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London: CHAPMAN & HALL, Limited 


Ey,o 71"?. 03 



Copyright, X903, 





This volume has been written in the hope of helping 
students of engineering and architecture to acquire a knowl- 
edge of Statics which will include the power to apply it cor- 
rectly in professional work. To this end an attempt has been 
made to carry out several specific purposes, prominent among 
which may be mentioned the following : 

1. To give much attention to the starting points of the sci- 
ence, and to make as clear as possible the course of deduction 

2. To point out the inherent mathematical limitations of 
pure Statics, and to show how all its important problems 
are solved. 

3. To develop algebraic and graphic methods of solution, 
or, if one prefer the terms, Analytical and Graphical Statics, 
side by side and with equal thoroughness. 

4. To present a graded set of problems illustrating not 
only universal principles but also how Statics is used in en- 
gineering practice. 

5. Finally, to keep the book of a size commensurate with 
the small amount of new matter which the reader, versed in the 
simplest operations of elementary mathematics, need master to 
gain the desired end. 

It may be pointed out that the phrases Analytical Statics 
and GrajJhical Statics are avoided. The ground for this is that 
there seems to be no necessity for using them, in this work at 



least, and that the terms seem objectionable as tending to 
obscure the unity of Statics, and to produce the impression 
that two merely alternative methods of procedure from identi- 
cal premises and of identical mathematical significance are 
loosely connected if not actually distinct branches of the 

The subject will be found developed in the following pages 
in such a way as to make it possible to solve problems from 
the outset by both methods in parallel (as in the plates), and 
the practice of making such double solution is believed to be 
of great value not only for the drill of checking one's own 
work, but also for the clearer light in which each method is 
seen by being kept in close relation to the other. Moreover, 
as the student checks the correctness of his own work, it is 
possible, even with classes of upwards of one hundred, to 
assign different problem data to each student, and still keep 
the labor of inspecting students' work at a minimum. After 
some experience with the two methods side by side, it is 
believed that practice should be had in the rapid solution of a 
large number of varied problems such as can be found in the 
familiar works on Statics, in which a single solution by either 
method is accepted and in which the student judges the 
correctness of his results not only by examination of his 
work, but by refkction upon their reasonableness under the 

Of course many different sources have been drawn upon 
freely for suggestions, methods, and material, but it will not 
be out of place to mention as especially prominent among 
them Rankine's Applied Mechanics and Hoskins' Graphic 
Siatics, and the writer takes pleasure in acknowledging his 
obligation accordingly. Grateful recognition is also due to 
Professor I. N. Hollis for valued criticism. 
Cambridge, Mass., July, 1902. 





1. Algebraic and Graphic Methods Compared I 

2. History of Graphic Methods in Statics 3 



3. Mechanics and its Subdivisions 4 

4. RigidBody. Particle. 5 

5. Rest and Motion 6 

6. Force 7 

7. Elements of Force 7 

8. Magnitude 8 

9. Direction and Sense 8 

10. Point of Application 8 

1 1. Coplanar and Non-Coplanar Forces. 9 

12. Concurrent and Non-Concurrent Forces 9 

13. Equilibrium 9 

14. Equivalence 10 

15. Resultant and Equilibrant 10 

16. Components 10 

17. Relation Between Two Components and their Resultant. Parallelogram 

of Forces 10 

Exercise 1 ii 

18. Couples and their Moments 12 

19. Center of Rotation for a Couple 13 

20. Single Force Always Replaceable by a Single Force and a Couple 13 

21. Moment of a Force 14 

Exercise 2 15 




22. Propositions Regarding the Moment of a Force 15 

23. Proposition Regarding the Moment of a Couple 16 

24. Sets of Forces Classified 16 

25. Conditions of Equivalence 17 



26. Notation. Conventions Regarding Elements of Forces 19 

27. Space Diagram and Magnitude Diagram 20 

28. Illustration of Scheme of Notation 20 



29. Demonstration of the Parallelogram of Forces 22 

30. Triangle of Forces 24 

31. Polygon of Forces. Location of Resultant of Inclined Forces. Magnitude 

Polygon 25 

32. String Polygon 26 

33. Additional Remarks on the String Polygon 30 



34. Nature of Statical Problems. 31 

35. Condition of Equilibrium 31 

36. Algebraic Statement of {A) and {B) (the Conditions of Equilibrium) for 

Coplanar Forces 32 

37. Graphic Interpretation of (A) and (B) 34 

38. Exercises in the Composition of Forces 35 

Exercises 3-7 38 

39. Generalization of the Three Classes of Resultants 39 

40. Establishment of Equilibrium by Use of Moments Alone 39 

41. Determination of Magnitudes by Single Moment Equations Alone 41 

42. Convention as to Algebraic Signs in Moment Equations 43 

43. Six Methods of Stating the Conditions of Equilibrium 43 



44. General Survey of the Scope of Pure Statics 45 

45. The Four Cases 47 



46. The Solution of Statical Problems 49 



47. Solution of Case i 50 

48. Solution of Case 2a 50 

48 a. Solution of Case 2b 51 

49. Solution of Case 3 53 

50. Solution of Case 4 55 

50 a. Remarks on Cases 3 and 4 58 

Exercises 8-15 58 



51. Graphic Representation of the Moment of a Force 62 

52. String Polygon for Parallel Forces a Diagram of Moments 63 

53. Remarks on the String Polygon as a Diagram of Moments 65 

54. To Pass a String Polygon Through Three Given Points 66 

Exercise 16 70 

55. An Alternative View of the Rays and the String Polygon 70 





56. Center of Gravity ' 72 

Exercise 17 77 



57. External and Internal Forces 78 

58. Stress 78 

59. Kinds of Stress 79 

60. Combined Stresses 79 

61. Further Particulars Relating to Stress 80 



62. Definitions 83 

63. Extent of Approximation to True Frames in Practice 84 

64. Loads Applied Elsewhere than at Joints 85 

65. Frames in General 86 

66. Loads 88 

67. Stresses in Structures 88 



68. Stresses in Non-Framed Structures 90 

Exercise 18 91 



69. Shear Diagrams 91 

7a Flexure Diagrams 92 

Exercise iq 93 

71. Coxmection Between Shear and Change in Flexure 93 

fjcercise 20 95 



72. Stresses in Framed Structures 96 

73. Method of Sections 97 

74. Method for Determining all the Stresses in a Frame under a Given 

Load 99 

75 . Example icx> 

76. Stress Diagrams 103 

77. General Instructions Regarding Exercises Involving Stress Diagrams. . • . Z04 

78. Special Instructions Regarding Exercises 21-23 • ^^ 

£biercises 21-23 i<>4 



79. Complications in Connection with the Analysis of Frames 106 

80. Reactions due to Non-vertical Forces 107 

81. The Fink Truss 108 

Exercise 24. 108 

82. Triangular Frame with Trussed Top Chord 108 

Exercise 25 109 

83. Counters 109 

Exercise 26 • no 

84. Bent of a Mill Building '. no 

Exercise 27 114 

85. Cantilever Bridge 115 

Exercise 28 115 

86. Three-Hinged Arch 116 

87. Line of Pressure 118 

Exercises 29, 30 119 

88. Hammer Beam Truss 120 

Exercise 31 , 122 

89. Stresses Due to Moving Loads 123 

90. Stability of a Masonry Dam 124 

Exercise 32 125 

91. Action and Reaction Not Necessarily Normal to the Surface of Contact ... 125 

92. Friction 126 

Appendix 130 




I. Algebraic and Graphic Methods Compared. — There 
are two methods of representing quantities for the purpose of 
computation, one in which numerals or letters are used — ^the 
algebraic or so-called analytical method — and the other in 
which quantities are represented by the lengths and relative 
positions of lines — the graphic method. 

The quantities of higher mathematics, such as imaginaries 
and infinitesimals, can be treated algebraically only. On the 
other hand, many complicated relations, such as profiles of rail- 
road lines and the varying pressure in a steam cylinder, can 
be comprehended and made subjects of computation or measure- 
ment only by graphic representation. 

The great bulk of ordinary arithmetic and trigonometric 
calculation can be done by either method, and which of the two 
should be selected is purely a matter of expediency. In the 
fundamental operations of arithmetic — addition, subtraction, 
multiplication, division, involution, and evolution — the graphic 
method is either so simple and obvious that it is rarely thought 


of as a formal method in connection with them, or else so dis- 
advantageous as to be little used. 

Problems involving much finding and summing of products 
or involving trigonometric or complicated geometric relations 
are found to be an advantageous field for the use of graphic 
methods, and in this field, graphic methods assume a place of 
relative importance. 

Problems in equilibrium are of precisely this sort, and 
graphic methods of dealing with them have been developed very 
extensively, and even named, as a body, Graphical Statics, in 
distinction from the body of algebraic methods in statical prob- 
lems — called Analytical or Algebraic Statics. All these 
terms, as pointed out in the preface, seem objectionable to the 
writer and will not be used further in this work. The terms 
algebraic method and graphic method will be used in their 

The underlying principles and the procedure in both meth- 
ods are, of course, scientifically correct, but they differ in the 
degree of precision attainable. The precision of results in 
algebraic computation depends upon the extent to which deci- 
mals or significant figures are carried out, and in graphical 
work upon care and skill in draughting. In the former case 
absolute precision may be approached as closely as the com- 
puter may care to go, but in the latter early bounds are set by 
the limitations of draughting appliances and human eyesight. * 
Sometimes it is worth while to use a combination of the two 
by sketching more or less roughly the desired diagrams and 
calculating trigonometrically instead of scaling the lengths of 
resulting lines. This of course is the algebraic method itself 
so far as precision goes. 

Whatever may be said about their relative precision and 
scope, it is certainly true that neither method can yield a result 
more accurate than the data from which it is derived. In 


structural design, the data are rarely or never known to a de- 
gree of accuracy greater than the degree of precision readily 
obtained in graphic computation, so that actually, from the 
point of view of the designing engineer, one method may 
be said to yield results as trustworthy as the other. As 
already stated, the two methods are of equal correctness in 

The choice between one method or the other for a given 
problem before an engineer is a matter of convenience, but in 
many cases both should be used for the sake of the check upon 
results thus obtained. 

The graphic method has the advantage, sometimes impor- 
tant, that it presupposes a knowledge of little or no mathe- 
matics beyond the elements of plane geometry, while the 
draughting appliances required in its use are of the simplest. 

2. History of Graphic Methods in Statics. — Knowl- 
edge of the graphic methods in statics and of their importance 
and scope is due mainly to the efforts of the late Professor 
Culmann of Zurich. Prior to 1866, when he published his great 
work, **Die Graphische Statik," the subject had attracted 
little attention and there existed only scattered and frag- 
mentary writings on it. Culmann died after completing the 
first, only, of his contemplated volumes, but he had thor- 
oughly established the subject, and interest in the graphic 
treatment of statical problems steadily spread through the 
engineering world. Among other names to be prominently 
associated with the development of the graphic methods are 
Bow, Maxwell, Mohr, W. Ritter, Cremona, Miiller-Breslau, 


3. Mechanics is the science which treats of rest and 
motion. It is subdivided into Kinematics and Dynamics. 

Kinematics treats of motion apart from its causes. Its 
scope is accordingly limited to the consideration of the paths 
of moving bodies and of that aspect of velocity and accelera- 
tion which involves only the relations between space and time. 

Dynamics treats of the effect of forces upon rest or motion. 
It is subdivided into Kinetics and Statics. Kinetics deals 
with the cases in which given forces produce a change in a 
body with respect to rest or motion, and Statics with the cases 
in which no such change is produced. 

Kinetics accordingly deals with the velocity and accelera- 
tion of given bodies resulting from given forces, and with such 
topics as work and energy, momentum, centrifugal force, 
impact, etc. Statics is concerned exclusively with the con- 
ditions under which a body under the action of forces will 
remain at rest or undergo no change of motion.* 

Statics — henceforth the exclusive subject of this book — is 
accordingly the science of equilibrium, of stability — the science 
by the aid of which are determined the forces necessary to 

* The four terms Kinematics, Dynamics, Statics, and Kinetics have an etymo- 
logical fitness which is worth noticing. Kinematics is from Gr. kituma, motion; 
Dynamics fh>m Gr. dynamis^ force; Kinetics from Gr. kinetikoSy putting in motion; 
and Statics from Gr. statikos, causing to stand. 



maintain a body undisturbed in its rest or motion in spite of 
disturbing tendencies, the forces which must interact between 
the various parts of the body to prevent its disruption under 
the action of given forces, and moreover the position in which 
a body or system of bodies under a given set of forces will be 
at rest. 

4. Rigid Body. Particle. — A rigid body is a body con- 
ceived to be incapable of change in shape or size under the 
action of forces. Such a body could be affected only as a 
whole by forces which act upon it. Herice to state that the 
body concerned in any problem is rigid or is to be so consid- 
ered is equivalent to stating that no thought is expected to be 
given to the possibility of the problem being complicated by 
the deformation or rupture of the body. 

Although no such thing as a rigid body exists in nature, 
all solids approximate rigidity to a greater or less extent. In 
the case of solids subject only to forces well within their capac- 
ity to withstand them, such as all properly designed engineer- 
ing structures, the approximation to rigidity is very close, and 
for the purposes of statics no material error results from as- 
suming complete rigidity in such cases. 

A complete examination into the safety of an engineering 
structure involves {a) the determination of the forces acting on 
the body or transmitted through its various parts ; and {b) the 
study as to whether the body and its parts are able to resist 
such forces without rupture or undue deformation. In the first 
step the body is assumed rigid, because the results of the step 
are practically the same as if the body were rigid, and in the 
second the lack of rigidity is clearly recognized and provided 
against. The first step falls within the domain of statics, and 
the second within that of the kindred science. Resistance of 

In statics it is the general practice to treat bodies as if they 


were rigid, unless distinctly stated to the contrary, — actual 
lack of rigidity being taken into account only as an aid in the 
solution of certain complex problems which would otherwise 
be indeterminate. 

A particle is a rigid body of the smallest conceivable di- 
mensions. It is also called a material pointy or, for short, 
simply a point. 

Every problem involving forces implies the existence of 
bodies or particles on which the forces act. 

5* Rest and Motion. — Rest is the relation existing be* 
tween two points when a line connecting them does not change 
either in length or in direction. If, on the other hand, the line 
does change in either length or direction, the relation between 
the points is motion. This line can change in one or both of 
these two particulars only, and motion can accordingly be 
divided into two corresponding classes only, translation and 

If the line between two points changes in length, one point 
is in translation with regard to the other. If the change is in 
direction, one point is in rotation about the other. 

Two bodies are at rest with regard to each other when 
every point of one is at rest with regard to every point of the 

Two bodies are in motion with regard to each other when 
any point of one is in motion with regard to any point of the 

The motion of a body is translation when any point of that 
body describes a straight line. If the points of the body de- 
scribe a series of straight lines, the motion is pure translation. 

The motion of a body is rotation when such points as 
change in position at all describe sets of concentric circles in 
parallel planes. 

The common normal to these planes which contains the 


centers of all the circles is called the axis of rotation. This 
axis may or may not traverse the body, but any points not 
changing in position must be contained in it. If the axis of 
rotation is fixed in position, the motion is pore rotation. 

Of course there can be compound translation, as that of a 
body sliding across the deck of a moving ship ; or compound 
rotation, as that of the moon about its own center, and also 
about the earth; or combined translation and rotation, as that 
of a ball rolling in a straight path. 

Two bodies at rest with reference to each other may be 
moving with respect to a third. Rest, then, far from meaning 
absolute motionlessness, means only motionlessness with 
regard to a definite body of reference, which in our work will 
usually be understood to be the earth. 

It may be added that translation can be regarded as the 
extreme case of rotation in which the axis is at an infinite 
distance, and instead of two subdivisions of motion, rotation 
alone might be regarded as covering the whole ground. This 
point of view, however, will not be convenient for the purpose 
now in hand. 

6. Force, — Force is an interaction between two bodies, 
either causing or tending to cause a change in their relative 
rest or motion. A force may always be conceived of as a 
push or a pull. 

Forces are designated by single capital letters as P, Q, R^ 
etc. In graphical work more elaborate designations are 
needed, which will be explained in .Chapter II. 

7. Elements of a Force. — In order that a force may be fully 
known, three characteristics of it, which will be spoken of as 
elements, are essential and sufficient, viz. : 

1. Magnitude. 

2. Direction (including sense). 

3. Point of Application. 


8. Magnitude. — The magnitude of a force is given by stat- 
ing numerically the ratio of its effectiveness in producing trans- 
lation to that of some arbitrarily selected standard force, 
usually the force of gravitation on a certain mass of standard 
material at a certain level and latitude; i.e., to the amount 
of force exerted by gravitation on the standard pound or other 
standard weight. In other words, the magnitude of a force 
is stated in units of weight. 

The absolute unit of force, that is the force which, acting 
for a unit length of time on a unit mass of material, will give 
rise to a unit velocity, is not used in statics. 

9. Direction and Sense The direction of a force is the 

direction of the line along which the force tends to produce 
motion, i.e., of the line of action of the force. Direction in- 
cludes the sense of the force as well as the slope of its line of 
action. By sense is to be understood the specification as to 
which of the two ways along the line of action the force 

Sense is the distinction between forward or back, up or 
down, to right or left on a given line. It is merely a matter 
of algebraic sign, requiring no separate equation for its deter- 
mination, and hence is not ranked as an element. 

10. Point of Application. — The point of application is the 
place (treated as a point) upon a body where the force is brought 
to bear. This point is, of course, upon the line of action and, 
together with the slope, it locates that line. 

In the case of rigid bodies, any point whatever on the line 
of action of the force may be taken as the point of application.* 

Change in the point of application of a given force affects 
the body only as to rotation, not at all as to translation. 

* This will be seen upon adding anywhere in the line of action of the force two 
opposite forces coincident with it and both equal to it in magnitude. The three, 
obviously equivalent to the original, amount simply to a new one just like it. 


11. Coplanar and Non-Coplanar Forces. — Forces are 
spoken of as coplanax and non-coplanar according as their 
lines of action do or do not lie in one plane. 

Except where distinctly stated to the contrary, and except 
where statements are evidently general (as in the whole of this 
chapter), coplanar forces will form the exclusive subject Oi 
these pages. This limitation conduces to simplicity without 
real loss of generality, for coplanar forces once understood, 
the treatment of non-coplanar forces will offer little difficulty 
to the unaided reader. 

12. Concurrent and Non-Concurrent Forces.— Forces are 
spoken of as concurrent or non-concurrent according as their 
lines of action do or do not meet in a point. Of course, either 
may or may not be coplanar. The study of concurrent forces 
is sometimes spoken of as statics of a particle, and of non- 
concurrent forces as statics of a rigid body. 

Whether a set of forces is concurrent or not is a matter 
affecting rotation only, the translatory effect of a force being 
independent of its point of application. 

13. Equilibrium. — Equilibrium is that condition of a set 
of two or more forces the result of which is that their combined 
effect on a body produces no change in the body with respect 
to rest or motion. 

A body is said to be in equilibrium when it is acted on by 
a set of forces in equilibrium. A body in equilibrium may be 
either at rest or moving with uniform speed (either of transla- 
tion or rotation) with regard to the body of reference. If it 
be moving in either of the ways just mentioned, it is the result 
of the previous application of an unbalanced force or of un- 
balanced forces, and any forces actually affecting it while in 
such motion are forces under the action of which a body could 
equally well be at rest. 

A set of forces, therefore, under which a body will be at 


rest is always a set in equilibrium ; or, in other words, condi* 
tions of rest are always conditions of equilibrium. It follows 
that, by establishing the conditions of rest, the familiar most 
usual phase of equilibrium, we establish the general, universal 
conditions of equilibrium. 

14. Equiyalence. — Two forces or sets of forces having 
identical effects on the rest or motion of a body both with re- 
spect to translation and to rotation are termed equivalent. 

15. Resultant and Equilibrant. Composition of Forces. 
— A single force equivalent to a set of forces is called the 
resultant of the set. A single force the addition of which 
to a set of forces produces equilibrium is called the equilibrant 
of that set of forces. The resultant and the equilibrant for a 
given set differ in sense only, i.e., one is simply the equal and 
opposite of the other. Of a set of forces in equilibrium any 
one of them is the equilibrant of all the rest. The operation 
of finding the resultant or equilibrant of a set of forces is called 
Composition of Forces. 

i6. Components. — Any one of a set of forces having a 
given resultant is a component of that resultant. Evidently 
there may be any number of components of a force. 

Rectangular components of a force are two components 
equivalent to it whose lines of action are at right angles to 
each other. 

A component in a given direction is understood to be the 
one of a pair of rectangular components whose line of action is 
parallel to the given direction. Such a component measures 
the total tendency of the given force to produce motion in 
the given direction. 

17. Relation between Two Components and their Result- 
ant. Parallelogram of Forces.* — If the adjacent sides of a 
parallelogram represent two concurrent forces in magnitude and 

* See also § 29 for proof and additional discussion of the parallelogram of forces. 


direction, the diagonal of this parallelogram starting at the in- 
tersection of these two sides will represent their resultant in 
magnitude and direction, provided the two forces act either 
both toward or both away from this intersection. The result- 
ant will then also act toward or away from the point of inter- 
section as the case may be. Ordinarily the components are 
regarded as acting away from their common point, and accord- 
ingly the resultant also. 

Corollary. — If a be the inclination of any force, P, to any 
given line, the components along and at right angles to this line 
will have the magnitudes P cos a and P sin a respectively. 

Exercise i.* Components. Find the components, both algebra- 
ically and graphically, of a force P whose magnitude, direction, and 
point of application are 20 lbs., 210'', and (2, 6) respectively, along the 
axes of A* and K,and also along five lines, M\ Nx^ M% N%^ M% N%, M^ N^ 
M% N%, inclined to the axis of -AT by 45', 6o', 120*. 45', and 30* respec- 
tively, and which pass through points (o, o), (-6, o), (6, o), (8, o), and 
ID. o) respectively. 

Suggestions as to Graphic Work. 
Use as scales i in. = 10 lbs. and 4 units of length. Plot the line of 
action of P in its proper relation to the preferably horizontal and vertical 
axes of coordinates and the given lines Mx Ni . . . M% A^». This will serve 
as a diagram of data for both methods. The problem can now well 
be stated in condensed form across the top of the sheet. 

Show P apart from this diagram with the proper direction and with 
the magnitude marked of! on it to scale. At a little distance from it, 
draw parallels to all the given lines. Construct the required com- 
ponents, scale, and show results in their proper places with dimension 

Suggestions as to Algebraic Work. 
Determine the angles between the line of action of P and each of the 
seven given lines. Proceed with the necessary computations using this 

For Mx Nx, 20 cos 45* = 14.14. 

" Mt Nt. etc. 
* For the solution of this and all the following exercises, sheets of paper about 
12 XiS inches, preferably ruled in inch and tenth-inch squares, are recommended 
as affording sufficient room, and as convenient in shape for the double solutions 
usually required. 


Finally collect results and show them in parallel columns thus : 

Line Gr. Al. 

Ml Nx 14. 1 14. 14 

Questions on Exercise i. 

How does the line MiNx differ from M^N^ ? 

What difference does this cause in the components along these two 
lines ? 

How would changing the point of application of P to say —3, 5, or 
elsewhere, affect the results of this exercise? 

From the seven components found, can two be selected which would 
fully replace P if properly located ? Is more than one such selection 
possible ? 

What would have to be the location of such components in order that 
they might replace P ? 

i8. Couples and their Moments. — ^Two equal and oppo- 
site forces with parallel lines of action constitute a couple. 
The sole tendency of a couple is to produce rotation. The 
measure of a couple, which can be nothing else than the 
measure of its tendency to produce rotation, is called the 
moment of the couple. 

It has been observed as one of the fundamental laws of 
matter that the moment of a couple is proportional alike to the 
common magnitude of the forces of the couple and to the 
shortest distance between their lines of action. This distance 
is called the arm of the couple. The product of such magni- 
tude and arm expresses the moment of the couple. 

The unit couple consists of two forces of unit magnitude 
with unit arm, and its moment is the unit for measuring 
couples. This unit is called, in the English system, inch- 
pound, foot-pound, foot-ton, etc. It must not be confounded 
with the units of work and energy with the same names. 

The sense of a couple will be called positive or negative 
according as its tendency is clockwise or not. 

The plane of a couple once known nothing further need be 


known of it but its moment, including, of course, the sense of 
the possible rotation. 

Either of the forces of a couple can be treated for computa- 
tion exactly like one of any other system of forces. 

The effect of a couple is the same wherever in its plane or 
in a parallel plane its forces may be applied. . 

Couples in the same plane can be summed by summing 
their moments, and a couple equivalent to the set can be thus 
evaluated, just as forces in the same straight line can be 
summed by summing their magnitudes with the effect of 
evaluating a force equivalent to the set. In such work sense 
must, of course, be duly regarded. 

19. Center of Rotation for a Couple. — The center of rota- 
tion for a couple may be anywhere in its plane, and is deter- 
mined only by the relative motions of the bodies from which 
come the forces constituting the couple. 

An interesting and important special case is that in which 
only one of the forces can be maintained in action at successive 
points of application. The center of rotation will then be the 
point of application of the other force, a familiar illustration 
being a pulley and belt. The point of application of the resist- 
ance of the shaft cannot change, while that of the force from 
the belt retreats freely as the pulley turns. The center of 
rotation is accordingly at the center of the shaft. 

20. Single Force Always Replaceable by a Single Force 
and a Couple. — Let P (Fig. i) be any force applied to a body 
and m be any point in the body. 

Suppose two additional forces, P' and 

P'\ opposite in sense, parallel to P 

and of the same magnitude as P, to 

be applied to the body at w, distant p ^"^^7^ 

from the line of action of P. Evidently Fig. i. 

the condition of the body is unchanged, but the three forces 


are seen to be equivalent to P with a new point of application 
nty and a couple of value /* X /• 

Hence it may be stated that any force P can be replaced 
by a force of the same magnitude and direction distant / from 
its original position, and a couple of the value P X P- In 
other words, points of application of forces can be shifted at 
will without effect on equilibrium, if^ at each change, a suit- 
able couple be added to the system. 

Any set of non-concurrent forces can thus be reduced to an 
equivalent set of couples and concurrent forces. 

31. Moment of a Force. — ^The moment of a force with re- 
spect to a point is the name given the moment of the couple 
consisting of the given force, and a force equal, opposite, and 
parallel to it conceived to be acting at the given point.* 

The arm of this couple is evidently the perpendicular dis- 
tance from the point to the line of action of the given force, a 
distance which may conveniently be called the arm of the force 
with respect to the point. 

The moment of a force with respect to a point is accordingly 
expressed by the product of the magnitude of the force into its 
arm, and is called positive or negative according as the con- 
ceived tendency to rotation is clockwise or not. 

The point to which the moment is referred is commonly 
called the center of moments, or, more briefly, the center. 

The moment of a force is sometimes referred to an axis, 
meaning thereby the axis of the conceived rotation. This axis 
will be the straight line through the center normal to the plane 
of the force and its arm. The moment with respect to this 

* Note that this is the same thing as saying that the moment o£ a force with 
respect to a point is the measure of its tendency to produce rotation about the 
point, assuming the point to be fixed. For assuming the point to be fixed is 
really assuming it to be always subject to a force equal, opposite, and parallel 
to the given one. This point of view gives rise to the phrase moment of a 
force about a point. 



axis is also spoken of as the moment with respect to the plane 

which contains this axis and is parallel to the force. 

Exercise 2. Moments of a Force. -^Find the moments in foot* 
pounds, both in magnitude and sign, of a force of 20 lbs., whose direction 
and point of application are respectively 60'' and (6, 2), about the five 
points (o, o), (—3, — 6), (8, o), (- 5. 2). (4, — 3). 

a. Obtain the results in the most direct way, by scaling arms from a 
carefully plotted diagram of data. Note that the result necessarily lacks 
precision and is affected by errors in draughting. Scale i in. = 2 ft. 
Show scaled dimensions in the diagram. 

b. Obtain the results by adding algebraically obtained moments of 
components of the given force — a purely algebraic process. * Note that 
the results are wholly independent of draughtsmanship and make a rigid 
check of results of (a) and can be made as precise as the nature of the 
case permits. 

Do the work in this form : 

= 20 X ? = ? 
20 X ? = ? 

0.500 0.866 

Ml = 20 cos 6o' X 2 — 20 sin 60® x 6 = 20.00 — 103.92 = — 83.92 

lo.o 5. 196 

Mt = etc 

Record results in this form : 




22. Propositions Regarding the Moment of a Force. — 

There follow at once from the definition of the moment of a 
force the three following important propositions. 

I. For a force greater than zero, the moment can be zero 
only with respect to a point in its own line of action. 

II. A zero moment with respect to a point off a stated line 
of action can result only from a zero force in that line. 

III. A zero force with an infinite arm can have a finite 


The ** zero force with an infinite arm " of the last proposi- 
tion is simply a way of expressing any couple, as will be seen 

23. Proposition. — The sum of the moments of two equal, 
opposite, and parallel forces (i.e., the moment ct a couple) is 
constant for all points in their plane. 

This the reader can readily prove for himself by taking any 
center and summing for it the moments of the two forces, using 
any convenient letters for their arms. 

24. Sets of Forces Classified.— It follows from § 5 that 
all possible conditions of a body with respect to rest and motion 
fall within one of four classes, viz. : 

1. Translation. 

2. Rotation. 

3. Both Translation and Rotation. 

4. Neither Translation nor Rotation. 

Any set of forces must accordingly be such as will produce 
one of these four conditions. We know by observation that 
a single force will produce translation and that a couple will 
produce rotation. Knowing also that translation and rotation 
can also result from sets of forces, we unhesitatingly conclude 
that for producing translation, a set could be replaced by some 
single force and for producing rotation, by some couple. 
Translation being the result of a single force or a set equiva- 
lent thereto, and rotation the result of a couple or a set 
equivalent thereto, it follows that any set of forces must fall 
into one of four classes corresponding to one of the four con- 
ditions of a body mentioned at the outset of this section. 

Any set of forces is therefore reducible to one of the follow- 
ing equivalents, viz. : 

1. Single Force, 

2. Couple, 

3. Single Force and Couple, 

4. Neither Single Force nor Couple, 


corresponding respectively to the four conditions of a body just 

Of course, the last named of these conditions is equilibrium. 

It should be pointed out, in passing, that a force and a 
couple in the same or parallel planes are reducible to a single 
force, and that the non-reducible case of a force and a couple is 
that in which the force is not parallel to the plane of the couple. 

25. Conditions of Equivalence.— Theorem. — If, for two 
sets of forces in a given plane, the sum of the moments of the 
forces about three points not in the same straight line be the 
same for each set, the two sets are equivalent. 

Proof. — As shown in the preceding section, each of the 
two sets must be reducible to a single force, a couple, or is in 
equilibrium, there being no possibility of the non-reducible case 
of a force and couple since the forces are coplanar. There can 
accordingly be but six different combinations of sets from this 
point of view. They and their respective relations to the 
theorem are as follows: 

{a) Each set rediuible to a single force, — Upon considera- 
tion of the moment of a force, as defined in § 2 1 , it is apparent 
that two forces can have the same moment about a point only 
when the point is distant from their lines of action inversely as 
the magnitudes of the forces. In a given case any point out- 
side a definite straight line cannot meet these conditions, 
except in the special case when the two forces are coincident, 
and of the same magnitude and direction, i.e., when the forces 
are equivalent. Therefore equality of moments for each of 
three points not in the same straight line demonstrates equiva- 
lence for this combination of sets. 

{b) Each set reducible to a couple, — The moment of a 
couple (§ 23) being a constant for all points in its plane, equal- 
ity of moments for the two sets for any one point demonstrates 
equivalence for this combination. 


{c) Each set in equilibrium. — The two sets are equivalent 
by definition, and the sums of their moments are bound to be 
equal for any and all points. The constant value of this sum is 
of course zero. 

{d) One set reducible to a single force^ the other to a 
couple, — necessarily non-equivalent. — As in combination I., 
the moments here could be equal only for points on a definite 
straight line. Their being equal for each of three points not in 
a straight line would exclude the possibility of this combination. 

{e) One set reducible to a simple force ^ the other in equi- 
librium, — necessarily non-equivalent. — Equality of moments 
could then arise only for points on a definite straight line — the 
line of action of the single force. Equality for each of three 
points not in a straight line would exclude the possibility of this 

(/) One set reducible to a couple^ the other to equilibrium^ 
— necessarily non-equivalent. — Here equality of moments 
could exist for no point whatever. Equality of moments for 
all points or any point excludes the possibility of this combina- 

Since in all of the possible combinations of sets of forces, 
the equality of the sums of the moments of the forces of the 
two sets for three points not in the same straight line is impos- 
sible without equivalence in the sets, the theorem is proved. 

Corollary. — If the sets are equivalent, their moments are 
equal for all points in the plane and conversely. 


26. Notation. Conventions Regarding Elements of 
Forces. (For illustrations see § 28.) — Magnitude is expressed 
in algebraic work by numerals or letters; in graphic work by 
lengths of lines. The capitals P^ Q, R, etc., are used both 
as general designations of forces, and also as the magnitudes 
of these forces when the magnitudes are not given numerically. 

Magnitudes may always be regarded as essentially posi- 
tive. If, in solving an equation, the value of a magnitude ap- 
pears to be negative, it is to be understood, nevertheless, that 
the force is positive as usual, but that its sense is opposite to 
that assumed in stating the equation. 

Direction can be expressed in algebraic work by the angle 
(a, 6, etc., or 30"*, 60°, etc.) which the line of action makes 
with, say, a horizontal line, and, by suitable convention, sense 
can readily be included in direction. The convention as to 
sense which will be used in this work will be the famih'ar one 
of trigonometry, in which the angles are measured anti- clock- 
wise from a horizontal axis continuously from 0° to 360°, the 
measurement always being made to that portion of the line of 
action on which the sense will be away from the horizontal 
axis. Accordingly, for example, a force with direction 30° 
would be one to the right and upwards; one of 210°, to the 
left and downwards. 

Sometimes when the line of action of a force is known but 


20 STj4T!CS, 

the sense as well as magnitude unknown, it will, as above in- 
timated, be convenient, for the algebraic determination of the 
sense, to permit it to be indicated by the sign of the root of the 
equation which determines the magnitude. The sense once 
determined, it should finally be associated with direction as 

Direction is expressed in graphic work by the slope of a 
visible line, and sense is distinguished by an arrow placed 
upon the line. 

The point of application is expressed in algebraic work by 
coordinates (one of which it will often be convenient to make 
zero) and in graphic work by points properly plotted. 

27. Space Diagram and Magnitude Diagram. — To com- 
pute graphically with forces, two distinct but intimately related 
diagrams are from the nature of the case necessary, viz., the 
Space Diagram and the Magnitude Diagram. 

In the space diagram only directions and points of appli- 
cation are plotted, and only points of application are determined. 

In the magnitude diagram only magnitudes and directions 
are plotted, and only magnitudes and directions determined. 

The scale of the former is that of lengfths, as inches, cen- 
timeters, etc., while the scale of the latter is the scale of force 
units, as pounds, kilograms, etc. 

Forces are designated in the space diagram by small let- 
ters on each Side of the line of action, and in the magnitude 
diagram by the corresponding capital letters at each end of the 
proper line. This particular style of lettering is Bow's sys- 
tem. It has great advantages over any other. 

Other convenient characters may be affixed to either dia- 
gram. See Figs. 2 and 3. 

28. Illustration of Scheme of Notation. — The scheme 
of notation enunciated in the two preceding sections may 
be illustrated fully by taking two forces, P^ and P^, their 



magnitudes, directions, and points of application being as- 
sumed respectively to be 20 lbs., 240°, (o, 4), and 40 lbs., 
150, (6, o). 

Putting them in their most condensed form algebraically, 
they may be written 

P^ = (20 lbs. 240** o, 4) 
P^ = (40 lbs. 150° 6, o). 

space Diafl^nun 
Scale, I in. = 8 ft. 

FlO. 2. 

Magnitude Diagram 
Scale, I in. s 60 lbs. 

Fig. 3. 

Graphically the same data would be shown as in Figs. 
2 and 3. 



29. Demonstration of the Parallelogram of Forces.* 

Theorem. — If two concurrent forces, P and (2» both taken as 
acting away from their common point My be represented in 
direction and magnitude by two sides of a parallelogram 
meeting at M^ their resultant, Ry is represented in all its 
elements by the diagonal starting at M. 

Proof. — From the corollary of § 25 it follows that the 
theorem will be proved if it can be shown that the sum of the 
moments of P and Q about all points O in the plane is equal to 
the moment of R about all such points O. 

Let (Fig. 4) /, q^ and r be the arms respectively of Py Q, 
and R with reference to O where O is any point whatever in 
their plane, and, noting the diversity in sign between the mo- 
ments of P and (2. we have only to prove that Rr = Pp — Qg. 

* The parallelogram of forces was first pointed out to the world in 1687 by 
Sir Isaac Newton and Varignon, probably independent of each other. The fore- 
going is only one of many known proofs. It is closely akin to that of Rankine, 
Applied Mechanics, 15th ed., p. 35. See Bowser*s Analytic Mechanics, p. 24, and 
Weisbach's Mechanik, I, p. 165. 

Many writers and teachers content themselves with the view that the whole 
proposition is a truth of observation and experience quite as much as the prem. 
ises which have to be resorted to for its proof, and so find little or no need 
for mathematical proofs. But when the premises are materially more simple and 
more a part of common experience than the proposition itself, the resulting proof 
is believed to be well worth the consideration of beginners. 



The point common to P and Q being M^ with coordinates 
X and y referred to axes, OX and OY parallel respectively to 

Fio. 4. 

P and Q, the values of the arms can be written directly from 
the figure, calling the inclination of P to Q and R, 00 and a 

p =z y sin a>, 
y = ;r sin CO, 
r =^ sin {go — a) ^ X sin a. 

From the figure, as a condition that R may be the diagonal 
of the parallelogram, by the theorem of sines 

^ R sxnico — a) , ^ R sin a 

P= > f- and Q——, . 

sin GO ^ s\n GO 

Substituting these five values in the equation Rr ^zPp— Qq 
as a means of testing its validity, there results 

Ry sin (a? — a) — Rx sin a = 

R sin {go — a) y sin go R sin a x sin go 

sm a> 

sin GO 



y sin {go —a) — xsma^= y sin {oo — a) — x sin a, • 

an identity,* and the theorem is proved. 

Corollary I. Three forces in equilibrium must meet in a 

Three forces may be parallel and still be in equilibrium — 
a fact not at variance with the preceding sentence if parallel 
forces be considered a limiting case with a common point at 

Corollary II. With rectangular axes, and with {x, y) as 
any point on the line of action of any force, it appears, 
from the value of r, that the arm of any force referred to the 
origin is 

a^=^ y cos a — X €\n a, 

where a is the direction of the force,. 

30. Triangle of Forces.— Observe in the parallelogram of 
forces of Fig. 4 that R divides the parallelogram into two equal 
triangles, MAB and MACy either of which shows fully the 
relations between two forces and their resultant as to magnitude 
and direction. Either serves the same purpose as the whole 
parallelogram so far as these two elements are concerned, 
and will be used in place of it under the name triangle of 

Observe that if the three forces are in equilibrium the 
arrows will follow one another around the triangle, and that 
in case all the arrows do not follow one another the two forces 
for which they do follow have the third for a resultant. 

* Observe that, whatever the location of M and the direction of the forces P 
and Q, the signs connecting I^ and Qg in /^r =z J^ -^ Qq, and those connecting y 
sin (oi— or) and j: sin (00— or) in the expression for r are always identical and that 
the identity here noted is not dependent upon special -conditions. 


Observe also in the triangle of forces that one of the three 
forces must always be shown in a line of action not its own ; 
in fact it will usually be found convenient to use parallels to 
the actual lines of action of all the forces. This gives rise to 
the magnitude diagram of § 27. 

From the triangle of forces it follows that if a be the incli- 
nation of a force -P to a given line, P cos a and P sin a 
measure the components of P parallel and normal to that line 
respectively (cf. Cor., § 17). 

31. Polygon of Forces. Location of Resultant of Inclined 
Forces. Magnitude Polygon. — The resultant of any set of 
forces can be determined in magnitude and direction by form- 
ing a polygon with its sides proportional to the magnitudes of 
the given forces, and parallel to them, with arrows following 
one another. The magnitude and direction of the resultant 
will be shown by the closing line of the polygon, the proper 
sense being away from the starting-point. 

This follows immediately from the triangle of forces, as 
will bo seen from Fig. 5, where the process of finding the mag- 
nitude and direction of the resultant of the four forces ab^ be, 
cd, and de is shown in detail. 

AB combined with BC by the triangle of forces. Fig. 5^, 
leads to their resultant ^ C, which is in position to be combined 
at once with CD to yield AD — the desired resultant of AB, 
BC, and CD. This process can obviously be continued indefi- 
nitely. The partial resultants are seen to be actually super- 
fluous in the construction, and the polygon ABCDE might have 
been built up without them. This polygon is commonly called 
the polygon of forces, but the more distinctive name, the 
magnitude polygon will be preferred in this book. 

Observe that while the magnitude polygon does not locate 
the points of application o{ AC, AD, or AE, it still furnishes 
important aid in finding them. We need only to observe that, 



in accordance with the parallelogram of forces, the point of 
application oi AC is at the intersection of ab and be. Then ac ' 

Fig. 5. 

can be drawn, and where it cuts cd is a point of ad. Drawing 
ady where it cuts ^j5f is a point of ae^ and so on for any number 
of forces. 

A method applicable to parallel forces will be developed 

Corollary. If the last point of the magnitude polygon co- 
incides with the first point, i.e. if the polygon closes, the result- 
ant is zero and the set of forces is equivalent to a couple or is 
in equilibrium, according to the relative positions of the given 

32. String Polygon. — As has been seen, the magnitude 
polygon takes .into account only magnitudes and directions of 
forces. The next step is to establish the general method by 
which points of application are taken into account and deter- 
mined. This method involves a construction in tlie space dia- 
gram in close relation with the magnitude polygon, which may 
be developed as follows. 

With any given set of coplanar forces, assume the addition 
of any force whatever, its magnitude, direction, and point of 
application being chosen entirely at random or to suit con- 


venience. Find, by aid of the magnitude polygon, the mag- 
nitude and direction of the second force that would have to be 
added to the set to close the magnitude polygon for the given 
set plus the assumed force. This second force would be the 
resultant of the new set, and if properly located in accordance 
with the method of § 31, would be a force which, together 
with the assumed force, would make a set of two forces which 
would be equivalent to or balance the given set according to 
the senses ascribed to them. 

The senses of the assumed force and its mate will be con- 
fluent in the magnitude polygon, and if they are also confluent 
with the rest of the set, the two would balance or equilibrate 
the given set ; if they are not so confluent, they would be two 
forces together equivalent to the given set. 

Thus two forces can always be determined, which will be 
equivalent to any given force or to any given set of coplanar 
forces, no matter how complicated, and they will furnish the 
desired means for taking account of points of application. 

It may be observed that these two forces must always fall 
within one of five cases which may be stated and interpreted as 
follows : 

{a) The two forces may be inclined to each other and either 
equal or unequal in magnitude. By the parallelogram of forces 
their resultant must pass through their point of intersection. 
But their resultant and that of the given set are identical. 
Therefore their intersection is a point on the line of action of 
the resultant of the given set. Their non-parallelism would 
show that there must be such a resultant. 

Observe that this method furnishes a means for locating the 
resultant of a set of parallel forces. 

(^') The two forces may be parallel ^ unequal, and opposite^ 
proving the given set to be reducible to a single force parallel 
to the two forces and acting at a point determinate only by re- 


placing the two forces in their turn by two others equivalent to 
them and inclined to each other as in {a). This may be re- 
garded as a special case of {a). This case can always be 
avoided in practice by judicious selection of the direction of the 
assumed force. 

(a") The two forces may be coincident^ unequal^ and opposite, 
proving the given set to be reducible to a single force, itself 
coincident with the t^vo coincident forces. This like {a') may 
be regarded as a special case of {a) and like (a') can always 
be avoided by proper selection of the assumed force. 

{b) The two forces may be parallel^ equals and opposite^ 
forming a couple, and proving the given set to be reducible to 
a couple. 

if) The two forces may be coincident y equals and opposite , 
that is in equilibrium, and proving the given set to be in 

The cases {a), {a\ and (tf") are evidently cases in which 
the magnitude polygon is not closed, and (V) and {c) are evi- 
dently cases in which that polygon is closed. 

In Figs. 6^, 6b y 6c are illustrated the important cases (a), 
(b), and (c) respectively. OA and oa are assumed outright and 
EO and eo or FO and fo determined just sls AE and ae were 
determined in Fig. 5. In Fig. da it appears that the given set 
amounts to a single force AE; in Fig. 6b to an anti-clockwise 
(negative) couple of value OA yp; and in Fig. 6c that they are 
in equilibrium, A and F coinciding and also ao and af In 
Fig. 6a the case {a^) would have resulted if O had been taken 
anywhere on AE; oa and oe would then of course be parallel, 
becoming coincident, case (a")> if ^^ should chance to contain 
the point common to ab and ae. 

Observe that, in Fig. 6a, oa and oe, and in Figs. 6b and 6c, 
oa and of, with the senses there shown» are pairs of forces 
equivalent to the given set. 


The forces oay ob^ oc^ od^ etc., form a polygon consisting 
of the lines of action of an arbitrarily chosen force oa^ and the 


9 y y 

Fig, 6. 

resultants respectively of oa and ab^ of oa^ aby and bc^ etc. , 
through the series of given forces. This polygon is called the 
string polygon, because it bears a certain relation to the form 


which a string would assume if secured at the ends and sub- 
jected to the given forces.* 

If the two added forces equivalent to or balancing the given 
set coincide, the string polygon forms a closed figure and is 
described as closed. 

Obviously, then, if the string polygon closes, the given set 
of forces cannot be reducible to a couple. 

Observe that, in Fig. 6a, ae was located where ad equal 
and opposite to oa would have given rise to a closed string 

33. Additional Remarks on the String Polygon. — The 
sides of the string polygon are called strings. The added force, 
OA, and the partial resultants, OB, OC, OD, etc., form a set 
of lines radiating from a common point (called the pole) and 
are hence known as rays. The set of rays and the magnitude 
polygon constitute the whole of the magnitude diagram. 

To construct a closed string polygon, one need only to see 
that the strings form a closed polygon having an apex on the 
line of action of each and every force of the system (and no- 
where else), — each apex to be formed by the pair of strings 
which represent components of the force on whose line of 
action the apex appears. 

Any one string is, of course, a closing line of a closed string 
polygon, but any string which it may be convenient to draw 
last will commonly be spoken of as the closing line. 

* Terms more strictly analogous to magnitude polygon would be point-of- 
application polygon or location polygon but they are not in use. Funicular polygon 
and equilibrium polygon are terms in common use as well as string polygon, 
the former being merely the Latin equivalent to string polygon, and the latter 
open to the objection that it might with equal appropriateoess be applied to the 
magnitude polygon. 



34. Statical Problems. — Problems in statics deal with 
bodies which may be conceived to be at rest under the action 
of a set offerees, some of which are not fully known, i.e., not 
known as to all their elements. The solution of such prob- 
lems consists of finding what must be the value of each of the 
unknown elements. This at once suggests the use of algebra, 
and the first and principal task is to find how to write equations 
which will truly represent the conditions — that is, which will 
be true only if equilibrium is established. The forces can be 
represented in such equations only by expressions for their ele- 
ments. Such elements as are unknown need only be expressed 
as unknowns, to be evaluated as usual by the ordinary process 
of solving the equations. To write such equations, one need 
only to consider carefully what must be true that a body may 
be at rest, i.e., what the conditions really are which are to be 

35. Conditions of Equilibrium. — As has been pointed out 
(§ I3)» conditions of rest are always conditions of equilibrium, 
and to arrive at the conditions of equilibrium it is only necessary 
to consider the conditions of rest. 

In order that a body may be at rest, two things must be 
true of the forces acting on it, viz. : 

(a) The total tendency of some of them to produce trans- 



lation in one direction must be met by an equal total tendency 
on the part of the others in the opposite direction. 

{b) The total tendency of any of them to produce rotation 
in one direction must be met by an equal total tendency on the 
part of the others. in the opposite direction. 

Accordingly a set of forces will be in equilibrium if they 
can result in 

{A) Neither translation 
{B) Nor rotation. 

Observe that this is the same thing as saying that there 
must be for a resultant neither (-^) a single force nor {B) a 

It remains to establish the algebraic equivalent of {A) and 
(^), and then the graphic equivalent. It will be best for the 
present to confine attention to coplanar forces. 

36. Algebraic Statement of (A) and (B) for Coplanar 
Forces. — Since, in a given problem, the directions of the forces 
may be very various, it is necessary in algebraic work to sub- 
divide translation in general into two component translations 
parallel tb a set of axes, preferably rectangular. For coplanar 
forces a pair of such axes will of course suffice. They will be 
most conveniently taken as horizontal and vertical and will be 
referred to as the axis of X and the axis of Y respectively, ac- 
cording to the usual convention. 

{A) will then be subdivided into two partial statements 
which will be distinguished by the proper subscripts, and the 
conditions can now be written 

{Aj) No translation to the right or left ; 
{A^ No translation up or down ; 
{B) No rotation. 

But the tendency of a force to produce translation to the 
right or left is simply the horizontal or JT-component of the force 
and can be expressed algebraically P^ cos a^^ P^ cos or,, etc. 


And the tendency of a force to produce translation up or 
down is simply the vertical or F-component of the force, and is 
expressed algebraically P^ sin a^, P^ sin cr,, etc. 

In order to express {B) algebraically, it will be convenient 
to use the principle of § 20 and consider the given set replaced 
by an equivalent set consisting of a series of concurrent forces 
and a series of couples. The point of concurrence may be 
chosen to suit convenience and be called m. The couples 
need to be considered further only in connection with rotation. 
The moment of each may be regarded as its contribution 
towards rotation, and the values of these moments are Pj^p 
P^a^j P^^y etc. If now m be taken as the origin of coordi- 
nates, and {x^y j/j), etc., be taken as coordinates of the points 
of application of the forces, it will be convenient to insert the 
equivalents for the ^'s, and the moments will appear in the form 
(P^y^ cos a^ — PyX^ sin a^ ), {P^^ cos a, — P^^ sin a^, etc., 
which may be looked upon indifferently as the sum of the 
moments of the components or as direct substitution of values 
of the tf's. Cf. Cor. II, § 29. 

{A) and {B) become, accordingly, expressed as equations: 


'/*, cos a, + /*j COS a-, + • • • + ■'*» cos a, = O. . A'J) 
P^ sin ar, + /*, sin »,+ ... + P„ sin », = o. . A',') 

{B") P.ijf, cos ai - X, sin a,) -\ 

+ -P-CA cos a, — X, sin a.) = o, 

or, more briefly, 


"SP cosa = o ^;) 

2Psina = o A',) 

(B') 2P{y cos a — X sin «) = o, 


or, representing all horizontal and vertical components by X 
and K respectively, these three equations take the form 2X = o, 
2Y= o, and ^{Xy — Yx) = o, or most simply of all, using 
^for moments, 

^^^ (:i^r=o Ay) 


{E) 2Jif=o. ...... B) 

This last is the usual form for the algebraic statement of 
conditions of equilibrium. These equations form an equipment 
for solving statical problems algebraically. 

Two other ways of insuring the satisfaction of (A) and {B) 
algebraically can be deduced, one of much practical impor- 
tance, the other of little. These will receive attention in due 
time (§§ 40, 41, 43), and will be shown to be mere transforma^ 
tions of the equations above stated. 

Note that the satisfaction of (^) precludes translation, and 
that the satisfaction of {B) precludes rotation. 

Show that the satisfaction of any one or any two of the 
A^f Ay^ and B equations will not insure equilibrium. 

Show that in case of concurrent forces (A) alone suffices, 
and gives the only two independent equations bearing on the 
case, i.e., that {B) will always hold in this case if {A) does. 

When then are both (^) and {E) always needed.^ 

Must three forces in equilibrium be concurrent } Why } 
How about three parallel forces ? How about more than three 
forces } Cf. Cor. I, § 29. 

37. Graphic Interpretation of (A) and (B).— That there 
may be no translation from a given set of forces it is evident 
that their resultant must be zero, and by § 31 it is seen that 


this can be only when the magnitude polygon for the set of 
forces closes. We can state then 

(A) A magnitude polygon must close. 

That there may be no rotation, the given set of forces must 
not be reducible to a couple, and, as has been shown in § 32, 
this condition will be fulfilled if a string polygon closes for the 
set of forces. We can state accordingly 

(B) A string polygon must close. 

The closed magnitude polygon precludes translation, and 
the closed string polygon precludes rotation. These two poly- 
gons constitute the full equipment for the solution of statica:l 
problems graphically. It remains only to get practice in using 

Here again, of course, for concurrent forces {A) alone 
suffices, and its satisfaction carries with it the certainty of sat- 
isfaction of (B). Hence with concurrent forces {B) is super- 

38. Exercises in the Composition of Forces. — The equip- 
ment for solving statical problems is now complete, except the 
transformation of the A^^ Ay^ and B equations referred to in 
§ 36, and developed in §§ 40, 41. Everything is, however, 
entirely ready for practice in the composition of forces, and 
practice in the use of what has already been developed will 
throw much light on what is to follow. 

Below are given explanations of the algebraic and graphic 
methods of procedure in a general case. In special cases more 
or less of both processes become superfluous and can be omit- 
ted afler a little practice. The plan for arrangement of work 
and for retention of memoranda at all steps is recommended 
as, in the long run, an economizer of time, both because it di- 


minishes the risk of numerical error, and because it aids in the 
location of such error when once committed. 

Let it be required to determine the equilibrant (and result- 
ant) of any set (say five in number) of non -parallel, non- 
concurrent coplanar forces. 

Graphic Solution, (Figs. 7a, 7b. Cf. PI. I.) Letter the 
five forces in the given set, preferably in the order of their 
occurrence from left to right, ab^ . , . ^ de. The required 
equilibrant will be ea and the resultant ae. 

Construct (Fig. 7b) the magnitude polygon (§ 31) ABCDE 
for the given forces, the sides having lengths proportional to 
the magnitudes of the forces to which they are parallel and so 
arranged that the arrows will follow one another. Evidently 
all that is needed to close this polygon now is to fill in the 
side EA . EA determines the magnitude and direction of the 
equilibrant, and AE tliose of the resultant. 

Construct the string polygon (Fig. 7^). The question 
now is how to locate ea so as to preclude rotation, i.e., how 
to locate ea so that the string polygon will close when this 
force is added to the set. 

Now ea must act at the intersection of strings oe and ao^ 
and to close the string polygon ao must coincide with oa. 
The position of ao thus being known and oe having already 
been located in the construction of the string polygon up to 
this point, it only remains to draw ea through their point of 
intersection. The string polygon is then seen to be closed, 
and any convenient point on ae may be taken as the required 
point of application.* 

Algebraic Solution. (Cf. PI. L) Let the unknown magni- 

* The short dash-lines in Fig. 'ja show ea in other positions than the correct 
one and the positions which ao would in those cases take, leaving the string 
polygon unclosed. Such forces ea and ao are there distinguished by subscripts. 



tude, direction, and point of application be designated E, or, 

and {x, y) respectively. Write the equations for horizontal 


Fig. 7. 

and vertical components — the A^ and A^ equations — the sec- 
ond considerably below the first. 

Insert in small figures, as memoranda, the values of the 


trigonometric functions just above each, taking care, by a 
moment's inspection, not to interchange the values of the sines 
and cosines. 

Perform the multiplications indicated and insert the prod- 
ucts in small figures as memoranda below each term. 

Proceed with the reduction as shown in PL I, and deter- 
mine the values of E and cr, observing that the a found will be 
that for the equilibrant and needs only to be changed by 180'' 
to give the a for the resultant. The quadrant indicated for a 
is th^ first, second, third, or fourth according as i? sin a and 
E cos a are both plus, the former plus and the latter minus, 
both minus, or the former minus and the latter plus. 

Write the equation of moments — the B equation — in ex- 
tended form, inserting the values of the given coordinates at 
once, striking out all terms in which a zero coordinate appears. 

Write as memoranda above each remaining P cos a and 
P sin a the value found for it in reducing the two preceding 

Perform the multiplications indicated and show products 
under each term as memoranda, and solve, assuming the value 
for either x or j/ as may be more convenient. 

Specifications for Exercises 3-7. In each exercise the object is to work out 
both graphically and algebraically, the equilibrant and resultant of the sets of co- 
planar forces there given. The results should be recorded side by side conspic- 
uously for comparison. Work will be arranged in general as in PI. I, and each 
exercise should be undertaken as a general case and so carried to completion. 

Magnitudes are expressed in poimds and lengths in feet. 

Angles will best be laid off by plotting from a table of natural tangents, and 
in favorable cases by the use of triangles and T-square. Small protractors will 
sometimes be found useful for checking. 

For scales, i in. = 20 lbs. and i in. = 4 or 5 feet are recommended. 

Exercise 3. Non-concurrent, non-parallel forces: (20 io5* 5,0). 
(25 263" II, o). (30 98^* IS, o), (10 80'' 17, o), (15 280* 2, o). 

Exercise 4. Parallel forces uniform in sense; five verticafly upward 
forces of 10, 20, 30. 40, and 50 lbs. respectively, with successive inter- 
vals between them of 3, 2, 6 and 5 ft. 


Exercise 5. Parallel forces varying in sense : the forces of the pre- 
ceding exercise but with the lo-lb. and 30-lb. forces reversed in sense. 

Exercise 6. Concurrent forces : five forces of 10, 20. 30. 40. and 50 lbs. 
respectively, with directions of 30*, 6o*, 135*, 220*, and 320* all applied at 
(o. o). 

Exercise 7. Three non-concurrent forces whose magnitude polygon 
closes: (50 330* — 6, o), (50 2io' o, 2). and (50 90* 8, o). 

Note. — This is a case similar to that of Fig. 66, § 32. The resiilts can be 
stated in foot-pounds with the proper algebraic sign to indicate sense, say -|- for 
clockwise and — for anti-clockwise rotation. The source of graphical results 
should be made clear by use of dimension lines, including scaled numerical re- 
sults and the expression of the necessary multiplication. There is no reason 
why OA may not be taken of some round magnitude, say 20, 30, etc., and it will 
be advantageous in this case. 

39. Generalization of the Three Classes of Resultants.— 

It may be observed in passing that the three classes of result- 
ants to one of which any set of coplanar forces may be re- 
duced can be expressed in generalized form in the notation 
of this book, as follows : 

Class I. (>o, 0° to 360°, unrestricted), i.e., a single force. 
In this case the magnitude polygon fails to close, and the string 
polygon may or may not close. 

Class 2. (o, indeterminate, at infinity), i.e., a couple. 
Here the magnitude polygon closes, but the string polygon 
does not close. 

Class 3. (o, indeterminate, indeterminate), i.e., a set in 
equilibrium. The magnitude polygon and the string polygon 
both close. 

Examples of the first two of these classes are contained in 
the data for Exercises 3-7. Inserting the equilibrant there 
required, the class then exemplified is in each case the last one, 
of course. 

40. Establishment of Equilibrium by Use of Moments 
Alone. — While the algebraic statement of equilibrium of § 36 
is, of course, of universal validity in cases of coplanar forces, 
there is an alternative algebraic statement which in many cases 


proves much more convenient in use. This statement takes 
into account moments only, and amounts simply to three 
equations of moments with certain limitations laid upon the 
choice of the centers of moments. This alternative is deduced 
as follows. 

To establish equilibrium, it must be made certain that the 
given set of forces will cause neither translation nor rotation, 
i.e., that they amount to neither a single force nor a couple. 
It has been shown that there can be no couple if 2M=o for 
any point whatever. It remains to be shown how moments 
can be used to prove the non-existence of a single force re- 
sultant as well. 

Suppose that for a given center of moments 2M =o for a 
given set of forces. This can arise from only one of two 
causes : either the set of forces is in equilibrium^ and would 
have no moment about any point, or the center of moments lies 
upon the resultant of the set. If, then, it can be shown that 
the center of moments does not lie on the resultant of the set, 
the set must be in equilibrium. 

Any one or any two points chosen at random might both 
chance to lie on the line of action of a resultant. In such a 
case 2M would be zero in spite of the resultant having a mag- 
nitude greater than zero, and if investigation went no further 
translation might exist undetected. If, however, three points 
not in the same straight line be chosen, one of the points at 
least will not lie on such line of action. If 2M = o for each 
of these points, 2M = o for at least one point not on the line 
of action of the resultant, and equilibrium is a certainty. 
Hence it can be stated that 

If, for a set of forces, 2M is zero for three points not in 
the same straight line, equilibrium is assured. 

It should be carefully observed that the three equations of 
moments implied in the preceding sentence are merely the A^, 


Ayi and B equations themselves, though transformed * in such 
a way as to be more convenient for use in certain problems. 
As mere variants or alternative forms of the equations of equi- 
librium first deduced they can solve no problems which the 
former could not be made to solve, and any general conclusions 
drawn from inspection of either of the two sets of equations 
could be drawn also from inspection of die other. The A^.^ 
Ay^ and B equations, as they are of simpler algebraic form, are 
more convenient for such inspection, but, nevertheless, they 
are a group of equations algebraically identical with the three 
moment equations. 

41. Determination of Magnitudes by Single Moment 
Equations Alone. — Complete determination of equilibrium 
requires in general three equations, A^f Ay^ and B^ or an equiva- 
lent, such as the three moment equations of § 40. Still it will 
be worth while to see under what circumstances, if at all, a 
single equation of moments will determine the magnitude of any 
one of three or fewer forces whose lines of action are known. 

Call this unknown force j2> and the others R and S. 
Then, since Q is to be the equilibrant of all the rest of the set, 
its line of action must be that of the resultant of the set. 

* The nature of this transformation appears upon writing one of these moment 
equations in its general form as follows. 

If the points of application of the forces be a series of (x, yysy and any one of 
the centers have the coordinates (Xay >^a) with reference to the same axes, the 
moment equation will take the form 

or, noting that Xa and>^a are constants and all the rest of the letters rariables, 
2{Xy - Yx) -y^X+ x^Y = o, 

which is obviously B minus Ax, multiplied through by a constant and plus Ay 
multiplied through by another constant. One or both of these constants might, 
of course, be zero. If the three centers be taken at the origin and elsewhere on 
each of the axes of x and y respectively, the three moment equations become 
simply Bj and B — AxX^k constant and B -^^ Ay X 9^ constant 


Realizing this it is easy to select a center which certainly does 
not lie on the line of action of the resultant of the set. K 
and 5, as well as j2> will in general produce moments about the 
point, and, appearing in the equation, will prevent Q from 
being the only unknown in it. If, however, the point can be 
selected so that it will be ofT the line of Q, and at the same 
time cause the arms of R and 5, or of 5 if 5 is the only other 
unknown, to be zero, determination of Q by one equation is 
possible. Such a point is to be found at the intersection of i? 
and Sy or anywhere on 5 if 5 is the only other unknown. 
Cases in which this cannot be done arise only when there are 
more than three unknowns, or when the lines of action of three 
are parallel or concurrent. 

An equation thus derived of course does not in general 
establish equilibrium, for the establishment of equilibrium re- 
quires the determination of two or three elements, and for this 
more than one equation is always necessary; but it does deter- 
mine one magnitude to which the other forces must inevitably 
conform in order to produce equilibrium. To get more, simply 
repeat the process. In general, then. 

To find the magnitude of a force whose line of action 
is given, solve an equation of moments for a center on the 
line of action of the other force whose magnitude is un- 
known ; or, if there are two others, for a center at their 
intersection. Repeat the process for the other one or two 

This method is of great use whenever the magnitude is 
sought for a force with a given line of action. Such problems 
are very common in practice. In such cases the method is 
oftin useful for determining senses of forces by mere inspec- 
tion without any calculation. 

Notice that solving Case 4 in this way is simply utilizing 


the general principles of the preceding section, choosing the 
centers so as to avoid simultaneous equations. 

The amount of it is that if there is equilibrium, 2M = o 
for any center, but the converse is not true. Hence the need 
of more than one moment equation. 

Suppose the condition requiring the center to be off the 
line of action of Q were violated, how would the equation be 
affected ? 

42. Convention as to Algebraic Signs in Moment Equa- 
tions. — In using A,^ A^y and B it has been convenient to conr 
sider magnitudes of forces as always positive and to associate 
sense exclusively witH slope. In using the method of mo- 
ments it will be more convenient to call arms universally posi- 
tive, and give magnitudes varying signs : -j- if for a given cen- 
ter they cause rotations in one sense, and — if the opposite. 
Throughout these pages clockwise rotations will be called 
positive, and anti-clockwise negative. 

To apply the method of moments, select the center, write 
the equation calling all moments positive that are not known 
to be negative. The sign found with the result will, in con- 
nection with the center for the equation, determine the sense 
of the force. The same force may be positive for one center 
and negative for another, hence the locations of the centers 
should be recorded near the work. 

43. Six Methods of Stating the Conditions of Equilib- 
rium. — In addition to the statements of the conditions of equi- 
librium already given, there is a third from the algebraic and 
two more from the graphic point of view. The six ways are 
here collected and stated for completeness, and for reference: 



Equilibrium will exist if 

(i) A string polygon closes 
for each of three poles not in 
the same straight line ; 
or if 

(2) Two string polygons 
close with the same pole ; 
or if 

Equilibrium will exist if 

(1) The sum of the mo- 
ments is zero for each of 
three points not in the same 
straight line; 
or if 

(2) The sum of the moments 
IS zero for each of two points, 
and the sum of the components 
is zero along a line not per- 
pendicular to the one joining 
these two points ; 
or if 

(3) The sum of the mo- 
ments is zero for one point 
and the sum of the compo- 
nents is zero for both of any 
two directions. 

The statements shown in bold face are the ones already 
developed at length and are the only ones of practical use. 
The critical reader, however, will assure himself of the correct- 
ness of the three others. 

* The first of these statements was shown in g 40 to be an al^braic identity 
with the third, and deducible directly from it, and the same might be prored 
of the second in a similar manner. 

(3) One string polygon and 
one magnitude polygon close. 


44* General Surrey of the Scope of Pure Statics. — The 
term Pure Statics is to be understood to mean the perfectly 
general science, in which problems can have no light thrown 
on them by considerations of size, shape, and elasticity of 
bodies. Its only resources are the conditions of equilibrium of 
rigid bodies. 

The first algebraic statement of these conditions which 
has been elaborated is of such a form as specially to invite 
thought as to the scope of the subject — as to how many and 
what coplanar statical problems are capable of solution. 

First may be considered 

(a) Non-concurrent Forces. There are only three funda- 
mental equations (§43), A^, Ay, and B. If, then, in any 
problem there are more than three forces of which elements 
are not known, there are more than three unknown quantities 
and the problem of establishing equilibrium is indeterminate. 
Hence all forces but three (at most) must be fully known. But 
three forces involve nine elements ; hence, unless six of these 
elements are known the problem is still indeterminate. 

These are a priori considerations regardless of tlie struc- 
ture of these particular equations. Taking this structure into 
account, the field becomes still more circumscribed. For 
example, since only one of the equations involves points of 



application, it is clear that there must be not more than one 
such point unknown. Even then the point is, strictly speak- 
ing, indeterminate, for the point appears as two unknown 
quantities, x and y. For the purposes of statics, however, any 
pair of values for x and y that will satisfy the equation will 
meet all requirements, and this actual indetermination is not a 
source of difficulty. In other words, B becomes simply the 
equation of the line of action of the force in terms of x and y, 
on which an indefinite number of points of application may be 

* Nine elements can be divided into groups of six known and 
three unknown in 9X8x7-7-3X2 = 84 different ways, 
which reduce to twenty separate cases at once (see Appendix 
for a complete statement of them), and of these, fifteen are 
more or less indeterminate, owing to the peculiar limitations of 
the equations above pointed out. Moreover, only three of 
the twenty cases are important. They are the following : 

a. When the three unknown elements all pertain to one 
force, i.e., one force is wholly wanting. This is the problem 
of finding the resultant and the equilibrant of a given set. — 
Composition of Forces. 

ft. When the three unknown elements pertain to two forces 
and the magnitude of one and the magnitude and slope of the 
other are wanting. 

y. When the three elements pertain to three forces, and 
three magnitudes are wanting. 

There are still to be considered 

{b) Concurrent {including Parallel^ Forces, Here there 
are only two fundamental equations, A^ and A^^* one of 

* The B equation becomes simply a mathematical identity with Ax and Ay in 
the case of concurrent forces, for the jt's and^^'s in 'SiXy — Kr)= o are then each 
constant, say Xa &nd>^a* The equation is then ya'SX— x^SV =: o, which is 



which must be replaced by B if the forces are parallel. They 
can determine only two unknown quantities. If there are 
more than two forces of which elements are unknown, there are 
more than two quantities unknown and the problem of estab- 
lishing equilibrium is indeterminate. Hence all forces but two 
(at most) in a set of concurrent or parallel forces must be 
fully known. But two forces involve six elements which may 
be unknown, and unless two of these elements are known, the 
problem is indeterminate. Six elements may be divided into 
groups of four known and two unknown in6x 5-5-2= 15 
different ways, which reduce to nine separate problems in the 
equilibrium of concurrent or parallel forces (see Appendix). 
Noting that composition of such forces is included in the gen- 
eral problem of composition of forces {a) as a special case, 
only one of the rest of the nine cases is important, viz. : 

d. When the two elements pertain to two forces, and the 
magnitudes of the two forces are wanting, — Resolution of 

45. The Four Cases. — Only four cases capable of solution 
by pure statics are of sufficient importance to require special 
study. These have been pointed out in § 44, and will here be 
restated and numbered for future reference. 

Case I. Magnitude, direction, and point of application of 
one force required, — Composition of Forces. 

Case 2. Magnitudes of two forces required, — Resolution of 

This case will be divided into two sub-cases, 2a and 2^. In 2a the 
two forces are non-parallel ; in 2b, parallel. 

Case 3. Magnitude of one force and magnitude and direc- 
tion of another required. 

Case 4. Magnitudes of three forces required.* 

* As will be seen ( § 50a), Case 4 may be regarded as a mere combination of 
Cases 3 and 2a, and Case 3 as combination of Cases 4 and i. 


This case is sometimes spoken of as Resolution in Three 

Case 4 is statically determinate only when the three forces 
are neither concurrent nor parallel. 

All these four cases or any of the few others which are 
capable of solution can, of course, be solved by applying A^^ 
Ayy and B — in some cases more conveniently in the form of 
the moment equations of §40 or §41 — or by closing the mag- 
nitude and string polygons. 

See Plates I-V respectively for a full treatment of these 
four cases. These plates are explained in the following 



46. The Solution of Statical Problems. — In general the 
first thing to be done in undertaking a problem in statics is to 
draw a sketch showing all that is known of all the forces, — 
their magnitudes, either as knowns or unknowns, being in- 
scribed near their lines of action. This operation is of funda- 
mental importance and is the step of greatest practical 
difficulty. It is sometimes called ** showing the forces." It 
consists really of accurately analyzing the situation and decid- 
ing just what forces are acting and just what is known of each. 
This once done, and the problem seen to be determinate, it is 
a matter of mere routine to proceed with the solution. 

The mastery of these solutions is what study pf this chap- 
ter is expected to bring about. The ability to show the forces 
will require a longer time to cultivate, and its importance may 
well be kept in mind. 

The methods of solution for the four important cases will 
now be taken up. It is well to note at the outset that the whole 
equipment available consists of the magnitude polygon and the 
string polygon for graphic solutions, and of the A^,, Ay, and 
B equations, or their alternative, the three moment equations of 
§ 40 for algebraic solutions. The task is now simply to become 
familiar with the manipulations by which this equipment can 
be made to fit the peculiarities of each case, and to yield most 
easily the desired results. 



In all cases, it should be noted, it will be found most con- 
venient in graphical work to letter the known forces in the 
order of their occurrence, going from left to right around the 
body on which they act, leaving the unknowns to be lettered 
consecutively after the knowns have all been provided for. 
The first letter will naturally be <z, and the last letter of the 
last unknown may properly be <z. Thus will be avoided double 
letters for apices of the magnitude diagram. 

Furthermore in algebraic work senses of forces given only 
in line of action will be assumed to be such as to make their 
directions less than i8o**, in writing the ^, and A, equations, 
and such as to give a positive moment in writing the non- 
simultaneous moment equations required for some of these 
cases. If the sense so assumed is the correct one, the fact 
will appear by the magnitudes proving to be plus quantities. 
If the magnitudes prove negative, of course the correct sense 
is the reverse of that assumed. 

After writing an equation of moments it will be well to 
check the lengths of all arms, whether given or not, by scal- 
ing them from a drawing, if one be at hand. Check also so 
far as possible the determinations of senses by inspection of the 

47. Solution of Case l. — Case i is mentioned here merely 
for completeness, and the account of its solution given in § 38, 
and presumably familiar to the reader by this time, need not 
be reprinted here. 

48. Solution of Case aa, (Cf. PI. Il^a:.)— Required the 
magnitudes (and senses) of two non-parallel forces in a set in 

Algebraic Solution. Let P and Q be the unknown mag- 
nitudes, and assume trial senses as per the next to the last 
paragraph of § 46. 

Solve, for P and Q, either {a) the A^ and Ay equations 


i .' 



PROBLEM. — Determine the Equilibrant of the following set of forces: (20 lbs. 

What is the Resultant of the set ? 








V'"^ ) 



^^ ^ 

/ ^ 
/ ^ 




















E A 18 THE REQUIRED 1 \ ^ 



1 9v^ 



«^-^| X in.- Z5iimt9 of length 


E 1 

>s. 45** 0,0), (15 lbs. 90** 13,0). (30 lbs. 6o*» 7.0). and (25 lbs. 315** 30,0). 


Let (E a x,y) be the force sought, then 
For no translation: 

.707 o .500 .707 

Az' E cosa-^- ao cos 45® + i$cos 90® + 30 cos 60® + 25 co5 31 5® — o 

+ X4-I4 o +15-00 


+ 17-68 

E cos a'^ —46.82 



Ay-. E sin a + 20 sin 45® + 15 sin 90** + 30 sin 6o* + 25 sin 315* — o 
+ 14.14 +15.00 +35.98 — X7.68 

J5 5f»a-— 37.44 

t^^^ ^sina -37.44 

tan a^m •" tist .000 

S cos a —46.82 


. ^ E sina ^ —37.44 
sin a —.625 "" 


For no rotation: 

Taking a point on the X-axis for the point of application of J? , y becomes 
f^ zero and x is determined as follows: 

o --S7'il4 

B:E cos ax y-^Esinax x + 20 cos 4$^ x 0—30 sin 4$^ x 0-^1$ cos 90^ xo 

o — 37.44« 000 

15-00 a^.98 —17.68 

-i55»»9o*XX3+3o^o*6o*xo-30 5»»6o"x7+ascoj 3x5^x0-85 J*«3XS**X3o-o 
—195.00 o —181.86 o +530^0 



Algebraic Graphic 

Eqtiilibrant (59.9 lbs. 218® 40' —4.10,0) Equilibrant (59.9 lbs. 218** 40' 
Resultant (59.9 lbs. 38^*40' —410,0) Resultant (59.9 lbs. 38*^40' 

.•: i ; 7/ D»> :'.:;/ c:'*j o.i.rm^JoIi 



• < \ 

./J- >' - . ;: "wOv* o.-. 


PROBLEM.-^In the foUowing set of forces in equilibrium, (ao lbs. zoo® o. 
determine the ujDknaimi. 



Scale I in.- 15 lbs. 



I 2a 

6o^ So** 

, (15 lbs. 150" 0,0), (P-? or 0,0), (10 lbs. 340** 0,0), and (Q-? or 0,0), 

340® 260** 


— .174 —.866 .500 .940 .174 

Axi ao cos 100® + 1 5 cos 1 50** +P cos 60® + 10 cos 340® +Q cos 80* —o 

-3.48 — xa.99 +0.S00P 4-9-40 +0.1740 

.SooP + .i74Q«7.o7 

.985 -soo .866 —•34a >985 

-4,: 20 5i» loo** + 15 sin 150® +P sin 60® + 10 5*n 340® H-Q sin 80* — ^ 

+ 19.70 +7.50 +0.866P — 3-4a +0.98SO 


P— 32.2 « (32.2 60*0 
G—— 52.6 -(52.6 260**) 

B: No couple being possible, the equation of moments is superfluous 


Algebraic Graphic 

P« (32. 2 lbs. 60") P- (32.3 lbs. 6o«) 

G--(52.6 1bs. 26o*») Q«(s2.5lbs. 260^ 


PROBLEM — In the following set of forces in equilibrium, (20 lbs. 2; 
determine the imknowns. 


Scales I ' l'^- - ^ 5 tinits of length 
( I in. — 15 lbs 


PLATE lib. 

B 2b. 

go® oo* 

P -? Of ©,o). (lo lbs. 90** 8,0). (16 lbs. 270** 14,0), and (0-? or 19.0), 
270* 370' 


Solution by Ax* A^. and B. 


. Axl 20 C05 270® + P C"05 90° + 10 C(?5 90" + 16 C05 270** +Q C<>J 90**— O 
o o o o o 

— 10 10 JO — 10 TO 

Agi 20 sin 270** +P sin 90^ + 10 sin 90^ + 16 sin 270® +(3 stn 90**— o 

— aoo +/* +IOO —160 +0 

P + (2-26o 

— aoo 

B: 20 cos 27o**Xo — 20 sin 270*'X4 +P cos 9o*X o — P jin 9o®X o + 10 cos 90*^0 

o +80 O O o 

10 O —16.0 T O 

— 10 sin 90®X 8 + 16 cos 270*^X0 — 16 sin 27o®X 14 +0 <^os go^Xo—Q 5i» oo*X 19 — o 

— 800 o +»d4.o o —19Q 

— 19(3 — —224.0 
G- II.8 

By A,. 

P —26.0 -C -26.0 —I i.s 

Alternative form of the preceding using moments alone. 
Center of moments at (0.0) 

80 -80 324 

PX0 + 20X4-10X8 + 16X14+QX19-0 

Q«^^--ii8-(ri.8 90^) 
Center of moments at (19.0) 

— 3CO no — 80 

PX19 -20X15 + 10X11- 16X5 +GXO-O 

P-«-i- — 14 2 — (14 2 90^ 

Check by AgZ 

14.2— 20 + 10 — 16 + 11. 8—0. 


Algebraic Graphic 

P-(r4 2Jbs 90**) P- (14.2 lbs. 90**) 

C-(ri.8)bs 90*^) P-(ii.8 1bs. 90**) 


(avoiding simultaneity, if desired, by taking one axis of refer- 
ence along one of the unknowns), or {b) their variants, the 
two moment equations of § 41. The assumption as to sense 
made at the outset is correct or must be reversed for either 
force according as its magnitude comes out plus or minus. 
Graphic Solution (Fig. 8 and PL lla). Close the mag- 

Fio. 8. 

nitude polygon (Fig. 8^). Suppose the first known of the 
given set be ab and the two unknowns de and ea. The mag- 
nitude polygon can be completed by familiar means from the 
apex A to D inclusive. Then DE drawn parallel to de and 
AE parallel to ae will locate the missing apex E, and the re- 
quired magnitudes and senses will be DE and EA. 

The string polygon is superfluous. It is bound to close if 
the magnitude polygon is closed. 

Note. — Case 2a nearly always occurs in practice for a set of concurrent 
forces, but it is capable of solution as well for a non-concurrent set if the resultant 
of all the knowns passes through the point of intersection of the two unknowns. 

48a. Solution of Case 2b. (Cf. PI. II*.)— Required the 
magnitudes (and senses) of two parallel forces in a set in 



Algebraic Solution. Let P and Q be the unknown mag- 
nitudes and assume trial senses as per the next to the last 
paragraph of § 46. 

As P and Q are parallel, the -A^axis may well be taken at 

right angles to them both, making their directions 90®, 

Write the A» and Ay equations as in Case 2a. Noting that the A^ 
equation vanishes, its place is made good by the ^equation (retaining the 
same senses as in the preceding equations) written as in Case i. Solving 
the simultaneous equations, P and Q are known, and the assumed senses 
are correct or to be reversed according as results come out plus or 

Or, better^ solve by moments alone by the method of § 41, 
taking a center on the line of action of P to find Q^ and on 

Fig. 9. 

the line of action of Q to find P. The assumed senses are cor- 
rect or to be reversed according as results come out plus or 
minus. Check results by seeing whether Ay is satisfied finally. 


:!/.;; i;:) c:x!::i 

-A'^ ^.•^ 






PROBLEM. — In the following set of forces in equilibrium. 

(P-? a-? 0,0) 

mine the unknowns. 


S^Im i I jo- — is units of length 
*'*^''^ (X in.— 30 lbs. 

PLATE 111 



bs. lao® 33,0), (G — ? x>r 40.0), (20 lbs. 75** 15, o). and (30 lbs. 60° 5,0) deter- 


Center of moments at the point of application of the force which is other- 
ise unknown, which, in this problem, is the point (0,0), 

: P cos aXo—P sin 0X0 + 10 cos 120^X0 — 10 sin 120^ X2^-\-Q cos 60® Xo 

000 —199.18 o 

.866 .966 

— Qjtn 60^X40 + 20 C05 75®xo— 20 5«n 75® X 15+30 (705 60^X0—30 ^tneo^'xs—o 
— 34-64 (? o —289.80 o —139.90 

— 34.64 Q =-618.88 

Q- -17.9 -(17.9 a40*») 

Using this value of Q in the next two equations, 

-.500 17^9 "'l^ -^^ « '3^ 

xi P COS a + 10 cos 120® -H(2 c^^ 240** + 20 cos 75+30 cos 60® — o 

— S.oo —8.8s +5-X8 4-iS-oo 

PC05 0— — 6.23 

.866 17.9 —.866 .966 .866 

i,: P sin a + 10 sin 1 20® + Q sin 240® + 20 sin 75^ + 30 sin 60** — o 

+8.66 —15.50 +19.32 +aS-98 

P sina— =38.46 

Pstna —38.46 ^ 
P cos a — 6.23 ' 


Pco5 0-6.23 


Algebraic Graphic 

P-38.9lbs. 260^48' P -38.9 lbs. a6i* 

0— zy.plbs. »4o* — 17.9 lbs. »4o* 

" A ' v-^ ''^^i .r- 'i '. ' 

'A jl 

,^ : n'u 'I- .'\* n :a 

-- :. '; r: •* !.. :^. f • . f' - :...r 
• . - ( •, > 

^ I i^v i :^^ 

I* I .\ 


Graphic Solution (Fig. 9 and PI. 11^). Let the knowns 
be lettered ab . . , cdy and the unknowns accordingly de and ea. 

Construct the magnitude polygon from A to D^ inclusive 
(Fig. 9^). Assume a convenient oa and construct the string 
polygon to ody inclusive (Fig. qd). In order that ao may coin- 
cide with oa and close the string polygon, oe^ the missing 
string, must intersect ea where oa does, and oe must start 
from the intersection of od and de. Hence oe is located, and 
the string polygon is closed. 

E must then be on a ray from O parallel to oe^ and on a 
line from A ox D parallel to ae or de. E is therefore located, 
the magnitude polygon is closed, and the required magnitudes 
and senses will be DE and EA. 

Note. — This case almost always occurs in practice for a set of parallel forces, 
but it is capable of solution as well for any set provided that the resultant of 
the knowns is parallel to the two unknowns. 

49. Solution of Case 3. (Cf. PL III.)-^Required the 
magnitude and direction, and the magnitude (and sense) re- 
spectively, of two forces in a set in equilibrium. 

Algebraic Solution. Let P and Q be the two unknown 
magnitudes and a the direction of P. Assume the sense of 
Q as per the next to the last paragraph of § 46. 

Write the B equation, taking the center on the line of 
action of P, and establish the magnitude and sense of G (§ 41)- 

One point only in the line of action of P being given, that 
point must, of course, be the center selected. 

In writing B^ memoranda may advantageously be inserted 
near each term, showing the value of each component of each 
force, as well as the moment of each component. 

Write the^^ and Ay equations, inserting the value of (2 just 
established, and get the horizontal and vertical components of 
P and establish P and a by routine like that for a similar pur- 
pose in Case i. 



Observe that the memoranda inserted near the terms of the 
B equation can be utilized in the A^ and Ay equations, just as 
the reverse was done in Case i. 

Observe also that Q once fully known, the problem is simply 
that of finding an equilibrant whose point of application is 
known in advance, and the work is thenceforth that of Case i 
for finding the magnitude and direction of the equilibrant. 

Graphic Solution (Fig. lo and PI. III). Let the knowns 
be lettered ab . . . cd^ and the unknowns accordingly de 

Fig. 10. 

and ea^ the former being unknown in magnitude (and sense) 
and the latter in magnitude and direction. 

Construct the magnitude polygon from A to Dy inclusive 
(Fig. 10^). Assume a convenient oa^ draw it through the 
giyen point of application of ea^ and construct tlie string 
polygon to ody inclusive (Fig. \od). In order that ao may co- 
incide with oa and close the string polygon, oe^ the missing 
string, must intersect ea where oa does, i.e., at the only known 


PROBLEM. — In the following set of forces in equilibrium, (15 lb 
and (R or 30,0). Hetermine P, Q, and i?. 



« , J I in. « 1 5 units of length 
Scales ],i^^y ^5 lbs. 


m 4. 

150** 30** 

^^^■5* 0.0). (-P or lo.o), (20 lbs. 90** 2X,o), Q or 50.0). (8 lbs, 240*35,0), 
330® aio** 


Solution by Ax, Ay, and B. 

.707 — 866 o .866 —.500 — 500 

Axi 15 COS 3i5®+P COS i5o*' + 2o cos 90^ + Q co5 30^ + 8 cos 240® +i? cos 120® «o 

XO.61 — 0.866P o +o.866(J —4.00 —050/? 

-0.866P + 0.866C -0.50/? « -6.61 

— .707 .500 100 .500 —.866 866 

Ay: 15 sin $i$°+P sin i ko^ -\- 20 sin go^ +Q sin so^ + S sin 240*^ -h R sin 120®— o 

— io.6x +0 sP +20.00 +0.50^ —6.93 +0.866/2 
O.50P +0.500+0.866-^- —2.46 

Center of Moments at (0,0) o.soP 

B: 15 C05 315^x0-15 sin sis^'xo+Pcos iso^xo-Psin JSo^xio+20 cos 9o**xo 
00 o +SO.P o 

ao.oo o.$oQ —6.93 

— 205*«90*'x2i+Qc(?5 3o°xo-G ^*« 30^x50+8 COS 240^x0-8 5tn 240**X35 

— 4ao.oo o —aS'OoQ o +343-55 


+R COS i20**Xo— /? sin i20°X3o-»o 

o — a5.98/?j 

— 5.00P--25.00Q — 25.98^^=177.45 

Combining the three equations, Ax, Ay, and B it appears that 
^ P-+6.82-(6.82 i5o'');Q--3.54-=(3.54 2io°);i?--4.73«(4.73 3oo') 

Alternative form of the preceding, using moments alone : 

To find P, take center of moments at (x^.y^ , the intersection of Q and R, 

Then since 

^-•577*= -28.87 and j' + 1.7322; « 5 1.96 

arc the equations of the lines of action of Q and /? respectively, it appears that 

*i*=3S-o s^^ >'i=— 8.66. 
^ Center of Moments at (35.0, —8.66), transforming co-ordinates: 
.707 -.707 -.866 .<oo 

15(8.66 COS 315** 35 stn 315°) +P(8.66 cos 150® 25 stn 150**) 

+ 6.ia -34-75 -7-50 +12.50 

o 1. 00 —.500 —.866 

+ 20(8.66 COS 90® 14 sin 90**) +8(8.66 cos 240® — o sin 240**) +QX o 

o 14.00 —433 o 


5.0P -279.45— 280.00 + 34.64 «34.o9. Whence P- +6.82 -(6.82 150®) 
To find Q, take center of moments at intersection of P and R and proceed 
as for P. 

Similarly, to find R, take center of moments at the intersection P and Q, 
In practical cases the algebraic work usually proves much simpler than 
for the general case used in this Plate. See Plate V. 


Algebraic Graphic 

P-(6.82lbs. 150°) P-(6. 8 lbs. 150°) 

G- (3-54 lbs. 210°) 0- (3.6 lbs. 2io<') 

;?- (4.73 lbs. 300°) /?-(4-7lbs. 300°) 


point of ea ; furthermore oe must start from the intersection of 
od and de. Hence oe is located and the string polygon is 

E must then lie on a ray from O parallel to oe^ and on a 
line from D parallel to de. E is thus located, and connect- 
ing E and A^ the magnitude polygon is closed. The required 
magnitude (and sense), and magnitude and direction, are DE 
and EA respectively. 

Observe that the key to this solution lies in locating oa in 
the only way possible, so that its point of intersection with ea 
may be known; i.e., by drawing it through the only point 
through which ea is known at the outset to pass. 

50. Solution of Case 4. (Cf. Pis. IV and V.) — Required 
the magnitudes (and senses) of three forces in a set in equi- 

Algebraic Solution. Let P, Q, and R be the required mag- 
nitudes. Assume senses as per the next to the last paragraph 
of §46. 

The Ax^ ^^and B equations can now be written, taking the center for 
B anywhere whatever, as in PI. IV, and there result three simultaneous 
equations, whence the desired quantities can be evaluated. This method 
is needlessly laborious, and the simultaneous equations can always be 
avoided by the aid of the three moment equations of § 41, as follows. 

Write three equations of moments, taking the centers at the 
intersections respectively of Q and R, R and P, and P and Q. 
These equations will yield the magnitudes (and senses) of P, 
(2, and R respectively. 

Obviously the solution is indeterminate if P, (2, and R 
are concurrent or parallel. 

In PI. IV is worked out a general case, and the deter- 
mination of the coordinates of the required center of moments 
is seen to involve considerable labor. Case 4, in its occur- 
rence in practice, however, almost invariably appears with the 


three unknowns so situated that the coordinates of the three 
centers are much more easily determined than in the problem 
of PI. IV. The arms of the forces can usually be read directly 
from the drawing or can be calculated with little labor. In 
such cases the equations will best be written with the arms 
inserted directly without resort to components except for the 
more inconveniently lying forces. 

A fairly typical example of Case 4 as it occurs in practice 
is fully worked out in PI. V. 

Sometimes two of the unknowns, say P and j2» are paral- 
lel. R can then be found by taking a center on the line of 
action of Q and off the line of P and /?, bringing into the 
equation the previously calculated value of P. 

Graphic Solution (Fig. 1 1 and cf. PI. IV). Let the knowns 
be lettered ab . . . . cdy and the unknowns de^ ef, and fa^ 
the three last being unknown in magnitude (and sense). 

Construct the magnitude polygon from A to Z>, inclusive 
(Fig. 11^). The two apices £ and /^remain to be located. 
That means that two strings oe and ^are to be located, while 
the direction of neither is known, nor is the point in which they 
will intersect ef known. One of these (directions might be 
assumed and the string polygon closed accordingly; the same 
result could be attained by assuming the point in which they 
intersect ef. The result of this process would in general be 
that the magnitude polygon would not close, and the labor 
would have been in vain. 

Now, if either of these two strings could be made to vanish, 
i.e., to be of zero length, it could be looked upon as having 
any direction whatever, including a direction consistent with 
the closure of the magnitude polygon ; then only one string re- 
maining to be located, there would be no further difficulty. But 
any string, om^ common to two forces, Itn and mn, will vanish 
if ol and on intersect at the point of intersection of Int and mn. 

Case 4 (jsccoMg 

PROBLEM. — A body in the shape of an isosceles triangle, base 20 feet, altitude ii^icet 
are known in line of action only. Determine P, Q, and R, 


a~t^^JS ^aRTMi 


o%.«uc» ^ J jj^ —300 1 





' V 

ob example). 

:^feet, is in equilibritim tinder the action of the forces shown; of these P, G, and R 


Center of moments at intersection of P and Q 
RX15 +6ooXio-65oX2Q+/^Xo+GXo«o 

6000 I 3000 
600 X 10 — 6<o X 20 7000 
^^^ oooAio 05 ^ .^ ^7^ ^+466.67 =(466.67 180^) 

Center of moments at intersection of Q and R 

PX15 cos 26' 34*+6ooXio-65oX20+QXo+i?Xo«o 

6000 13000 

„ 600X10—650X20 7000 , 

P^ -— = =-522.0 = (522.0 3S3 26») 

I5X.894 13.41 

Center of moments at intersection of P and R 
GX30 -600X20 +650 X 10 +FXo+/^Xo.=o 
3000 6500 

600X20—650X10 5500 




« + 183.33 = (183.33 90°) 



P« (522.0 lbs. 333''26') 

G- (183.3 lbs. 90° o) 

i?- (466.7 lbs. i8o° o") 

P = (522 lbs. 330*26') 
0- (183 lbs, 9o*» oO 
7? -(467 lbs, 180"* oO 




Accordingly if oa be drawn (Fig. 1 1^) at the outset through 
the intersection of ef and /a, on proceeding with the string 

Fig. II. 

polygon oe will fall into place, connecting the point where od 
cuts de with the intersection of of (whatever its direction) and 
ef. The string polygon thus being closed, E is located at the 
intersection of lines from D and O parallel respectively to de and 
oe. F can now be located at the intersection of lines from E 
and A parallel respectively to ^and af The magnitude poly- 
gon is now closed, the direction of the ray OF is, of course, 
perfectly consistent with the string polygon being closed, and 
the required magnitudes and senses are DE^ EFy and FA. 


Observe that the key to this solution lies in drawing the first 
string through the point of intersection of two properly selected 
consecutively lettered unknown forces, and that the problem 
could also have been solved by starting od at the intersection 
of de and </*, causing oe to vanish, F to be located by aid of 
the string of, and E by parallels to de and ef from D and F 

Observe also that the graphic, unlike the algebraic, solu- 
tion, is not essentially more laborious in an instance like that 
of PI. IV than in PI. V. 

50a. Remarks on Cases 3 and 4. Case 3 may be looked upon 
as a combination of Cases 4 and i , for any two convenient com- 
ponents of the force unknown in magnitude and direction might 
be substituted for this force. Then they and the other un- 
known could be handled by Case 4, and the two components 
combined into the single required force by Case i . In fact it 
not uncommonly happens that the two components are as ac- 
ceptable a result in practice as their resultant and some labor 
is saved by substituting them. 

In a similar way Case 4 may be looked upon as a combina- 
tion of Cases 3 and 2a. The intersection of the two unknowns 
may be regarded as the given point of application of the re- 
sultant of the two unknowns, which can then be found by Case 
3, and resolved into its components, the required forces, by 
Case 2a. This view is of little value in practice, unless per- 
haps in throwing additional light on the graphic method for 
solving Case 4 given in the preceding section. 

Exercise 8. Sketch carefully free hand the graphic solution of each 
of t)ie four cases, assuming the necessary data for each. 

Exercise 9. Same as Exercise 8, but done to scale as usual. 

Exercise 10. A derrick mast, supported by the usual socket and 
guys, is 40 ft. high and the boom is 60 ft. lonf;^. The boom is held at an 
inclination to the vertical of 30 degrees by a stay running from its upper 
end to the top of the mast. If a weight of 5000 lbs. be suspended from 



the end of the boom, what would be the forces transmitted through the 
boom and stay ? 

Solve graphically and algebraically, in the latter case using the Ax and 
Ay equations, and also by two moment equations of § 41. 

Exercise 11. A uniform beam rests upon supports at the ends at the 
same level. The length of the beam is 20 ft. It carries vertical loads of 
10, 20, 30, 40 cwt. at 5, 8, 13, and x6 ft. respectively from the left ends. 
Its own weight is 5 cwt. Determine the reactions ^1 and R% at the ends. 
Solve by both methods. 

Suggestions. Draw the data diagram to the scale i in. = 5 ft. One 
data diagram serves as usual for both methods of computation. A 
single heavy line will suffice for representing the beam. 

Letter forces in the space diagram in continuous circuit around the bar. 

For the algebraic solution apply § 41, determining Rx and ^s each by 
an independent equation in the form given in PI. 11^. 

-^1 -and R% once determined, check by seeing if Ax and Ay are 

Exercise 12. A body in the shape of an isosceles triangle is supported 
by a smooth hinge at one of the equal angles and by a smooth horizontal 
plane (at the same level as the hinge) at the other. Assume three un- 
equal parallel forces normal to the slope on the side next the hinge to be 
given, and determine fully the pressures on the hinge and plane. Solve 
by both methods. 

Exercise 13. P\, P%, Pt, P*, and P* (Fig. 12) are known forces acting 
on the body shown, and Q, R, and S are known only in line of action. If 

Fig. is. 

the magnitudes of the given forces are 4000, 3000, 2000, 1200, and isoolba. 
respectively, determine the magnitudes (and senses) of Q, R, and 5 by 
both methods. 

Suggestions. Construct the data diagram carefully. Scale i in. = 5 ft. 


See PI. IV for solution by the usual Axf Ay, and B, Observe what labor 
this solution involves. 

Use the method of § 41 and solve, scaling arms from the drawing 
only for checking. In writing the equation it will be convenient to 
write some of the moments in the form Pxa, and some in the form 
P cos ay — P sin a x. (Cf. Plate V.) 

Exercise 14. A square plate (Fig. 13) weighing 900 lbs. is held in a 
vertical plane, with its edges inclined 30** to the horizontal and vertical. 

Fig. 13. 
by a horizontal force and forces along two adjacent sides as shown. 
Determine these three forces graphically only. 

Suggestions. Though this is clearly Case 4, note and use the short 
cut possible in this special case where there are two pairs of forces con- 
cerned whose resultants are necessarily equal, opposite, and coincident. 
Putting in this common resultant — which will naturally be lettered ac— 
there are two groups of concurrent forces, ab, be, and ac, and ac, cd, and 
da. In the first ab is known, whence be and ca follow by Case 2a ; uc 
can then be resolved similarly into ^//and da. 

Note that what makes this exercise a special case is that there is only 
one known force. The method just pointed out may be regarded as the 
standard method of resolving a force graphically into three components. 

The algebraic method if called for would have employed the three 
moments equations as usual in Case 4. 

What could be specified as to the position of the pole in the general 
method of Case 4 which would lead to a solution identical with this 
short cut? How are all four sides of the string polygon then ac- 
counted for? 

Exercise 15. Resolve (both graphically and algebraically) a given 
force of any convenient assumed magnitude into two parallel com- 
ponents whose points of application are 

{a) On opposite sides of the given foroe. 

{b) On the same side of that force. 


Suggestions. Avoid a small scale for the space diagram. As is usual 
in this Case, the equations of moments are to be preferred to the 
Ax* Ay, and B equations for the algebraic work. 

Note. — For further general problems in statics, the reader is referred to such 
works as Loney's Statics, Bowser's Analytical Mechanics, Minchin's Statics, Wal- 
ton's Problems, etc. The reader seeking practice among the problems there given 
must be prepared in many of those problems for a large amount of geometrical 
analysis and trigonometric reduction extraneous to the purely statical solution. 


51. Graphic Representation of the Moment of a Force. 

— ^Theorem. If, through any point, a line be drawn parallel 
to a given force, /*, the distance, 7, intercepted from this line 
by the two strings belonging to the force is a length such that 
when multiplied by the force H^ measured by the perpendic- 
ular dropped from O upon the given force in the magnitude 
diagram, the result will be numerically equivalent to the mo- 
ment of P about the given point. 

Proof. — Let m (Fig. 14) be any point whose perpendic- 

Fig. 14. 
ular distance from the. line of action of P is p. If P be lettered 
ab, and the pole be tajcen and the string polygon be con- 
structed as shown, j^ is a length such that 




For, noting the similarity, by construction, of the triangles 
OAB and the one bounded by oa, od, and the line through m^ 
it appears that 

AB:H = y:p, 

or 2J& AB represents the magnitude of jP, 

Hy = Pp. Q. E. D. 

The intercept 7 may accordingly be said to represent the 
moment of P about w, it being understood that, to get the 
numerical value of this moment, y measured in the proper 
scale of lengths must be multiplied by H measured in the 
proper scale of force magnitudes. 

So far H and y are simply a force and a distance whose 
product is the same as that of P and /. There are of course 
an indefinite number of forces and distances whose products 
will have this value, and the substitution if desired might be 
made in simpler ways even than by this theorem, if that were 
all that is desired. With a series of parallel forces, however, 
treated as usual with the magnitude and string polygons, H 
will be constant for them all, and aided by this fact the theorem 
leads to a convenient graphical method for the treatment of 
moments which will be developed in the next section. 

Observe that the strings oa and ob are simply any two com- 
ponents of /*, and that H is merely their common component 
normal to P. 

52. String Polygon for Parallel Forces a Diagram of 
Moments. — As a corollary to the theorem of the preceding 
section, it may be stated that with parallel forces any two 
sides of the string polygon, extended if necessary, will cut from 
a line parallel to the forces an intercept which will represent 
the sum of the moments (about any point in that line) of all 
the forces at the apices of the polygon included between the 
two sides. 



For example, suppose, in Fig. 15, the string polygon be 
drawn for any set of parallel forces (magnitude polygon not 
shown) which for the sake of generality are not taken in equi- 
librium and which are lettered preferably, though not necessa- 
rily, in the order of their occurrence. Lengths are intercepted 
on the line MNy parallel to the forces, which represent the 

Fig. 15. 

moments of groups of these forces about any point m on MN 
as follows. 

The intercept a^a^ bounded by oa and og represents the 
moment of ag by direct application of § 51; that is, it repre- 
sents the moment of the resultant of the whole set, and hence 
the sum of the moments of all the forces of the set. 

Similarly, a^a^ represents the moment oiad\ that is, the sum 
of the moments of ab^ bCy and cd. Likewise a^a^ represents 
the moment ol dg\ that is, oi de^ ef^ and^ combined. 

Moreover, the intercepts are the actual summations of other 
intercepts each representing the moment of one of the indi- 
vidual given forces. Take, for instance, the last case, that of 
de^ efy and^. Extending the strings oe and of till they cut 
MN^ the resulting intercepts a^a^ , afi^ , and a^a^ are seen by 


§ 5 1 to represent the moments of de^ ef, and fg respectively 
about ntj and what is more, a^a^ is their algebraic sum (—^3^4+ 
■^5^4— ^6^2)* ^^ ^^ should be. 

The force H^ it may be repeated, by which the lengths of 
these various intercepts must be multiplied to give the numeri- 
cal values of the respective moments, is the force represented 
by the perpendicular (measured in the same scale as the rest 
of the forces) dropped from O upon the straight line ABCD^ 
etc., upon which fall the parallel forces ab, be, cd, etc., when 
they appear in the magnitude polygon. H being thus a con- 
stant force, the moments are proportional to the intercepts as 
above stated, and the string polygon for the parallel forces is a 
diagram of moments. 

53. Remarks on the String Polygon as a Diagram of 
Moments. — Referring still to Fig. 15, it will be useful to note 
the following facts and deductions. 

If the forces had been in equilibrium, og and oa^ the strings 
including the whole set, would have coincided and there would 
have been no difference between a^a^ and a^^^ except in sign, 
and their sum would be found to be zero, as it should be. 

If the forces had been reducible to a couple, oa and og 
would have been parallel and a^a^ constant, as it should be, for 
the constant moment of a couple. 

Where oa and og intersect, a^a^ would vanish, as it should 
do, for that intersection is known to be on the line of action of 
the resultant of the set of forces. 

If the forces in the given set are not parallel, the string poly- 
gon can still be made to yield the sum of the moments about 
a point of any consecutively lettered group of these forces. It 
is only necessary to draw a line through the point parallel to 
their resultant, scale the intercept from this line by the strings 
inclosing the group, and multiply it by the H for this particu- 
lar resultant. The process is a straightforward application of 


§51. Since in general no two resultants will have the same Hy 
or be in the same direction, the string polygon for non-parallel 
forces is not a diagram of moments in the same useful sense as 
when the forces are parallel. 

Hence the string polygon is not only an equivalent to an 
equation of moments in its scope and uses, but also an exact 
parallel to it in that it readily gives the value of any term of 
that equation and is simply a graphical means of finding and 
summing the values of those terms. 

If any set of forces whose magnitude polygon closes is not 
reducible to a couple, the intercept from any line whatever in 
the plane by the two extreme strings must be zero. This could 
not be unless these two strings coincide, or, in other words, 
unless the string polygon close, a fact the converse of which 
could be taken as a demonstration of the preclusion of rotation 
by the closure of a string polygon alternative to that of § 32. 

If the set of forces is reducible to a couple, the resultant 
looked upon as a single force is indeterminate in direction (cf. 
Exercise 7). Nevertheless its moment can still be found from 
the string polygon by assigning it any convenient direction, 
taking the intercept by the extreme strings (lettered, say, oa and 
on) from a parallel to it and multiplying this intercept by the 
H measured by the perpendicular from dropped upon a par- 
allel to the intercept drawn through the two coinciding points 
A and -A^in the magnitude polygon. The process by which 
the result was evaluated in Exercise 7 may be regarded as 
really an application of this method, in which a direction normal 
to the extreme strings was assigned to the resultant, making 
-^identical with OA. 

54* To Pass a String Polygon Through Three Given 
Points. — A process called passing a string polygon through 
three given points is sometimes useful in the study of the equi- 
librium of structures. It consists in locating a pole for a set 



of two or more forces so that three specified strings of the result- 
ing string polygon will each (prolonged if necessary) pass 
through one of three given points not in the same straight line.* 
Problem. — Suppose, for example, that the forces in Fig. 16, 

ab^ bcy cdi and de^ their magnitude polygon ABCDE^ and the 
three points m^, m^, and m^he all given, and that it is required 
to select a pole O so that oa, oc, and oe will pass respectively 
through Wj, m^f and m^ 

Solution. — Taking a pole (7 at random, determine (Case 
3) two forces CC^ and A C^y the former acting at m^ in any 

* If there is only one force in the given set, the string polygon can still be 
made to pass through three points not in a straight line, but two of the points will 
of course have to be on one of the strings. 


convenient direction, and the latter at m^^ in a direction deter- 
mined accordingly, such that they and the given forces a6 and 
ic, intervening between m^ and w,, will constitute a set in 
equilibrium. Now, when oc passes through m^ as required, oc^ 
will have to pass through m^ and m^ in order to close the string 
polygon for this balanced set. It follows, moreover, for the 
same reason, that if oc^ passes through m^ and m^, oc will have 
to pass through w,, and working backwards, oa will have to 
intersect oc^ again at m^. Therefore an O taken anywhere 
on a line through Cj parallel to m^^ will be one for which, if 
oa be drawn through w^, oc will pass through w,. 

Repeat this process, determining an ^^^ and an EE^ at m^ 
and Wj respectively, which would balance /7^, bc^ cdy ditidde, — 
AE^ and EE^ corresponding exactly to the AC^ and CC^ 
respectively of the preceding step. Then reasoning as before, 
an O taken anywhere on a line through E^ parallel to mytn^ will 
be one for which, if oa be drawn through m^, oe will pass 
through niy 

Therefore the intersection, d?, of lines from C^ and Ey^ par- 
allel respectively to m^^ and m^^ will be the required pole 
for which if oa be passed through w^, oc will pass through m^ 
and oe through m^ as required. The string polygon oa^ ob^ ocy 
ody oe can now be constructed at leisure. 

The foregoing process may be described in general terms 
as follows. 

If for a set of given forces, ab^ be, cd , . . yz^ it be required 
to pass any three specified strings, ody oly and oSy of their string 
polygon respectively through the given points m^y m^y and m^ 
not in one straight line, draw a string polygon at random 
from a pole Cy selected at random, beginning with the string 
o'dy making that pass through m^y and by working both ways 
complete this random polygon at least as far as &s. Use this 
string polygon to determine (Case 3) //^ and dl^ and ss^ and 


ds^ (//^ and ss^ being assumed to act at m^ and m^ respectively 
in any convenient direction, and the other two forces at w, 
and determined accordingly), which would balance de . . . >&/, 
and de . . . rs respectively. DL^^ and DS^ do not need to be 
drawn. Parallels to m^m^ and m^m^ through L^ and S^ re- 
spectively will locate the required pole O at their intersection. 
To draw the required string polygon, od will be drawn first and 
made to pass through m^ Then working both ways, the 
polygon is completed. 

The problem obviously does not admit of solution when the 
three points are in a straight line. 

It is advisable to take (7 so that the random string poly- 
gon will lie as well as possible out of the way of the final 
required string polygon. 

The random string polygon can be obviated if desired by 
an algebraic computation of the magnitude and direction of such 
a force as OD of the preceding general problem. D being 
given, O can then be plotted directly. 

To determine OD we need to observe that its point of appli- 
cation m^ is given, and that the resultant of it and de , , , kl, 
and of it and de . . . rs must pass through m^ and m^ respec- 

It will be well to consider od replaced by its horizontal and 
vertical components meeting at m^, whose magnitudes and 
senses * can be determined from two simultaneous equations of 
moments for centers respectively at m^ and m^^ the first one for 
the series of forces de . . , kl and the unknowns, the latter for 
the series de . . , rs and the unknowns. 

Combining the two components (by Case 2d)y the required 
OD is found, and O can be plotted. 

*It will be found advisable, in order to avoid error in such simultaneous 
equations, to assume senses of the components arbitrarily and show them in the 
sketch of data subject to subsequent correction on the solution of the equation. 


It will be highly desirable in practice to check this loca- 
tion of O by determining in a similar manner the magnitude 
and direction of SO. The centers will be at m^ and m^ and in- 
cluded in the equations, besides the two unknown components 
will be the sets sr . . . mly and sr . . . ed respectively. 

The object of the whole process is sometimes merely the 
determination of forces at two of the points, such as the OD and 
OS of the preceding work. The foregoing section constitutes 
therefore a complete description of the solution of such prob- 
lems by both the algebraic and graphic methods. (Cf. § 86.) 

Exercise i6. Assume any set of seven vertical forces uniform in sense 
and letter them in the order of their occurrence ad, . . , hi. Assume 
three points mi,m%, and m^ not in the same straight liiie and at the left 
of ad, between de and e/, and at the right of At respectively. Pass a 
string polygon for the assumed forces through the three assumed points 
so that oa, oe, and oi will go through mx, m%, and m% respectively. 
Solve graphically only. 

Remark. Assuming the set of forces as specified does not interfere 
with the generality of the method required and will save some labor. 

55. An Altematiye View of the Says and the String 
Polygon. — Each two successive rays may be regarded as two 
components of the force with which they form a closed force 
triangle, and, moreover, by reversing them in sense, as com- 
ponents respectively of the forces next preceding and follow- 
ing that force in the magnitude polygon. Inserting these 
pairs of components in place of their respective resultants, there 
will result a set equivalent to the given set. If by moving the 
assumed point of intersection of the components along the lines 
of action of their resultants the pairs of equal and opposite com- 
ponents of the consecutively lettered forces can all be made 
coincident, the components must be a set in equilibrium, and 
hence the original set must have been as well. 

The fact that the components were pairs of equals as well 
as opposites could only be if the magnitude polygon closed. 


The pairs all coinciding form the closed string polygon. The 
lines of action of pairs of coinciding forces constitute the 

If the pairs of equal and opposite components all coincide 
and neutralize one another but one, that pair of components 
constitutes a couple to which the given set is reducible. 

If the set is not reducible to a complete set of pairs of equal 
and opposite components (a case which can only arise when the 
magnitude polygon fails to close), there will be only one pair 
not of this character. The pairs of equals and opposites could 
be made to coincide throughout in the space diagram as usual, 
leaving the other two to locate their resultant passing through 
their intersection, a resultant likewise that of the given system. 

From the preceding three paragraphs could be deduced 
anew the principle that the closure of the magnitude polygon 
and the string polygon insures equilibrium. 




56. Center of Gravity.— The center of gravity of a body 
is that point through which the resultant of the gravity forces 
acting upon all parts of the body will pass, whatever the atti- 
tude of the body, or what amounts to the same thing, whatever 
we may regard the slope of the gravity forces, so long as they 
remain parallel.* The center of gravity may be looked upon, 
then, as the invariable point of application of the force which 
represents the weight of the body. 

Since the moment of the weight of the body, referred to any 
plane, is evidently equal to that of the weights of all the parts 

* Strictly speaking, the center of gravity is a point in which the whole mass of 
a body might be concentrated without affecting gravity attractions existing between 
the body and all other bodies, whatever may be their relative distance and 

Very few bodies have a true center of gravity. What is commonly called the 
center of gravity would more properly be called center of mass or center of area. 

All centers of gravity are also centers of mass, but ail centers of mass are not, 
strictly speaking, centers of gravity. 

The term center of gravity is so thoroughly fixed by general usage in the sense 
above given, and the need for making the distinction is so rare, that there is no 
great need for urging the substitution of the more accurate terms. For further 
discussion of this distinction see Du Bois, Mechanics, voL 2, chap. 4. 



referred to the plane, we can write (conceiving gravity acting 
parallel to the plane) : 

x2w = w^x^ + w^^ + . . . + w^^, 

that is x2w = ^{wx) 

likewise y2w = ^{wy) - , . . . . (i') 

also li2w = 2{wz) 

where x^y^'z and x^ y, z are the coordinates of the centers of 
gravity of the whole and of parts of the body respectively, and 
w the weights of these parts. 

It is often required to find ^, J, 5, and to do it we need 
only to divide the body into parts whose x^ y^ z can readily be 
determined, and apply from (i') 

Equations (i) show that for all planes or axes of reference 
passing through this center of gravity, i.e. for x =z o, y = o, 
and z= o, 

2(wx) = o; 2(wy) = o; 2{wz) = o; . . . (2) 

as might have been foreseen. 

It is often convenient to apply the term center of gravity 
to bodies irrespective of their weights or to those having no 
weights, as geometrical shapes, and planes and lines, meaning 
thereby the point which would be the center of gravity of the 
body if the body were of uniform density or had weight pro- 
portional to volume, area, or length. Any factor introduced 
to express weights in terms of volumes, areas, and lengths 
would cancel out of equations (i) and (2) and leave simply 
volumes and areas and lengths in place of weights, to be treated 
exactly like weights. In (i) and (2), then, might be substi- 
tuted V, a, or / for w. In the following work attention will be 
devoted mainly to the center of gravity of areas, and they like 


lines will preferably be referred to two rectangular axes, and 2 
will disappear. 

In the foregoing the term ** system of bodies ** (if the bodies 
be understood to be fixed relatively to one another) might be 
substituted for * ' body. * ' 

Statically the location of a center of gravity is the location 
of the point of application of the resultant of a set of parallel 
forces, a simple special case under Case i requiring only the 
familiar equations (of which (i) above is, of course, a deriva- 
tive) and polygons. 

In the case of simple and symmetrical bodies the result of 
applying equations or polygons can, of course, be seen at once 
by mere inspection, just as can sometimes be done with other 
statical problems. 

The center of gravity of a straight line is evidently its 
middle; point; of a rectangle is the intersection of normals to 
the centers of its sides ; of any parallelogram the intersection 
of its diagonals ; of a circle or ellipse its geometrical center; of 
a triangle the intersection of its medial lines. This opens the 
way for proof by the general method that the center of gravity 
of a trapezoid is the intersection of the line connecting the 
middle points of its parallel sides with the diagonal connecting 
points found by extending each of the parallel sides in opposite 
directions by an amount equal to the length of the opposite 
side, or at the intersection of two such diagonals. (Fig. 17.) 

Y -iSa k b - -^— 6--3|f ;?a-*--— X 

Fig. 17. 
The center of gravity of a surface bounded by a curved line 
is, of course, treated by equation (i), but for strict solutions 


in such cases subdivisions w are made infinitesimally small and 
the problem falls into the field of the integral calculus rather 
than arithmetic, without, however, introducing any new stati- 
cal principle. The solution may be approximated with any 
desired degree of exactness by arithmetical methods by mak- 
ing the parts w smaller and smaller. 

In the case of a segment of a parabola cut off by a chord 
normal to the axis, if b be the length of such chord and h its 
distance from the vertex, integration will show that its area is 
\bh, and that its center of gravity is on the axis distant \h 
from the verte^c (Fig. i8). The center of gravity of a circular 

Fig. i8. 
sector is shown by similar means to be distant fi-om the center 

of the circle on the central radius by \r ^ , where r is the 

radius of the circle and is half the central angle of the sec- 
tor in circular measure. 

Any plane surface which can be divided with sufficient 
exactness into triangles, rectangles, trapezoids, parallelo- 
grams, segments of parabolas, or sectors of circles can accord- 
ingly be treated by the general method. For a surface it is 
necessary in general to assume the gravity forces acting first 
in one direction, then in another, it may well be, at right angles 
to the first. The point common to the two resultants for the 
two different directions for. the forces will be the center of 

Numerical example. Required the location of the center 
of gravity of the irregular surface shown in Fig. 19. 

Solution. Taking the origin at the lower left-hand corner, 
and dividing the figure into two rectangles and a semicircle, the 
areas and the coordinates of the centers of gravity of each of 





these divisions are worked out and recorded in the figure for 
convenience in reference. The center of gravity of the semi- 

FiG. 19. 
circle is, according to what has been said just above as to the 
center of gravity of a sector, 

2 X 4 X sin - 




= 1.70 m. 

above the center of the circle, making the ordinate of the 

center of gravity 2.0 + 4-0 + 1.70 = 7.70 in. as shown. 

Algebraic method. Calling x and J? the coordinates of the 

center of gravity of the whole figure, the algebraic solution can 

conveniently be arranged as follows : 

25.1 X 4-0 = 100.40 25.1 X 7^7 = 193-27 
32.0 X 4-0 = 128.00 32.0 X 4.0 = 128.00 
20.0 X 5.0 = 100.00 20.0 X i.o = 20.00 

77.1 X ^ = 328.40 
- 328.40 

X = 


= 4.26 in. y = 

77.1 X 7 = 341.27 

- 341.27 


= 4.43 m. 


Graphic method. Considering forces proportional to the 
areas of the divisions acting in two directions at right angles to 
each other from the centers of gravity of their respective divi- 
sions, and locating the resultant of each of these sets of parallel 
forces by the usual method, the required center of gravity is 
found at the intersection of the two resultants. This work is 
done in Fig. 19, with results agreeing with those found above. 
One magnitude polygon only is required. By transferring nor- 
mals to its rays as well as parallels, both string polygons can 
be constructed. 

The reader interested in a wider range of applications of 
the principles of this chapter is referred to works such as those 
mentioned at the close of Chapter VI, in addition to which 
might well be mentioned Rankine's Applied Mechanics, or 
Weisbach's Mechanics. 

Exercise 17. Find the center of gravity of any irr^;ular figure 
bounded by curves and straight lines. Solve both graphically and alge- 

Suggestion. Select the figure from among the rolled-steel sections, 
such as the channels, unequal-legged angles, bulb angles, deck-beams, 
etc., given in the Carnegie, Cambria, Pencoyd, or other manufacturers* 
handbooks. Results may then be compared with those given in the 


57. External and Internal Forces.— External forces upon 
a body are forces exerted upon that body by or through 
another body. 

Internal forces in a body are those transmitted to one part 
of that body through another part of the same body. 

Investigation of external forces determines whether a body 
as a whole is stable, without regard to whether all parts are 
strong enough to do what is required of them. Investigation 
of internal forces determines the strength required of the parts 
to insure the stability of the whole. A knowledge of both 
external and internal forces is therefore indispensable. Both 
internal and external forces are, of course, subject to the gen- 
eral laws governing forces. 

The internal forces at any section of the body hold in equi- 
librium the external forces on either side of section. 

58. Stress. — Stress is the tendency to distortion or rupture 
in a body due to the action of the external forces. It is a 
measure of the responsibility of the body or of a specified part 
of the body in service. 

Stress may vary greatly in the various parts of the body, 
and it is necessary in any study of it to compute the stresses 
in existence at one or more plane sections of the body. In 
such cases the plane of the section will divide the body into 
two segments, and the external forces into two sets, one acting 



directly on one segment, the other on the other. The two 
equal, opposite, and coincident resultants of these two sets of 
forces may be looked upon as the immediate cause of the stress 
and its measure as well. The magnitude of the stress depends 
upon the common magnitude of these two resultants, and its 
nature upon their other elements. 

The determination of stresses, then, is simply the deter- 
mination of elements of forces — a statical process pure and 
simple and one of the most important fields of application of 
statics. It will receive much attention accordingly. 

59. Kinds of Stress. — The resultants mentioned in the 
preceding section may, of course, be either single forces or 
couples, and four special cases of stress are distinguished, each 
corresponding to a special direction for these single forces and 
to, a special position for the planes of the couples. 

With the single force resultants there is B'ormal Stress or 
Tangential Stress (more commonly called Shear) at the sec- 
tion according as the resultants are normal or parallel to the 
section. Their common magnitude measures each stress. 

With the couples there is Flexure or Torsion at the sec- 
tion according as the couples are in a plane normal to the sec- 
tion or in planes parallel to it. The common magnitude of 
the couples measures the stress. 

60. Combined Stresses. — As a matter of fact, however, 
the resultants are very frequently not in any one of these four 
simple relations to the section, and there results some combi- 
nation of the four stresses accordingly. If the single force re- 
sultants are inclined to the section there exists both normal 
stress and tangential stress, each measured by the common 
magnitudes of the components normal and parallel to the sec- 
tion respectively. If such resultants fail to pass through the 
center of gravity of the section it will be convenient to observe 
that by § 20 each resultant is equivalent to a force acting at 


that point* and a couple, and in such cases accordingly there 
would be combined normal stress, shear, and flexure. If 
the forces are non-coplanar, there may be still another 
couple in a plane parallel to the section, and torsion will be 
present in addition to the other stresses. 

If there is normal stress or shear at the section as well as 
flexure, as is usually the case, the resultant couple which will 
measure the flexure is the resultant of the set made up of all the 
forces external to the body acting on the segment and also a 
force, internal to the body but external to the segment, equal, 
opposite, and parallel to these external forces conceived to be 
applied at the center of gravity of the section. t 

In other words, flexure may be looked upon as measured 
by the couple remaining after providing the forces which will 
prevent motion of the segment normal or parallel to the section. 

6i. Further Partictilars Relating to Stress. — Normal 
stress is divided into its more familiar subdivisions, compres- 
sion and tension, according as the resultants act from their 
respective segments towards or away from the section. 

Shear and Flexure are similarly subdivided according to the 
relative senses of the resultants, but their subdivisions have no 
established names, and are distinguished from each other by 
the adjectives positive and negative, arbitrarily applied. These 
distinctions are rarely needed except in the case of horizon- 
tal beams subject to vertical loads, in which the sections 

* Any other convenient point might be selected, but, as a principle of Resistance 
of Materials worth noting in passing, it may be said that there is no advantage in 
making the substitution of the force and the couple unless the center of gravity be 
the point of application of the force. It is known how to provide for a force so 
applied, but not for one elsewhere, except by this substitution. 

■f If this force be taken into account .the value of the flexure would be independ. 
ent of the center chosen, as the value of a couple is cotistant for all centers in 
the plane (§ 23). But taking the center of moments at the center of gravity the 
moment of this force vanishes and the force need not then be determined for the 
calculation of the flexure. 



are taken normal to the axis of the beam, i.e. vertical. In 
such cases it is natural and customary to call shear and flexure 
positive when the left-hand segment is subject to an upward 
force or clockwise couple respectively, and vice versa. 

Under normal stress a body simply tends to lengthen or 
shorten (according as the stress is compression or tension) 
along a line at right angles to the section. Compression is 
the stress typical of columns, posts, struts, and pedestals. 
Tension is typical of ropes and chains in service, or tie-rods. 
Anything through which a push is transmitted is in compres- 
sion, and anything through which a pull is transmitted is in 

In Shear one segment of the body tends to slide by the 
other with no tendency to rotate about an axis normal to the 
section. Shear is the stress typical of rivets; in fact, their 
main purpose is usually to resist this kind of stress. The 
shearing resistance of the rivet in Fig. 20 prevents it from 


Fig. 20. 

being separated into three parts by shearing on the planes of 
contact of the plates which it connects. Another illustration 
of shear is to be found (Fig. 21) in the plane indicated by 
the dotted line in the timber receiving thrust from the rafter. 

Flexure is simply a tendency to bend. Familiar examples 
are a stick bent over the knee, or a loaded floor joist. 

Torsion is a twisting tendency due to the two segments of 


the body tending to rotate in opposite directions about an axis 
normal to the section. Torsion is the stress typical of shaft- 
ing for the transmission of power. It is rarely permitted else- 
where if it can be avoided. 

When it is said that Flexure or Torsion exists at a given 
section it means that the segment of the body on one side of 

Fig. 21. 

the section is likely to bend or twist respectively with respect 
to the other segment. These two stresses may be looked upon 
as complex cases of normal stress and shear respectively, i.e. 
cases in which the tendency to failure under the normal stress 
or shear varies in different parts of the section. It is familiar 
fact that the upper and lower surfaces of a bent beam and the 
surface portions of a shaft are subject to severer stress than the 
interior portions. 

Familiar examples of combined stresses occur in beams 
subject to transverse loads in which shear and flexure appear 
together; in shafting in which shear, flexure, and torsion ap- 
pear together, to which may be added compression if the shaft 
is vertical, 

A stress is fully described as soon as its magnitude and 
nature are stated, nature being here understood to include 
algebraic sign as well as kind (§ 59). 



63. Structures. Definitions. — Structures are simply artifi- 
cial contrivances for supporting loads or resisting the active 
forces of nature. They must resist the stresses due to the 
action of these loads or forces on the one hand and the reaction 
of the earth's surface op the other. They may be very simple 
(as mere posts, pedestals, tension rods), acting under com- 
pression alone or under tension alone, and such may be desig- 
nated by the term elemental structures. In general, however, 
the term structure will be understood to mean only the 
complex kind, such as are subject as a whole to bending and 

As regards their composition, structures may be divided 
into two broad types, (a) framed) and {6) non-framed. 

A framed structure, or frame, is one composed of a series 
of straight bars fastened together by their ends only, so as a 
whole to make substantially one rigid body. 

Ideally (that is, if the joints could be made frictionless 
hinges, if the weights of the bars could be made to act only 
at their ends, and if all other loads are applied only at the 
joints) the stress in each bar of a frame would be purely axial 
(i.e., pure tension or pure compression), making each member 
an elemental structure. 

Such ideal frames are also called true frames, and are what 
is meant when frames are referred to without further description. 

The nearest actual approximation to a true frame is a pin- 



connected truss. Indeed pin-connections have been used very 
widely in truss construction largely because the assumed condi- 
tions of computation can thus be most nearly realized in prac- 
tice (cf. § 63). 

The resultant of the external forces at any section of a true 
frame is resisted by elemental members into which the section 
can easily be analyzed (for the center lines of the bars must 
be the lines of action of the forces internal to the structure) and 
each pin or joint is in equilibrium under the action of a set of 
concurrent forces. 

A non-framed stntcture is one consisting of one continu- 
ous member or of a number of members so fastened together 
throughout their lengths as to make one solid piece. 

A non-framed structure cannot be analyzed into separate 
elemental members. The flexure and shear at any section is 
resisted by the whole section under stress varying in intensity 
from point to point of the section according to more or less 
complex laws. 

The relations between the external forces are entirely in- 
dependent of whether the structure is framed or non-framed. 
The difference in the types is wholly one of internal make-up, 
and effects the method of dealing with internal forces only. 

The periphery of a frame constitutes what are called the 
two chords of the frame, the portion of the periphery bounded 
by the end joints on the upper side of the frame being further 
designated as the top, or upper, chord and the one on the 
lower side as the bottom, or lower, chord. 

63. Extent of Approximation to True Frames in Practice. 
— Outside of the fact that joints cannot possibly be made fric- 
tionless, and that the weights of the bars cannot possibly be 
made to act only at their ends, true frames are very uncommon. 

Compression chords, that is, the large part of the periphery 
of every frame which is subject to compression, are usually 


made stiff at the joints. Other members may be hinged to 
them, but the chords themselves are not broken at the joints, 
or are so rigidly spliced as to be practically continuous and 
hence stiff. Moreover, trusses are frequently made with no 
hinge joints whatever, but with tightly riveted connections. 

The conception of the true or ideal frame as above defined 
is, however, fundamental to the design of all these structures, 
and is the hypothesis upon which the normal stresses in the 
members are determined. The difference between this ideal 
frame and actual ones can, by proper design and construction, 
be rendered negligible. When, however, the difference is 
permitted to be considerable, means are available for estimat- 
ing and providing for the so-called secondary stresses thereby 

If the axes of the various bars intersect at common points. 
I.e., if the forces at a joint are really concurrent, secondary 
stresses are considered negligible in spite of the joint being far 
from frictionless. This is due to the materials being so nearly 
rigid that only very slight changes of shape in the structure 
occur and hence there occurs only a very slight tendency for 
the bars to turn about the joints. The forces themselves by 
meeting at the joint produce no tendency of this kind. 

64. Loads Applied Elsewhere than at Joints. — It occa- 
sionally happens that loads must be brought to bear on frames 
at points where there is no joint, and where it is not practicable 
to make one by the addition of more bars to the frame. The 
result is that the bar to which the force is applied has to do 
double duty — that of any frame member subject to tension or 
compression and also that of a beam or girder subject to shear 
and flexure. 

In its capacity as a beam the bar transmits its transverse 
loads to the frame in the form of reactions upon the joints by 
which the bar is incorporated into the frame. These reactions 



are determined as in any case of a beam with given loads and 
supports. Reversing their senses, they are the forces acting 
on the joints of the frame. Making these determinations and 
substitutions the frame can be analyzed as usual. The shear 
and flexure for which (in addition to their tension or compres- 
sion as part of the frame) the bars subject to transverse loads 
must be designed are determined as for any beam. 

This form ot construction is uneconomical of material, and 
is usually to be avoided. 

Strictly speaking, all bars of a material frame are in the 
condition just described owing to their own weight, but in 
frames of moderate size the resulting shear and flexure in indi- 
vidual bars is neglected even though the frame itself is analyzed 
for tensions and compressions due to its own weight considered 
concentrated at the joints. 

For example of a frame illustrating this section see § 84 
and Exercise 27. 

65. Frames in General. — Frames may be {a) complete, 
{b) incomplete, or {c) redundant. A complete frame is one 
composed of just enough bars to insure its keeping its shape 
under all conditions of loading. If it has fewer bars than this 

FlO. 33. 

Ftg» 91^ 

Flo. 34. 
requires, but nevertheless is able to carry a load if properly 
distributed, it is incomplete, and if more, it is redundant. 
Examples of each are shown in Figs. 22, 23, and 24 re- 


Since the triangle is the only geometrical figure in which 
a change of shape is precluded unless the lengths of its sides 
are changed, the triangle is necessarily the basis of arrange- 
ment of the bars in a frame* 

A complete frame must accordingly be made up of the 
minimum number of bars consistent with its being composed 
wholly of triangles. Removing a bar common to two triangles 
would render it incomplete. Adding bars to a complete frame, 
as by adding the second diagonal in one or more quadrilaterals, 
renders it redundant, but if such bars are capable of resisting 
one kind of stress only, as in the case of counters, the redun- 
dancy may be only apparent. See § 83 and Exercise 26. 

Complete frames are the type which is usually closely ap- 
proximated, and which accordingly receives most attention. 

Their analysis involves only a straightforward application 
of the familiar principles of statics. 

Incomplete frames are stable only under symmetrical or 
other specially arranged loads. Under such loads they are 
analyzed with as much ease and certainty as Complete 
Frames, and require no further explanation. 

Structures having outlines of incomplete frames may resist 
loads of any distribution by virtue of the flexural strength of 
members continuous through several joints, but such structures 
are not true frames and require special treatment, which takes 
note of the ability of some of their bars to resist bending. 

Redundant frames* will resist loads of any distribution, and 
some forms are not uncommonly found advantageous in use. 
Their analysis involves statically indeterminate problems and 

♦ A test for redundancy can be worked out as follows. Beginning with a tri- 
angle, each two bars added establishes a new joint. Then if 3 -|- x equals the 
number of joints, the number of bars for the complete structure will be 3 -|- 2jp, i.e., 
twice the number of joints minus 3. Therefore, for complete frames, if m equals 
the number of bars and n equals the number of joints, m equab 2ff— 3. If m is 
less than (211—3), the frame is incomplete; if m is more than (211—3), the frame 
is redimdant. 


their stresses therefore cannot be worked out by purely statical 
methods, but with special data as to the material, or as to the 
perfectness of workmanship, or by the aid of some outright 
assumption as to probable action of the bars, a more or less 
satisfactory estimate of stresses can be made. 

66. Loads. — Loads are {a) permanent or dead, those which 
are always present; or {b) moying or live, those which are 
only occasionally present. 

The permanent load for roofs is the weight of the roof cov- 
ering, purlins, trusses, etc. 

The live load for roofs is wind pressure, snow, etc. 

The permanent load on bridges is the weight of the floor, 
trusses, etc. 

The live load on bridges is the weight of trains, carts, 
crowds of people, wind pressure, etc. 

Loads which act simultaneously on a structure may or may 
not be considered all at once in the determination of stresses. 
If they are not considered all at once the total effect is ob- 
tained by simply taking the algebraic sum of all the stresses 
caused by the partial loads. It is usually desirable in practice 
to follow this course for the sake of avoiding very serious com- 
plications in the work. 

The way in which stresses are provided for falls within the 
domain of Resistance of Materials. 

67, Stresses in Structures. — As has been stated, stresses 
result directly from external forces. External forces are either 
loads or reactions. Loads are always known or assumed and 
the reactions determined accordingly by the methods already 
developed, commonly Case 2b or Case 3.* 

* It may be noted that certain cases of statically indeterminate problems occa- 
sionally arise in connection with reactions. An example would be found in a 
beam resting on three or more supports, giving rise to' three or more parallel 
reactions. The reactions can be worked out in such cases, by the aid of the laws 
of elasticity, involving methods outside the scope of this book. The need of re- 
sorting to such methods is usually avoided in the design of the structure. 


As soon as the reactions are known, the external forces are 
all known and the stresses can be determined as a purely statical 
matter by methods to be inferred from the discussion of stresses 
in §§ 58-60. As there stated the stress at any section is really 
a question of the resultant of all the forces external to the 
body on one of the segments into which the section divides the 
body. This resultant is a measure of what may be looked 
upon as normal stress, shear, flexure, or torsion whctlicr 
the body is framed or non-framed. Before either kind o 
structure can be designed, however, the statical analysis must 
go a step further still. In the case of non-framed structures 
this further analysis is statically indeterminate and requires 
the aid of the elementary principles of Resistance of Materials 
for its accomplishment. Stresses in a non-framed structure arc 
accordingly considered determined as soon as each of the four 
kinds of stress is fully known at a requisite number of sections. 

In framed structures this further analysis may be statically 
determinate. It consists in determining just how much tension 
or compression exists in each bar as a consequence of a given set 
of loads. The determination of stresses in a framed structure 
does not stop with the determination of normal stress, shears, 
flexures, and torsions in the body as a whole. In fact, in 
many cases it proceeds directly to the determinations of the 
tensions and compressions in the bars consequent upon the four 
fundamental stresses without stopping to find those stresses in 
their unanalyzed state at all. 

The next two chapters will state the statical processes ap- 
plied to each of the two kinds of structures after the external 
forces are all known. 


68, Stresses in Non-framed Structures. — In a non-framed 
structure, stresses are found at any section by finding the re- 
sultant of all the forces on one side of the section. Its com- 
ponents normal and parallel to the section will measure the 
normal stress and shear respectively, and the senses of the 
components will decide whether the former is compression or 
tension and the latter positive or negative. Taking the inter- 
section of the force with, the section plane as its point of ap- 
plication, the moment of the normal component and of the 
parallel component about the center of gravity will measure 
the flexure and torsion respectively, and the signs of the mo- 
ments will distinguish between positive and negative values of 
the stresses. This process is simply Cases i and 2a combined. 

A process statically identical with the preceding is fre- 
quently more convenient in practice. In this process the com- 
ponents of the individual forces normal and parallel to the 
section are summed to get the normal stress and shear respec- 
tively, and the moments of the individual forces about the 
center of gravity of the section and about an axis through the 
center of gravity normal to the section are summed to get the 
flexure and torsion respectively. 

In non-framed structures, stresses have to be worked out 
at only a relatively small number of critical sections. It is 
sufficient at other sections to be sure that the stresses do not 



exceed certain amounts. Stresses existing at all sections can 
easily be shown, if required, as ordinates of properly con- 
structed curves. (Cf. Exercise i8.) 

Exercise i8. A horizontal bar is acted on by (ao 240* 5, o), (30 300* 
8, o), (40 90*" 20,0), and by a force at (o, o) whose line of action is inclined 
30"* to the horizon, and one at (12, o) unknown both in magnitude and 
direction. Establish equilibrium and determine all the stresses at a 
section through <io, o). Solve graphically only. Show clearly on the 
drawing how results were found and where they were scaled. 

69. Shear Diagrams. — ^A diagram showing the value of 
the shear at all points of a vertically loaded beam can easily 
be constructed by selecting a horizontal base line and drawing 
parallels across the intervals between the adjacent forces at dis- 
tances above or below the base line proportional to the magni- 
tude of the sum of the forces on either side of the interval. 
Resultants of forces at the left of any segment indicate positive 
or negative shears and are set off upward or downward irom 
the base line according as they are upward or downward. 

Thus, in Fig. 25, the values of the shears in proceeding 
from left to right are in the four intervals respectively, +51 .yy^ 
—48.23, —18.23, and +31.77. The shear diagram is con- 
sequently as shown in the lower shaded diagram. 

The four numerical values just given are evidently the values 
of the resultants AB, ACy AD^ and AE^ and could have been 
projected across into their respective intervals from the magni- 
tude diagram. 

70. Flexure Diagrams. — Since the measure of flexure at 
any section is the sum of the moments about the section of all 
the forces on one side of the section, all that is necessary for 
obtaining a diagram showing the flexure at all sections of a beam 
under a set of parallel forces is to letter the forces in the order 
of their occurrence along the beam and draw their closed string 
polygon (§ 52). The intercept by this polygon from a line 
parallel to the forces is proportional to the flexure in the sec- 



tion of the beam traversed by that line. The intercept needs 
only to be multiplied by the common If (§ Si)of the forces to 
give the numerical value of that flexure. 

An example of a flexure diagram for a beam subject to a 
series of vertical forces is shown in the upper of the two shaded 
diagrams in Fig. 25. To find the flexure at any section all 

Fig. 25. 

that is needed is to drop a vertical line from that section, meas- 
ure the portion of this vertical within the diagram and multiply 
the resulting length by 30 lbs., which was selected as a con- 
venient value for If in constructing the diagram. 


Thus the intercept directly below. the load dc scales 5.2 
ft. The flexure in the section directly under 6c is therefore 
5.2 X 30= 156 ft. -lbs. Algebraically the value is found to 
be more precisely 51.77x3= I55-3I ft- lbs. Values can be 
obtained in like manner for any other section. It should be 
observed that the percentage accuracy of the graphical result 
is no greater than the accuracy of the scaling of the intercept. 
With small intercepts therefore the percentage of inaccuracy in 
the graphical result may be considerable. 

Intercepts above ao indicate positive flexure and those 
below ao negative flexure. 

Observe that the diagram shows at a glance that the flex- 
ure in the beam increases steadily to a positive maximum under 
6c, then decreases to zero at a point a little nearer to cd than 
to 6c and continues to increase negatively, though more slowly, 
after passing cd to a negative maximum under de, when it neg- 
atively decreases rapidly again to zero under ea. The nega- 
tive maximum is found to be numerically slightly larger than 
the positive maximum, and the dangerous section of the beam^ 
so far as flexure is concerned, is under de. 

Observe that the sections where the flexure reaches its 
greatest values, whether positive or negative, are those in 
which the shear passes through zero — a phenomenon of in- 
evitable occurrence, as will be shown in the next section. 

Exercise 19. A horizontal beam of 30 ft. span, supported at each end, 
carries loads of 300, 600, 1800, and 1200 lbs. at points 6, 10, 18, and 25 ft. 
respectively from the left end. Neglecting the weight of the beam itself, 
determine by both methods the numerical value of the flexure at sec- 
tions 8, 12, 18, and 30 ft. from the left end. Record results side by side 
for comparison. 

Suggestions. Take the scale of lengths as great as i in. = 4 ft. Take 
some convenient round number for the magnitude of H, 

71. Connection Between Shear and Change in Flexure. — 

It will be useful to see if there is a simple relation between the 


flexure at the end of an interval and the shear and flexure at 
the beginning of the interval. 

Accordingly, let ACy Fig. 26, 

**« be any segment of a beam, C^^ and 

T ^> -: ^ ^«9 ^2' 2Lnd M^ and J/g the ends of the 

k - ^r^-lVjj^?*^ interval and values of the flexures 

Fig. 26. existing there respectively, P^ the 

resultant of all the forces at the left 

of Cj, P^ of those in the interval C^C^. Dimensions as shown. 


and M^ = P^a^. 

Subtracting the latter from the former, 

M,-M, = P, (a, -\-a^^ P^^ 

But P^ is the measure of the shear at the beginning of the 
interval and {a^ + a,) is the length of the interval. Calling 
the former V^ the latter « , giving P^a^ the more convenient 
characterization + M^, the last equation takes the form 
M.^M^JrVa + M,. 

That is, the flexure at the end of an interval is measured 
by the algebraic sum of the flexure at the beginning of the 
interval, the product of the shear at that section by the 
length of the interval, and the sum of the moments of the 
forces in the interval about the center of gravity of the end 
of the interval,* 

Corollary. If dr be of infinitesimal length, dxy M^ will van- 
ish, Jfj — M^ may be written dM^ and the equation takes the 

* The data of Fig. 26 are of the simple sort usual in the practical examples of 
this problem. The reader should satisfy himself that the same conclusion would 
have been reached whatever the directions of Px and P,, or if one or both of them 
had been couples. 


wnence it appears that the shear at any section is the ;r-deriva- 
tive ot the flexure at that section, and it follows that where 
the shear passes through zero the flexure is at a maximum 
or minimum. 

Exercise 2q. Shears and Flexures from fixed loads. Draw diagrams 
of shears and flexures for each of the seven following cases.* 

(a) Cantilever with IV concentrated at the outer end ; (3) cantilever 
with fV uniformly distributed ; (c) simple beam with IV concentrated at 
the middle ; (d) simple beam with IV uniformly distributed ; (e) simple 
beam' under several concentrated loads; (/) beam overhanging one 
support, concentrated load at the end of the overhang, two others be- 
tween the supports ; (g) beam with equal loads at each end, supports at 
equal distances from the ends. 

Compute algebraically and compare the maximum shears and flex- 
ures in cases {a), (d), (c), {{i). 

Suggestions. Note that in (a), (c\ and (^) the diagram of shear can be 
transferred directly from the magnitude diagram, — strictly also in (d) 
and (d). 

In dealing with (d) and (d) construct approximate diagrams by divid- 
ing the load into short portions, and treat as a set of equivalent loads 
concentrated at the centers of gravity of these portions. Points in the 
true diagrams will lie at the intersections of the extended vertical 
boundaries of the portions with the approximate diagrams thus found 
and in the case of shear will be the straight line through these points; 
in the case of flexure will be the parabola inscribed in the approximate 
string polygon and tangent to it at these points. 

* Here the term cantilever is used to designate a beam supported at one end 
only (by being built rigidly into a wall, for example) or in any way overhanging a 
support ; a simple beam is understood to mean one resting upon supports at each 
end without constraint 


72. Stresses in Framed Structures. — In a framed structure 
the resultant of all the loads on one side of a section can pass 
the section, so as to hold the other segment in equilibrium, 
only in the form of simple pushes and pulls which must act 
along the center lines of the bars cut. These bars should be 
imagined to be replaced by forces acting along their center 
lines, which are a set of forces into which the resultant can be 
resolved — a problem which is determinate if the number of 
forces does not exceed two or three. This resolution accom- 
plished (Case 2 or Case 4), the compression or tension in 
each bar is known and the requisite stresses determined. 

Here as with non-framed structures it is usually unneces- 
sary actually to evaluate the resultant of the external forces 
on one side of the section. The individual external forces are 
treated as the given forces in Case 2 or Case 4, and the un- 
knowns are the two or three components of the resultant which 
measure the required stresses. 

This method could be applied repeatedly until every bar 
in the structure had been cut and the stress in it determined, 
but when the stresses in a large number or in all of the bars in 
the frame are to be found, a less laborious method suggests 
itself, when it is observed that each and every joint is in equi- 
librium under a set of concurrent forces. Selecting a joint if 
possible where only two bars concur with one or more external 
forces, the stress in these two bars can be found by Case 2a. 



The joint at the other end of one of these bars will then com- 
monly be found to be subject to forces, only two of which are 
now unknown — one of those previously unknown being the 
one just established, for the force which a bar exerts upon a 
joint at one of its ends must be the equal and opposite of that 
exerted upon the joint at the other end. Proceeding thus, 
applying Case 2a to joint after joint till all have been treated, 
the stresses in all the bars are known. 

If there are no joints where so few as two bars concur, the 
problem is of course statically indeterminate unless the frame 
is so composed as to permit the determination of one of three 
concurring bars by using method of sections, § 73. (Cf. 
Exercise 25.) 

Sometimes a similar step has to be taken to deal with joints 
in the interior frames of where three forces concur, all incapable 
of determination by the ordinary method of Case 2a. (Cf. 
Exercise 24.) 

73. Method of Sections. — Method of Sections * is the name 
applied to the method of determining stresses in a framework 
by dividing the structure into two segments by means of a sec- 
tion plane, treating the bars cut as mere lines of action of 
forces external to the segment, and finding magnitudes accord- 
ingly. The problem presented is Case 4, when three bars are 
cut by the section, or Case 2, in the rarer case, when only two 
are cut. The calculations can be made for whichever segment 
the work will be easier, or, if a check be important, for both 

* This method solved algebraically is sometimes more explicitly called Ritter's 
Method of Sections, after Professor August Ritter, of Aix-la.Chapelle, who used it 
freely in his DacA- u. Braeken-Constructumen, Solved graphically the process 
is also called Culmann's Method of Sections, after Professor Culmann, of Zurich. 

It must be observed that the methods used in this book for finding stresses are 
really all methods of sections, and the limitations imposed in this section must be 
seen to be arbitrary or conventional, with a view to the establishment of a con- 
venient technical term. 



Wherever we can divide a structure without cutting more 
than three bars, if these three bars do not meet in a point, the 
stresses in the bars can be determined. 

If stress in a single bar only is desired, one simply selects 
a convenient section plane which will cut this bar and only two 
other bars. 

The algebraic method of solving Case 4 will usually be 
found more convenient than the graphic for these problems. 

Lever-arms, however, are sometimes so troublesome to 
calculate that scaling them from a carefully made large scale 
drawing is the best way to get them. Writing moments in 
form P {y cos at — jr sin a) is always an available resource. 

The reasoning underlying this method is still further ex- 
plained by Fig. 27. Fig. 27^1 represents any frame in equililj- 

FiG. 27. 

rium under the external forces P, Q, R, 5, and T. It is re- 
quired to find the stress in V. Intersecting the frame by a 
section cutting V and two other bars, U and W^ we have as a 


result what can be regarded as two rigid bodies shown shaded 
in Figs. 2y6 and 2yc in equilibrium under the forces (external 
to them) r,P, 0. C/, F, fTand R,S, C/, V, W, in either of which 
Kcan be established by an equation of moments with J/^, 
the intersection of U and W^ as a center. If stresses in U 
and PFbe also required, they can be found by taking centers at 
AI^ and M „ respectively. 

Observe that this solution is entirely independent of the 
number or inclination of bars in the frame other than those cut, 
provided P, Q, i?, 5, and T remain unchanged in all their 

Plate V may be looked upon as giving a complete graphic 
as well as algebraic solution by this method, if the shaded 
body be considered the segment of a frame, and the three 
forces, jP, (2» and i?, the forces acting through the cut bars. 

.74. Method for Determining All the Stresses In a Frame 
Under a Given Load. — To determine all the stresses in a non- 
redundant frame, since it is made up of a series of sets of con- 
current forces, we need only to 

(i) See that the external forces are in equilibrium, and 

(2) Work out case 2 a for each joint. 

In the former of these two processes the algebraic method 
is generally to be preferred. It usually is less laborious, and 
the superior precision of its results is welcome, as the accuracy 
of all the succeeding work depends upon them. 

In the latter process, however, the graphic method is 
greatly to be preferred, as saving much troublesome labor with 
a minimum risk of serious error with a degree of accuracy 
amply sufficient for the needs of engineering design. This 
work consists simply of beginning with a joint where two bars 
concur with one or more external forces, and working out a 
closed magnitude polygon for the set of concurrent forces so 
composed. Two of the sides of this polygon will determine 


the stresses m the two bars — not only in magnitude, but also 
in nature, the former being shown by the lengths of the sides, 
the latter by the arrows upon them, for these arrows must 
follow one another around the polygon, and when transferred 
to the bars to which they belong they indicate compression or 
tension according as they point toward or away from the joint. 
Proceeding to the next joint, which, with the force acting 
through the first two bars known, will in turn have only two 
unknowns concurring upon it, a second polygon is constructed 
and so on through the whole series of joints. Thus would 
result a series of magnitude polygons, one for each joint. 

75. Example. — An example is worked out in Fig. 28. 
The frame there shown. Fig. 2%a, is known to be in equilib- 
rium under the five external forces, 20, 200, 30, 140, and 
no lbs., through the first three having been given at the 
outset and the last two having been determined (Case 2b) 
from the dimensions of the frame and the positions of the sup- 

It is most convenient to letter the external forces in the 
order of their occurrence around the outside of the figure and 
have the letter common to two adjacent forces apply also to 
the bar or bars in the periphery of the frame between their 
points of application. Letters added inside each triangle of 
the frame complete the lettering of the forces. 

The work can begin at either of the two end joints. Tak- 
ing the left end joint, we lay off EAB as the beginning of its 
magnitude polygon ; then a line from B parallel to bf and one 
from E parallel to ^^ would locate -Fas in Fig. 28^. The 
polygon is in the order of the arrows, EABFE, and the stresses 
in ^/ and ^ are consequently 127 lbs. compression and in 
lbs. tension respectively. Since ^ and ^are now known, the 
polygon for the lower of the two joints next on the right can * 
be completed by drawing from F and E parallels respectively 



to^ and e^^ locating G, and closing the polygon in which the 
arrows are in the order EFGE, showing the stresses in^ and 
^^ to be 173 lbs. compression and 130 lbs. tension respectively. 




Fig. 28. 

Only two forces are now unknown at the joint where dc is 
applied, and. this polygon can be closed. The bar ^ acted 
downward upon the preceding joint; it must therefore act up- 
ward at this joint. GF and FB of the polygon are already 


drawn, adding the known BC^ drawing CH and GH parallel 
respectively to ch and gh, H is located and the polygon is 
closed and consists, following the arrows, of BCHGFBy show- 
ing stresses in ch 2JiA gh to be 132 lbs. compression and no 
lbs. tension respectively. 

For the joint at the lower end of ^A only eh is unknown, 
and to close this polygon we have only to connect E and H, 
The arrows are in the order EGHE^ and M is 133 lbs. tension. 

For the joint at the extreme right two of the forces, ch 
and ehy have been determined and cd and de were known at 
the outset. It is important as a check to see if they are in 
equilibrium. Laying off CD downward and DE upward, equal 
to 30 and 1 10 lbs. respectively, E is found to iall where it was 
before located, and the work checks with CDEHC^ read with 
the arrows as the last magnitude polygon. 

Examining Fig. 28^, it appears that it consists not only of 
a closed magnitude polygon for each joint, bpt also a complete 
magnitude polygon, ABCDEA^ for the external forces. In 
fact, the ordinary method of determining all the stresses in a 
given frame under given fixed loads is, after equilibrium is 
established, to see that the external forces are lettered * in the 
order of their occurrence in passing around the outside of the 
frame, and complete their magnitude polygon before construct- 
ing any of the minor ones for the internal forces at the joints. 
Then adding two new lines to the diagram will establish the 
first joint, two lines more the next, and so on, each two lines 
completing a new magnitude polygon until the last joint is 
reached, where one bar only remains to be determined, and let- 
ters are already in the diagram which only need to be con- 

* It should be pointed out that a letter which is common to two adjacent forces 
may be written once midway between them just as well as closely adjoining each, 
and is so written in practice. 


nected to show this last stress. If the line so drawn is parallel 
to its corresponding bar the work checks, and the results can 
be scaled off as required. If the work does not so check, it 
shows that there is an error somewhere which must be found, 
and corrected before the result for any bar can be looked upon 
as trustworthy. 

76. Stress Diagrams, — A diagram such as just described, 

(i). A closed magnitude polygon for the external forces, 

(2). A closed magnitude polygon for each joint of the 
frame, is called a stress diagram — sometimes also a Maxwell, 
or Cremona diagram. 

In constructing stress diagrams as large a scale as is con- 
venient should be used throughout, especially in the diagram 
of the frame, where otherwise short lines may give rise to in- 
accuracies when long lines have to be drawn parallel to them. 

The equivalent of very large scale for frame diagram, with- 
out some of the worst disadvantages of such large scale, can 
be produced by drawing in a group long parallels to the bars 
through calculated points. These can then be used with a frame 
diagram of moderate scale or even a mere sketch for a guide. 

If a number of stress diagrams are to be drawn from one 
frame diagram, natures of stresses from different loadings may 
well be recorded on small freehand diagrams accompanying 
each stress diagram. 

To avoid confusion, the lines of action of the external forces 
should be shown entirely outside the periphery frame as in Fig. 

Each line of the stress diagram, except those representing 
the external forces, represents the two equal and opposite 
forces which are in action at each end of the bar bearing the 
same letters as the line. 


An error outside of common blunders in the use of draught- 
ing instruments, which is frequently the cause of failure of stress 
diagrams to close, is an incorrect determination of the reactions 
which gives rise to a set of external forces amounting to a 
couple instead of being in equilibrium, and such an error is not 
exposed by the magnitude polygon for these forces. The 
reactions should therefore be figured out independently, i.e., 
the correctness of one should not be allowed to depend upon 
the correctness of the other. 

77. General Instructions Regarding Exercises Involying 
Stress Diagrams. — Determine reactions, whenever possible, 
by inspection, otherwise algebraically. 

After completion of the stress diagram see that the work 
checks before proceeding further. 

Show upon the diagram of truss, by means of algebraic 
signs, the nature of stress in each bar, using sign -|- for com- 
pression and — for tension. 

As a guide in selecting a place for magnitude polygon of 
the external forces with a view to preventing work from run- 
ning off the paper, it will be well to compute in advance one 
or two of the largest stresses, where the bars in which they 
will exist are easily discernible, as in Exercises 21-23. 

78. Special Instructions Regarding Exercises 21-23 

State for comparison the graphic, algebraic, and semi-algebraic 
results for the bar specified, having used for the algebraic work 
the method of sections, and understanding by semi-algebraic 
work a similar algebraic solution in which lever arms are scaled 
from the drawing instead of being computed. 

Note that checking semi-algebraically does not check the 
accuracy of the frame diagram. 

Exercise 21. Truss shown in Fig. 29. Span 64 ft. Rise one fourth 
of the span. Eight equal panels; 1000 lbs. vertical load at each of the 
joints o and 8, and 2000 lbs. at each of the joints 1-7, inclusive. Scale 



as large as i in. s= 8 ft. Check the results for the t>ar 2-3 algebraically 
and semi-algebraically. 

Exercise 22. Same truss and loads as in Exercise i6» but with addi- 
tional loads of 4000 lbs. at joints 11 and 13, and 2000 lbs. at joints 9 and 
15. Check the result for bar lo-ii algebraically and semi-algebraically. 

Exercise 23. Same conditions as in Exercise 17, except that all loads 
at right of the center are doubled and 2000 lbs. added at joint 14, 

Check the result for bar 3-12 algebraically and semi-algebraically. 


79* Complications in Connection with the Analysis of 
Frames. — ^The methods of the preceding chapter will be found 
easy of application to any simple frame under vertical loads. 
But many cases are unavoidable in practice where the con- 
ditions are not so simple, and a number of the most important 
and illustrative of them will be taken up in this chapter. The 
sources of difficulties are various. Those treated in the follow- 
ing pages fall into four classes, more than one of which may, 
of course, be found exemplified simultaneously in connection 
with one structure. These classes of the sources of difficulty 

1. Indeterminateness, apparent or real, as to the reactions. 

2. Indeterminateness in the analysis of a frame, apparent 
but not real,— due to special systems of grouping bars. 

3 . Apparent but not real redundancy in framing. 

4. Structure being partly framed, partly non-framed. 

The really indeterminate cases here taken up are such only 
as can be brought within the scope of statical methods by mak- 
ing assumptions of such reasonableness and of such general 
acceptability in practice that the fact that the cases are statically 
indeterminate is usually ignored. 

Structures generally recognized as redundant do not fall 
within the scope of this book. 



80. Reactions Due to Non-Vertical Forces. — Indeter- 
minateness as to the reactions occurs in the case of structures 
subject to sidewise or any non-vertical forces, such as roof- 
trusses, towers, etc., exposed to the wind. These reactions 
are, in general, two forces of which only the points of applica- 
tion are known, hence involving four elements unknown (two 
slopes and two magnitudes), and the problem of finding them 
is indeterminate. Results satisfactory in practice can, how- 
ever, be obtained in any one of these three ways : 

(i) By supporting one end upon rollers, assuming them fric- 
tionless, thus making one of the slopes known. 

Rollers are frequently present anyway in large roofs to pro- 
vide for expansion and contraction, and such roofs at once 
come under this method. 

(2) By assuming that each wall resists half the horizontal 
component of forces, thus indirectly assuming two slopes. 

(3) By assuming that both reactions will be parallel to the 
resultant of all non-vertical loads. 

In (i) we will call to our aid a mechanical contrivance and 
make an assumption, and in (2) and (3) we make assumptions 

Each of these ways, experience has justified as of sufficient 

This kind of indeterminateness arises whenever the resultant 
of the loads has a component parallel to the supporting surface 
or, if the supports are in different planes, to one or both of the 
supporting surfaces. For example, in the case of a door or 
gate hinged to a vertical jamb, the door may be in equilibrium 
however its weight be divided between the hinges. In such a 
case, of course, if analysis were necessary, the assumption 
would be that either hinge may have to furnish the whole ver- 
tical support, or else assume it all on one or the other of the 


hinges, and by setting this hinge a little high, make the 
assumption a certainty. 

8i. The Fink Truss. — The form of roof-truss shown in 
Fig- 30f and known in this country as the Fink truss, has some 

Fig. 30. 

practical merits which cause it to be widely used. Though 
statically determinate it cannot be analyzed by the stress- 
diagram method without some special manipulation. On reach- 
ing either of the joints M or iV, the three bars P, Q^ and /?, 
or R, 8y and T^ respectively are unknown, and the stress- 
diagram appears to be blocked. If any one of the five forces 
P, <2> R^ ^9 or T can be evaluated the difficulty will be over- 
come. This suggests the method of sections, and it appears 
that a section can be taken through T either side of the ridge, 
which will cut only two other bars. T is determined accord- 

In general in dealing with Fink trusses, the value of this 
stress might well be looked upon as something always to be 
computed in advance, as the reactions are, and to be laid off 
in place as soon as the magnitude diagram for the external 
forces is completed. Cf. Exercise 24. 

Exercise 24. A Fink truss, which in this case will be made up of hor- 
izontal bars and bars inclined 30 degrees and 60 degrees to the horizon, 
as shown. Eight equal divisions in the upper chord. Load at each 
upper chord joint 2000 lbs. Construct the stress diagram. 

82. Triangular Frame with Trussed Top Chord.— A 

fr?ime somewhat similar to the Fink truss is that shown in 



Fig. 31. It is a construction developed from the ordinary 
triangular frame by a simple system of 
trussing applied to the rafters. The re- 
sult is, however, that there is no joint at 
which a stress-diagram can be started 
immediately. The way out of the diffi- 
culty is, as in the case of the Fink truss, 
the previous determination of the stress ^iq, 31. 

in the horizontal tie by the method of sections. 

The Fink truss may, in fact, be regarded as a special case 
of this construction, in which the trussing of the top chord is 
more elaborate, and in which the horizontal tie and the lower 
chord of that trussing are coincident to some extent. Cf. 
Exercise 25. 

Exercise 25. A triangular frame of 24 ft. span and 20 ft. rise is made 
with its top chord trussed as shown (Fig. 32), by means of struts, nor- 
mal to them, 3 ft. 6 in. in length. The frame is 
supported by a hinge at the left end of the hori- 
zontal side and a set of rollers at the right end 
of the same bar. Loads Pu A, Pt, P*, and Pt 
are applied at the joints and with the directions 
shown. Taking these forces at 2Gbo, 10,600, 
4066, 3600, and 4000 lbs. respectively construct 
the stress diagram. 

Suggestion. Here we can treat all external 
forces as acting outside the frame, by dotting 
Fio. 32. such lines of action as may be needed and treat- 

ing such parts of these lines as lie inside the frame periphery as if they 
were actual bars of the frame. 

83, Counters. — Frames if made of a simple triangulation 
may be called upon to resist, in some of their members, both 
tension and compression according to variations in loading con- 
stantly occurring. Such reversal of stresses is to be avoided 
in the interests of economy and simplicity of design. If a 
diagonal in a given quadrilateral is replaced by its mate, if the 
loads remain unchanged, the stresses produced in the two 


diagonals will be opposite in character. Hence, if in a quad- 
rilateral where a diagonal might be called upon to resist com- 
pression, a mate to it, if present, would resist the stress in 
tension, and no compression could exist in either diagonal. A 
diagonal added for such a purpose is called a counter. 

Counters in bridge trusses are idle when the truss carries a 
full symmetrical load. They are in action only under certain 
partial or unsymmetrical loads. Reversal of stress may be 
unpreventable in some members of certain kinds of trusses. 

The greater the permanent load compared with the live, 
the less the likelihood of reversals, and the less there is for 
counters to do. 

The presence of counters gives the frame an appearance of 
redundancy, but if the counters are of such a character or 
secured at the joints in such a way that they are incapable of 
resisting both tension and compression, the bars which must 
inevitably be out of action under the given loads can simply be 
Ignored, and the analysis can proceed as usual. 

Cf. Exercise 26 and Fig. 33. 

Exercise 26. A four-sided framed tower ^o ft. 
high and 14 ft. wide at the base, and whose other 
dimensions are as shown (Fig. 33). is made with a 
double set of slender diagonals throughout. In a 
gale each half of it is supposed to be subject to the 
forces shown. Assuming half the horizontal thrust 
J taken up at each column base, construct the stress 
~£~ diagram. 
-r- — — j-^^ Suggestion. In drawing the diagram of the frame 
t*"^* 1 dot the set of diagonals which are out of action. 
Fig. ii. Cf. § 83. 

84. Bent of a Mill Building. — A common and important 
type of structure subject to loads elsewhere than at the joints 
and hence exemplifying the partly framed, partly non-framed 
type of structure is the combination of columns with a truss 
constituting the bents of a mill-building and shown in Fig. 



34. They are usually of steel throughout. The forces which 
give rise to their peculiarities are sidewise forces due to wind, 
tension of belts, shocks from traveling cranes, etc. The 
general relation between such structures and true frames is 
explained in §64 {q-v,)y but it will be worth while to amplify 
that explanation by taking as an example the bent of Fig. 34 

Fig. 34. 
and discussing in detail the peculiar steps in its statical analysis. 

The steps peculiar to the statical analysis of a mill-building 
bent are : 

{a) The assumption of the points of application of the foun- 
dation reactions, R^ and R^. 

{b) Tlie ^xi^\ys\s (by inspection) of the bent into its con- 
stituent true frame and ilexural members. 

{c) The determination of the forces brought to bear upon 
the joints of the frame by the flexural members. 

The step {a) is one made necessary by the practical con- 
sideration that the columns are members of considerable 
breadth in the plane of the bent even at their bases. More- 
over, they have to be anchored to the foundations so as to resist 
horizontal displacement as well as an actual lift. The result 
is an anchorage which offers considerable resistance to the rota- 


tion of the column about its base. This resistance is of course a 
couple in the plane of the bent which combined with the requi- 
site horizontal and vertical resistances amounts to the single 
forces R^ and R^ intersecting the axes of the columns at points 
at some distance y above their bases. The evaluation ofy^ 
which is dependent upon the evaluation of the couples, is a 
statically indeterminate problem. Consideration of the elas- 
ticity of the material* would lead to the assumption that^ is 
half the distance between the column bases and the attach- 
ment of the knee-braces — ^the name given to the lowest of the 
bars secured directly to the column. This would minimize the 
stress in the columns, but would necessitate careful attention 
to the design and execution of the anchorages and foundations. 
In small structures in which such attention to the anchorage of 
the columns is not considered worth while, the existence of the 
couple may be ignored outright, and the structure treated as if 
y were zero and the bases of the columns hinged to the foun- 
dations. Then the couple actually materializing in the life of 
the structure is simply so much addition to the factor of safety. 
Cases might arise in which the designer would feel justified in 
assuming other values of ^ and proportioning the anchorages 
and other parts of the structure accordingly. 

The points of application once decided upon, the distribution 
of the horizontal components between the two reactions (§80) 
would be made, the vertical components calculated (Case 2b) 
accordingly, and R^ and R^ established. 

Step {b) is not difficult, for structures of the kind in question 
are simple triangulations with one or more bars subject to loads 
between joints, or extended beyond joints so as to receive loads 
on the extended part, usually at its end. The simple triangu- 
lation constitutes the true frame, and the bars just described 

*Cf. Johnson, Bryan, and Tumeaure's Modem Framed Structures y Art. 151. 


are the flexural members, — ^the bars spoken of in §64 as doing 
double duty. 

Step {c) consists of finding the reactions from the frame 
upon the flexural members necessary to hold them in place in 
opposition to the transverse loads to which they are subject. 
Reversing these reactions in sense, they become the forces 
transmitted to the joints of the frame by the flexural members. 

The analysis of the frame then goes on as usual, establish- 
ing the tension and compression in all the members, including 
such as are the parts of the flexural members common to the 
frame. The flexural members have also shears and flexures, 
arising from their beam-action, to be determined, and when 
this is done the statical analysis of the structure is complete. 

The whole process is one proceeding on a number of 
uncertainties, but there is no reason why it may not surely be 
kept on the side of safety and that, too, without serious lack of 

The bent of Fig. 34 is accordingly worked out as follows. 

Assuming the columns hinged at their bases (step a), and 
the horizontal components H^ and H^ of the foundation re- 
actions to be equal (§80), the vertical components, F and 
Fj, of these reactions follow at once by Case 2b, 

In Fig. 35 are shown (step b) the true frame, including the 
whole of the triangulation of bars in the bent, and, separate 
from it, the four flexural members, — the two top-chord bars 
subject to transverse loads, and the two columns from eaves to 

The two top-chord bars will naturally be regarded (step c) 
as two centrally loaded beams requiring reactions of ^fF from 
the frame joints at their ends and hence transmitting that 
amount to those joints. The two columns differ from the top- 
chord bars in the immaterial particular that their supporting 
joints are both on the same side of all the loads. V^ and V^ 



being regarded as transmitted directly up the column to the 
nearest joint of the true frame, the loads for the windward 
column are P and H^, and for the leeward column Hy The 

.8, 8 

Fio. S5. 

reactions 5p 5,, S,, and S^ are determined by Case 2b. Re- 
versing them in sense, they furnish the only remaining external 
forces in action on the true frame. The frame can now be 
analyzed as desired by familiar methods. Cf. Exercise 27. 


Fig. 3& 

Exercise 27. A bent of a mill building is framed and subject to loads 
as shown in FiG. 36. The truss proper consists of four panels of equal 
length. The columns are to be assumed hinged at the base and the 
horizontal thrust divided equally between them. (Cf. §§ 80, 84.) 



Construct the stress diagram and, preferably working with original 
frame as above shown, check algebraically the stresses in Q and H re- 
spectively, and record them side by side with the corresponding graphi- 
cal results for comparison. 

Suggestions. Determine the ITs and Vs algebraically. The deter- 
mination of J^ will require the previous determination of one other bar. 
Show the bent completely analyzed into its constituent parts, giving the 
numerical magnitudes of all the external forces. (Cf. Fig. 35.) 

The 200o-lb. force may be assumed to be normal to the top chord and 
applied midway between the two nearest joints. The knee-braces of this 
bent are seen to be horizontal. 

85. Cantileyer Bridge. — ^A cantilever bridge consists of 
one or more trusses or girders supported at one or both ends 
by ends of other trusses or girders which overhang their sup- 
ports. The overhanging part is the cantilever whose promi- 
nence in this style of bridge gives it its name. 

A cantilever bridge always requires more than two points 
of support, and the loads and reactions constituting usually a 
set of parallel forces, the reactions would seem at first glance 
to be indeterminate. Noticing, however, that the structure is 
composed of at least two or three structures, the reactions 
prove to be determinate. Cf. Exercise 28. 

Exercise 28. A cantilever bridge proportioned and loaded as shown 
is supported by vertical reactions at A, B, C, and /?. Determine these 
reactions graphically only. 

Suggestions. Observe that the hinges at E and F divide the bridge 

Fig. 37. 

into three separate bodies, one resting upon two others, that none of 
these three bodies is in a statically indeterminate condition, and that 
therefore the whole structure is determinate. 

Letter as usual around the figure and construct the string polygon 
for the given forces. One of the three strings then missing can be 


located from the knowledge that the flexure at E and F must be zero 
(§ 70), and the other two follow at once. 

86. Three-hinged Arch. — The structure typified in Fig. 38, 
consisting of two ribs (fiamed or non-framed, straight or 
curved, but usually curved) which are hinged together at their 
upper ends C and rest on hinges at their lower ends A and 


Fig. 38. 
B, is called a three-hinged arch. It is used for roofs where the 
greatest spans are required, and for bridges as well. 

It is a structure which evidently cannot stand up even under 
vertical loads without horizontal resistances at the supports. 
In this respect it is like an arch, and it easily lends itself to the 
pleasing curved form of an arch if desired. Moreover, it is the 
only statically determinate method of arch construction, and an 
important and fruitful subject of study accordingly. 

There appears to be an indetermination in connection with 
the reactions at A and B. Here two forces are required of 
which only the points of application are given. Two magni- 
tudes and two slopes must be determined, or else, if each 
reaction be replaced by two convenient components, four 

Now, the hinge at C (assumed like the other hinges to be 
frictionless) cannot resist any force which does not pass through 
it, and this fact in connection with the three general conditions 
of equilibrium furnishes the basis for the necessary four equa- 
tions and the problem is seen to be determinate. 

These four conditions would be satisfied if two forces were 


determined which, if applied at A and B respectively, would 
combine with all the forces between those points and C and 
yield two equal, opposite, and coincident resultants passing 
through C, These two forces would be the required reactions. 
Their intersection in the magnitude polygon would be the pole 
which (§54) would direct a string polygon for the given loads 
through the three given points A^ B^ and C in such a way 
that the strings through A and B would be the extreme strings 
of the whole given set, and the one through C the string com- 
mon to the resultants of the two groups on each side of C. 

Conversely, such a pole once determined, the two lines 
from it to the ends of the magnitude polygon for the given 
forces would give the magnitudes and directions required. 

The methods of § 54 will therefore furnish the graphic and 
algebraic methods for the determination of the reactions. 

Moreover, the string polygon through these three points for 
forces lettered in the order of their occurrence is the locus of 
the intersection (with any section) of the resultant of all the ex- 
ternal forces on either side of the section. What is more, 
each string is the resultant of all the external forces on either 
side of it and its magnitude and sense can be found from the 
corresponding ray. 

This string polygon once drawn, the stress in any bar under 
the most complicated system of loading can be determined by 
the method of sections (§ 73) with an equation of moments 
involving only four quantities, the known magnitude of the re- 
sultant, the required magnitude and their respective lever arms. 
These last can, of course, be scaled or calculated, but as a 
careful drawing is necessary in any event for the construction 
of the string polygon, scaling would commonly be far more 

An alternative solution of the three-hinged arch is some- 
times given which is based upon the fact that it may be 


regarded as two separate structures, and the groups of loads 
coming on each assumed to act separately. Thus (Fig. 38) if 
the portion ^C be unloaded while CB is loaded, the direction 
of the reaction at A must be through AC, and this reaction on 
CB and the one at B are determinate, falling under Case 3. 
Repeating the process for AC^ two more partial reactions are 
determined. The resultants of the two reactions at each 
hinge are the reactions required for the structure. 

It should be observed that the hinges A and B may be put 
at different levels without affecting the determinateness of the 
problem or its method of treatment. 

The horizontal components at A and B may be furnished 
by abutments as with any arch, or, in cases where circum- 
stances permit, by tie-rods connecting A and B, together 
with such anchorage as would be required with the given 
load to keep any truss from moving as a whole. With the 
tie-rod the structure becomes what might be regarded as a 
triangular truss, two of whose members are more or less curved 
and more or less frameworks themselves. 

87^ Line of Pressure. — The line of pressure in any struc- 
ture is the locus of the intersections with successive section 
planes throughout the structure of the resultants of all the 
external forces on either side of those sections. ^ 

In case the loads are non-continuous, the line of pressure 
will be a broken line; with continuous loads, it is a cu^e 
which must be plotted point by point at sections taken at short 

An example of the former case arises in connection witfi 
the three-hinged arch under a series of concentrated loads. 
The string polygon including the end reactions as extreme 
strings and passing through the three hinges is tlie line of pres- 
sure of that arch for the given loads, if the loads be lettered 
in the order of their occurreacc* 



An example of the latter case is a masonry dam exposed 
as it is to uninterrupted hydrostatic pressure. Points in the 
line of pressure are located in this case by taking successive 
horizontal sections and finding the resultants of the water 
pressure and weight of the masonry above that section. Where 
these resultants pierce the sections which define and limit their 
components are points in the line of pressure of the section of 
the dam. 

Exercise 29. A three-hinged arch of 240 ft. span is subject to loads 
as shown in Fig. 39. The top chord is divided into eight equal bays, and 

the bays of the bottom chord are each equal to AE in length, DE being 
horizontal. The joints of the bottom chords are on the arc of a circle 
of 145 ft. radius through A and B. Other dimensions as shown. 
Required : 
I. Reactions graphically by the method of § $4. 

Reactions algebraically (§ 54), showing all lever arms dimen- 
sioned except such as are given directly by the main dimen- 
Line of pressure. 
Stress diagram.* 
Note that the stress diagrams of frames ^C* and ^C* will 
each have its own check. 
Pressure on the hinge at C graphically. 

6. Stress in the bars marked Q, R, and S with the aid of the line 
of pressure and scaled arms. 

7. Stress in Q, R, and S from the stress diagram. 

8. Tabulated record of results of i, 2, 5, 6, and 7. 





Suggestions . Use a large scale for the frame diagram, say i in. = 30 
or 40 ft., and a moderate scale for the magnitude diagram. 
Is the line of pressure a closed string polygon ? 

Exercise 30. Suppose the coordinates of the three hinges. A, B, 
and C of a three-hinged arch to be expressed in feet as (o, o), (60, 30), 
and (30, 50) respectively, and the ribs to be of any curvature and either 
framed or non-framed. Suppose a vertical load of 20 tons acting 20 ft. 
horizontally from A, and a horizontal force of 10 tons acting towards 
the right and applied between B and C and 10 ft. vertically above B. 
Determine the reactions algebraically. 

Suggestion. The horizontal and vertical components of the two 
reactions would usually be a more convenient form for the expression 
of the required answer than the resultant reactions themselves. 

Use the method of § 54, and check the results carefully. 

88. Hammer-beam Truss.— Like air other roof trusses, 
the hammer-beam truss, Fig. 40, has vertical loads to resist 

Fig. 40. 

and also non-vertical ones. Unlike most trusses, some of its 
main members are curved. This curvature alone would not 
call for any special treatment here, for curved members may, 
for purposes of analysis, be considered replaced by straight 
ones connecting the same joints and the curvature left out of 
account till the design of the piece itself is taken up. The 
truss would then be a complete frame and would be treated 
like any other. 

The combined effect of two circumstances, however, ren- 


ders it necessary to approach the hammer-beam truss in a dif- 
ferent way. These two circumstances are, (i) curved bars in 
frames at best work under very unfavorable conditions, and (2) 
in these trusses they are made of wood, — a material specially 
weak and difficult to connect. Moreover, even at best the 
shape of this truss is in itself unfavorable to rigidity. The 
result is that even under perfectly vertical loads the hammer- 
beam truss must be expected to produce thrust upon each sup- 
porting wall, as an arch would do, or as rafters without any 
collar-beam (or with the collar-beam too close to the ridge) 
would do. That means that the walls must be prepared to 
resist the spreading at the base of the truss which the curved 
ties are unable to prevent. 

To estimate this thrust from vertical loads, a good way is 
to place no reliance whatever upon the upper curved members 
— to consider them absent. This reduces the truss to an in- 
complete frame — the parts below the joints 3 and 7 are mere 
framed inclined struts, reaching from their foothold on the 
wall to joints 3 and 7 of the truss 3, 4, 5, 6, 7, 12 which they 
support. The loads on this truss are known and hence the 
vertical load at the top of each strut. The vertical reaction at 
the base of each strut is, of course, equal and opposite, but as 
it is not coincident with it, the two form a couple, and the 
struts would fall into the building but for horizontal resistances 
at their upper ends along the bar 3, 12, 7. This horizontal 
resistance calls for an equal and opposite one at the base of each 
strut. These horizontal forces constitute the second couple 
which balances the first. The common magnitude of the hori- 
zontal forces may be taken as the thrust of the roof, and with 
the vertical reactions, which are easily found, there is no 
further obstacle to the construction of the stress diagram. 

Finding thrusts as just described and making no claim on 
the upper curved members, and not an unduly severe one on 


lower ones, leaves the former ready to act at their full value 
when the wind blows, and the latter in as favorable a condition 
as possible. 

In determining wind stresses, the only thing to be done 
(except as described below) is to treat the truss as a complete 
frame, as explained above. The results thus obtained will be 
combined with those from the vertical loads. 

The only difficulty, then, is to provide for the stresses. The 
stiffiiess of the rafters at 7 may have to be relied upon to assist 
the upper curved members, and the buttresses may have to rise 
to the level of i and 9 (as shown in the figure) in order to 
relieve the lower curved members. 

This last would raise real points of support of the truss to I 
and 9, and 9, 10 and o, i would become mere posts prevented 
from overturning by the buttresses, and 0,11 and 10, 13 mere 

Of course in a large roof it may be necessary to rely upon 
the stiffness of the rafters, and even then higher stresses than 
would ordinarily be regarded as satisfactory may have to be 
tolerated. It may be advisable even to neglect all curved 
members and rely entirely on high buttresses and the stiffiiess 
of the rafters. 

The hammer-Leam truss is a very imperfect structure as 
regards the economical application of material. The less the 
curvature of members the better from this point of view. 

Exercise 31. With a vertical load of iocx> lbs. per bay of the top 
chord of a hammer-beam truss such as is shown in Fig. 40, determine the 
thrust on the walls on the assumption that the upper pair of curved 
bars is wholly inoperative, and construct the stress diagram. All bars 
of the truss are to be taken as horizontal, vertical, or inclined at 45"* as 
shown, and o. i and 9, 10 each of the same length as i, 1 1 or 9, 13. 

Suggestion. Draw a separate diagram of one of the framed struts 
o, f, 2, 3, II, showing all the loads upon it, and determine the reactions 
upon it as an independent structure. The way is then dear for the 
stress diagram as usual. 


89. Stresses Due to Moving Loads.— Some structures, 
as bridges and viaducts, are subject to loads varying greatly 
in magnitude and taking widely varying positions on the 

The special steps involved in the determination of the 
stresses in such a structure are 

(i) The assumption of the maximum reasonable load or 
system of loads. 

(2) The determination of the locations of this load or sys- 
tem of loads which will produce the extremes of stress in each 
and every part of the structure. 

(3) The determination successively of the extremes of stress 
in each and every part of the structure due to loads assumed 
to be stationed at all the various points decided by the preced- 
ing step. 

The first of these steps is an exercise of judgment in the 
light of experience of what the structure may reasonably be 
expected to carry, leaning of course towards too high rather 
than too low an estimate. 

The second of the steps opens a large field in the study of 
special structures and systems of loading. In general it may 
be said to be a process based upon the observation of the stress 
in a given part of a structure due to a single force followed by 
a study of how and to what extent other forces may be grouped 
with it to reinforce its effect. The methods are a simple 
development of the general principles of statics as treated in 
Part I with the addition of various devices, largely graphical, 
for saving time and labor. 

The third step is a straightforward solution of the simple 
statical problem of finding the stress in a certain part of any 
structure subject to a given set of loads. 

The reader interested further in this subject will find it 
treated at length in Burr's Stresses in Bridge and Roof 


Trusses, Johnson, Bryan, and Turncaure's Modem Framed 
Structures^ Merriman and Jacoby's Roofs and Bridges^ 
Parts I and II, Hoskins' Graphic Statics, DuBois' Stresses 
in Framed Structures, etc. 

The suddenness of the application of the moving load or 
impact is not considered at all in this anal}^is. Allowance is 
made for that by certain empirical methods in connection with 
the proportioning of the parts to resist the extreme stresses de- 
termined as just described. 

90. Stability of a Masonry Dam. — The lines of pressure 
for the various conditions to which a masonry dam is exposed 
afTord a very important aid in the study of the stability and de- 
sign of such a structure. 

To construct such a line of pressure the cross-section of the 
dam is drawn to a large scale and divided into a suitable num- 
ber of strips by a series of horizontal planes. The line of 
pressure will be located when there have been found the inter- 
sections with each of these planes of the resultant of all the 
forces acting upon the dam above that plane. 

The gravity and hydrostatic forces on each strip are then 
determined (assuming the thickness of the strip normal to the 
paper to be one foot) and shown in their proper places. 

This being done the successive resultants can be located by 
Case I. The forces can sometimes be most conveniently let- 
tered by beginning with the top strip and lettering the gravity 
forces, in the order of occurrence of their strips, ab, be, etc., 
and lettering downward, giving the hydrostatic pressures in a 
similar order the letters ab', Vc* , etc. 

A general string polygon will locate the successive result- 
ants bV , cc\ etc., and their intersections with the lower limits 
of the successive portions of the profile to which they belong 
will be points of the line of pressure. 

An alternative lettering suitable to some cases would be to 


letter the gravity and hydrostatic forces on the strips respec- 
tively ab and be, cd diVid de, etc., beginning at the top. Tak- 
ing the pole coincident with A, the strings oc, oe, etc., would 
be the required resultants. 

Sometimes it may be more satisfactory to determine the 
locations of the resultants for the successive groups of gravity 
forces algebraically, and use the string polygon only for the 
hydrostatic forces. The intersections of the pairs of partial 
resultants will be points of the series of required resultants 
which can then be shown by transference from the magni- 
tude diagram. 

The location of the line of pressure in masonry arches 
differs from the preceding process only in the greater uncer- 
tainty about the elements of the external forces on the struc- 
ture and the lack of a determinate point for beginning the line 
of pressure within the structure. 

Exercise 32. A masonry dam with vertical upstream face has a cross- 
section determined by the following coordinates (origin at the top of 
the upstream face) ; (20,0), (20, — 17), (21, — 30), (31. — 50), (43, — 70), 
(581 - 90). (73» - "o). (o, - no). 

Required the lines of pressure {a) when the reservoir is empty, and 
(b) when the reservoir is full. 

Weight of masonry to be taken at 150 lbs. per ca. ft. and of water at 
62.5 lbs. 

Statical work to be done graphically only. 

91, Action and Reaction not ITecessarily Nonnal to the 
Surface of the Contact of the Bodies. — ^The function of 
statics is to deal with the equilibrium of bodies without regard 
to the origin of the forces acting on them. In order, however, 
to deal with questions of equilibrium it is necessary to be 
familiar with the various possible sources of forces and the 
conditions under which they are effective in order correctly to 
prepare a problem for the application of purely statical methods. 

In the exercises prescribed in the foregoing work, the 


sources of all the forces dealt witli might be classified broadly 
as either directly or indirectly the attraction of gravitation, 
which needs no further comment, and to an interaction be- 
tween masses most naturally thought of as normal to their 
surface of contact. Accordingly special attention will now 
briefly be given to the possibility and causes of interactions 
between masses inclined to the normal to their surface of 

92. Friction. — It is a property of all bodies that they 
offer more or less resistance to being moved over one an- 
other. This resistance is a force in the plane of contact 
always to be considered when one body rests upon another 
and the other forces acting on the body have an unbalanced 
component parallel to that plane. This kind of resistance is 
called friction. The resultant of the friction with the normal 
resistance is a force acting on the body inclined to the normal 
to their surface of contact. 

The elements of a force due to friction have the following 

The point of application may always be taken in the sur- 
face of contact. 

The direction is always opposite to any unbalanced com- 
ponent of the other forces parallel to the surface of contact. 

The magnitude, like that of any other passive resistance, is 
variable. It is as large as it has to be (up to a certain limit) to 
maintain equilibrium and no larger. In this respect, friction, 
or tangential resistance, is precisely like the normal resistance. 
The limit in the latter case is the ultimate compressive strength 
of the weaker of the t)vo bodies and, in the former, the limit 
is a certain percentage (called the coefficient of friction) of 
the normal pressure existing at the time in question. If the 
former be overstepped penetration or crushing will occur, if. 
the latter, sliding or rolling. Both limits are experimentally 


determined for various materials and conditions, and data re- 
garding them can be found on record in the standard reference 

The coefficient of friction is commonly designated by the 
letter ;i. Its values vary greatly according to the materials in 
contact, the smoothness of the surface of contact, the degree of 
lubrication, whether the motion is likely to be sliding or rolling. 

When the limit of friction is reached and the body is about 
to move, the reaction upon this body from the one under it is 
at a limiting inclination to the normal to the surface of con- 
tact called the angle of frictioii. The tangent of this limiting 
angle is evidently the ratio of the limiting frictional resistance 
to the normal pressure with which it is associated. That is, 
the coefficient of frictioii is the tangent of the angle of 

If greater resistance to motion on the surface of contact be 
required than friction can be trusted to furnish, recourse can be 
had to various devices, prominent among them the use of one 
or more bodies penetrating each of the two given ones, pass- 
ing through their surface of contact and furnishing the required 
resistance in that surface by virtue of their shearing strength. 
Nails, bolts, rivets, and dowels are examples of the bodies 
used for such purpose. 

What is of most importance with regard to friction from 
the point of view of statics is that it is a source of passive force 
tangential to a surface of contact between, two bodies just as 
normal resistance is a source of passive force normal to that 
surface. Of course a force due to friction is treated statically 
just like any other force. 

Example i. A body of weight W rests upon a rough 
plane inclined a to the horizon and is subject to a horizontal 
force P. The coefficient of friction being ;i, between what 
two limits may P vary while the body remains at rest i 



Solution. The forces acting on the body are the active 
forces P and W, and the passive ones iVand T=zfJiN, When 
Tacts downward as shown, Fig. 41, P is at its upper limiting^ 

value and the body is about to move 
up the plane. 

Statically the problem is one to 

determine two magnitudes in a set 

of concurrent forces whose lines of 

action are all given (Case 2d). The 

Fig. 41. unknowns are P and Ny T being 

known as soon as N is. The regular solution of Case 2a 

can now be applied. No numerical data being given the 

algebraic method will be used. 

Resolving the four forces along and at right angles to the 
plane for the ^, and Ay equations there results 

P cos a — )wiV — W^ sin or = o, 

P sin a— N^ Wqos a = o. 

Eliminating N^ it appears that 

^_ sin or + jn cos a ^^^ 
"" cos a — )[i sin a 

Reversing the direction of T the other limiting value of P is 
found by similar means to be 

^_ sin « — /i cos a ^ 
"" cos a -^ pi sin a 

The same result might have been reached by reversing the 
sign of pi in the other value of P. 

Example 2. A uniform ladder rests upon a horizontal floor 
and against a vertical wall. The floor and wall are of differ- 
ent materials and the coefficients of friction between them and 
the material of the ladder are respectively .30 and .20. What 
is the least inclination to the horizon at which the ladder can 
rest ? 



Solution. The ladder is to be in equilibrium under three 
forces : W its weight acting at its center 
of gravity, P and Q at its upper and 
lower ends respectively with directions 
inclined to the normals to the wall and 
floor by their respective angles of fric- 
tion, and in such a way as to give P an 
upward component and Q a component 
towards the wall. The part that pure 
statics plays in this problem is to furnish 
the condition that W must pass through 
the intersection of P and Q, To this 
end the middle of the ladder must be 
in the same vertical with the intersec- 
tion of P and Q. The inclination, 0, of the ladder when this 
condition is satisfied will be the limiting inclination required. 

The determination of this inclination is henceforth purely 
a geometric and trigonometric process. Thus can be written 
(Fig. 42), 

cos (0 + g,) sin (0 + e,) 

sin 6^ 

cos e^ 

. ,. , , ctn ^, - tan ^, 
whence can be established tan = -, and 6^ and 

6^ being given respectively as tan-^ 0.30 and tan~^ 0.20, it 
follows that 

= tan-^ 1.567 = 57** 27'. Ans. 

Additional problems involving friction may be found in 
abundance in such works as those mentioned at the end of 
Chapter VI. 


In this Appendix, as in Chapter V, the term Pure Statics is 
used to designate the treatment of problems in which the data 
consist exclusively of elements of forces, the desiderata in- 
clude only other elements of forces, and the conditions of equili- 
brium of rigid bodies form the only determining conditions. 

A somewhat different class of statical problems is some- 
times met, especially in text-books, in which the object is to 
determine the position in which a given body will rest under 
given forces and with given conditions of support. The de- 
siderata here are geometrical relations between bodies con- 
sistent with equilibrium, and not elements of forces. Such 
problems might not improperly be called geometrico-statical 
problems to distinguish them from purely statical problems. 
An illustration of such is to be found in Example 2 of §92. 
Their solution consists in using the conditions of equilibrium 
to identify the position of the body for which the forces acting 
would be interrelated in a manner consistent with equilibrium, 
and then in using sufficient geometric insight to derive a simple 
definition of that position, — the latter of the two steps often 
proving the more puzzling of the two. 

Moreover, as stated in §46, a large part and often the 
most difficult part of an ordinary statical problem in practice 
is the discernment of the data in proper shape for insertion into 
the purely routine processes of pure statics. 



Though pure statics must accordingly be understood to 
have a restricted meaning, nevertheless within its domain will 
be found nearly all of the statical problems of engineering. 
The mathematical extent of this domain offers an inviting field 
of inquiry with a view to finding out how many and what 
problems might arise, which would not be obviously beyond 
its purview by involving more unknowns than there are deter- 
mining conditions. 

As a simple matter of permutations and combinations an 
early limit is set to the number of such problems. It is merely 
a question of the number of ways in which groups of elements, 
each group containing as many elements as there are determin- 
ing equations, can be selected without repetition from the total 
number of elements which pertain to forces not exceeding 
in number the determining equations. It will be convenient 
to treat forces in two groups, {a) non-concurrent (excluding 
parallel) forces and {b) concurrent (including parallel) forces, 
in which the determining equations are three and two re- 
spectively, and the limiting number of combinations eighty- 
four and fifteen as deduced in §44. As implied in §44, many 
of these combinations are, after all, statical identities. For 
example, a magnitude, direction, and point of application can 
be selected from a group containing three each of these ele- 
ments in three different ways, but in each case the statical 
significance is the same, viz., that the magnitude, direction, 
and point of application of one of a group of forces are to be 
evaluated. Collecting the statical identities as separate cases, 
there are found for non-concurrent forces only twenty such 
cases, numbered in the following table I-XX, and for concur- 
rent only nine, numbered XXI-XXIX. 

It should be borne in mind that for non-concurrence, as 
here understood, it is necessary and sufficient that the resultant 
of all the known forces shall not include a point common to 


three unknowns nor be parallel to tliem, and for concurrence 
that this resultant shall intersect the point common to two 
unknowns or be parallel to them. It hardly need be added 
that with non-concurrent forces if three unknowns should have 
a point in common or be parallel, the problem would always 
be incapable of solution. 

The following table gives a list of all the twenty-nine 
cases, with the number of statical identities included in each 
case. Py a^ and m are used to designate magnitude, direc- 
tion, and point of application respectively of any of three 
forces, (2> ^i 2md 5, which may involve unknowns. The 
sense may be understood to be unknown whenever either the 
magnitude or the direction is unknown. 

The reader who has mastered the four cases of §45 will 
find the additional ones an interesting field for further practice 
with the same methods, adjusting them of course to the 
peculiarities of each case. The graphical method is recom- 
mended as especially convenient for this investigation. The 
problem in the first twenty cases is then uniformly to close a 
magnitude polygon and a string polygon, observing that in 
the last nine cases, insuring the concurrence of the forces may 
replace the closure of the string polygon. The number of 
forces involved need never exceed four, any number of given 
forces being considered to be represented by any single force 
as their resultant. Some of these new cases will be found in- 
capable of solution, or capable of one or more real solutions, 
according to the relations between the data in each case. 
Others are always indeterminate. Very few are always capa- 
ble of only a single solution. 

Some of the last nine cases might be regarded as special 
cases already included in the first twenty. 




No. of 










Case 1 of f 45 







Case 3 of S 45 














































Case 4 of 145 











































































Case 2 of (45 














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The World's Columbian Exposition of 1893 4to, 1 00 

Worcester and Atkinson. Small Hospitals, ESstablishment and 
Maintenance, and Suggestions for Hospital Architecture, 

with Flans for a Small Hospital 12mo, 1 26 


Green's Grammar of the Hebrew Language 8yo, 3 00 

" Elementary Hebrew Grammar 12mo, 1 26 

" Hebrew Chrestomathy 8yo, 2 08 

Geeenius's Hebrew and Chaldee Lexicon to the Old Testament 

Scriptures. (Tregelles.) Small 4to, half morocco, 6 (10 

Letteris's Hebrew Bible 8yo, 2 2fr 



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the Library on or before the last date 
stamped below. 

A fine of Ave cents a day ia incurred 
by retalhing it beyond the specified 

Please return promptly. 

OCT 25 Bt H 

- mj 'Gill 

Eng 719.03 

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