 



77 



the funicular polygon, and draw the closing string of the poly- 
gon. Then the dividing ray, drawn through the pole O paral- 
lel to this closing string, will determine the magnitudes of the 
two reactions, HM being the magnitude of the right reaction 
R2, and MA that of the left reaction Rj (§ 43). The lines of 









s; 










s; 


i8 


s; 




s 





^ 








Jp 


^^ 


/ 


\ 


v^S 


in 


m M 


^^Y 


/ 


\ 


v*^ 


(VJ 


CM 


< 


/ 


/ 


\ 


\ 


^v. 12 


i 


■ 




I 


1 










1 


1 






R. 


^--^ 


— i — ^"^ 


■"— "-^.' 








k-*^ 




^^ 


k 

*fc 
*fc 
*fc 






0" 


10 


20f 














J 





xtA- 



y 



^-^-^ 



^4C 






^Xv 

X 






0' 



sqoor ioop(>>^j-i. 



D 



R. 



Ri 



Fio. 49. Dead Load Reactions. 

action of these reactions will be parallel to the load line — 
which is vertical. By scaling the lines HM and MA, it is found 
that these reactions are each equal to 7950 pounds. 

The reactions might have been found, algebraically, by tak- 
ing moments about the supports. Thus, to find the value of 
the left reaction, take moments about the right support. Then 
the equation of moments is +Ri X 48 — 1325 X 48 — 2650 X 
40 — 2650 X 32 — 2650 X 24 — 2650 X 16 — 2650 X 8 = o. 
Solving this equation gives Rj = 7950. Since the two reac- 
tions must hold the given loads in equilibrium, the right reac- 
tion is therefore equal to the total load minus the left reaction 
= 15 900 — 7950 = 7950 pounds. 

If the loads are symmetrical with respect to the center line 
of the truss, it is evident that the reactions are equal, and that 
each is equal to 15 900 •— 2 = 7950 pounds. 

95. Snow Load Reactions. The reactions and the stresses 
for snow loads are usually considered separately from those 
due to dead loads. The same methods may be used for find- 
ing the snow load reactions as have been explained for finding 
the dead load reactions, and need no further explanation. 



I 



78 BEACTIONS. C?/iap. IX, 

96. Effective Reactions. It is seen from Fig. 49 that the 
half joint loads at the ends are transferred directly to the 
supports without causing any stress in the truss. These half 
loads act through the same lines as the total reactions, and 
subtract from these reactions. The resulting forces are called 
the effective reactions. If the effective reactions are used, the 
half loads at the ends of the truss are neglected in computing 
the stresses in the truss. The effective reactions for the truss 
loaded as shown in Fig. 49 are each equal to 7950 — 1325 = 
6625 pounds. 



Art. 2. Reactions for Wind Loads. 

The magnitudes and lines of action of the wind load reac- 
tions for any given truss depend upon the condition of the 
ends of the truss. If the span is small, or if the truss is made 
of wood, the ends are usually fixed to the supports by anchor 
bolts. For large spans, the changes of temperature and the 
deflections due to the loads cause the truss to expand or con- 
tract a considerable amount. If the ends are fixed, the tem- 
perature changes and the loads may produce large stresses in 
the truss. The usual method of providing for the changes in 
length is to place one end of the truss upon a planed base 
plate or upon rollers. 

97. Wind Load Reactions. The wind load reactions will 
be determined for each of the following assumptions^ (i) that 
both ends of the truss are fixed ; and (2) that one end of the 
truss is supported upon rollers. In the example that will be 
given, a triangular form of truss will be used, although the 
methods employed are applicable to any type of simple truss. 

98. (i) Truss Fixed at Both Supports. The problem of 
finding the lines of action of the wind load reactions for a truss 
fixed at both ends is indeterminate; but assumptions may be 
made which will give approximately correct results. The ver- 
tical components of the reactions are independent of any 
assumptions ; but the horizontal components depend upon the 



Art, S. 



WIND LOAD REACTIONS. 



79 



condition of the ends of the truss. If the roof is comparatively 
flat, i. e., if the resultant of the normal components of the wind 
loads is nearly vertical, the reactions may be taken parallel to 
this resultant; as each reaction will then be approximately 
vertical. The above assumption gives erroneous results when 
the roof makes a large angle with the horizontal. The assump- 
tion that the horizontal components of the reactions are equal 
more nearly approximates actual conditions; and this is par- 
ticularly true if the ends of the truss are fastened to the sup- 
ports in the same manner and the supports are equally elastic. 
The method of finding the reactions of a truss with fixed ends 
will now be shown by a problem, assuming (a) that the reac- 
tions are parallel, and (b) that the horizontal components of 
the reactions are equal. 

(a) Reactions Parallel, It is required to find the wind load 
reactions for the truss shown in Fig. 48. This truss has a span 
of 48 feet and a rise of 12 feet, the distance between trusses 
being 14 feet. The wind pressure on a vertical surface will be 
taken at 30 pounds per square foot, and Duchemin's formula 
will be used for finding its normal component. Referring to 
Fig. 47, it is seen that, for P = 30 pounds and for a pitch of 
one-fourth, the component normal to the roof is about 22 
pounds. The length of a panel of the upper chord is 8.94 feet 
(see § 94, a). The panel load is therefore equal to 8.94 X 14 X 
22 = 2753, or say 2750 pounds. The half panel loads at the 
ends of the truss and at the apex are each equal to 1375 pounds. 




FiQ. 50. Wind Load Reactions — Ends Fixed. 



80 REACTIONS. Chap. IX. 

The truss with its loads is shown in Fig. 50. To determine 
the reactions, construct the force polygon or load line ABCDE, 
assume a pole, and draw the funicular polygon. The dividing 
ray OF, drawn through the pole O parallel to the closing 
string of the funicular polygon, will give the magnitudes of 
the two reactions, FA being the magnitude of the left reaction, 
and EF that of the right. The lines of action of these reac- 
tions are given by drawing, through the supports, lines paral- 
lel to the load line AE. By scaHng the lines FA and EF, it is 
seen that the left reaction is equal to 5670 pounds, and the right 
reaction to 2580 pounds. The values given for the reactions 
were actually taken from a diagram four times as large as that 
shown in Fig. 50; as the scale used here is too small to give 
accurate results. 

(b) Horizontal Components of Reactions Equal, The reac- 
tions will be found for the same truss and loads as in Fig. 50, 
assuming that the horizontal components of the reactions are 
equal. First find the reactions, as in § 98 (a), assuming that 
their lines of action are parallel. The vertical components of 
these reactions are correct ; since they are independent of the 
ends of the truss. Now draw a horizontal line through the 
point F (Fig. 50), and make mn equal to the horizontal com- 
ponent of all the loads on the truss. This may be done by 
drawing vertical lines through A and E, intersecting the hori- 
zontal line at the points m and n, respectively. Bisect mn, 
making mF' = F'n, and draw the lines F'A and EF', which 
represent the magnitudes of the required reactions. The lines 
of action of these reactions are given by drawing lines through 
the left and also through the right points of support of the 
truss parallel respectively to F'A and EF'. By scaling the 
lines F'A and EF', it is seen that the left reaction is equal to 
5400 pounds, and the right reaction to 2960 pounds. 

99. (2) One End of Truss Supported on Rollers. If one 
end of the truss is supported on rollers, then the roller end can 
take no horizontal component of the wind loads (neglecting 
the friction of the rollers) ; for there would be a continued 
movement of the rollers if a horizontal force was applied at 



Art, iS, 



WIND LOAD REACTIONS. 



81 



this end of the truss. The reaction at the roller end is there- 
fore vertical, and hence all the horizontal component of the 
wind loads must be taken up at the fixed end of the truss. 
Since the wind may come from either of two opposite direc- 
tions, the rollers may be under the leeward side, or under the 
windward side of the truss. The method for finding the reac- 
tions will now be shown by a problem, considering (a) that 
the rollers are under the leeward end of the truss, and (b) that 
the rollers are under the windward end. 

(a) Rollers Under Leezvard End of Truss, It is required 
to find the wind load reactions for the truss shown in Fig. 48, 
the leeward end being on rollers. The wind loads are the 
same as those shown in Fig. 50. In this problem, the line of 
action of the reaction at the roller end is vertical ; while the 
line of action of the reaction at the fixed end is unknown, its 
point of application, only, being known. To determine the 
reactions for this case, apply the method explained in § 44. 



Pv 












i 




l/T -^ 


K^ — 


x\. 


.a\]\^ 


1 *V' 


■^^^^^ ^^--'''' ■ 


H''\ 


^1 ^^-^"^ ^*^*^^^-^^^*^^^ ^2 




/ ^- 


'" "' — "^^^-^ 




fc.^*^ 


0* - lO* 20' ^^ 




^ 2 / ^ > 



o'*'---::::::^5 

0* ZOOO^AOOO*" 



F 
Rf 



Fig. 51. Wind. Load Reactions — One End on Rollers. 



Draw the load line ABCDE (Fig. 51) for the given loads, 
assume any pole O, and draw the funicular polygon (shown by 
full lines), starting the polygon at the only known point on the 
left reaction — its point of application at the left end of the 
truss. From the pole O, draw the dividing ray OF parallel 
to the closing string of the funicular polygon to meet a vertical 
line drawn through E parallel to the known direction of the 
right reaction. Connect the points A and F; then EF repre- 
sents the magnitude of the right reaction, and FA that of the 



82 REACTIONS. Chap. IX. 

left reaction. The lines of action of these reactions, which are 
represented by Rj and Rg, are given by drawing lines through 
the ends of the truss parallel respectively to EF and FA (§ 44). 
By scaling the lines EF and FA, it is seen that the right reac- 
tion is equal to 2310 pounds, and the left reaction to 6270 
pounds. 

The reactions might also have been found by a slightly dif- 
ferent method, as follows: Find the reactions for the truss 
fixed at both ends, assuming that their lines of action are par- 
allel (see § 98 (a), and Fig. 50). EF and FA represent the 
magnitudes of these reactions, their vertical components being 
independent of the condition of the ends of the truss. Now 
since the right reaction can have no horizontal component, 
draw the line En through E to meet the horizontal line through 
F; then En represents the magnitude of this reaction. Now 
all tlie horizontal component of the wind loads must be taken 
up at the left end of the truss, this component being repre- 
sented by the line mn. The vertical component of the left 
reaction is represented by the line mA; therefore the result- 
ant of these two components, or the line nA (not drawn), 
represents the magnitude of the left reaction. 

(b) Rollers Under Windzvard End of Truss. If the rollers- 
are under the windward, instead of the leeward end of the 
truss, then the left reaction is vertical and the right is inclined, 
the only known point on the right reaction being its point of 
application at the right support. For this case, the funicular 
polygon must start at the right support, the remainder of the 
solution being similar to that explained in § 99 (a). The con- 
struction for this solution is shown by the dotted lines in Fig. 
51, EF' and F'A representing the magnitudes of the two reac- 
tions, their lines of action Rg' and R/ passing through the 
right point of support and the left point, respectively. By 
scaling the lines EF' and F'A, it is seen that the right reaction 
is equal to 4350 pounds, and the left reaction to 5070 pounds. 

The reactions might also have been found by a method 
similar to that explained in the last paragraph of § 99 (a), mA 
and Em (not drawn) representing these reactions (see Fig. 50)- 



Art. IS. 



COMPABISON OF REACTIONS. 



83 



ICO. Fig. 52 shows to scale the wind loads and the wind 
load reactions for different assumptions as to the condition of 
the ends of the truss. It is seen from the figure that the verti- 




Fig. 52. 



S^ Both ends fixed. reactions parallel, shown by ^ 

'^ Both ends fixed, hor. com'ps equal, shown by 

^•. Windward end of truss on rollers, shown fay 

Leeword end of truss on rollers. shown by —• — ••— 

Wind Load Beactions fob Different Assumptions as to Condition 

OF Ends of Tbuss. 



cal components of the reactions are independent of, and that 
the horizontal components are dependent upon, the assump- 
tions as to the condition of the ends of the truss. If the roof 
had made a greater angle with the horizontal than that shown 
in the figure, then the differences in the reactions, due to the 
assumptions as to the condition of the ends of the truss, would 
have been more apparent. 

loi. In determining the reactions for a truss, great care 
should be exercised in drawing the truss, and in locating the 
panel points accurately. The truss and diagrams should be 
drawn to a large scale to insure good results. Care should be 
taken in laying out the load line, and the pole should be chosen 
in such a position that acute mtersections are avoided. 



CHAPTER X. 

STBESSES IN BOOF TRUSSES. 

The determination of the external forces acting upon a truss 
has been taken up in Chapter VIII and Chapter IX. The dead 
and wind loads acting upon the roof are transferred to the 
supports through the members of the truss, and this chapter 
will treat of the determination of the stresses in the members 
due to these external forces. 



Art. I. Definitions and General Methods for 

Determining Stresses. 

102. Definitions. The external loads tend to distort the 
truss, i.e., to shorten some of the members, and to lengthen 
others ; and since the materials used in the construction of the 
truss are not entirely inelastic, the members are actually dis- 
torted. The deformation in any member caused by the loads 
is called strain, and the internal force which is developed in 
the member and which tends to resist the deformation, or 
strain, is called stress. There are three kinds of stress which 
may be developed by the external forces, viz. : (a) tension, (b) 
compression, and (c) shear. 

(a) Tension, A member is subjected to a tensile stress, or 
is in tension, if there is a tendency for the particles of the mem- 
ber to be pulled apart in a direction normal to the surface of 
separation. Iti this case, the external forces causing the ten- 
sion act in a direction from the center toward the ends; and 
therefore the internal forces, or stresses, act from the ends 
toward the center. 

(b) Compression. A member is subjected to a compressive 

84 



Art. I. METHODS FOR DETERMINING STRESSES. 85 

Stress, or is in compression, if there is a tendency for the par- 
ticles of the member to move toward each other. If the mem- 
ber is in compression, the external forces causing the compres- 
sion act in a direction from the ends toward the center; and 
therefore the internal forces, or stresses, act from the center 
toward the ends. 

(c) Shear. A member is subjected to a shearing stress, 
or is in shear, if there is a tendency for the particles of the 
member to slide past each other. 

Since the loads are generally applied at the joints, the mem- 
bers are usually subjected to a longitudinal stress only, and 
are either in tension or in compression; although a shearing 
stress may be developed at the connections of the members. 

103. General Methods for Determining Stresses. It has 
been shown in the preceding chapter that the reactions of a 
truss may be determined, graphically, by means of the force 
and the funicular polygons. They may also be determined, 
algebraically, by the three fundamental equations of equilibr 
rium: 

S horizontal components of forces =^0, (i) 

S vertical components of forces = 0, (2) 

2 moment of forces about any point = 0. (3) 

Having found the reactions (see Chapter IX), the stresses 
may be determined either by equations (i) and (2) or by 
equation (3). The first two equations involve the resolution 
of forces, and they may be solved either algebraically or 
graphically. The third equation involves the moments of 
forces, and it may also be solved either algebraically or graph- 
ically. There are, therefore, four methods for determining the 
stresses in a truss. 



Moment of Forces: 



Resolution of Forces: 



Algebraic Method, (a) 
Graphic Method. (b) 



Algebraic Method, (c) 
Graphic Method. (d) 

It is possible to solve the stresses in the members of any 



86 



STRESSES IN ROOF TRUSSES. 



Chap. X. 




Simple truss by using any one of the above four methods; but 
all are not equally well suited to any particular case. It is 
usually the case that a certain one of the four methods is bet- 
ter suited to a particular problem than are any of the other 
three ; and in solving stresses, the problem should be studied 
in order that the simplest solution may be found. 

104. Notation. The notation that will be used to desig- 
nate the members of a truss is 
known as Bow's notation, and 
is shown in Fig. 53. Referring 
to this figure, it is seen that the 

FIG. 53. Notation. Upper chord members, begin- 

ning at the left end of the truss, may be designated by X-i, X-2, 
X-4, X-4', X-2', and X-i'; the lower chord members by Y-i, 
Y-3, Y-5, Y-5', Y-3', and Y-i'; and the web members by 1-2, 

2-3, 3-4» 4-5> 5-5'» 5'-4', 4'-3'> 3'-2', and 2'-i'. The numerical 
value of the stress will generally be written directly on the 
member. 

A tensile stress will be denoted by prefixing a plus ( + ) sign, 
and a compressive stress by prefixing a minus ( — ) sign before 
the number representing the stress. This use of these signs is 
not universal, but the above designation was adopted as it is more 
often employed. 

105. It is the object of this work to deal with graphic — 
rather than with algebraic methods; and since the method of 
graphic moments is not well suited to the determination of 
the stresses in a truss, except to explain other methods, the 
first three methods will be treated only briefly, while this text 
will deal chiefly with the determination of stresses by the 
fourth method — graphic resolution. These four methods will 
now be taken up in the order shown in § 103. 



Art. 2. Stresses by Algebraic Moments. 

The method of algebraic moments furnishes a convenient 
means of finding stresses, especially when the upper and lower 
chords of the truss are not parallel. 



Art. S. 



BTBESSES BY ALGEBBAIC MOMENTS. 



87 



io6. Method of Computing Stresses by Algebraic Mo- 
ments. To obtain the stress in any particular member, cut 
the truss by a section ; and replace the stresses in the mem- 
bers cut, by external forces. These forces are equal to the 
stresses in the members cut, but act in an opposite direction. 
The section should be so taken that the member whose stress 
is required is cut by it; and if possible, so that the other 
members cut by the section (excepting the one whose stress 
is required) pass through a common point, which point is 
taken as the center of moments. To determine the sign of 
the moment and of the resulting stress, assume the unknown 
external force to act away from the cut section, i. e., to cause 
tension. Write the equation of moments, considering the 
external forces which replace the members cut and those on 
one side of the section only, equate to zero (see § 103, equa- 
tion 3), and solve for the unknown stress. The sign of the 
result will then indicate the kind of stress. If it is positive, 
the assumed direction of the unknown force is correct, and 
the stress is tension. If the result is negative, the assumed 
direction i'S incorrect, and the stress is compression, i. e,, the 







X 


p 


p 

Us 




1 ■" 




^ 


p ^ 




hi 


W^\ 


> 


p 
r 


\ 


u 


V 


^\Ay. J AU A\ 


V 


R! 




S 


IV 


p 

f 




Fia. S4, 8tb£si 



88 STRESSES IN ROOF TRUSSES. ^^P- ^' 

equation of moments is not equal to zero unless the assumed 
direction of the external force is reversed. 

The application of the method of algebraic moments to the 
determination of the stresses in a truss will now be shown. 

107. Problem. It is required to find the stresses, due to 
the dead load, in the members of the truss shown in Fig. 54. 

To determine the stress in the member X-i, cut the mem- 
bers X-i and Y-i by the section m-m (see Fig. 54, a), replace 
these members by external forces, assume that the unknown 
external force replacing the required stress X-i acts away 
from the cut section, and take moments about the joint L^ 
(Fig. 54, a). Then 

+ RiXd + X-iXa = o, 

or X-I = — — I (compression). (i) 

a 

Since the sign of the result is negative, the equation shows 
that the external force acts in an opposite direction to that 
assumed, i. e., that it acts in the direction shown in Fig. 54, b, 
and causes compression. 

To find the stress in the member Y-i, use the same section 
and take moments about Uj, assuming that the external force 
replacing the stress in Y-i acts away from the cut section. 
Then 

+ RiXd — Y-iXe=o, 

or Y-i = +.^i211. (tension). (2) 

e 

Since the result is positive, the equation shows that the 
external force acts in the direction assumed, i. e., that it acts 
in the direction shown in Fig. 54, b. 

To find the stress in the member 1-2, cut the members 
X-2, 1-2, and Y-I by the section s-s (see Fig. 54, c), and take 
moments about L^, assuming that the external force which 
replaces the stress in 1-2 acts away from the cut section. Then 

+ PXd + l-2Xd=:o, 

P X d 
or 1-2 = — = — P (compression) . (3) 



'^rt, g. STRESSES BY ALGEBRAIC MOMENTS. 89 

To find the stress in X-2, cut the members X-2, 2-3, and 
Y-3 by the section n-n (see Fig. 54, a and Fig. 54, d), and 
take the moments about Lj, the intersection of 2-3 and Y-3. 
Then 

+ RiXd + X-2Xa=o, 

or X-2 = — — • (compression). (4) 

a 

To determine the stress in 2-3, cut the members by the 
section n-n, assume the external force replacing the stress in 
2-3 to act away from the cut section, and take the moments 
about Lq. Then 

+ PXd — 2-3Xc = o, 

P X d 

or 2-3 = + (tension). (5) 

c 

This equation shows that the assumed direction of the 
external force replacing 2-3 was correct, i. e., that this force 
acts away from the cut section, as shown in Fig. 54, d. 

For the stress in Y-3, cut the members by the section n-n, 
and take moments about Ug. Then 

+ Ri X 2d — P X d — Y-3 X f =0, 
or Y-3=_ 1 (tension). (6) 

For the stress in 3-4, take a section (not shown in the 

figure) cutting X-4, 3-4, and Y-3, and take moments about Lo- 

Then 

-f P X d -f P X 2d + 3-4 X 2d =0, 

— PXd — PX2d , 

or 3-4= J = — ~P (compression). (7) 

For the stress in X-4, take the section p-p, and the center 
of moments at Lg. Then 

-f R, X 2d — P X d ^- X-4 X b = o, 

— Ri X 2d -f P X d 

or X-4== r (compression). (8) 

b 

For the stress in 4-5, take the section p-p, and the center 
of moment at Lo. Then 



90 STRESSES IN ROOF TRUSSES. Cf"*P- ^- 

+ P X d + P X 2d — 4-5 X g=o, 

+ PXd + PX2d +3Pd 
or 4-5 = = — - — (tension) . (9) 

For the stress in Y-5, take the section p-p, and the center 
of moments at Uj. Then 

+ RiX3d — PX2d — PXd — Y-sXh = o. or ¥-5 = 

+ R. X 3d - P X 2d - P X d _ + 3d (R.-P) ^^_,„^ (^^^ 

h h 

The stress in the member 5-5' =o, as may be shown by 
taking a circular section cutting Y-S, 5-5', and Y-s', and the 
center of moments at Lg. 

Since the truss and the loads are symmetrical about the 
center line, it is evident that it is necessary to find the stresses 
for only one-half of the truss. 

In some trusses, it is impossible to take a section so that 
all of the members except the one whose stress is required will 
pass through the center of the moments. If another mem- 
ber cut by the section does not pass through the center of the 
moments, it is necessary to first solve for the stress in this 
member, and then replace this stress by an external force. If 
the stress in this member is tension, the external force is 
taken as acting away from the cut section; and if it is com- 
pression, toward the section. 

If the moment arms for the forces are computed alge- 
braically, considerable work is required ; and these arms may 
generally be most easily found by drawing the truss to a 
large scale, and scaling the moment arms from the diagram. 

One of the most important advantages of the method of 
moments is that the stress in any particular member may be 
found independently of that in any other member. 

io8. By noting the results obtained for the stresses in the 
preceding problem, several conclusions may be drawn as to 
the nature of the stresses in the different members of the truss. 
Referring to equations (i) and (4), § 107, it is seen that the 
stress in X-i is equal to that in X-2 ; and referring to equa- 
tion (3), it is seen that 1-2 is an auxiliary member whose func- 



^^' ^' STRESSES BY QRAPUIC MOMENTS. 91 

tion is to transfer the load P to the joint L^ (by compression). 
Further, if there is no load at Ui, the stress in 1-2 is zero. 
Since the load is transferred to L^ by the member 1-2, the 
stress in X-i must be equal to that in X-2; as the load has 
no horizontal component and all of its vertical component is 
taken up by the member 1-2. There can be no stress in the 
member 5-5' unless there is a load at Lg; and if there is a 
load at that point, the stress in 5-5' is tension, and is equal to 
the load. The member 5-5' is usually put in, its functions 
being merely to support the lower chord and prevent deflec- 
tion. 

Art. 3. Stresses by Graphic Moments. 

The stresses in the members of a truss may be found by 
graphic moments, although this method is not generally the 
simplest that may be used. 

109. Stresses by Graphic Moments. This method is some- 
what similar to that of algebraic moments, as explained in 
Art. 2 ; except that in this method, the moment of the exter- 
nal forces is found graphically instead of algebraically. Since 
the structure is in equilibrium, the moment of the known 
external forces must be equal to the moment of the forces 
which replace the stresses in the members cut by the section. 
If all the members cut, except one, pass through the center of 
moments; then the moment of the external force replacing 
the stress in this member must be equal to the moment of 
the known external forces. To determine the stress in any 
particular member, cut the member by a section, and take 
moments about the intersection of the other members cut. 
To find the moment of the known external forces, construct 
the force and the funicular polygons for these forces. Then 
the moment is equal to the pole distance multiplied by the inter- 
cept. This intercept is measured on a line drawn through the 
center of moments parallel to the resultant of the external 
forces on one side of the section ; and is the distance on this 
line cut off by the strings drawn parallel to the rays meeting 



STRESSES IN ROOF TRUSSES. 



Chap. X.' 



on the extremities of the line representing the magnitude of 
the resnitant in the force polygon. If all the members cut by 
the section, except the one whose stress is required, pass 
through the center of moments, the algebraic sum of the 
moments of the unknown force replacing the stress in this 
member and of the known external forces on one side of the 
section must be equal lo zero. 

The solution of the following problem shows the applica- 
tion of the above method to the determination of the stresses 
in the members of a truss. 

no. Problem. It is required to find the stresses in the 
members of the truss loaded as shown in Fig. 55. 




To determine the stress in the member X-i, cut the mem- 
ber by the section m-m, and take the center of moments at L^, 
Then the moment of the left reaction is equal to + H X ya. 



^rt. 3. STRESSES BY GRAPHIC MOMENTS. 93 

and the equation of moments (assuming the unknown external 
force to act away from the cut section) is 

+ HXy8 + X-iXa=o, 

or X-i=— . (i) 

a ^ ^ 

A plus sign placed before the stress indicates tension, and 
a minus sign indicates compression. 

To find the stress in Y-i, cut the member by the section 
m-m, and take the center of moments at Ui. Then 

+ HXy8 — Y-iXe=o, 

V ^ HXys 

or Y-i = + ■ . (2) 

e ^ ^ 

To find the stress in 1-2, take a section (not shown in the 
figure) cutting X-2, 1-2, and Y-i, and take the center of 
moments at L^. Then 

+ HXyi + i-2Xd=o, 

HXyx 
or 1-2 ^— . (3) 

To find the stress in X-2, take the section n-n, and the 
center of moments at Lj. Then 

+ HXy8 + X-2Xa = o, 

HXy, 
or X-2 = ^—. (4) 

To find the stress in 2-3, take the section n-n, and the 
center of moments at Lo- Then 

+ HXyi — 2-3Xc=o, 

. HXy. 
or 2-3 = H ^— . (5) 

To find the stress in Y-3, take the section n-n, and the 
center of moments at Ug. Then 

+ HXy4 — Y-3Xf=o, 
or Y-3 = + ^^. (6) 



94 STRESSES IN ROOF TRUSSES. Chap. X. 

To find the stress in 3-4, take a section (not shown in the 
figure), and the center of moments at Lj. Then 

+ HXy» + 3-4X2d = o, 

HXy, 
or 3-4 ^3—. (7) 

To find the stress in X-4, take the section p-p, and the 
center of moments at Lj. Then 

+ HXy. + X-4Xb = o, 

• ■ H X y* 
or X-4 = ^. (8) 

To find the stress in 4-5, take section p-p, and the center 
of moments at Lo. Then 

+ HXy2 — 4-5Xg=o, 

. HX.y. 
or 4-5 = H —. (9) 

To find the stress in Y-5, take the section p-p, and the 
center of moments at U3. Then 

+ HXy5 — Y-5Xh=o, 

H Xy, 
or Y-5 = + j^. (ID) 

Referring to the equations of moments given in § no, it 
is seen that the sign of the moment of the external forces to 
the left of any section is always positive. Keeping this in 
mind, it is possible to write the value for the stress in any 
member, together with its proper sign, without first writing 
the equation of moments. The stress in any particular mem- 
ber — irrespective of whether it is tension or compression — is 
equal to the moment of the external forces to the left of the 
section cutting the member divided by the arm of the force 
which replaces the unknown stress in the member. The sign 
of this stress is opposite to that of the moment of the unknown 
external force replacing the stress in the member, assuming 
this stress to act away from the cut section, i. e., if the sign 
of the moment of this external force is negative, the stress is 



Art, 4. 



STBESSES BT ALGEBRAIC RESOLUTION. 



95 



tension ; and if the sign is positive, the stress is compression. 
It is seen that this follows directly from equations of moments. 

Art. 4. Stresses by ALGEBRi\ic Resolution. 



III. Stresses by Algebraic Resolution. The stresses in 
the members of a truss may be found by the application of 
equations (i) and (2), § 103, which are 

2 horizontal components of forces = 0, ( i ) 
2 vertical components of forces =0. (2) 

The above equations may be applied either (a) to the 
forces at a joint, or (b) to the forces on one side of a section, 
including those replacing the stresses in the members cut by 
the section. The method of algebraic resolution is not applica- 
ble if more than two of the forces at a joint or at a section are 
unknown; since there are but two fundamental equations of 
resolution. As it is necessary to find the stresses in some of 
the members before those in other members may be found, 
this method may be most easily explained by solving a par- 
ticular problem rather than a general one. 

In writing the equations, forces acting upward and to the 
right will be considered positive, and those acting downward 
and to the left, negative. 




L 



L. Y 

Stn 60'» 0666 Co5.60*-a500 

Fig. 56. 



(a) Forces at a Joint, To find the stresses in the members 
meeting at a joint, apply equations (i) and (2), §111, assum- 



96 STRESSES IN ROOF TRUSSES. Chap. X. 

ing that the unknown forces replacing the stresses act away 
from the joint, i. e., that they cause tension. If all the forces 
are known except two, these may be found by solving the 
above equations. The signs of the results will determine the 
kind of stress, i. e., a plus sign indicates that the assumed 
direction is correct and that the stress is tension, while a 
minus sign indicates that the assumed direction is incorrect 
and that the stress is compression. 

Problem, It is required to find the stresses in the members 
of the truss loaded as shown in Fig. 56, all loads being given 
in pounds. 

To find the stresses in X-i and Y-i, apply equations (i) 
and (2) to the forces at the joint Lo, assuming that these 
forces act away from the joint. Then 

+ X-isin6o° + Y-i=o, 
and + X-i cos 60° + 3000 = o. 

Substituting the values of sin 60° (0.866) and cos 60° 
(0.500), and solving the equations for X-i and Y-i, we have 

X-I = — 6000 (compression), and Y-i = -j-5i96 (tension). 

To find the stresses in X-2 and 1-2, apply equations (i) 
and (2) to the forces at the joint Uj. Then 

+ X-I sin 60° + X-2 sin 60° + 1-2 cos 60** = o, 

and + X-I cos 60° + X-2 cos 60° — 1-2 sin 60° — 2000 = o. 

Substituting the values of sin 60° and cos 60°, also the 
value already found for X-i, and solving these equations for 
1-2 and X-2, we have 

1-2 = — 1732, and X-2 = — 5000. 

To find the stresses in 2-3 and Y-3, apply equations (i) 
and (2) to the forces at the joint L^. Then 

— Y-I + 1-2 cos 60° + 2-3 cos 60** + Y-3 =* o, 

and — 1-2 sin 60° + 2-3 sin 60° = o. 

Substituting the values of sin 60° and cos 60**, also the 
values of Y-i and 1-2, and solving these equations for 2-3 
and Y-3, we have 

2-3 = + 1 732, and Y-3 = + 3464. 



Art. 4. 



STRESSES BT ALGEBRAIC RESOLUTION. 



97 



Fig. 56, a, shows the truss just solved, together with its 
loads and stresses. This figure illustrates the method of writ-> 
ing the stresses on the members of the truss. 




o 
o 

5? 



+ 5196 

Sin 60'«0.866 



+ 3464 
Y 

Fig. 56, a. 



+5196 
C<»60*-a500 



8 
8 



(b) Forces on One Side of a Section. The method of 
algebraic resolution may also be applied to the forces on one 
side of a section, including those replacing the stresses in the 
members cut by the section. Since the forces replacing the 
stresses in the members cut and the external forces on one 
side of a section must be in equilibrium, we may apply the 
two fundamental equations (i) and (2), § 103, and find the 
unknown stresses, provided not more than two of the stresses 
are unknown. The kind of stress may be determined by 
assuming the unknown external forces replacing the stresses 
in the members cut to act away from the section. A plus 




m n 
Sin 60**.0t666 



P Y 

FiQ. 57. 



Cos 60^0.500 



98 STRESSES IN ROOF TRUSSES. Chap, X. 

sign for the result indicates tension, and a minus sign, com- 
pression. 

Problem. It is required to find the stresses in the mem- 
bers of the truss loaded as shown in Fig. 57, all loads being 
given in pounds. 

To find the stresses in the members X-i and Y-i, take the 
section m-m (Fig. 57), and assume that the unknown forces 
replacing the stresses in X-i and Y-i act away from the cut 
section. If a known stress is considered in any equation, the 
external force replacing it is taken as acting in its true direc- 
tion. Then 

+ X-I sin 60^ + Y-I =0, 

and + X-I cos 60° -f 6000 = 0. 

Substituting the values of sin 60® (0.866) and cos 60® 
(0.500) in these equations, and solving for X-i and Y-i, we 
have 

X-I = — 12 000, and Y-I = -f 10 392. 

To find the stresses in 1-2 and X-2, take the section n-n, 
and assume that the unknown external forces replacing the 
stresses in X-2 and 1-2 act away from the section (the direc- 
tion of the force replacing the stress in Y-i has already been 
found to act away from the section). Then 

-f- X-2 sin 60° -f 1-2 cos 60° + Y-I = o, 

and + X-2 cos 60° — 1-2 sin 60° + 6000 — 4000 = o. 

Substituting the values of sin 60° and cos 60°, also the 
value already found for Y-i, and solving these equations for 
1-2 and X-2, we have 

1-2 = — 3464, and X-2 = — 10 000. 

To find the stresses in 2-3 and Y-3, take the section p-p, 
and assume that the unknown external forces replacing the 
stresses in 2-3 and Y-3 act away from the cut section (the 
force replacing the stress in X-2 has already been found to act 
towards the section). Then 

— X-2 sin 60° + 2-3 cos 60** -+- Y-3 == o, 

and — X-2 cos 60° -f 2-3 sin 60° +6000 — 400Osssso 



-4»'*- ^- STRESSES BY GRAPHIC RESOLUTION. 99 

Substituting the values for sin 60® and cos 60°, also the 
value already found for X-2, and solving for 2-3 and Y-3, we 
have 

2-3 = + 3464, and Y-3 = + 6928. 



Art. 5. Stresses by Graphic Resolution. 

The method of graphic resolution is usually the most con- 
venient one for finding the stresses in roof trusses ; since it is 
rapid and has the advantage of furnishing a check on the 
work. It consists of the application of the principle of the 
force polygon to the external forces and stresses acting at 
each joint of the truss. Since the external forces and stresses 
at each joint are in equilibrium, the force polygon must close, 
and the forces must act in the same direction around the 
polygon (see § 28). As the lines of action of all the forces 
are known, it follows that if a sufficient number of the forces 
at a joint are completely known to permit of the drawing of a 
closed polygon, the magnitudes and directions of the unknown 
stresses may be determined. 

The reactions may be found by means of the force and 
funicular polygons, as explained in Chap. II. As soon as 
these reactions are determined, all the external forces acting 
on the truss are known, and the stresses may then be found. 

112. Stresses by Graphic Resolution. Loads on Upper 
Chord, It is required to find the stresses in the truss loaded 
as shown in Fig. 58. 

The reactions Rj and Rg are first found by constructing the 
force polygon (Fig. 58, b) for the given loads (see § 94, b). 
If both the truss and the loads are symmetrical about the 
center line, it is unnecessary to construct the funicular polygon; 
as each reaction is equal to one-half of the total load. The 
stresses may then be determined by applying the principle 
of the force polygon to each joint of the truss. Referring to 
the joint L^ (Fig. 58, a), of the three forces acting at this 
joint, it is seen that the reaction R^ is completely known. 



536085 



aiBESSES I}f ROOF TBUSSEB. 



while the internal forces or stresses X-i and Y-i are unknown 
in magnitude and direction. These forces are shown in Fig. 
58, c. The force polygon for these forces is constructed by 




(r) Stns33 Diaqnam 



Fia. SS. SIBBSSES BI Gbafbic Bbsoldiion. 



^r^' ^« STRESSES BY GRAPHIC RESOLUTION. 101 

drawing Ri (Fig. 58, c), acting upward, equal and parallel to 
the known left reaction; and, from the extremities X and Y 
of this line, drawing the lines X-i and Y-i parallel respect- 
ively to the members of the truss X-i and Y-i, meeting at 
the point i. Then X-i and Y-i represent to scale the magni- 
tudes of the stresses in the members X-i and Y-i. These 
forces are in equilibrium, so they must act around the polygon 
in the same direction ; and by applying the forces to the joint 
Lq, it is seen that the stress X-i acts towards the joint and is 
compression, while the stress Y-i acts away from the joint 
and is tension. 

The forces at the joint Uj are next taken, instead of those 
at Lj ; since at the former there are but two unknown forces, 
while at the latter there are three. The forces at Ui are shown 
in Fig. 58, d. Since the stress X-i has been found to be com- 
pression, it must act towards the joint Ui. The force P is also 
known, while the internal forces X-2 and 1-2 are known in 
lines of action only. The force polygon for these forces (Fig. 
58, d) is constructed by drawing X-i, acting upward and to 
the right, equal and parallel to the known force X-i, and the 
force P, acting downward, equal and parallel to the known 
load P. The force polygon is then closed by drawing from 
the point X the line X-2 parallel to the member X-2, and 
from the point i, by drawing 1-2 parallel to the member 1-2. 
Then X-2 and 1-2 represent the magnitudes of the unknown 
forces, X-2 acting downward and to the left, and 1-2 acting 
upward. Both X-2 and 1-2 are compression; as may be seen 
by applying these forces to the joint Ui (Fig. 58, a). 

The forces acting at the joint Lj are next considered. 
These forces are shown in Fig. 58, e, the forces 2-3 and Y-3 
being unknown. The unknown forces are found by construct- 
ing the force polygon, as shown in Fig. 58, e. Applying the 
forces in this polygon to the joint Lj, it is seen that both 2-3 
and Y-3 act away from the joint, and therefore the stresses 
are tension. 

The forces acting at the joint Ug are shown in Fig. 58, f. 
Of these, the forces P, X-2, and 2-3 are known, while 3-4 and 



102 STKKSSKS IN KOOF TRUSSES. Chap. X. 

X-4 are unknown. The unknown forces are found by con- 
structing the force polygon shown in Fig. 58, f. Applying the 
unknown forces to the joint Uo, it is seen that both 3-4 and 
X-4 are compression. 

The forces acting at Lg are shown in Fig. 58, g. The 
unknown forces 4-5 and ¥-5 are found by constructing the 
force polygon shown in Fig. 58, g. Both forces act away from 
the joint, and therefore the stresses are tension. 

The forces acting at Ug, together with their force polygon, 
are shown in Fig. 58, h. The stress in X-4' is compression, 
while that in 4^-5' is tension. The stress in 5-5' is zero. 

Since the truss and loads are symmetrical about the center 
line, the force polygons for the joints to the right of the center 
need not be drawn ; as they are the same as those for the left. 

Stress Diagram. Referring to the separate force polygons 
shown in Fig. 58, it is seen that some of the forces in one 
polygon are repeated in another, i. e., it is necessary to find 
some of the stresses in one polygon and use these stresses in 
drawing the next polygon. It is thus seen that these separate 
force polygons may be grouped together in such a manner 
that none of the lines are drawn twice. Again referring to the 
separate diagrams, it is seen that the stress in any member 
acts in one direction in a particular polygon and in an oppo- 
site direction when repeated in another polygon. If these 
force polygons are combined, the line representing the stress 
will therefore have two arrows pointing in opposite direc- 
tions. The force polygon for all the joints of the truss are 
grouped together into the stress diagram shown in Fig. 58, i. 
If the diagram is drawn for the forces at all the joints, as in 
this figure, the last polygon in the stress diagram must check 
with one side on the line representing the known right 
reaction. 

As the stress diagram is being drawn, it is usually most 
convenient to put the arrows representing the directions of 
the stresses at any joint directly upon the diagram of the 
truss; as shown in Fig. 58, a. If this is done, they may be 
omitted in the stress diagram. 



Art. 5. 



STRESSES BY (JUAPHIC UKSOLrXIOX. 



103 



The student should follow through the separate force 
polygons in the stress diagram shown in Fig. 58, i, paying 
particular attention to the fact that the forces at a joint, 
whose magnitudes are represented by a closed polygon in the 
stress diagram, must act in the same direction around the 
polygon. Great care should be used in drawing the truss, and 
in transferring the lines from the truss to the corresponding 
lines in the stress diagram to secure accuracy. 

113. Stresses by Graphic Resolution. Loads on Longer 
Chord, In the problem given in § 112, and in the preceding 
problems in this chapter, the loads have all been applied at 
the upper chord panel points; as it is the usual practice to 
consider all the dead load on the upper chord. In some build- 
ings, however, the ceiling loads and other miscellaneous loads 
are supported at the lower chord panel points ; and the follow- 




Fio. 50. Stress Diagram — Loads on Lower Chord. 



104 8TKE88E8 IN ROOF TRU88ES. ^^«P- ^' 

ing problem will show the methods used in drawing the stress 
diagram for a truss loaded at the lower chord panel points 
only. 

Problem. It is required to find the stresses due to ceiling 
loads in all the members of the truss sho^n in Fig. 59. The 
trusses are spaced 15 ft. apart, each truss having a span of 40 
ft. and a rise of 10 ft. The ceiling load is 10 lbs. per sq. ft. 

40 

The panel load P is equal to ^ X IS X 10 = 1000 lbs., 

and the effective reactions are each equal to 2j4 X 1000 = 
2500 lbs. 

To draw the stress diagram, lay off the reactions Rj and 
R2 on the load line, as shown in Fig. 59, and start the diagram 
with the forces acting at the left end of the truss. The stress 
diagram is drawn considering the forces at the joints in the 
following order : Lo, Uj, L^, Uj, Lg, Lg, U,,, L/, Ug', L/, U/, and 
and L„'. The arrows indicating the kind of stress are placed at 
each panel point of the truss, as the portion of the stress 
diagram for that panel point is drawn. The arrows, with the 
exception of those in the load line, are omitted in the stress 
diagram. Since the truss and the loads are symmetrical about 
the center line, it is seen that the stresses are all determined 
as soon as the diagram is drawn for the forces up to and 
including those acting at L3. It is therefore unnecessary to 
draw the diagram for the right half of the truss. 



CHAPTER XI. 



WIND LOAD STRESSES. 



This chapter will treat of the stresses in rcx)f trusses due 
to wind loads for different conditions of the ends of the 
trusses. The method for finding the wind load stresses in roof 
trusses by graphic resolution will be shown for the four fol- 
lowing cases: (a) Both Ends of Truss Fixed — Reactions 
Parallel ; (b) Both Ends of Truss Fixed — Horizontal Com- 
ponents of Reactions Equal; (c) Leeward End of Truss on 
Rollers ; (d) Windward End of Truss on Rollers. 



Art. I. Both Ends of Truss Fixed — Reactions Parallel. 

When the roof truss is comparatively flat, it is usually 
customary to assume that both reactions are parallel to the 
resultant of the wind loads. The wind load stresses in the 
members of a roof truss will be found by the method of 
graphic resolution, assuming that the reactions are parallel. 

114. Problem. It is required to find the wind load stresses 
in all the members of the truss shown in Fig. 60. The truss 
has a span of 60 ft., a rise of 15 ft., and the trusses are spaced 
15 ft. apart, center to center. The lower chord has a camber 
of 2 ft., and the normal component of the wind is taken at 23 
lbs. per sq. ft. (§ 93). 

After computing the loads, the reactions are found by 
means of the force and the funicular polygons (see § 98, a). 
The left reaction is represented by R^, and the right reaction 
by Ro (Fig. 60). 

105 



106 



WIND LOAD STRESSES. 



Chap. XI. 



The stress dia^am is started by drawing the f^rce polygon 
for the forces at the joint L„ (Fig, 60). The stresses in the 
members X-i and Y-i are unknown, white the wind load at 
Lj and the left reaction are known. The unknown stresses at 

this joint are determined by drawing the force polygon — , 
X, I, Y, R,, noting particularly that the polygon closes, and 
that the forces act in the same direction around the polygon. 




The lines X-l and Y-r in the force polygon represent the 
stresses in the members X-i and Y-i. By placing the arrows 
on the members at the joint Lo, corresponding to the direc- 
tions of the forces in the force polygon, it is seen that X-i is 
compression and that Y-i is tension. 

The force polygon, or stress diagram, for the forces at the 
joint Ui is X, X, 2, I, X, the sequence of the letters and figures 
denoting the directions of the forces. The kind of stress in 



ArLl. ENDS FIXED — REACTIONS PARALLEL. 107 

the unknown members X-2 and 1-2 is determined by placing 
arrows on the members meeting at the joint Uj. The stresses 
in both these members are compression. 

The stress diagram for the forces at the joint Lj is Y, i, 
2, 3, Y, the sequence of the letters and figures denoting the 
directions of the forces, which directions are shown by arrows 
placed on the members at the joint L^. The stresses in both 
'2-3 and Y-3 are thus found to be tension. 

The unknown stresses at the joint Uj are X-4 and 3-4, 
and the stress diagram for the forces at the joint Ug is X» X, 

4, 3> 2, X. 

The unknown stresses at the joint- Lg are 4-5 and Y-s, 
and the stress diagram for the forces at Lg is Y, 3, 4, 5, Y. 

The unknown stresses at U3 are X-6 and 5-6, and the stress 
diagram for the forces at this joint is X, X, 6, 5, 4, X. 

At the joint Lg, the unknown stresses are 6-7 and Y-7, 
and the stress diagram for the forces at this joint is Y, 

5, 6, 7, Y. 

At the joint U4, the unknown stresses are X-(6'-i') and 
6'-7, and the stress diagram for the forces at this joint is X, 
X, 6'-I^ 7, 6, X. 

The unknown stress at the joint L, is Y-(6'-i'), and the 
stress diagram for the forces at this joint is Y, 7, 6'-i', Y. 

The stress diagram for the forces at the joint Lq' 5s Y, 
(6'-i0, X, Y. 

The numerical values for the stresses in all the members 
of the truss may be found by scaling the corresponding lines 
in the stress diagram to the given scale. 

It is seen from the stress diagram that there are no stresses 
in the members 6'-5', 5'-4', 4'-3^ 3'-2^ and 2'-i'. This fol- 
lows, since there are no intermediate loads between the joints 
U4 and Lo', and the members X-(6'-iO, 7-6', and Y-(6'-i') 
form a triangle. It is further seen that it is necessary to draw 
the stress diagram for the complete truss, and that the mem- 
bers X-(6'-i') and Y-(6'-i') must form a triangle, one side 



of which is the known reaction Rj, 
the work. 



WIND LOAD 8TBE88E8. Chap. XI. 

hich furnishes a check on 



Art, 2. Both Ends of Truss Fixed — Horizontal Components 
OF Reactions Equal. 



When the horizontal component of the wind loads is large 
and the supports are equally elastic, actual conditions are 
most nearly approximated by taking the horizontal components 
of the reactions equal. The horizontal component of the wind 
loads may be large if the roof is steep, or if a ventilator such 
as shown in Fig. 6i is used. The wind load stresses in a 
roof truss having a "monitor" ventilator will now be found,. 
assuming that the reactions have equal horizontal components. 

115. Problem, It is required to find the wind load stresses 
in all the members of the truss shown in Fig. 61. The truss 
has a span of 50 ft., a pitch of one-fourth, and has a monitor 




Load 

Stress Diogram 
)" 3000* 6000* 



B Fixed — Hok. Ccup. of Reactions Bqcui. 



-^r*. ^. ENDS FIXED — HOR. COMPS. EQUAL. 109 

ventilator, as shown in Fig. 6i. The trusses are spaced i6 ft. 
apart, center to center. The wind load on the vertical surface 
of the ventilator is taken at 30 lbs. per sq. ft., and the com- 
ponent of the wind, normal to the roof surface, is taken at 23 
lbs. per sq. ft. 

The wind loads at the joints U3 and U4 are found by taking 
4:he resultants of the horizontal and inclined loads acting at 
these joints (see Fig. 61). The reactions Rj and Rj are 
determined by the methods explained in § 98, b, assuming first 
that the reactions are parallel and finding these reactions by 
means of the force and the funicular polygons ; and then mak- 
ing their horizontal components equal. 

The stresses in the members of the truss are found by 
drawing the force polygons for the forces at each joint and 
then combining these polvgons, taking the joints in the fol- 
lowing order: Lo, Ui, lii, ^2, I^2» Ug, L3, U4, U5, U/, U„, 
U3', Lg^ and Lo'. The complete stress diagram is shown in 
Fig. 61. When the stress diagram has been drawn for the 
' forces up to those acting at U3, it is seen that there are three 
unknown stresses at this joint, viz.: 5-6, 6-7, and X-7. The 
stress in X-7 is taken as zero, and the load acting at U4 is held 
in equilibrium by the stresses in the two members X-8 and 
7-8. There are no stresses in the members S'-4', 4'-3', 
3^-2', and 2'-i'. The numerical value of the stresses in the 
members may be found by scaling the corresponding lines in 
the stress diagram to the given scale. The kind of stress, 
whether tension or compression, is given by the arrows placed 
on the members of the truss, as shown in Fig. 61. If the 
arrows act awa^ from the center of the member, i. e., toward 
the joints, the stress is compression; and if they act toward 
the center of the member, i. e.j away from the joints, the stress 
is tension. ' 

The student should carefully follow through the construc- 
tion of the stress diagram, placing the arrows, which show the 
directions of the forces at the joint, on the members, as the 
force polygon for that joint is drawn. 



110 



WIND LOAD STRESSES. 



Chap, XI. 



Art. 3. Leeward End of Truss on Rollers. 

If the span of the truss is large, the change in its length 
due to the loads and to the temperature variations is usually 
adjusted by placing one end of the truss on rollers. The 
stresses in a roof truss with its leeward end supported on 
rollers will now be determined. 

116. Problem. It is required to find the wind load 
stresses in all the members of the "Fink" truss shown in Fig. 
62, the leeward end of the truss being supported on rollers. 
The span of the truss is 60 ft., the rise 20 ft., and the lower 
chord is cambered 3 ft. The trusses are spaced 16 ft. apart, 
and the normal component of the wind is taken at 26 lbs. per 
sq. ft. 




Wind Load 
Stress Diaqram 
5000* 10000* 



J. 



Fig. G2. Leeward End of Truss ox Rollers. 



The reactions Rj and R^ (Fig. 62) are found by means of 
the force and funicular polygons, the funicular polygon being 



^rt. 3. LEEWARD EXD ON ROLLERS. Ill 

Started at the left point of support of the truss (see § 99, a). 
The stresses in the members intersecting at the joints Lq, Uj, 
and Lj are found by drawing the stress diagram for these 
forces, as in the preceding articles. Referring to the forces at 
the joint Uj, it is seen that there are three unknown stresses 
at this joint, viz. : X-5, 4-5, and 3-4 ; and therefore the stress 
diagram cannot be drawn for this joint. There are also three 
unknown stresses at Lj. The unknown stresses at U2 may, 
however, be found as follows : Replace the members 4-5 and 
5-6 by the auxiliary dotted member connecting the joints Lj 
and Ug (Fig. 62). Now draw the stress diagram for the forces 
at the joint Ug, this diagram being X,X,4',3,2,X. Next draw 
the stress diagram for the forces at U3, which is X,X,64^X. 
Now the line X-6 in the stress diagram represents the true 
stress in the member X-6. After this stress is determined, 
remove the dotted auxiliary member, and replace it by the 
original members 4-5 and 5-6. Two of the forces (P, and 
X-6) at Ug are now known, and the stress diagram may be 
drawn for this joint, this diagram being X,X,6,5,X. Four of 
the forces at Ug are now known, the unknown forces being 
3-4 and 4-5. The stress diagram for the forces at this joint is 
X,X,5,4,3,2,X. The stress diagram for the forces at the remain- 
ing joints may be drawn in the following order: Lg, M, U^, 
Lg', Lo'. There are no stresses in the members 6'-5', 5'-4', 
4^-3^ 3'-2^ and 2'-i'. The complete stress diagram for all 
the members of the truss is shown in Fig. 62. The kind of 
stress, whether tension or compression, is denoted by the 
arrows placed on the members of the truss ; and the numerical 
values of the stresses may be found by scaling the lines in the 
stress diagram to the given scale. 



Art. 4. Windward End of Truss on Rollers. 

Since the wind may act from either of two opposite direc- 
tions, it is necessary to find the stresses in the members of the 
truss when the rollers are under the leeward end of the truss 



112 



WIND LOAD STRESSES. 



Chap. XI. 



and also when the rollers are under the windward end. The 
stresses in a roof truss with its windward end on rollers will 
now be found. 

117. Problem. It is required to find the wind load 
stresses in all the members of the "Camels Back" truss shown 
in Fig. 63, the windward end of the truss being supported on 
rollers. The span of the truss is 60 ft, the rise 14 ft., and, the 
trusses are spaced 15 ft. apart. 




Wind Load 
Stress Diaqram 
0* 3000* 



Fig. 63. 



Windward End of Truss on Rollers. 



The normal components of the wind loads are found from 
the diagram based on Duchemin's formula (see Fig. 47), 
assuming that P equals 30 lbs. per sq. ft. The reactions Rj 
and R2 (Fig. 63) are found by means of the force and funicular 
polygons (see § 99, b). 

The complete stress diagram for all the members of the 
truss is shown in Fig. 63. The numerical values of the stresses 
may be found by scaling the lines in the stress diagram to the 
given scale. The kind of stress in each member, whether 
tension or compression, is shown by the arrows placed on the 
members of the truss. Referring to the arrows on the truss, it 
is seen that the left half of the lower chord is in tension, while 
the right half is in compression. It is also seen that the mem- 



Art. 4. WINDWARD EXD ON ROLLERS. 113 

bers 1-2, 3-4, and 5-5' are in tension, while the corresponding 
members i'-2' and 3'-4' are in compression; and further that 
the members 2-3 and 4-5 are in compression, while 2'-3' and 
4'-5' are in tension. Since the wind may act from either 
direction, it is seen that these members will be subjected to 
reversals of stress. 



} 



CHAPTER XII. 

STRESSES IN CANTILEVER AND UNSYMMETRICAL 
TRUSSES— MAXIMUM STRESSES. 

■ 

This chapter will be divided into two articles. The first 
article will treat of the application of the method of graphic 
resolution to the solution of stresses in cantilever and unsym- 
metrical trusses, showing a method for drawing a combined 
stress diagram, and the second will treat of the determination 
of the maximum stresses in trusses. 

Art. I. Stresses in Cantilever and Unsym metrical Trusses. 

A cantilever truss is one which is supported at one end 
only, the other end being entirely free. Such a truss is often 
used to project over platforms and entrances to buildings. 
The cantilever truss may be fastened to the walls of the build- 
ing or to the columns supporting the main trusses. 

ii8. Stresses in a Cantilever Truss. Problem, It is 
required to find the dead load stresses in the cantilever truss 
shown in Fig. 64. The span of the truss is 24 ft. The trusses 
are spaced 15 ft. apart, center to center, and support a dead 
load of 12 lbs. per sq. ft. of the horizontal projection of the 
roof. 

The point of application A (Fig. 64) and the line of action 
of the reaction R, are known; while the point of application, 
only, of the reaction R^ is known. The line of action of Ri 
may be found by applying the principle that, if a body is in 
equilibrium under the action of three external forces, these 

114 



Art. 1. 



STRESSES IN A CANTILEVER TRUSS. 



115 



forces must all intersect at a common point. One of these 
three forces is the resultant of the loads acting on the truss, 
the other two being the two reactions. 



El 








5 1 

Stress Diaqram 

0* 1000* zooo* 
I i I 

Fig. 64. Stress Diagram — Cantilever Truss. 



The resultant R of the loads acting on the truss is found 
by drawing the force and funicular polygons for these loads, 
as shown in Fig. 64. Now produce the lines of action of R and 
R2 until they intersect at the point C. Then the line of action 
of Rj must pass through the points B and C. The magnitudes 
of these reactions are found by drawing the force triangle for 
the loads and the two reactions (see Fig. 64). Then Y-7 
represents the magnitude of the reaction Rg, and X-7 that of Rj. 

The stresses in the members of the truss are found by 
drawing the stress diagram, starting the diagram with the 
forces acting at the point B. The forces acting at D are then 
taken, noting that the -stress in the member Y-7 is equal to 



116 CANTILEVER AND UN8YMMETRICAL TRUSSES. Chap. XII. 

the known reaction Rg. The complete stress diagram is shown 
in Fig. 64. The numerical values of the stresses may be found 
by scaling the lines in the stress diagram to the given scale. 

The stress diagram might also have been drawn without 
first finding the reactions by starting the diagram with the 
forces acting at E, the left end of the truss. 

Referring to the truss and to the stress diagram (Fig. 64), 
it is seen that the stress in the member 6-7 would be zero if 
the inclination of the member Y-7 was changed so that the 
lines of action of R and Rj would intersect on the member 
X-2 ; and further, that the stress in 6-7 would be compression 
if the lines of action of R and Rg intersected above the mem- 
ber X-2. 

iig. Unsymmetrical Truss — Combined Stress Diagram. 

In the problems that have been given (excepting that given in 
§ 113), the loads have all been applied at the panel points of the 
upper chord of the truss ; and separate stress diagrams have been 
drawn for the dead and for the wind loads. This section will 
treat of the loads on both the upper and lower chords; and the 
stresses due to the dead loads and to the wind loads will be found 
by drawing a single diagram. The method for drawing the com- 
bined stress diagram will now be shown by the following problem. 

Problem. It is required to find the stresses in the members 
of the truss shown in Fig. 65, the leeward end of the truss 
being supported upon rollers. The span of the truss is 50 ft., 
the rise i6f ft., and the adjacent trusses are spaced 14 ft. apart, 
center to center. The dead load is taken at 12 lbs. per sq. ft. 
of horizontal projection, the ceiling load at 10 lbs. per sq. ft. 
of horizontal projection, and the wind load at 26 lbs. per sq. 
ft. of roof surface. 

The resultants of the dead and wind loads at each panel 
point of the upper chord are first found. These resultants are 

• 

shown on the truss diagram (Fig. 65). The reactions Rj anc 
Ro, considering the right end of the truss on rollers, are founc 
for these loads by means of the force and funicular polygons. 
These reactions are shown in the force polygon, but are not 
shown on the diagram of the truss. The reactions due to the 



COMBINED STRESS DIAORAM. 



loads on the lower chord are then found, these reactions heing 
each numerically equal to two lower chord panel loads. 




FlO. SB. UNBTMHBtBICAI. THOBS— COMBIIIKD 



The left reaction due to all the loads is then determined by 
finding the resultant of the reactions due to the dead and 
wind loads on the upper chord and to the ceiling loads on the 
lower chord. This reaction is represented by R,' (Fig. 65), 
The right reaction is determined in the same manner, and is 
represented by R^'. 

The stresses in the members of the truss are now found 
by drawing the stress diagram, as shown in Fig. 65. The 
numerical values of the stresses may be found by scaling the 
lines in the stress diagram to the given scale. 

To determine which condition gives the largest stresses, it 
is necessary to take the wind as acting both from the right 
and from the left; and further, to assume that the rollers may 
be under either end of the truss. 

The above method of laying oiT the loads in the load line, 



118 MAXIMUM STRESSES. (^^P- -^^^• 

and of determining the reactions, is applicable to all trusses 
carrying loads on both the upper and lower chords. It is seen 
that this method places the loads in their proper order in the 
load line. 

Art. 2. Maximum Stresses. 

120. Maximum Stresses. In the preceding articles, 
methods have been given for finding the stresses in various 
types of trusses, due separately to the dead load, to the snow 
load, and to the wind load. In this article, the maximum 
stresses that may c/ccur in the members of a truss due to the 
combined effect of the different loadings will be found. The 
stresses will first be found separately for the different load- 
ings, and then combined to determine the maximum stresses. 

If the truss and also the loads are symmetrical about the 
center line, each reaction is equal to one-half of the total load 
on the truss (neglecting the half loads at the ends). In this 
case, it is unnecessary to draw the funicular polygon to deter- 
mine the reactions. The dead load stresses may then be found 
by drawing the stress diagram for the dead loads. 

Since the snow load is always taken at so much per square 
foot of the horizontal projection of the roof, it is seen that it is 
unnecessary to draw a stress diagram for the snow loads ; but 
that the snow load stresses may be determined directly by 
proportion from the known dead load stresses. The minimum 
snow load in this work will be taken at lo pounds and the 
maximum snow load at 20 pounds per square foot of the hori- 
zontal projection of the roof. 

If the truss is symmetrical and is fixed at both ends, the 
wind need only be taken as acting from one side; since the 
stresses in the corresponding members would be the same as 
if the wind was taken from the other side. If the truss has 
rollers under one end, the wind must be taken as acting both 
from the roller side and from the fixed side of the truss. The 
condition which gives the larger stresses is then considered in 
making the combinations for maximum stresses. 



^^' ^' MAXIMUM STRESSES. 119 

The dead load is always acting upon the truss, and there- 
fore it must be used in all the combinations for maximum 
stresses. If the maximum snow load is taken, then the wind 
load will be neglected; as it is improbable that the maximum 
wind and the maximum snow will ever occur at the same 
time. If the maximum wind is considered, then the minimum 
snow load will be used. 

From the above discussion, it is seen that the following 
combinations should be made to obtain the maximum stresses 
in the members of a truss when there are no reversals of stress : 

(a) Dead load stress plus maximum snow load stress. 

(b) Dead load stress plus minimum snow load stress plus 
wind load stress (rollers under leeward end of truss). 

(c) Dead load stress plus minimum snow load stress plus 
wind load stress (rollers under windward end of truss). 

The method for finding the maximum stresses in the mem- 
bers of a truss will now be shown by the solution of a prob- 
lem. 

121. Problem. It is required to find the maximum 
stresses in the members of the Fink truss shown in Fig. 66. 
The dimensions of the truss, together with the loadings taken, 
are shown in the following table : 
Span of truss = 80 ft. 
Rise of truss = 20 ft. 
Distance between trusses = 16 ft. 
Dead load taken at 12 lbs. per sq. ft. hor. proj. 
Minimum snow load taken at 10 lbs. per sq. ft. hor. proj. 
Maximum snow load taken at 20 lbs. per sq. ft. hor. proj.. 
Wind load taken at 23 lbs. per sq. ft. of roof surface. 
Since one-half of the upper chord is divided by the web 
members into eight equal parts, preliminary computations 
give the following panel loads : 

Dead panel load 960 lbs. 

Minimum snow panel load 800 lbs. 

Maximum snow panel load 1600 lbs. 

Wind panel load 2060 lbs. 

Wind loads at end and apex 1030 lbs. 



120 



MASIUUM STRESSES. 



Chap. zn. 



Since the dead loads are symmetrical about the center line, 
the effective dead load reactions are each equal to one-half of 
the total dead load (neglecting the half loads at the ends of 
the truss). The dead load stresses are found by drawing the 
stress diagram, as shown in Fig. 66, b. 




DiAGRAUa — Pine Truhg (No Cambeb) 



The minimum and maximum snow load stresses are 
obtained from the dead load stresses without drawing another 
stress diagram. The ratio of the minimum snow load to the 



^^' ^- MAXIMUM STRESSES. 121 

dead load is as 5 is to 6, and that of the maximum snow load 
to the dead load is as 5 is to 3. The minimum and maximum 
snow load stresses are obtained by multiplying the correspond- 
ing dead load stresses by these ratios. 

The wind load reactions, considering the leeward end of 
the truss on rollers, are found by means of the force and 
funicular polygons. These reactions are represented by Rj 
and R2 (Fig. 66, c). The wind load stress diagram is shown 
in Fig. 66, c. The apparent ambiguity at some of the joints 
is overcome by substituting the dotted members in the truss 
diagram and the corresponding ones in the stress diagram, as 
is shown in Fig. 66, a.and Fig. 66, c (see § 116). 

The wind load reactions, considering the windward end of 
the truss on rollers, are represented by R/ and Kg' (Fig. 66, 
c). The wind load stress diagram for this case is also shown 
in Fig. 66, c. It is seen that the stresses in all the upper chord 
and web members for this case are the same, and tliat the 
stresses in the lower chord members are smaller than they 
were when the rollers were considered under the leeward end. 

The stresses in the members of this truss due to the differ- 
ent loadings, together with the maximum stresses, are shown 
in tabular form in the upper half of the table given in Fig. 68. 
It is seen that, for this particular truss, there are no reversals 
of stress in any of the members. 

122. Problem, It is required to find the maximum 
stresses in the members of the truss shown in Fig. 67, the lower 
chord being cambered two feet. The other dimensions of the 
truss, together with the dead, snow, and wind loads, are the same 
as for the problem given in the preceding section. 

The minimum and maximum snow load stresses are obtained 
by proportion from the dead load stresses (see § 121). 

The dead load stress diagram is shown in Fig. 67, b. 

The wind load stress diagram, considering the leeward end 
of the truss on rollers, is shown by the full lines in Fig. 67, c. 
The wind load stress diagram, considering the windward end 
of the truss on rollers, is shown by the broken lines in Fig. 
67, c. 



122 



HAXIHUU BTKE88BS. 



Chap. XIJ. 



The stresses in the members of the truss due to the dif- 
ferent loadings, together with the maximum stresses, are 
shown in the lower half of the table given in Fig. 68. It is 
seen that, for this truss, there are no reversals of stress in any 
of the members. 

By comparing the stresses given in Fig. 68, it is seen that 
even a small camber in the lower chord increases the stresses 



^' Stress Diagram 

O" 5000* kXXX)* 




Fta 67. STBess DiAQRAua — fine: Truss (With Caubeb). 

in most of the members a considerable amount. A small 
camber, however, improves the appearance of the truss, and 
also increases the clearance in the building. If no camber is 



Art, 2. 



MAXIMUM STRESSES. 



123 



STRESSES IN A FINK TRUSS (No Comber ). 



Truss 
Member 



Dead 

Load 

Stress 



Snow Load 
Stress 



Wind Load Stress 



Min. 



Max. 



Rollers 
Leeward 



Rollers 
Windward 



Max- 
imum 
Stress 



X-l 
X-2 
X-5 
X-6 
X-9 

x-io 

X-13 

X-14 

X-(14'- 

Y-l 

Y-3 

Y-7 

Y-15 

Y-(I5- 

1-2,5 

2-3.4 

3-4,11 

4-7,8 

6-7,8 

7-8 

8-15 

12-15 

14-15 



1') 



r) 

6,9-10,13-14 
•5,10-11,12-13 
-12 
-II 
9 



-16100 
- 1 5700 
- 1 5300 
-14800 
-14400 
- 14000 
- 1 3600 
-13100 

■1-14400 
+13500 
+11600 
+ 7800 



600 
900 
1700 
1400 
2800 
3400 
3800 
5700 
6700 



+ 
+ 

+ 
+ 



-13400 
-13100 

- 12700 
-12300 

- 12000 
-I 1700 
-I 1300 
-10900 

+ 12000 
+ 11300 
+ 9700 
+ 6500 



700 
800 
1400 
1200 
2300 
2800 
3200 
4500 
5600 



+ 
+ 

+ 
+ 



-26800 
-26200 
-25500 

- 24700 
-24000 
-23300 

- 22700 
-21800 

+ 24000 
+ 22500 
+ 19300 
+ 13000 



1300 
1500 
2800 
2300 
4700 
5700 
6300 
9500 
I 1200 



+ 
+ 

+ 
+ 
+ 



-20600 

- 20600 

- 20600 

- 20600 

- 20600 
-20600 

- 20600 

- 20600 

- 10300 
+ 25400 
+ 23000 



+ 
+ 
+ 



+ 
+ 

+ 
+ 
+ 



18400 
9200 
9200 
2100 
2300 
4100 
4600 
6900 
8Z00 
9200 
13600 
16100 



-20600 
-20600 

- 20600 

- 20600 

- 20600 

- 20600 
-20600 

- 20600 

- 10300 
18000 
15700 
II 100 

1900 

1900 

2100 

2300 

4100 

4600 

6900 

8200 

9200 

13800 

16100 



+ 
+ 

+ 
+ 
+ 



+ 
+ 



+ 
+ 
+ 



-50100 
-49400 
-48600 
-47700 
- 47000 
-46300 
-45500 
-44600 

+ 51800 
+ 47800 
+ 39700 
+ 23500 



3600 
4000 
7200 
7200 
12000 
14400 
16200 



+ 
+ 



+ 24300 
+ 28400 



STRESSES IN A FINK TRUSS (With CamberV 


X-l 


-19300 


-16100 


-32200 


-265*00 


-24500 


-61600 


X-2 


-18800 


- 1 5700 


-31400 


-26200 


-24500 


-60700 


X-5 


- 1 8400 


- 1 5300 


-30700 


-26200 


-24500 


-59900 


X-6 


-18000 


-15000 


-30000 


-26200 


-24500 


- 59200 


X-9 


-17600 


-14700 


-29300 


-26200 


-24500 


- 58500 


X-IO 


-17200 


-14300 


-28600 


-26200 


- 24500 


- 57700 


X-13 


- 1 6700 


-13900 


- 27800 


-26200 


-24500 


- 56800 


X-14 


-16300 


-13600 


-27200 


-26200 


- 24500 


-56100 


X-(I4'-I') 








-12300 


- 1 0700 




Y-l 


+17300 


+ 14400 


+ 28800 


+ 30400 


+ 21600 


+ 62100 


Y-3 


+ 1 6200 


+13500 


+ 27000 


+27700 


+18800 


+ 57400 


Y-7 


+13900 


+1 1600 


+ 23200 


+22200 


+13300 


+ 47700 


Y-15 


+ 8600 


+ 7200 


+14300 


+10200 


+ 2000 


+ 26000 


Y-(7'-l') 








+ 1 1 100 


+ 2200 




1-2,5-6,9-10,13-14 


- 800 


- 700 


- 1300 


-2100 


- 2100 


- 3600 


2-3,4-5,10-11,12-13 


+ UOO 


+ 900 


+ 1800 


+ 2800 


+ 2800 


+ 4800 


3-4.H-I2 


- 1700 


- 1400 


- 2800 


- 4100 


- 4100 


- 7200 


4-7,8-11 


+ 2300 


+ 1900 


+ 3800 


+ 5500 


+ 5500 


+ 9700 


6-7.8-9 


+ 3400 


+ 2800 


+ 5700 


+ 8300 


+ 8300 


+ 14500 


7-8 


- 3400 


- 2800 


- 5700 


- 8200 


- 8200 


-14400 


8-15 


+ 5600 


+ 4700 


+ 9300 


+12300 


+ 11300 


+ 22600 


12-15 


+ 7900 


+ 6600 


+13200 


+17800 


+16800 


+ 32300 


14-15 


+ 9000 


+ 7500 


+15000 


+20600 


+ 19600 


+ 37100 


l5-(l4-8'/ 








+ 1200 


+ 200 





Fia. 68. Table of Stresses — Maximum Stresses. 



124 MAXIMUM STRESSES. (^^P- ^II» 

used and the span is long, the lower chord has the appearance 
of sagging. 

123. Maximum and Minimum Stresses. By the term? 
"maximum and minimum stresses'* are meant the greatest 
ranges of stress that may occur in any member due to the 
different loadings. If there is no reversal of stress in any 
member, the maximum stress is the greatest numerical stress, 
and the minimum is the smallest numerical stress that may 
ever occur in any member. If there is a reversal of stress, the 
maximum is the greatest numerical stress, and the minimum 
is the greatest numerical stress of an opposite kind that may 
ever occur. 

In finding maximum and minimum stresses, it must be 
borne in mind that the dead load is always acting, and that 
there is no reversal of stress in any member unless the wind 
load stress, or stress due to another condition of loading, in 
that member is greater in amount and is of an opposite kind 
to that caused by the dead load. 

In designing, there is no need of finding minimum stresses 
unless there are reversals of stress ; since if there are no rever- 
sals, the members must be designed for their maximum 
stresses, while if there are reversals, they must be designed for 
each kind of stress. 

If both the maximum and minimum stresses are required, 
the following combinations should be made : 

(a) Dead load stress alone. 

(b) Dead load stress plus wind load stress (rollers lee- 
ward). 

(c) Dead load stress plus wind load stress (rollers wind- 
ward). 

(d) Dead load stress plus maximum snow load stress. 

(e) Dead load stress plus minimum snow load stress 
plus wind load stress (rollers leeward). 

(f) Dead load stress plus minimum snow load stress plus 
wind load stress (rollers windward). 



CHAPTER XIII. 



COUXTEEBEACIXG. 



The use of counterbracing adds considerably to the work 
required to find the maximum stresses in a truss; since the 
same diagonals are not always stressed. In many cases, how- 
ever, its use is more economical than to design the member for 
a reversal of stress. There are two methods which may be 
used to determine the stresses in a truss with counterbracing, 
viz. : (a) by the use of separate stress diagrams, and (b) by 
the use of combined stress diagrams. The former method 
may be used to advantage when several different combinations 
must be made to determine the maximum stresses, while the 
latter is useful when but few combinations are required. 

The determination of stresses in trusses with counterbrac- 
ing will now be taken up in three articles, as follows: Art. i. 
Definitions and Notation; Art. 2, Stresses in Trusses with 
Counterbracing — Separate Stress Diagrams; and Art. 3, 
Stresses in Trusses with Counterbracing — Combined Stress 
Diagrams. 

Art. I. Definitions and Notation. 

124. Definitions. The triangle is the only geometrical 
figure which is incapable of any change in shape without a 
change in the length of one or more of its sides. The triangle 
is therefore the elementary form of truss; and the trusses in 
common use consist of a series of triangles so arranged as to 
form a rigid body. 

A polygonal figure which is composed of more than three 

125 



126 



COUNTERBRAOINO. 



Chap. Xni, 



sides and which is free to turn at its joints may be distorted 
without changing the length of any of its sides. For example, 
take the quadrilateral frame shown in Fig. 69, a, the frame 
being acted upon by an external force. This figure may be 
distorted without changing the Icfngth of any of its sides, as 
is shown by the dotted lines in the figure. 



i 


^ 


B 




>A^-- 


i 




/ 


/ 




/ 


/ 




/ 


/ 




/ 


/ 




/ 


/ 




' 


1 





(a) 



CD C 

( b) (c) 

Fig. 69. 




Now suppose a diagonal, connecting the points B and D, 
is added to the frame ABCD, as shown in Fig. 69, b. The 
figure is then composed of two triangles, and any force acting 
towards the right will tend to distort the frame, and will pro- 
duce tension in the diagonal BD. The tension produced in this 
diagonal will prevent any distortion of the frame. Now sup- 
pose that the external force acts towards the left, as shown in 
Fig- 69, c. The distortion of the frame will be prevented by 
the diagonal BD, which will be in compression. 

It is thus seen that the diagonal BD, which is capable of 
resisting both tension and compression, will prevent any dis- 
tortion in the frame ABCD. The same is true of the diagonal 
AC. If the diagonal member is capable of resisting tension 
only, or compression only; then two diagonals will be re- 
quired to prevent distortion, as shown in Fig. 69, d. If two 
diagonals are used, and both are designed to take the same 
kind of stress, it is evident that only one acts at a time. 
Referring to Fig. 69, d, if both the diagonals are rods and 
therefore can resist tension only, it is seen that BD is in 
tension when the external force acts towards the right and 
that there is no stress in AC. It is further seen that AC is in 



Art. 1. 



DEFINITIONS AND NOTATION. 



127 




FiQ. 70. 



tension when the force acts towards the left and that there is 
no stress in BD. If two diagonals are used, and each can 
resist both compression and tension, the problem is indeter- 
minate by static methods. This case will not be considered 
in this work. 

Let the quadrilateral ABCD (Fig. 70) be one panel of a Pratt 
truss loaded with dead 
load, as shown in the fig- 
ure. In this type of truss, 
the intermediate diagonals 
are tension members. The 
members AC and A'C, 
which are in tension under 
the dead load, are called main diagonals or main braces; while 
the members BD and BD', which are not stressed by the dead 
load, are called counterhraces or counters. The counters may be 
stressed either under the action of wind loads or of unsymmetrical 
loads. 

125. Counterbracing. The two tension or two compres- 
sion diagonals in the same panel cannot act at the same time 
unless they are subjected to initial stress. A truss will not be 
in equilibrium under the action of initial stress unless external 
forces are applied, or unless the initial stress is held in equi- 
librium by the resisting moment of some of the members of 
the truss. Since initial stress will not be considered in this 
work, it is evident that if the diagonals can resist only one 
kind of stress, but one diagonal in each panel will act at a 
time. 

The method for determining whether the main member or 
the counter is under stress due to any system of loading, 
together with the magnitude of the resulting stress, will now 
be given. Referring to Fig. 70, and noting the fact that the 
dead load is always acting upon the truss, it may be readily 
shown that there will be a tensile stress in the diagonal AC 
when the dead load alone is acting. The kind of stress in thi* 
member may be determined by drawing the stress diagram, os 



128 COUNTERBRACING. (^^^P- ^^^^* 

by a method which will be explained in the following article. 
Now suppose that the snow load covers the right half of the 
truss only, or that the wind is acting from the right. Either 
of these conditions will tend to cause a compressive stress in 
the diagonal A'C. If this compression is less than the dead 
load tension in the member, then the tension already in the 
member will be reduced by an amount equal to the magnitude 
of the compressive stress. The resultant of these stresses will 
be the actual stress in the member due to the combined loads. 
If this compression is exactly equal to the dead load tension, 
the resulting stress in the member is zero. Now suppose that 
the compression caused by the unsymmetrical loading is 
greater than the dead load tension. In this case, the member 
A'C can only take enough compression to neutralize the dead 
load tension. If more than this amount of compression is 
thrown into A'C, the member will tend to be distorted, and 
the counter BD' will be thrown into tension to resist this 
distortion. 

If the chords are parallel and the diagonals have the same 
inclination, as in the truss shown in Fig. 70, the magnitude of 
the tension in the counter BD' is equal to the difference 
between the stress in A'C caused by the wind or unsym- 
metrical load and that caused by the dead load. If the chords 
are not parallel, i. e., if the main member and the counter in 
any panel have different inclinations, the stress in the counter 
IS equal to the difference in shears in the panel resolved in the 
direction of the counter. 

126. Notation. When counterbracing is used, the system 
of notation must be slightly modified ; since either diagonal in 
a panel may be stressed. A convenient system of notation is 
shown in Fig. 71. In this system, one diagonal in each panel 
is shown as a dotted line. The diagonals shown as full lines 
are designated by the figures 2-3 and 4-5 ; while those shown 
as dotted lines are designated by the same figures accented, 
2'-3' and 4'-5'. 

To illustrate this system of notation, if the diagonals 



Art, IB, 



SEPARATE STRESS DIAGRAMS. 



129 



stressed are 2^-3' and 4-5, then the upper chord members are 
X-i, X-3', X-5, and X-6; the 
lower chord members are Y-i, 
Y-2', Y-4, and Y-6 ; and the web 
members are 1-3', 2'-3', 2^-5, 4-5, 
and 4-6. If the dead load alone is 
acting, the diagonals 2-3 and 4-5 
are stressed, and the web mem- 
bers are 1-2, 2-3, 3-5, 4-5, and 4-6. 




F16. 71. Notation. 



Art. 2. Stresses in Trusses with Counterbracing — 

Separate Stress Diagrams. 



127. Determination of Stresses in Trusses with Counter- 
bracing — Separate Stress Diagrams. In the explanations that 
follow, the diagonals will be assumed to be tension members, 
as this is the more common case; although the same general 
principles will apply if they are compression members. The 
method for determining the maximum stresses in trusses with 
counterbracing, when separate stress diagrams are used, in- 
volves the following steps : 

(i). Construct the dead and snow load stress diagrams, 
assuming that the diagonals all slope in the same direction. If 
the truss is symmetrical about its center line, it will only be 
necessary to construct the diagrams for one-half of the truss. 
The snow load stresses may be found by direct proportion from 
the dead load stresses without the use of another diagram. 

(2). From these diagrams, determine in which panels, if 
any, the diagonals will be subjected to compression, and draw 
in the second diagonals in these panels. Revise the stress 
diagrams to include the added diagonals. The revised diagrams 
will now contain the actual stresses in all the members due to 
vertical loads. The main members (those stressed by the dead 
load) are represented by full lines, and the counters by dotted 
lines. 

(3). Construct the wind load stress diagrams, using those 



130 COUNTEKBRACING. Chap. XIII. 

diagonals which have been found to be in tension due to the 
dead load. If the truss is symmetrical, it will only be necessary 
to consider the wind as acting from one direction ; while if the 
truss is unsymmetrical, it will be necessary to consider the 
wind as acting from both directions. 

(4). P>om the wind load stress diagram, determine which 
diagonals, if any, will be in compression. Draw counters in 
these panels. Revise the stress diagrams to include the added 
diagonals. 

(5). From the stress diagrams, determine the stresses due 
to the different loadings, combine these stresses to determine 
which diagonals are stressed, and then find the maximum 
stresses in all the members of the truss. In making any par- 
ticular combination to determine the maximum stress in any 
member, it is necessary to first find which diagonals are acting 
at that time, and then to combine the stresses found in that 
particular member when these diagonals are acting. In mak- 
ing the combinations for maximum stresses, it is necessary to 
consider, not only the member itself, but also the corresponding 
member on the other side of the center of the truss. 

The following combinations will be considered in this work 
in determining maximum stresses in roof trusses. 

(a) Dead load plus maximum snow load. 

(b) Dead load plus minimum snow load plus wind load 
(wind acting from either direction). 

(c) Dead load plus wind load (wind acting from either 
direction). 

The method outlined above will now be explained by the 
solution of two problems. The first problem will be to deter- 
mine the maximum stresses in a truss with parallel chords, and 
the second to determine the maximum stresses in a truss with 
non-parallel chords. 

128. Problem i. Truss With Parallel Chords. It is 
required to find the maximum stresses in all the members of 
the Pratt truss with counterbracing shown in Fig. 72. The 
span of the truss is 40 ft. ; the height, 7.5 ft. ; and the trusses 
are spaced 15 ft. apart. The dead load will be taken at 10 lbs. 



Art. S. 



SEPARATE STRESS DIAGRAMS. 



131 



per sq. ft. of hor. proj.; the minimum snow load, at lo lbs. per 
sq. ft. of hor. proj.; the maximum snow load, at 20 lbs. per sq. 
ft. of hor. proj. ; and the component of the wind normal to the 
roof surface, at 27 lbs. per sq. ft. of roof surface. The wind 
load reactions will be assumed parallel to the resultant of all 
the wind loads. 




KXX) ZOOO 3000 

1 I I I 



Fig. 72. Stbess Diagbams — Tbuss with Paballel Chords. 



The dead load stress diagram (see Fig. y2) is first drawn, 
assuming that all the diagonals slope in the same direction, i. e., 
that 2-3 and 4^-5' are acting. From this diagram it is found 
that 4'-5', if acting, would be in compression ; therefore, the 
other diagonal, 4-5, in the same panel is acting due to the dead 
load. The stress diagram is now revised to include the diagonal 
4-5, also the diagonal 2'-3'. The dead load stress diagram for 
the entire truss is shown in Fig. y2. The dead load stresses 
when the different diagonals are acting are shown in the table 
in Fig. 73. 

The minimum and maximum snow load stresses are deter- 



132 



OOUNTEBBRACING. 



Chap, XIJL 



mined by direct proportion from the dead load stresses without 
constructing any new diagrams, and are shown in Fig. 73. 



Truss 


Dead 


Snow Load 


Wind 


Maxi- 


Mem- 
ber 


Load 
Stress 


Stn 


ess 


Load 
Stress 


mum'^ 
Stress 


Min. 


Max. 


X-l 


- 3750 


- 3750 


- 7500 


— 


2070 


-II250* 


rx-3' 

IX-3 


- 3000 


- 3000 


- 6000 


— 


3160 




- 4000 


- 4000 


- 8000 


— 


2110 


-I2000» 


fX-5 
lX-5' 


- 4000 


- 4000 


- 8000 


— 


2110 


-12000 


- 3000 


- 3000 


- 6000 


— 


1060 




X-6 


- 3750 


- 3750 


- 7500 


— 


1320 


-11250 


Y-l 


+ 3000 


t 3000 


+ 6000 


+ 


2580 


-t 9000* 


/Y-2' 
I Y-2 


•h 4000 


-f- 4000 


-h 8000 


+ 


1530 




+ 3000 


+ 3000 


+ 6000 


+ 


2580 


-h 9000* 


/Y-4? 
1 Y-4 


+ 4000 


-h 4000 


+ 8000 


+ 


1530 




-h 3000 


+ 3000 


+ 6000 


+ 


480 


+ 9000 


Y-6 


+ 3000 


+ 3000 


+ 6000 


+ 


480 


+ 9000 


(1-3' 
I 1-2 


+ 750 


+ 750 


-1- 1500 


— 


790 


- 40* 



















1 2-3 


- 1250 


- 1250 


- 2500 


+ 


1320 


+ 70* 


+ 1250 


t 1250 


+ 2500 


— 


1320 


-h 3750 


(3-5 
2'-5 


- 1500 


- 1500 


- 3000 







- 4f>00* 


- 750 


- 750 


- 1500 


— 


790 




(4'-5' 
I 4-5 


- 1250 


- 1250 


- 2500 


— 


1320 




+ 1250 


+ 1250 


4 2500 


+ 


^20 


+ 3820* 


r4-6 
I 5'-6 



















+ 750 


+ 750 


-h 1500 


- 


790 





Fio. 73. Table of Stbesses-t-Tbuss with Paballel Chobds. 

The wind load stress diagram for the wind acting from the 
left is also shown in Fig. 72. The diagram is first drawn using 
the diagonals which are found to be in tension for dead 
load, VIZ. : 2-3 and 4-5. The stresses due to the wind load 
are shown in Fig. 73. By comparing the dead and wind load 
stresses, it is seen that the wind tends to cause compression 
in 2-3, and further that the wind load compression in this 
member is greater than the dead load tension; therefore, the 
counter 2^-3' will act. The stress diagram (Fig. 72) is now 
revised to include the counter 2'-3', also the counter 4'-5'; 
since the latter member will be stressed when the wind acts 
from the right. Since all the members which may ever act 
are included in the stress diagram and the truss is symmetri- 



^^' -• SEPARATE STRESS DIAGRAMS. 133 

cal, it is evident that it is unnecessary to construct the wind 
load stress diagram for the wind acting from the right. The 
wind load stresses when the different diagonals are acting are 
given in Fig. 73. The maximum stresses in all the mem- 
bers are determined by making the combinations indicated 
in § 127, and are shown in the last column of Fig. 73. From 
this table, it is seen that the counters 2'-3' and 4^-5' are 
required. These counters are shown by dotted lines in the 
truss and stress diagrams. In determining the maximum 
stresses, it is seen that not only the member itself, but also 
the corresponding member on the other side of the truss must 
be considered. The stresses shown with a star after them in 
Fig. 73 are maximum stresses. 

For a truss with parallel chords, it is evident that, if the 
diagonals 2-3 and 2'-3' act separately, the stresses in them are 
numerically equal but have opposite signs, i. e., if one is ten- 
sion, the other will be compression. It is therefore seen that 
it would not be necessary to actually use the member 2^-3' 
in the stress diagrams. 

129. Problem 2. Truss With Non-parallel Chords. It is 
required to find the maximum stresses in all the members of 
the truss with counterbracing shown in Fig. 74. The span of 
the truss is 100 ft;^ ; the total height, 37.5 ft. ; the height of 
the vertical sides, 12.5 ft.; the pitch of the roof, one-fourth; 
and the trusses are spaced 15 ft. apart. The panel points of 
the lower chord lay on the circumference of a circle of 106.25 
ft. radius. The dead load is taken at 10 lbs. per sq. ft. of hor. 
proj.; the minimum snow load, at 10 lbs. per sq. ft. of hor. 
proj.; the maximum snow load, at 20 lbs. per sq. ft. of hor. 
proj.; the wind on the vertical sides of the truss, at 30 lbs. 
per sq. ft. of surface ; and the component of the wind normal 
to the roof surface, at 23 lbs. per sq. ft. of roof surface. The 
wind load reactions are assumed parallel to the resultant of 
all the wind loads. 

The dead load stress diagram for one-half of the truss is 
shown in Fig. 74. In constructing this diagram, it was first 
assumed that all the diagonals sloped downward toward the 



134 



COUN'XEBB&ACINQ. 



Chap. XIII. 



right. From the stress diagram, it is seen that the dead load 
tends to cause compression in 5'-6' and 7'-8' ; therefore, 5-6 
and 7-8 are stressed by the dead load. The stress diagram 
was then revised to include these members. 




De^j) Load Stbebs 'Dii.asjkU. 



The minimum and maximum snow load stresses are found by 
proportion from the dead load stresses without constructing addi- 
tional diagrams. 

The wind load stress diagram, assuming the wind to act 
upon the left side of the truss, is shown in Fig. 75. This dia- 



^"- f- SEPABATE STBEBB DIAOBAMS. 135 

gram was constructed by first using the diagonals which are 
in tension due to the dead load. 




Pl8. 76, NOU- 



The table of stresses is shown in Fig. 76, the stresses being 
determined from the diagrams in Fig. 74 and Fig. 75. This 
table is constructed as follows: First record the stresses in 
the diagonals, starting with the member 1-2. The dead and 
wind load stresses are obtained from the stress diagrams, and 
the minimuni and maximum snow load stresses are obtained 
by direct proportion from the dead load stresses. The dead 



136 COUNTERBllACIXG. <^^<^P' ^UL 

and wind loads produce the same kind of stress in the diag- 
onals 1-2, 3-4, 5-6, and 7-8; therefore no counters are required 
in the left half of the truss when the wind acts towards the 
right. The dead load stress in 9-10 is + 3500> 2ind the wind 
load stress in that member is — 2400; therefore no counter 
is required in that panel. The dead load stress in 11-12 is 
+ 300, and the wind load stress in that member is — 4300. 
Since the member 11-12 can only take enough wind load com- 
pression to neutralize the dead load tension already in it, it is 
seen that the counter ii'-i2' will be thrown into action. 
Revise the wind load stress diagram to include the counter 
11'- 1 2', as shown in Fig. 75. Also revise the dead load stress 
diagram to include the corresponding counter 5'-6' (if not 
already included). Record the member ii'-i2' in the table, 
and tabulate the dead, snow, and wind load stresses. Also 
record the stresses in the diagonals 13-14 and 15-16. It is 
seen that the dead and wind loads produce the same kind of 
stress in these members, therefore no counters are required in 
these panels. It is thus seen that ii'-i2' is the only counter 
required in the truss when the wind acts towards the right. 

Likewise, tabulate the stresses in the verticals. The 
stresses in the verticals adjacent to the diagonal 11-12 should 
be considered, both when the main member 11-12 and when 
the counter ii'-i2' act, i. e., when 11-12 acts, the verticals 
are lo-ii and 12-13, and when ii'-i2' acts, the verticals are 
10-12' and ii'-i3. 

Also, tabulate the strCvSses in the upper and lower chord 
members. The stresses in the chord members in the same 
panel as 11-12 should be considered, both when 11-12 and 
when 11'- 1 2' act. 

Combine the stresses to determine the maximum stress in 
each member, as shown in Fig. 76. In making the combina- 
tions, it must be borne in mind that, to get the stress in any 
member, it is necessary to first determine which diagonals are 
acting for each combination, and to combine the stresses in 
that particular member when these diagonals are acting. The 
^combinations considered are those indicated in § 127. Since 



SEPABATE STBE85 DIAQRAMS. 



Truss 


Dead 


Snow Load 


Wind 


Max- 


Mem- 
ber 


Load 
Stress 


Stress 


Load 

Stress 


imum 
Stress 


Min- 


Max. 


\-z 




+ 7200 


+ 7200 


+ 14400 


+ 9700 


+ 24100" 


3-4 




+ 3600 


+ 5600 


+ 7200 


+ 1200 


+ 10600 


5-6 




+ 300 


+ 300 


+ 600 


+ 8400 


t 9000* 


7-8 




+ 3500 


+ 3500 


+ 7000 


+ 1 3700 


+ 20700' 


9-10 


t 3500 


+ 3500 


+ 7000 


- 2400 


+ 10500 


11-121 




+ 300 


+ 300 


+ 600 


- 4300 


+ 900 


ir-12') 


- 200 


- 200 


- 400 


+ 3200 


+ 5000* 


13-14 




+ 3600 


+ 5600 


+ 7200 


+ 4200 


+ 1 I400» 


15-16 




+ 7200 


+ 7200 


+ 14400 


+ 4500 


+ EI600 


A-l 




- 7600 


- 7600 


- 15200 


- 13100 


-26300' 


2-3 




- 5500 


- 5500 


- 11000 


- 6600 


- 17600 • 


4-6 




- 1900 


- 1900 


- 3800 


- 5400 


- 9200* 


5-8 




- 2100 


- 2100 


- 4200 


-1060O 


- 14600* 


7-9 


X 


+ 1000 


+ 1000 


-1- 2000 


+ 500 


+ 3000 • 


10-11 1 


a 


- 2100 


- 2100 


- 4200 


+ 2700 


- 6300 


10-12' 


t 


- 1900 


- 1900 


- 3600 





- 5700 


12-13 


^ 


- 1900 


- 1900 


- 3800 





~ 5700 


ll'-13. 




- 1700 


- 1700 


- 3400 


- 3400 


- 6600 


14-15 




- 5500 


- 5500 


- 1 1000 


- 4200 


- 16500 


A- 16 




- 7600 


- 7600 


- 15200 


- 4200 


- 22800 


X-2 




- 7100 


- 7100 


- I4ZO0 


- 14000 


-28200" 


X-4 




-10300 


- 10500 


- 20600 


- 17400 


-380O0' 


X-6 


1 


- 10500 


- 10500 


-21000 


-19800 


-40800" 


X-8 


-10300 


-10300 


-20600 


-17500 


-36100' 


X-IO 




-10300 


-10300 


- 20600 


- 10900 


-31500 


X-121 


1 


-10500 


-10500 


-21000 


- 8400 


-31500 


X-!2'J 


- 10500 


-10300 


- 20600 


-10900 


-31500 


X-13 




- 10300 


- 10300 


- 20600 


- 8400 


- 50900 


X-15 




- 7100 


- 7100 


- 14200 


- 4500 


- 21 300 


Y-1 













+ 6300 


+ 6300* 


Y-3 




+ 6700 


+ 6700 


+ 13400 


+ 15100 


+ 28500' 


Y-5 


1 


+ 9300 


+ 9300 


+ 18600 


+ 1 1 300 


+ 29900* 


Y-7 


+ 7600 


-1- 7600 


+ 15200 


+ 5200 


+ 2280O' 


Y-9 


. 


+ 7600 


t 7600 


+ 15200 


+ 5200 


+ 22600 


Y-Ul 


5 


t 9300 


+ 9300 


t 16600 


+ 4100 


+ 27900 


Y-n'I 




+ 9500 


+ 9500 


+ 19000 


i 1900 


+ 20900 


Y-14 




+ 6700 


+ 6700 


+ 13400 


- noo 


+ 20100 


Y-t6 













- 6300 


- 6300* 




.^^ 


^^ 






^ -( 




^W 


V"^ 


P^ 










o\lO 


^ 


-v^X 




4 


3/7 


§g\ 


> 
.''11 


Ji|/|4 ' 


> 

16 A 


A § 




$^ 


29900 


+Z2S00 







h. 7a. Tablb o 



138 COUNTERBRACING. C^<^P' ^HI' 

the wind is taken as acting towards the right, only, it is nec- 
essary to consider the member itself and also the correspond- 
ing member on the other side of the center line. The maximum 
stresses are indicated by stars in Fig. 76. 

Referring to the table of stresses, it is seen that the maxi- 
mum stress in 1-2 is given by the combination of dead, mini- 
mum snow, and wind loads; that the maximum stress in 7-9 
is given by the combination of dead and maximum snow loads ; 
and that the maximum stress in the counter ii'-i2' is given 
by the combination of dead and wind loads. The attention of 
the student is called to the stress Y-ii', which is the stress 
in the lower chord when the counter ii'-i2' is acting. Unless 
care is taken to determine which diagonal is acting, the stu- 
dent is liable to make the mistake of combining the dead and 
maximum snow load stresses, which would seem to give a 
stress of -f 28 500 in Y-ii'. However, this stress can never 
occur, as the main diagonal 11-12 is acting for this combina- 
tion. When the dead, minimum snow, and wind loads are on 
the truss, the counter Ii'-I2' is acting, and the stress in 
Y-ii' is then -{"20900, which is the maximum stress in that 
member when the counter is acting. When the main diagonal 
11-12 is acting, the combination of dead and maximum snow 
loads gives a stress of + 27900 in Y-ii. However, the maxi- 
mum stress occurs in the corresponding member Y-5 when 
the dead, minimum snow, and wind loads are acting, and is 
-f- 29 900. It is seen that the stress in Y-ii is also +29900 
when the wind acts towards the left. Referring to Y-i and 
Y-16, it is seen that the stresses in these members are -f 6300 
and — 6300, respectively. 

If the direction of the wind is changed and is made to act 
towards the left, it is seen that Y-i and Y-16 are subjected to 
reversals of stress. It is further seen that the counter 5'-6' 
is then thrown into action. The diagonals 5'-6' and ii'-i2' 
are the only counters required, and Y-i and Y-16 are the 
only members which have reversals of stress. 

The maximum stresses are shown on the truss diagram in 
Fig. 76. 



^^*- ^- COMBINED STRESS DIAGRAM. 139 

In this problem, it has been shown that the only counter 
required when the wind acts towards the right is Ii'-I2'. In 
some trusses, it might happen that when the wind acts towards 
the right a counter is required in some panel on the left of the 
center and also in another panel (not the corresponding one) on 
the right of the center of the truss. In this case, it would be more 
convenient and would give less cause for errors in making the 
combinations for maximum stresses if the dead load stress dia- 
gram was drawn for the entire truss. The diagram should also 
be revised to include both the. main diagonal and the counter in 
each panel in which a counter is required when the wind acts 
towards the right. 

Art. 3. Stresses in Trusses With Counterbracing — 

Combined Stress Diagram. 

130. Determination of Stresses in Trusses With Counter- 
bracing — Combined Stress Diagram. In determining the 
maximum stresses in a truss with counterbracing by means of 
the combined stress diagram, it is necessary to first find which 
diagonal in each panel is stressed by the combined loadings. 
The stress diagram is then constructed, using only those diag- 
onals which are found to be stressed. Two methods will now 
be given for finding which diagonals are stressed. The first 
method may be used to advantage for trusses with parallel 
chords, and the second, for trusses with non-parallel chords. 

(i) Trusses with Parallel Chords. If the truss has parallel 
chords, the simplest method for determining which diagonal 
in each panel is stressed consists of an application of the con- 
dition of equilibrium that 2 V = o to the external forces on 
one side of a section and the members cut by the section. 
This method will now be explained by means of a problem. 

It is required to find which diagonals are stressed in the 
truss loaded as shown in Fig. yy. The load line, together with 
the reactions, is shown in Fig. yy, b. To determine whether 2-3 
or 2'-3' is acting, cut the members X-3, 2-3, 2^-3', and Y-2 by 



140 



COUNTERBRACING. 



Chap, XIII, 



1000 




(b 



Fig. 77. 



the section m-m, and apply the condition that 2 V = o to the 
forces acting upon the shaded portion of the truss. Since the 

chords are horizontal 
members and can resist 
jio vertical force, it is 
seen that the vertical 
component of the stress 
in the diagonal 2-3, or 
in the diagonal 2^-3', 
must be equal and oppo- 
site in direction to the re- 
sultant of the external 
forces to the left of the 
section m-m. Now the 
resultant of these external forces is represented in the load 
line (Fig. y^y b) by EB, and acts downward; therefore, the 
vertical component of the stress in the diagonal must act upward 
for equilibrium. Referring to Fig. ^7, it is seen that if a mem- 
ber is in tension, the force it exerts upon the shaded portion of 
the truss acts away from that portion ; and if it is in compres- 
sion, the force it exerts acts toward the shaded portion. There- 
fore, if the diagonals are tension members, as they are in the 
Pratt truss shown in Fig. yy, the member 2-3 will act in tension. 
In like manner, it may be shown that the diagonal 4-5 is also in 
action. From the above discussion, it is seen that by observing 
whether the resultant of the external forces to the left of any 
panel acts upward or downward and knowing whether the diag- 
onals are tension or compression members, it may be determined 
at once which diagonal in any panel is stressed. The method 
described above is called the method of shears. 

The stresses may now be found by constructing the stress 
diagrams for the combined loads, using only those diagonals 
which have been found to be stressed. It is seen that this 
method requires considerable work if several combinations 
must be made to determine the maximum stresses; as it is 
necessary to determine for each combination which diagonals are 
stressed, and then to draw a stress diagram for each combination. 



Art. S. 



COMBINED 8TBESS DIAOBAH. 



141 



(z) Tntsses with Non-parallel Chords. If the chords of the 
truss are not parallel, the condition of equilibrium that S M^o 
may be used to determine which diagonals are stressed. The 
apphcation of this method will now be shown by means of a 
problem. 

It is required to determine which diagonals are stressed in 
the truss loaded with dead and wind loads, as shown in 
Fig. 78. 




The reactions (assumed to be parallel) are first determined 
by means of the force and funicular polygons, as shown in 
Fig. 78. To determine whether 2-3 or 2' -3' is in action due 
to the combined loads, cut these members, together with the 
chord members in the same panel, by the section m-m ; and 
consider the members cut and the external forces to the left of 
the section. Prolong the chord members X-3 and Y-2 until 
they intersect at P, and take this point as the center of 
moments. Now from the condition that S M :^ o, the moment 
of the stresses in the members cut by the section must balance 
the moment of the external forces to the left of the section 
m-m. But the moment of each chord stress is zero, since its 
line of action passes through the center of moments; there- 
fore, the moment of the stress in the diagonal 2-3, or the 
diagonal 2'-%', must balance the moment of the external forces 
to the left of the section m-m. The next step is to determine 



142 COUNTERBRACINO. Chap, XIIL 

in which direction the moment of the external forces to the 
left of m-m tends to produce rotation. Now the resultant of 
these external forces is represented in the force polygon by 
GC, acting in the direction from G towards C; and its line 
of action is through the intersection of the strings oc and og. 
Although the intersection of these strings falls outside of the 
limits of the drawing, it is evident that the moment of the 
resultant of the external forces is clockwise; therefore, the 
moment of the stress in the diagonal acting at this time must 
be counter-clockwise. In the truss shown in Fig. 78, the diag- 
onals are tension members; therefore, the diagonal 2-3 is 
stressed. 

In like manner, by using the section n-n and taking the 
center of moments at P', it is found that the moment of the 
resultant GE of the external forces to the left of the section 
is counter-clockwise ; hence the moment of the stresses in the 
diagonal 6-7, or 6'-7', is clockwise. The diagonal 6-7 is 
therefore stressed. 

To determine whether 4-5 or 4'-5' is stressed, cut the mem- 
bers by the section p-p. Instead of using the method of 
moments for this case, which would necessitate first finding 
the kind of stress in X-5 or Y-4, the method of shears, 
described in § 130 (i), will be used. From the force polygon, 
it is seen that the vertical component of the resultant GD of 
the external forces to the left of the section p'-p acts down- 
ward; therefore, the vertical component of the stress in the 
diagonal must act upward. The diagonal 4-5 is therefore 
stressed. 

The stresses may now be found by constructing the stress 
diagrams for the combined loads, using only those diagonals 
which have been found to be in action due to the combined 
loading. 

It is seen that this method may be used to advantage if 
only a single combination is required to determine the maxi- 
mum stresses. 



CHAPTER XIV. 



THREEHINGED ARCH. 



131. Definition. An arch is a structure which has inclined 
reactions for vertical loads. The only type of arch which will 
be considered here, and the one commonly used for long span 
roof trusses, is the three-hinged arch. A three-hinged arch is 
composed of two simple beams or trusses, hinged together at 
the crown, and also hinged at their points of support. This is 
the only form of arch construction which is statically determi- 
nate. An example of a three-hinged arch is shown in Fig. 
79, a. This structure is composed of two simple trusses, 
hinged together at the crown C and also hinged at the supports 
A and B. 

The reactions may be obtained by applying the principles 
which have already been explained. The method of finding 
these reactions may best be explained by first determining the 
reactions due to a single load. 





Fia. 79. Thbee-hinged Arch — Reactions for Single Load. 



132. Reactions Due to a Single Load. It is required to 
find the reactions in the three-hinged arch shown in Fig. 79, a, 
due to the single load P. 

143 



144 THREE-IIINOED ARCH. Chap. XIV, 

This arch is composed of the two segments AC and BC, 
and is hinged at the points A, B, and C. The load P is sup- 
ported by the segment AC, the other segment being unloaded. 
There are only two forces acting upon the segment BC, the force 
exerted by the segment AC against the segment BC (acting 
downward), and the reaction Rg of the segment BC against 
its support. Since this segment is held in equilibrium by 
these two forces, they must have the same line of action and 
must act in opposite directions. 

Now consider the segment AC. This segment is held in 
equilibrium by three forces, viz.: the reaction Rj at A, the 
load P, and the force (acting upward) exerted by the segment 
BC against the segment AC. This last force is equal in mag- 
nitude, but acts in an opposite direction to the force exerted 
by the segment AC against the segment BC; and is also equal 
to the reaction Ro. Since the segment is held in equilibrium 
by these forces, they must intersect at a common point. This 
point is determined by prolonging the line of action of the 
reaction, which acts through the points B and C, until it inter- 
sects the line of action of the force P at D (Fig. 79, a). The 
reactions at C will neutralize each other, and the three-hinged 
arch, taken as a whole, is in equilibrium under the action of 
the three forces, Ri, P, and Rg. The lines of action of the two 
reactions Rj and Ro being known, the magnitudes of these 
reactions may be determined by drawing their force polygon. 
This polygon is shown in Fig. 79, b, the reactions being 
represented by R^ and Rg. 

The reaction at either support due to any number of loads 
may be found by first determining that due to each load sepa- 
rately, and then combining these separate reactions. 

The method for finding the reactions and stresses due to a 
number of vertical loads will now be shown. 

133. Reactions and Stresses for Dead Load. It is required 
to find the dead load reactions and stresses for the three- 
hinged arch, loaded as shown in Fig. 80, a. The span of the 
arch is 125 ft., and its rise is 50 ft. Both the segments AC 
and BC are alike, and are symmetrically loaded. 



DEAD LOAD STRESSES. 



145 



To find the reactions, lay off the load line, as shown in 
Fig. 80, b, choose any pole O, and draw the funicular poly- 
gons, one for each segment, as shown in the figure. The ver- 
tical reactions at supports A and B and at the hinge C are 







Dead Load 
Stress Diagram 
0* 10000^ 20000* 



Fig. 80. Stbess Diagram for a Three-hinged Arch. 

found by drawing the rays from the pole O parallel to the 
closing strings of these funicular polygons. These reactions 
are represented by R^, Rb, and (Ro + R'o), respectively. 
Since the truss and the loads are symmetrical about the center 
line, half of the load at the crown, or Ro, will be transferred 



\ 



146 TIIREE-HIXGED ARCH. Chap, XIV. 

to the left support, and the other half, or R^.', will be trans- 
ferred to the right support. To determine the reaction at A 
due to all the loads, consider first the reaction at this point 
due to the single load Rjj acting at C. Since there can be no 
resisting moment at the hinge C, it follows that the reaction 
at A caused by this load must also pass through the hinge C, 
which gives its line of action. Now the vertical component 
of this reaction is equal to Ro, and its line of action is AC; 
therefore, its magnitude is given by the line Rj' (Fig. 80, b), 
drawn from the point E parallel to the line joining the points 
A and C. The total reaction at A due to all the loads is now 
determined by combining the vertical reaction Rj^ with the 
reaction R/. This total reaction is represented by Rj (Fig. 
80, a and Fig. 80, b). The reaction at B is determined in like 
manner, and is represented by Rg. Taking the equilibrant of 
the reaction R^ and of the resultant of the loads acting upon 
the segment AC, it is seen that the reaction at C is horizontal 
and acts towards the left. This reaction is represented by FY 
(Fig. 80, b). In like manner, it may be shown that the reac- 
tion at C, due to the loads on the segment BC, is horizontal 
and acts toward the right, and is represented by YF. 

The reactions having been determined, the stresses in the 
members of the three-hinged arch may be found by drawing 
the stress diagram, starting the diagram with the forces acting 
at A. Since the truss and the loads are symmetrical about the 
center hinge, it is only necessary to draw the stress diagram 
for one segment. The stress diagram for the segment AC is 
shown in Fig. 80, b. The magnitudes of the stresses may be 
determined from the stress diagram, and the kind of stress is 
indicated by the arrows placed on the members of the truss, 
as shown in Fig. 80, a. 

134. Wind Load Stresses for Windward Segment of 
Truss. It is required to find the wind load stresses in the 
windward segment of the three-hinged arch, loaded as shown 
in Fig. 81, a. 

To find the reactions for this segment, consider it as a sim- 
ple truss supported at the hinges. Construct the force polygon 



WIND LOAD 



147 



(Fig. 81, b) for the wind loads, choose any pole O, and draw 
the funicular polygon, as shown in Fig. 81, a. The reactions 
are then determined by drawing the ray through the pole O 
parallel to the closing string of the funicular polygon. These 




FlQ. 81. 8TREB3 DlAOaAU 



reactions are parallel to the resultant of the wind loads, and 
are represented by R, and R,, (Fig. 81, b). Now consider the 
three-hinged arch as a whole. Since there are no loads on the 
segment BC, for equilibrium, the right reaction Rj, acting at 
B, must also pass through the center hinge C. The reaction 
R„ which was found by considering the segment AC as a 
simple truss, will now be resolved into the two reactions R, 
and R/ (Fig. 81, b), drawn parallel to R, and R/ (Fig. 81, a), 
respectively. The reaction R^ is found by drawing the closing 
line of the force polygon. The arch is held in equilibrium by 



148 THREE-HINGED AEOH. ^^<^P' ^^^^ 

the three following forces, viz. : the resultant R of all the wind 
loads, the reaction Rj at A, and the reaction Rg at B. Since 
the line of action of Rg is known, that of Rj may be determined 
by prolonging R and Rg until they intersect, and connecting 
this point of intersection with A. In this problem, the point 
of intersection of R and Rj falls outside the limits of the drawing. 

The reactions may also be determined, as follows: Since 
there are no loads on the right segment, the line of action of 
R2 must pass through the hinges B and C. Therefore, choose 
any pole O, and, starting at A, the only known point on the 
left reaction Ri, draw the funicular polygon for the given truss 
and loads, closing on the line of action of Rg. The dividing 
ray, drawn parallel to the closing string, will then determine 
the two reactions. 

The latter method is somewhat simpler than the one 
shown in Fig. 81. 

The reactions having been determined, the stresses in the 
windward segment may be found by drawing the stress dia- 
gram, starting the diagram with the forces acting at A. This 
diagram is shown in Fig. 81, b. The magnitudes of the stresses 
may be determined from the stress diagram, and the kind of 
stress is indicated by the arrows placed on the members of the 
segment AC. 

To determine the maximum and minimum stresses, it is 
also necessary to find the stresses in the members of the lee- 
ward segment BC, and these stresses will be determined in the 
following section. 

135. Wind Load Stresses for Leeward Segment of Truss. 
It is required to find the wind load stresses in the leeward 
segment of the three-hinged arch shown in Fig 82, a, the arch 
and the loads being the same shown as in Fig. 81, a. 

To facilitate a comparison of stresses in the windward and 
leeward segments, the stresses will be found in the same 
segment AC as in the preceding section. 

The wind load reactions are the same as those found in 
§ 134, and are represented in magnitude in Fig. 82, b and in 
line of action in Fig. 82, a. The wind load stresses are found 



WIND LOAD 8IK]iSaE& 149 

by drawing the stress diagram, starting the diagram with the 
forces acting at A. This diagram is shown in Fig. 82, b. The 
magnitude of the stresses may be determined from the stress 
diagram, and the kind of stress is indicated by the arrows 
placed on the segment AC. 




Wind Load 

Stress Dioqrom 

For 

Leeword Side 

0* lOOOO" ?0000 

(b) 



Fin. 82. Srataa Lil. 




By comparing the wind load stress diagrams and also the 
kind of stress indicated by the arrows (see Fig. 81 and Fig, 
82), it is seen that there are many reversals of stress. 

The maximum stresses may be determined by making the 
combinations indicated in § 127. 



CHAPTER XV. 

STRESSES IN A TRANSVEBSE BENT OF A BUILDING. 

136. Construction of a Transverse Bent. It has been 
assumed in the preceding discussion of^ roof trusses that the 
trusses were supported upon walls or pilasters. However, in many 
types of mill building construction, the trusses are supported by 
columns, to which they are rigidly connected, thus forming a 
transverse bent. The columns carry not only the roof trusses 
and the loads on them, but may also support the side covering 
and resist the pressure of the wind against the side of the build- 
ing. In such buildings, the trusses are usually riveted to the 
columns and are braced by members called knee-braces. The 
side covering is fastened to longitudinal girts, which are con- 
nected to the columns. Additional rigidity is secured by means 
of a system of wind bracing. This bracing may be placed in the 
planes of the sides gf the building and in the planes of the upper 
and lower chords ^f the trusses. 

The intermediate transverse bents support a full panel load; 
while those at the end carry but half a panel load, and are some- 
times made of lighter construction. The end bent may be built 
by running end posts up to the rafters, or the same construction 
may be used as for the intermediate transverse bents. The latter 
method is preferable when an extension in the length of the 
building is contemplated. 

137. Condition of Ends of Columns. The stresses in a 
transverse bent depend to a considerable extent upon the condi- 
tion of the ends of the columns. Several assumptions may be 
made, although it is difficult, if not impossible, to exactly realize 
any of the assumed conditions. The columns may be taken as 
(i) hinged at the top and base, (2) hinged at the top and fixed at 
the base, or (3) rigidly fixed at the top and base. 

150 



/ 



CONDITION OF ENDS OF COLUMNS. 151 

(i) Columns Hinged at Top and Base. If the columns 
merely rest upon masonry piers, or if no effectual attempt is 
made to fix them at the base by embedding the columns in con- 
crete or by fastening them with anchor bolts, the columns should 
be taken as hinged at the top and base. The common assumption, 
and the one which will be made in this text, is that the horizontal 
components of the reactions due to the wind are each equal to 
one-half of the horizontal component of the total external wind 
load acting upon the structure. The vertical components may be 
found by the method of moments, or by means of the graphic 
construction shown in § 139. 

The maximum bending moment in the column is at the foot 
of the knee-brace of the leeward column, and is equal to the 
horizontal component of the reaction multiplied by the distance 
from the foot of the knee-brace to the foot of the column, i. e., = 
Hgd (see Fig. 83). 

(2) Columns Hinged at Top and Fixed at Base, If the 
deflections at B, the foot of the knee-brace (Fig. 83), and at C, 
the top of the column, are assumed to be equal, it may be shown 
that the vertical components of the reactions are the same as if 
the columns were hinged at D, the point 
of contra-flexure. The distance y from / /"^v. 
the base of the column to the point of / \/ 
contra-flexure depends upon the relative ^ ^x,^^ 
lengths of h and d. It may be shown 
that y varies from f d when d = ^h, to ^d 
when d = h. As soon as the position 
of the point of contra-flexure has been 
found, the column may be taken as 
hinged at that point, the wind acting 
upon the portion below the point of 
contra-flexure being neglected. The ver- fig. 83. ^ 

tical components of the reactions may 

now be found, and, since the horizontal components are assumed 
to be equal, the reactions themselves may be determined. 

The maximum positive bending moment in the column is at 
the foot of the knee-brace of the leeward column, and is equal 



152 STRESSES IN A TRAXSVERSE BENT. Chap, XV. 

to H2 (d — y). The maximum negative moment is at the foot 
of the column, and is equal to Hjy (see Fig. 83). 

(3) Columns Fixed at Top and Base, If the columns are 
fixed at the top and base, the point of contra-flexure is at a dis- 
tance y=— from the base (see Fig. 83). In this case the 

2 

column may be taken as hinged at the point of contra-flexure, 
the external wind below this point being neglected. The maxi- 
mum positive moment is at the foot of the knee-brace, and equals 

Hod ... 

+ — =— ; and the maximum negative moment is at the base of 

H^d 
the column, and equals . 

mi 

When an attempt is made to fix the columns, the resulting con- 
dition probably lies between that shown in Case 2 and that shown 
in Case 3. It is seen from Case 2 that y varies but slightly with a 
considerable difference in the ratio of d to h, and that it has its 

minimum value of — when h = d. In Case 3 it is seen that 

2 

y = -. The assumption commonly made when some effective 
means are used to fix the columns at the base is that 7 = — , and 

2 

this assumption will be made, in this text. The horizontal com- 
ponents of the reactions will be taken equal, and the vertical com- 
ponents may be found by moments, or by the method shown 

in § 139- 

138. Dead and Snow Load Stresses. The dead and snow 
load stresses in the truss of a transverse bent are the same as for 
a truss supported upon masonry walls. If the columns are hinged 
at the top, the stresses in them are direct compressive stresses, 
caused by the load? upon the truss and by the weight of the sides 
of the building supported by the columns. If the columns are 
fixed at the top, the deflection of the truss will produce bending 
moments in the columns and corresponding stresses in the knee- 
braces. Since the deflection of the truss is usually quite small, 



DEAD AND SNOW LOAD STRESSES. 



153 



the bending moments in the columns and the stresses in the knee- 
braces due to vertical loads will be neglected. 

The dead load stress diagram for a transverse bent is shown 
i.i rig. 84. The general dimensions of the building and of the 
transverse bent, together with the loads used, are shown in the 
figure. 




i 



Spa 

Length of euilding - 90-0". 
Distance Between Trusses « 15-0" 
Height of Colunnns - 20-0". 
Dead Load « 12 , Min.Snow Load«/0,and 
Max-5now Load = 20 lbs- per sq-f t- hen proj. fjj^ 
6pl0" J 



vJ o' 



LI 



Dead Load 

ZOOO 4000 

' ' ' I 

Max- Snow Load 

^ClDO 4000 6000 
L-J-J l_J 

Min.Snow Load 

1000 2000 3000 

« ■ I I 




Dead and Snow Load 
Stress Diaqrann 

FiQ. 84. Stsesses in a Tbansyebse Bent. 



The snow load stresses may be determined by proportion from 
the dead load stresses. In this problem the maximum snow, which 
is used when the wind load is not considered, is taken at 20 lbs. 
per sq. ft. of horizontal projection ; and the minimum snow load, 
which is used in connection with the wind load, is taken at 10 lbs. 
per sq. ft. of horizontal projection. 



154 



STRESSES IN A TRANSVERSE BENT. 



Chap. XV. 



Since the deflection of the truss is neglected, there are no 
stresses in the knee-braces due to vertical loads. 

139. Graphic Method for Determining Wind Load Reac- 
tions.* The following is a very convenient graphic method 
for determining the wind load reactions for a transverse bent. 

Lay off 2Wj^ (I'ig- 85) equal to the total normal wind load 
acting upon the roof area supported by the transverse bent, its 
point of application being at the center of the length RN. Also, 
lay off SW^ equal to the total horizontal wind acting upon the 
side area supported by the column of the bent, its point of appli- 




■^. ,nz 






FIG. 85. 



_L _\>,.,-' i 

Graphic Method for Determininq Rbac^ons. 



cation being at the center of the column length AR. Find the 
resultant LS = 5W of the normal and horizontal wind forces, 
and produce the line of action of this resultant until it intersects 
at B a vertical line through N, the apex of the truss. Lay oflP 
DB = BG = ^LS (since the horizontal components of the reac- 
tions are equal, by hypothesis). Join the points A and B, also 
the points B and C, and from the points D and G, draw the ver- 



*This graphic method of determining the reactions is that given in 
Ketchum 's ' * Steel MiU Buildings. " 



GRAPHIC DETERMINATION OF REACTIONS. 155 

tical lines DE and GF, respectively. Then ED represents the 
vertical component V^ of the right reaction Rj, and GF repre- 
sents the vertical component Vi of the left reaction Rj. 

This construction may be proved, as follows: Taking the 
moments of the external forces about the point C, and solving 
for Vi, we have 

or, since BG = (by construction), 

4 (area triangle BGC) . v 

\*C^ \y T 

But area triangle BGC = 

4 

For, area triangle BGC = area triangle BGF + area triangle 

CGF 

FG X c FG X d FG X L 

= — ^-^ + — ^ — = — : — • (3) 

224 
Substituting this value of the area of the triangle BGC in equa- 
tion (2), we have 

Vi = GF, which proves the construction. 

In like manner, it may be shown that ED = Vo. 

The left reaction R^ = GM is now determined by finding the 
resultant of Hi and Vi, as shown in the figure. The right reac- 
tion R2 = MD is equal to the resultant of Hg and Vg. 

In the case shown, the columns are taken as hinged at the base. 
If the columns are fixed at the base, the above construction 
should be modified by taking the columns as hinged at the point 
of contfa-flexure and neglecting the wind load below this point. 

140. Wind Load Stresses — Columns Hinged at Base. The 
wind load stress diagram for a transverse bent having the columns 
hinged at the base is shown in Fig. 86. The dimensions of the 
bent are the same as those shown in Fig. 84. The wind load on 
the roof is taken at 22 lbs. per sq. ft., which is the normal com- 
ponent of a horizontal wind load of 30 lbs. per sq. ft. (see Fig. 



156 



STRESSES IN A TBANSVEBSE BENT. 



Chap, XV, 



47, Duchemin's formula). The horizontal wind load on the side 
of the building is taken at 20 lbs. per sq. ft. The distribution of 
the loads is shown on the truss diagram, and the reactions are 
determined by the method shown in § 139. Since the reactions act 
at the bases of the columns, they produce a bending moment at 




Nor. Wind, 22 Ibs.sqft. ^' 
Hor. » 20 



II 11 It 



^ Columns Hinged at Base. ^ Max- Moment- Hzd. 




Wind Load 
Stress Diagram 

5000 loooa 
■ »'■'■ I 

FiQ. 86. Stresses in a Transvebsb Bent — Columns Hinged. 

the foot of the knee-brace. The difficulty of drawing .the stress 
diagram due to this moment is overcome by trussing the columns 
as shown in the figure. The stresses obtained for the members of 
the auxiliary trusses are not used, and the resulting stresses in 
the columns are not true stresses. The stresses in all the mem- 
bers of the truss and in the knee-braces are, however, true stresses. 



COLUMNS HINGED AT BASE. 157 

The c»mplete stress diagram is shown in Fig. 86, the kind of 
stress in each member being indicated by arrows in the truss 
diagram. The members shown by dotted lines in the truss dia- 
gram are not stressed when the wind acts upon the left side of 
the truss. The true direct stresses in the windward and leeward 
columns are respectively equal to Vi and Vj. 

The maximum bending moment occurs at the foot of the 
knee-brace of the leeward column, and is equal to Hjd. 

The unit stress in the extreme fiber of the column due to the 
wind moment may be found from the formula. 



S = 



My .^ 



I± 



cE 

where 

M = maximum bending moment in inch-pounds ; 
y = distance from neutral axis to extreme fiber, the axis 
being perpendicular to the external force causing the 
bending moment; 
I = moment of inertia of the section of the member about an 
axis perpendicular to the direction of the force causing 
moment ; 
P = total direct loading in the member in pounds ; 
L = length of member in inches. 

c = constant depending upon condition of ends of member. 

For a member hinged at both ends, use c=io; for 

member fixed at one end and hinged at the other, use 

c = 24 ; and for member fixed at both ends, use c = 32 ; 

E = modulus of elasticity of the member. For steel, it may 

be taken at 29000000. 
The sign is to be minus if P causes compression, and plus if 
P causes tension. 

141. Wind Load Stresses — Columns Fixed at Base. The 
wind load stress diagram for a transverse bent with the columns 
fixed at the base is shown in Fig. 87. The dimensions of the bent 



*See * * Theory and Practice of Modern Framed Structures, ' ' by John- 
soii; Bryan, and Turneaure. 



15S 



BTIIE»SE3 IN A TBAN8TEB8E BENT. 



Chap. XV. 



are the same as shown in Fig. 84. The wind load on the roof is 
taken at 22 lbs. per sq. ft., and on the sides, at 20 lbs, per sq, ft. 

It should be noted that in this case the columns are considered 
as fixed at the points of contra-flexure, and that the wind below 




Fio. ST. Stresses in i. TBANavEnsE Bent — CoLmus Fixed. 



these points is neglected. The remaining solution is similar to 
that shown in § 140. 

The maximum positive bending moment is at the foot of the 

knee-brace of the leeward column, and is +_J_-. The maxi- 

2 
mum negative bending moment is at the foot of the leeward 

column, and is — '—: 



COLUMNS FIXED AT BASE. 159 

ThcL advantages of fixing the columns are shown by a com- 
parison of the stress diagrams given in Fig. 86 and in Fig. 87 
and of the maximum bending moments for the two cases. Both 
stress diagrams are drawn to the same scale. 

The unit stress in the column caused by the bending moment 
may be found by applying the formula shown in § 140. 

Since the wind may act from either side, it is seen that many 
of the members are subjected to reversals of stress. 

The maximum and minimum stresses caused by the different 
loadings may be found by making the combinations indicated 
in § 127. 



CHAPTER XVI. 



MISCELLANEOUS PEOBLEMS. 



142. Stresses in a Grand Stand Truss. It is required to 
determine the stresses in the grand stand truss shown in Fig. 
88, a. In this structure, the knee-brace and the seat beam meet 
at the point M. Since the left column is braced by the seat beam, 
all the horizontal component of the wind will be taken by this 
beam. The right reaction due to the wind loads will therefore be 
vertical, and there will be no bending moment in this column due 
to the wind. The wind on the vertical side of the building will 
not be considered, as it will be resisted by the seat beam and will 
cause no stresses in the truss. The general dimensions of the 
truss, together with the loads used, are shown in Fig. 88, a. 

The dead load stress diagram is shown in Fig. 88, b. The 
reactions Rg and R3' may be obtained by drawing the force and 
funicular polygons, or by the method of moments. The kind of 
stress is indicated by the arrows in the stress diagram, compres- 
sion being denoted by arrows acting away from each other, and 
tension by arrows acting toward each other. In this problem, the 
arrows are placed on the stress diagram, instead of the truss 
diagram, to avoid confusion; since the same truss diagram is 
used for all the stress diagrams. 

The snow load stress, if considered, may be determined from 
the dead load stress diagram. 

The wind load stress diagram, considering the wind as acting 
towards the left, is shown in Fig. 88, c. The reactions R2 and Rg' 
are determined by means of the force and funicular polygons. 
In finding these reactions, the resultant 2Wr of the wind load is 
used instead of the separate loads to simplify the solution. It is 
seen that the left reaction has a downward vertical component 

160 



BTassass in a obamd stand tbubs. 



161 




'"'^^ h 






Dead Load -10 Ibspersqfthorproj 
Nor. comp. of wind - 17 lbs- per sq-ft- 

Distance between trusses « 20 ft. 



o/ 



13 



o, 

Oi 



0' 5' 

''■■■■ 



10' 



15' 20' 

_i I 



^hn 



ost 



Of 



^/i 



w 




Dead Load 
Stress Diagram 
(b) 




Wind Load 
Stress Diaqrajn ) 
Wind Right 
(c) 



4000 8000 
I 1 I I I I 

Scale -All Diagranns 




Wind Load 
Stress Diagram 

Wind Left '^' 8 

FIO. 88. 8TBB8BB8 IN ▲ GBAND StAND TRUSS. 



162 



MISCELLANEOUS PROBLEMS. 



Chap. XVI. 



due to the wind on the cantilever side of the truss. The stresses 
may be obtained from the stress diagram shown in Fig. 88, c. 

The wind load stress diagram, considering the wind as acting 
towards the right, is shown in Fig. 88, d. The reactions Rj and 
Rj' are determined by means of the force and funicular polygons 
as shown, the resultant of the wind loads on the left being used 
instead of the separate loads. The stresses may be obtained from 
the stress diagram shown in Fig. 88, d. 

If the maximum and minimum stresses are required, they may 




FiQ. 89. Stbesses in a Tbbstle Bbitt. 



be determined by combinmg the stresses due to the different 
loadings. ^ 

143. Stresses in a Trestle Bent. It is required to find the 
stresses in the trestle bent, loaded as shown in Fig. 89, a. Since 
the same detail is used at the bases of the columns, it will be 
assumed that the horizontal components of the reactions are 
equal. 

' The -stress diagram for the bent, shown in Fig. 89, b, is drawn 



STBESBEfl IJ{ A TRESTLE BEXT. 



ins 



by starting with the force I', acting at the top of the bent. Tbt- 
diagram is completed by taking each of the horizontal components 
of the reactions at the base equal to one-half of the total wind 
load upon the structure. The stresses may be obtained from the 
stress diagram, the kind of stress being indicated by the arrows. 




Momenl = 18000 K J.4'61200 inlfas 
Direct 5heor.5-ia000;5-3600lb5 
To get o , the shear due to momen* 
at a units disfonce from the 
center of qravKy. we have 
Moment, M,= a(dhdUdlrdUdl) 
Or,6l200 = a(.9V?66'.l64',|.54'f;6fe.'J 
Therefore, □ > E6Z0 







RC3U 


LTS 






Rivet 


d 


d^ 


5« 


s 


R 


z 


90 
2 66 


-8! 
7.01 


2530 
7500 


3600 
3600 


6100 
9300 


3 


1.54 


3 39 


5190 


3600 


3500 


4 


1-84 


3 39 


5190 


3600 


3500 


5 


?66 


7.07 


7500 


3600 


9300 


5.^= 3heorduetoMoment = axd. 
5 = Shear due to direct load, P. 
R = Resultant 5heor 




Fra. 00. Eccentric Riveted Connection. 



164 MISCELLANEOUS PROBLEMS. Chap. XVL 

The vertical components of the reactions are equal, but act in 

opposite directions. 

144. Eccentric Riveted Connection. It is required to find 

the shearing stress in each of the five rivets in the connection 

shown in Fig. 90, b. The spacing of the rivets is that used in 

a six-inch angle for a standard channel connection. The load P 

is transferred to its connecting member at the left edge of the 

angle. Since the load is not transferred at the center of gravity 

of the connecting rivets, it is seen that there is a tendency for 

rotation, which causes additional shearing stresses in the rivets. 

The total shearing stress in each rivet may be found by the 

following method. 

Replace the force P by an equal force acting through the 

center of gravity of all the rivets and a couple whose moment M 
is equal to P multiplied by the distance from its line of action 
to the center of gravity of the rivets. In this example, M = 18 000 
X 3.4=61 200 in. lbs. Now the force P acting through the cen- 
ter of gravity will cause a direct shearing stress in each rivet 
equal to P divided by the number of rivets. In this case, the 
direct shearing stress is 18 000 -f- 5 = 3 600 lbs. The moment M 
of the couple will cause a shearing stress in each rivet, which may 
be computed as follows : Since the shear in each rivet due to the 
moment will vary as the distance of that rivet from the center 
of gravity of all the rivets, it follows that the resisting moment of 
each rivet (which is equal to the shear in the rivet multiplied by 
^.ts moment arm) will vary as the square of its distance from the 
center of gravity of all the rivets. Now if a represents the shear 
at a unites distance from the center of gravity due to the moment 
M ; and di, dg, etc., represent the distances of the respective rivets 
from the center of gravity, the following relation is true : 

M = a (d,^ + d,^ + d- + d,^ + d,2), 

M 
and ^=2d^- 

Since a is the shear at a unit's distance due to the moment, the 
shear on any rivet is equal to a multiplied by its distance from the 
center of gravity of all the rivets. The total shear in each rivet 
may now be determined by finding the resultant of the shear due 



ECCENTRIC BIVETED CONNECTION. 1G5 

to the direct load P and that due to the moment of the couple. 
This resultant may be found by drawing the parallelogram of 
forces, as shown in Fig. 90, b. 

A table showing the important data for this problem, together 
with the resultant shear on each rivet, is shown in Fig. 90, a. 
This table shows a convenient form for recording results. 

The force polygon for the shearing forces is shown in Fig. 
90, c, and the funicular polygon in Fig. 90, d. Since both poly- 
gons close, the forces are shown to be in equilibrium. 

Referring to the resultant shears given in the last column of 
the table shown in Fig. 90, a, it is seen that the eccentric connec- 
tion causes large shearing stresses in some of the rivets. It 
therefore follows that all eccentric connections should be carefullv 
investigated. 



PART III. 



BEAMS. 



CHAPTER XVII. 

BENDING MOMENTS, SHEARS, AND DEFLECTIONS IN BEAMS 

FOR FIXED LOADS. 

This chapter will treat of graphic methods for determining 
bending moments, shears, and deflections in beams for fixed 
loads. It will be divided into three articles, as follows: Art. i. 
Bending Moments and Shears in Cantilever, Simple, and Over- 
hanging Beams ; Art. 2r, Graphic Method for Determining Deflec- 
tions in Beams; Art. 3, Bending Moments, Shears, and Deflec- 
tions in Restrained Beams. 

Graphic methods may be readily applied to the determination 
of bending moments, shears, and deflections, and in many cases 
afford a simpler and more comprehensive solution than algebraic 
processes, especially when these functions are required at several 
points along the beam. 



Art. I. Bending Moments and Shears in Cantilever, Sim- 
ple AND Overhanging Beams. 

145. Definitions. Vertical forces, only, will be considered 
in this article ; as the forces acting upon a beam are usually ver- 
tical loads. 

Bending Moment. The bending moment at any point, or at 

167 



168 



BENDING MOMENTS AND SHEARS IN BEAMS. Chap, XVII. 



any section, of a beam is the algebraic summation of the moments 
of all the forces on one side of the point, or section. The bending 
moment will be considered positive if there is a tendency for the 
beam to bend convexly downward, and will be considered nega- 
tive of there is a tendency for the beam to bend convexly upward. 

A bending moment diagram is a diagram representing the 
bending moments at points along the beam due to the given 
loading. 

Shear. The shear at any section of a beam is the algebraic 
summation of all the vertical forces on one side of the section. 
The shear is positive if the portion of the beam to the left of the 
section tends to move upward with reference to the portion to 
the right of the section. The shear is negative if the portion to 
the left tends to move downward with reference to the portion 
to the right of the section. 

A shear diagram is a diagram representing the shears at points 
along the beam due to the given loading. 

146. Bending Moment and Shear Diagrams for a Canti- 
lever Beam. Two conditions of loading will be considered for 
the cantilever beam, viz.: (a) beam loaded with concentrated 
loads, and (b) beam loaded with a uniform load. 

(a) Cantilever Beam zvith Concentrated Loads. It is 




i_ 



Fig. 91. Cantilever Beam — Concentrated Loads. 



^^^' ^' CANTILEVEB BEAM. 169 

required to draw the bending moment and shear diagrams for 
the beam MN (Fig. 91), loaded as shown with the three con- 
centrated loads AB, BC, and CD. 

Bending Moment Diagram, To construct the bending moment 
diagram, draw the force polygon (Fig. 91, a), assume any 
pole O, and draw the funicular polygon (Fig. 91, b) for the 
given loads. The diagram shown in Fig. 91, b is the bending 
moment diagram for the beam loaded as shown. For the bend- 
ing moment at any point along the beam is equal to the intercept 
under that point, cut off by the funicular polygon and the hori- 
zontal line m-n, multiplied by the pole distance H. (See § 64.) 

Referring to Fig. 91, it is seen that the maximum bending 
moment occurs at the fixed end of the beam. 

Shear Diagram. To construct the shear diagram, lay off the 
reaction R = AD (Fig. 91, c), and draw the horizontal line DD. 
Since there are no loads between the left reaction and the load 
AB, the shear is constant between the points of application of 
these forces, and is represented by the intercept between the lines 
DD and AA. At the point of application of the load BC, the 
shear is reduced by the amount of that load. The shear is con- 
stant between the points of application of the loads AB and BC, 
and is represented by the intercept between DD and BB. Like- 
wise, the shear between the loads BC and CD is constant, and is 
represented by the intercept between DD and CC. 

Referring to the cantilever beam shown in Fig. 91, it is seen 
that the maximum bending moment and the maximum shear both 
occur at the same point — the fixed end of the beam. 

(b) Cantilever Beam zvith Uniform Load. It is required 
to draw the bending moment and shear diagrams for the beam 
MN (Fig. 93), loaded as shown with a uniform load. 

Bending Moment Diagram. Referring to Fig. 92, it is seen 

^ w- lbs- per ft- 



Fig. 92. 



170 BENDING MOMENTS AND 8HBAKS IN BEAMS. ^^P- ^V^^* 

that a couple is required to fix the beam at the left end. The 
magnitude of this couple may be found as follows : 

w(l — x)2 
Moment of forces to right of A= , (i) 

wx* 

and moment of forces to left of A = — Pa + Rx (2) 

which are the equations of a parabola. 

For equilibrium, the sum of the moments of the forces on both 
sides of A must be equal to zero, and the magnitude of the couple, 
which is Pa, may be found by equating (i) and (2) to zero and 
solving for Pa, noting that R = wl. Then 

wx^ w (1 — x)* 
-Pa + Rx — + ' =0, 

2 2 

wP 
or Pa= , which is the magnitude of the couple required to 

2 
fix the beam at the left end. It is seen that the value of P 
depends upon the arm of the couple. 

wP 

If x = o, M = — Pa = , 

2 

and 

wP 

if x = l, M=— PaH = 0. 

2 

To construct the moment diagram for the beam MN (Fig. 93), 
divide the load area into any number of equal parts (in this case 
eight) by verticals, and take the weight of each part as a load 
acting through its center of gravity. Draw the force polygon 
(Fig. 93, a) and the funicular polygon (Fig. 93, b) for these 
loads; and trace a curve (not shown) tangent to the funicular 
polygon at its ends and at the middle points of its sides. The 
bending moment at any point along the beam is equal to the inter- 
cept, measured between the horizontal line m-n of the funicular 
polygon and the curve, multiplied by the pole distance H. The 
greater the number of parts into which the load area is divided, 
the more nearly will the funicular polygon approach the bending 
moment parabola. 



Art. J. 



CANTILEVEIl BEAM. 



171 



Shear Diagram. The shear at any point whose distance from 
the left end of the cantilever beam is x may be represented by 
the equation, S = R — wx. If x»=o, S = R; and if x = l, 
S = o. Since the load is uniform, the shear decreases by a con- 

j Uniform ioad of w- lbs, p er lin- f t- 



y. 



v/. 



.J-L4 



f 



.N 



Moment Diagram 




Shear Diagram 
(c) 





Force Polygon 
(a) 



Fig. 03. Cantilever Beam — Uniform Load. 



stant amount towards the right end of the beam. Therefore, to 
draw the shear diagram, lay off AB (Fig. 93, c) == R, and draw 
the horizontal line BC. Join the points A and C, completing the 
triangle ABC, which is the required shear diagram. 

Referring to the bending moment and shear diagrams, it is 
seen that neither the bending moment nor the shear changes sign 
throughout the length of the beam. 

147. Bending Moment and Shear Diagrams for a Simple 
Beam. The bending moment and shear diagram for a simple 
beam will be constructed for two conditions of loading, viz. : (a) 
beam loaded with concentrated loads, and (b) beam loaded with 
a uniform load. 

(a) Simple Beam with Concentrated Loads. It is required 
to draw the bending moment diagram and the shear diagram 



172 BENDING MOMENTS AND SHEARS IN BEAMS. ^^^P- ^V^J- 



for the beam MN (Fig. 94), supported at its ends, and loaded 
with concentrated loads as shown. 

Bending Moment Diagram, To construct the bending moment 
diagram, draw the force polygon (Fig. 94, a), assume any pole O, 
and draw the funicular polygon (Fig. 94, b), which is the required 
bending moment diagram. For, the bending moment at any 



r- 




±_ 



B 



, (b) , 

Moment Diaqram 
I « • 



I 




(a) 
Force Polygon 



(c) C 
Shear Diagram 



Ri 



I 
I 
I 



Fig. 94. 



C E 

SiMiPLB Beam — Concentrated Loads. 



point along the beam is equal to the intercept under the point of 
moments multiplied by the pole distance H. Referring to the 
bending moment diagram (Fig. 94, b), it is seen that the moment 
of the forces to the left of any point along a simple beam is posi- 
tive, and does not change its sign throughout the length of the 
beam. In this particular example, it is seen that the maximum 
bending moment occurs at the point of application of the load BC. 
Shear Diagram, To construct the shear diagram, lay off 
FA=^ Ri (Fig. 94, c), and draw the horizontal line FF. Between 
the left reaction and the load AB, the shear is equal to Ri ; there- 
fore from A, draw the horizontal line AA. Then the intercept 
measured between FF and AA represents to scale the shear 



SIMPLE BKAlf. 



173 



between the left end of the beam and the load AB. Between the 
loads AB and BC, the shear is equal to Rj — AB ; therefore from 
A, draw the vertical line AB, representing to scale the force AB, 
and from B, draw the horizontal line BB. Then the intercept 
between BB and FF represents to scale the shear between the 
loads AB and BC. It is seen, by referring to Fig. 94, c. that the 
shear between the left end of the beam and the load BC is posi- 
tive. Between the loads CD and DE, the shear equals R, — AB — 
BC — CD, and is represented by the intercept between FF and 
DD. Between the load DE and the reaction R,, the shear equals 
Ri — AB — BC — CD — DE, and is represented by the intercept 
between FF and EE. The shear between the load BC and the 
right end of the beam is negative. 

Referring to Fig, 94. b and Fig. 94, c, it is seen that the maxi- 
mum bending moment occurs at the point of zero shear, i. e., at 
the load BC. 

(b) Simple Beam with Uniform Load. It is required to 
draw the bending moment and shear diagrams for the beam MN 
(Fig. 95), supported at its ends, and loaded with the uniform 
load, as shown. 



Uniform load of w Ibs-per Irn- ft- 

-It 1 1 'T ' f ' 1 ' T '' ' ' 




Force Fblyqon 



c Beam — Unifobji Load. 



174 BENDING MOMENTS AND SHEARS IN BEAMS. ^^P- -3lF//. 

Bending Moment Diagram. To construct the bending moment 
diagram, divide the load area into any number of equal parts 
(in this case eight) by verticals, and take the weight of each part 
as a force acting through its center of gravity. Draw the force 
polygon (Fig. 95, a) for these loads, assume any pole O, and 
draw the funicular polygon (Fig. 95, b). Trace a curve (not 
shown in the figure) tangent to the funicular polygon at its ends 
and at the middle points of its sides. Then the bending moment 
at any point along the beam will be equal to the intercept under 
that point, between the curve and the closing line of the funicular 
polygon, multiplied by the pole distance H. 

The curve drawn tangent to the middle points of the sides of 
the funicular polygon is a parabola, and the greater number of 
parts into which the load area is divided, the more nearly will the 
funicular polygon approaching the bending moment parabola. It 
will now be shown that the bending moment curve for a uniform 
load is a parabola. 

Let L = span of beam, w = weight of uniform load per linear 
foot, and x = distance from the left support to the point of 
moments. Then the bending moment at any point whose distance 

wx* w 

from the left end is x is, M = R^x = — (Lx — x^), which 

is the equation of a parabola. The moment is a maximum when 

x= — , and is equal to ^wL^. 
2 

The parabola may be drawn without constructing the force 
and funicular polygons, as follows: In Fig. 95, b, lay off the 
ordinate mn = nr = JwL^ = moment at the center of the beam; 
and connect the point r with the points i and 5. Divide the lines 
ir and rs into the same number of equal parts, and number them 
as shown in Fig. 95, b. Join like numbered points by lines, which 
will be tangents to the required parabola. 

Shear Diagram, To construct the shear diagram, lay off 
AC = Ri (Fig. 95, c), and from C, draw the horizontal line CC. 
Also, lay off CB = Rg downward from C, and connect the points 
A and B, which gives the required shear diagram. 

It is seen from Fig. 95, b and Fig. 95 c that the shear is zero 



Art 1. 



OVEBHANGING BE.VM. 



175 



at the center of the beam, and that the moment is a maximum at 
the point of zero shear, i. e;, at the center of the beam. 

The equation expressing the shear at any point is S = Rj — 

wx = iwL — wx = w ( x), which is the equation of the 

inclined line AB (Fig. 95, c). 

It will now be shown that the bending moment at any point 
along a simple beam is the definite integral of the shear between 
the point in question and either point of support. For, 



Jo Jo 



w I — 
V 2 



x)=— (Lx — x2) =M. 



The above equation shows that the bending moment at any 
point in a simple beam uniformly loaded is equal to the area of 
the shear diagram on either side of the point. 

148. Bending Moment and Shear Diagrams for an Over- 
hanging Beam. Two conditions of loading will be considered 
for the overhanging beam, viz.: (a) beam loaded with concen- 
trated loads, and (b) beam loaded with a uniform load. 

(a) Overhanging Beam with Concentrated Loads, 



It 



IS 




Moment Diaqrams 




pi Force Polygons 




Pio. 96. Overhanging BsA:\f — Concentrated Loads. 



176 BENDING MOMENTS AND SHEARS IN BEAMS. ^^P- ^^11' 

required to draw the bending moment and shear diagrams for 
the beam MPN (Fig. 96), loaded with concentrated loads. 

Bending Moment Diagram. To construct the bending mo- 
ment diagram, draw the force polygon (Fig. 96, a), assume any 
pole O, and draw the funicular polygon (Fig. 96, b), which is 
the required bending moment diagram. The moment at any 
point along the beam is equal to the intercept under the point 
multiplied by the pole distance H. 

The bending moment diagram shown in Fig. 96, b, is not in 
a convenient form for comparing moments at different points, 
and an equivalent diagram will now be drawn which shall have 
all intercepts measured from a horizontal line. If the line mn 
(Fig. 96, d), common to the two outside forces R^ and EF, is 
made horizontal, it is seen that the intercepts will all be meas- 
ured from a horizontal line. To draw the bending moment 
diagram (Fig. 96, d), construct the new force polygon (Fig. 96, 
c), as follows: Since mn is to be horizontal and common to the 
forces Ri and EF, draw the ray O'F (Fig. 96, c)=H (Fig. 
96, a), corresponding to the string mn; and from F draw Ri = 
FA (already found) upward. From A, draw in succession the 
loads AB, BC, and CD downward from D; and from D, draw 
DD' = R2 upward. Then from D', draw the loads D'E = DE 
and EF downward, closing the force polygon at F. Construct 
the funicular polygon shown in Fig. 96, d. It is readily seen, 
since Rj and EF have a common point F in the force polygon 
and since FO' is horizontal, that* the funicular polygon, or bend- 
ing moment diagram (Fig. 96, d), will have intercepts measured 
from a horizontal line equal to those in Fig. 96, b. 

Referring to Fig. 96, b and Fig. 96, d, it is seen that the 
bending moment for this problem has a maximum positive value 
at the load BC, that it passes through zero at the point p, and 
that it has its maximum negative value at P, the point of applica- 
tion of the reaction Rg. 

Shear Diagram. The shear diagram for the overhanging 
beam MN loaded with concentrated loads is shown in Fig. 96, e. 
Referring to this diagram, it is seen that the shear is positive at 
the left end of the beam, that it passes through zero at the load 



Art 1, 



OVERHAXQING BEAM. 



177 



BC, and that it has its maximum negative value between the 
load CD and the reaction R^. It is further seen that the shear 
passes through zero at P, the point of application of the reaction 
Rg, and that it has its maximum positive value between the reac- 
tion R2 and the load DE. 

The maximum moment occurs at the point of zero shear, and 
since the shear passes through zero at two points, the moment 
will have maximum values at these points — one of which is the 
maximum positive moment, and the other, the maximum negative 
moment. 

Uniform load of w lbs- per linft- 




Moment Diagrams 





Force Polygons 




Fig. 97. Ovebhanging Beam — Uniform Load. 



(b) Overhanging Beam with Uniform Load, It is required 
to draw the bending moment and shear diagrams for the beam 
MN (Fig. 97). The beam overhangs both supports, and is loaded 
with a uniform load. 

Bending Moment Diagram. To draw the bending moment 
diagram, divide the load area into any number of equal parts (in 
this case nine), and assume the weight of each part as a load 
acting through its center of gravity. Draw the force polygon 
(Fig. 97, a), and the funicular polygon (Fig. 97, b). Trace a 
curve (not shown in the figure) tangent to this funicular polygon 



178 DEFLECTIONS IN BEAMS. ^/lap. XVII. 

at its ends and at the middle points of its sides, which will be the 
required bending moment diagram! 

The equivalent funicular polygon showing the intercepts meas- 
ured from a horizontal line is shown in Fig. 97, d, and the force 
polygon is shown in Fig. 97, c. If a curve is drawn tangent 
to this funicular polygon at its ends and at the middle points of 
its sides, the diagram will be the, required bending moment dia- 
gram for the given los^ds and beam. The bending moment at 
any point is equal to the intercept under the point of moments 
multiplied by the pole distance H = H'. 

Referring to Fig. 97, d, it is seen that the bending moment is 
negative at both supports, and that it passes through zero between 
the supports and becomes positive. 

Shear Diagram. The shear diagram for the overhanging 
beam loaded with a uniform load is shown in Fig. 97, e. It is 
seen that the shear between the left end of the beam and the left 
support is negative, and that it passes through zero at the left 
support and becomes positive. The shear passes through zero at 
p and becomes negative. At the right support it again passes 
through zero and becomes positive, and is positive between the 
right support and the right end of the beam. 

The shear passes through zero at three points, therefore the 
bending moment has maxima at these points. The maximum 
negative moments are at the supports, and the maximum positive 
moment is between the supports. 



Art. 2. Graphic Method for Determining Deflections in 

Beams. . 

149. Explanation of Graphic Method — Constant Moment 
of Inertia. The deflection at any point along a beam may be 
readily determined graphically, and a graphic method for finding 
the deflections will now be explained. The graphic method is 
especially useful when the deflections are required' at several 
points along the beam. 

Let MN (Fig. 98, a) be a horizontal beam, supported at its 



Art. 2. 



GRAPHIC DETEKMIXATIOX OF DEFLECTIONS. 



179 



ends, and let the beam be divided into a sufficient number of parts 
(in this case four) that the polygon representing its neutral sur- 

M N 



id) 




'*-sU 




(e) 



Fig. 98. Graphic Deflections. 



face may very closely approximate the elastic curve. By the 

elastic curve is meant the curve assumed by the neutral surface 

of the beam when the elastic limit of the material is not exceeded. 

Let ab (Fig. 98, c) be the position taken after flexure by the 

neutral surface of the segment at the 

left end of the beam. Also let Fig. 99 

represent a portion of a beam which 

has deflected as shown. The angle a 

(Fig. 98, c) between be and bCi (the 

projection of ab) represents the angle 

of rotation of the section CD (Fig. 

99) to CD', originally parallel to AB. 

The angle a (Fig. 98, c) also equals 

the angle a between mn and mo 

(Fig. 99). 

Let ab = dl (Fig. 98, c). Then the angle a may be found from 

Mdl 
the formula tan a =-pY~' 




where 



M = bending moment at b, in inch-pounds, 
dl = aborbCi, 



180 DEFLECTIONS IN BEAMS. Chap. XVII . 

E — modulus of elasticity of the material, in pounds, 
I = moment of inertia of the cross section of the beam, 
which in this case is assumed to be constant throughout its length. 

This formula may be deduced as follows : The stresses at 
any point in the beam shown in Fig. 99 will vary as the distance 
of the point from the neutral axis. From the similar triangles 
Omn and DnD' (Fig. 99), we have 

R : c : : dl : A, 
or, RA = cdl. (i) 

Now let S = stress on extreme fiber, and let E = modulus of 
elasticity of the material. Then 

A : S : : dl : E, 

A Sdl 
or, A=^-. (2) 

Substituting this value of A in equation (i), and solving for R, 
we have 

Ec 

R=-s-. (3) 

But from the common theory of flexure, we have 

Mc 
S = ^p. (4) 

Substituting this value of S in equation (3), we have 

EI y . 

M 

dl 
From Fig. 99, it is seen that tan a = — p-. (6) 

Substituting the value of R found in equation (5) in equation (6), 
we have 

Mdl. 
tana=-^j-. (7) 

Calculate the value of the angle a from equation (7), and 
draw be. The angles aj and ag may be found in like manner. Cal- 
culate the values of these angles, lay them o& at c and d (Fig. 
98), respectively, and draw the lines cd and de. Draw the closing 
line ae, and the vertical ordinates will have their true values in 



Art. IB. DEFLECTIONS IN A SIMPLE BEAM. 181 

either Fig. 98, b or Fig. 98, c. For in both Fig. 98, b and I'ig. 
98, c, we have 

at B, Bb = Bb, 

at C, Cc = Ccj-cCj ^ 2 Bb-cCj, 

at D, Dd = Ddg-didg-ddi = 3 Bb-2 cCi-dd^, 

and at e, o = ee8-e2e8-eie2-eei = 4 Bb-3 cCi-2 ddj-eej. 

Since cCj, ddj, and eCj are equal in both Fig. 98, b and Fig. 
98, c ; therefore Bb, Cc, and Dd must also be equal in both figures. 
It is thus seen that the closing line ae may be horizontal or 
inclined without changing the values of the deflection intercepts. 

Referring to Fig. 98, it is seen that the successive triangles 
aBb, bcCi, cddj, and dee^ have one side equal in each successive 
pair of triangles. These triangles may therefore be placed in 
contact, as shown in Fig. 98, d and Fig. 98, e, Fig. 98, d cor- 
responding to Fig. 98, b, and Fig. 98, e to Fig. 98, c. The tri- 
angles shown in Fig. 98, d and Fig. 98, e suggest another method 
of drawing the polygons representing the neutral surface of the 
beam. Thus with a pole distance dl equal to one division of the 
span, lay off in succession the distances cCj, ddj, and eCi, com- 

Mdl 
puted from the formula dl tana = dl-=rj-. Draw the rays, and 

construct the funicular polygons shown in Fig. 98, b and Fig. 
98, c. 

150. Practical Application. The method explained in 
§ 149 will now be applied to a practical problem. 

Let MN (Fig. 100) represent a 12 in. X 31.5 lb. X 40 ft. I 
beam, supported at its ends, and sustaining a load of 4 000 pounds 
at a point 15 feet from the left end of the beam. It is required to 
draw the elastic curve representing its neutral surface and find the 
magnitude of its maximum deflection. 

Draw the force polygon shown in Fig. 100, a and the funicular 
polygon shown in Fig. 100, b. Divide the span into any number 
of equal parts (in this case eight) by verticals in the moment 
diagram, and bisect each of these laminae by a vertical, drawn 
between the closing line cd and the funicular polygon ced. Con- 
sider these verticals as loads, and lay them off successively on the 



182 



DEFLECTIONS IN BEAMS. 



Chap. XVIL 



vertical load line XY (Fig. lOO, c). Assume any pole O', draw 
rays, and construct a new funicular polygon fgh (Fig. lOO, d) 
with the closing line fg. Draw a curve tangent to the middle 





^ '&\rT:. .^--J99o . , , A 




^ id) 
Deflection Diagram 



Fig. 100. Deflections in a Simple Beam — Concentrated Load. 

points of the sides of this funicular polygon. This curve will 
then represent the elastic curve of the beam in an exaggerated 
form, and the intercepts between the funicular polygon and its 
closing line, which represent the deflections of the beam, are 
each to be divided by a constant to obtain their true values. 

. The value of this constant will now be determined. Referring 
to Fig. 98, b and Fig. 98, c, it is seen that the ordinates cci, ddi, 

Mdx2 
and eCi are each equal to dx tan a, which may be written -^y — J 

since tan a has been shown equal to , and dl may be taken 

equal to dx for the elastic curve with a pole distance dx. Now in 
Fig. loi, a, lay off successive distances on the deflection load line 

M 

equal to tpr* where H = first pole distance ; and then construct 
Jri 

Fig. loi, b. It is now seen that Fig. loi, a corresponds to Fig. 
100, c, and Fig. loi, b to Fig. 100, d. From the similar triangles 



Art, f. 



DEFLECTIONS IN A SIMPLE BEAM. 



183 



bcci and 0'i2, cq : — : : dx : H', from which cci =.: ^■. A sim- 

ilar relation is true of the intercepts ddi and eej. Comparing the 
values obtained for the verticals cCi, ddj, and tt^ in Fig. loi, a 




Fig. 101. 



with the true verticals cCj, dd,, and ecj in Fig. 98, b and Fig. 
98, c, we find the following relation, 

Mdx2 Mdx HH'dx , . 

^ (i) 

EI • HH'— EI 

Therefore the true verticals may be found from those in Fig. 10 1 

by multiplying the intercepts measured in this figure by 

HHMx - Mdx2 Mdx HHMx 

; for, = X 

EI EI HH' EI 

The same relation is true for the actual deflection intercepts 
Bb, Cc, and Dd, which correspond to those in Fig. 100, d ; since 
these are each composed of the elements cCj, dd^, and ee^, as has 
been previously shown. 

In the problem given, H = 7 000 pounds, H' = 20 feet = 240 

480 . 

inches, dx =-:5— = 60 inches, the maximum intercept = 6.8 feet 

o 

= 81.6 inches, 1 = 215.8, and E = 29 000 000. Therefore the 

A a 4.' A /• • u \ ^1-6 X 7 000 X 240 X 60 
maximum deflection A (in inches) = 1 Z 

215.8 X 29000000 

= 1.31- 

If the constant by which each measured intercept is to be mul- 



184 DEFLECTIONS IX BEAMS. Chap, XVII. 

HH'dx 
tiplied is used in the form . , then the measured intercept, 

H', and dx must all be expressed in inches. 

If the measured intercept, H', and dx are expressed in feet, 
then the constant should be 

1728 HH'dx 

Ei • (") 

The method explained above may be used to find the deflec- 
tion at any point of a simple beam, a cantilever beam, or a beam 
overhanging the supports, sustainmg either concentrated or uni- 
form loads. If the length of the parts into which the beam is 
divided is small, this method gives results sufficiently accurate 
for practical purposes. 

151. Deflection Diagram — Variable Moment of Inertia. 

The method of determining deflections, described in § 149 and § 
150, is applicable to beams having a constant moment of inertia. 
For a simple beam, the bending moment increases towards the 
center, and for an economical design of built-up beams, it is 
necessary to increase the moment of inertia of the beam cor- 
respondingly. When the section of the beam is not constant, the 
method of determining deflections should be modified; as will 
now be shown by the solution of a practical problem. The solu- 
tion which will now be shown gives accurate results. 

It is required to draw the deflection diagram for the plate 
girder shown in Fig. 102. The span of the girder is 62 feet 4 
inches, center to center; the distance, back to back of angles, is 
6 feet 2 inches ; and the girder is loaded with a uniform load of 
3 600 pounds per linear foot. The section of the beam at the 
different points is as shown in the lower part of Fig. 102. The 
moment of inertia is increased towards the center of the girder 
by the addition of cover plates. 

Draw the force polygon (Fig. 102, a) for the given uniform 
load. In this case, one-half the total uniform load is assumed 
to be divided into five parts, and these partial loads are assumed 
to act through the center of gravity of each part. Assume any 
pole O and pole distance H, and draw the funicular polygon 



Art. H. 



DEFLECTIONS IK A PLATE GIBDfR. 



185 



(Fig. 102, b). A curve tangent to this polygon will give the 
moment diagram. Since the girder is symmetrical about its cen- 
ter line, the moment and deflection diagrams need be drawn for 
only one-half of the span. 

Uniform load of 3600 Ib5« oer lln« ft- of 




FLANee 

SeCTIOH 

WebPu-- rfrxg 
Max. DerLECTtON , ^ , 



I4»l5-i20560 
r-*- Is. 100080 
A/-^I«- 80130 
^ "^1^1 1 • 60700 



ITZaxHxHVv - 1738x240000x150x106 A ,7. 
TTT '" ?90S>000K607dd "°-^' 



Fig. 102. Deflections in a Plate Gibdeb. 



To . draw the deflection diagram, instead of dividing the 
moment area by equidistant verticals as was done in § 150, divide 
it into laminae by verticals dropped from the ends of the cover 



186 DEFLECTIONS IN BEAMS. Chap. XVII. 

plates. The moments of inertia between these verticals will then 
be constant. Compute the area of each lamina in square feet 
(using scale of beam), locate its center of gravity, and assume 
a force numerically equal to the area of the lamina to act through 
its center of gravity. Compute the moments of inertia of the 
different sections of the beam. The values of these moments of 
inertia for the given girder are shown in the lower part of Fig. 
102. Determine the ratio of the moment of inertia of each sec- 
tion to that at the end of the beam. These ratios (reading from 
the end of the beam) are equal to i.oo, 1.32, 1.65, and 1.99, 
respectively. Lay off the moment area to any convenient scale 
on the load line CD (Fig. 102, c). From D, draw a horizontal 
line DO4, and on this horizontal line, lay off the pole distance Hi, 
using the same scale as for the load line CD. This pole distance 
may be taken of any convenient length. Now the deflection at 
any point varies inversely as the moment of inertia of the section 
at that point. Therefore to draw a deflection diagram whose 
intercepts shall bear a constant ratio to the true deflections, it is 
necessary to increase the pole distances in the same ratio that the 
moments of inertia are increased. Lay off Hg (Fig. 102, c) = 

.— XH^; H3=-fxHi; and H, = H,= ^ X H,. 

Join the point 4 with the point O4. Also, since the moment 
of inertia is constant for the two center moment areas, join the 
point 3 with the same point O4. From the point 7, draw 7-O3 
vertical, to intersect 3-O4 at O3 ; and connect the points 2 and O3. 
From the point 6, draw 6-O2 vertical, to intersect 2-O3 at O2; 
and join the points i and Og. Also, from the point 5, draw 5-O1 
vertical, to intersect 1-O2 at O^; and join the points C and Oi. 
Using the lines C-O^, i-Oi, 2-O2, 3-O3, 4-O4, and D-O5 as rays, 
draw the funicular polygon (Fig. 102, d). Trace a curve tangent 
to the polygon, which will give the required deflection diagram. 
To get the maximum deflection A (at the center), multiply 

,, , . , , , 1728 X H X Hi X dx T .t,- 

the measured intercept y by J- - In this case 

dx = I ; since the area was taken instead of the intercept in the 



-^rt. S. KE8TRAINBD BEAMS. 187 

moment diagram. H^ is always to be taken as the least value for the 
moment of inertia. Therefore 

1728 X 240 000 X 150 X 10.6 

A = ^y 4z ^ = 0.37 inch. 

29 000 000 X 60 700 ^^ 

The deflection Ai at any other point along the girder, where 
the intercept is yi, may be found by proportion, i. e., if \\ repre- 
sents the intercept at the center of the beam, then 

Vi :y :: Ai :o.37, 

0.37 X y'l 
or Ai = ] — '—• This requires less work 

than a second substitution in the formula. 



Art. 3. Bending Moments, Shears, and Deflections in 

Restrained Beams. 

The following examples of restrained beams will be taken up 
in this article, viz.: (i) cantilever beam — a beam fixed at one 
end and free at. the other; (2) beam fixed at one end and sup- 
ported at the other; (3) beam fixed at both ends. 

152. Definitions. A restrained beam is a beam fastened 
at one or more points in such a manner that the beam is not free 
to deflect at these points. 

A beam is fixed at any point if its neutral surface at that point 
is horizontal. 

The bending moment and shear diagrams for a cantilever 
beam may be drawn without first finding the moment of the 
fixing couple. In other restrained beams, it is necessary to first 
find the deflections at the ends, considering the beam as sup- 
ported at both ends, and then to determine the value of the fixing 
moment that will make the neutral surface horizontal at the fixed 
points. 

Referring to the overhanging beam shown in Fig.~97, it is 
seen that the overhanging portions of the beam may be made 
of such lengths that the beam will be fixed at its supports. 

153* (i) Bending Moment, Shear, and Deflection Dia- 



188 



MOMENTS, 8HEABS, AND DEFLECTIONS. Chap. XVIL 



grams for a Cantilever Beam. It is required to draw the bend- 
ing moment, shear, and deflection diagrams for the cantilever 
beam MN (Fig. 103). 

^ alb b|c 




Moment Dioqrarn 1 n 





Sheap Diagram 
fc) 





Deflection Diagram 
(o) 



Fig. 103. Cantileteb Beam — Co^centbated Loads. 

The bending moment and shear diagrams are constructed by 
the methods explained in Art. I. The bending moment diagram 
is shown in Fig. 103, b, and the shear diagram in Fig. 103, c. 

To draw the deflection diagram, use the method explained in 
Art. 2. Divide the moment area (Fig. 103, b) into segments by 
verticals, and lay off the lengths of these intercepts on a vertical 
load line (Fig. 103, d). Take any pole O', and construct the 
funicular polygon shown in Fig. 103, e. Trace a curve tangent 
to this funicular polygon, which is the required deflection dia- 
gram. 

To get the actual deflection at any point, multiply the intercept 
between the curve and the horizontal line by the constant deter- 
mined by equation (i), or equation (2), § 150. 

154. (2) Bending Moment, Shear, and Deflection Dia- 
grams for Beam Fixed at One End and Supported at the 
Other. The diagrams will be drawn for the beam loaded (a) 



Art. S, 



CANTILEVER BEAM. 



189 



with concentrated loads, and (b) with a uniform load. The con- 
struction will first be taken up in detail, and the proof given; 
after which simplified constructions will be shown. 




[)ef lection Diagram -Fixed and Supports 

(i) 
Fig. 104. Beam Fixed and Supported — Concentrated Loads. 



(a) Beam Fixed and Supported — Concentrated Loads. It 
is required to draw the bending moment, shear, and deflection 



190 MOMENTS, SHEARS, AND DEFLECTIONS. C^P- ^^^^. 

diagrams for the beam MN (Fig. 104). The beam is fixed at 
the right end and is supported at the left, and is loaded with the 
three concentrated loads, as shown. 

Assume that the beam is supported at both ends, and draw 
its bending moment, shear, and deflection diagrams, as in § 150. 
Also draw the dividing ray O'm (Fig. 104, d) parallel to the 
closing string eh (Fig. 104, e). The force polygon is shown in 
Fig. 104, a; the bending moment diagram in Fig. 104, b; the 
shear diagram in Fig. 104, c ; the deflection force polygon in Fig. 
104, d ; and the deflection diagram in Fig. 104, e. 

Since the beam is to be fixed at the right end, assume that a 
couple whose moment is sufficient to make the neutral surface 
horizontal is applied at the right end of the beam. Let the tri- 
angle F'G'K' (Fig. 104, f) be the moment area of the couple 
required to fix the beam at this point. The moment of this fixing 
couple is equal to wH ; where H is the first pole distance, and w 
is the intercept under the fixed end of the beam, this intercept 
being the altitude of the moment area for the fixing couple. The 

value of w may be computed from the formula w = jLjl. ; where 

S = span of beam in feet, 1' = length of one division of the 
moment area in feet, and V2 = length m-7 (Fig. 104, d). It 
is seen that w may be found from the above formula as soon as 
the deflection diagram for the beam treated as supported at both 
ends has been drawn. Compute w, and construct the bending 
moment triangle F'G'K' (Fig. 104, f). Then wH is the moment 
of the fixing couple at N, and the moment at any point along the 
beam due to the fixing couple is equal to the intercept under that 
point multiplied by H. 

Construct the deflection diagram for the fixing couple shown 
in Fig. 104, h. The force polygon, containing the intercepts in 
the moment triangle (Fig. 104, f), is shown in Fig. 104, g, the 
pole distance Hj being taken equal to H'; and the deflection 
diagram is shown in Fig. 104, h. Since the moment area shown 
in Fig. 104, f is that for the fixing couple, it is seen that fg (Fig. 
104, e) must be equal to fg' (Fig. 104, h), i.e., the deflection 



^rt. S. BEAM FIXED AND SUPPORTED. 191 

caused by the fixing couple must be equal and opposite to that 
at the end of the beam supported at both ends. 

It will now be shown that the altitude G'K' (Fig. 104, f) 

3V2I' 
= w = — F-. Assume that the verticals in the triangle F'G'K' 

were drawn one foot apart. The sum of all the verticals would 

then be equal to the area of the triangle. Hence, 1-7 (Fig. 104, g) 

area of triangle Sw 2, . ,— . 

= -r,. But m-7 = V2=-(i-7) (Fig- i04, 



— V — zV ' ^ 3 

g) ; as will now be shown. From the similar triangles efg (Fig. 
104, e) and 0'm7 (Fig. 104, d), ef : O'm ::fg :m-7. Also, 
from the similar triangles e'f'g' (Fig. 104, h) and 02n7 (Fig. 
104, g), e'f : Ojn ::f'g' = fg :n-7. But ef = e'f'; therefore 
m-7 = n-7. Since the moment area from which the intercepts 
in Fig. 104, g are taken is a triangle, it is seen that m-7 = n-7 = 

2 22 Sw Sw 

- ( 1-7) . Now m-7 = V. = -( 1-7) = >. _ from which 

3 3 3 2I' 31' 

3V'.l' / X 

w= i-^-. (i) 

To construct the deflection diagram for the beam fixed at 
the right end and supported at the left, draw TU (Fig. 104, i) 
horizontal, and from this horizontal line, lay off ordinates equal 
to the differences between the ordinates in Fig. 104, e and Fig. 
104, h. Draw the polygon as shown, and trace a curve tangent 
to this polygon at the middle points of its sides, which will be the 
required deflection diagram. 

To get the actual deflections, multiply the intercepts in the 
deflection diagram by the constant given in equation (i), or 
equation (2), § 150. 

Simplified Construction. A simpler construction for the above 
will now be given. The bending moment diagram, considering 
the beam supported at both ends, is FrpGF (Fig. 104, b). To 
draw the diagram for the beam fixed at the right end and sup- 
ported at the left, from G, lay off GK = w (computed from 
equation i), and connect the points F and K, completing the 
fixing moment triangle FGK. Now the fixing moment tends to 



192 MOMENTS, SHEARS, AND DEFLECTIONS. ^^^P- ^^^L 

cause rotation in an opposite direction to that caused by the loads 
on the beam. Therefore the differences between the ordinates 
in the polygon FrpGF and the triangle GFK will give the ordi- 
nates of the bending moment diagram for the beam fixed at 
one end and supported at the other. This bending moment dia- 
gram is FrpGKF. The moment at any point along the beam is 
equal to the intercept in the diagram under the point multiplied 
by the pole distance. The point p is the point of contra-flexure, 
i. e., it is the point where the beam changes curvature and has zero 
moment. Since the change in curvature of the neutral surface is 
proportional to the bending moment, it is evident that the deflec- 
tion diagram for the beam fixed at one end and supported at the 
other may be drawn directly from its bending moment diagram. 
To draw the deflection diagram by the method suggested, lay off 
the intercepts between the line FK (Fig. 104, b) and the broken 
line FrpG on the deflection load line (Fig. 104, j), noting that 
intercepts representing positive moments are laid off in one direc- 
tion, and those representing negative moments, in the opposite 
direction. From the point 7, draw the horizontal line 7-O3, and 
with the pole distance H3 = H', construct the funicular polygon 
(Fig. 104, i). Trace a curve tangent to this polygon, which will 
give the required deflection diagram. 

The latter method requires fewer constructions, and is more 
accurate than the former. 

It is seen that the fixing moment at N will decrease the reac- 
tion at M and increase that at N by an amount z, which is the 
force represented by the distance between the shear axes PQ 
(considering the beam supported at both ends) and the shear 
axis RS (considering the beam fixed at the right end and sup- 
ported at the left). 

To find z, let M^ =the moment of the fixing couple at N. 

Then M^ =wH, and z=— ^-=— ^-, where H = the pole dis- 
tance and S = the span of the beam. Compute z, and lay it off 
upward from P. Draw the shear axis RS, which is the shear axis 
for the beam fixed at the right end and supported at the left. 



Art.S. 



BEAM FIXED AND SUPPORTED. 



193 



(b) Beam Fixed and Supported — Uniform Load. It is re- 
quired to draw the bending moment, shear, and deflection dia- 
grams for the beam MN (Fig. 105), fixed at the right end and 
supported at the left, and loaded with a uniform load. 




Moment Dioc^roms 
(b) 



i^ P 



A _ 




Shear Diagram 
(c) 



Deflection Dioqram- Simple Beami 

le) , 





Deflection Diagram -Rxed and Supported 

(g) 

Fig. 105. Beam Fixed and Supported — Uniform Load. 

Divide the uniform load into segments, and assume the weight 
of each segment as a force acting through its center of gravity. 
Apply the method explained in the preceding article, and con- 
struct the bending moment, shear, and deflection diagrams. Trace 
curves (not shown in the diagram) tangent to the moment and 
deflection diagrams, and measure intercepts between the closing 
lines and these curves. The bending moment diagram is shown 
in Fig. 105, b; the shear diagram in Fig. 105, c; the deflection 
diagram, considering the beam as supported at both ends, in Fig. 
105, e; and the deflection diagram, for the beam fixed at the right 



194 



BEXDIXO MOMENTS. SHEARS, DEFLECTIONS. Chap. XVII, 



end and supported at the left, in Fig. 105, g. The intercepts laid 
bff in Fig. 105, f are the distances between the line FK and the 



zrz, 




Deflecl-ion Diagram -Both Ends F\xed 

Fig. 106. Beam Fixed at Both Ends — Concbntbatisd Loads. 



curve (not shown), drawn tangent to the funicular polygon. The 
point of contra-flexure is at p (Fig. 105, b). 

To determine the actual deflections, multiply the intercepts in 



Art.S. BR.VM FIXED AT BOTH ENDS. 195 

Fig. 105, S ^y ^^^ constant shown in equation (i), or equation 

(2), § 150. 

155* (3) Bending Moment, Shear, and Deflection Dia- 
grams for a Beam Fixed at Both Ends. It is required to draw 
the bending moment, shear, and deflection diagrams for the beam 
MN (Fig. 106), fixed at both ends, and loaded with concentrated 
loads, as shown. 

Construct the bending moment, shear, and deflection dia- 
grams, considering the beam as supported at both ends, and 
draw the ray O'm (Fig. io6, d), dividing the deflection load line 
at m into Vj and Vj. 

Since the beam is fixed at both ends, assume a couple to act 
at each end whose moment is suflicient to make the neutral sur- 
face horizontal at these points. The moment M^, which fixes the 
right end of the beam, is represented by the moment area triangle 
FGK (Fig. 106, b) ; and the moment Mg, which fixes the left 
end, is represented by the moment area triangle GFL (see pre- 
ceding section). Draw the force polygon (Fig. 106, f), for the 
moment triangle FGK, and construct its deflection diagram (Fig. 
106, g). Draw the dividing ray Ogn parallel to the closing 
string, which divides the load line into Vg and V4. Also draw 
the force polygon (Fig. 106, h) for the moment triangle GFL, 
and construct its deflection diagram (Fig. 106, i). Draw the 
dividing ray Ogn, which divides the load line into Vg and v^. 

The angles lO'm and mO'7 (Fig. 106, d) and their corre- 
sponding angles in the deflection diagram represent the deflec- 
tions at the ends of the beam, considered as supported at both 
ends. Likewise, the angles lOgn and nOg 7 (Fig. 106, f) and 
their corresponding angles in the deflection diagram represent the 
deflections at the ends of the beam for the right fixing couple; 
and lOgn and n037 ^^^ their corresponding angles in the 
deflection diagram represent the deflections for the left fixing 
couple. Since the pole distances H', Hg and H3 are all equal, it 
is seen that Vj, Vg, V3, V4, V5, and v^ are respectively proportional 
to these angles. To fix the beam at the ends, 1. e., to make the 
neutral surface horizontal at the ends, v^ must equal Vj -f- Vg, and 
V2 must equal v^ + v^. 



196 BEXDINO MOMENTS, SHEARS, DEFLECTIONS. Chap. XVIL 

The values of w^ and Wj (Fig. io6, b) may be found, as 
follows : As has been shown in § 154, the length of the deflection 
load line 1-7 (Fig. 106, f) for the moment triangle FGK = 

i ; and this line 1-7 is divided by the ray Ojn into one-third 



2V 
and two-thirds its length. Therefore v,, which equals i (1-7),= 

_i: and v., which equals J (1-7),= — ^. In like manner, 

considering the deflection load line 1-7 (Fig. 106, h) for the 
moment triangle GFL, it may be shown that 

Sw, J Sw. (j\ 

T,= -^,andVe= ^- CU 

Now, v, = V3 + v,=|^^ + |^, (2) 

and v, = v, + ve=|p+^- (3) 

Solving equations (2) and (3) for w^ and Wj, we have 

Wi= -g-(2rvi — vj, (4) 

and W2 = -^(2V2 — vj. (5) 

To get the bending moment at any point for the beam fixed 
at both ends, compute Wi and Wj from the above formulae, and 
lay them off downward from F and G (Fig. 106, b), respectively. 
Join the points L and K. Then the moment area FGKL repre- 
sents the bending moment of the fixing couples at both ends of 
the beam. Since the polygon Fprp'GF represents the bending 
moment, considering the beam supported at both ends, it is seen 
that the bending moment for the beam fixed at both ends is repre- 
sented by the moment area Fprp'GKLF. The moment at any 
point is equal to the intercept in this diagram multiplied by the 
pole distance H. 

The points p and p' are points of contra-flexure. 



Arts, BEAM FIXED AT BOTH EXDS. 197 

To draw the deflection diagram (Fig. io6, k) for the beam 
fixed at both ends, construct the deflection load line (Fig. io6, j) 
by laying off on this load line distances equal to the intercepts 
between the line LK (Fig. io6, b) and the broken line Fprp'G. 
The intercepts 6-i and 1-2 are laid off in an upward direction, 
while 2-3, 3-4, 4-5, and 5-6 are laid off in a downward direction. 
From the point i, with the pole distance H^ = H', draw the hori- 
zontal line 1-O4. Construct the funicular polygon (Fig. 106, 
k), and draw a curve tangent to the polygon at the middle 
points of its sides, which will give the required deflection dia- 
gram. It is seen that the neutral surface is horizontal at the ends 
of the beam. To obtain the actual deflections, multiply the inter- 
cepts in Fig. 106, k by the constant shown in equation (i), or 
equation (2), § 150. 

The deflection diagram might also have been drawn by sub- 
tracting from the intercepts in Fig. 106, e the sum of the corre- 
sponding intercepts in Fig. 106, g and Fig. 106, i; and laying 
off their differences from the horizontal line TU. 

The former method is preferable, and is subject to less error 
than the latter. When the former described method is used, the 
diagrams shown in Fig. 106, f. Fig. 106, g, Fig. 106, h, and Fig. 
106, i need not be drawn except to prove the constructions. 

To construct the shear diagram, it is necessary to find the 
effect upon the reactions due to the fixing moment at each end. 

Let Z.2 = decrease in the reaction at M due to the fixing 
moment at N, 

and Zi = decrease in the reaction at N due to the fixing moment 
at M. 

Then Z2 — Zi = distance (to scale of forces) between shear axes 
PQandRS (Fig. 106, c). 

No^ 2^ = .^, and z, = ^ (see § 154 for proof). 

H 
Therefore, Zg — Zi=-^(w. — wj. (6) 

If Zo is greater than z^ (as in this problem), Zo — z^ is to be 



198 



HKXDIXG MOMENTS, SJIEARS, DEFLECTION'S. Chap. XV 11. 



measured upward from PQ (the shear axis for the beam sup- 
ported at both ends) ; and if z^ is greater than Zj, then z^ — z^ i« 
to be measured downward from PQ. 

Draw the shear diagram and shear axis PQ, considering the 
beam as supported at both ends, lay off Z2 — z^ upward from PQ, 
and draw the shear axis RS, which is the shear axis for the beam 
fixed at both ends. 



Beam a Lhading 



ip 



f 



W-wL 



f 



k. 



W-wL 



iP 



W-wL 



f 
I 



t 



ip 



W-wL 



1 



Max. Moment 



-PL 
-WL 

2 



. WL , .WL 



Max- Deflection 



PL? 
3 El 

Wl? 
6 El 

4dE:i 

5WL^ 
384 LI 

0.0093 P\} 
El 

a0054 WL^ 

tr 

PL* 



192 El 

WL* 
384 CI 



Fig. 107. Fobmulae for Maximum Moments and Deflections. 

156. Algebraic Formulae. In Fig. 107 are given several 
algebraic formulae for determining the maximum bending mo- 
ments and maximum deflections in beams for some of the simpler 
forms of loading. 



CHAPTER XVIII. 

MAXIMUM BENDING MOMENTS AND SHEAItS IN BEAMS FOR 

MOVING LOADS. 



This chapter will treat of the determination of the maximum 
bending moment and maximum shear at any point in a simple 
beam loaded with moving loads ; and of the position of the mov- 
ing loads for a maximum bending moment or maximum shear. 




Maximum Moment Diagram 

Fig. 108. Simplb Beam — Uniform Load. 

157. Beam Loaded with a Uniform Load, (a) Maximum 
Bending Moment. It is required to find the maximum bending 
moment at any point O (Fig. 108) of a beam loaded with a 
uniform load. First assume the beam to be unloaded, and then 
assume a uniform load to move onto the beam. The bending 
moment at every point along the beam is increased with each addi- 
tion of the uniform load until it reaches its maximum value at 
every point when the beam is fully loaded. It has been shown in 
§ 147, b that the bending moment diagram for a beam fully 
loaded with a uniform load is a parabola with a maximum ordi- 
nate at the center of the beam equal to ipL^. The maximum 
bending moment diagram for the beam MN is shown in Fig. 108. 

199 



2>: 



x->iZJL*iL J^y^cs^ xv 



S AXD SHEARS. Chap. XT III. 



. :ti : 



sz srj pccrt O {Fig. loS), at a distance 






2 • 



(I) 



. rt* i>--^* *:'is.rj re r 



M 



It* 



=f^T-'=»=f(T^')(-^-')- 



(2) 



..J 






:rnfcv><-i — mv*TCs is: The bending moment at 
: 31 i ztiirz z=:i>:r=h- koded is equal to one-half the load 
rr^— Ijfi !:> ibe rrcoact of the two s^ments into which 
:: If iii-.i*"! "bv tbf ^rnren pocnt. 

-r^x--n:urr. Ser^iiris: rnccnent at any point along a beam 
!y !.:,iif-i riiy r* reacilY found from equation (2). 
.VjLrrwciT 5i<-j'^. It is required to find the maximum 
ir.y r*::r.: O >: ±e beans MX 1 Fig. 109), loaded with a 






>:<i:L 




i._a 



Fig. 109. Siscplb Bulv — Uxifouc Loaix 

It has been shown in § 147, b that the equation express- 
ing the shear in a beam fully loaded with a uniform load is 

>> = p I— X 1 where p is the uniform load per foot, L the 

length of the beam, and x the distance from the left end of the 



SIMPLE BEAM — UNIFORM LOAD. 201 

beam to the point where the shear is to be determined. It is seen 
from this equation that when the beam is fully loaded the shear 

has its maximum positive value -\- when x = o; that it has 



its maximum negative value — — when x — L ; and is zero 
when x = 

2 

Now it is seen that the load to the left of any point O (Fig. 
109) decreases the positive shear at that point by the amount of 
the load, while it increases the left reaction by a less amount. 
Since the shear is equal to the reaction minus the load to the left, 
it will have its maximum value at any point when there is no load 
to the left of the point. Therefore, to get a maximum positive 
shear at any point in a beam due to a uniform load, the segment 
to the left of the point should be unloaded and that to the right 
fully loaded. 

The equation for a maximum positive shear is 

S = R,= PiL=iO(Lzi£L»= P (L-x)». (3) 

2L 2L 

This is the equation of the parabola CB, which has its vertex 
at the right end of the beam. The maximum ordinate is at x = o, 

T 

the left end of the beam, and is equal to J_ . 

2 

To get the maximum negative shear at any point, the portion 
to the left of the point should be fully loaded and that to the 
right unloaded. The equation expressing the maximum negative 
shear is 

S=-R,= -^\ (4) 

^ 2L 

which is the equation of the parabola AD. 

158. Beam Loaded with a Single Concentrated Load, (a) 

Maximum Bending Moment. Let the beam MN (Fig. no) be 



202 



MAXIMUM BENDING MOMENTS AND SHEARS. Chap, XV III, 



loaded with a single moving load P. The maximum bending 
moment at any point O occurs when the load is at that point ; for 
a movement of the load to either side of the point decreases the 




(a) Maximum Moment Diagram 




{b Maximum 5hear Diagram 
FiQ. 110. SxMPLE Beam — Concxntbated Moyino Load. 

opposite reaction and hence the bending moment. The equation 
expressing the maximum bending moment at any point due to a 
single moving load is 



M = 



Hk*' ) e-») 



L 

2 



(s) 



This is the equation of the parabola (Fig. no, a), which has 

PL 
its maximum ordinate equal to when x = o. 

4 
If ^is substituted for P in equation (l), it is seen that we 

have equation (s). Therefore, the bending moment due to a sin- 
gle concentrated load is the same as for twice that load uniformly 
distributed over the beam. 

(b) Maximum Shear, To get the maximum positive shear 



SIMPLE BEAM-CONCENTRATED MOVING LOADS. 203 

at any point O (Fig. no) due to a single moving load, the load 
should be placed an infinitesimal distance to the right of the 
point. (For all practical purposes it may be considered at the 
point.) The maximum shear may be expressed by the equation 

'^^ L • 

which is the equation of the straight line CB. The positive shear 

is a maximum when x = — , i. e., at the left end of the beam. 

2 

The maximum negative shear occurs when the load is just to 
the left of the point, and may be expressed by the equation 



-^(^-0 



<;_ . .__ (7) 

which is the equation of the straight line AD. 

159. Beam Loaded with Concentrated Moving Loads, (a) 

Position for Maximum Moment. Let MN (Fig. in) be a beam 
loaded with concentrated moving loads at fixed distances apart. 
(For simplicity, three loads are here taken, although any number 



R 



C^ce-^CO X 



R. ^ ' ^q.--4.A-j 



L-x 



R 



1 



Fio. 111. Position for Maximum Moment — Moving Loads. 

might have been considered.) It is required to find the posi- 
tion of the loads for a maximum bending moment in the beam, 
together with the value of this moment. 

The determination of the position of any number of moving 
loads for a maximum bending moment at any point in a beam will 
be treated in § 208. 

Let X be the distance of one of the loads Po from the left end 



204 MAXIMUM BENDING MOMENT? AND SHEARS. Chap. XV III. 

of the beam when the loads are so placed that they produce a 
maximum moment under Pg. Now if the bending moment is 
found, and its first derivative placed equal to zero, the position of 
the loads for a maximum bending moment may be determined. 
To determine the bending moment, first find the reaction R^ by 
taking moments about the right end of the beam. Thus, 

_ P, (L-X + a) + P, (L-x) + P3 (L-x — b) 
R,_ j^ 



(P, + F2 + P3) (L-x) + Pia— Pab 
"" L 

and the bending moment under P2 is 

M = RiX — P^a 

X (Pi + P2 + P3) (L-x) + X (P,a- Psb) 



(8) 



P,a. (9) 



DiflFerentiating equation (9) and placing it equal to zero, we have 

dM (P, + P, + PJ (L - 2x) + P.a- P3b 
-dx= L =°- ('°) 

Solving equation (lo) for x, we have 

__L P.a — Psb 

^- 2+2(P, + P, + P3)- ^"^ 

The position of the wheel Pg for a maximum bending moment 
in the beam is determined by equation (11). For, P^a — Pgb = the 

P^a — Pgb 
moment of the loads on the beam about Pg, and -^ — ^-^5 — t—^ = 

^1 + ^2 + ^3 

distance from P2 to the center of gravity of all the loads ; hence 

L I 
x = — h — (distance from Pg to c. g. of all the loads) . 

Therefore, the criterion for a maximum bending moment under 
the load P2 is : The load Pg w«^f be as far from one end of the 
beam as the center of gravity of all the loads is from the other end. 



SIMPLE BEAM-CONCENTRATED MOVING LOADS. 



205 



Since Pj is any one of the moving loads, it is seen that theo- 
retically this criterion must be applied to, and the bending mo- 
ment found for, each of the loads ; and the greatest value taken 
as the maximum bending moment. However, it may often be 
determined by inspection which load will give a maximum mo- 
ment. If some of the loads are heavier than others, the maximum 
moment will occur under one of the heavier loads. 

If X had been measured from the center of the beam instead 
of from the ends, the following equivalent criterion for the maxi- 
mum bending moment would have been found : For a maximum 
bending moment under the load Pg, this load must be as far to 
one side of the center of the beam as the center of gravity of all 
the loads is to the other side of the center. 

The methods employed for determining the bending moment 
for fixed loads may be applied to moving loads as soon as their 
position for a maximum bending moment has been found. 




Fig. 112. Position tor Maximum Moment — Two Equal Loads. 

A Special case of the above, which often occurs, is that of a 
beam loaded with two equal loads at a fixed distance a apart 
(Fig. 112). Applying the criterion for a maximum moment to 

L a 

this case, it is seen that x (measured from the end) =— 4. — , 

2 4 

or if measured from the center = — . 

4 

Placing one of the loads P at a distance —from the center of the 

beam, and taking moments about the left end (noting that the 
loads are equal), we have 



206 MAXIMUM BENDING MOMENTS AND SHEARS. Chap, XVIII, 



P(^-f) 



R2= j- 

and the maximum moment M is 



' (t-f ) (t-t) 

2 4 



"-R'(t-t) = 



C-D' 



= p - 2 / . (I2) 

2L 

This equation is true for the ordinary values of a and L, but 
does not give a maximum bending moment when a > 0.586L. 
When a > 0.586L, one of the wheels placed at the center of the 
beam (the other being then off of the beam) will give a maximum 
bending moment, as will now be shown. 

Assume that one of the two equal loads P is placed at the 

center of the beam, and that the distance a between the loads is 

greater than 0.5 L. The maximum bending moment is then equal 

PL 
to Equating this to the value found for the maximum 

4 

bending moment in equation (12), we have 

4 2L 

Solving for a, we have 

a^o.586L. (13) 

Therefore, when a = 0.586L, the moment due to a single load 
P at the center of the beam is the same as the moment given by 
placing the loads according to the criterion for maximum mo- 
ments. When a > 0.586L, the maximum bending moment for 
two equal loads at a fixed distance apart is given by placing one 
of the loads at the center of the beam ; and the criterion for a 
maximum bending moment does not apply. 

If the two loads are unequal, the maximum moment will 
always occur under the heavier load. 



SIMPLE BEAM-COXCENTRATED IMOVIXO LOADS. 207 

(b) Position for Maximum Shear. For two equal loads, 
the maximum end shear will occur when both loads are on the 
span and when one of the loads is at an infinitesimal distance 
from the end of the beam ; since the reaction will be a maximum 
for this condition. For a maximum shear at any point along a 
beam due to two equal loads, one of the loads must be at the point. 

For two unequal loads, the maximum end shear will occur 
when both loads are on the span and when the heavier load is 
at an infinitesimal distance from the end. For a maximum shear 
at any point along the beam due to two unequal loads, the heavier 
load must be at the point. 

For a maximum shear at any point along a beam due to any 
number of loads, one of the loads must be at the point. The 
criterion for determining which load placed at the point will give 
a maximum shear will now be determined. 

Let MN (Fig. 113) be a beam loaded with any number (in this 
case four) concentrated loads, and let O be any point along the 
beam whose distance from the left end is x. Also let 2P = the 
sum of all the loads on the beam when there is a maximum shear. 
It is required to determine which load placed at O will give a 
maximum shear at that point. 



M 






L 

Fio. 113. Position for Maximum Shear — Moving Loads. 

When Pi is placed at O, the shear Si at O is equal to the left 
reaction Ri, i. e., 

P, (L — x) + P,(L — X — a) 
Sx = Rx = L 

P, [L — X— (a + b)] + P, [L — x--(a + b + c)] 
+ L 

= 5P(L — x)— P,a — PaCa + b)— P,(a + b + c). 



208 MAXIMUM SHEAK. Clmp. XVIII. 

When Pj is at O, the shear S^ at O is 

P, (L-x + a)+P,(L-x) 
bj = Ki — r*! = J- 

P,(L-x-b)+P;[L-x-(b + c)] 
+ j- P, 

_ SP(L — x)+Pia — P«b — P, (b + c)— P^L 
_ _ . 

Subtracting S, from S,, we get the difference in the shear for the 
two cases, or 

P,L — 5Pa 

Si — S, = J- . (14) 

From the above equation, it is seen that Si will be greater than 

S, if P,L > SPa, i. e., if 

PiL 

-^>5P. (15) 

The above equation expressed in words is : The maximum pos- 
itive shear at any point along a beam occurs when the foremost 

load ts at the point if is greater than SP. // is less than 

2Pi, the greatest shear will occur when some succeeding load 
(usually the second) is at the point. 

The methods employed to determine the shear due to fixed 
loads may be applied to moving loads as soon as their position 
for a maximum shear has been determined. 



PART IV. 



BRIDGES. 



CHAPTER XIX. 

• TYPES OF BRIDGE TRUSSES. 

Bridge trusses are comparatively recent structures, the ancient 
bridges being pile trestles or arches. Somewhat later, a combina- 
tion of arch and truss was used, although the principles govern- 
ing the design were not understood. It was not until 1847 that 
the stresses in bridge trusses were fully analyzed, although trusses 
were constructed according to the judgment of the builder before 
this date. In 1847, Squire Whipple issued a book upon bridge 
building, and he was the first to correctly analyze the stresses 
in a truss. Soon afterward, the solution of stresses became very 
generally understood, wooden trusses were discarded for iron 
ones, and still later, steel replaced iron as a bridge-truss mate- 
rial. From this time, the development of bridge building was 
very rapid, culminating in its present high state of efficiency. 

160. Through and Deck Bridges. Bridges may be grouped 
into two general classes, viz. : through bridges and deck bridges. 

A through bridge is one in which the floor is supported at, or 
near, the plane of the lower chords of the trusses (see Fig. 
114, e). The traffic moves through the space between the two 
trusses. Except in the case of a pony truss (one in which there 
is no overhead bracing), a system of overhead lateral bracing is 
used. 

209 



210 



TYPES OF BRIDGE TRUSSES. 



Ctiap, AIX. 



A deck bridge is one in which the floor is supported directly 
upon the upper chords of the trusses. In this type, the trusses 
are below the floor (see Fig. 114, d). 

161. Types of Bridge Trusses. In Fig. 114 are shown 
several types of bridge trusses that have been very generally 
used. 



AAA/ W\ A 




(a) Warren 



(b) Howe 



(c) Pratt 



^^E^ap^ ^aiffii^ 



(d) Baltimore ( Deck) 



(e) Baltimore (Thru) 



M 



r^N 




NN^ 



(f) Whipple 




(g) Camels BacU 




(h) Parabolic Bowstring 




(i) Parabolic Bowstring 




ij) Petit 

Fig. 114. Ttfes of Bhidob Trusses. 

Fig. 114, a shows a Warren truss. This truss is still used foi 
short spans, but has the disadvantage that the intermediate web 
members are subjected to reversals of stress. 

Fig. 114, b shows a Howe truss. This truss was in favor 



TYPES OF BRIDGE TRISSES. 211 

when wood was extensively used as a building material for 
trusses, but is little used at present. It has the disadvantage of 
having long compression web members. 

Fig. 114, c shows a Pratt truss. This form of truss is exten- 
sively used, both for highway and railroad bridges, up to about a 
200-foot span. It is economical and permits of good details. 

Fig. 114, d shows a Baltimore deck truss, and Fig. 114, e, a 
Baltimore through truss. These trusses are used for compara- 
tively long spans, and have short compression members. 

Fig. 1 14, f shows a Whipple truss. This truss was quite exten- 
sively used, but is now seldom employed. It is a double intersec- 
tion truss, and has a redundancy of web members. The stresses 
are indeterminate by ordinary graphic methods. 

Fig. 114, g shows a Camels-Back truss. This truss is used 
both for short and long spans. 

Fig. 114, h and Fig. 114, i show Parabolic Bowstring trusses. 
The upper chord panel points are on the arc of a parabola. A 
great disadvantage of these types is that the upper chord changes 
direction at each panel point and that the web members change 
both their angle of inclination and length at the panel points. 

This type is sometimes modified by placing the panel points 
on the arc of a circle. 

Fig. 114, j shows a Petit truss. This truss is quite exten- 
sively used for long spans, and is economical. 

162. Members of a Truss. The general arrangement of 
members is given in the through Pratt railroad truss shown in 
Fig. 115. The arrangement of members in the various types of 
trusses is somewhat similar to that shown. In this truss, the ten- 
sion members are shown by light lines, and the compression mem- 
bers by heavy lines. 

Main Trusses, Each truss consists of a top chord, a bottom 
chord, an end post, and web members. The web members may 
be further subdivided into hip-verticals, intermediate posts, and 
diagonals. The diagonals may be divided into main members and 
counters, the main members being those stressed under a dead 
load, and the counters those stressed only under a live load. In 
Fig. 115, LqUi is an end post; U1U2, a panel length of the upper 



212 



TYPES OF BRIDGE TBUS3E3. 



Chap. XIX. 



chord; LoL„ a panel length of the lower chord; U,L„ a hip 
vertical ; UjLj and U^Lj, main diagonals ; and LjU^, a counter. 
Lateral Bracing. The bracing in the plane of the upper 

■o^^^l/Jop Loreral Stait 
X^p Lateral Ties 




Stringers- 
Ftedestal 



Mom Ties 
Intermediate Rast 
rioorbeoms 
Boltom Lateral Ties 
Through Pratt Truss 

T&uss Membbbs. 



chord (Fig. 115) is called the top lateral bracing; and that in 
the plane of the lower chord, the bottom lateral bracing. The 
members of the lateral systems are stressed by wind loads and by 
the vibrations due to live loads. The top lateral system is com- 
posed of top lateral Struts and ties. The floorbeams act as the 
struts in the lower lateral system. 

Portals. In through bridges, the trusses are held in position 
and the bridge made rigid by a system of bracing in the planes 
of the end posts. This system of bracing is called the portal 
. bracing, or portal. 



MEMBERS OF A TRI'SS. 213 

Knee-braces and Sway Bracing, The braces connecting the 
top lateral struts and intermediate posts (see Fig. 115), in the 
plane of the intermediate posts, are called kncc-braces. When 
greater rigidity is required, a system of bracing somewhat similar 
to the portal bracing is used instead of the knee-braces. This 
bracing is called the sway bracing. The top lateral strut is also 
the top strut of the sway bracing. Knee-braces and sway bracing 
are often omitted on small span highway bridges. 

Floor System, The floor systems of ordinary highway bridges 
differ considerably from those of railroad bridges. Both types, 
however, have cross-beams running from one hip vertical or inter- 
mediate post to the opposite one. These beams are called floor- 
beams. The beams at the ends of the bridge are called the end 
floorbeams, and those at the intermediate posts, intermediate floor- 
beams. The end floorbeams are usually omitted in highway 
bridges, and an end strut, or joist raiser, is substituted. 

In railroad bridges, there are beams which run parallel to the 
chords and are connected at their ends to the floorbeams. These 
beams are called stringers. 

In highway bridges, there are several lines of beams which 
run parallel to the chords and which rest upon the floorbeams. 
These beams are called joists. 

In railroad bridges, the ties, which support the rails, rest 
directly upon the stringers; and in highway bridges, the floor 
surface is supported directly by the joists. 

Pedestals. The supports for the ends of the trusses are called 
pedestals. For spans over about 70 feet, the pedestals at one end 
of the bridge are provided with rollers, to allow for expansion 
and contraction. 

Connections. The members of the truss may be either riveted 
together or connected by pins. In the former case, the truss is 
said to be a riveted truss, and in the latter, a pin-connected truss. 
Riveted trusses are often used for short spans and are very rigid. 
Pin-connected trusses are easy to erect, and are used for both 
short and long spans. 



CHAPTER XX. 



LOADS. 



The loads for which a bridge must be designed may be classi- 
fied, as follows: dead load, live load, and wind load; and these 
loads will be discussed in the following three articles. 



Art. I. Dead Load. 

The dead load is the weight of the entire bridge, and includes 
the weights of the trusses, bracing, floor, etc. It is, of course, 
necessary to determine the weight of the trusses before the dead 
load stresses in them may be determined; therefore an assump- 
tion must be made as to the dead load. If the weights of similar 
bridges are available, then these weights may be used to determine 
the dead load ; but it is customary to use formulae for finding the 
approximate dead load. It should be borne in mind that the dead 
load for a single-track bridge is carried by two trusses, each sup- 
porting one-half the total load. 

163. Weights of Highway Bridges. The total dead load 
per linear foot of span for a highway bridge not carrying inter- 
urban cars may be very closely approximated by the formula* 

w= 140 -f i2b + 0-2l>L — 0.4 L, (i) 

where w = weight of bridge in pounds per linear foot, 

b^ width of bridge in feet (including sidewalks, if any), 
L = span of bridge in feet. 



*Merriman and Jacoby's "Roofs and Bridges," Part II. 

2U 



Art,l. DEAD LOAD. 215 

The weight of a heavy interurban riveted bridge may be 
closely approximated by the formulaf 

w = 6oo+ i.8L + 27bH bL(iH L), (2) 

I '12 1000 

where, 

w = weight of bridge in pounds per linear foot, 
b = width of roadway (including sidewalks), 
L == span of bridge in feet. 

164. Weights of Railroad Bridges. The weights of rail- 
road bridge trusses per linear foot of span are given very closely 
by the following formulae :$ 

ForEsc w= (650 + 7L), (3) 

E40, w = J(65o + 7L), (4) 

E30, w = H65o + 7L). (5) 



(t 



tt 



The above formulae do not include the weights of the ties and 
rails, which may be assumed at 400 pounds per linear foot of 
track. If solid steel floors are used, the weight of the track 
should be taken at 700 pounds per linear foot. 

E50, E40, and E30 refer to the live load for which the 
bridge is designed ; as given by Theodore Cooper in his "General 
Specifications for Steel Railroad Bridges and Viaducts.'' This 
live loading will be explained in § 167 (b). 

These formulae give the weights of single-track spans. Double- 
track spans are about 95 per cent heavier. 

Art. 2. Live Load. 

The live load consists of the traffic moving across the bridge. 
For highway bridges, the live load consists of vehicles, foot pas- 
sengers, and interurban cars ; and for railroad bridges it consists 



tE. S. Shaw. 

JF. E. Turneaure. 



216 LOADS. Chap, XX, 

of trains. The standard highway bridge specifications of Theo- 
dore Cooper and J. A. L. Waddell give the live loads to be used 
for different classes of highway bridges, and are recommended by 
the writer for general use. 

165. Live Load for Light Highway Bridges. The live 
load for highway bridges is usually specified in pounds per square 
foot of floor surface. The weights given in the following table 
are recommended for use when the bridge does not carry inter- 
urban traflfic. 



Live Loads for Highway Bridge Trusses. 



Spans up to 100 feet 100 lbs. per sq. ft. 



125 to 150 " 90 " " " " 



100 to 125 95 

125 to 150 *' 90 

150 to 200 " 85 

" over 200 " 80 " " " " 

In some states, the law requires that highway bridges be 
designed for a live load of 100 pounds per square foot of floor 
surface, but the law as usually stated is defective in that the 
allowable unit stresses are not given. 

The floor systems of light highway bridges should be designed 
to carry a live load of 100 pounds per square foot of floor space, 
and to be of sufficient strength to carry a heavy traction engine. 
The total load is obtained by multiplying the weight per square 
foot by the clear width of the roadway and sidewalks; and this 
load is to be so placed as to give the maximum stress in each truss 
member. 

166. Live Load for Interurban Bridges. For the live load 
on interurban bridges, the student is referred to the highway 
bridge specifications of Theodore Cooper and J. A. L. Waddell. 

167. Live Load for Railroad Bridges. The live load for 
railroad bridges varies on account of the great variations in spans 
and wheel spacings of engines. The bridges for main tracks are 
usually designed for the heaviest engines now in usef, or that may 
reasonably be expected to be built in the near future. 



Srt.ff. LIVE LOAD. 217 

The live load may be treated either (a) as a uniform load, (b) 
as a number of concentrated wheel loads followed by a uniform 
train load, or (c) as an equivalent uniform load. The second 
loading is preferable and gives more accurate results, but requires 
more work to determine the stresses. 

(a) Uniform Load, When train loads were light, it was 
customary to design the bridges for a uniform load instead of 
considering actual wheel concentrations; and a uniform load is 
still often used in practice. The same load is generally taken for 
both the chord and the web numbers. This method is simpler 
than that involving wheel loads, and gives fair results if intelli- 
gently used. Care should be exercised, however, in determining 
the load to be used for each truss. 

(b) Concentrated Wheel Loads. The method of using con- 
centrated wheel loads is complicated by the great variation in the 
weights and spacings of engine wheel loads. Most railroad com- 
panies specify that the stresses shall be computed for two engines 
and tenders followed by a uniform train load. 

The present practice among many railroad companies is to use 
a conventional wheel loading, one of the best examples of which 
being that given by Theodore Cooper in his "Specifications for 
Steel Railroad Bridges and Viaducts*'. The three common classes 

o oooo oooo o oooo ^ooo 

o oooo oooo o oo^o oooo 

o oooo mminio o oooo loinmin c^^,. ,. 

in oooo ojfoMCvJ in oooo (vjcj(vi(^ 5000 lbs. 

CJ lototnm roiororo OJ mmiOiO roroioro ,. ^ 

Live Load per Track for Cooper's Er 50 Loading 

Pio. 116. Cooper's E50 Loading. 

recommended by Theodore Cooper are his E50, E40, and E30. 
Fig. 116 shows the wheel loads and spacings for Cooper's E50 
loading, which corresponds to the heaviest engines in common 
use, although a live load corresponding to an E60 loading has 
been used. 

An E40 loading has the same wheel spacings, the weight of 



218 



LOADS. 



Chap, XX, 



each wheel being four-fifths of that for the E50 loading, followed 
by a uniform train load of 4000 pounds per linear foot of track. 

An E30 loading has the same wheel spacings, the weight of 
each wheel being three-fifths of that for the E50, followed by a 
uniform train load of 3000 pounds per linear foot of track. 





Equivalent Uniform Loads - 


- Cooper's. E-40. 


Span 

(Ft.) 


Equivalent Uniform Load 


Span 

(Ft.) 


Equivalent Uniform Load 


Chords 


Webs 


ri-Bms. 


Chords 


Webs 


FIBms. 


10 


9000 


12000 


8200 


46 


6330 


7240 


5240 


II 


9310 


1)640 


7960 


48 


6220 


7140 


5200 


12 


9340 


11330 


7830 


50 


6110 


7060 


5140 


13 


9340 


11080 


7600 


52 


6040 


6940 


5130 


14 


9210 


10860 


7460 


54 


5960 


6820 


5120 


15 


9030 


10670 


7330 


56 


5880 


6720 


5110 


16 


8850 


10500 


7120 


58 


5800 


6620 


5090 


17 


8650 


10350 


6940 


60 


5730 


6530 


5080 


18 


8430 


10240 


6780 


62 


5690 


6490 


5080 


19 


8220 


lOlOO 


6630 


64 


5700 


6450 


5070 


20 


8000 


10000 


6500 


66 


5620 


6450 


5070 


21 


8040 


9780 


6390 


68 


5560 


6380 


5060 


22 


8040 


9580 


6290 


70 


5510 


6340 


5060 


23 


8010 


9400 


6200 


72 


5490 


6320 


5030 


24 


7960 


9230 


6120 


74 


5460 


6300 


5010 


25 


7890 


9080 


6040 


76 


5440 


6290 


4990 


26 


7780 


8930 


5970 


78 


5420 


6270 


4970 


27 


7660 


8790 


5900 


80 


5400 


6250 


4950 


28 


7540 


8660 


5830 


82 


5370 


6230 


4930 


29 


7420 


8540 


5770 


84 


5340 


6200 


4310 


30 


7300 


8430 


5720 


86 


5310 


6180 


4890 


31 


7220 


8320 


5680 


86 


5270 


6150 


4870 


32 


7140 


8190 


5650 


90 


5250 


6130 


4860 


33 


7050 


8080 


5620 


92 


5250 


6110 


4850 


34 


6960 


7980 


5600 


94 


5210 


6090 


4810 


35 


6870 


7890 


5570 


96 


5170 


6060 


4780 


36 


6820 


7820 


5530 


98 


5150 


6040 


4760 


37 


6760 


7750 


5500 


100 . 


5140 


6020 


4740 


38 


6700 


7690 


5460 


125 


5100 


5770 


4720 


39 


6630 


7630 


5430 


150 


5010 


5570 


4700 


40 


6560 


7570 


5400 


175 


4890 


5350 


4680 


42 


6530 


7450 


5340 


200 


4740 


5240 


4660 


44 


6470 


7340 


5300 


250 


4510 


5030 


4640 



Fia. 117. Equivalent Uniform Loads. 



The great advantage of Cooper's loadings is that the spacing 
of wheels is the same for all loadings. It is thus seen that the 



Art.S. WIND LOAD. 219 

moments and shears for the different loadings are proportional 
to the class of loading. 

The actual determination of moments, shears, and stresses 
due to concentrated wheel loads will be given in Chapter XXIII. 

(c) Equivalent Uniform Load. An equivalent uniform load 
is one which will give results closely approximating those for 
actual wheel concentrations. To secure accuracy, it is necessary 
to use a different uniform load for each span ; also a different load 
for the chords, the webs, and the floorbeams. The table in Fig. 
117 gives the equivalent uniform loads for Cooper's E40 loading. 
For E50 loading use 125 per cent of these values, and for E30 
use 75 per cent. 

The stresses in trusses due to equivalent uniform loads may be 
determined by the methods given in Chapter XXI. 



Art. 3. Wind Loads. 

The wind load is usually expressed in one of the two follow- 
ing ways : (a) in pounds per square foot of actual truss surface ; 
or (b) in pounds per linear foot, treated as a dead load acting 
upon the upper and lower chords and as a live load acting upon 
the traffic as it moves over the bridge. 

168. Wind Load, (a) The method of stating the wind 
load in pounds per square foot of actual truss surface has the 
disadvantage that an assumption must be made as to the area of 
the truss surface. If this method is used, the wind load is usually 
taken at 30 pounds per square foot of truss surface. The load 
may be considered to act against the windward truss only, or to 
be equally resisted by the two trusses. 

169. (b) The method of specifying the wind load in 
pounds per linear foot is more logical, and is usually employed. 

For highway bridges, the usual practice is to take a wind load 
of 150 pounds per linear foot, treated as a dead load acting upon 
both the upper and lower chords; and a load of 150 pounds per 
linear foot, treated as a live load acting upon the portion of the 
bridge covered by the traffic. 



220 LOADS. Chap. XX, 

For railroad bridges, a higher value is used for that portion 
of the wind which is treated as a live load. This is to provide, 
not only for the wind loads, but also for stresses caused by the 
vibrations due to trains. The following wind loads are recom- 
mended as conforming to good practice: 150 pounds per linear 
foot, acting upon both the upper and lower chords; and 450 
pounds per linear foot of live load upon the bridge, the latter 
force being assumed to act six feet above the base of the rail. 
The portion of the wind load considered as live load will be re- 
sisted by the bottom laterals in through bridges, and by the top 
laterals in deck bridges. 



CHAPTER XXI. 

STEESSES IN TRUSSES DUE TO UNIFORM LOADS. 

In this chapter there will be given several graphic methods for 
determining the stresses in various types of bridge trusses under 
uniform loads, together with the algebraic method of coefficients. 
These methods will be explained by the solution of particular 
problems; as they may be more easily understood than general 
ones. Most of the methods are of general application to the 
various types of bridge trusses; but the problem should first be 
studied to determine which method may be employed to the best 
advantage. 

The chapter will be divided in seven articles, as follows : Art. 
I, Stresses in a Warren Truss by Graphic Resolution; Art. 2, 
Stresses in a Pratt Truss by Graphic Resolution ; Art. 3, Stresses 
by Graphic Moments and Shears ; Art. 4, Stresses in a Bowstring 
Truss — Triangular Web Bracing; Art. 5, Stresses in a Parabolic 
Bowstring Truss; Art. 6, Wind Load Stresses in Lateral Sys- 
tems ; and Art. 7, Stresses in Trusses with Parallel Chords by the 
Method of Coefficients. 



Art. I. Stresses in a Warren Truss by Gr.\phic Resolution. 

The application of the method of graphic resolution to the 
solution of the stresses in a Warren truss will now be explained. 

170. Problem. It is required to find the maximum and 
minimum dead and live load stresses in all the members of the 
highway Warren truss shown in Fig. 118, a. The truss has a 
span L of 70 feet ; a panel length 1 of 10 feet ; and a depth D of 
10 feet. The bridge has a width of 14 feet. 

221 



222 



STRESSES IX BRIDGE TRUSSES. 



Chap. XXL 



171. Dead Load Stresses. The dead load may be obtained 
ty applying equation (i), § 163. This load is found to be 476 lbs. 
per ft. of span, or say 240 lbs. per ft. of truss. The panel load 
\V = 240 X 10 = 2400, all of which will be assumed to act at 
the lower chord. The effective reaction R,=3X 2400=7200. 

X 



V, 


Uz Us 


U4 Us Uz 


u! 


A 


2 /\ 4. /\ 6 


/\ 6'/\ 4-'/\ 2" 


\ 



Lb Y 1^3 (a) \lz 



(c) 



3 (b) 

0* 3000** 4000* \ 
I I I I I 



i load on upper chord , | on lower. Load all on lower chord . 
D-IOft.j I =10 ft, L=70ft-.,w-240lbs.perft. 

Fro. 118. Dead Load Stresses — Warren Truss. 

To determine the dead load stress, lay off XY = 7200 (Fig. 118, 
b), and draw the dead load stress diagram for one-half of the 
truss. Since the truss and loads are symmetrical and the full dead 
load is always on the bridge, it is necessary to draw the diagram 
for only one-half of the truss. The kind of stress is determined 
by placing the arrows on the truss diagram as the stress diagram 
is drawn. The dead load stresses for both the chord and web 
members are shown in Fig. 119, a and Fig. 119, b. By a com- 
parison of these stresses, it is seen that, for this truss, the two web 
members meeting upon the unloaded chord have the same nu- 
merical stress. It is also seen that, for an odd-panel truss, the 
center web members are not stressed. 

The assumption is sometimes made that one-third of the dead 
load acts at the upper chord and two-thirds at the lower chord. 



Art, 1. 



WARREN TRUSS GRAPHIC RESOLl'TIOX. 



223 



The dead load stress diagram for this assumption is shown in 
Fig. ii8, c. For highway bridges, it is customary to assume all 
the dead load on the lower chord. 



Maximum and Minimum Stresses in Chord Mejmbers I 




Upper Chord 


Lower Chord | 


Chord Member 


X-2 


X-4 


X-6 


Y-l 


Y-3 


Y-5 


Y-7 


Dead Load 
Live Load 
Maximum 
Minimum 


- 7200 
-21000 
-28200 

- 7200 


-12000 
-55000 
-47000 
-12000 


-14400 
-42000 
-56400 
-14400 


•*■ 5600 
+10500 
+14100 
+ 5600 


+ 9600 
+28000 
+57600 
+ 9600 


+ 15200 
+58500 
+51700 
+15200 


+14400 
+42000 
+56400 
+14400 



(a) 



Dead Load Stresses in Web Members 



Web Member 



X-l 



1-2 



2-3 



3-4 



4.-5 



5-6 



6-7 



Dead Load 



- 8060 



+ 8060 



-5570 



+ 5570 



-2690 



+ 2690 



(b) 


Live Load Stresses in Web Members 


Web Member 


X-l 


1-2 


2-3 


3-4 


4-5 


5-6 


6-7 


Live Load af \1\ 


- 1120 


+ 1120 


-1120 


+ 1120 


-1120 


+ 1120 


-nzo 


tl »f M L^ 


- 2240 


+ 2240 


-2240 


+ 2240 


-2240 


+ 2240 


-2240 


m •' •» Ls 


-5360 


+ 5560 


-3360 


+ 3360 


-5360 


+ 3360 


-3360 


M M •• Ls 


-4480 


+ 4480 


-4480 


+ 4480 


-4480 


+ 4480 


+ 3360 


». •» •• L? 


-5600 


+ 5600 


-5600 


+ 5600 


+ 2240 


-2240 


+ 2240 


•f n i. Jji 


-6720 


+ 6720 


+ 1120 


-1120 


+ 1120 


-1120 


+ 1120 


Max* Live Load 


-23520 


+25520 


-16800 


+16800 


-11200 


+11200 


-6720 


Min- » »• 








+ 1120 


- 1120 


+ 5360- 


-3360 


+ 6720 


Uniform » "^ 


-23520 


+25520 


-15680 


+15680 


-7840 


+ 7840 






(C) 



Maximum and Minimum Stresses in Web Members 



Web Member 



X-l 



1-2 



2-3 



3-4 



4-5 



5-6 



6-7 



Dead Load 
Max- Live Load 
Min- Live Load 
Max- Stress 
Min- Stress 



- 8060 
-25520 

r3l580 
-8060 



• 8060 
+23520 


+51580 
+ 8060 



- 5570 
-16800 
+ 1120 
-22170 
-4250 



+ 5370 
+ 16800 
- 1120 
-22170 
+ 4250 



- 2690 
-11200 
+ 3360 
-13890 
+ 670 



+ 2690 
I+II200 
-3360 
+13890 
- 670 




-6720 
+ 6720 
-6720 
+6720 



Id) 
Fio. 119. Table op Stresses — Warren Truss. 

172. Live Load Stresses. The live load will be taken at 
1400 lbs. per linear ft. of bridge, or 700 lbs. per ft. per truss. 

Since the upper and lower chords are parallel, the chords will 
resist the bending moment due to the loads, and the web members 
will resist the shear. 



224 



8TUK8SKS IN BKIDGE TUUSSES. 



Chap. XXL 



(a) Chord Stresses. It is seen that each increment of live 
load brought upon the bridge increases the bending moment and 
therefore increases the stresses in each chord member. The maxi- 
mum live load stress in each member of the upper and lower 
chord is therefore given when the bridge is fully loaded with 
the live load. The minimum stress occurs when there is no live 
load on the bridge. 

Since the maximum live load chord stresses are obtained when 
the bridge is fully loaded, it is seen that it is unnecessary to draw 
a new stress diagram; as the live load chord stresses may be 
obtained from the corresponding dead load stresses by multi- 
plying each stress by the ratio of the live load to the dead load. 

The maximum live load chord stresses are shown in Fig. 
119, a. The minimum live load chord stresses are zero. 

X 




0* 1000*2000* 



a r 

V 




Y D» lOff.-, l^lOfr... L- 10 ft. 
p« 700 Ibsperft-iP- 7000 lbs. 

Stress Diagram 
for 

Live Load at Lli only 
(b) 



S 



5' 7 5 3 » 

Fig. 120. Live Load Stresses — Warren Truss. 



(b) Web Stresses. By partially loading the truss with live 
load, it is possible to obtain stresses which are larger than those 
obtained when the bridge is fully loaded ; or it is even possible to 
get stresses of an opposite kind to those due to the dead load. 



Art.l. WARBEX TRUSS — GRAPHIC RESOLUTIOX. 225 

The maximum and minimum live load web stresses will now be 
found by applying one load at a time, and noting the effect of 
that load upon the stress on each web member. 

Fig. 1 20, b shows the stress diagram for a single live panel 

P 

load placed at L/, the reaction at L© being - . Since the truss 

7 

is sjTnmetrical about the center line, a load placed at L/ will pro- 
duce the same stresses in the web members to the left of the cen- 
ter as a load at Li will in the web members to the right of the 
center. It will therefore be necessary to record only one-half of 
the web members, thus simplifying the table of stresses. From the 
stress diagram (Fig, 120, b), it is seen that the stresses in all of 
the members to the left of L^' due to a load at Lj' are constant. 

The stresses in the web members to the left of the center line 
of the truss due to a live load at L/ are shown in the first line 
of Fig. 119, c, these stresses being alternately tension and com- 
pression. Now place a live load at La', the remainder of the truss 

. 2 • 

being tmloaded. The reaction at L© equals — P, and it is seen 

without drawing another stress diagram that the stresses in the 
web members to the left of the center line are twice those due to 
the live load at L^'. The stresses due to a load at L2' are shown 
in the second line of Fig. 1 19, c. It is also seen that the stresses 
for a load at h/ are three times those for a load at L/, The 
stresses for a load at h/ are shown in the third line of Fig. 119, c. 
The stresses for a live load at L3 are shown in the fourth line ; for 
a live load at Lg in the fifth line ; and for a load at Lj ni the sixth 
line. 

Maximum and Minimum Live Load Wch Stresses. By the 
maximum live load stress is meant the greatest live load stress 
that ever occurs in the member ; and by the minimum live load 
stress is meant the smallest live load stress if there is no reversal 
of stress, or the greatest stress of an opposite kind if there is a 
reversal of stress in the member. 

From Fig. 1 19, c, it is seen that the addition of each load pro- 
duces a compressive stress in X-i and a tensile stress in 1-2, the 
maximum stress being — 23 520 for X-i and + 23 520 for 1-2 



226 STRESSES IN BRIDGE TRUSSES. Chap, XXI. 

when the truss is fully loaded. The minimum live load stresses 
in X-i and 1-2 are zero when there is no live load on the truss. 

It is also seen that loads at L^', Lg', h/, Lg, and Lj all produce 
compression in 2-3 and tension in 3-4; and that the load at Li 
produces tension in 2-3 and compression in 3-4. The maximum 
live load stresses in 2-3 and 3-4 are therefore obtained by adding 
the stresses due to the loads at L/, Lg', L3', Lg, and Lg, and are 

— 16800 and +16800, respectively, (see Fig. 119, c). The 
minimum live load stresses in 2-3 and 3-4 are obtained when 
there is a live load at Li only, and are + 1120 and — 1120, 
respectively. 

The maximum live load stresses in 4-5 and 5-6 are obtained 
when the loads at L/, Lg', Lg', and Lg are on the bridge, and are 

— II 200 and +11 200, respectively. The minimum live load 
stresses in 4-5 and 5-6 are obtained when the loads at Lj and Lg 
are on the bridge, and are + 3360 and — 3360, respectively. 

The maximum live load stress in 6-7 is obtained when the 
loads at L/, Lg', and Lg' are on the bridge, and is — 6720. The 
minimum live load stress in 6-7 is obtained when the loads at 
Li, L;,, and Lg are on the bridge, and is -f- 6720. 

A comparison of the corresponding stresses in line 7 and line 
9 shows the difference in the stresses in each member for the 
bridge loaded for maximum live load stresses, and fully loaded 
with live load. 

173. Maximum and Minimum Dead and Live Load 
Stresses, (a) Chord Stresses, The maximum and minimum 
chord stresses due to dead and live loads are shown in Fig. 119, a. 
The maximum chord stresses are obtained when the bridge is 
fully loaded with dead and live loads, and are equal to the sum of 
the dead and live load stresses. The minimum chord stresses are 
obtained when there is no live load on the bridge, and are the dead 
load stresses. 

(b) Web Stresses. The maximum and minimum web stresses 
due to dead and live loads are shown in Fig. 119, d. The maxi- 
mum web stresses are obtained by adding the dead and maximum 
live load stresses ; since they have the same signs. 



Art.l. WABREN TRUSS — GRAPHIC RESOLUTION. 227 

The minimum web stresses are obtained by adding, alge- 
braically, the dead and minimum live load stresses. It is seen 
that the dead and minimum live load stresses have opposite signs. 

By comparing the dead and the minimum web stresses, it is 
seen that the members 4-5, 5-6, and 6-7 have reversals of stress. 

174. Loadings for Maximum and Minimum Stresses. 
From a comparison of the stresses in Fig. 119, the following con- 
clusions may be drawn for the maximum and minimum dead and 
live load stresses in a Warren truss. 

(a) For maximum chord stresses, load the bridge fully unth 
dead and live loads, 

(b) For minimum chord stresses, load the bridge u^ith dead 
load only, 

(c) For maximum web stress in any member, load the longer 
segment of the bridge with live load, the shorter segment being 
unloaded, 

(d) For minimum web stress in any member, load the 
shorter segment of the bridge with live load, the longer segment 
being unloaded, 

175. Simplified Construction for Live Load Web Stresses. 
Referring to Fig. 120, b, it is seen that the numerical stresses in 
all the web members to the left of any lower chord panel point 
are constant when the live load extends to the right of that panel 
point. As the stresses are alternately tension and compression, 
it is seen that it is only necessary to draw the triangle YaY (Fig. 
120, b). This triangle is constructed by making YY equal to 
one live panel load, and drawing Ya parallel to be to meet a 
horizontal line Ya. The inclination of be is determined by mak- 
ing Yb equal to (or some multiple of) the half panel length, and 
Yc equal to (or the same multiple of) the depth of the truss. The 
stresses in the members to the left of L^' when there is a load 

at L/ are equal to — Ya. When there is a load at Lo', the stresses 

to the left of L^' are equal to — Ya, etc. It is therefore unneces- 

7 

sary to draw the entire diagram shown in Fig. 120, b. 



228 



STRESSES IX BBIDQE TRUSSES. 



Chap. XXL 



Art. 2. Stresses in a Pratt Truss by Graphic Resolution. 

176. The stresses in a through Pratt truss will be deter- 
mined in this article. It was shown in the preceding article that 
some of the web members of the Warren truss were subjected to 
reversals of stress. In the Pratt truss shown in Fig. 121, a, the 
web members are so constructed that they can not resist reversals 
of stress, the intermediate posts taking compression only, and the 
intermediate diagonals, tension only. This involves the use of 
another set of diagonals, called counters, in those panels where 
there is a tendency for reversals. 

Ui Uz U3 X ui uJ u; 




Y I 

3 load on upper chord, | on lower. Load all on lower chord. 
D • 28 ft-i 1 - 20 fr.j L-140 ft-, w-710 lbs.per ft. 

Fig. 121. Dead Load Stbesses — Pbatt Truss. 

The two center diagonals of the seven-panel truss shown in 
Fig. 121, a are both counters; as there is no dead load shear in 
the center panel of an odd-panel truss. The counter JJ^L/ acts 
when the live load extends to the right of Lg', and the counter 
Uj'La, when the live load extends to the left of L3. 

Care must be taken in determining the stresses in the verticals 
adjacent to the counters; as they differ greatly from those which 
would exist in these members if the main ties could resist com- 
pression. 



Art.S 



PRATT TRUSS — GRAnilC RKSOLUTIOX. 



229 



177. Problem. It is required to find the maximum and 
minimum dead and live load stresses in the through Pratt truss 



Maximum and Minimum Stresses in Chord Members - 


\ 


Upper Chord 


Lower Chord | 


Chord Member 


X-3 


Xr5 


X-6 


Y-l 


Y-2 


Y-4 


Y-6 


Dead Load 
Live Load 
Maximum 
Minimum 


-50.7 
-182.8 
-233.5 
- 50.7 


-60.8 
-219.2 

-28ao 

- 60.8 


- 60.8 
-219.2 
-280.0 

- 60.8 


+ 30.4 
+ 109.6 
+ 140.0 
+ 30.4 


+ 30.4 
+ 109.6 
+ 140.0 
+ 30.4 


+ 50.7 
+182.8 
+233.5 
+ 50.7 


+ 60.8 
+219.2 
+280.0 
+ 60.8 



(a) 



Dead Load Stresses in Web Members 


Web Member 


X-l 


1-2 


Z-3 


3-4 


4-5 


5-6 


6-6' 


Dead Load 


-5Z.A 


+ 14.2 


+ 34.9 


-14.2 


+ 17.5 


0.0^ 


0.0 








(b) 











Live Load Stresses in Web Members 


Web Member 


X-l 


1-2 


2-3 


3-4 


4-5 


5-6 


6-6' 


Live Load at l!i 


- 9.0 


0.0 


+ 9.0 


- 7.3 


+ 9.0 


- 7.3 


+ 9.0 


II 11 II L*2 


- 18.0 


0.0 


+ 18.0 


- 14.6 


+ 18.0 


-14.6 


+ 18.0 


» m * \JS 


- 27.0 


0.0 


+ 27.0 


-21.9 


+ 27.0 


-21.9 


+ 27.0 


n n ** Lb 


-36.0 


0.0 


+ 36.0 


-29.2 


+ 360 


+ 21.9 


-27.0 


m H M L2 


-450 


0.0 


+ 45.0 


+ U.6 


- 18.0 


+ 14.6 


- 18.0 


•• »» • Li 


-540 


+ 51.2 


- 9.0 


+ 7.3 


- 9.0 


+ 7.3 


- 9.0 


Max. Live Load 


-189.0 


+ 51.2 


+135.0 


-73.0 


+ 90.0 


-43.8 


+ 54.0 


Min. f 


0.0 


0.0 


- 9.0 


+ 21.9 


-27.0 


+43.8 


-54.0 


Uniform »» » 


-189.0 


+ 51.2 


+126.0 


- 51.2 


+ 63.0 


0.0 


0.0 









(C) 












Maximum and Minimum Stresses in Web Members 


Web Member 


X-l 


1-2 


2-3 


3-4 


4-5 


4-5 
Counter 


5-6 


6-6' 
Counter 


Dead Load 
Max- Live Load 
Min. Live Load 
Max- Stress 
Min. Stress 


- 524 
-189.0 
0.0 
-24U 
-524 


+ 14.2 
+ 51.2 
0.0 
+ 65.4 
+ 14.2 


+ 349 
+ 135.0 
- 9.0 
+1699 
+ 259 


-14.2 
-73.0 
+ 21.9 
-87.2 
00 


+ 17.5 
+90.0 
-27.0 
+I0T5 
0.0 


+ 95 
0.0 


0.0 
-43.8 
+43.8 
-43.8 

0.0 


0.0 
+ 54.0 
-54.0 
+ 54.0 

0.0 



All stresses are in thousands of pounds 

Fig. 122. Table of Stresses — Pratt Truss. 

shown in Fig. 121, a. The truss has a span of 140 feet, a panel 
length of 20 feet, and a depth of 28 feet. 

178. Dead Load Stresses. The dead load will be taken at 
710 lbs. per ft. per truss. The panel load W = 710 X 20 = 14 200. 



230 STRESSES IN BRIDGE TRUSSES. Chap, XXL 

The effective reaction Ri = 14200 X 3 = 42600. In this prob- 
lem, all the dead load will be taken on the lower chord of the 
truss. The stresses will be expressed in thousands of pounds, 
being carried to the nearest 100 pounds; thus, 14200 will be 
written 14.2. 

The dead load stress diagram for the left half of the truss is 
shown in Fig. 121, b. Since the truss and loads are symmetrical 
with respect to the center line, it is only necessary to draw the 
diagram for one-half of the truss. 

The dead load chord stresses are shown in Fig. 122, a, and 
the dead load web stresses in Fig. 122, b. The stresses for the 
left half of the truss only are shown ; as those for the right half 
are equal to the stresses in the corresponding members shown. 
It is seen that the upper and lower chord stresses increase from 
the end toward the center of the truss, and that the web stresses 
decrease from the end toward the center. It is further seen that 
the stresses in the web members in the center panel of this truss 
are zero, and that the stresses in the three center panels of the 
upper chord are equal. 

Fig. 121, c shows the dead load stress diagram for the left 
half of the truss, assuming that one-third of the dead load is on 
the upper chord and two-thirds on the lower chord. By compar- 
ing the stress diagrams shown in Fig. 121, b and Fig. 121, c, it is 
seen that the chord and inclined web stresses are the same for both 
cases, and that the stresses in the vertical posts, only, are changed. 
When one-third of the dead load i^ taken on the upper chord, 
the stresses in the intermediate posts are greater by the amount 
of the upper chord panel load than when all the load is taken on 
the lower chord; while the stress in the hip vertical is smaller 
by the amount of the upper chord panel load. 

179. Live Load Stresses. The live load will be taken at 
2560 lbs. per ft. per truss. The panel load P = 2560 X 20 = 
51 200, or say 51.2. 

(a) Chord Stresses. Since the addition of each increment of 
live load increases the bending moment, and since the upper and 
lower chords in trusses with parallel chords must resist this bend- 
ing moment, it is seen that the maximum stresses in all the chord 



Art, jf. 



PRATT TRUSS — GRAPHIC RESOLUTION'. 



231 



members are obtained when the bridge is fully loaded with live 
load. 

The minimum live load chord stresses are zero, when there is 
no live load on the bridge. 

The chord stresses may be obtained either by loading the 
bridge fully with live load and drawing a stress diagram similar 
to that shown in Fig. 12 1, b, or they may be obtained from the 
dead load chord stresses by direct proportion. Each chord stress 
is equal to the corresponding dead load stress multiplied by the 



Uj X '^i 





L^ Y^i:3 (a) l:^ 

' D = 28fr.a.20ft.-, L-I40ft. 
p=2560 Ib5.perft-.,P.5I2(X) lbs- 



Stress Diagram 
for 

Live Load at Li only 
(b) 



Fig. 123. Live Load Stresses — Pratt Truss. 



ratio of the live load to the dead load. The latter method requires 
less work and was used in this problem. The maximum live load 
chord stresses are shown in Fig. 122, a. 



232 STRESSES IX BKIDGE TRUSSES. Chap, XXL 

(b) IV eb Stresses, The web stresses, and the positions of 
the loads for maximum and minimum live load web stresses, will 
now be found by applying one load at a time and obtaining the 
stress in each member due to that load. 

The stress diagram for a single live load at L^' is shown in 
Fig. 123, b. Since there are seven panels in this truss, the re- 
action at Lo will be equal to— P, and the reaction at L^', to 

—P. This diagram is constructed by laying off Ri = — P, and 

drawing the stress diagram for the entire truss (see Fig. 123, b). 
In drawing this diagram, it is assumed that the diagonals which 
are stressed by the dead load, and the diagonal 6-6' (U3L3') are 
acting. This set of diagonals will act unless the dead load stresses 
in some of the members are reduced to zero by the live load, thus 
throwing some of the counters into action. 

Since the stresses in 3^-4', 5-6, and 3-4 are numerically equal, 
as are also those in 2'-3', 4'-5', 6' -6, 5-4, 3-2, and X-i, for a 
single load at L/, it is unnecessary to draw the entire diagram 
shown in Fig. 123, b. The stresses in these members may be 
obtained from the triangle YaYj. This triangle is constructed by 
laying off YYi equal to the live panel load P, and drawing Ya 
parallel to be to meet a horizontal line through Y^. The inclina- 
tion of be is determined by laying off Yjb equal to some multiple 
(in this case the multiple is 2) of the panel length, and YjC equal 
to the same multiple of the depth of the truss. For a single live 

load at L/, the reaction at L© is equal to — P, and the stresses in 

the inclined diagonals are each equal to — Ya; while those in the 
verticals, with the exception of the hip vertical, are each equal to 

iYYj. The stress in the hip vertical i'-2' is not influenced by 

any of the loads on the truss except that at L/, and therefore the 
stress in this member is either o or P. 

The stresses in the web members to the left of the center of the 



Art.g. PRATT TRUSS — GRAPHIC RESOLUTION. 233 

truss when there is a live load at Lj' are shown in the first line 
of Fig. 122, c. It is seen that the stress in 1-2 is zero. 

When there is a single live load at Lj', the reaction at L^ is 

equal to — P, and the stresses in the inclined and vertical web 

2 2 

members (1-2 excepted) are each equal to - Ya and- YYi, 

7 7 

respectively. The stresses in the web members to the left of the 

center when there is a live load at L^' are shown in the second 
line of Fig. 122, c. 

When there is a single live load at L3', the reaction at L^ is 

equal to 3. p, and the stresses in the web members to the left of 

the center for this loading are shown in the third line of Fig. 
122, c. 

The stresses in the web members to the left of the center 
when there is a single live load at L3 are shown in the fourth 
line of Fig. 122, c ; when there is a single live load at Lj, in the 
fifth line ; and when there is a single live load at L^, in the sixth 
line. 

Maximum and Minimum Live Load Web Stresses, By com- 
paring the web stresses shown in Fig. 122, b with those shown in 
Fig. 122, c, it is seen that the live load stresses in the members to 
the left of the center due to loads at L/, Lg', and h/ have the 
same signs as the corresponding dead load stresses. It is also 
seen that loads placed at L3, Lj, and Lj, successively, produce the 
same kinds of stress in the members to the left of these points 
as does the dead load. 

A live load placed at L^ tends to produce an opposite kind of 
stress in the members 2-3, 3-4, and 4-5 to that caused by the 
dead load ; and a load placed at Lg tends to produce an opposite 
kind of stress 3-4 and 4-5. 

Since in this truss the intermediate diagonals are tension mem- 
bers, the loads at and to the right of h/ cause the counter U3L3' 
to act ; and loads at and to the left of L3 cause the counter U3'L8 
to act. 

From Fig. 122, c, it is seen that the addition of each live load 



234 STKE8SES IN BKIDGE TKU8SES. Chap. XXI. 

produces a compressive stress in X-i, the maximum stress being 

— 189.0, which is the sum of the stresses caused by the separate 
loads. The minimum live load stress in X-i is o, when there is 
no live load on the bridge. 

It is also seen from Fig. 122, c that the maximum live load 
stress in the hip vertical 1-2 is obtained when there is a live 
load at Li, and is equal to the load at that point. This member 
simply transfers the load to the joint Ui, and the stress in it is 
not influenced by any other loads on the truss. The minimum live 
load stress is o, when there is no live load on the bridge. 

The maximum live load stress in the member 2-3 is obtained 
when the loads at and to the right of Lg are on the truss, and 
is + 13SA the sum of the separate stresses caused by these loads. 
The minimum live load stress in 2-3 is obtained when the single 
live load at L^ is on the truss, and is — 9.0 (provided there is 
already that much tension in the member due to the dead load). 

The maximum live load stress in the vertical 3-4 is obtained 
when the loads at and to the right of L3 are on the truss, and is 

— 73.0. The minimum live load stress in 3-4 is obtained when 
the loads at Li and Lg are on the truss, and is +21.9 (provided 
the counter in the panel Lg Lg does not act for this loading). 

The maximum live load stress in 4-5 is obtained when the 
loads at and to the right of Lg are on the truss, and is + 90.0. 
The minimum live load stress in 4-5 is obtained when the loads 
at Li and Lg are on the truss, and is — 27.0 (provided there is 
that much dead load tension already in it). 

The maximum live load stress in 5-6 is obtained when the 
loads at Lj', Lg', and Lg' are on the truss, and is — 43.8. The 
minimum live load stress in 5-6 is obtained when the loads at 
Lj, L2, and Lg are on the truss, also when the truss is fully loaded. 

The maximum live load stress in 6-6' (U3L3') is obtained 
when the loads at Li', Lg', and Lg' are on the truss, and is -j- 54.0. 
The minimum live load stress in 6-6' is obtained when the loads 
at Li, L2 and Lg are on the truss, also when the truss is fully 
loaded. 

By comparing the above stresses, the following conclusions 
may be drawn: (a) there is no stress in the hip vertical 1-2 



A^'S. PRATT TBl'SS — GIIAPHK' RESOLITIOX. 235 

unless there is a load at Lj, and when there is a load at this point, 
the stress in the hip vertical is equal to that load; (b) the web 
members meeting on the unloaded chord have their maximum 
and minimum live load stresses under the same loading; (c) the 
maximum live load web stresses are obtained when the longer 
segment of the truss is loaded with live load; (d) the minimum 
live load web stresses are obtained when the shorter segment of 
the truss is loaded. 

A comparison of the stresses shown in line 7 and line 9 (Fig. 
122, c) shows the differences in the corresponding stresses for 
the truss loaded for maximum live load stresses and fully loaded 
with live load. 

180. Maximum and Minimum Dead and Live Load 
Stresses, (a) Chord Stresses. The maximum and minimum 
chord stresses due to dead and live loads are shown in 
Fig. 122, a. The maximum chord stresses are obtained when 
the truss is fully loaded with dead and live loads, and are equal 
to the sums of the corresponding dead and live load stresses. The 
minimum chord stresses are obtained when there is no live load 
on the bridge, and are the dead load stresses. 

(b) Web Stresses, The maximum and minimum web 
stresses due to live and dead loads are shown in Fig. 122, d. Since 
the intermediate posts take compression only, and the intermediate 
diagonals, tension only, care must be used in making the com- 
binations for maximum and minimum stresses. It should be borne 
in mind that the counter and main tie in any panel cannot act at 
the same time, and that the counter does not act until the dead 
load tension in the main tie in that panel has been reduced to 
zero by the live load. It is thus seen that a main tie can resist 
as much live load compression as there is dead load tension 
already in it, and no more, the remainder of the live load stress, if 
any, being taken by the counter. Since the dead load always acts, 
it must be considered in all combinations. 

Referring to Fig. 122, c and Fig. 122, d, it is seen that the 
maximum stress in X-i is obtained when the bridge is fully 
loaded with live and dead loads, and is — 241.4. The minimum 



236 STRESSES IN BRIDGE TRUSSES. Chap, XXI. 

Stress in X-i is obtained when there is no live load on the bridge, 
arid is — 524. 

The maximum stress in the hip vertical 1-2 is obtained when 
there is a live load at Lj, and is + 65.4, the sum of the dead and 
live panel loads. The minimum stress in 1-2 is obtained when 
there is no live load on the bridge, and is + 14.2. 

The maximum stress in 2-3 (see Fig. 122, c and Fig. 122, d) 
is obtained when there are live loads at and to the right of Lg, 
and is + 169.9. The minimum stress in 2-3 is obtained when 
there is a live load at L^ only, and is + 25.9. In this case, it 
is seen that the minimum live load stress has an opposite sign 
from that of the dead load, and subtracts from it. No counter 
is required in this panel, as the live load compression is less than 
the dead load tension. 

The stress in the member 4-5 will be determined before that 
in 3-4, as it is necessary to determine which diagonal is acting 
before the stress in the post can be found. 

The maximum stress in 4-5 is obtained when the loads at 
and to the right of Lj are on the bridge, and is + 107.5. The 
minimum stress in 4-5 is obtained when the live loads at Lj and 
Lj are on the bridge, and is o; as will now be shown. Referring 
to Fig. 122, c, it is seen that the live load tends to cause a com- 
pression of 27.0 in 4-5 when the loads at Li and Lg are on the 
bridge. The member can only resist a compression of 17.5 (this 
being the dead load tension already in it), the resulting stress in 
4-5 being o. The remainder of the live load stress tends to dis- 
tort the member 4-5, and throws the counter in this panel into 
action. The stress in the counter is then + 9-5« 

The maximum stress in 3-4 is obtained when the live loads 
at and to the right of L3 are on the bridge, and is — 87.2. The 
minimum stress in 3-4 is obtained when the live loads at Li 
and L2 are on the bridge, and is o. It is seen that the stress in 
3-4 must be o; as the stress in 4-5 has been shown to be o for 
this loading, and for equilibrium at U2, the stress in 3-4 must 
also be o. 

The maximum stress in the counter 6-6' (UgLg') is obtained 
when there are live loads at L3', L/, and h/, and is + S4-o* 



Art.g. PRATT TRUSS — GRAPHIC RESOLUTIOX. 237 

The minimum stress in 6-6' is obtained when there is no live 
load on the bridge, when the bridge is fully loaded, or when the 
live loads at Li, Lg, and Lg are on the span, and is o. There are 
no dead load stresses in the center diagonals of this truss. 

The maximum stress in 5-6 is obtained when there are live 
loads at L3', Lj', and L/, and is — 43.8. The minimum stress 
in 5-6 is obtained when there is no live load on the bridge, also 
when the live loads at L^, Lj, and L, are on the bridge, and is o. 

The maximum stress in the other counter L3U3' is obtained 
when the loads at Lj, Lj, and Lj are on the bridge, and is + 54.0. 
The minimum stress is obtained when there are no live loads on 
the bridge, when the bridge is fully loaded, or when there are 
live loads at L3', Lj', and L/. 

It is unnecessary to determine the stresses in the members 
beyond the center line ; as the truss is symmetrical. 

i8i. Loadings for Maximum and Minimum Stresses. 
Conclusions. From a comparison of the stresses shown in Fig. 
122, the following conclusions may be drawn for maximum and 
minimum dead and live load stresses in a Pratt truss. 

(a) For maximum chord stresses, load the bridge fully unth 
live and dead loads. 

(b) For minimum chord stresses, load the bridge unth dead 
load only. 

(c) For maximum web stress in any member (except the hip 
vertical), load the longer segment of the truss zvith live load. For 
maximum stress in the hip vertical, load the bridge so that there 
will be a live load at L^. 

(d) For minimum web stress in any member (except the hip 
vertical), load the shorter segment of the bridge with live load. 
For minimum stress in the hip vertical, load the bridge zvith dead 
load only. 

In making the combinations for maximum and minimum 
stresses, it should be borne in mind that the counter in any panel 
does not act until the dead load stress in the main diagonal in that 
panel has been reduced to zero by the live load. 

182. The Stresses in the members of a Howe truss may be 
determined in a similar manner to that used in finding those in a 



238 STBES8E8 IN BHIDQE TRUSSES. Chap.IXl 

Pratt truss. In the Howe truss, the vertical members take tension 
only, and llie diagonals, compression only. 



Art. 3. Stresses by Graphic Moments and Shears. 

The method of graphic moments and shears, as applied to 
the solution of the stresses in a bri(^e truss, will be explained in 
this article, 

183. Problem. It is required to find the stresses in the 
six-panel Warren truss shown in Fig. J24, a. The truss has a 
span of 130 feet, a panel length of 20 feet, and a depth of 20 feet. 
The dead load will be taken at 800 lbs. per ft. per truss, and the 
live load at 1600 lbs. per ft, per truss. 



2 


r-^ 


w 


■^' 


[^ 


^u 


p-E 


I 


W 




« , t 


5^ 


^ -^ 


? 




« 




't 


? 


■n 


y^y ' 






















3r---. "^ 


k- 








,'. 


y^ 




.--- ' 


._^....^^ 


*(b) 


i 


z ij 


if. 


:^,.: 


% ^ 


: 


q- 


h 










Ir 


-V/. '■ k9. ,c, ,1 j, 


1 




i__ 


• JOMO' BOX 


»•«■! ja 1 


L. 




2 




40 






L 1 



D«20ft.il-20ff.iL-l20 ft., w. 800 per ft., W- 16000 lbs. 
1. 124. De^ap Txjaii Rtbesses bt Gkafhic Mombstr i>D Sboars. 



184. Dead Load Chord Stresses. Draw the truss diagram 
(Fig. 124, a) to scale, and load it fully with dead load. The dead 



Art,S. GRAPHIC MOMENTS AND SHEARS. 239 

panel load is equal to 800X20=16000. Draw the load line 
shown in Fig. 124, c for the dead load. Take the pole O with 
a pole distance H, draw the rays, and construct the funicular 
polygon shown in Fig. 124, b. The pole distance H, expressed 
in thousands of pounds, should be numerically equal to the depth 
of the truss, or to some multiple of the depth, i. e., if D = 20, 
then H should be equal to 20 000, or to some multiple of 20 000. 
The vertices of the funicular polygon shown in Fig. 124, b lie on 
the arc of the bending moment polygon for the given truss and 
loads. The bending moment in any upper chord member is equal 
to the intercept under the center of moments multiplied by the 
pole distance H ; and the stress in the member is equal to the 
bending moment divided by the depth of the truss. Now if the 
pole distance in thousands of pounds is taken equal to the depth 
of the truss or to some multiple of the depth, then the intercept 
will represent to scale the stress in the member ; provided a scale, 
which may be determined in the following manner, is used in 
measuring the intercept. In this problem, D = 20; and suppose 
that H = 40 000, and that the linear scale of the truss is i inch = 
40 feet. Then the intercepts should be measured to a scale of 

40 X =80000 pounds to the inch. If H had been taken 

equal to 20 000 pounds, then the scale to be used in measuring the 
intercepts should have been i inch = 40 000 pounds. 

The stresses in the lower chord members are mean propor- 
tionals between those in the adjacent upper chord members, and 
the intercepts are to be taken as shown in Fig. 124, b. It is thus 
seen that the stress in any chord member may be found directly 
by scaling the intercept under the center of moments. 

The ordinates to the stress polygon shown in Fig. 124, b may 
be obtained without drawing the force and funicular polygons, 
in the following manner. The stress in the center member X-6 
of the upper chord is equal to the bending moment at the center 
of the truss due to the uniform load of w lbs. per linear ft. of 

wL^ 
truss divided by the depth of the truss, i. e., to . Lay off the 



240 STRESSES IN BRIDGE TRUSSES. Chap. XXL 

•JJ1 J- . V /r 1 . 800 X 120 X 120 

middle ordinate X-o equal to = 72000, to any 

8 X 20 

given scale. Draw the horizontal line 4-4 equal to one-half the 
span of the truss; also draw the line 1-4 parallel to the middle 
ordinate X-6. Divide the horizontal line 4-4 into three equal 
parts (one-half the number of panels in the truss) ; also divide 
the line 1-4 into the same number of equal parts, and number as 
shown. Draw the radial lines 1-4, 2-4, and 3-4 to meet verticals 
through the lower chord panel points. Then these points of inter- 
section will give points on the required stress polygon. 

If the truss has an odd number of panels, the above method 
should be modified as follows: Divide the horizontal line 4-4 
(one-half the span of the truss) into as many parts as there are 
panels in the entire truss; and use only the alternate points of 
division. 

It is seen that the method of graphic moments and shears does 
not give the kind of stress. If the kind of stress is not evident 
by inspection, then it may be found by algebraic methods. 

185. Live Load Chord Stresses. The live load is equal to 
1600 lbs. per ft. per truss, and the live panel load is 1600 X 20 = 
32000. The live load chord stresses may be obtained by con- 
structing a diagram similar to that used for the dead load chord 
stresses, or they may be more easily obtained from the dead load 
stresses by direct proportion. 

186. Dead Load Web Stresses. The reactions for the 
truss loaded with the dead load are represented by Rj and Rg 
(Fig. 124, c). In a bridge with parallel chords, the web members 
resist the shear. The shear in panel LqLj is equal to + Ri ; in 
L1L2, to -f (Ri — W) ; and in L2L3, to + (R^ — 2W). At Lj, the 
shear passes through o. The dead load shear line is shown by 
the stepped line in Fig. 124, c. Now the stresses in the web mem- 
bers are equal to the shears in the members multiplied by the 
secant of the angle that the members make with the vertical. The 
shears are graphically multiplied by the secant of the angle by 
drawing the lines cc^, ee^, and ggi parallel to 1-2, 3-4, and 5-6, 
respectively. The stresses in 1-2, 3-4, and 5-6 are tension. The 



QHAPHIC MOMENTS AND 8HGAHS. 



241 



Stresses in X-l, 2-3, and 4-5 are numericall)' equal to those in 
1-2, 3-4, and 5-^, respectively, and are compression. Since the 
truss and loads are symmetrical about the center, it is only neces- 
sary to draw the diagram for one-half of the truss. 

187. Live Load Web Stresses. Another diagram must be 
drawn for the live load web stresses. This diagram may be con- 
structed in the following manner : 

Assume that the Warren truss shown in Fig. 125, a instead 
of being a simple span is a cantiliver truss, the right end being 




__H-L_ 

O-e0ft.,l.Z0fT-iL-IZ0ft.i W-I6000lbs-:P-32000II». 
Pia. ISS. LiTB Load Wib Stbessbs — Uaxihum Wbb Stbihsbs. 

fixed and the left end free. Then with the cantilever truss fully 
loaded with live load, lay off the load line (Fig. 125, c), assume a 
pole O, and with a pole distance H = span L, draw the funicular 
polygon, starting the polygon at a, the top of the load line. Now 



242 STRESSES IN BRIDGE TRUSSES. Chap. XXL 

the bending moment at the right support is equal to the intercept 
y\ multiplied by the pole distance H. But the given truss instead 
of being a cantilever truss is a simple span ; and is held in posi- 
tion by the reaction Rj at the left end, instead of being fixed at 
the right end. Therefore the moment at the right end, which is 
equal to Hy^, must be equal to the moment of the left reaction. 
Now the moment of the left reaction is equal to RiL, where L is 
the span of the truss ; therefore Hyi = RjL, and since H =^ L by 
construction, y^ = Rj. 

The maximum live load shear in X-i and 1-2 is obtained 
when the truss is fully loaded with live load, and is yi = Ri. The 
stress in X-i, which is numerically equal to that in 1-2, is 
obtained by drawing a line through a point a parallel to the 
member X-i. 

Now move the truss one panel to the left, but let the loads 
remain stationary. The new position of the truss, together with 
the new loading, is shown in Fig. 125, b. With the same pole O 
and pole distance H, draw a new funicular polygon. This funicular 
polygon will coincide with a part of the first polygon, the part to 
the left of the point d (Fig. 125, c), being identical in both poly- 
gons. As above, the bending moment at the right end for this 
loading is equal to the moment of the left reaction, i. e., Hyg = 
R3L, or y2 = R3. Now the loading shown in Fig. 125, c is that 
for a maximum shear in the members 2-3 and 3-4; and since 
y2 = R3 is the shear in these members, the stresses are obtained 
by drawing through d a line parallel to the member 2-3. 

In like manner, gh may be shown to be the shear in the mem- 
bers 4-5 and 5-6 when there is a maximum stress in 4-5 and 5-6 ; 
and the stresses in these members are obtained by drawing a line 
through g parallel to the member 4-5. 

Likewise, jm is equal to the shear in 6-5' and 5'-4' when 
there is a minimum shear in 6-5' and 5'-4', and the stresses in 
these members are obtained by drawing a line through j parallel 
to the member 6-5'. 

Similarly, the minimum stresses in 4^-3' and 3'-2' are obtained 
by drawing a line through o parallel to 4^-3'. 



Art. 4. BOWSTRING TRUSS. 243 

The minimum live load stresses in 2'-i' and X'-i' are zero, 
when there is no live load on the truss. 

The kind of stress is not given by this graphic construction, 
and if not evident by inspection, it may be obtained by algebraic 
methods. 

i88. Maximum and Minimiun Dead and Live Load 
Stresses, (a) Chord Stresses. The maximum chord stresses 
may be obtained by adding the dead and live load stresses. The 
minimum chord stresses are equal to the dead load stresses. 

(b) Web Stresses. The maximum and minimum web stresses 
may be obtained by graphically combining the dead and live load 
shears, provided the dead and live load shear diagrams are placed 
as shown in Fig. 125, c. In this figure, the dead load shear in 

X-i is be =— W, the live load shear is Ri=yi = ab, and the 

maximum shear is ac. Likewise, df = maximum shear in 2-3, gi = 
maximum shear in 4-5, jk = minimum shear in 6-5', no = mini- 
mum shear in 4'-3', and rs = minimum shear in 2'-i'. The 
stresses in the members are obtained by graphically resolving these 
shears. 



Art. 4. Stresses in a Bowstring Truss — Triangular Web 

Bracing. 

The solution of the maximum and minimum stresses in a bow- 
string truss will be taken up in this article. The dead load stresses 
will be determined by graphic resolution, and the live load stresses 
by a special application of the method of graphic resolution. The 
method used is general in its application to trusses with triangular 
web bracing, and the upper chord panel points may lie on the arc 
of any curve. 

189. Problem. It is required to find the maximum and 
minimum stresses in the bowstring truss shown in Fig. 126, a. 
The upper chord panel points lie on the arc of a parabola with a 
middle ordinate of 24 ft. The truss has a span of 160 ft., and a 



244 



STRESSES IN BRIDGE TRUSSES. 



Chap. XXL 



panel length of 20 ft. The dead load will be taken at 500 lbs. per 
ft. per truss, and the live load at 1000 lbs. per ft. per truss. 



-665 



-67.3 




(b) 

Dead Load. 
Stress Diaqram 







10000 20000 
I I 



Ranel Load -10000 lbs. 



FiQ. 126. Dead Load Stresses — Bowstring Truss. 



190. Chord Stresses, (a) Dead Load Chord Stresses. The 
dead panel load is equal to 500 X 20 = 10 000 lbs. The dead load 
chord stresses are obtained by drawing the dead load stress dia- 
gram for one-half of the truss, as shown in Fig. 126, b. The 
stresses, expressed in thousands of pounds, are shown on the 
members of the truss in Fig. 126, a. 

(b) Live Load Chord Stresses. The live panel load is equal 
to 1000 X 20 = 20 000 lbs. The live load chord stresses are 
obtained by proportion from the dead load chord stresses. These 
stresses are double the corresponding dead load stresses. 

191. Web Stresses, (a) Dead Load Web Stresses. The 
dead load web stresses are obtained from the dead load stress 
diagram shown in Fig. 126, b. These stresses are shown on the 
members of the truss in Fig. 126, a. It is seen that the web 
stresses in this type of truss are small, and that the stresses in 



Art. 4. 



BOWSTBIXG TRUSS. 



245 



all the web members are tension. These members act as auxiliary 
members to transfer the load to the upper chord panel points. 

(b) Live Load Web Stresses. The live load web stresses 
are obtained from the stress diagram shown in Fig. 127, b. This 
diagram is constructed by assuming a value (in this case 10 000 
lbs.) for the left reaction R,, and drawing the stress diagram 
with no loads on the truss. 

X 




2' 



r-^ 



Panel Load -20000 lbs. 



Member 


True Ri 
Assuniecl Ri 


1-2 


7.00 


2-3. 3-4 


5.25 


4-5.5-6 


3.75 


6-7, 7-8 


2.50 


75-8. 6'-r 


1.50 


5^6*. 4-5' 


0.75 


3'-4'. 2^3' 


0.25 


r-2' 






Stress Diagnam for ^^ 

Max. and Min.Live Load Stresses in Web Members. 
Ri assumed-No loads on truss. 

To get stress in any web member, 

. measured stress by J^^^ ^' 

' Assumed Ri 

20000^ 40000* 

— \ I 




(C) 



3' 



(b) 



5' 



7' 7 531 



Fia. 127. Live Load Web Stresses — Bowstring Truss. 

It has already been shown that, for a maximum live load stress 
in any web member, the longer segment of the truss should be 
loaded ; and for a minimum live load stress, the shorter segment 
should be loaded. Therefore, for a maximum stress in any web 
member to the left of the center of the truss, there should be no 
loads to the left of that member; and for a minimum stress in 
the corresponding member on the right of the center, there should 
be no loads to the left of that member. The shear in any member 
is therefore equal to the reaction ; and the stress in the member 
may be found by multiplying the stress in that member, obtained 



2-46 STRESSES IN BRIDGE TRUSSES. Chap, XXL 

from the stress diagram shown in Fig. 127, b, by the ratio of the 
actual live load reaction (for a maximum or minimum stress in 
that member) to the assumed reaction. For example, to get a 
maximum stress in 1-2, the bridge is fully loaded with live load, 
and the reaction Ri=7oooo. The maximum stress in 1-2 is there- 
fore equal to the stress scaled from Fig. 127, b multiplied by 

7^^=+ 15.8. 

lOOOO 

For stresses in 2-3 and 3-4, load all joints except L^; for 
stresses in 4-5 and 5-6, load all joints except L^ and 1^; for 
stresses in 6-7 and 7-8, load all joints except Lj, Lj, and L3 ; for 
stresses in 8-7' and 7'-6', load the joints h/, Lg', and h^ ; for 
stresses in 6'-5' and 5'-4', load joints Lg' and Lj' ; for stresses in 
4'-3' and 3'-2', load the joint L/ only; and for stress in 2'-i', 
there should be no loads on the truss. 

The ratios of the actual reactions to the assumed reactions for 
the above loadings are shown in Fig. 127, c. 

The live load stresses are shown on the members of the truss 
in Fig. 127, a, and are obtained by scaling the stresses from the 
diagram in Fig. 127, b and multiplying these stresses by the ratios 
shown in Fig. 127, c. 

192. Maximum and Minimum Dead and Live Load 
Stresses, (a) Chord Stresses. The maximum chord stresses are 




+ 186.3 -I-I94.I +195.9 +196.5 

+ 62.1 + 64.7 + 65.3 + 655 

Fio. 128. Maximum and Minimum Stresses — Bowstring Truss. 

equal to the sums of the corresponding dead and live load stresses, 
and the minimum chord stresses are the dead load stresses. These 
stresses are shown on the truss diagram in Fig. 128. 



Art, 5. PARABOLIC BOWSTRING TRUSS. 247 

(b) Web Stresses. The maximum and minimum web stresses 
are obtained by combining the dead and live load stresses shown 
in Fig. 126, a and Fig. 127, a, respectively. In making the com- 
binations, both the stresses in the member and those in the corre- 
sponding member on the other side of the center line of the truss 
must be considered. 

The maximum stress in 1-2 is +7-9+ i5-8 = + 23.7, and 
the minimum stress is + 7.9, the dead load stress. 

The maximum stress in 2-3 is + 7-3 + 21.6= + 28.9, and the 
minimum stress is + 7.3 — 6.8 = + 0.5. 

The maximum stress in 3-4 is + 5.5 + 20.5 = + 26.0, and the 
minimum stress is + 5.5 — 9.1 = — 3-6. 

The maximum stress in 4-5 is + 6.0 + 25.3 = -\- 31-3, and the 
minimum stress is -f- 6.0 — 13.2 = — 7.2. 

The maximum stress in 5-6 is + 5.3 + 24.7 = + 30-o> and 
the minimum stress is + 5.3 — 13.7 = — 8.4. 

The maximum stress in 6-7 is + S- 2 + 27.0 = + 32.2, and 
the minimum stress is + 5.2 — 16.2 = — 11. o. 

The maximum stress in 7-8 is + 5.4 + 27.0 = + 32.4, and the 
minimum stress is + 5.4 — 16.2 = — 10.8. 



Art. 5. Stresses in a Parabolic Bowstring Truss. 

The maximum and minimum stresses in a parabolic bowstring 
truss will be found in this article by a special graphic method. 
The method given applies only to trusses whose upper chords 
panel points lie on the arc of a parabola. 

193. Problem. It is required to find the maximum and 
minimum stresses in the parabolic bowstring truss, half of which 
is shown in Fig. 129. The truss has a span of 160 ft., a panel 
length of 20 ft., and a depth at the center of 24 ft. The dead load 
will be taken at 500 lbs. per ft. per truss, and the live load at 1000 
lbs. per ft. per truss. 

194. Chord Stresses, (a) Dead Load Chord Stresses. The 
dead panel load = 500 X 20 = 10 000 lbs., or say lo.o. Now the 
bending moment at any point in a truss loaded with a uniform load 



248 8TBEBSB8 IN BRIDGE TBUSBES. Chap. XXI. 

varies as the ordinates to a parabola ; and the stress in any chord 
member is equal to the bending moment divided by the depth of 
the truss. Since the moment at any point varies as the ordinates 
to a parabola, and the moment arm for the stress in any lower 



Fia. 120. Maxiuuu anu M[M[uuh Btbesbbh— Pababolic Bow9thin(i Tbuss. 

chord member is an ordinate to the parabola, it is seen that the 
stress in the lower chord is constant. 

y/JJ 

For this truss, the moment at the center ^ , and the dead 

8 

wL= 500 X 160 X j6o 



load stress - in the lower chord ; 



8D ~ 8X24 



= 66 700. Since the stress in the lower chord is constant when the 
bridge is fully loaded, it is seen that the diagonals are not stressed, 
and that the horizontal components of the stresses in the upper 
and lower chords are equal. The dead load stresses in the upper 
chord members may be found graphically, as follows : Lay off ab 
(Fig. 129) equal to the stress in the lower chord, and draw the 
vertical lines aa' and bb'. Draw lines parallel to the upper chord 
members, and the distances intercepted on these lines between the 
vertical lines aa' and bb' will represent the stresses in the corre- 
sponding upper chord members. 



Art.S. PARABOLIC BOWSTRING TRUSS. 249 

(b) Live Load Chord Stresses, The maximum live load chord 
stresses are obtained when the bridge is fully loaded with live load, 
and are found in the same manner as the dead load chord stresses. 
To find these stresses, lay off be (Fig. 129) equal to the live load 

1000 X 160 X 160 

stress in the lower chord = 5—- = i33 vx>. The 

8 X 24 *^*^ *^ 

upper chord stresses are represented by the distances intercepted 
between the vertical lines bb' and cc' on lines drawn parallel to 
the chord members. 

The minimum live load chord stresses are zero. 

195. Web Stresses, (a) Dead Load Web Stresses. Since 
the stress in the lower chord is constant when the bridge is loaded 
with a uniform load, it is seen that the dead load stresses in the 
diagonals are zero. The dead load stresses in the verticals are 
tensile stresses, and are each equal to the dead panel load 

= + lO.O. 

(b) Live Load Web Stresses. The maximum live load 
stresses in the diagonals and the minimum live load stresses in 
the verticals may be obtained directly from the diagram shown in 
Fig. 129. The maximum live load stresses in the verticals, and 
the minimum live load stresses in the diagonals are obtained when 
the bridge is fully loaded with live load. The maximum live load 
stresses in the verticals are + 20.0 (the live panel load), and the 
minimum live load stresses in the diagonals are o. 

The diagram shown in Fig. 129 is constructed, as follows: 

1000 X 160 X 160 
Lay off be to any scale= ^-— = 133 300. Construct 

the half truss diagram, making the half span equal to b'd' = ^bc. 

Since the span is 160 ft. and the depth of the truss is 24 ft., make 

24 
dd'=— ^bc. With b'd' equal to the half span and dd' equal to 

the depth at the center, draw the outline of the given truss. 

The maximum live load stresses in the diagonals sloping upward 
to the left are obtained when the longer segment of the bridge is 
loaded, and the maximum stresses in the diagonals sloping upward 



250 STBE88ES IN BRIDGE TRUSSES. Chap, XXL 

to the right, when the shorter segment of the bridge is loaded with 
live load. These stresses are obtained directly by scaling the mem- 
bers to the given scale. The minimum live load stress in the first 
vertical from the left end is zero. The minimum live load stresses 
in the other verticals are compression, and are represented by the 
vertical distances between the first upper chord panel point and 
each succeeding upper chord panel point. 

The proof of the construction shown in Fig. 129 is given 
in § 197. 

196. Maximum and Minimum Dead and Live Load 
Stresses, (a) Chord Stresses. The maximum stress in the 
lower chord is represented by ac (Fig. 129), which represents 
the sum of the dead and live load chord stresses. The maximum 
stress in the upper chord members are represented by the dis- 
tances intercepted between the vertical lines aa' and cc' by the 
upper chord members produced. The minimum chord stresses are 
the dead load stresses. The maximum and minimum chord 
stresses are shown on the members produced. 

(b) Web Stresses. The maximum stresses in the diagonals 
are represented by the lengths of the diagonals, the dead load 
stresses in these members being zero. These stresses are shown 
on the members of the truss diagram. The minimum stresses in 
the diagonals are zero. The maximum stresses in the verticals 
are equal to the sums of the dead and live panel loads == + lo.o + 
20.0 = + 30-0. The minimum stresses in the verticals are shown in 
Fig. 129. These stresses are obtained by laying off to scale the dead 
panel load, W = lo.o, above the first upper chord panel point, 
and drawing the horizontal dot and dash line shown in Fig. 129. 
The dead and live load stresses are of an opposite kind, and the 
resulting stress is thus obtained. The minimum stresses in the 
verticals are represented by the distances between this line and 
each upper chord panel point. Distances measured below the line 
represent tensile stresses, and those above, represent compressive 
stresses. 

197. Proof of Construction Shown in Fig. 129. It will 
now be proved that for the construction shown in Fig. 129, (a) 
the maximum live load stresses in the diagonals are represented by 



Art, 5. 



PAR.\BOLIC BOWSTRING TRUSS. 



251 



the lengths of the diagonals, and (b) that the minimum live load 
stresses in the verticals, i. e., the maximum live load compressive 
stresses, are represented by the vertical distances from the first 
upper chord panel point to each upper chord panel point. 




Fio. 130. Pababolic Bowstring Truss. 



(a) Stresses in Diagonals. Let Fig. 130 represent a parabolic bow- 
string truss in which the span = E-_, and the depth = _ X -E = 

oJj Li 8D 

E — = panel load P. 

Also, let x = abscissa and y = ordinate of any point O on the parabola, 
p = live load per foot, 
n = number of panels in truss, 
m = number of loads on truss, 
L = span of truss, 
D = depth of truss. 

It is required to prove that the maximum live load stress in any 
diagonal is represented by the length of that diagonal. 
The equation of the parabola about the left end Lq is 



2 



4D 



(D-y), 



from which, 



_4Dx.L_x^ 



N6w the horizontal component of the maximum live load stress in 
any diagonal member is equal to the difference of stress in the two adja- 
cent lower chord members when the truss is loaded for a maximum stress 
in that diagonal. 

Consider any diagonal, say U2L3 in panel LjLs, whose maximum stress 
occurs when there are m loads on the truss. 

Tor this loading, R, = i^±i>. ^ J^ . 

2n ^ n 

Moment of Ri about the panel point to right of diagonal ^ 



(n\+l)m pL ,„ ^xL 
-^^ — TT- y — V (n — m)_, 



2n ^ n ^ 



n 



252 STRESSES IN BRIDGE TRUSSES. Chap. XXL 

and stress in lower chord member to right of diagonal = 

£n n n 

pL*(ni + 1) (n — m)m 
"= 8n^Dx(L — x) 

In above equation, x^— (n — m), and substituting this value of x, 

n 
we have 

pL* (m + 1) (n — m)m 



S = 



8n»Di^(n — m) [L— i^(n— m)] 
n n 

pL» (m + 1) 



- 8nD 

Also moment of B^ about panel point to left of diagonal = 

(m4- l)m pL ,„ ^ in L 

2n ^ n ^ n 

and stress in lower chord member in same panel as diagonal = 

(m -f l)m pL , , L 

2S— XlfX ("-'"-D n 

8 — . 

_ pL*(m + l) (n — m — l)m 
= 8n»Dx(L — x) 

In above equation, x^— (n — m — 1), and substituting this value 

of X, we have 

g/^ pL* (m + 1) (n — m — l)m 



8n»D-t(n — m — 1) [L — t(n — m — 1)] 
n n 



pL'm 



8nD 



Now S-S' = Jl^!i^±lI IEB. 

8nD ~" 8nD 

-. JB = horizontal component of stress in diagonal. 

t)L' 
Now the span of the truss was laid off equal to tL-. . Therefore 

the horizontal component of the maximum stress in the diagonal, which is 
±. — , is represented by one panel length of the lower chord. Since one 

panel length represents the horizontal component of the stress in the 



Art, 5. PARABOLIC BOWSTRING TRUSS. 253 

diagonal, the length of the diagonal itself will represent the maximum 
stress in it. 

(b) Stresses in Verticals. It is required to prove that the minimum 
live load stresses in the verticals, i. e., the maximum compressive stresses, 
are represented by the vertical distances from the first upper chord 
panel point to each upper chord panel point. 

Consider any vertical, say UgLs (Fig. 130), whose maximum com- 
pressive stress occurs when there are m loads on the truss. Now the 
stress in U2L2 when there are m loads on the truss is equal to the 
difference in stress between L2L, and LiL, for this loading multiplied 
by the tangent of the angle that UiLs makes with LiL,. 

The stress in the lower chord member to the right of the vertical is 

pL'm 

The stress in the lower chord member to the left of the vertical is 



Ri(n — m — 2) i^ 

n 



8" = 



(m 4- l)m pL /„ ^ o\ L 

pL*(m + 1) (n — m — 2)m 
= 8n»Dx(L — x) 

In above equation x= — (n — m — 2), and substituting this value 

n 

of X, we have 

pL*(m + l)(n — ni — 2)m 



8'' = 



8n«I>- <n — m — 2)[L— It (n — m — 2)] 
n n 



pL'(m + l)m 
- 8nD(m + 2) * 

g/ s'' = pL*m pL'(m -f l)m 

SnD 8nD(m + 2) 

pL'm 
" 8nD(m + 2) ' 

The stress in any vertical member is 

^ tan 9, where 9 is the angle that the diagonal 



8nD(m + 2) 



which meets the required vertical at the lower chord makes with the 
lower chord. 



254 STRESSES IN BRIDGE TRUSSES. Chap. XXL 

An expression will now be found for tan 6. 

Tan e=:L =57. 



Now y= — r~"'(^ — ^)> ^" which x= ^^ (n — m — 2), 
Substituting values of x and y, we have 

i^X ~(°-"™-2)[L- h .(n — m — 2)] 
tan e = 



L 

4D(n — m — 2)(m-i-2) 
En 



o^ *• 1— pL'm 4D(n — ra — 2)(m + 2) 

Stress in vertical = 1- v ^ — ■ — 

8nD(m + 2) ^ Ln 

- pLm(n- m^2)_ _ pnlm(n-m-2) ^.j^^^^ impanel length, 
■" 2n' - 2n=' f s * 

Pm(n — m — 2) 



2n 



, where P = panel load. 



For the problem given in § 193, where n = 8; 

when m = 6, then stress in vertical = o; 
*' m = 5, '' '' " '* =^P 
'' m=:4, '' ** ** *' =iP 
'' m = 3, '* ^' '* *' =:ftP 

Since, by construction, D = — v ? = P» it is seen that the above 

stresses in the verticals are represented by the ordinates to a parabola 
whose ordinates are ^P less than those to the first parabola. 

The above relation may be proved in a different manner, as 
follows: Produce each upper chord member of the eight-panel 
truss (Fig. 130) to intersect the lower chord produced. Load 
the bridge for the maximum compressive stress in each vertical, 
and find the stresses in the verticals by moments. If 1 represents 
a panel length, the moment arms for the members UiLj, UgLg, 
UgLg, and U4L4 are 1, 2 f 1, 5 1, and 16 1, respectively ; the reactions 
for maximum compressive stresses are ^ P, -^ P, -^P, and f P; 
and the stresses are o, -^Pj^ P, and^P. Now the above ordi- 
nates are equal to the ordinates to the given parabola minus the 
first ordinate -^F, 



Art 6. 



WIND LOAD STRESSES IN LATERALS. 



255 



Art. 6. Wind Load Stresses in Lateral Systems. 



The wind load stresses in the upper and lower laterals of a 
bridge will be determined in this article. All the wind load will be 
assumed to act on the windward truss ; although the assumption 
is sometimes made that half of the wind acts on each truss. An 
initial tension is sometimes put into the laterals to give additional 
rigidity to the truss, but will not be considered in this article. 
The upper chord members are a part of the top lateral system, and 
the lower chord members and floorbeams are a part of the bottom 
lateral system. Most specifications state that the wind load 
stresses need not be considered unless they are 25, or, in some 
cases, 30 per cent of the live and dead load stresses. 

198. Upper Laterals. It is required to find the stresses in 
the upper lateral system shown in Fig. 131, a. The panel length 
is 20 ft., and the width is 16 ft. The wind load will be taken at 
150 lbs. per ft., and will be treated as a fixed load. The panel load 
W= 150 X 20 = 3000 lbs. The end strut of the upper lateral 
system is also a part of the portal system, and its stress will not be 
considered at this time. 

X 




Y Hxed Load 

Panel Length »E0', Width* 1 6*. 
W- 3000 lbs. 
Fig. 131. ' Wind Load Strbsses in Upper Laterals. 



(a) Chord Stresses. The fixed wind load stress diagram for 
one-half of the truss is shown in Fig. 131, b. When the wind acts 
in the direction shown, the diagonals represented by full lines are 
in action; and the stress diagram was drawn using these diago- 
nals. When the direction of the wind is reversed, the diagonals 
shown by dotted lines are in action. The stresses in the chord 



256 



STRESSES IX BRIDGE TRUSSES. 



Chap. XXL 



members for the wind acting in the direction shown in Fig. 131, a 
are given in the first line of Fig. 132, a. The stresses in these 
members when the direction of the wind is reversed are given in 
the second line of Fig. 132, a. It is seen that the chord members 
are subjected to reversals of stress due to the wind load. Since 



Stresses in Chord Members 


Chord Member 


Y-l 


Y-2 


Y-3 


x-r 


X-2' 


X-3' 


Wind as shown 
Opposite direction 


- 7.50 
0.00 


-11.25 
+ 7.50 


- 11.25 
-^ 11.25 


0.00 
- 7.50 


+ 7.50 
- 11.25 


-h n.25 

-11.25 



(a) 



Stresses in Web Members | 


Web Member 


I'-l 


1-2' 


2'-2 


2-3' 


3'-3 


Wind as shown 


•f 9.60 


-6.00 


+ 4.S0 


-3.00 


0.00 


When wind comes from opposite direction, other set of diagonals acts. | 



(b) 
FiQ. 132. Table of Stresses — Upper Laterals. 

the chord members are already stressed, it is seen that there are 
actually no reversals of stress unless the wind load stresses are of 
an opposite kind and are larger than the stresses already in the 
chord members. 

(b) Web Stresses. The wind load stresses in the web mem- 
bers may be obtained from the stress diagram shown in Fig. 
131, b. These stresses are shown in Fig. 132, b. When the wind 
acts from the opposite direction, the set of diagonals shown by the 
dotted lines is in action ; and the stresses in them are equal to the 
stresses in the corresponding members when the wind acts in the 
direction shown. 

199. Lower Laterals. It is required to find the stresses in 
the lower lateral system shown in Fig. 133, a. The panel length 
is 20 ft., and the width is 16 ft. The wind load will be taken at 
600 lbs. per ft. Of this load, 150 lbs. per ft. will be treated as a 
fixed load, and 450 lbs. per ft. as a moving load. The fixed panel 
load is 150 X 20 = 3000 lbs., and the moving panel load is 450 X 
20 = 9000 lbs. 

(a) Chord Stresses, (i) Fixed Load. When the wind acts 
in the direction shown in Fig. 133, a, the diagonals shown by full 



Art. 6, 



WIND LO.VD STRESSES IN LATERALS. 



257 



lines are in action. The fixed wind load stress diagram is shown 
in Fig. 133, b; and the stresses in the chord members are shown 
in Fig. 134, a. 

X 




B B' (c) 

Pixed Load Moving Locd 

Fbnel Lenqth - 20'-, Width- 16', W = 3000 lbs-, P« 9000 lbs- 
FiQ. 133. Wind Load Stresses in Lower Laterals. 

(2) Moving Wind Load, The maximum stresses in the chord 
members due to the moving load are obtained when the load 
covers the entire span. Since the moving panel load is three times 
that of the fixed load, these stresses are obtained by multiplying 
the corresponding fixed wind load stresses by three. The stresses 
due to the moving load are given in the second line of Fig. 134, a. 

(b) Web Stresses, (i) Fixed Load. The stresses in the 
lower chord web members for the fixed load of 150 lbs. per ft. are 
obtained from the stress diagram shown in Fig. 133, b. These 
stresses are given in Fig. 134, b. 

(2) Moving Load. The stresses in the lower chord web mem- 
bers for the moving load of 450 lbs. per ft. are obtained from the 
stress triangle shown in Fig. 133, c. This triangle is constructed 
as follows : Lay off A'C equal to twice the width, and B'C equal 
to twice the panel length ; and draw the line A'B'. Then lay off 
AC equal to the moving panel load P, and draw AB parallel to 
A'B' to meet the horizontal line BC. Then AC and AB repre- 
sent the stresses in the struts and diagonals, respectively, due to 
a reaction equal to P. The stresses in these members due to any 



258 



STRESSES IX BRIDGE TRUSSES. 



Chap. XXL 



reaction are obtained by proportion. The stresses in the web 
members to the left of the center of the truss for a moving load 
at each of the lower chord panel points are shown in Fig. 134, c. 
The maximum stress in each member due to the moving load is 
shown in the last line of Fig. 134, c. 



Stresses in Chord Members 



Chord Member 



Fixed WincJ Load 
Movinq Wind Load 
Maximum Stress 
Opposite dinection 



Y-l 



-11.25 
"35.75 
-45.00 
• 0.00 



Y-Z 



-18.75 
-56.25 
-75.00 
+45X)0 



Y-5 



-22.50 
-67.50 
-90.00 
+75.00 



Y-4 



-22.50 
-67.50 
-90.00 
+90.00 



x-r 



0.00 

0.00 

0.00 

-45.00 



X-2' 



+ 11.25 
+33.75 
+45.00 
-75.00 



X-3* 



+16.75 
+56.25 
+75.00 
-90.00 



X-4' 



+22.50 
+67.50 
+90.00 
-9000 



(a) 



Stresses in Web Members -Fixed Lcad 


Web l^ember 


r-i 


1-2' 


2'-2 


2-3" 


3'-3 


5-4' 


4'-4 


Rxed Wind Load 


+ 14.40 


-9.00 


+ 960 


-6.00 


+-4.80 


-300 


000 



(b) 



Stresses in Web Members - Moving Load 


Web Member 


r-i 


1-2' 


2-Z 


2-3* 


3'-3 


3-4' 


4' -4 


Moving Load at Lli 


+ 2.06 


- 1.29 


+ 2.06 


- L29 


+-2J06 


- 1.29 


+ 2.06 


., n \i^ 


+ 4.12 


- 2.58 


+ 4.12 


- 2.58 


+ 4.12 


-2.58 


+ 4.12 


n . [^3 


+ 6.18 


- 3.87 


+ 6.18 


- 3.87 


+ 6.18 


-3.67 


+ 6.18 


• " L^ 


+ 8.24 


- 5.16 


+ 8.24 


-SI6 


+ 8.24 


-5.16 




" " U 


+ IO.30 


-6.45 


+ 10.30 


-6.45 








m n n 1 , 


+ 12.36 


-7.74 












Max Moving Load 


+43.26 


-27.09 


+30.90 


-19.35 


+20.60 


-12.90 


+ 12.36 



(C) 



Maximum Stresses in Web Members 



Web Mennber 



r-i 



1-2' 



2*-2 



Z-V 



S'l 



3-4 



4-4 



Fixed Wind Load 
Movinq ^'nd Load 
Maximum Stress 



+ 14.40 
+43.26 
+ 57.66 



•9.00 
•2T09 
56.09 



+ 9.60 
+30.90 
+40.50 



-6.00 
•19.35 
•25.35 



+ 4.80 
+ 20.60 
+ 25.40 



3.00 
12.90 
15.90 



0.00 
1+12.36 
+ 12.36 



When wind conaes from opposite direction, otlTer set of diaqonais acts 

(d) 



Fig. 134. Table of Stresses — Lower Laterals. 



200. Maximum and Minimum Stresses in Upper Laterals. 

The upper laterals are loaded with a fixed load only, and the maxi- 
mum chord stresses are shown in the first line of Fig. 132, a. The 



Art. 7. 



METHOD OF COEFFICIENTS. 



259 



minimum chord stresses are obtained when the wind acts from 
the opposite direction, and are given in the second line of Fig. 
132, a. 

The maximum web stresses are shown in Fig. 132, b. The 
minimum web stresses are zero. 

201. Maximtun and Minimum Stresses in Lower Laterals. 
The maximum stresses in the lower chord members for fixed and 
moving loads, when the wind acts iij the direction shown, are 
given in the third line of Fig. 134, a. The minimum chord stresses 
are obtained when the wind acts from the opposite direction, and 
are shown in the fourth line of Fig. 134, a. 

The maximum web stresses for fixed and moving loads are 
given in Fig. 134, d. When the wind acts from the other direc- 
tion, the other set of diagonals is thrown into action, and the 
stresses in the first set are zero. 



Art. 7. Stresses in Trusses with Parallel Chords by the 

Method of Coefficients. 



In trusses with parallel chords, the bending moment is resisted 
by the chords, and the shear by the web members. For such 
trusses, the method explained in this article may be used to advan- 
tage for determining the stresses. 

202. Algebraic Resolution — Method of Coefficients. To 



U. 






ut "i Us - 



6 



-^ 




6 '6 "6 '^ P 

Chord Stresses « Coeificients X P tan 6 
Web Stresses « Coefficients x P sec 
Coefficients for a single load 
FiQ. 135. Dead Load Coefficients — Warben Tbuss. 



explain this method, let it be required to find the stresses in the 
members of the truss shown in Fig. 135, the truss being first 



260 STRESSES IX BRIDGE TRUSSES. Chap. XXL 

loaded with a single load P. For this loading, the left reaction is 
equal to ^ P, and the right reaction to f P. For equilibrium, 
the summation of both the horizontal and vertical forces at any 
point must be equal to zero. Resolving the forces at L©, it is seen 
that the stress in the member X-i = — ^ P sec 6, and the stress 
in Y-i = + |. P tan 6; where 6 is the angle that the inclined web 
member makes with the vertical. Using the stress in X-i already 
found, and resolving the forces at Ui, the stress in X-2 = — |P 
tan 6, and the stress in 1-2 = + i P sec 6. Using the stresses in 
Y-i and 1-2 already found, and resolving the forces at Lj, the 
stress in 2-3 = — ^ P sec 0, and the stress in Y-3 = + f P tan 6. 
In like manner, the stresses in the remaining members may be 
determined. Referring to the stresses found above, it is seen that 
all the chord stresses have the common factor P tan 6, and that 
all the web members have the common factor P sec 6. It is 
further seen that these factors are multiplied by coefficients which 
are expressed in terms of the number of panels in the truss. 

These coefficients may be readily found in the following man- 
ner : In this discussion, it should be borne in mind that the chord 
coefficients are to be multiplied by P tan 6, and the web coefficients 
by P sec 6 to get the stresses in the members. Considering the 
forces at Lq, it is seen that R^ = -J- and acts upward, therefore, for 
equilibrium, X-i = ^ and acts downward; as indicated by the 
arrow. Since X-i = ^ and acts toward the left, Y-i = -J- and 
acts toward the right. It is seen from the arrows that X-i is 
compression ( — ), and that Y-i is tension (+). 

Now consider the forces at Uj. It has already been shown 
that X-I is compression, therefore the arrow on X-i will act 
toward the joint. Since X-i = 1. and acts upward, for equilib- 
rium, 1-2 = -J- and acts downward. Likewise, since both X-i and 
1-2 act toward the right, for equilibrium, X-2=^ + ^ = |- and 
acts toward the left. 

Next consider the forces at Lj. Since 1-2= ^ and acts 
upward, 2-3 = -J- and acts downward. Likewise, since Y-i, 1-2, 
and 2-3 all act toward the left, for equilibrium, Y-3 == |- -)- |- + 
^ = {^ and acts toward the right. 



Art, 7. 



METHOD OP COEFFICIENTS. 



261 



In like manner, the coefficient and the kind of stress may be 
found for each member (see Fig. 135). 

The chord stresses may be obtained by multiplying the coeffi- 
cients by P tan 6, and the web stresses, by multiplying the coeffi- 
cients by P sec $. 

The coefficients for any loading may be found in the manner 
indicated above. 

203. Loading for Maximum and Minimtun Live Load 
Stresses. The coefficients for all the members for a load 
applied at each lower chord joint in turn are shown in Fig. 136. 
The coefficients shown in the top line are those for a load on the 
left of the truss. The stresses in the chord and web members may 
be obtained by multiplying the coefficients by P tan 6 and P sec 6, 
respectively. 




Chord Stresses - Coefficients x P Tan 6 
Web Stresses = Coefficients X P sec 6 
Coefficients for load at each panel point 

Fia. 136. Live Load Cobppicibnts — Warren Truss. 



The Stress in any number due to any loading may be readily 
found from Fig. 136. It is seen that each load produces compres- 
sion in the upper chord and tension in the lower chord. The max- 
imum stresses in the upper and lower chords are therefore 
obtained when the truss is fully loaded. Referring to Fig. 136, 
it is seen that the maximum live load stresses in both X-i and 1-2 



262 STRESSES IX BRIDGE TRUSSES. Chap. XXI. 

are obtained when the truss is fully loaded ; since each load pro- 
duces compression in X-i and tension in 1-2. The minimum live 
load stresses are zero, when there are no loads on the truss. The 
maximum live load stresses in 2-3 and 3-4 are — ^^^ P sec 6 and 
-f "VP sec By respectively, when the loads at Lg, L3, Lg', and Lj' 
are on the truss. The minimum Uve load stresses in 2-3 and 3-4 
^^^ + i P sec ^ and — -JP sec 6, respectively, when the load at Lj 
is on the truss. When the truss is fully loaded, the stress in 2-3 
is ( — V + i ) P sec 6 = — -j-P sec 0, and the stress in 3-4 is 
(-f-i^ — ^) p sec = -\-^P sec 6. The maximum stresses in 
4-5 and 5-6 are — f P sec ^ and + 1" P sec 0, respectively, when 
the loads at Lg, L/ and Li' are on the truss. The minimum 
stresses in 4-5 and 5-6 are + f P sec ^ and — f P sec 6, respect- 
ively, when the loads at L^ and Lj are on the truss. When the 
truss is fully loaded, the stress in 4-5 is ( — 1--{-|-) P sec 6 = 
— f P sec 6f and the stress in 5-6 is (+ i — | ) P sec ^ = + f 
P sec 6. 

Conclusions. The following conclusions may be drawn from 
the live load coefficients shown in Fig. 136. 

( 1 ) For maximum live load chord stresses, load the truss 
fully tvith live load. For minimum live load chord stresses, there 
should he no live load on the truss. 

(2) For maximum live load iveh stresses, load the longer 
segment of the truss only. 

(3) -P^^ minimum live load web stresses, load the shorter 
segment of the truss only. 

Simplified Method. Instead of considering the coefficients for 
the member due to each separate load, it is usually more convenient 
to find the coefficients directly for all the loads. The dead load 
coefficients for the truss may be readily found by first finding the 
effective reactions in terms of the panel load, and then determining 
the coefficients from the end toward the center of the truss (see 
Fig. 137, a). The live load coefficients for the chords are the 
same as those for the dead load. The live load coefficient for the 
maximum or minimum stress in any web member may be readily 
found from the formula : 



UBTUOD OF COEyFICIEXTS, 

Live load web coefficient;^ , where 



(^)" 



m = number of loads on truss for a 

stress in any web member, and 
n = number of panels in truss. 
_-9 . 



or mmimum 




Chord Stresses -Coefficients x W tan6 
Web Stresses =Coefficier>t5 x W sec9 

DEAD UOAD COEFFICIENTS 




Chord Stresses -Coefficients X P tan 9 
Web Stresses = Coefficients x P sec 9 

UVE LOAD COEFFICIENTS 



D.L - 6500 

LL -20000 ^_ „,-.. 

Max- 26500 MaK^4ZOO0 

I Min-- b'bOO Uz Min- 10000 



10000 DL -12500 

■3ZDD0 L-L -4000D 

Max- 5? 500 

12500 




- 5100 Li Ol + 8100 L2 DL +10600 Li 

L.I; +10000 LL +260OO LL +54000 

Max + 13100 Moxt MlOO Max+44fciO0 

Min + 3100 Mm* StOO (;;) Min+i06O0 

Leno|ih= 60 fti Heiqht = 10 ft-. WiOth - 16 ft- 
W- 2500 lbsiP-&000 Ibs-iTon 9-0500; Sec 6 - II 15 
FiQ. 137. Maximum ahd Minimum STBBSSEia — Wabbem Tbdss. 



264 STRESSES IX BRIDGE TRUSSES. Chap. XXI. 

The second differences of the maximum coefficients in the web 
members are constant, which furnishes a check on the work. The 
numerators of these coefficients are i, 3, 6, 10, etc., while the 
denominator of each is the number of panels (see Fig. 136). 

204. Coefficients and Stresses in a Warren Truss. It is 
required to find the maximum and minimum stresses in all the 
members of the Warren truss shown in Fig. 137. The truss has 
a span of 60 ft. ; a height of 10 ft. ; and a panel length of 10 ft. 
The dead panel load is 2500 lbs., and the live panel load is 
8000 lbs. 

The dead load coefficients are shown in Fig. 137, a. The dead 
load chord stresses are obtained by multiplying the chord coeffi- 
cients by W tan 6, and the dead load web stresses, by multiplying 
the web coefficients by W sec 6. These stresses are shown on 
the truss diagram in Fig. 137, c. 

The live load coefficients are shown in Fig. 137, b. The chord 
and web stresses are obtained by multiplying the coefficients by 
P tan 6 and P sec 6, respectively. These stresses are shown in 
Fig. 137, c. Instead of placing both the maximum and minimum 
coefficients on the same member, the maximum coefficients are 
placed on the member and the minimum coefficients on the corre- 
sponding member on the other side of the center line. 

The maximum and minimum stresses are obtained by combin- 
ing the corresponding dead and live load stresses. These stresses 
are shown in Fig. 137, c. 

205. Coefficients for a Pratt Truss. The dead load co- 
efficients for a Pratt truss are shown in Fig. 138, a. The hip 
vertical acts as a hanger to support a single panel load, and its 
coefficient is unity. The dead load chord and web stresses may 
be obtained by multiplying the coefficients by W tan $ and W 
sec 6y respectively. 

The live load coefficients are shown in Fig. 138, b. The coeffi- 
cients for maximum live load stresses are shown on the left, and 
those for minimum stresses, on the right of the center line. The 
live load chord and web stresses may be obtained by multiplying 
the coefficients by P tan and P sec Oy respectively. 

206. Coefficients for a Baltimore Truss. The dead load 



AH.7. 



METHOD OF COEFFICIENTS. 



265 



coefficients for a through Baltimore truss are shown in Fig. 139, a. 
The signs placed before the coefficients and the arrows on the 
members indicate the kind of stress. The sub-members in this 
truss carry the loads to the main members and are not a part of 
the main truss system. Care must be used in determining the 
coefficients, as some of the sub-members carry the loads toward 
the center of the truss, and the loads are then carried back to the 
abutment by the main members. The dead load chord and web 
stresses may be obtained by multiplying the coefficients by W tan B 
and W sec tf, respectively. 



-5 _ -6 . -6 -6 




\0 Q,' 



\ / 

X 



/ \ 




W '"^ W -^5 y^^^t ^ +5 w •-» w 

Chord Stresses = Coef tic lents x W tan 
Web Stresses = Coef fiaents xWsec0 

DEAD LOAD COEFFICIENTS 



Rj* 




Chord Stresses = Coefficients x P tan 8 
Web Stresses = Coefficients x P sec 6 

UVE LOAD COEFFICIENTS 
FiQ. 138. Dead and Live Load Coefficients — Pratt Truss. 



The live load coefficients are shown in Fig. 139, b. The live 
load chord coefficients are the same as those for the dead load. 
The coefficients for maximum live load stresses are shown on the 
left, and those for minimum live load stresses, on the right of 
the truss. The maximum live load stresses in the diagonal web 
members whose coefficients are +ff ^"^ +f|2ir^ b^^h obtained 
when the loads extend to the right of the member whose coeffi- 
cient is +f}. Likewise, the maximum live load stresses in the 



8TEE88ES IK BRIDGE TRUSSES. 



Chap. XXI. 



members whose coefficients are — ti; + fij ^^'^ + ti ^''^ ^' 
obtained when the loads extend to the right of the member whose 
coefficient is+ f|. The former bading in this particular case 
will give the same coefficient. The apparent peculiarity of the 
-lOt -\Zi -IZt -IZi _ -IQj 




Chord Stressea - Coefficients X W tan 9 

Web Stresses - Coefficients X W sec 9 

DEAD LQ^O COEFFICIENTS 

-12i 



y 


'^W 


^ 
"W^ 




^ 


)r 


yk 


^ +6i 
R. P 


' P 


♦10 *10 
> P F 


IWP p p P P P...P 1 



""" Chord Stresses » Coefficients X P tan 9 '^'"' 
Web Stresses - Coefficients X P sec 6 

LIVE LOAD COEFFICIENTS 
Fio, 13S. Dud and Livg Load CoBPPicniHTS — Ualtiuobe Tbusb. 

loading for maximum stresses is due to the sub-members. Care 
should be used in getting the coefficients for minimum web 
stresses in this type of truss. The live load chord and web stresses 
may be obtained by multiplying the coefficients by P tan and 
P sec d, respectively. In making the combinations for maximum 
and minimum dead and live load web stresses, care must be used 
in determining whether or not the counters are acting. 



CHAPTER XXII. 

INFLUENCE DIAGRAMS, AND POSITIONS OF ENGINE AND 
TRAIN LOADS FOR MAXIMUM MOMENTS, SHEARS, 

AND STRESSES. 

This chapter will treat of the construction of influence dia- 
grams, and of their use in determining the positions of wheel 
loads for maximum moments, shears, and stresses in girders and 
trusses. When a series of concentrated loads moves across a 
girder-bridge, the maximum moments and shears at various 
points are usually given by different positions of the wheel loads, 
and in many cases by a different wheel load at each point. Like- 
wise in the case of trusses,, the maximum moments in chord 
members and maximum shears in web members are usually given 
by different positions of the loads, and often by a different wheel 
load at the various panel points. It is possible to derive criteria 
for the positions of wheel loads for maximum moments and 
shears in girders and trusses, and this chapter will take up the 
determination of such criteria by means of influence diagrams. 

207. Influence Diagrams.* Definitions. An influence dia- 
gram is a diagram which shows the variation of the effect at any 
particular point, or in any particular member, of a system of 
loads moving over the structure. Influence diagrams representing 
the variations of bending moments and shears in trusses and 
beams are commonly used, and will be taken up in this chapter. 
The difference between an influence diagram and a bending mo- 
ment or shear diagram is that the former shows the variation of 
bending moments or shears at a particular point, or in a particu- 



*For a full discussion of influence diagrams, see a paper by G. F. Swain, 
Trans. Am. Soc. C. E., July, 1887. 

267 



268 



INFLUKNCE DIAGRAMS — POSITION OF LOADS. 



Chap, XXII. 



lar member, for a system of moving loads ; while the latter shows 
the bending moments or shears at different points for a system of 
fixed loads. 

The influence diagram is usually drawn for a load unity, and 
in this chapter only unit influence diagrams will be considered. 
The moments, shears, or stresses for any system of moving loads 
may be obtained from the intercepts in the unit influence diagram 
by multiplying them by the given loads. 

The equation representing the influence diagram at any point 
may be derived by writing the equation for the function due to a 
load unity at the point. 

The principal use of influence diagrams is to find the position 
of a system of moving loads which will give maximum moments, 
shears, or stresses ; although they may also be used to determine 
the values of these functions. 

208. Position of Loads for a Maximum Moment at Any 
Point in a Beam, or at Any Joint of the Loaded Chord of a 
Truss with Parallel or Inclined Chords (a) Concentrated 




p per ft. |"0! 
^^ C" ' 



dx b 



—4 



Fig. 140. Influence Diagram — Moment at Loaded Chord Joint b'. 



Moving Loads, Let 2 Pi (Fig. 140, a and Fig. 140, b) be the 
summation of the moving loads to the left of the point b' of the 



MAX. MOMENT — LOADED CHORD JOINT. 269 

truss or beam, and let S Pj be the summation of the moving loads 
to the right of b'. It is required to draw the influence diagram 
for the bending moment at b' ; also to determine the position of 
the moving loads for a maximum bending moment at b'.J 

To construct the influence diagram (Fig. 140, c) for the 
bending moment at b', compute the moment at b^ for a load unity 

d(L — d) 
at that point. This moment = z = ordinate be. Draw 

the horizontal line ac, lay off the ordinate be, and draw the lines 
ae and ce. Now when the load unity is to the left of b', the bend- 
ing moments at b' are represented by the ordinates to the line ae ; 
and when the load unity is to the right of b', by the ordinates to 
the line ce. 

A convenient method for drawing the influence diagram with- 
out actually computing the moment will now be shown. Let y 
represent the ordinate to the line ae, and x, the distance from the 
load unity to the left end of the span. Then the equation of the 
line ae (i. e., for the load unity to the left of b') is 

(L-x )^ , . , x(L-d) 

y= — j—d— (d— x)= — J- , (i) 

and the equation of the line ce (i. e., for the load unity to the 
right of b') is 

d(L-x) 

y= — L • (^) 

When x = d, these two equations have a common ordinate = 

d(L — d) 
= . When x = L, the ordinate to the line ae is L — d; 

and when x = o, the ordinate to the line ce is d. Therefore, to 
construct the influence diagram (Fig. 140, c), lay off at a the 
distance am = d, and at c, lay off en = L — d. The intersection 
of an and cm will then locate the point e of the influence diagram. 



tTbe criterion determined in this section applies to the unloaded chord 
joints if they are on the same vertical lines as those in the loaded chords. 



270 INFLUENCE DIAGRAMS — POSITION OF LOADS. Chap. XXII. 

The moment at any point along the truss or beam due to any 

number of moving loads may be found from the unit influence 

diagram by multiplying the ordinate under each load by the load 
and taking the sum of these products. 

The position of the loads for a maximum bending moment at 
any point b' will now be determined. Since 2 Pi and 2 Pg repre- 
sent the summation of all the loads to the left and to the right of 
b', respectively, they will have the same effect as the separate 
loads. The bending moment at b' due to the loads 2 Pi and 
2P2is 

M = 2Piyi + 2P,y,. (3) 

Now let the loads be moved a small distance dx to the left, 
the movement being so small that none of the loads pass a', b', 
or c'. Then the bending moment is 

M + dM = 2 Pi ( yi - dyi ) + 2 P, (y, + dy, ) . (4) 

Subtracting equation (3) from equation (4), we have 

dM= — 2Pidyi + 5P2dy2. (5) 

Referring to Fig. 140, c, it is seen that dyi = dxtan cx^, 

be L— d 

and that dyo = dxtan oc.,. But tan ocj = — —= — , and tan 

ab L 

be f\ 
oc 2 = T" ="■" • Substituting these values of dyi and dyg in equa- 

dM 
tion (5), dividing through by dx, and placing— j — =0 for a maxi- 
mum, we have 

-^=5P,(^)+2P,(d)=o. (6) 

Solving equation (6), we have 

5P.L=(2P, + 2P,)d, 

2 P. 2 Pi + 2 P, 
or — p= = . (7) 



MAX. MOMENT — LOADED CHORD JOINT. 271 

Equation (7) is the criterion for the maximum bending mo- 
ment at b' in a truss or beam. Since b' is any point along the 
truss or beam, this criterion expressed in words is — the maximum 
bending moment occurs zvhen the average load to the left of the 
point is equal to the average load on the entire span. 

For a given system of moving loads, it may happen that two 
or more positions of the loads will satisfy the criterion. In this 
case, the moment for each position must be found. As soon as 
the position of the loads has been determined, the bending moment 
may be found, either algebraically or graphically, by the methods 
given for fixed loads. 

If the truss has equal panels, the most convenient unit of 
length to use is a panel length. For a truss with unequal panels, 
or for a beam, the common unit of length is the foot. 

(b) Uniform Load. The moment at b' (Fig. 140) for any 
length of uniform load of p pounds per linear foot is equal to 
the area of the influence diagram included between the extreme 
ordinates of the uniform load multiplied by the load per linear 
foot. For, the bending moment due to a length dx of the uni- 
form load = pydx, which is the area under that load. If the 

length of the uniform load=l, the moment at b'= j^^^pydx 

= p X area of influence diagram under the uniform load. For a 
maximum bending moment at any point, it is seen that the span 
must be entirely covered with the uniform load. The maximum 
bending moment at b' = p X area of influence diagram aec. 

209. Position of Loads for a Maximum Moment at Any 
Joint of the Unloaded Chord of a Truss with Parallel or 
Inclined Chords. Let 2 P be the total moving load on the 
truss ; let 2 Pi be the summation of the moving loads to the left 
of b' (Fig. 141, a) ; let 2 Pj be the summation of the moving 
loads in the panel b'c' ; and let 2 P3 be the summation of the 
moving loads to the right of c' in any truss with horizontal or 
inclined chords in which the unloaded chord joints are not ver- 
tically over those of the loaded chord. Also, let r = the horizontal 
distance from e' to the next loaded chord joint toward the left 
end of the truss; let s = the horizontal distance from e' to the 



272 



IXFLL'ENCE DIAOHAMS — POSITION OF L0AD8. Chap. XXIL 



left end of the truss ; and let L = the span of the truss. It is 
required to draw the influence diagram for the moment at e', and 
to determine the position of the moving loads for a maximum 
moment at e'. 

Now it is seen that the moment at e' for a load unity moving 
from b' to a' and from f ' to c' is the same as for a simple truss. 
The portions of the influence diagram for these parts of the 
truss are ag and fh, respectively, which are portions of the 
influence diagram aof. The load in the panel b'c' is carried to 
the panel points b' and c' by the stringer; and the influence line 
for this part of the truss is therefore the straight line gh. 

The criterion for a maximum moment at e' may be deter- 
mined in the following manner : Let the loads be moved a small 





• Influence Diaqram 
for 
h Moment at Joint e' 

(b) 



<0i 
I 



LA 



Fig. 141. INFLUENCB Diagram — Moment at Unloaded Chobd Joint e'. 



distance dx toward the left end of the truss from the position 
shown in Fig. 141, a. The increase in the moment at e' will be 

dM = S Pgdx tan a 3 — 2 Pgdx tan a 2 — 2 Pidx tan oc ^, 
For a maximum, 

=SP8tan cxg — SPgtan CX2 — SPitan oci = o. (i) 

dx 



MAX. MOMEXT AT UXLOADED CHOBD JOIXT. 273 

T» 8 he — gb L — S 

But tan «,=- ; tan <^2= 3 ; and tan ax=s — 

Also, 

he = (L — s + r — d) tan oc 3, and gb = (s — r) tan oc ^. 

Substituting these values of he and gb in the expression for 
tan 0C2, and then substituting the values of tan a, and tan ccj, 
we have 

rL — sd 



tan ««= — . 

qL. 

Substituting the values of tan 0C3, tan ocj, and tan cc^ in 
equation (i), we have 

_ s ^ rL — sd L — s 

Now substituting 2 P for 5 Pi + S Pj + 2 Pj, and reducing, 
we have 

5 Pj 4 + 2 Pi 
2P d 

-^ : =0, (2) 



2P SP,-^ + 2P, 
o'- il= (3) 



which is the criterion for a maximum moment at any joint of the 
unloaded chord. 

2P SP,;+2P. 



Referring to equation (2), it is seen that-y- 

must become o by passing from positive to negative, and that 
this can only occur when some load passes c' or b'. 

210. Position of Loads for a Maximum Moment at a 
Panel Point of a Truss with Subordinate Bracing. It is 
required to determine the position of the wheel loads for a 



274 INFLUENCE DlACiHAMS — POSITION OF LOADS. Chap. XXII. 



maximum moment at the panel point g' of the truss with subor- 
dinate bracing shown in Fig. 142, a. A truss with subordinate 
bracing is one which has points of support for the floor system 
between the main panel points. It is seen that g' is the center 
of moments for determining the stress in the lower chord member 
b'e'. The portion of the load in the panel c'e', carried to c', is 
on the left of the section p-p; and its moment must therefore 
be considered in getting the moment at g'. 




* 1. tV ♦ 



/0C2 ! 




Influence Diagrxam 

for 
Moment at Joint q' 

(b) 



a b e i 

Fig. 142. Influence Diagram — Truss with Subordinate Bracing. 



The influence diagram for the moment at g' (Fig. 142^, b) 
is drawn as follows: For a load unity moving from a' to b', 

the moment at g' increases from o to ^^ Ij and ag is the 

influence line for the segment a'b'. Likewise, the influence 
line for a load unity moving from f to e' is fh. As the 
load unity moves from b' to c', the moment at g' of the 
forces to the left of section p-p is increased at the same rate 
as from a' to b'; and gm is the influence line for the segment 
b'c'. The influence line for the segment c'e' is the straight 
line mh. 

To determine the position of the wheel loads for a maximum 
moment at g', let 2 Pj, 2 Pj, 2 Pg, and 2 P be the loads in the 



MAX. MOMENT — SUBORDINATE BRACING. 



275 



segment a'b', in the panel c'e', in the segment e'f, and on the 

entire span, respectively. Now if the loads are moved a small 

distance dx to the left, the moment at g' will be increased by 

dM = 2 Padx tan « , -f 5 P^dx tan oc ^ — J P^dx tan oc ^. 

For a maximum moment, 

dM 
-J — =SP3tan ocj + SPgtan cxj — SPjtan oc^=o, (i) 

But, 



L — s 



tan oc 3 == — ; tan oc ^ = — — 



and tan oc„== 



nr + rh d tan a ^ -f 2d tan a 3 L + s 



Substituting these values in equation ( i ) and reducing, we have 

2P 2Pi — 2P2 

— = s ■ (^) 

which is the criterion for a maximum moment at g\ This 
criterion will be satisfied when some wheel load is at the panel 
point c'. 

211. Position of Loads for a Maximum Shear at Any 
Point in a Beam, (a) Concentrated Moving Loads, It is 
required to find the position of the loads for a maximum shear 
at any point O (Fig. 143, a). 

M (^ 000 n o o N 



\-.-.-.-A 



^0 



L 




ll 



1. 



Influence Diagram 

Shear otO 
(b) 




- — 1-1. 



Fio. 143. Shear in a Beam. 



The unit influence diagram (Fig. 143, b) will first be drawn. 
For a load unity entering the beam at N and moving along the 



276 INFLUENCB DIAGRAMS — POSITION OF LOADS. Chap, XXII. 

L — S 

span, it is seen that the shear increases from zero at N to -j z — 

when the load is just to the right of O. When the load passes 
O, the shear at that point is suddenly decreased by the amount 

of the load unity, and is equal to — r- , and then increases to 

zero at M. For a system of moving loads, it is seen that the 
positive shear increases as the loads move toward the left until 
a load passes O, when it is suddenly decreased by the amount 
of the load. The shear therefore reaches a maximum every time 
a load reaches O. Now it may happen that, even though the 
loads are moved to the left and a load passes O, the reaction 
will be increased sufficiently to increase the shear at O. 

A method will now be shown for determining which of the 
two consecutive loads, Pj or Pj, when placed just to the right 
of O, will give a maximum shear. 

Let 2 Pi = total load on the beam when P^ is at O. Now 
move the loads to the left until Pg is just to the right of O. If 
no additional load moves on the beam and no load moves off, 
the shear will at first be suddenly decreased by Pj, and then 

SPib 
gradually increased by an amount equal to 2 Pib tan oc = — - — . 

SP,b 
(see Fig. 143, b). The total increase in shear will be -^ Pi. 

Pi SPi 
If this expression is negative, i. e., if -r- > ^j: — , then P^ placed 

at O will give the maximum shear; and if it is positive, i.e., if 

Pi 5 Pi , Pi SPj 

-j— < -r= — , then Pg will give the maximum shear. IfT--= -y— ^, 

then both wheels give equal shears. 

If an additional load moves on the beam and no load moves 
off when Pg moves up to O, the expression representing the 

S Pib P„x 
increase in shear will be —7 h -7 Pi ; where P^ is the 



MAX. SHEAR AT ANY POINT IX A BEAM. 277 

load which enters the span, and x is the distance from the right 
end of the beam to the load P^. If 5 Pg is the total load on the 
beam when wheel Pj is at O, then the increase in shear will lie 

S Pjb 2 P^b 
between — = P^ and — =: P^. WTien the former expres- 
sion is negative and the latter positive, then both positions of 
the loads should be tried. This condition will occur for only a 
short distance, to the right of which both expressions are nega- 
tive, and to the left, both are positive. Wheel i at the point 

, Pi_5Po 
will give a maximum shear when-^— — -=:— , and wheel 2 will 

b > L 

Pi_SPi 
give a maximum shear when-r—"— r — . 

(b) Uniform Load, From Fig. 143, b, it is seen that the 
maximum shear at any point O due to a uniform load will occur 
when the load extends from the right end of the beam to the 
point O. The minimum shear will occur when the load extends 
from the left end of the beam to the point O. 

212. Position of Loads for a Maximum Shear in Any 
Panel of a Truss with Parallel or Inclined Chords, (a) Con- 
centrated Moving Loads. Let Fig. 144, a represent a truss 
loaded witli concentrated loads. It is required to determine the 
position of the loads for a maximum shear in any panel, say 
b'c'. Let 2 Pi represent the total load on the left of this panel, 
S P2 that on the panel, and 5 ?« the total load on the right of the 
panel b'c'. Let m = number of panels on the left of the panel 
b'c', and n = total number of panels in the truss. 

The influence diagram (Fig. 144, b) is constructed, as fol- 
lows: For a load unity moving from a' to b', the shear in the 

m 

panel b'c^ increases from zero when the load is at a' to 

'^ n 

when the load is at b' ; and ab is the influence line for the load 
to the left of the panel b'c'. For a load unity moving from iV 

n — m — I 

to c', the shear increases from zero to + ; and dc 

n 



278 INFLUENCE DIAORAKS — POSITION OP LOADS. Chap. XXII. 



is the influence line for the loads to the right of the panel b'c'. 

m 
For a load unity on the panel b'c', the shear varies from 



for the load at b' to + 



n 



m — I 



n 



for the load at c' ; and be is 



the influence line for the load on this panel. The load in the 
panel b'c' is carried to the joints b' and c' by the stringers, the 
amount of load transferred to each varying inversely as the dis- 
tance from the load to the joint. 

The influence diagram for the entire span is abed, and it is 
seen that the lines ab and be are parallel, and that the vertical 
distance between them is unity. 

Referring to the influence diagram (Fig. 144, b), it is seen 
that for a maximum positive shear in b'c' due to a moving load, 







*_, 



Influence Diagram 
Shear in b'c' 
(b) 




* ^ ^. 



Fio. 144. Shear in a Tbuss. 



---J-i 



the load should be placed at c'; and for a maximum negative 
shear, the load should be placed at b'. 

The maximum positive shear in the panel b'c' is 

S = SP3y, + SP^,-SP,y,. (i) 

Now move the loads a small distance dx to the left, the dis- 



ICAX* SHEAR IN ANY PANEL OF A TBUSS. 279 

tribution of the loads remaining the same as before. The shear 
is then 

S + dS = SP. (y.+dy.) + SP, (y,-dy,) -SP, (y,-dy,) (2). 
Subtracting equation (i) from equation (2), we have 

dS = S P,dy, - S P,dy, + S P,dy,. (3) 

But dy3 = dxtan oc3 = dx— 5. 

nd 

■> 

n — I 



dy2 = dxtan oc2 = dx — -r— , 



I 

and dyi = dxtan oc- = dx — -. 

•^* nd 

Substituting these values of dyg, dyj, and dyi in equation (3), 
and dividing through by dx, we have 

ds 
Placing -j— = o for a maximum, and multiplying through by 

nd, we have 

5P3 — SP^Cn — i)+2Pi = o, (5) 

S Pi + 2 P2 + 2 P3 

from which 2 P^ = —^ — * (6) 

* n 

which is the criterion for the maximum shear in any panel. This 

criterion expressed in words is — the maximum shear in any 

panel of a truss will occur when the load in the panel is equal to 

the total load on the truss divided by the number of panels. This 

is equivalent to saying that the average load in the panel must be 

equal to the average load on the entire bridge. 

This criterion requires that some wheel near the head of the 

train shall be at the panel point to the right of the panel in 

which the shear is required. Any particular wheel P placed at 

the right-hand panel point will give a maximum shear in that 

panel if the entire load on the bridge lies between 2 Pon and 



280 INFLUENCE DIAGRAMS — rOSITION OF LOADS. Chap. XXII. 

(2 Pj + P) n ; where 2 P^ is the load in the panel other than P, 
and n is the number of panels in the truss. 

(b) Uniform Load. Referring to Fig. 144, b, it is seen that 
the maximum shear in the panel b'c' due to a uniform load will 
occur when the load extends from the right end of the truss to 
the point h, i. e., to the point where the shear changes sign. The 
minimum shear will occur when the uniform load extends from 
the left end of the truss to the point h. By minimum shear is 
meant the greatest shear of an opposite kind. 

213. Position of Loads for a Maximum Stress in any Web 
Member of a Truss with Inclined Chords. The criterion for 
maximum stresses in chord members is the same for trusses 
with parallel chords as for those with inclined chords. The 
criterion for maximum stresses in web members for trusses with 
inclined chords differs from that for those with parallel chords; 
since in the former case, the inclined chord members take a part 
of the shear. 

Let g'c'^ (Fig- I45> a) be any web member of a truss with 
inclined chords ; let p-p be a section cutting g'h', g'c', and b'c' ; 
and let O be the point of intersection of g'h' and b'c', which is 
the center of moments for determining the stress in g'c'. Also 
let 2 P be the entire moving load on the truss when there is a 
maximum stress in g'c' ; let S Pi be the summation of the mov- 
ing loads in the panel b'c' ; and let S Pj be the summation of 
the moving loads to the right of c'. It is required to draw the 
influence diagram for the moment at O, and to determine the 
position of the moving loads for a maximum stress in g'c'. 

The stress in g'c' is equal to the moment of the external 
forces to the left of the section p-p about the point O divided 
by the moment arm r; and the stress in this member is a 
maximum when the moment at O is a maximum. 

L — s — d 
The moment at O for a load unity at c'=i =^ k, 

L — s 
and for a load unity at b' = -j ^ — k — k — s^ The mo- 



MAX. WEB STRESS — INCLINED CHORDS. 



281 



nient passes through zero when the load is at some point between 
b' and c'. The influence diagram for the moment at O may 
therefore be constructed by laying off ch (Fig. 145, b) = 

L — s — d L — s 
1 k (downward from c), bg = — z — k — k — s (up- 
ward from b), and drawing ag, gh, and hf. 



<y 




Fio. 145. Stress in Web Member — Truss with Inclined Chords. 

The maximum stress in g'c' occurs when some of the wheels 
near the head of the train are in the panel b'c'. It will be 
assumed in this case that there are no loads to the left of b'; 
as this is the usual condition. 

The criterion for a maximum stress in g'c' may be deter- 



282 INFLUENCE DIAGRAMS — POSITION OF LOADS. Chap, XXIL 

mined in the following manner: Let tlie loads be moved a 
small distance dx toward the left end of the truss from the 
position shown in Fig. 145, a. The increase in the moment at 
O will be 

dM ^ S Pjdx tan ex , — 2 P^dx tan oc ^. 
For a maximum, 

dM 

-j — = SP2tan 0C3 — 2 Pi tan a 2 = 0. (i) 

ch k L — s — d 
But tan oc 3 = := = — — , since ch = = k ; 



and tan 0^2 = 



ch + gb 



/L — s — d,v /L — s V 

-(^j— k) + (-j_k-k-s) 



k k + s 

Substituting these values of tan 0C3 and tan ocg in equation (i), 
and putting 2 P instead of 2 Pi + 2 P2, we have 

— =0, 



or 



2P 



d 



2P 



■0+1) 



(2) 



L d 

which is the criterion for a maximum moment at O and therefore 
for a maximum stress in the member g'c'. 

By comparing the above criterion with that given for a 
maximum shear in § 212, it is seen that they are very similar, 
the only difference being that in this case the load in the panel 

is to be increased by the — th part of itself before dividing by 
the panel length. 



MAXIMUM FLOORBEAM REACTION. 



283 



To get the maximum stress in the member g'b' (Fig. 145, a), 
the same panel b'c' is partially loaded and k is replaced by k', 
the other quantities remaining the same as for g'c'. 

214. Position of Loads for a Maximum Floorbeam Re- 
action. It is required to find the maximum load on the floor- 
beam at O (Fig. 146, a) due to loads on the panels MO and ON. 
The loads are carried to M, O, and N by the floor stringers. 




e a' e' 

(Q) Influence Diagram for jjjj 

Shear at Moment at 0' 

FiQ. 146. Floobbeam Reaction. 



b 



Fig. 146, a shows the influence diagram for the floorbeam 
load at O. The influence line for the load in the panel MO 
is ac ; and that for the load in the panel ON is cb. The ordinate 
ce is equal to unity. 

Let M'N' (Fig. 146, b) be a beam whose length di + ^2 is 
equal to the length of the two panels MN of the truss, and let 
a'b'c' be the influence diagram for the bending moment at O', 
whose distance from the left end of the beam equals dj. The 

djd, 

ordinate under O' equals -z — — V • 

di + d^ 

By comparing the diagrams in Fig. 146, a and Fig. 146, b, 
it is seen that they differ only in the value of the ordinates ce 
and c'e'. It is also seen that the maximum floorbeam reaction 
occurs for the same position of the loads as does the maximum 
bending moment. The ratio of any two corresponding ordinates 



i 



284 INFLUENCE DIAGRAMS — POSITION OF LOADS. Chap. XXII. 

djdj di -j- dg 

Vi and ya is as ce is to c'e' = i -7— ^ — p^-=- , , — . 

dj -f- dj didj 

Therefore, the maximum floorbeam reaction may be obtained by 

finding the maximum bending moment at a distance d^ from 

the end of a beam whose length is equal to the sum of the two 

di + d., 
panel lengths dj + dj, and multiplying this moment by — -r-^ 

If the two panel lengths are equal, the maximum moment at 

the center of the beam should be multiplied by _ where d is 

equal to the panel length. 



CHAPTER XXIII. 

MAXIMUM MOMENTS, SHEARS, AND STRESSES DUE TO ENGINE 

AND TRAIN LOADS. 

This chapter will treat of the determination of maximum 
moments, shears, and stresses due to engine and train loads. 
Graphic methods of applying the criteria derived in Chapter XXII 
will also be shown. The dead load moments, shears, and stresses 
will not be considered; as they may be found by the methods 
already explained. 

The chapter will be divided into two articles, as follows: 
Art. I, Maximum Moments, Shears, and Stresses in Any Par- 
ticular Girder or Truss ; and Art. 2, Maximum Moments, Shears, 
and Stresses in Girders and Trusses of Various Types and Spans. 



Art. I. Maximum Moments, Shears, and Stresses in Any 

Particular Girder or Truss. 

When it is required to determine the maximum moments, 
shears, and stresses in any particular span, the methods which 
will now be given will be found convenient. These methods will 
be explained by the solution of two problems, involving (i) the 
plate girder, and (2) the Pratt truss. 

215. (i) Maximum Flange Stresses and Shears in a 
Plate Girder. Let Fig. 147, a represent one girder of a single- 
track deck plate girder-bridge whose span, center to center of end 
bearings, is 60 ft., and whose effective depth is 6 ft. It is required 
to find the maximum flange stresses and shears at the tenth points 

285 



36 MAX. MOMENTS AND SHEARS — ENGINE LOADS. Chap. XXIII. 

abng the girder due to the engine and train loading shown in 
Fig. 147, c, the loading being that for one girder. 

Flange Stresses. The diagram for flange stresses is shown 
in Fig. 147, b. To construct this diagram, lay off the engine dia- 








Plate Oirdcr unkr Ensinc and Train Load - — . . ____ 

2par,.W-0-Depthteff).e-0-,Loadrngas5ho«n. ^'« "' ^°°'^'' °^ ^^^^ 
All loods and srmses are In rhoiaands of pounds- 

FlO. 147. MlXIHUU FlAHQE STSHBSEB AKt S)IE4BS IN 1 PI.ATK OlRDEB. 



Art, t. PLATE GIRDER — MAXIMUM FLANGE STRESSES. 287 

gram (Fig. 147, c) to any convenient scale, and mark the begin- 
ning, middle point, and end of the uniform load. Since the span 
of the girder is 60 ft., it will be necessary to consider only about 
20 ft. of uniform load. Lay off the load line (Fig. 147, e) for 
the given engine and 20 ft. of train load, and with a pole distance 
equal to some multiple of the effective depth of the girder (in this 
case four times the depth), construct the funicular polygon 
ABGKO (Fig. 147, b). The portion of the funicular polygon 
KO for the uniform train load is an arc of a parabola, tangent to 
KL at K and to LO at O, the method used in constructing the 
parabola being that shown in Fig. 148, b. Prolong the line AB 
as far as necessary to complete the construction. Now divide 
the span of the girder into ten equal parts, and drop verticals 
from these points of division to meet the funicular polygon. 
The line bT)' (Fig. 147, b), which connects the points of inter- 
section of verticals 60 feet apart with the funicular polygon, 
is the closing string of the funicular polygon for the loads that 
come upon the girder when it is in the position shown in Fig. 
147, a. For this position of the girder, wheel 2 is at section 2. 
The ordinates under the several points of division, intercepted 
between the closing string b'b' and the funicular polygon, repre- 
sent to scale the flange stresses at the several points along the 
girder for this position of the load. These ordinates represent 
actual flange stresses, and the scale used in measuring them is 
four times that used for the loads in the load line ; since the pole 
distance was taken equal to four times the depth of the truss 
(see § 184). Now consider the girder to be shifted one division, 
or 6 feet, toward the right from the position shown in Fig. 147, a. 
(This has the same effect as moving the loads 6 feet to the left.) 
The closing string of the funicular polygon for this new position 
of the loads and truss is c'c'. The ordinate directly above wheel 
2, intercepted between the closing string c'c' and the funicular 
polygon, represents the flange stress at section i of the girder; 
and the other ordinates represent the flange stresses at various 
points along the girder. Likewise, d'd' and e'e' are closing 
strings for the girder moved two and three divisions to the right, 
respectively, from its original position; and the flange stresses 



288 MAX. MOMENTS AND SHEARS — ENGINE LOADS. C^^^P- ^^HL 

at different points along the girder are represented by the ordi- 
nates between these closing strings and the funicular polygon. 
Also, a'a' is the closing string for the funicular polygon when 
the girder is moved one division to the left from the position 
shown in Fig. 147, a. From an inspection of the diagram shown 
in Fig. 147, b, it is seen that it is unnecessary to draw any other 
closing strings for this problem. Now if the curve MM is drawn 
through the points distant each one division horizontally from 
b', c', d', and e', respectively, it is seen that the ordinates between 
this curve and the funicular polygon will represent the successive 
flange stresses at section i as the girder is moved to the right, 
i. e., when the load moves along the girder toward the left. Like- 
wise, the curve NN, drawn through the points distant each two 
divisions horizontally from a', b', c', d', and e', respectively, will 
represent the successive flange stresses at section 2 as the load 
moves along the girder. Also, the curves RR, SS, and TT repre- 
sent the successive flange stresses at sections 3, 4, and 5, respect- 
ively, as the load moves along the girder. By scaling the ordi- 
nates at the different points, it is seen that the maximum flange 
stress at section i occurs when wheel 2 is at section i, and is 
82000 lbs. Likewise, it is seen that the maximum flange stress 
at section 2 occurs when wheel 3 is at section 2, and is 143 000 
lbs. ; that the maximum flange stress at section 3 occurs when 
wheel 3 is at section 3, and is 186000 lbs.; that the maximum 
flange stress at section 4 occurs when wheel 4 is at section 4, and 
is 210000 lbs.; and that the maximum flange stress at section 5 
occurs when wheel 5 is at section 5, and is 212 000 lbs. It should 
be noted that the maximum moment and flange stress at any 
point always occurs when some wheel is at the point, i. e., the 
maximum ordinate is always at some vertex of the funicular poly- 
gon. If the ordinate is not at one of the division points through 
which the curves MM, NN, etc., were drawn, its length may be 
more accurately determined by drawing the closing line for the 
required position of the loads. It is seen that the construction 
shown in Fig. 147, b not only gives the maximum stresses, but 
also the wheels which cause these stresses. Referring to § 159, 
it is seen that the maximum moment, and therefore the maximum 



'^^*^' PLATE GIRDER — MAXIMUM SHEARS. 289 

flange stress in the girder, occurs at some wheel near the center 
of the span when that wheel is as far to one side of the center 
as the center of gravity of all the loads is to the other side of the 
center. The position of the center of gravity of all the wheels on 
the girder at this time is given by producing the extreme strings 
AB and KL (Fig. 147, b) until they intersect. It is seen that the 
center of gravity of all the wheels is 0.4 ft. to the right of wheel 
5 ; therefore the maximum flange stress in the girder occurs when 
wheel 5 is 0.2 ft. to the left of the center. A part of the closing 
string for this position of the wheels is shown by the dotted line 
near b' (Fig. 147, b) ; and the maximum flange stress in the 
girder is 2:13 000 lbs. 

Shears, The maximum live load shears at the tenth points 
may be obtained from the diagram shown in Fig. 147, d. In con- 
structing this diagram, use the same load line as for flange 
stresses, but take the pole at P'. The pole P' is located on a hori- 
zontal line through the top of the load line with a pole distance 
equal to the span of the girder, 60 feet. Draw the funicular polygon 
A'B'CG'K'O' (Fig. 147, d) for the given loading, the portion 
K'O' being the arc of a parabola tangent to K'L' at K' and to 
L'O' at O'. With the first driver, wheel 2, at the left end of the 
girder, lay oflF the span to the right, and divide it into tenths, as 
shown in Fig. 147, f. The ordinate dd" above the right end of 
the girder, intercepted between the horizontal line A'B'N' and 
the funicular polygon A'B'C'G'K'O', then represents the reac- 
tion at the left end of the span (see § 187) ; provided none of the 
loads used in the construction of the funicular polygon are oflf of 
the girder. Now when wheel 2 is at the left end of the girder, 
wheel I is oflf of the span, and should not be considered in draw- 
ing the funicular polygon, 1. e., the horizontal line A'B'N' should 
be replaced by the line B'C'rs. Since there are no loads on the 
girder to the left of wheel 2, the maximum shear at the left end 
of the span is equal to the left reaction, which is represented by 
the ordinate sd" = 98ooo lbs. When wheel 2 is at section i, 
which is 6 feet from the left end of the girder, wheel i is still off 
of the girder, and the maximum shear at section i is represented 
by vc" = 82 400 lbs. The scale used in measuring these ordi- 



290 MAXIMUM STRESSES — ENGINE LOADS. Chap. XXIIL 

nates is the same as that used for the loads. Now if sections 2, 3, 
and 4 (Fig. 147, f) are successively placed under wheel 2, then 
the intercepts above the right end of the girder will represent the 
left reactions. The shears at these points are equal to the reac- 
tions minus the weight of the pilot wheel i ; and are represented 
by the intercepts between the funicular polygon and the line mr, 
the vertical distance between A'B'N' and mr representing the 
weight of wheel i. The shears at sections 2, 3, and 4 are 67 400, 
53 000, and 39 700 lbs., respectively. When wheel 2 is at section 5, 
the center of the girder (see Fig. 147, h), the shear at section 5 is 
27900 lbs.; when wheel i is at section 5 (Fig. 147, g), the shear 
at section 5 is 24 500 pounds ; and when wheel 3 is at section 5 
(Fig. 147, i), the shear at that section is 17800 lbs. For maxi- 
mum shears, it is unnecessary to consider any points to the right 
of the center of the girder. It is seen that wheel 2 gives the 
maximum shears at all points from the left end to the center of 
the span. It may be easily determined by trial, from the diagram 
shown in Fig. 147, d, which wheel — wheel i or wheel 2 — gives 
the maximum shear at any point along the girder, or it may be 
determined directly by the criterion given in § 211. 

Another method of obtaining the maximum shears will now 
be given, in which the shears are determined from the diagram 
for flange stresses. When wheel 2 is at the left end of the 
girder, wheel i is off the span, the closing string of the funicular 
polygon is d'd' (Fig. 147, b), and the ray, drawn through P 
parallel to d'd' to intersect the load line, will determine the two 
reactions (Fig. 147, e), the left reaction being 98000 lbs., which 
is the shear at the left end of the girder. Likewise, when wheel 2 
is at section i, wheel i is still off the girder, the closing string 
of the funicular polygon is c'c' ; and the ray, drawn through P 
parallel to c'c', will determine the two reactions, the left reaction 
being 82400 lbs., which is the shear at section 2 of the girder. 
Likewise, when wheel 2 is at section 2, wheel i is on the girder, 
the closing string is b'b', the left reaction is yy 400 lbs., and the 
shear is 77 400 — 10 000 = 67 400 lbs. In a similar manner, the 
shears at sections 3, 4, and 5 of the girder are found to be 53 000, 
39 700, and 27 900 lbs., respectively. In this problem, wheel 2 is 



Art, 1. PRATT TRUSS — MAXIMUM CHORD STRKSSES. 291 

at one of the sections. If the wheel causing maximum shears is 
not at one of the sections, then another set of verticals and closing 
lines should be drawn to determine the maximum shears. In 
some cases, this method gives the simpler solution; while in 
others, the first method described is more efficient. 

In the problem given, the loading consisted of one engine and 
tender followed by a uniform train load. If the loading consists 
of two engines and tenders followed by a uniform train load, the 
maximum shears will be somewhat larger. This is evident from 
the fact that the straight line AB (Fig. 147, b) would then be 
replaced by a broken line, thus increasing the length of the ordi- 
nates between the closing lines and the funicular polygon. 

The process explained is more efficient if the equidistant ver- 
ticals, which represent the divisions of the girder, are drawn upon 
a separate sheet of tracing cloth or transparent paper; as the 
position of the girder may then be readily shifted. 

The method explained in this section affords an efficient solu- 
tion for any particular span and loading; and if a large scale is 
used and care is exercised in making the constructions very good 
results may be obtained. 

216. (2) Maximum Chord and Web Stresses in a Pratt 
Truss. Let Fig. 148, a, represent one truss of a single-track 
through Pratt truss-bridge, whose span is 120 ft., and whose 
depth is 28 ft. It is required to find the maximum chord and 
web stresses due to the engine and train loading shown in Fig. 
148, c, the loading shown being that for one truss. 

Chord Stresses. In this problem, the chords are parallel, and 
the upper chord panel points are directly over those of the lower 
chord, which simplifies the solution somewhat. The construc- 
tions and methods used for this problem are similar to those used 
for the plate girder (see Fig. 147, and § 215) ; and only the 
points in which the methods differ for the two cases need be 
explained. The diagram for chord stresses is shown in Fig. 
148, b. By an inspection of the truss and loading, it was seen 
that about 80 feet of uniform train load would be sufficient. The 
load line is shown in Fig. 148, f ; and since the pole distance was 
taken equal to twice the depth of the truss, the ordinates in the 




Pfjatt Truss under Engine and Train Loaj) 
Span.ieo-0"i Depth - Se'-O', Loading os shown 
All loods and Stresses are in thousands of pounds 



Scoie of Web Stresses 
Pia. 148. Maxiudu Chogd and Web Btbebsbs in a Pbatt Tbuss. 



Art. 1, PRATT TRUSS — MAXIMUM WEB STUESSKS. 29:i 

chord stress diagram should be measured by a scale equal to twice 
that used for the loads. The chord stresses, together with the 
ordinates which represent them, are shown in Fig. 148, b. From 
an inspection of this diagram, it is seen that the maximum stress 
in L^Lo is 88 000 lbs. when wheel 3 is at Lj ; that the maximum 
stresses in U1U2 and LoLj are each 135000 lbs. when wheel 5 is 
at Lo ; and that the maximum stress in U0U3 is 145 000 lbs. when 
wheel 8 is at L3. It is also seen that wheel 7 at L3 gives almost 
as large a stress in U2U3 as does wheel 8. 

If a truss with an odd number of panels had been given, then 
the maximum stress in the center panel of the lower chord would 
have occurred when the loads are so placed that there would be 
zero shear in the center panel. 

If the live load had consisted of two engines and tenders fol- 
lowed by a uniform train load, then the stresses caused by the 
loads of the second engine should also have been considered. 

Weh Stresses, The diagram for all the web stresses, except 
the hip vertical, is shown in Fig. 148, d, and for the hip vertical, 
in Fig. 148, e. For the diagram shown in Fig. 148, d, the pole 
is taken at P^ with a pole distance equal to the span of the truss. 
When wheel 3 is placed at L^ (Fig. 148, g). the left reaction, 
which is represented by the intercept above L/ (Fig. 148, g),= 
136 000 lbs. ; the shear in the panel LqL^ = 136 000 — 11 500 (the 
portion of the loads in the panel L^Li that is carried to Lq by the 
stringer) = 124 500 lbs.; and the stress in L^Uj, which is ob- 
tained by drawing a line parallel to the member LoUi,= 153000 
lbs. (Fig. 148, d). This is the maximum stress for LqUj ; as 
may be shown by successively placing wheels 2 and 4 at h^, and 
finding the stresses in this member caused by these positions of 
the loads. Likewise, the maximum stress in U^Lo is obtained 
when wheel 3 is at Lo (Fig. 148, h), and is 103 500 lbs. : the maxi- 
mum stresses in L\Lo and U0L3 are both obtained when wheel 2 
is at Lo (Fig. 148, i), and are 50 700 and 62 300 lbs., respectively : 
the maximum live load stresses in UX, and L^.Lo' are both 
obtained when wheel 2 is placed at h/ (Fig. 148, j), and are 
25 100 and 30900 lbs., respectively Cif there is no dead load shear 
in the panel) ; and the maximum live load stresses in Uo'L.,' 



294 MAXIMUM STRESSES — ENGINE LOADS. Chap, XXIIL 

and Ug'L/ are both obtained when wheel 2 is at h^ (Fig. 148, k), 
and are 6000 and 7300 lbs., respectively (if there is no dead load 
shear in the panel Lg'Li'). When wheel 2 is at any panel point, 
the part of the load carried to the panel point ahead by the 
stringer is 4000 lbs. The constructions when wheel 2 and wheel 3 
are successively at L3 are shown in Fig. 148, d, wheel 2 at Lg 
giving the larger stress. 

The diagram for determining the maximum stress in the hip 
vertical is shown in Fig. 148, e. The stress in this member is 
equal to that portion of the total load in the two panels LoL^ and 
LjLj which is carried to Lj by the stringers. It is seen that this 
is a maximum when the heavy wheels are near Lj. To determine 
the exact position of these wheels for a maximum stress in UjLi, 
and also the maximum stress in the member, place the center of 
gravity of the first six wheels (those in the panels LqLi and LiLo) 
at Li ; and draw the funicular polygon shown in Fig. 148, e, with 
a pole distance equal to one panel length. If the pole distance is 
taken equal to a panel length, then the intercepts in the funicular 
polygon will represent the reactions at L© and L2 caused by the 
loads in the panels L^L^ and L^Lj, respectively. This is true 
because in the funicular polygon intercept X pole distance (one 
panel length) == moment; and the reactions = intercept X pole 
distance (one panel length) -f- moment arm (one panel length) = 
intercept. When the center of gravity of the six wheels is placed 
at Lj, the reactions at Lo and Lj are each equal to 19200 lbs. 
(Fig. 148, e), and the stress in UiLi= 103 000 (the total weight 
of the six wheels) — 38400 = 64600 lbs. Now place wheel 3 at 
Lj. For this position of the loads, the reactions at L© and L, are 
1 1 500 and 2y 400 lbs., respectively ; and the stress in UiLj = 
103 000 — II 500 — 27 400 = 64 100 lbs. Next place wheel 4 at 
Lj. For this position of the load, an additional wheel — wheel 7 — 
is brought upon the length LoLj ; the reactions at Lo and Lg are 
24000 and 26500 lbs., respectively; and the stress in UiLi = 
1 16 000 (the total weight of the seven wheels) — 24 000 — 26 500 
= 65 500 lbs. It is thus seen that the maximum tension in UiL^ 
= 65 500 lbs. when wheel 4 is at Lj. 

Referring to Fig. 148, e, it is seen that the ordinates under 



'^rt.g. LOAD LINE AND MOMENT DIAGRAM. 295 

wheels 2 and 3 represent the loads that were deducted from the 
reactions in Fig. 148, d to obtain the shears. 

The stress in the hip vertical may be obtained in a different 
manner, as follows: Place wheel 4 at Li, and draw a vertical 
through wheel 4 to intersect the funicular polygon at v (Fig. 
148, b). Also, mark the points m and n on the funicular polygon, 
distant horizontally one panel length from v. The lines mv and 
nv are the closing strings of the funicular polygons for the loads 
in the panels LqLi and L^Lj, respectively. Now draw rays 
through P parallel to these closing strings, and these rays will cut 
off on the load line the total load transferred to Li, which is the 
stress in UiLi. 



Art. 2. Maximum Moments, Shears, and Stresses in 
Girders and Trusses of Various Types and Spans. 

When it is required to determine the moments and shears in 
girders and trusses of different spans, subjected to the same load- 
ing, the work may be greatly facilitated by constructing a load 
line and a moment diagram for the load on one rail. If the dia- 
grams are drawn to a large scale, very good results may be 
obtained by their use. The construction of the diagrams will be 
explained, and then their application to the determination of 
moments and shears in girders and trusses will be shown. 

217. Load Line and Moment Diagram. The load line and 
moment diagram for O)oper's E40 loading are shown in Fig. 
149. The diagrams shown in this figure were originally drawn 
upon cross-section paper, divided into one-tenth inch squares, 
using a scale of four times that shown in Fig. 149. The engine 
diagram was laid off to a scale of ten feet to the inch ; the loads, 
to a scale of 50 000 pounds to the inch ; and the moments, to a 
scale of 2 500 000 foot-pounds to the inch. The engine and uni- 
form train load diagram is shown in the lower part of the figure. 
The engine wheels are numbered, the load on each wheel shown, 
and the spacing of the wheels given. 



296 MOJIBN'TS AND SllEAltS — ENGINE LOADS. Chap. SXIII. 

I I i I I i § 



g5^wgw ¥Wffii 




FiQ. 149. MoMsy-c \ 



UB DiAOBAus — Cooper's B40. 



The load line, which is the heavy stepped line 1-2-3-4-5, etc., 
is a diagram whose ordinates measured from the line 0-0 repre- 



Art.e, LOAD LINE AND MOMENT DIAGRAM. 297 

sent the summation of the loads to the left. Each step in the 
load line represents to scale the load directly under it. The por- 
tion of the diagram above the uniform load is a straight line 
having a uniform slope of 2000 pounds per foot. 

The moment lines, which are numbered i, 2, 3, 4, 5, etc., near 
the right edge of the figure, are constructed as follows : Starting 
at o, the right end of the horizontal reference line 0-0, lay off 
successively on a vertical line the moment of each wheel load 
about that point, beginning with wheel i. The moment line i is 
then drawn frc«n a point in the reference line, directly over 
wheel I, to the first point at the right of the diagram; moment 
line 2 is drawn from a point on moment line i, directly over 
wheel 2, to the second point at the right of the diagram ; moment 
line 3, from a point on moment line 2, directly over wheel 3, to 
the third point, etc. The moment line BC is a parabolic curve; 
and may be easily constructed by computing the moments of por- 
tions of the uniform load between B and several points to the 
right about these points, and laying off these moments above the 
moment line 18. It is seen that the broken and curved line ABC 
is an equilibrium polygon for the given loads. The ordinate at 
any point, measured between the reference line 0-0 and any 
moment line, represents the sum of the moments about this point 
of all the loads up to and including the wheel load corresponding 
to the moment line under consideration. To illustrate, the sum 
of the moments of the wheel loads i, 2, 3, and 4 about wheel 18 
is represented by the ordinate over wheel 18, measured between 
the reference line 0-0 and moment line 4. Also, the moment at 
any point, measured between any two moment lines — as between 
2 and 6— represents the sum of the moments of wheel loads 3 to 
6, inclusive, about that point. The line of equal shears for wheels 

I and 2, shown in Fig. 149, has an upward slope of -7-; where 

Pi is the first wheel load, and b is the distance between wheels 
I and 2. The use of this line will be explained in § 220 ( i ) . 

218. Application of Diagrams in Fig. 149 to Determining 
Maximum Moments in Plate Girders or at Joints of the 



298 MAX. MOMENTS AND SHEARS — ^ENGINE LOADS. C^^^P- XXIII. 

Loaded Chord of a Truss with Parallel or Inclined Chords. 

The diagrams shown in Fig. 149 may be used for finding the 
position of the wheels for a maximum moment, and also the value 
of the moment. To illustrate the use of these diagrams, let it be 
required to determine the position of the wheels for a maximum 
moment at Lg (Fig. 150), this point being either a panel point of 

F 



u. ut ,^V J^' 



/ 






^ 






7 • . • [ \ 

/I / ^ 



— t^h: /I / I \ 

X ^ * \ / I \ 

t /v « ' 



» ^•'^ "Cv*^ I /^^v I Maximum Moment at Li 



a/-^ I' i E*L ^(^^ i, ^, 

U L Li L K ili go 

Fig. 150. Maximum Moment at Loaded Chord Joint. 

the truss whose span is AB, or any point along a girder of the 
same span. To avoid confusion, a portion of the load line shown 
in Fig. 149 is reproduced to a larger scale in Fig. 150. Since the 
girder, or truss, must be shifted, the points along the girder, or 
the panel points of the truss, should be marked on the edge of 
a separate slip of paper or upon a piece of tracing cloth ; and it 
will be assumed that the points marked on the span AB (Fig. 
150) are on a separate slip of paper. The criterion for a maxi- 
mum moment at Lj is — the average load S Pi to the left of Lj 
must be equal to the average load S P on the entire span (see 
§ 208). Try wheel 4 at L2, i. e., shift the truss until Lg is under 
this wheel (Fig. 150). The total load on the span is represented 

2P 
by the ordinate BC = S P; while the average load is -^ — , which 

is represented by the slope of the line AC. The load 2 Pi to the 
left of Lg is represented either by the ordinate FE or the ordinate 
FD, depending upon whether wheel 4 is at Lo or just to the left 



'Art.g. MAX. MOMENT AT LOADED CHORD JOINT. 299 

5 Pi 

of Lj; and the average load -r-rrto the left of Lo is represented 

either by the slope of the line AE or the line AD. Since the slope 

of the line AC is less than that of AD and greater than that of 

AE, it is seen that wheel 4 at Lj gives a maximum moment at 

this point If the line AC lies above the line AD, the loads must 

be moved to the left; and if the line AC is below the line AE, 

then the loads must be moved to the right. It is thus seen that 

2P 
if the line whose slope is — r— cuts the vertical line representing 

the load at the point, this position of the load gives a maximum 

moment. Instead of actually drawing the line AC, its position is 

usually determined by stretching a thread. In the illustration just 

given, none of the wheels were off the bridge to the left, and the 

line AC starts from the zero shear Une at L©. If some of the 

loads had passed the left end of the bridge Lo, then AC should 

start in the load line vertically over L^. 

The position of the wheels having been determined, the 

moment itself may be easily found from the moment diagram 

(Fig. 149), as follows: With wheel 4 at. Lj, read the ordinate 

at the right end of the span Lo' between the reference line 0-0 

and the moment line 10, which is the moment of all the loads on 

the span about L^'. The moment at Lj is obtained from the 

AF 
moment at Lo' by multiplying it by ^ — and subtracting the 

moment of the loads to the left of L2, this latter moment being 
given by reading the ordinate at Lj between the reference line 
0-0 and moment line 4. 

Since the line ABC (Fig. 149) formed by the segments of 
the moment lines is a funicular polygon for the given loads, the 
moment at Lg is also equal to the ordinate at that point inter- 
cepted between the funicular polygon and the closing line. The 
extremities of this closing line are on the verticals which pass 
through the ends of the bridge. 

In finding the maximum moment in a girder — which occurs 
near its center — it is necessary to locate the center of gravity of 



300 



MAX. MOMENTS AND SHEARS — ENGINE LOADS. G^P- XXUl. 



all the loads on the girder (see § 159 (a)) ; and the center of 
gravity of any number of loads may be found from Fig. 149 by 
producing the extreme strings of the funicular polygon until they 
intersect. 

a 19. Application of Diagrams in Fig. 149 to Determining 
Maximum Moments at Panel Points in the Unloaded Chord 
of a Truss with Parallel or Inclined Chords. The use of the 
diagrams in Fig. 149 for finding the maximum moment at any 
unloaded chord panel point will be shown by the following prob- 
lem : It is required to determine the position of the engine load 
for a maximum moment at the panel point U, of the truss shown 

I — ^ 

Maximum Moment at Us 




Fig. 151. Maximum Moment at Unloaded Chord Joint. 



ill Fig. 151, together with value of the maximum moment. Let 
the heavy stepped line 1-2-3-4, etc. (Fig. 151), be a portion of 
the load line, and let L be the span of the truss. It has been 
shown (§ 209) that the criterion for a maximum moment at U3 is 



2 p S P, ^ + 2 P, 



; where S P is the total load on the span, 



2 Pj is the load to the left of Lg, S Pg is the load in the panel 
LoLg', d is the panel length, r is the horizontal distance from Ug 
to L2, and s is the horizontal distance from Ug to L©. Try 
wheel 6 at L/, which is the position of the load in Fig. 151. Now 
the total load S P on the span is represented by BC, and the aver- 



Art, e, MAX. MOMENT AT UNLOADED CHORD JOINT. 301 

2P 

age load — p, by the slope of the line AC. The load 2 Pi to the 

left of Lj is represented by GK, and that in the panel LjLj^', by 
HD or HE (depending upon whether wheel 6 is just to the left 
or to the right of Lg'). The average load in the panel LjLj' is 
represented by the slope of the line GD, or the line GE. The 

r 
term SPo— 7-(see above criterion) is represented by JN, or 

r 
JM ; the term 2 Po -r+ 2 Pj, by RN, or RM ; and the term 

2P,^ + 2P, 

by the slope of the line AN, or the line AM (not 

s 

drawn in the figure). Since the line AC cuts the vertical through 
Ug between the points N and M, it is seen that the wheel 6 at L/ 
satisfies the criterion for a maximum moment at Ug. If it had 
been impossible to satisfy the criterion by placing some wheel at 
h/, then the criterion should have been tested by placing a wheel 
at Lg. In using the diagrams shown in Fig. 149, the upper and 
lower chord panel points should be marked on a separate slip of 
paper. The slip should then be shifted until the wheel which 
gives a maximum moment is at the panel point about which the 
moment is required. The auxiliary lines shown in Fig. 151 need 
not be drawn ; as the positions of the Hnes GD and GE may be 
determined by stretching a thread, the points N and M being 
marked, and the thread then moved to the position AC. 

The position of the engine load (wheel 6 at h/) having been 
determined, the left reaction may be found as in § 218. The 
determination of the moment at the unloaded chord joint U3 
differs from that at a loaded chord joint, in that only the portion 
of the load in the panel LJ^/, transferred to L, by the stringers, 
should be considered. The moment at U3 is equal to that of the 
left reaction minus the moment of the portion of the loads in 
L2L2' transferred to Lj together with the moment of all other 
loads to the left of U3. 



302 MAX. MOMENTS AND SHEARS — ENGINE LOADS. ^^«P- XXIII. 



The moment at Ug may also be readily found, as follows: 
With wheel 6 at Lo', find the moments at Lj and Lj', as in § 218. 
Then if Mi, and M^ represent the moments at Lj and L/ respect- 

r 
ively, the moment at U8 = Ml+ (Me — Ml) -t- 

220. Application of Diagrams in Fig. 149 to Detennining 
Maximum Shears. Two cases will be considered, viz: 
(i) maximum shears in beams and girders; and (2) maximum 
shears in trusses. 

(i) Maximum Shears in Beams and Girders. It has been 
shown (§ 211) that wheel i at any point will give a maximum 

Pi_5P2 
shear when -r- — -7 — » ^^^ that wheel 2 .will give a maximum 

D > Lr 

Pi 5 Pi 
shear when -r-'^^ — r- ; where Pj is the first wheel load, S Pi is 

the total load on the span when Pj is at the point, S Pg is the total 
load when Pj is at the point, b is the distance between wheels i 
and 2, and L is the span of the beam or girder. 

J 





fio 




|6 : 




f\ 


s : 


c 


I 


2 


3 


1 

4 


.^^0 • 1 
^^ • ^^^^ 1 

Maximum Shears in Girder 


w 


m|!_.---^ 


• i 1 
1 I 


N 


"^ ^ -.5 U B 



Fig. 152. Maximum Shears in a Girded. 

Let 1-2-3-4, etc. (Fig. 152) be a portion of the load line, 
and let AB == L be the length of a beam or girder. It is required 
to determine the segment of the beam in which wheel i gives the 
maximum shear, and also the segment in which wheel 2 gives the 
maximum shear. Place wheel i at the left end of the beam ; and 



^^'^- MAXIMUM SHEARS IN A GIRDEH. 303 

from A, draw the line AC having a slope of -r^ to intersect a ver- 

b 

tical through the right end of the beam. Now the ordinate BC 

Pi 
represents the load which divided by L equals -r—. It therefore 

represents the load 2 Pi, or 2 Pg. Draw the line CD parallel to 
AB to intersect the load line at D; and also draw the vertical 
line DE. It is then seen that the average load on the entire span 

Pt 
equals — , when wheel 5 is at the right end of the beam. Now lay 

off the span AB on a separate slip of paper, and mark the points 
E' and F' on this slip to correspond with the points E and F 
(Fig. 152). Place the point B (on the slip) at E, i.e., directly 
under wheel 5, and mark the positions of wheels i and 2, calling 
these points i' and 2'. Then any point in the segment 2'B 
(between wheel 2 and the right end of the span) has its maxi- 
mum shear when wheel i is at the point; and any point to the 
left of i' (wheel i) has its maximum shear when wheel 2 is at 
the point. Between i' and 2' (the distance between wheels i 
and 2), both positions should be tried. 

The line marked equal shears, wheels i and 2 (Fig. 149), 
corresponds to the line AC (Fig. 152). Since Fig. 149 has cross- 
section lines, it is unnecessary to draw any additional lines. 

(2) Maximum Shears in Trusses. It has been shown (§212) 
that the criterion for a maximum shear in any panel of a truss 
is — the load in the panel must be equal to the total load on the 
span divided by the number of panels. If 2 P is the total load 
on the span, and 2 Pj is the load in the panel other than the 
wheel load at the panel point to the right, then any load P at 

. SP 
this panel point will give a maximum shear in the panel if-= — 

2 P, 2 Pi + P 

lies between — -z — and 5 . 

d d 

Let L (Fig. 153) be the span of any truss, and 1-2-3-4, etc., 
a portion of the load line. It is required to determine which 



304 WAX. MOMKNTS AND bllEAUS — ENGINE LOADS. Chap, XXIII. 

wheel load placed at the panel point to the right will give a maxi- 
mum shear in each panel of the truss. Place wheel i at the first 
panel point L^, as shown in Fig. 153. Draw the lines AC, AD, 



• E 



[S^ i 



zF^ 



/ 









Maximum Shears in Truss 



/ ^-^ I 



v^--lr 



f^/ 



I / N 






wt L ^---i-j 

^-.^ — r ^i 



?j Nk: Li 5^ I 



t 



Lo d L. L2 H , L't Hi F to 

Fig. 158. Maximum Shears in a Truss. 



and AK with ordinates at L^ representing the wheel loads Pi, Po, 

P Pi + Po 

and P3, respectively. These lines will have slopes of -^,— ^ ^' 

d d 

P, + P, + P3 
and -T , respectively. Wheel i placed at any panel 

point will give a maximum shear in the panel to the left when 

\\ S P 2 P 

-r- = -Y— • Now-r — is represented by BC, the ordinate over the 
cl > L. i- 

Pi 2P 
right end of the truss L/; and -7-= -y— as soon as wheel 3 

enters the span and until wheel 4 enters the span at Lo'. There- 
fore wheel I will give a maximum shear in the panels to the left 
of each panel point passed by it in moving the loads to the left 
until wheel 4 enters the span at L^', i. e., in the panel Lo'Lj'. 
Likewise, wheel 2 will give a maximum shear in any panel whtn 

2P Pi P1 + P2 . 

-^r-Hes between -r- and — -l \ 1. e., for values of 2 P between 

L d d 

Pi + P2 2 P 

BC and BD. Now — 1 = —. — • when wheel 10 is at Lo'. 

d L 

Therefore wheel 2 will give a maximum shear in the panels to 



ArL2, MAXIMUxM SIIEAIIS IX TRUSSES. 305 

the left of each panel point passed by it in moving the loads from 
the position with wheel 4 at Lq' to the position with wheel 10 at 
Lo', i. e., in the panels Lj'L^', LgLj', and L^Lg. Likewise, wheel 3 
will give a maximum shear in the panels to the left of each panel 
point passed by it in moving the loads from the position with 
wheel 10 at L^' to the left end of the truss, i. e., in the panels 
LqLi and LiLg. This is determined from the slopes of the lines 
AD and AK. From the above discussion, it is seen that both 
wheel I and wheel 2 satisfy the criterion for a maximum shear 
in the panel Lg^L^' ; and that both wheel 2 and wheel 3 sat- 
isfy the criterion for a maximum shear in the panel L^Lj. 
Since the diagram in Fig. 149 has cross-section lines, the 
auxiliary lines shown in Fig. 153 need not be drawn, the 
actual process being as follows: Lay off the span AB = L 
on the edge of a separate slip of paper, and mark the panel 
points Lo, Lj, L2, Lo', L/, and L^'. Place the edge of the 
slip on the reference line 0-0 (Fig. 149) with the right end L^ 
of the truss under wheel 4, and mark the positions of wheel i 
and wheel 2 on the slip. Now move the slip to the right until 
Lo' is directly under wheel 10, and mark the positions of wheel 2 
and wheel 3. The marking of the slip will now correspond to 
that of the line AB (Fig. 153). It is now seen that wheel i will 
satisfy the criterion for a maximum shear in the segment Lq'i'; 
that wheel 2 will satisfy the criterion for a maximum shear in 
the segment 2'2' ; and that wheel 3 will satisfy the criterion, in 
the segment 3'Lo. It is also seen that the segment in which 
wheel I satisfies the criterion for a maximum shear is overlapped 
by that in which wheel 2 satisfies the criterion, by the distance 
I '2', which is the distance between wheels i and 2. It is further 
seen that the segment in which wheel 2 satisfies the criterion for 
a maximum shear is overlapped by that in which wheel 3 satisfies 
the criterion, by the distance between wheels 2 and 3. 

The position of the wheels having been determined, the shear 
itself may be easily found from the moment lines (Fig. 149). 
To determine the shear in any panel, say in LjLg' (Fig. 153), 
place the right panel point L2' under wheel 2 (this position of 



306 



MAXIMUM STKESSKS — ENGINE LOADS. 



Chap. XXIIL 



the load being the one which satisfies the criterion for a maximum 
shear) ; and read the ordinate to the funicular polygon ABC 
at the right end of the truss. The moment represented by this 
ordinate divided by the span is equal to the left reaction. Since 
the loads can only come upon the truss at the panel points, being 
transferred by the stringers, the shear in the panel LgLg' is equal 
to the left reaction minus the portion of the load in this panel 
which is carried to the left panel point L2. The load which is 
carried to this panel point is equal to the moment of the load to 
the left of Lg' about Lj' divided by the panel length d ; and is 
represented by the ordinate to the funicular polygon above the 
panel point Lg'. 

221. Application of Diagrams in Fig. 149 to Determining 
Maximum Web Stresses in Trusses with Inclined Chords. 
The following problem will show the application of the diagrams 
in Fig. 149 to determining the maximum stress in any web mem- 



,''\ 












,.-v 






;^ 



■^ 






Maximum Stress in UiLz 



U-Eo.. 5 lL:.__.d_VjL2 __L ^^ lI; Co 

Ir; \p' "Ir* 



Fig. 154. Maximum Stress in Web Member — Truss with Inclined Chords. 



ber of a truss with inclined chords. It is required to find the 
maximum live load stress in the member UiLg of the truss shown 
in Fig. 154. A portion of the load line shown in Fig. 149 is 
reproduced to a larger scale in Fig. 154. It has been shown 
(§ 213) that the criterion for a maximum stress in the member 



SP 
U1L2 is = 



2P 



.(■+-D. 



where 2 P is the total load on the 



Art,^, MAX. WEB STRESSES — INCLINED CHORDS. 307 

span, 2 Pi is the load in the panel LiLj, s is the distance from L^ 
to the left end of the truss, and k is the distance from the left end 
of the truss to the point of intersection O of the members UiUg 
and LjLg, this point being the center of moments for determining 
the stress in UiLj. It is seen that this criterion is similar to that 
for a maximum shear in the panel, indicating that some wheel 
near the head of the train placed at Lg will give a maximum 
stress in UiLg. Try wheel 2 at Lg (Fig. 154). The load S P^ 
in the panel L^Lo is either P^ or Pi + P2> depending upon 
whether wheel 2 is to the right or to the left of Lo. The term 

2 Pj(i + y) ^f ^^^ criterion is represented by GF, or by GE; 

and the term , by the slope of the line AF, or the 

line AE. The total load on the span is represented by BC, and 
the average load, by the slope of the line AC. Since AC lays 
between AF and AE, it is seen that wheel 2 satisfies the criterion 
for a maximum stress in UiLg. 

To determine the stress in the member U1L2 place wheel 2 
at Lo, and read the ordinate to the funicular polygon (Fig. 149) 
at the right end of the truss. The moment represented by this 
ordinate divided by the span L is equal to the left reaction Rj. 
If Gi represents the portion of the load to the left of the section 
p-p (i. e., the portion of the load carried to L^ by the stringer), 

-R,k + Gi (k + s) . 
then the stress in U,Lo = r . The part of the 

load in L1L2 which is carried to L^ may be found by reading the 
ordinate to the funicular polygon at Lg, and dividing by the panel 
length. 

222. Determination of Maximum Stresses in a Truss with 
Subordinate Bracing Petit Truss. A truss with subordinate 
bracing has been defined as one which has points of support for 
the floor system between the main panel points. The Baltimore 
trusses shown in Fig. 1 14, d and Fig. 1 14, e are examples of such 



308 



MAXIMUM STUESSES — ENGINE LOADS. Chap. XXIII. 



a truss, in which the chords are parallel ; while the Petit trusses 
shown in Fig. 114, j and Fig. 155 are examples of this type, in 
which the chords are not parallel. The effect of the subordinate 
bracing upon the stresses in the main members of the truss will 
be shown by the following problem : It is required to determine 
the maximum stresses in the members in the panel L4Le of the 
truss shown in Fig. 155. The members L^M, UjM, and L5M 
are tension members; while L4M is in compression when the 
counter U3M is not acting, and in tension when the counter is 
acting. It is seen that UgM does not act when the main members 
have their maximum stresses. 



PIU3 




Fig. l.jo. Maximum Stresses — Truss with Subordinate Bracing. 



To determine the maximum stress in U2U3, cut the members 
U2U3, LeM, and hJ^f^ by the section p-p, and take the center of 
moments at Lg. Since the counter U3M is not acting, the posi- 
tion of the loads for a maximum moment at Lg, together with the 
value of this moment, may be obtained by the methods shown 
in § 218. The stress in U2U3 is equal to this moment divided by 
the perpendicular distance from Lq to U2U3. It is thus seen that 
the stress in this chord member is the same as for a truss with 
the subordinate bracing omitted. 

To find the maximum stress in the lower chord member L^Lg, 
the same section should be taken with the center of moment at Uo. 
The position of the wheel loads for a maximum moment at U2 
may be determined from the criterion deduced in § 210. By the 
use of the diagrams in Fig. 149, it is easy to apply this criterion, 
and to determine which wheel placed at L5 will give a maximum 
moment at Ug. The moment at this panel point is equal to the 



drug. TRUSS WITH SUBORDINATE BRACING. 309 

reaction at Lo multiplied by the distance LqL^ minus the moment 
of the loads to the left of L4 about L4 plus the moment of the 
portion of the loads in the panels L4L5 and LgLg which is carried 
to Lg by the stringers. It is necessary to consider the joint load 
at Lg ; since it is to the left of the section p-p. 

The maximum stress in L^M is obtained when there is a maxi- 
mum joint load at L5, and is equal to that load. The stress in 
this member may be obtained as shown in Fig. 148, e, and ex- 
plained in § 216. 

The position of the wheel loads for a maximum stress in UjM 
and the stress itself are the same as if the subordinate members 
LgM and L4M were omitted. The stress in UoM may therefore 
be determined as shown in § 221, the subordinate members being 
omitted, and the panel length taken as L^L^. It is seen that this 
is true, since the small triangular truss L^ML^ merely acts as a 
trussed stringer to transfer the loads to L^ and Lg. 

The stress in L(jM is influenced by the subordinate members 
if there are any loads between L4 and L^. The stress in this 
member may be determined by taking the center of moments at 
the point of intersection of U2U3 and L^L^. 

The maximum stress in U2L4 may be determined as in § 221, 
considering the members LgM and L4M removed. The section 
r-r should be cut, and the center of moments taken at the point 
of intersection of UiUg and hj^^. 

The stress in L4M when the counter is not acting, 1. e., when 
L4M acts as a compression member, may be readily found by 
graphic resolution. Since UgM and LqM are collinear, the re- 
solved components in L4M and L5M perpendicular to UgLg must 
be equal. 

The maximum tensile stress in L4M, which occurs when the 
counter U3M is acting, may be found in the following manner: 
Consider the numbers L4M and UgM replaced by a straight mem- 
ber L4U3. Now the maximum stress in this member may be 
found by the same method used for that in U2M, i. e., by consid- 
ering the members L.5M and UoM removed. The member LqM 
is not considered ; as it does not act for this loading. After find- 
ing the stress in L4U3, the maximum stress in L4M may be deter- 



310 MAXIMUM STRESSES — ENGINE LOADS. Chap. XXIIL 

mined, as follows: Lay off on a line parallel to the member 
L4U3 a length L^U,, representing to scale the stress in that mem- 
ber. Through U,, draw a line parallel to the member LeM ; and 
through L4, draw lines parallel respectively to L4M and UgM to 
intersect the line parallel to L^M. This construction gives the 
maximum stress in the member L4M. 

The maximum stress in the counter UgM may be found by 
taking the section p-p, remembering that the member LeM is not 
acting when the truss is loaded for a maximum stress in the 
counter. The load should be brought upon the truss at L^, and 
the left segment loaded for a maximum in the member. The 
kanger L5M should be considered; as the loads carried by it 
increase the stress in U3M. 



I 

d 



INDEX. 



A 

PAGE. 

Accurate method, Moment of In- 
ertia 61 

Action, known lines of 23 

Line of, defined 5 

Algebraic formulse for beams 198 

methods for beams 167 

moments, stresses by 86 

resolution, stresses by 95 

Application of diagrams 297 

web stress 'SOii 

Point of, defined 5 

Approximate method, moment of 

inertia 59 

Arch, defined 143 

Line of pressure in 35, 37 

Area, center of gravity of, 47 ; ir- 
regular 50 

Irregular 49 

Geometrical 47 

Moment 39 

Moment of inertia of 58 

about parallel axes 64 

of parallelogram 47 

of quadrilateral 48 

of sector 48 

of segment 48 

of triangle 48 

Radius of gyration of 60 

Axes, parallel, moment of inertia 

of 56 



B 

Baltimore bridge truss 210 

Coefficients for 264 

Base of column 151 

wind load stress, hinged column 155 

Beam, cantilever, moment, deflec- 
tion and shear 188 

fixed both ends 195 

Floor, maximum reaction 279 

Overhanging, concentrated loads 175 

Overhanging, uniform load 177 

one end fixed 189 

Beams 167 

Deflection in 178 

Restrained 187 

Bending moment, see also Moment. 

in beams 167 

in simple beam 171 

Bent, stresses in transverse 150 

Trestle 162 

Body, rigid, deflned 3 

311 



PAOB. 

Bow's notation 86 

Bowstring bridge truss 210 

truss, stresses in 243, 247 

Braces, main 127 

Bracing, bridge 212 

Subordinate 273, 308 

Sway 70 

Weight of 70 

Bridge, see also Trusses. 

Bridge, floor system 213 

Joists 213 

loads 214 

stringers 213 

truss members 211 

trusses, see names of trusses, 

trusses, types of 210 

Bridges 209 

Live loads for 216 

Wind load on 219 

Weights of 214 

Building, transverse bent. 150 

C 

Camel's back bridge truss 210 

Cantilever beam 168 

beam, moment, deflection, shear 188 

roof truss 67 

trusses 114 

Center of gravity 47 

irregular area 50 

Centroid, 47; of parallel forces. . . 49 

Chord, defined 68 

Loads on upper, 99 ; on lower . . 103 

Maximum stress in, Pratt truss. 291 
Chords, counterbraced, parallel, 

139 ; non-parallel 141 

Inclined, 306 ; truss with 280 

Loaded 298 

Stresses in trusses with parallel 259 

Circular chord truss 67 

Closed polygon li 

Closing funicular polygon 23 

Coefficients, method of for stresses 259 

for Baltimore truss 264 

for Pratt truss 264 

for Warren truss 264 

Columns, conditions of ends. ..... 150 

fixed at base 151, 152, 157 

fixed at top 152 

hinged top and base 151 

hinged at base, stresses in 155 

Combined stress diagram 116 

counterbraced truss 139 

Complete frame structure 65 



312 



INDBX. 



PAGE. 

Components, defined, 6 ; horizon- 
tal 80 

liorizontul, oi* reactions equal.. 108 
Non-parallel, non-concurrent . 23, 24 
Composition of forces, 6 ; concur- 
rent forces 8 

Compression, defined 84 

members, long 68 

Compressive stress, sign of 86 

Concentrated loads, cantilever 

beam 168 

one end fixed 180 

overhansing beam 175 

8imi)le beam 171 

moving load 201 

wheel loads 217 

Concurrent forces 5, 8 

Equilibrium of 12 

Resolution of 11 

Resultant of 9 

Conditions for equilibrium 26 

of ends of columns 150 

Connections 213 

Eccentric riveted 164 

Constant moment of inertia 178 

Construction, special, for funicular 

polygon 34 

of a roof 60 

Simple, beam one end fixed.... 191 

Contraction and expansion 70 

In bridges 213 

Cooper's Class E-40 296 

Copianar forces defined 5 

Corrugated steel, 69; weight of.. 70 
Counterbraced truss, combined 

stress diagram 139 

with parallel chords 130 

with non-parallel chords 133 

Stresses in 129 

Counter bracing, 'i25,' 127;' notation 128 
Couple, defined, 6 ; a resultant, 

21 ; moment of 40 

Culman's method 53 



Dead load 69, 70 

on bridges 214 

reactions, roof 75 

reactions and stresses, arch .... 144 

stresses, transverse bent. ...... 152 

Deck bridges 210 

Deflection in beams 178 

Curve, elastic • 179 

one end fixed, beam 190 

beam two ends fixed 195 

diagram 184 

formulae, beams 198 

problem, plate girder 184 

Determining stresses 85 

Diagonals, importance of 126 

Main 127 

Diagram, bending moment 168 

Deflection, 184; plate girder... 184 

Force, defined 6 

Moment and shear, simple beam 171 

for shears 302 

Shear, beams 168 

Space, defined 6 

Stress 102 

stress in uusymmetrical truss.. 116 



PAQB. 

for web stresses 306 

Wind load 74 

Diagrams, application of 297 

Influence 267 

Different polygons for same forces 34 

Direction, defined 5 

Distance pole, defined 20 

Duchcmln's formula for wind pres- 
sure 73 

Dynamics, definition 3 

E 

Eccentric riveted connection 164 

Economical trusses 68 

BIffective reactions, roof 78 

Elastic curve 179 

End, beam with one fixed 189 

Leeward, on rollers 110 

Windward, on rollers 112 

Ends, beam with two fixed 195 

Fixed, parallel reactions 105 

Fixed, horizontal, components, 

reactions equal 108 

of columns, conditions of 150 

Engine loads 267 

and train loads 285 

Equillbrant, defined 6, 13 

Equilibrium, defined 5 

of concurrent forces 12 

Conditions for 26 

of non-concurrent forces 26 

Proolems in 13, 27 

polygon 20 

of system of forces 42 

Equivalence, defined 6 

Equivalent uniform bridge load. . 

-< 218, 219 

Exact method, moment of inertia. 61 

Expansion in bridges 213 

and contraction 70 



F 



Figure, polygonal 125 

Fink trusses 67 

Maximum stresses in 119 

Fixed beam, one end 189 

columns 151 

columns, stresses in 157 

ends, horizontal components of 

reactions equal 108 

ends, parallel, reactions 105 

truss, reactions 78 

Flange stresses, plate girder 285 

Floor beam, maximum reaction... 279 

system, bridge 213 

Force, definition. . . , 4 

diagram, definition 6 

Moment of 38 

polygon, described 9 

triangle, described 9 

Forces at a Joint 95 

Centroid of parallrt 49 

Composition of 6 

Concurrent, resolution of 11 

Concurrent 8 

Concurrent, equilibrium of.. 12, 26 

Different polygons for same. ... 34 

Kinds of, defined 5 

Moment of system of 42, 44 

Moments of 85 

Moment of parallel 45 



INDEX. 



313 



PAOK. 

Forces, non-concurrcut 10 

Non-concurrent, resolution of... 23 

Non-concurrent, non-parallel. . . . 1(» 

Forces, one side of section 97 

Parallel 22 

parallel, moment of inertia of. 53 

Resolution of G, 83 

Resultant of concurrent 9 

System of 41 

Formulse for beams 198 

roof weights 71 

wind pressure 73, 74 

weight of bridges 214, 215 

Frame, triangle 65 

Framed structures 65 

Funicular polygon 20 

Closing of 23 

through two points, 35; three. 30 



G 



General method for determining 

stresses 85 

Geometrical areas 47 

Girder, plate, problem 184 

stresses and shears 285 

Grand stand truss, stresses 100 

Graphic determination, radius of 

gyration 58 

methods for beams 107 

methods for beam deflections... 178 

transverse bent 154 

moments 43 

moments, stresses by 91 

resolution, stresses by 99 

statics, defined 3 

Gravity, center of 47 

center of, irregular area 50 

Gyration, radius of 57, 60 



II 



Highway bridges, live loads for. . . 210 

Weights of 214 

Hinged arch 143 

Columns 151 

Columns, stresses in 155 

Horizontal components 80 

ot reactions equal 108 

Howe bridge truss 210 

roof truss 67 

Ilutton's formula for wind pres- 
sure 73 



Inaccessible points of intersection. 31 

Inclined chords, truss with 280 

Inclined chords 306 

surface, wind pressure on 73 

Incomplete framed structure 66 

Influence diagrams 267 

Inertia, see Moment of Inertia. 
Irregular areas, 49 ; center of 

gravity of 50 

Intersection, inaccessible points of 31 

Interurban bridges, live loads for. 216 



J 

PAGE. 

Jack rafters 69 

Joint, forces at 95 

Position of loads for maximum 

moment at 268, 271 

Joists, bridge 213 

K 

Ketchum's formula for roof 

weights 71 

Kinetics, defined 3 

Knee-braces 213 

Known lines of action 23 

L 

Lateral bracing ; 212 

systems, wind load stresses in. 255 

Leeward end of truss on rollers.. 110 

rollers 81 

segment of arch, wind load 148 

Line of action, defined 5 

of pressure of arch 35, 37 

Lines of action, known 23 

Line, load and moment diagram. 295 

Live loads for bridges 216 

Load, dead 69 

Dead, arch, reactions and 

stresses 144 

line and moment diagram 295 

reactions, Wind 78 

Snow, on roof 72 

Uniform, on bridge 217 

Uniform, on cantilever beam... 169 

Uniform, overhanging beam.... 177 

Uniform, simple beam 173 

Uniform, beam one end fixed . . . 193 

Wind 73 

Wind, stresses 105 

Wind, stress on arch 146, 148 

Wind, stresses in lateral systems 255 

Loaded chords 298 

Loads on bridge 214 

Concentrated I68 

Concentrated, simple beam 171 

Concentrated, one fixed end 189 

Concentrated, overhanging beam 175 

Engine and train 267 

Maximum, for floor beam reac- 

^ tion 279 

Moving 199 

Position of, for maximum mo- 
ment 208, 271 

Position of, for maximum shear. 

275, 277 

Position of, for maximum stress 

in web 280 

on roofs 69 

on lower chord 103 

on trusses 72 

on upper chord 99 

Wind, on bridges 219 

Long compression members 68 

Lower chord Qg 

M 

Magnitude, definition 5 

Main braces '. 127 



814 



INDEX. 



PAGE. 

Main dlagonalH 127 

trusses, bridge 211 

Mazlmum and minimum stresses. 124 

stresses 114, 118 

chord stresses, Pratt truss 201 

flange stresses, plate girder.... 285 

floor beam reactions 279 

moment, concentrated moving 

load 201 

moment diagrams 297 

moment, position of loads for . . 

268, 271 

moment, uniform load 199 

■hear 302 

shear, plate girder 285 

■hear, position of load for. 275, 277 

■hear, two loads 207 

■hear, uniform load 200 

■tress. Inclined chords 280 

stresses, with subordinate brac- 
ing 308 

stress In web members 280 

web stresses, Pratt truss 291 

web stresses 306 

Members of bridge truss 211 

Web, defined 08 

Merriman's formula for roof 

weight 71 

Methods for moment of inertia, 

Accurate, 61 ; Approximate. . . 59 

Method of coefficients for stresses. 259 

Cuimann*s 53 

for determining stresses 85 

Graphic, for deflection 178 

Mohr's 55, 63 

Minimum and maximum stresses. . 124 

Moment area 39 

(See also Bending Moment.) 

beam with both ends fixed 195 

Bending 167 

Bending, simple beam 171 

Constant, of Inertia 178 

of couple 40 

diagram and load line 295 

formulse for beams 198 

of Inertia 52 

of inertia of areas, 58 ; about 

parallel axis 63 

of Inertia, table 63 

of inertia, variable 184 

Maximum, tables 297 

Maximum, concentrated moving 

load 201 

Maximum, uniform load 199 

of parallel forces 45 

Position of loads for maximum. 

268, 271 

of resultant 39 

of system of forces 42, 44 

Moments, defined 38 

Algebraic, stresses by 86 

of forces 85 

Graphic, 43 ; stresses by 

graphic 91 

Stresses by, in bridge trusses . . 238 

Motion, defined 4 

Moving loads 199 



N 
Negative moments 38 



PAGB. 

Non-concurrent forces 5, 16 

Equilibrium of 26 

Resolution of 23 

Non-coplanar forces, defined 5 

Non-parallel chords, counterbraced 141 

Non-concurrent forces 16 

forces, problem 29 

Notation, Bow's 86 

Counterbraclng 128 

descrllied 7 



O 

One end of truss on rollers 80 

Overhanging beam, concentrated 

loads 175 

uniform load 177 



Parabolic bowstring truss. .. .210, 247 
Parallel axes, moment of inertia . . 

56, 64 

chords, counterbraced 139 

chords, stresses in trusses with. 259 

forces 22 

forces, centroid of 49 

forces, moment of 45 

forces, moment of inertia of . . . . 53 

forces, problem 27 

reactions 79 

reactions, fixed ends 105 

Parallelogram, area of 47 

l*article, definition 4 

Pedestals 213 

Permanent loads on trusses 72 

Petit bridge truss 210 

Pin connections 213 

connected trusses 68 

Pitch of roof, defined 68 

Plate girders 297 

Problem 184 

Stresses and shears 285 

Point, defined, 4 ; of application, 

defined 5 

Position of loads for maximum 

moment at 268, 271 

Points, inaccessible intersection. . . 31 
Polygon, through two, 35 ; 

tnrough three 36 

Pole, defined, 20 ; pole distance, 

defined 20 

Polygon, closed 11 

Closing funicular 23 

Equilibrium 20 

Force, described 9 

Funicular 20 

through two points, 35 ; 

through three 36 

Polygonal figure 125 

Polygons, different for same fig- 
ures 34 

Portals 212 

Position of loads for maximum 

fioor beam reactions 279 

for maximum moment, concen- 
trated moving loads 203 

of loads for maximum moment. 
268, 271 



IND£X. 



315 



PAGE. 

Position of loads for maximum 

8ii6ar •••••••••••••■•••^1 Of mil 

of loads for maximum stress, 
web members 280 

Positive moments 38 

Pratt roof truss 67 

truss, stresses in 228 

bridge truss 210 

truss, coefficients for 204 

truss, maximum cord and web 
stresses 201 

Pressure line of arch 3o, 37 

Wind 73 

Problems, general 32, 40, 75 

in equilibrium 13, 27 

Problem, Algebraic moments 88 

Graphic method for deflections. 181 

Graphic moments 92 

Graphic resolution 104 

Leeward end on rollers 110 

Maximum stresses in Fink truss 119 

Plate girder 184 

Stresses by algebraic resolution. 96 

Stresses in cantilever truss 114 

Truss with non-parallel chords, 

counterbraced 133 

Truss with counterbraced paral- 
lel chords 130 

Unsymmetrical truss 116 

Wind load stresses 105, 108 

Windward end on rollers 112 

Purlins, 69 ; weight of 70 

Q 

Quadrangular truss 67 

Quadrilateral, area of 48 

R 

Radius of gyration 57 

Table 63 

of area 60 

Rafter, jack 69 

Rafters, weight of 70 

Railroad bridges, live loads for.. 216 

Weights of 215 

Rays, deflned 21 

Reactions for arch, dead load.... 144 

Effective, roof 78 

fixed ends 108 

by graphic resolution 99 

Horizontal components, equal... 108 

Maximum, floor beam 279 

Parallel 79 

Parallel, fixed ends 105 

for roof loads 75 

for single load, arch 143 

for transverse bent 154 

Wind loads, roof 78 

Redundant frame 66 

Relation between different poly- 

f:ons for same forces 34 

ution of forces 6. 85 

of concurrent forces 11 

of non-concurrent forces 23 

Resolution, stresses by graphic... 99 

Algebraic 95 

Rest, definition 4 

Restrained beams 187 

Resultant of parallel forces 22 



PAGE. 

a couple 21 

defined 6 

Moment of 39 

of non-parallel, non-concurrent 

forces 16, 18, 21 

of several concurrent forces 9 

of two concurrent forces 8 

Rigid body, definition 3 

Rise, defined 68 

Riveted connections 213 

Eccentric 164 

trusses 68 

Rollers, leeward end on 110 

under truss 80 

Windward end on 112 

Roof (see also Trusses). 

construction 69 

Effective reactions 78 

loads 69 

Reactions 75 

trusses, 65 ; types, 67 ; weights 

of 71 

truss stresses 84 

Rotation 4, 38 

S 

Saw tooth roof 67 

Section, forces on side of 97 

Moment of Inertia of, table .... 63 

Radius of gyration, table 63 

Segment, area of 48 

Leeward, of arch 148 

Windward of arch truss 146 

Sector, area of 48 

Shear, defined 85 

beams I68 

beams, with both ends fixed 195 

diagrams 296 

diagrams, simple beam 171 

Maximum, two loads 207 

Maximum, for uniform moving 

load 200 

Maximum, position of loads for. 

^^ 275, 277 

Shears 302 

plate girder 285 

Stresses by, In bridge trusses. . . 238 

Sheathing 69 

Shingles, weight of 70 

Signs of stresses 86 

Simple beam 171 

construction, beam with one 

fixed end 191 

Slate, weight of 70 

Snow load 72 

reactions, roof 77 

stresses, transverse bent 152 

Space diagram, definition 6 

Span, denned 67 

Special construction for funicular 

polygons 34 

Stand, grand, truss stresses IGO 

Steel (see also Corrugated Steel). 

Weight of 70 

Statics, defined 3 

Straight line formula for wind 

pressure 73 

Strain, deflned 84 

Stress, defined 34 

diagram 102 

diagram, unsymmetrical truss.. 116 



316 



INDEX. 



PAGE. 

Stress iu trestle bent 1G2 

Wind, on arch 146 

Stresses by algebraic moments. . . 86 

by algebraic resolution 05 

for arch, dead load 144 

in bridge trusses (sec names of 
trusses). 

in cantilever trusses 144 

and coefficients, Warren truss. 264 

in columns 155 

in counterbraced trusses.. . .120, 130 

in flange, plate girder 285 

in grand stand fruss 160 

by graphic moments 01 

by graphic resolution 00 

Alaximum 114, 118 

Maximum and minimum 124 

Maximum, with inclined chords. 280 

by moments and shears 238 

in roof trusses 84 

with subordinate bracing 308 

in transverse bent 150, 151 

in trusses, method of coefficients 250 

in trusses with parallel chords. 250 

in unsymmetrical trusses 114 

Web 306 

Wind load 105 

Wind load, in lateral systems. . . 255 

Strings, defined 21 

Stringers, bridge 213 

Structures, framed 65 

Strut, defined 68 

Sway bracing 70, 213 

Subordinate bracing 273, 308 

System of forces, moment by.. 42, 44 

Moment of resultant 41 

Floor, of bridges 213 

of parallel forces, moment of 

inertia 53 



PAGE. 

Grand stand 160 

with Inclined chords 280, 306 

on rollers 80, 112 

Roof, stresses 84 

with subordinate bracing 808 

triangle 3 25 

Unsymmetrical, stress diagram 

for 116 

Trusses, bridge, types of 210 

Cantilever 114 

Counterbraced, stresses in 139 

Economical 68' 

Loads on 72 

Roof 65 

Weight of 71 

Stresses in, with parallel chords 259 
with parallel chords, counter- 
braced 139 

Stresses in counterbraced 129 

Unsymmetrical 114 

Two non-parallel, non-concurrent 

components 24 

Types of roof trusses 67 

of bridge trusses 210 

IU 

T^nlform load, beam fixed one end 193 

on bridge 217 

on cantilever beam 169 

overhanging beam 177 

loads, simple beam 173 

T'nloaded chords 300 

Upper chord 68 

Loads on 09 

Unsymmetrical trusses 114 

truss, stress diagram 116 



Table of moment of Inertia and 

radius of gyration. 63 

Tar and gravel roof, weight of . . . 70 

Tensile stress, sign of 86 

Tension, defined 84 

Throe hinged arch 143 

non-parallel, non-current com- 
ponents 23 

Through bridges 200 

Tie, defined 68 

Ties for tracks 213 

Tiles, weight of 70 

Tin, weight of 70 

Top of column 151 

Tooth, saw, roof 67 

Train loads 267, 285 

Transformation of moment area.. 30 

Transverse bent, stresses In 150 

Translation, defined 4 

Trestle bent 162 

Triangle, area of 48 

frame, shaped 65 

Force, described 

as a truss 125 

Truss (see also Bridges and Roofs 
and name of each truss). 

Arch 143 

Fink, maximum stresses in 110 

fixed both supports 78 



Variable moment of inertia 184 



W 

Warren bridge truss 210 

Coefficients and stresses In ... . 264 

Stresses in 306 

Web members, defined 68 

Maximum stress in.. 280 

Maximum stress in, Pratt truss. 201 

Stresses 306 

Weights (see article). 

of highways bridges 214 

of railroad bridges 215 

Wheel loads on bridges 217 

Whipple bridge truss 210 

Wind load 73 

on bridges 210 

reactions 78 

stresses in arch truss 146 

stresses fixed column 157 

stresses hinged column 155 

stresses in lateral systems 255 

Leeward segment of arch .... 148 
on roof 105 

f)ressure 73 
ndward end of truss on rollers 

82, 112 

segment, stress on arch 146 

Wooden shingles, weight of 70 



-.C 



r.'