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>l 



THE PRINCIPLES 



OF 



GRAPHIC STATICS 



BY 

LT.-COL. SIR G. S. CLARKE, K.C.M.G. 



i 

BOTAL KNGINKER8 




THIRD EDITION 



bonbon: 
E. & F. N. SPON, Limited, 125 STRAND 

JJeto f>orfc: 
SPON & CHAMBERLAIN, 12 CORTLANDT STREET 

1897. 



G 4 G 2. 2. 
AUG I 'A 1J02 

.C5S 



t»ll^ 




PBEFACE TO SECOND EDITION. 



Since this book was published, nearly nine years have elapsed, during which the 
subject has made steady progress in this country. Systematic teaching has to a great 
extent taken the place of the desultory adoption of a method here and there in a 
course of engineering study. Graphic Statics tends more and more to be read on the 
ground of its educational value, and not merely for what it can accomplish ; while it 
is now generally recognised that no real grasp of the subject, no certainty of handling, 
can be attained without a mastery of principles. 

Under these circumstances, and to meet the demand which exists both here and 
in America, it has been decided to issue a cheaper edition. Pressure of other work 
has rendered it impossible for the writer to make additions which would perhaps add 
to such usefulness as the book may claim. He has been obliged, therefore, to content 
himself with revision and the minor amplifications suggested by small difficulties which 
have been experienced by various correspondents. The proofs have been read by 
Mr. T. H. Eagles, M.A., Instructor in Geometrical Drawing and Lecturer in 
Architecture at the B. L E. College, Coopers Hill, to whose kind assistance, generously 
proffered, the writer is deeply indebted. 

IiONDOW, 

May, 1888. 



a 2 



PKEFACE. 



The study of Graphic Statics, as a subject sui generis, has made but little progress in 
England, though the great value of numerous Graphic methods has long been fully 
acknowledged. While in many of the great Engineering Schools of the Continent 
the subject is deemed worthy of a professorial chair, in England it is left to be 
gleaned almost hap-hazard. A method more or less is thrown into a course of 
teaching, according to the predilection or dislike of the teacher for Graphic modes 
of procedure. 

And yet the subject is valuable, not merely as a means to an end, but as a part 
of mental training. A mind brought up only on mathematical symbols is but half 
trained : the Graphic is the complement of the Analytical process. And the power 
conferred by Graphic method is to a large extent at the disposal of those who have had 
but little mathematical training. The writer once had occasion to explain a practical 
application of the triangle of forces to a class of working men, who seemed at once to 
grasp and appreciate it. 

The present work is an attempt to steer a middle course between the too pro- 
nouncedly abstract character of many of the numerous foreign treatises, and the too 
narrowly practical treatment the subject has received in England. Though con- 
fessedly incomplete, it will, it is hopted, prove suggestive, and may perhaps lead some 
of its readers to a further prosecution of a really fascinating study — one, too, in which 
very much still remains to be done. 

In order to make the work self-contained, an Appendix is added, giving tables of 
the weights and strength of materials. By the help of these tables it will be found 
possible to apply the various constructions to actual practice. 

The scale to which the figures are drawn is necessarily contracted, but it has been 
endeavoured to make these figures clear and easy to follow. 

It is perhaps necessary to apologize for the employment of a few words not yet 
quite naturalized in the sense here given to them. Thus " pressure " has been adopted for 
compressive stress, which has usually been termed " compression." Using " pressure " 
in this sense, " compression " is available for the alteration of length due to " pressure." 
Tension and extension are universally so used, and analogy seems to demand pressure 



vi PEEPACE. 

and compression. The words " vector " and " stress centre " are also adopted, while 
" kern " has been borrowed bodily from the German. The latter word has no English 
equivalent which possesses a sufficiently suggestive meaning, and the thing meant — 
the locus of the stress centre under certain conditions — has hitherto received no 
English name. 

Besides the ordinary English text books, the following works have been put 
in requisition, viz. ; — 

* Wrought-iron Bridges and Roofs/ — Unwin. 
4 La Statique Graphique.' — Levy. 

1 Der Constructeur/ — Reuleaux. 

' Die Graphische Statik.' — Culmann. 

1 Die Grundziige des Graphischen Rechnens/ — Von Ott 

* Elemente der Graphischen Statik/ — Bauschinger. 
'Taschenbuch der Hutte/ 

The writer is, moreover, much indebted to his colleague, Professor W. 0. Unwin, 
B.Sc., for many valuable hints, and to his late and present colleagues, Professor 
A. G. Greenhill, M.A., and Mr. T. H. Eagles, M.A., for kindly reading proofs at 
different times. 

Coofebb Hill, 

November, 1879. 



CONTENTS. 



CHAPTEE L 

GRAPHIC ARITHMETIC. 

PABA. PAGE 

1. Multiplication of Lines •• 1 

2. Raising a Line to any Power 2 

8. Extraction of Roots 2 

4. Division . . • . • • • • • • . • • . • . . . • • • • • • 5 

6. Fractions 5 

6. Rectilinear Figures 7 



CHAPTEE II. 

COMPOSITION AND RESOLUTION OF FORCES, &o. 

7. Forces acting at a Point 10 

8. Forces in any Direction .. . . 12 

9. Properties of the Funicular Polygon 14 

10. Resolution of a Force in three directions . • . 17 

11. Resolution of a Force in three directions without the direct employment of the Funicular Polygon 19 

Parallel Forces. 

12. Resultant of a System of Parallel Forces 20 

18. Resolution of a Force into two parallel Components haying given Lines of Aotion . . . . 22 

14. Determination of the Reactions of the Supports of a Loaded Beam 28 

15. Constant Component of Stress in the Sides of the Funicular Polygon 24 

Moments op Forces, 

16. Graphic Representation of the Moments of Forces 24 

17. Reduction of Moments to a Common Base 25 

18. Moment of a Single Force 26 

19. Moment of the Resultant of a System of Forces 27 

20. Moment of the Resultant of Parallel Forces 28 

Couples. 

21. Moment of a Couple 29 

22. Resultant Couple 30 

23. Interior Forces, or Stresses 31 



• • fl 



vm CONTENTS. 



CHAPTER III. 

RECIPROCAL FIGURES. 

PARA. PAQB 

24. Definition of Reciprocity . , • . 31 

25. Classification of Figures 32 

26. Conditions of possible Reciprocity . . .... . . • . • • • • • • • . 33 

27. Examples of Reciprocal Figures 33 

28. Exceptional Figures 36 



CHAPTER IV. 

STRESS DIAGRAMS. 

29. Mechanical Property of Reciprocal Figures .. .. •• •• •• •• •• 88 

30. Determination of Stresses in the Bars of a Polygonal Frame 38 

81. Distinction between Ties and Struts 40 

32. Roof Trusses with Symmetrical Vertical Loads . . . • • • 41 

33. Roof Trusses with Unsymmetrical Vertical Loads • . . . . . . . 44 

34. Pent Roof: Vertical Load 44 

86. Wind Pressure . . . • . . . . . . . . . . . . . . • • . . 45 

36. Stress Diagrams for Normal Wind Pressure . . . . . . . . . . . . . . 46 

87. Diagrams of Horizontal and Vertical Components of Wind Pressure . . 47 

38. Bowstring Roof with Normal Wind Pressure 48 

89. Warren Girder 60 

40. Bollman Girder • .. # . 61 



CHAPTER V. 

ACTION OF STATIONARY LOADS. 

Biams Fixed at Onb End. 

41. Single Concentrated Load 52 

42. Several Oonoentrated Loads • • . . . # 54 

43. Uniformly Distributed Load .. . . 54 

Bbams Supported at Both Ends. 

44. Single Oonoentrated Load 65 

45. Any Number of Oonoentrated Loads . . . . . . . . . . . . . . . . 56 

46. Uniformly Distributed Load 58 

47. Load with any Fixed Distribution . . . . . . . . . . 62 

48. Forces in one Plane, but not parallel • 63 

49. Beam Resting on Three Supports 64 



CONTENTS. 



IX 







CHAPTER VL 

TEAVELLING LOAD. 

PABA * # PAGE 

50. Single Concentrated Load 65 

51. Any Number of Concentrated Loads 67 

52. Uniformly Distributed Load . . . . . . . . . . # . 70 

53. Combined Stationary and Travelling Load 71 

54. Curve of Total Stress in Booms • • 72 

CHAPTER VII. 

BRACED GIRDERS. 

55. General Considerations • • . . . . . . . . . . # . . . . . . . 72 

56. Nature of the Loads on a Railway Girder 73 

57. Warren Girder. — Load on Lower Joints 73 

68. Warren Girder, Loaded at all Joints 77 

59. Lattice, or Trellis Girder 79 

60. Bowstring Suspension Girder 81 

61. Bowstring Girder . . . . . . 84 

CHAPTER VIII. 

CENTRE OF PARALLEL FORCES.— CENTRE OF GRAVITY. 

62. Parallel Forces acting at Points in one Plane . • • • . . . . . . • • . . 85 

63. Parallel Forces in Space 87 

64. Centre of Gravity of Lines and Curves . . . . . . . . . . 89 

65. Centre of Gravity of Plane Figures . . 90 

66. Curved Surfaces and Solids . . • • • • 94 



OHAPTEE IX. 



• a 



MOMENT OF INERTIA— CENTRAL ELLIPSE, &o. 

67. Moments of Parallel Forces 

68. Reduction of Moments to a Common Base 

69. Moment of Inertia of a System of Parallel Forces 

70. Radius of Gyration 

71. Curve of Inertia and Central Curve 

72. Properties of the Central Ellipse, Ac. 

73. Moment of Inertia of a System of Forces determined by means of the Central Ellipses of its 

XDriTOUUB a. a. • m .a a. a* • . m • 

74. Ellipse of Inertia of a System of five Parallel Forces 



96 

99 

99 

101 

101 

103 

105 
106 



x x CONTENTS. 

CHAPTER X. 

MOMENT OP BESISTANCE.— OENTEAL ELLIPSE, &o. 

PABA. PAOK 

75. Bending Stress 107 

76. Besistance Area 110 

77. Examples of Besistance Areas . . . . . . . . . . Ill 

78. Moment of Inertia of a Section . . . . . . . . . . . . . . . . . . 116 

79. Moment of Inertia by means of Funicular Polygon 116 

80. Moment of Inertia by means of Besistance Area. Table of Moments of Inertia of various 

OUCUOUS .. •• •• a. » • «. ss « # a . •• a • •• XXI 

81. Central Ellipse and Kern of a Section . . 121 

82. Examples : Central Ellipse of Simple Figures 122 

83. Central Ellipse and Kern of an I Section . • . . • . . . • . . • • . . . 128 

84. Central Ellipse and Kern of an Angle Iron . • . . . • . . . • . . . . 130 

85. Besistance to Shearing . . . . . . . • . . . . . . . • . . • • 132 

86. Intensity of Stress at Neutral Axis • . . • • • • • . . • . • • . . • • 132 

87. Intensity of Stress at any part of a Section. . 133 

88. Curve of Shearing Stress 135 

89. Besistance of Long Struts 138 



APPENDIX. 



TABLES. 

Weight of a cubic foot of various substances . . , . 140 

Weight of Roof Coverings 140 

Weight of Boof Framing 141 

Weight of Platform : Boad Bridges 141 

Weight of Platform : Bailway Bridges 141 

Weights of Locomotives and Tenders 142 

Working Strength of Materials . . ' 142 



GKAPHIC STATICS. 



CHAPTER L 



Pig. 1. 



GBAPHIC AEITHMETIO. 

By Graphic Arithmetic is meant the employment of lines to represent numbers, and 
the performance, by means of construction, of the ordinary processes of arithmetic. 
The results obtained are lines which represent the required product, quotient, &c, on 
the same scale as that of the lines dealt with. 

Thus by the square root of a given line is meant a second line which represents 
the square root of the number which the first line is taken to mean, and which 
represents that square root on the same 
scale as that on which the given line 
represents the number. It will be 
seen, therefore, that if a line is said to 
represent a number, a scale is always 
implied. In dealing with all questions, 
the scale must never be lost sight of. 

Although in the following pages 
lines only are spoken of, it must be 
understood that numbers represented by 
lines on some convenient scale are meant, 
and by unity is implied a unit length on 
this scale. 

1. Multiplication of Lines. — To 
determine the product of any number 
of lines 

o 1 Xo l Xa,X..,,Xfl, 

draw any line X (Fig. 1), and set 

off from a length 1 equal to unity. At 1 erect a perpendicular T, and set off 
from 1 the distances 1 a x = a if 1 Oj = a* and so on ; join the points a n a % . . . . to O, 
and produce the lines a l9 a a • • • • indefinitely. 

From set off 2 = a x along X, and raise a perpendicular 2 2' at 2, cutting 
A a in 2'. 

Then 22' i a % 




02 01 



; or, 22' = Oj x «* 



B 



GBAPHIO STATICS, 



From set off 3 = 2 2' along X, and at 3 erect a perpendicular 3 3', cutting 



A s in 3'. 

Then 



33' 

03 



rTB - or or > 33 - 22 X «» = «i X a, X <H. 



Proceeding thus, the last ordinate 5 5' gives the product a x x a* x (h X a A x Oj, 
which must of course be read off on the same scale as that used for the lines and for 
the unity length 1. 

2. Raising a Line to, any Power. — Draw two axes X X, Y Y (Fig. 2), at right 
angles to each other, intersecting in 0. From set off 1 along X equal to unity 

and a x along Y equal to a, the line 
to be raised to any power, say 5. 

Draw a x a* at right angles to 1 a x , 
*+ cutting X X in a 2 ; a, a 3 at right angles 

to a x a^ cutting YY in a 3 ; and so on, 
cutting each axis in turn, and drawing 
each line at right angles to the last. 
Then 



Fig. 2. 




q, _ a, 

O a! ~" 01 



; or, Ooj, = a". 



Similarly, 

Oa, Oa. 

Chi = o- 1 1 ;°r,Oa 8 = Oa a x« = a\ 

And finally, 

0<h-tt. 

3. Extraction of Boots. — The value of y/a can be obtained by determining a 
mean proportion (x) between a and unity ; thus x = y/ a x 1 = V^oT By repeating 
this process the values of y'a, v^a, y/a, &c, can be determined ; other roots can, 
however, be obtained by means of the logarithmic spiral or the logarithmic curve. 

To draw the former, make a (Fig. 4) equal to unity, and taking as pole, 
draw any number of radii, r u r 2l r 3 . . . . at equal angular distances, a. Now the required 
curve cute every radius vector at the same angle. Hence, if 1, 2, 3 are the points 
in which the curve cuts r u r a , r 3 , the triangles 3 2, 2 1, 10 a, &c, will, if the 
angles a are supposed to diminish indefinitely, be similar. Therefore 



1 












or 



fj = fi ; f B — r ! ; r 4 — r x . • . • r„ = r t • 



Hence successive radii at equal angular distances form a series in geometric progression 



GRAPHIC ARITHMETIC. 



of which r x is the ratio. This enables the curve to be drawn for any assumed value 
of this ratio r x . With (Fig. 3) as centre describe an arc a a' with radius unity; 
on this arc set off a chord a a! equal to the assumed value of r l9 produce a, a', 
describe an arc b V with radius equal to r x ; with as centre and the chord r a of the 
arc b V as radius, describe an arc c d ; proceeding in this way the successively obtained 
chords r a , r 3 . . . .form the required geometric series. Set off the lengths r lf r a , r 3 . . . ., 
&c. (Fig. 3), along the successive radii in Fig. 4; thus the points 1, 2, 3 .... are 
obtained, and the curve drawn through them is part pf the required logarithmic 



Fig, 3. 




Fro. 4. 




Fro. 5. 



spiral. To continue this curve the other way, i. e. below 
the unity radius a, draw the radii 01', 2' .... at 
equal angular intervals a, and obtain their lengths by 
constructing an isosceles triangle cH a (Fig. 5), of which 
the two sides a, a' are both equal to r l9 and the 
third side a a' equal to unity. Having constructed this 
triangle, describe an arc with centre and radius o b equal 
to a a' (Fig. 4), the chord r x of this arc is the required 
radius 1' ; draw a second arc with radius c equal to 
r lf then the chord r 2 f is the required radius 2' (Fig. 4) ; 
by continuing this process the spiral may be extended at 
pleasure. 

The spiral so drawn serves to determine any root of any line. Thus, suppose the 
5th root (v7) of any line I is required. With as centre and radii equal to unity 
and l 9 cut the spiral in a and L respectively, join L, and divide the angle L a into 

b 2 




GRAPHIC STATICS. 



Fig. 6. 



five equal parts : make the angle a x equal to one-fifth the angle L a, and on the 

same side of a as O L. Then 

Ox = V(5L - ^T- 

In the same way the spiral can be used to raise a line to any power. Suppose 

the value of a line s 9 raised to the power 7, or s\ to be required. Cut the spiral, with 

O as centre and radii equal to unity and s, in a and 1. Join 1 0, and draw a radius 

7 such that the angle 7 a is equal to 7 x angle 10 a. Then 

In the figure, to avoid confusion of lines, s has been taken equal to r x ; the 
construction is, however, quite general. 

Instead of the logarithmic spiral the logarithmic curve may be used. 
Take any line as axis, from any point on this axis set off abscissas equal to a 
series of numbers sufficiently near to each other read off from any convenient scale. 
At the extremity of each abscissa draw ordinates equal to the logarithms of the 
respective numbers taken from a table of logarithms and read off from the same scale, 
and through the extremities of these ordinates draw a curve. This curve can then be 
used in place of a table of logarithms. 

Thus, to obtain the product of a series of lines, set them off from the origin along 

I the axis, and draw the corresponding ordi- 

nates ; the sum of these ordinates is the 
logarithm of the product, and the abscissa 
corresponding to an ordinate equal to this 
sum is the required product. 

To raise a line I to any power n, set I 
off along the axis from the origin, and draw 
the corresponding ordinate, and obtain the 
abscissa corresponding to an ordinate equal to n times the ordinate of the abscissa L 

Conversely, to extract the n* root of a line Z, set off I from the origin along the 
axis, and obtain the abscissa corresponding to an ordinate of length \j* of the ordinate 
of the abscissa I 

The cube root of a given line may be found by means of a special curve drawn 
as follows : 

Make A (Fig. 6) equal to unity, and on A describe a semicircle. At A erect 
a perpendicular to A. Take any point on this perpendicular ; join C 0, cutting 
the semicircle in D. Make F = D, and erect a second perpendicular at F. Make 
H = 0. Then H is a point on the curve, and any number of points can be 
similarly obtained. 

Draw H M at right angles to H ; 

Then OH 1 = 0F.0M. 




GKAPHIO ARITHMETIC. 



Now the triangles A C and A D are similar. 
Hence 



Thus 



Therefore 



OA OD 1 OD 

OC = OA ; or 'OC = T ; or * OD - 00 = 1 - 

OF.OH= 1, andOF = -JL. 

O Jo. 

OH 8 = J-.O M, and H 8 = M. 
OM = ^OH". 



Fig. 7. 



The curve can be used as follows : — 

From 0, with radius equal to O X = I the line of which the cube root is required, 
cut the curve in X ; draw X B perpendicular to X. 
Then as above : — 

4. Division. — To divide one line by another, or to obtain the value of -?, it is 

merely necessary to obtain a fourth proportional x, to — , and unity. 
Then 

ma a 

r=-6 ;or '* = 6- 

6. Fractions. — If it is required to add a number of fractions together, or to 

obtain the value of r + t- 1 + t- 2 + . . . . , 

the fractions must first be reduced to 
a common denominator, d. Draw two 
axes, X, Y (Fig. 7), at right angles 
to each other. From O set off D along 
X, equal to the assumed common de- 
nominator d\ on the same axis set off 
O b, b x , b % . . . . , equal to the respective 
denominators 6, b Xy b 2 . . . . of the fractions 
to be added, and along Y set off a, 
a x , a,. . . . respectively equal to a, a X9 
a, .... , the numerators. Join a 6, a x b x , 
a 2 b % . . . . , and draw through D the lines 
Dy, Dy lf Dy, .... respectively parallel to ab, a x b u a 2 b 2 . . . , 

Thus 

Oo _ a _ Oy 

6 " b "OD' ° r b 




a Oy 



Oo, _ a, _ Oy, ftr «i_ Oft 
OTT" ^'OTV b ~ d 



GRAPHIC STATICS. 



and so on ; hence 



b + \ + b % ''^ 



= Oy + Oyi + Oy, .... 

d 



Therefore, if S is the sum of the lines Oy, 0^. . . . a fourth proportional to d, S, 
and unity will be a line representing the required sum of the given fractions. 

If any of the fractions of the given series have a negative sign, they should be 
separately reduced to the same common denominator d, and the operation can 
conveniently be performed by setting the numerators along Y from O in the 
opposite direction to the numerators of the positive fractions. Then the sum of the 
numerators of the reduced negative fractions must be subtracted from the sum of 
the numerators of the reduced positive fractions. 

To obtain the product of a series of fractions -r x - x r- x . . . . the method of § 1 
can be employed in a slightly modified form. From O set off along any axis * 



Fig. 8. 




\ *** 



(Fig. 8) the distance b x , 6 a , b z . . . . respectively equal to the denominators 
b i9 b u b 9 . . . . of the given fractions; at b x b 2 > 6 a • • • • draw ordinates b x a ly b 2 a 2 , b z a s 
respectively equal to a x , a, , a* . . . . the numerators ; through a x , a* a* . . . . draw the 
lines A X9 A a , A 3 . . . . 

Along X set off x x , equal to unity, and draw an ordinate x x y x cutting A x in 
y x ; set off x 2 along X equal to x x y x > and draw an ordinate x % y 2 , cutting A a in y a . 

Then *iyi __ 

Hence 



5 and ** 









and similarly 



«i <h <h 

** = *, x s x - 



Proceeding in this way the product of any number of fractions can be obtained. 



GEAPH1C ARITHMETIC. 



To raise a fraction to any power, or to determine the length of a line representing 
the expression (^ j , draw two lines, X, Y (Fig. 9), at any angle, with as centre 
describe two arcs, a a', b b' with radii respectively equal to a and 6, join a 6', a' b. 



Fig. 9. 




From set off along Y any arbitrary length r, and, starting from r, draw a series 
of lines r 1, 1 2, 2 3, 3 4 ... . alternately parallel to V a and b a!. 

01 Oa _ a 

o7 = Ob' '~ &••••(*)» 



O^ _ a 
01 "6 



(ft. 



Multiplying the equations a and ft together 



02 



Again, 
Multiplying y and 8 



o;-(J) — -Cr>- 

03 - a ^ 

03 



and finally, 



O 3 /a\» 

07 " W f 

06 _ /a\« 
Or" W 



06 



a 



Thus a fraction ^- has been obtained equal to f r J , and having any assigned denomi- 
nator Or. If r is made equal to unity, then 

°" {t) ': 

If b = 1 and the radius b b 1 is taken as unity, then 6 = a\ and .the construction 
may be used in place of that of § 2. 

6. Rectilinear Figures. — The areas of any rectilinear figures can be expressed 
by lines. Thus in the triangle a b c (Fig. 10), with any angle b as centre, describe an 



f 



8 



GBAPHIC STATICS. 



Pig. 10. 




arc with radius be equal to 2 units on the scale to which the triangle is drawn. 
Through c draw a line cf touching the arc, and through a a line parallel to c b, cutting 
cf in d. Then the triangle b d c is equal to the triangle b a c 9 and the area of b d c 
ia%bey>dc = dc 9 hence d c represents the area of the triangle a b c on the same 

5 scale as that to which it is drawn. Thus the 

scale taken was \ full size, and the length 
of d c read off on that scale gives 2*95 square 
inches as the required area. This process is 
termed the reduction of a triangle to a given 
altitude : in the above case the altitude is 
2 units. If the absolute area of a triangle is 
^ ^ not required, but only a line representing that 
area, any convenient length can be taken for 
the radius b e ; thus if a number of triangles 
are to be represented by lines proportional to their areas, it is necessary to reduce each 
separately, by the method of Fig. 10, to a triangle having any the same height b e 9 
then the bases of the reduced triangles are proportional to their areas. The sum of 
the bases of the reduced triangles multiplied by half the assumed height gives the sum 

of the areas of all the triangles. 
Pw. n. ^fcv Any rectilinear polygon can be 

similarly treated, having first reduced 
it in the usual way to a triangle with 
vertex at one of its angles and base 
on a side produced ; or it can be re- 
duced directly to a triangle having 
any given height. Thustheirregular 
seven-sided polygon a. . . .g (Fig. 
11) is reduced to a triangle AiB, 
which has a height equal to 2 A. 

In the case of a trapezium, if 
m is the arithmetic mean of the 
parallel sides, d the perpendicular distance between those sides, h the base to which the 
area of the trapezium is to be reduced ; then 

X d 

x.K = m.d, or - = t» 

in a 

whence x can be obtained by construction, then x represents the area of the trapezium 
reduced to the base h. 

The above construction for the reduction of areas to a given base is required in 
obtaining the centres of gravity of plane figures. 




*; 



GRAPHIC AEITHMETIC. 



NOTE. 

The foregoing constructions are only directly applicable when the numbers (lines) to be multiplied, 
divided, &c, are 60 small that the unit can be taken of workable length, as otherwise the triangles 
become of such " ill-conditioned " forms that accuracy is not attainable. Well-formed figures are 
essential in all graphic work. If, for example, in Fig. 1 the lines Oj, a, ... represent large numbers, 
they would be of unmanageable length unless the unit length 1 were made extremely small. In this 
case the length 1 instead of being made of unit length should be taken some convenient multiple of 
the unit— say 10, 20, 40, 100, or 1000 times the unit. 

Suppose 1 is made equal to m times the unit, 



Then 



And similarly the last ordinate = 



^ = * or 22' = *** 
a x m m 

2^ = m° r33 rf 

«i X <hX . .. X a, 
m % ~ 1 



The result must therefore be read off on a scale the unit of which is — . of the original scale used. 

As a numerical example, suppose that the length 1 is ten times the unit length employed in 
setting off the lengths c^, a* &c. 

Then 2 2' the product of Oj and a* must be read on a scale -^th smaller than the original one, i. e. 
each unit of the original scale represents 10 units in the product. 

3 3' the product of a 1} Og, and a* must be read on a scale rhn^ 1 smaller than the original one, i. e. 
each unit of the original scale represents 100 units in the second (continued) product, and so on. 



( io ) 



CHAPTER II. 



COMPOSITION AND BESOLUTION OP FOECES, Bra 

Forces having any Directions in one plane. 

7. Forces acting at a Point. — In order that a line may represent a force 
correctly, it is necessary, 1st, that its length should be proportional to the magnitude 
of the force ; 2nd, that its position on the paper should correspond with the line of 
action of the force ; 3rd, that the direction of the force along its line of action, i. e. its 
" sense? should be indicated by an arrow, or by lettering. 

Forces which do not act in one plane have to be represented by their projections 
on two rectangular co-ordinate planes. 

If the lines a, b (Fig. 12) represent in magnitude and direction two forces 
P x and P a acting at a point 0, then the diagonal r of the completed parallelogram 

Oar b represents the resultant of P x and P a in 
magnitude and direction. Now the force which, 
acting in conjunction with P A and P 2 , would maintain 
the point in equilibrium, is equal to R, and has 
the same line of action, but opposite sense. Hence if 
three forces, P lf P 2 , P s , acting at a point (Fig. 13), 
are in equilibrium, lines drawn parallel to and in the same direction as the forces 
must form a triangle, and the direction arrows of the forces point the same way round 




Fro. 13. 




Fio 14. 




this triangle. The order in which the forces are taken in drawing the triangle is 
immaterial, hence either b a c or b a d is -obtained. 

The triangle bac or bad can evidently be drawn if the magnitude of one of the 
forces and the directions of the other two are known* Thus a force P (Fig. 14) can 



COMPOSITION AND EESOLUTION OF POECBS IN A PLANE. 



11 



Fig. 15. 



be resolved in two directions a, Ob; draw c d parallel to the direction of P, and 
representing its magnitude on any scale, and from c and d draw lines parallel to b, 
a meeting in e> then c e and e d are the magnitudes of the components of P on the 
same scale. To obtain the direction arrows of these components affix to e d an arrow 
indicating the reverse direction to that of P, then c d represents the force which would 
maintain equilibrium with the two forces ce, ed, and the arrows must point the same 
way all round the triangle ced. This determines the sense of the two components 
of P. 

To obtain the resultant of any number of forces P! . . . . P 6 acting at a point 
(Fig. 15), determine the resultant r x of P x and P a ; combine r x and P 8 , thus obtaining 
the resultant r a of P n P a , P s ; con- 
tinuing the process, the last resultant 
R is the resultant of all the forces 
acting at 0. Thus to determine the 
resultant of any number of forces 
acting at a point, it is necessary to 
draw a polygon Oabcdef formed of 
lines drawn successively parallel to, 
proportional to, and in the same 
direction as the forces, then the closing 
line 0/ of this polygon of forces gives 
the magnitude and line of action of 
the resultant 

Since in the triangle of forces 
Oef, Of is the resultant of e and/e, 
the direction arrow of / will point 
in the opposite way round the triangle 
to the arrows of e and/ e. This fixes the sense of the resultant R of P x . . . . P 6 . 

Since the force which, acting in conjunction with P x ... . P 6 , would maintain the 
point in equilibrium, is equal in magnitude to R, has the same line of action but the 
opposite sense, it follows that if any number of forces acting at a point are in equili- 
brium, then lines drawn successively parallel to the lines of action of those forces, in 
the same direction as the sense of the forces and having lengths proportional to the 
magnitudes of the forces, must form a closed polygon. Thus if the forces P x . . . . P 5 
(Fig. 16) form a system in equilibrium acting at the point 0, the polygon of forces 
a .... / must close, the direction arrows pointing the same way all round the polygon. 

The order in which the forces are taken is immaterial, thus the polygon of forces 
(Fig. 16) may take the form abcdef or abhde€. 

If n forces acting at a point form a system in equilibrium, the directions of all 

a 2 




12 



GRAPHIC STATICS. 



and the magnitudes of n — 2 of them being known, then the polygon of forces can be 
drawn and the two unknown magnitudes obtained. 

In the case of a system of forces acting at a point, the sole condition of equilibrium 
is that the polygon of forces should close. This condition corresponds to the analytical 
condition of equilibrium of forces acting at a point. 

If Pu P a ... . are any such system of forces, and a 1? cuj . . . . the angles made by 
their directions with any axis, then counting forces acting in one direction as positive 
and in the other direction as negative 



Pi COS. 04 + P2 COfl » °« + • • • • 

P* sin. at + P 8 sin. a, + .... 



= 

= 



i. e. the sums of the projections on any two perpendicular axes of all the forces acting 



Fig. 16. 




in one direction are respectively equal to the sums of the projections on the same axes 
of the forces acting in the other direction. Thus the sum of the projections of c d, d e, 
and ef (Fig. 16) on the line X is equal to the sum of the projections of/ a, aft, and 
b c y while the sum of the projections of be, c d, de, and fa on Y is equal to the sum 
of the projections of ef and a b. 

8. Forces acting in any Direction. — If three forces only maintain a body in 
equilibrium, then their directions must pass through a point, hence the sole condition of 
equilibrium is that stated in the preceding section. If, however, more than three forces 
act, a further condition must be satisfied. 



COMPOSITION AND RESOLUTION OF FORCES IN A PLANE. 



13 



Suppose the forces P x . . . . P 6 (Fig. 17) to act on a body and to maintain that 
body in equilibrium, and suppose the forces connected by a system of rectilinear 
bars Aj A a , A a A s . . . . jointed at A x , A a . . . ., and that the system remains in equili- 
brium, the body having been removed and the jointed bars alone remaining. As a 
condition of equilibrium of the system, each joint must be separately in equilibrium 
under the forces acting at it. Thus the exterior force P u together with the 
interior forces or stresses Si and S fl in the bars A x A 2 and A x A 6 maintain the point 




A! in equilibrium. Hence Pi , Si , and S* must form a triangle a ft (Pig. 1 7a) ; 
similarly P a , S a , and S x form a triangle ft c (Fig. 17a) which has one side ft common 
to the triangle a ft 0, and so on. Finally P B , S 6 , and S< form a triangle eaO which has 
one side a common to the first triangle a ft 0. Thus each joint of the polygonal 
frame Ai . . . . A* (Fig. 17) furnishes a triangle, each triangle has one side common to 
only two other triangles, and all the triangles make up a closed polygon abcde, the 
polygon of the exterior forces P x . . . . P 6 . Further, the vectors drawn from to the 



14 GEAPHIO STATICS. 

angles of this polygon of forces determine in magnitude, direction, and sense the 
stresses in the several sides of the polygonal frame A x . . . . A 6 . 

The polygon A x . . . . A 6 (Fig. 17), the sides of which are respectively parallel to 
the vectors drawn from a pole (Fig. 17a) to the angles of the polygon of forces 
a .... e, is termed the funicular polygon of the forces P x . . . . P $ with respect to the pole 0. 

Reasoning conversely to the above, it will be evident that if a system of forces 
acting on a body is in equilibrium, then the polygon of forces a . . . e must close, and 
if the angles of this polygon are joined to a point 0, and starting at any point t a 
polygon is traced whose angles A x . . . A B are on the lines of action of the forces, and 
whose sides are respectively parallel to the vectors drawn from 0, this second polygon 
must close also. 

If the funicular polygon A x . . . A B of the forces P x . . P 6 with respect to any pole 
closes, then the funicular polygon B & . . . B 6 of the same forces with respect to any other 
pole 0' closes also. 

The conditions of equilibrium of a system of forces acting in any directions on a 
rigid body are, therefore, 

1st. The polygon of forces must close. 
2nd. The funicular polygon of the forces with respect to any pole must close. 

This second condition corresponds to the analytical condition, that the sum of the 
statical moments of a system of forces in equilibrium, about any point in their plane, is 
zero ; i. e. if B lf Ra • • • • form a system of forces in equilibrium, then r x r 2 . . . . being 
the perpendicular distances of their lines of action from any point in their plane, 

Bt.n + R, r, + =0 

reckoning moments which tend to cause rotation in one direction as positive, and in 
the other direction as negative. 

9. Properties of the Funicular Polygon. — The funicular polygon (Fig. 17) 
was drawn starting from the point t ; any other point might, however, have been 
taken as origin. Hence there is an infinite number of funicular polygons having the 
same pole but different starting points. Similarly there is an infinite number of 
funicular polygons having the same origin but different poles. 

If one of the forces P x (Fig. 17) be supposed removed, and a force — P x having 
the same magnitude and line of action but the opposite sense to "P x , this new force will 
evidently be the resultant of the remaining forces P„ P 3 , P 4 , P 6 . 

Now suppose that the magnitudes and directions of the four forces P a . . P 6 were 
known, the magnitude and direction of their resultant — P x being unknown. Draw 
the polygon bcdea (Fig. 17a) of the four known forces, then the closing side a b gives 
(§ 7, p. 11) the magnitude of the resultant of P a . . P,. The direction arrow of this 



COMPOSITION AND BESOLUTION OF FORCES IN A PLANE. 



15 



resultant will point the opposite way round the polygon to the arrows of the four 
known forces. All that is required, therefore, is a point on the line of action of the 
resultant, then a line through this point parallel to ba gives the position of the 
resultant, its magnitude and direction being already known. Taking any point r 
(Fig. 17) as origin, draw the funicular polygon r A a A 3 A 4 A 6 t ; if then the two last 
sides A*r, A5* of this funicular polygon coincide, the forces P a . . P B are in equilibrium, 
and do not admit of a resultant; if A a r, A 5 f are parallel, the forces P a . . P 5 do not 



Fig. 18. 




Fig. 18a. 




admit of a resultant, but may be reduced to a couple in an infinity of ways ; if, 
however, A a r, A 6 1 produced intersect in A x , then A x is a point on the line of action of 
the resultant of P a . . P 6 , and that resultant is fully determined. 

Thus, to obtain the resultant of a number of forces it is necessary : 1st, to draw 
the polygon of those forces in order to obtain the magnitude, direction, and sense of the 
resultant ; 2nd, to draw the funicular polygon of the forces with respect to any pole in 
order to obtain a point on the line of action of the resultant. 



16 GEAPHIC ST A11CS. 

The above process has been carried out in the case of the forces Pj P t 

(Fig. 18), and the magnitude and direction of their resultant R thence obtained. 

In Fig. 17 any number of the forces P x . . . P 6 can be replaced by their resultant 
without disturbing equilibrium. Suppose the. funicular polygon to be cut by the section 
plane xy, then to restore equilibrium a force must be applied equal to the resultant of 
the stresses S 5 S 3 in the sides A x A 5 , A 3 A 4 . The line of action of this resultant must 
pass through I, the intersection of A x A B , A 3 A 4 produced, and since the forces R, S 6 
and S 3 acting at I must be in equilibrium, these forces must form the triangle d a 
(Fig. 17a), of which the side a d gives the magnitude and direction of R. The arrow 
indicating the sense of R must (§ 7, p. 10) point in the opposite way round the triangle 
d a 0, to the arrows of the sides a 0, d, the direction of which can be ascertained by 
inspection, from the triangles a e, and e d corresponding to the joints A 5 and A* 
respectively : hence R is completely determined. 

Thus the funicular and force polygons can be employed to determine the partial, 
resultant of any number of adjacent forces in a system, the polygon of forces furnishing 
the magnitude, direction, and sense of this partial resultant, while the intersection of 
those sides of the funicular polygon between which the lines of action of the forces lie, 
gives a point on its line of action. 

Suppose that in Fig. 17 the magnitudes of two of the forces, e. g. P 4 and P 5 , are 
unknown, their lines of action being given. Starting from a (Fig. 17a) the polygon 
(abed) of the remaining forces P lf P 3 , P 8 is drawn, then lines from d and a parallel 
to the given lines of action of the unknown forces determine e ; and hence the magni- 
tudes, directions, and sense of these forces are given by the lines de and ea. Starting 
from t (Fig. 17) draw the funicular polygon t A x A^ A 3 n with respect to any pole O; 
produce the extreme sides of this polygon to cut the given lines of action in A 6 and A*, 
join A 5 A* ; then A« A4 will be parallel to O e. Now if all the forces are parallel, the 
polygon of forces becomes a straight line, and since in that case d e and a e would not 
intersect, the only way of obtaining e is by drawing a vector from parallel to A* A*, 
the closing line of the funicular polygon. This will be noticed hereafter in determining 
the reactions of supports. 

One other remarkable property of funicular polygons remains to be noticed. 

If the pole describes a straight line, then the corresponding sides of the successive 
funicular polygons with respect to successive positions of the pole will all intersect in 
a straight line, and this straight line is parallel to the locus of the pole. 

This is shown in Fig. 19, where the corresponding sides 1 2, l l 2 1 ; 2 3, 2 1 3 l , &c, 
of two funicular polygons 1 . . . 5, l l . . . 5 1 of the forces P x . . . P 6 drawn with respect to 
two different poles 0, 1 intersect on the straight line X Y, and this line is parallel to 
the line O l . 

A weightless string suspended at its two extremities to fixed points, and strained 



COMPOSITION AND RESOLUTION OF FORCES IN A PLANE. 



17 



by forces distributed at intervals along its length, takes the form of the funicular 
polygon ; the term is, however, used independently of its derivation, and may be 
equally applied, whether the sides of the polygon to which it refers are in tension or 
pressure, provided that the polygon conforms to the conditions above described. In 
the latter case the funicular polygon is sometimes termed a linear arch. 



PM.1& 




10. Resolution of a Force in Three Directions. — Suppose that the force P 
(Fig. 20) is to be resolved in three given directions, X X, Y Y, Z Z. If, then, the 
direction of P is supposed to be reversed, then the reversed force P and the three 
required components form a system in equilibrium, and the polygon of forces as well 
as the funicular polygon of this system must close (§ 9). In Fig. 20 the polygon 
of forces is commenced by drawing a b parallel to the direction of the given force P, 
and joining a and b to an assumed pole 0. From a and b draw ax y bx parallel to the 
given directions X X, Z Z ; join x. It is required to obtain in the polygon of forces 
a side d c parallel to the given direction Y Y, and such that the funicular polygon 
traced by lines parallel to the four vectors a, d, c, b will close. 

Suppose the required component Y Y to be resolved into two components Y x , Y 2 , 
of which Y x coincides in direction with X X, and Y a is parallel to Z Z. Then evidently 
the component Y u together with the required component XX, make up the length ax 



18 



GRAPHIC STATICS. 



in the polygon of forces ; and similarly Y a , together with the required component Z Z, 
make up the length b x. The component Y a must act along til drawn from the 
intersection t of X X and T T parallel to Z Z. 

Starting from o draw the funicular polygon for the four forces X X, Y^ Y a , and 
Z Z ; o 1 parallel to a is the first side, and since X X and Y x have the same line of 
action, the second side becomes nil, the third side is 1 8 parallel to x and terminated 
at 8 on the line of action 1 1! of the component Y a . But the five forces P, X X f Y lf Y a , 
Z Z, are in equilibrium, hence their funicular polygon must close. Produce o 1 to cut 
the line of action of P in 4, and draw 4 5 parallel to b 0, cutting Z Z in 3, then 8 3 is 

Fig. 20. 




the closing line of the funicular polygon. In the polygon of forces draw c parallel 
to 8 3, c d parallel to Y Y, and join d. In the funicular polygon draw 1 2 parallel 
to d, and produce 8 3 ; this amounts to replacing the two components Y lf Y a by their 
resultant Y Y, and the lines 8 3 produced and 1 2 will intersect at 2 on the line of 
action of Y Y. Thus the complete funicular polygon of the forces XX, YY,ZZ, and 
P reversed, is 1 2 3 4. The polygon of forces determines the sense of the components : 
affix to a b an arrow indicating the reverse sense to that of P, then the arrows point 
the same way all round the polygon of forces. 

In Fig. 21 a special case of the same problem is dealt with. Two of the three 
directions X X, Y Y are parallel to each other, and at right angles to the direction of 



COMPOSITION AND RESOLUTION OF FORCES IN A PLANE. 



19 



the given force, while the third, Z Z, is parallel to the direction of the given force. If 
the construction above explained is carried out in principle, it will be found that in 
the polygon of forces the two parallel components a d, dc 9 are equal in magnitude but 
of opposite sense, forming therefore a couple, while the third, c 6, is equal in magnitude 
and has the same sense as a 6, the given force. The figuring of the funicular polygon 
in Fig. 21 has been made to correspond with that in Fig. 20. 

The resolution of a force in mare than three directions is always indeterminate. 

The resolution of a force in three directions is impossible if two of the given 



Fig. 21 



Fig. 22. 




directions intersect on the line of action of the given force, and indeterminate if the 
three given directions intersect on the line of action of the force, or if the three given 
directions are all parallel to that of the force. It is also obviously impossible if the 
three given directions are all at right angles to that of the force. 

11. Resolution of a Force in three directions without the direct employ- 
ment of the Funicular Polygon. — The construction explained in the above paragraph 
has been given first as being the most general possible ; a simpler method can, 
however, sometimes be adopted. Thus, in Fig. 22, P is to be resolved in three 

d 2 



20 



GEAPHIO STATICS. 



directions, X, T, Z. Produce one of the given directions X to cut the direction of P 
in 1 ; join 1 and 2, the intersection of the other two directions, Y and Z. Resolve P 
first along X and 1 2, and then resolve the component 1 2 along Y and Z. Thus in 
the polygon of forces draw a b parallel and equal to P, and ac>bc respectively parallel 
to X and 1 2 ; affix to a b an arrow in the reverse direction to that of P, then the 
arrows in the triangle of forces b ac point the same way round; reverse the arrow of 
b c, and draw cd,bd parallel to Y and Z respectively, the arrows of cd 9 db point the 
same way round as that of c b reversed. 

This construction is really a special case of § 10, in which c, one of the angles of 
the polygon of forces, is taken as pole, which allows the lines of action o£ the forces to 
be used as the sides of the funicular polygon. 

Parallel Forces. 

12. Resultant of a System of Parallel Forces. — In the case of a system of 
parallel forces the polygon of forces becomes a straight line, which is sometimes 
termed the line of loads; the principles above laid down are, however, absolutely 
unchanged. 

To obtain R, the resultant of any system of parallel forces P x . . . . P 6 (Fig. 23), 



Fig. 23. 





draw the polygon offerees abedefg starting from a; this polygon is a straight line, 
and ga 9 its closing line, gives the magnitude and sense of the required resultant 

Take* any pole 0, and draw the vectors Oo, 06 Og ; starting at any point 0, 

draw the funicular polygon o, 1, .... 6 7, its sides 1, 1 2, &c,, being respectively 



COMPOSITION AND EESOLUTION OF FOBCES IN A PLANE. 



21 



parallel to the vectors a, J, &c. Then, as in § 9, the first and last sides #1,76 
of the funicular polygon produced, intersect on the line of action of R : R is therefore 
fully determined. 

Similarly 2 3 and 6 7 produced intersect in s, and 8 is (§ 9) a point on the line of 
action of the partial resultant r of P 3 P 4 P 5 P 6 . 

Fig. 24. 




Tht>s, given any system of parallel forces, the resultant of the entire system, or 
the partial resultant of any number of the forces can be obtained. 

If the number of parallel forces is infinite, and their lines of action are 
infinitely near together, the funicular polygon becomes a curve, termed tho funicular 



curve. 



The property of the funicular polygon stated in § 9, p. 16, deserves special notice in 
the case of parallel forces. In Fig. 24, P x . . . . P 6 are a system of vertical forces whose 
lines of action are equidistant. On a load line o 6, o 1 is made equal to P M 1 2 to P 2 , 
and so on. Suppose that it is required to combine these six forces with a horizontal 



22 GRAPHIC STATICS. 

force H : in the polygon of forces draw o equal to H, and at right angles to the load 
line ; take as pole, and draw the vectors 01, 2, &c. Starting from I, the inter- 
section of the lines of actibn of V x and H, draw the funicular polygon I, II ... . VII. Let 
a new horizontal force H' acting at I' be substituted for H, the vertical forces remaining 
as before ; make o 0' equal to H' : then (§ 9, p. 16), since the pole has moved from to 0', 
the sides of any funicular polygon of the forces with respect to the new pole will 

severally intersect the corresponding sides of the funicular polygon I VII on some 

line parallel to 0' 0. To determine this line, draw I' II' parallel to 0' 1, and produce 
I' II' and I II to meet : the line through their meeting point parallel to 0' is the 
required line. If, however, the intersection of I' II' and I II is at an inconvenient 
distance, as in Fig. 24, draw lines from I parallel to any vector 4 of the first force 
polygon, and from I' parallel to 0'4 the corresponding vector of the second force 
polygon. These lines intersect in a, then A B drawn through a parallel to 0' is 
the required line, and corresponding sides of the two funicular polygons intersect on 
A B as shown. 

The above construction has an application in the case of the arch. If P,. . . .P f 
represent the weights of strips of equal breadth of the arch and its surcharge, and H 
the assumed, or calculated horizontal thrust at the crown, the funicular polygon I . . . .VII 
is termed the line of pressure of the arch. By means of the construction of Fig. 24 a 
new line of pressure corresponding to any new thrust H' can at once be derived from 
the first; or on the other hand, if the line of action of the thrust is known, its magni- 
tude can be so determined that the line of pressure shall pass through any given point. 
Suppose H", a thrust of unknown magnitude, to act at I" (Fig. 24a), and that the line 
of pressure is to pass through X. Starting at I", draw the funicular polygon of the 
forces with respect to any pole 0'. Produce the line of action of H" ; then evidently, 
if different values are assigned to H", or different poles taken along o 0, the corre- 
sponding sides of the resulting funicular polygons will all intersect on this produced 
line. Produce VII", VI", the last side of the already drawn funicular polygon to meet 
this line in x. Join X #, and draw 6 0" in the polygon of forces parallel to X a?. Then 
o 0" gives the magnitude of H" such that the resulting line of pressure passes through 
X. This line of pressure is then completed as in Fig. 24a. 

13. Resolution of a Force into two parallel components having given lines 
of action. — In Fig. 25, b a is drawn equal and parallel to P, and b and a are joined to 
any pole 0. From any point 1 on the line of action of P draw 1 2, 1 3 respectively 
parallel to 6 0, a 0, cutting tue given directions X and Y in 2 and 3 respectively ; join 
2 3, then 2 3 is the closing line of the funicular polygon 12 3. Draw c parallel to 
2 3, then b c is the magnitude of the component X, and c a that of Y. If it is remem- 
bered that the component X and the stresses in the two sides 2 3 and 1 2 of the 



COMPOSITION AND KESOLUTION OF FOKCES IN A PLANE. 



23 



funicular polygon must in the polygon of forces form a triangle b c 9 no doubt can 
arise as to which of the components b c represents. 

In Fig, 26 the same construction with identical lettering is shown for the case in 
which the two given directions are on the same side of P. In this case the components 
cannot both have the same sense. 

Fig. 25. Fig. 26. 






In either case the resolution can be effected without the funicular polygon, as 
follows : Draw a line cutting the three directions at any angle. Divide b a in c so that 

be 1/ 

— = — , or in the second case produce b a to a point c such that this proportion holds. 

C Cv OX 

If the scale at which ba is taken is such that ba is not less than the perpendicular 
distance between X and Y, it will only be necessary to place b a in between X and Y, 
producing it in the second case to meet P. 

Fig, 27. & 




14. Determination of the reactions of the supports at the extremities of 
a Loaded Beam. — If a beam A B (Fig. 27) is supported at its two ends, and loaded 
with the weights W! . . . W 5 , then, in order that equilibrium may be maintained, the 



24 GEAPHIO STATICS. 

two supports A and B must furnish reactions BiR 2 , whose resultant is equal to the 
sum of the weights W x . . . W 6 . These reactions having been obtained, the supports 
can be supposed removed, and the forces R M R a , W l . . . W 6 can then be dealt with as 
forming a purely statical system. 

It is required, therefore, to determine two forces Rj Ra , having given lines of 
action, and such that their resultant is equal to the resultant of the given forces 
W x . . . W 6 . The polygon of forces and the funicular polygons with respect to any 
pole, of the whole system R x Ra^Wj . . . W 5 must close. Draw the load line a V 2' 3' 4' 5', 
making a V equal to W l9 V 2' equal to W a , and so on, then the whole line a 5' is equal 
to Rx + R^. Assume any pole 0, and draw the vectors a, 1\ 2\ . . , then starting 
at any point o draw the funicular polygon o 2 ... 6 8 corresponding to the pole 0. The 
sides o 2, 6 8 of this polygon cut the given directions R x and Ra in 1 and 7 respec- 
tively, hence 1 7 is the closing side of the funicular polygon. Draw the vector x 
parallel to 1 7, then a? a is the magnitude of B. x and x 5' that of R a . The resultants of 
R x and R a and of W x . . . W 5 both pass through 9, the intersection of 1 2 and 7 6 
produced, and these resultants are of equal magnitude but opposite sense. 

It will be evident that, if the forces are symmetrical and symmetrically distributed 
about the centre of the beam, the vector x will bisect a b 9 and Rx is then equal to Ra. 

15. Constant component of stress in the sides of the Funicular Polygon. — 

In the case of a system of parallel forces, the component perpendicular to the direction of 
the forces of the stress in the sides of any funicular polygon is constant. In Fig. 27 the 
stresses t x , t % . . . in the sides of the funicular polygon are given by the vectors t x , t 2 . . . 
in the polygon of forces. Resolve one of these stresses t x into a horizontal and vertical 
component by drawing h horizontal and therefore perpendicular to a b 9 then h 
gives the magnitude of the horizontal and a h that of the vertical component. 
Similarly the horizontal component of any other of the stresses is A, and hence the 
horizontal component of all the stresses is the, same. If with vertical loads the sides of 
the funicular polygon are in pressure, as is the case in Fig. 24, the horizontal 
component is termed the constant horizontal thrust The distance A is called the 
polar distance. 

Moments of Forces. 

10. Graphic representation of the Moments of Forces. — By the moment of a 
force about a point is meant the product of the force and the perpendicular let fall on its 
line of action from the point. 

If a force P act on a rigid body in the plane of the paper, and an axis perpendicular 
to this plane is imagined to pass through the body at any point 0, it is evident that the 
tendency of the force will be to rotate the body round such axis. Hence, although it 



MOMENTS OF FORCES. 



25 




is customary to speak of the moment of a force about a point, it should always be 
remembered that moments have relation to some axis, and that it is only in the case 
of forces acting in one plane that the point can be taken ^iq. 27a. 

instead of that axis. The point is the trace of that axis 
on the plane of the forces, or in the case of a single force ^* — o 

the point about which the moment is taken is the trace / \P 

of the axis on the plane through the force and perpen- 
dicular to the axis. 

The moment of a force about any point in its plane is equal to double the area of 
a triangle whose base represents the magnitude of the force, and whose height is equal 
to the perpendicular distance of the point from the direction of the force. The 
moments of a system of forces can, therefore, be represented by a series of rectangles, 
and if these rectangles are reduced to equivalent rectangles, all of which have equal 
bases, then evidently a scale of moments can be constructed from which the heights of 
the rectangles read off give the values of the several moments in terms of any unit of 
moment adopted. The moment unit is the moment of a unit of force acting at a unit 
of perpendicular distance, and may be a foot pound, a foot ton, an inch tan, &o, 
according to the linear and force units adopted. 



Fig. 28. 




Seal* of Morruerua 



r-r-r-r-r-t-r-ir-r-T-r 



KtCOO 



tooooFoct lbs, 

-1 



17. Reduction of Moments to a common base. — In Fig. 28 the lines a x Y ly 
a, P s , . . • represent in magnitude, direction, and position forces whose moments about 
any point in their plane are to be reduced to a common base. 



26 GRAPHIC STATICS. 

The linear scale is 80 feet to 1 inch. 
The scale of force is 100 pounds to 1 inch. 

Thus— OiPi= 76lb8. 

«,P t = 120 * 
0,P, = 242 n 

aiP 4 = 88 „ 

Suppose that toe common base to which the moments are to be reduced is 
H = 100 feet. 

Draw any line X Y through O, and from <h 9 <h • • draw the lines a^ b u a, b 29 . . . . 
parallel to X Y, and equal to the given base H. Join 61P1, & S P S • • • • Through O 
draw O c O Ca . . . parallel to b x P lf 6 a P a . . . . and meeting a x P x , a, P, . . . . produced, if 
necessary, in c l9 c,, . • From c l9 c % . . drop perpendiculars c x d ly e % d %9 . ... on the line 
X Y. Then these perpendiculars c x d x = m l9 c 2 d 2 = m t9 &c., are the required heights 
of the reduced rectangles, having H as base. 

From O drop a perpendicular O x 9 on the direction of ^ P M and let s be the point 
in which a x P x produced cuts X Y. Then the triangles a x P x b x and 8c x O are similar. 

Hence «i p i _ * fr 

H ""#0 

But since the triangles 8 d x c x and sxO are also similar, 

• O ~ O*' 
Thus •i^i ^.^^p n.--_ tt 

-jj- « g- ; or a, Pi . U« a flh . Jtt. 

Similarly the moment of <h Pa about O is m, . H, and so on. 

The scale of moments must be so drawn that y units on the scale of force equals 
y x 100 units on the scale of moments ; thus on the latter scale 1 inch represents 
10,000 foot-pounds. 

Beading off the distances f^ ^ m, m 4 on a scale of moments so constructed, we 

obtain — Moment of Oi P t = 12,000 foot-pounds. 



^P,* 5,000 „ 
M 0, P s = 4,500 

<*.P 4 = 9,400 



Moments tending to effect rotation in the direction of the hands of a watch are 
usually reckoned as positive; hence m l9 m*, and m 4 are positive, and m, is a negative 
moment. The sum of the moments, or the total moment, is 21,900 foot-pounds. 

18, Application of the Funicular Polygon to the Determination of the 
Moment of a Single Force.— In Fig. 29 P is the force, the moment of which about 
the point A is required. 

Draw b c parallel to the direction of P, and equal to it in magnitude. Take a 
Dole at a distance from b e equal to H (any convenient multiple of the linear unit). 



MOMENTS OF FOBCES. 



2* 



Fig. 29. 




Feo. 80. 



and draw the funicular polygon 1 2 3 of the force P with respect to the pole 0. 
Through A draw a line parallel to the direction of P, and cutting the sides 1 2, 
2 3 of the funicular polygon in n and 
m respectively. Draw 2 d perpendicular 
tomn. 

Then, since the triangles m 2 n 9 e b 
are similar — 

be tnn 
or, p . A a m m n . H. 

Hence, if a line is drawn through any 
point parallel to the direction of a force, 
then the portion of this line intercepted 
between the two sides of the funicular poly- 
gon which meet on the line of action of 
the force, multiplied by the polar distance, 
is the moment of the force about the point. 

If the polar distance H is made unity on the linear scale, then the length m n read 
off on the scale of force gives the num- 
ber of moment units. If H is not unity, 
then the length m n read off on the Q 
scale of forces must be multiplied by 
the number of linear units which H 
represents; the product will be the 
number of moment units. It will 
therefore be convenient, if the moments 
of a system of forces have to be dealt 
with, to take H a round number of 
linear units. The multiplication can 
be performed arithmetically, or by 
drawing a scale of moments as in the 
preceding paragraph. 

19. Moment of the Besultant 
of any System of Forces. — In accord- 
ance with the principles of the pre 
ceding section, the moment about any 
point of the resultant of a system 
of forces P x . . . P § (Pig. 30) can be 




28 



GEAPHIO STATICS. 



obtained. Draw the polygon a . . . / of the five forces, then the resultant of the forces 
is parallel to / a, the closing line of this polygon. Take a point 0, whose distance h 
from a f is equal to the base H of the required moment, and draw the funicular polygon 
0123456 of the forces P x . . . P 6 with respect to 0. Then the first and last sides 
1, 6 5 of the funicular polygon produced will intersect on the line of action of the 
resultant. Draw through C a line parallel to a/, cutting the produced sides 1, 6 5 in 
x and y respectively. 

Then the moment of the resultant of the whole system is equal to x y . H, and this 
moment is equal to the sum of the moments of the separate forces about C. 

If the funicular polygon closes, the sides 1, 6 5 coincide, and the intercepted 
distance x y disappears. The moment of the resultant is therefore zero, and the 
forces are in equilibrium. Hence, as has been noticed in § 8, the closing of the 
funicular polygon corresponds to the analytical condition of equilibrium, that the 
algebraic sum of the moments about any point is equal to zero. 

20. Moment of the Resultant of Parallel Forces. — The moment about the 
point C of the resultant of the five parallel forces Pj . . . P 5 (Fig. 31) can be obtained 



Fig. 31. 





in a precisely similar manner. Draw the line of loads af, and also a funicular polygon 
01 ... 6 with respect to a pole at a distance H from af Then the moment of the 
resultant of the system about G is equal to x y . H. 

The moment of the partial resultant of any number of the forces can be similarly 
obtained. Thus the moment of the resultant of P 4 and P s is equal to y'y . H, and the 
moment of P 3 is equal to a/ y'. H. 

The above construction gives the moment of any force or combination of forces in 
the form of a product which has one common factor H. This factor is termed the 



MOMENT OF PARALLEL FORCES. 



29 



" moment base " ; and since the length of H in the polygon of forces may be taken at 
pleasure, the moments of any number of forces, or the moments of the resultants of 
any groups of those forces, can, by means of the above construction, be reduced to any 
common base. The moments of any number of forces having been thus reduced to a 
common base, their summation reduces itself to the summation of the lengths of 
certain intercepts by the sides of the funicular polygon on a line through the centre of 
moments, and parallel to the direction of the forces. 

The above has a direct practical application in the determination of the bending 
moments at the various sections of a loaded beam, and also in obtaining the centre of 
gravity and moment of inertia of plane figures. 



Couples. 



-P x and P, (Fig. 32) are two equal parallel forces 
Starting from any point a, draw the polygon of 



Fro. 32. 



21. Moment of a Couple, &a 

of opposite sense forming a couple, 
forces a b y be of the couple, the 
two sides of which coincide, since 
the forces are equal and parallel. 
Taking any point as pole, draw 
the funicular polygon 012 3, of 
which the two sides 1, 2 3 are 
parallel to each other and to the 
vector a 9 while the side 1 2 is 
parallel to b. Suppose P x to be 
replaced by its two components 
a 0, 6 acting along the lines 1, 
1 2 respectively, and P a to be re- 
placed by b and e acting 
along the lines 1 2 and 2 3, then 
the two components acting along 
the line 1 2 evidently balance 
each other, and the remaining two 
components a0 9 Oc acting along 1 and 2 3 form a new couple Q 2 Q a , which would 
replace the original couple P x P a . 

Now, the moment of a couple about any axis perpendicular to its plane is equal 
to the product of one of its equal forces into the perpendicular distance between their 
directions. Make 1 a equal to a b or P lf and join a 2, then the moment of the couple 
Pj Pj is equal to double the area of the triangle 1 a 2. From 1 set off 1 y on 1 equal 
to a or Q! , then the moment of the new couple Q x Qa is equal to double the area of 




30 



GRAPHIC STATICS. 



the triangle 1 y 2. Join y a, then the angle 1 ay is equal to the angle abO and ya is 
parallel to 1 2. Hence the triangle 1 y 2 is equal to the triangle 1 a 2, or the moment 
of the couple Qi Q, is equal to that of Px P s . 

It is evident, therefore, that a given couple can be replaced by an equivalent 
couple having any given lines of action, provided only that the area of a triangle 
having a base on the line of action of one of the forces and equal to that force, while 
its vertex is on the line of action of the other force, remains constant. 

Thus, if the original couple P x P 3 is to be replaced by an equivalent couple having 
any lines of action 8 X S, cutting P x and P a in 1' and 2' respectively ; make 1' a' .equal to 
P u join 1' 2', and draw a'y' parallel to 1' 2', then y* V is the required magnitude of the 
equal forces S x S,. The sense of these forces must of course be such that the couple 
tends to effect rotation in the same direction as the original couple. 



22. Resultant Couple. — A number of couples acting in a plane can be replaced 
by a resultant couple having given lines of action. P x P a and Q x Q, (Fig. 33) are two 

*■* * * couples ; it is required to replace them 

by a resultant couple having given lines 
of action Ri,Rj. 

Draw the polygon of forces abed, 
making ab parallel and equal to Pi, 6 c 
parallel and equal to Qi, and so on. The 
figure obtained will be, of course, a 
parallelogram. Take any point as 
pole, and draw the vectors a, b 9 c 9 
d. Starting at any point 0, draw the 
funicular polygon 1 2 3 4 t of the four 
forces Pi, Qi, P a , <& relative to the pole 
0. Prom a in the polygon of forces 
draw a x parallel to the given directions 
Rx and Rj, and in the funicular polygon 
produce the two last sides 1 0, 4 X to cut 
these given directions in 5 and 6 respec- 
tively. Join 5 and 6, and from the pole 
draw e parallel to the last side 5 6 of 
the funicular polygon, then a e gives the 
magnitude of the equal forces R^ R,, 
forming the required resultant couple. 
The polygon of forces abedaea will 
§u give the sense of the required couple. 




BECIPBOCAL HOUSES. 31 

Attach direction arrows to the side a e as shown, the arrows then point the same way 
all round the polygon, and as both the polygon of forces and the funicular polygon 
close, the system is in equilibrium ; hence the direction arrows of the resultant couple 
must be the reverse of those appended to the side a e of the polygon of forces, that of 
Bx pointing down, and that of R, pointing up. 

23. Interior Forces or Stresses. — In Fig. 34, A, A B, B D are three sides of 
a jointed polygonal frame strained by two forces P ly P, acting at the joints A and B 
respectively; resolve Pi along the 

directions of the bars A, A B, and „ Pio. w. 

P, along the directions of A B, B D. 
The resolution is effected by means of 
the triangles of forces A Pj a, BP 2 i, 
and the directions of the components 
are those shown by the arrows (§ 7) ; 
then the stresses in the bars will be 
equal in magnitude but opposite in 

direction to the resolved components of the exterior forces P, . P a . The directions 
of these stresses are also shown by arrows, and the bar A B is evidently in tension. 
Hence, for tension in any bar of a structure the arrow-combination — > < — obtains, 
and for pressure 




CHAPTER in. 

BECIPBOCAL FIGUBES. 



24 Definition of Reciprocity. — Two plane figures are said to be reciprocal to 
each other when both fulfil the following conditions : — 

a. To every side of one figure there is one corresponding side in the other 

figure. 

b. Corresponding sides are parallel, perpendicular, or inclined at some 

constant angle to each other. 
e. To every system of lines meeting in a point of one figure there corresponds 
a closed polygon in the other figure. 

Thus the first of two reciprocal figures can be deduced from the second, or the 
second from the first, in precisely the same way. The number of sides in two 
reciprocal figures is the same ; and since the question of size does not enter into the 
above conditions, if one figure is reciprocal to another, all figures similar to the first 
will be reciprocal to the second, and vice versd. 



32 GEAPHIC STATICS. 

25. Classification of Figures. — In order that it may be possible for a given 
figure to fulfil the above conditions, it is necessary that — 

1st. The figure can be decomposed into a number of closed polygons such 
that each side of the original figure is a side of two only of these 
polygons. 

2nd. At least three lines of the original figure meet at each of its angular 
points, or vertices. 

3rd. Every side of the original figure passes through at least two vertices. 

In any figure fulfilling these three conditions, if s is the number of sides, tj the 
number of vertices, and p the number of closed polygons into which it is possible to 
decompose it — 

Then it can be shown that p + v - * = 2.* 

Plane figures can be divided into the following classes : — 

Deformable Figures, whose angles admit of variation without alteration in the 
length of the sides. Such figures are not therefore sufficiently determined if their 
sides only are known. 

Indeformable Figures, whose angles are determined if the lengths of their sides 
are known. 

The latter class can be subdivided into Strictly Indeformable Figures and Figures 
having surplus lines. 

A figure is strictly indeformable if it contains only the exact number of sides 
sufficient to determine it. Such a figure therefore becomes deformable if any one 
side is suppressed. A figure is said to have one, two, or three surplus lines if it contains 
one, two, or three more sides than are absolutely necessary to determine it. 

If s is the number of sides, v the number of vertices of a strictly indeformable 
figure, then, since every vertex is determined by two sides, and two only, there will 
be, exclusive of any side a b> 2(v — 2) sides ; the total number of sides is therefore 
2(v — 2) + 1. Draw n new lines on the figure joining vertices already determined. 
These new lines will be surplus lines, that is to say, they are not required to determine 
the figure, and it must be possible to express them in terms of the other sides of the 
figure. Hence, in the case of a figure having n surplus lines, the total number of 
sides is 2{v — 2) + 1 + n. Again, in the original strictly indeformable figure, suppose 
n' of the lines removed, then the total number of sides remaining will thus be 
2 (v — 2) + 1 — n 1 ; and it is impossible to construct the figure if only these remaining 
sides are known, for there will be an infinite number of figures having these sides but 
different angles. Hence, in order that the new figure should become strictly indeform- 

* Levy, ' La Statique Graphiqne/ Chap. III. § 27 



KECIPBOCAL PIGUEES. 33 

able, it is necessary that it should fulfil n' geometrical conditions. The results of the 
above reasoning may be expressed as follows . — 

In any deformable figure, s < 2 v — 3. 
„ strictly indej *or unable figure, 8 = 2 v — 3, 
» figure having surplus lines, s > 2 v — 3. 

26. Conditions of possible Reciprocity. — A figure containing one surplus line 
admits of only one reciprocal. 

A strictly indeformable figure admits a reciprocal only when it satisfies one 
geometrical condition. 

A deformable figure admits a reciprocal only when it satisfies one more 
geometrical condition than the number of lines required to render it strictly 
indeformable. 

A figure containing n surplus lines admits of n — 1 series of reciprocals, each 
series infinite. 

The above may be expressed as follows : — 

If v is the number of vertices in a figure, p the number of closed polygons into 
which it can be decomposed, the figure admits (a) one reciprocal; (p) an infinity of 
reciprocals ; (c) none, according as v is (a) equal to ; (b) greater than ; (c) less than p. 

The figures with which it is required to deal in practice are for the most part 
strictly indeformable. 

27. Examples of Reciprocal Figures. — Fig. 35 consists of two five-sided 
polygons A . . . E, a . . . e, whose vertices A .... a, ... are joined by the lines 
A a 9 B b . . . . The total number of sides (15) is less than 2 v — 3 ( = 17), the figure is 
therefore deformable. If, however (as in Fig. 37), the figure is formed of two triangles 
A B C, a b c, whose angles are joined, then 5 = 9 = 2x6 — 3, and the figure is strictly 
indeformable. Hence a figure of the form shown in Fig. 35 is deformable only so long 
as the number of sides in each of the two polygons A ... E, a ... e is greater than 3. 

Moreover (§ 26), Fig. 35 does not admit a reciprocal unless it satisfies three 
geometrical conditions, i. e. one more condition than the number of additional sides 
required to render it strictly indeformable. 

Similarly Fig. 36 does not admit a reciprocal unless it satisfies one geometrical 
condition. 

On the other hand any figure formed by joining the angles of any closed polygon 
to any point in its plane, admits of one reciprocal. 

For example, to draw the reciprocal of Fig. 35. From any point 0, Fig. 35a, 
draw vectors A', B', &c, respectively parallel to the sides A B, B C, &c, of the 
polygon A . . . E (Fig. 35). Now to every vertex A, B . . . (Fig. 35) there is a 

F 



34 



GRAPHIC STATICS. 



corresponding triangle in Pig, 35a, and in every such triangle two of the sides will be 
the already drawn vectors from 0. Starting from A' draw A' B' parallel to B 6, BV 
parallel to C c y C D' parallel to D d> and D' E' parallel to B e. In order to obtain the 
reciprocal of the vertex A, E' A' must be drawn parallel to A a, but E' and A' are 
already fixed, and hence it is evident that if the polygon A . . . E (Fig. 35) is 
arbitrarily assumed in the first instance, it does not follow that the line joining E' and * 
A' is parallel to A a, and if it is not parallel Fig. 35 does not admit of a reciprocal. 
Thus four out of the five lines A a, B b ... . can be drawn arbitrarily, but the fifth is 
determined, since it must be parallel to a line passing through two points previously 



Fig. 85a. 



Fig. 85. 




fixed. In other words, as a first condition that Fig. 35 should admit a reciprocal, the 

sides A a, B b must be parallel to the sides of a closed polygon ; this constitutes 

one geometrical condition. Suppose this condition fulfilled : it now remains to 
determine the reciprocals of the vertices a . . . e (Fig. 35) These reciprocals will all 
be triangles, and one side of each of these triangles is already determined. Hence, to 
draw the reciprocal of the vertex a (Fig. 35), draw A'O', E'O' in Fig. 35a, respec- 
tively parallel to a ft and a c, thus fixing the position of a point 0'. Passing to the 
vertex b, two sides A' 0/ A' B' of its reciprocal triangle are already fixed, hence the 



EECIPBOCAL FIOUBES. 



35 



third side 0' B' must be parallel to be, or Fig. 35 will not admit of a reciprocal. In 
fact any two sides a b 9 a e of the polygon a . . . e can be assumed arbitrarily, but the 
remaining three are then fixed. Hence the conditions that Fig. 35 should admit of a 
reciprocal are — 1st, that the lines A a, B b ... . should be drawn parallel to the sides of 
any closed polygon ; 2nd, that the sides of the polygon A . . . E should be parallel to the 
vectors drawn to the angles of the polygon A' . . . E' from some fixed point in its 
plane ; and, 3rd, that the sides of the polygon a . . . e should be parallel to the vectors 
drawn to the angles of the polygon A' . . . E' from some other fixed point 0' in its plane. 
In other words, the two polygons A ... E, a ... e must be funicular polygons of the 
lines A a, B6, &a, with respect to any two poles. But if this is the case, every pair 
of corresponding sides of the two funicular polygons must (§ 9) intersect on a straight 
line. It is easy to see that this involves only three geometrical conditions in the 
construction of Fig. 35. Draw any two pairs of corresponding sides at pleasure ; e. g. 
A a A 8 in the polygon A . . . E and a a, a 8 in the polygon a . . . e 9 the intersections 
a and 8 of these assumed pairs of sides determine a line X Y. Draw another side E D 
of the polygon A . . . E at pleasure, the line E D cuts X Y in c, then the side 
corresponding to E D in the polygon a . . .d must be drawn through a fixed point c. 
This constitutes one geometrical condition. Similarly assuming the two remaining 



Fig. 86 



Fig. 87. 



Fig. 87a. 



r 





lines B C and C D, their corresponding lines bc 9 cd must be 
drawn through fixed points y, /J respectively, thus making 
three geometrical conditions in all. 

By similar reasoning it can be shown that, in order 
that Fig. 36 may admit of a reciprocal, one geometrical 
condition must be satisfied. Fig. 36a is the reciprocal of 
Fig. 36. 
The construction of Fig. 37a, the reciprocal of Fig. 37, needs no explanation. 

f 2 



36 



GBAPHIO STATICS. 



28. Exceptional Figures. — Fig. 38 is formed of a figure A . . . E, through the 
vertices of which the lines A a, B /J, . . . are drawn, and the points a, /J, ... on these 
lines are joined, forming a polygon a .... c. Call / the original figure A . . . E and P 



Fio. 38. 



/ 
\ 




Fio. 38a. 



the whole figure made up of/, the lines A a, B /? . . . . and the polygon a . . • c. Let s 
be the number of sides and v the number of vertices in the original figure. Then if 

f=2t>-8-fr [a] 

the figure / will (§ 25) be deformable in r different ways if r is negative, strictly 
indeformable if r is zero, and it will have r surplus lines if r is positive. 

Suppose that through n of its vertices lines have been drawn, their extremities 
being joined by lines forming a closed polygon, then if d is the number of sides in the 
complete figure F, and t/ the number of its vertices, 

**= • + 2nand t>'= v + fk 



BECIPBOCAL FIGURES. 37 

Hence from equation [a] — 

* + 2ti = 2(* + »)-8 + r 
or f 

Thus the complete figure F will be deformable, strictly indeformable, or will have 
surplus lines, according as the original figure / is deformable, indeformable, or has 
surplus lines. 

Now, in Fig. 38, reciprocals of the vertices a . . . c must be triangles, and if these 
triangles are drawn commencing with the vertex a, the last triangle would have one 
side determined by the first triangle, and another side determined by the last triangle 
but one, its third side would therefore be fixed, and if this third side were not parallel 
to the corresponding side of the original figure the latter would not admit of a 
reciprocal. 

Hence, in order that a figure of the form shown in Fig. 38 may admit of a 
reciprocal, there is a condition with respect to the drawing of the lines A a, B /J . . . . 
and a /J, fi y . . . , which is perfectly independent of the form of the original figure 
A . . . E. Thus if the original figure A . . . E had one surplus line, it should (§ 26) be 
possible to construct a reciprocal to the complete figure ; but this will evidently be 
impossible unless the new condition above indicated is satisfied. 

To satisfy this condition it is necessary that the polygon a .... c should be the 

funicular polygon of the lines A a, B /3, Thus if the lines A a, B /3 are regarded 

as forces, the forces must either be in equilibrium, or one of them must be the resultant 
of the others. 

If this condition is fulfilled, a figure of the form of Fig. 38 will admit one reciprocal 
if the original figure A ... E is strictly indeformable ; none if it has any surplus lines ; 
while if it is deformable in r ways, it will not admit a reciprocal unless it satisfies r new 
conditions. 

The above is of importance in passing from the conception of purely geometrical 
figures to that of jointed polygonal frames with forces acting at their vertices, or joints. 

[There are several other known exceptions to the rule stated in § 26, and new 
ones doubtless remain to be discovered : the exception above noticed alone occurs in 
the frameworks used in roof and bridge construction,] 



38 GEAPHIC STATICS. 



CHAPTER IV. 

STRESS DIAGRAMS. 

29. Mechanical Property of Reciprocal Figures. — Suppose abco (Fig. 36) to 
be a polygonal frame formed of six bars jointed at the vertices a, b> c, o, and that at 
the extremities of each bar two equal and opposite forces are applied acting along the 
axis of that bar. Each pair of equal and opposite forces will be separately in equi- 
librium, and the whole frame will therefore be in equilibrium, whatever are the 
magnitudes of the equal and opposite pairs of forces. If the frame is in equilibrium, 
each separate vertex or joint must be in equilibrium, and the forces acting at that joint 
must therefore (§ 7) form a closed polygon. Hence there must be a figure such that 
to every vertex of the original figure abco there corresponds a triangle ; but if such 
a figure exists, it will (§ 24) be the reciprocal of a b c o. Hence the problem of deter- 
mining the stresses in the bars of such a figure resolves itself into the drawing of a 
reciprocal to the figure, and is possible or not in accordance with the conditions of 
possible reciprocity indicated in the preceding chapter. 

The present figure has one surplus line, since «=2v— 3 + 1, and a 1 V d <f 
(Fig. 36a) is its reciprocal. The sides of the new figure will be proportional to the 
stresses in the corresponding bars of the given figure, and conversely a system of pairs 
of equal forces severally proportional to the sides of the figure a 1 V d o f , and applied to 
the extremities of the corresponding bars of the original figure abco, would maintain 
the latter in equilibrium. 

30. Determination of the Stresses in the Bars of a Polygonal Frame. — 
ABODE (Fig. 38) is a polygonal frame acted upon by five exterior forces Pi, P 2 , 
P 3 , P 4 , P s applied at its several joints, and maintaining it in equilibrium. Whatever 
is the form of the frame, the exterior forces must satisfy the graphic conditions of 
equilibrium stated in § 8 ; that is to say, the polygon of forces and all funicular 
polygons of the system must close. Draw one of the funicular polygons a/3ySe of 
the system corresponding to any pole (Fig. 38a), and suppose that at every vertex 
of this funicular ♦polygon a force equal and opposite to the exterior force passing 
through that vertex is applied. Then each vertex of the funicular polygon will be 
separately in equilibrium, and hence to every vertex of the complete figure (Fig. 38) 
made up of the original frame and the funicular polygon, there must correspond a 
closed polygon in Fig. 38a. Thus, if three lines meet at any vertex in Fig. 38, there 
must be a corresponding triangle in Fig. 38a ; if four lines meet at any vertex in 
Fig. 38, there must be a corresponding quadrilateral in Fig. 38a, and so on. Of two 



STRESS DIAGRAMS. 39 

figures thus related, each is the reciprocal of the other, and the reciprocal of the figures 
made up of the bars of a polygonal frame and the lines of external forces acting at its 
joint* is termed the stress diagram of that frame. 

The sides of the stress diagram are proportional to the exterior forces and to the 
stresses or interior forces in the corresponding bars of the frame. 

Thus the conception of a jointed polygonal frame acted upon by exterior forces 
which produce stresses in its several bars is reducible to that of a purely geometrical 
linear figure conforming to the three conditions of § 25, and the statical problem of 
determining the stresses in the bars of a framework becomes the geometrical problem 
of determining the reciprocal of a given figure, and is possible, impossible, or indeter- 
minate, according as the construction of the reciprocal is possible, impossible, or 
indeterminate. 

Starting with the necessary hypothesis that the exterior forces Y x . . . P, (Fig. 38) 
maintain the given frame A ... E in equilibrium, then the complete figure admits of 
one reciprocal if the original figure is strictly indeformable 9 of an infinite number of 
reciprocals if the figure has surplus lines 9 while if the figure required n new lines to 
render it strictly indeformable, it would not admit a reciprocal unless it satisfied n 
geometrical conditions. 

The problem of determining the stresses in the bars of a given framework is 
therefore, strictly speaking, determinate only when the latter is strictly indeformable, 
and this is the case with most of the frameworks with which it is necessary to deal in 
practice. If the framework has surplus bars the distribution of stress is indeterminate, 
and in dealing with such a framework it will be necessary to assume that one of the 
possible modes of distribution is the actual one. On the other hand, if the framework 
is deformable, it must have some special form ; that is to say, there must be some 
special relations existing among the lengths of its bars, or the determination of the 
stresses is impossible. 

To proceed with the drawing of the reciprocal in the present case. The reciprocals 
of the vertices a ... c are already obtained, since they are the triangles, each made 
up of one of the sides of the polygon of forces and of the vectors drawn to from its 
extremities. To obtain the reciprocal of the vertex A, draw a line from b parallel to 
B A, and from a parallel to E A. These lines intersect in /, then a bf is the reciprocal 
of the vertex A. Passing to E, draw eg parallel to E D and/^r parallel to E B, thus 
<* e fff is the reciprocal of the vertex E. For the vertex B, draw c h parallel to C B 
and g h parallel to D B, thus the five-sided figure bchgf is the reciprocal of B, 

For the vertex D, join d h ; then d h should be parallel to C D, and d e g h is the 
reciprocal of D. The reciprocal of C is c d h 9 all the lines of which have been already 
drawn. 

It follows from the principle of the triangle and polygon of forces (§ 7) that the 



40 GBAPHIC STATICS. 

lines of the reciprocal figure give the stresses in the bars of the framework on the 
same scale as that adopted in drawing the polygon of forces. The lines of the 
reciprocal are figured similarly to the bars to which they correspond. The sides of 
the funicular polygon are similarly treated. 

The following points in connection with the above should be noticed :— 

1st. The polygon of forces or line of loads must first be drawn, and should close 
properly. In drawing this polygon, the forces should be taken one after the other 
right round the figure, as in Fig. 38. 

2nd. Having drawn the reciprocal of any vertex, it is always necessary to proceed 
next to that one of the adjacent vertices at which the least number of unknown forces 
meet. Thus the vertex E, and not B, is dealt with after A. If this rule is observed, 
there will never be more than two unknown forces at any vertex, provided that the 
framework is strictly indeformable. If, however, the framework had surplus lines, 
there would be more than two unknown forces acting at some of the vertices, and the 
distribution of the stresses at those vertices would be (§ 7), strictly speaking, 
indeterminate. 

3rd. It is evident that if the five forces P x . . . P 6 are given, arid if they are known 
to form a system in equilibrium, the reciprocals of the vertices A . . . E can be drawn, 
and the required stresses obtained without the use of the funicular polygon. It should 
be remembered, however, that the figure thus obtained is not the reciprocal of the 
framework, but is only a portion of the reciprocal of the complete figure made up of 
the framework, the lines of action of the forces, and a funicular polygon. 

4th. On arriving at the vertex D, the only unknown force is 6, and the line join- 
ing k and d (points previously obtained) must be parallel to 6. This closing up of the 
reciprocal figure serves as a verification of the accuracy of the preceding construction. 

[Note. — In future the skeleton drawing of a framework will be called the " frame 
diagram," and the reciprocal the " stress diagram/'] 

31. Distinction between Ties and Struts. — From the stress diagram it can 
always be ascertained whether the stress in any bar of a framework is a tension or a 
pressure ; i. e. whether the bar is a tie or a strut Thus in the stress diagram (Fig. 38a) 
the triangle a bf represents the three forces acting at the joint A, and these forces 
being in equilibrium, the arrows point the same way round the triangle (§ 7). The 
direction arrow of P x is known, and this therefore fixes the other direction arrows. 
Transferring these arrows to the corresponding bars of the frame diagram, it will be 
seen that the direction arrow of 1 points towards the joint A, and that of 2 points away 
from it. Hence A B is in pressure, and A E in tension. 

In the figures following, pressure bars are shown in strong lines, tension bars in 
fine lines. 



STRESS DIAGRAMS. 41 

32. Roof Trusses with Symmetrical Vertical Loading. Figs. 39, 40, PI. I. ; 

41, 42, PL II. — In the preceding case all the exterior forces were supposed to be 
given ; in practice it is required to deal with frameworks having known loads and 
resting on supports, the reactions of which must be treated as exterior forces forming 
with the known loads a system in equilibrium. In all stress diagrams, therefore, the 
reactions of the supports must first be determined, and then the polygon of the exterior 
forces (including those reactions) should be drawn. 

When the loading is vertical and symmetrical, the reactions will be vertical, and 
each reaction will be equal to half the total load. Moreover the polygon of forces 
becomes a straight line, the " line of loads." 

Fig. 39 is the frame diagram of a roof truss largely used for moderate spans up to 
about 60 feet. It consists of two rafters A C, C E, a polygonal tie A F G E, two struts 
B F, D G, and two braces C F, C G. The five upper joints ABODE are taken as the 
loading points, the purlins carrying the adjacent bays of roof being supposed to pass 
over the joints B and D. If R t and Ej are the respective reactions at A and B, then 



Rj = }&2 == 



2 w 1 + 2 w 2 + w z 



2 

Starting from a on the line of loads (Fig. 39a), set off ac = w l9 cd = w u de = «7 8 , 
ef = w 2 ,fb = i0 x , successively from a to 6, reading off their magnitudes in pounds or 
tons from any convenient scale. Bisect a b in o, then a o and o b represent the equal 
reactions R x and R 3 . 

From c and o draw eg and og respectively parallel to B A and FA, then the 
figure acgo a is the reciprocal of the joint A, and eg and og give the stresses in the 
bars 1 and 2. 

Proceeding next to the joint B (at which joint there are only two unknown forces), 
draw g h and d h respectively parallel to B F and C B. Then the figure cdhg\B the 
reciprocal of the joint B, and g k and d h give the stresses in the bars B F and B 0. 

Proceeding to the joint F, draw h t and o t parallel to F C and F G respectively, 
then the figure othgo is the reciprocal of the joint F, and ht and ot give the stresses 
in the bars F C and F G. 

Proceeding to the joint C, draw t i and e i parallel to G C and D C, then the figure 
dhtied is the reciprocal of the joint C, and the lines ti and e i give the stresses in the 
bars GC and DC. 

The whole of the stresses are now obtained, since- by the symmetry of the structure 
and of its load, the stress in D G is equal to that in B F, and so on. It will be best, 
however, to complete the figure, proceeding next to the joint D and then to G. The 
figure ti h o t is the reciprocal of the joint G, and in drawing this reciprocal the point h 
is fixed. If now the previous construction has been correctly carried out, the line 

o 



42 GRAPHIC STATICS. 

joining k and / should be parallel to D E. By § 31 it can be ascertained which of the 
bars composing the frame diagram are in pressure and which in tension. In Fig. 39 
and the succeeding figures the corresponding lines of the stress and frame diagrams are 
similarly numbered. 

Fig. 39b is the stress diagram of the roof shown in Fig. 39, the loading being the 
same as before, with the addition of two equal loads to if w+ suspended from the joints 

F and Or. Each reaction is now equal to w x + w % + w 4 + -£ • 

Starting from a, draw the polygon of the exterior forces (or load line), taking 
these forces in the following order: w l9 w %f m? 8 , w 2 , w l9 Ea, w? 4 , w? 4 , Rj. The reactions 
Rj and B a will overlap each other on the load line as shown, and the whole length a b 
of the load line will be unaltered by the addition of the two forces w A . 

The forces acting at the joint A are w lf 1, 2, and R. From c and s on the load 
line draw lines intersecting in g 9 and respectively parallel to B A and F A. Then the 
five-sided figure a eg s a is the reciprocal of the joint A, and the lengths eg, sg represent 
the stresses in the bars A B, FA, The completion of the stress diagram needs no 
explanation. 

Figs. 40, PI. I., and 41, PL II., are the frame diagrams representing two forms of 
truss frequently used for spans of from 60 to 100 feet. Figs. 40a, PL L, and 41a, 
PL II., are the stress diagrams corresponding to the loading indicated. The drawing 
of these stress diagrams should present no difficulty. 

Fig. 42, PL II., represents a modification of the truss shown in Fig. 39, which 
has been adopted for roofs up to 90 feet span, but which Professor Unwin recommends 
should be restricted to spans not exceeding 60 feet. In drawing the stress diagram 
(Fig. 42a, PL II.) a difficulty presents itself in dealing with the joint D. If the joints 
are taken in the order A, B, C, D, then on arriving at D there are three known forces 
w z , 4> 5, and three unknown forces 7, 9, 8 ; the distribution of stress in the three bars 

7, 9, 8 is therefore, strictly speaking, indeterminate. In fact it will be necessary to 
assume that the stresses 5 and 9 are equal,* so that 7 is equal to the resultant of the 

* The assumption that the stresses 5 and 9 (Fig. 42, PI. II.) may perhaps be deemed hardly 
satisfactory. A direct solution may be obtained without difficulty by the " method of sections." If a 
section be imagined by a plane perpendicular to the plane of the truss cutting not more than 3 bars, the 
stresses on which are unknown, the resultant of the external forces acting on one side of this section 
plane can be resolved (§ 10) in the 3 directions of these bars, and the resolved parts so obtained are the 
stresses in the respective bars. In the figure a vertical section plane may be taken a little to the left of 
the ridge cutting the bars 12, 13, and 14, and the total external force w x + to a + w, + w 4 — 1^ resolved 
along the 3 directions 12, 13, and 14, so determining the stresses in those bars. The stresses 8 and 11 
at the junction of 8, 11, 12, and w A can then be determined in the usual way, and when 8 is known (as 
also the stresses in the ban meeting at B and O) we shall have only two unknown stresses, viz. 7 and 9, 
acting at D. If a section plane is imagined passing a little to the right of D, and cutting the bars 6, 7, 

8, and 9, the stress acting in 6 {determined from the joint C) would have to be compounded with w k , w v 
w, , and Hi as the external force acting on one side of the section plane. 



STRESS DIAGRAMS. 



43 



equal tensions 5 and 9, and. of the component of w z resolved along D C. In the stress 
diagram Fig. 42a, de represents to Z9 and ef drawn at right angles to d9 gives the 
magnitude of the component of w z resolved in the direction of the bar D B. Draw g i 
parallel to the bar 9, and make g i equal to g A, since g h represents the stress 5. Then 
h i is the resultant of the two equal stresses 5 and 9 ; produce h i to l> making i I equal 
to ef 9 the whole line h I will therefore represent the resultant of 5 and 9 added to the 
component of w %y hence hi gives the magnitude of the stress 7 in the bar DE. 





i 


i 


1 


i 






Fig. 






X 


< » 


> 

• ^^ 

« .x 

\^X 

.x* 

x * 

x * 








\ / 

\/ 

A 

X 
* X 


% A 

% X 

• x 

» X 

% X 

%^x 

x» 

x » 

X * 

X « 

X * 


vs. 

«■ -- 
^ ' \ 1 


cY 


/2^k — - 






' 1 » 




Proceeding to the joint E, the four forces acting at this joint are 6, 7, 10, 14 : draw I r 
parallel to 10 and or parallel to 14, then the figure ohlr is the reciprocal of the 
joint E. The remaining construction will require no explanation. In Fig. 42a, 
PI. II., only half the stress diagram has been drawn, the remaining half will be 
symmetrical about the line o r. 

Fig. 44, PL III., shows another modification of the truss in Fig. 39, PI. I., which 

G 2 



44 GBAPHIC STATICS. 

can also be used for spans up to about 60 feet. Fig. 44a is the stress diagram for a 
symmetrical vertical loading applied at tbe upper joints. 

Pig. 46 sbows a bowstring roof, the form used for large spans. In roofs of this 
form the verticals are usually constructed as struts, and the diagonals as ties. If the 
loading is symmetrical, the single system of diagonals shown in continuous lines will 
be all in tension. If, however, the loading is unsymmetrical, as in the case of wind 
pressure, some of these diagonals would be placed in pressure, which pressure would 
be taken up as a tension by the diagonals shown in dotted lines. It would clearly be 
impossible to draw a stress diagram for the complete roof with its crossed diagonals, as 
the distribution of stress at every joint, except the end joints, is indeterminate. Before 
proceeding to draw the stress diagram for a bowstring roof it is necessary therefore 
to reduce the frame diagram to a single system of triangulation. Moreover, if the 
diagonals are to be all tension bars, then in deciding which system of triangulation to 
deal with, any bar which the stress diagram shows to be in pressure must be omitted, 
those only which are proved to be in tension being retained. 

Fig. 46a is the stress diagram for Fig. 46. It will be found that the segments of the 
arched rib are the only bars in pressure. 

33. Roof Trusses with Unsymmetrical Vertical Loading. — If the load is 
unsymmetrical, the reactions of the supports will be unequal, and the stresses in bars 
similarly situated with respect to the centre line of the truss will be unequal also. 
Suppose that the truss (Fig. 39, PI. I.) is loaded only by the loads w x% w 2j w z applied at 
the joints A, B, C respectively. To determine the reaction of the supports, set off 
w u w 29 w z successively from a to b on the line of loads (Fig. 39c), take any point as 
pole, and draw the funicular polygon afiyS with respect to ; from draw the line 
o parallel to the closing line a 8 of the funicular polygon, then (§ 14) a o, o b represent 
the magnitudes of the reactions R x , R, at the supports A and E respectively. In 
drawing this funicular polygon it should be noticed that since the lines of action of 
w x and Ex coincide, the side of the funicular polygon parallel to the ray a vanishes. 
The reciprocal can now be drawn, and it will be found, as might have been expected, 
that the strut G D is unstrained, and that the stresses 8 and 1 1 are equal. 

34. Pent Hoof. Vertical Load. — In this roof (Fig. 43, PI. II.) the exterior 
forces are the five loads w 1 . . . w s , and the forces supplied by the wall at the points of 
support A, B. The latter may be resolved into two forces P, Q acting in the direction of 
the bars which are fixed in the wall, and two vertical reactions r 2 and r x acting at A and 
B respectively. Set off the loads successively along a b (Fig. 43a), then a b represents 
their resultant B in magnitude, and since the loading has been taken symmetrical the 
line of action of this resultant passes through e, the centre point of B C. Join e A, and 



STRESS DIAGRAMS. 



45 



from a (Fig. 43a) draw a c horizontal to cut b c parallel to £ A in c. From c and b draw 
lines respectively parallel to A B, A C, and intersecting in d. Then (§ 1 1 ) a c, c d, d b 
represent the components of R resolved along, B 0, B A, A C. But r a , the reaction at 
A, is equal to w 6 . Make df equal to w 6 , then abdfc is the complete polygon of the 
exterior forces. The drawing of the stress diagram (Fig. 43a) presents no difficulty. 



Wind Pressure. 

35. Wind Pressure. — The maximum pressure of the wind in England has been 
variously taken at 40 and 50 lbs. per square foot of surface perpendicular to its direction. 
The usual direction of the wind is probably nearly horizontal, though it is possible 
that, in occasional gusts, this direction may make a considerable angle with the 
horizontal, becoming in fact normal to roof surfaces of high pitch. 

For determining P n , the normal pressure of the wind on any plane surface in 

terms of P, the pressure on a plane surface perpendicular to its direction, and i the 

angle of inclination of that direction to the plane of the surface, the following formula 

was deduced by Hutton from experiment : — 

P. = Pain.* 18 *"* 1 - 1 . 

Supposing the direction of the wind to be horizontal, i is equal to the pitch of the roof 
surface. 

The following table gives the value of P n in lbs. per square foot of surface for 
horizontal winds acting with forces of 40 and 50 lbs. per square foot of vertical surface 
exposed to them, on roofs of various pitch : — 



* 






Angle of Roof. 


P = 40. 

m 


P = 50. 


5° 


5-0 


6-3 


10° 


9-7 


12-1 


20° 


18-1 


22-6 


30° 


26-4 


33 


40° 


33-3 


41-6 


50° 


38-1 


47-6 


60° 


40-0 


50*0 


70° 


41-0 


51-3 


80° 


40-4 


50-5 


90° 


40-0 


500 



In order therefore to determine the total stresses in the bars of a roof truss, it 

will be necessary — 

1st. To draw a stress diagram for the dead or permanent load, i. e. the weight of 
the rafters, purlins, and roof covering,* together with the weight of the maximum snow- 

* The weights per unit of area of different kinds of roof covering are given in the Appendix. 



46 GEAPHIO STATICS. 

fall. The latter may be taken in England at from 5 to 6 lbs. pet square foot of the 
plan of the area covered. Inasmuch as the maximum wind pressure is never likely to 
act simultaneously with the maximum snowfall, it is probable that, in England at all 
events, it will be quite sufficient to provide for the former only. 

2nd. Assuming the direction of the wind to be horizontal, to draw either (a) a 
single diagram, corresponding to the normal pressure taken from the above table, or 
(b) to resolve the normal pressure into a horizontal and vertical component, and to 
draw a stress diagram for each component separately. Now if the direction of the wind 
is supposed normal to the roof surface, it is clear that the stresses due to this normal 
wind can be read off from the diagram drawn for the normal pressure of a horizontal 
wind by merely varying the scale. Similarly a diagram giving the stresses due to the 
horizontal component of a horizontal wind can be used to give the stresses due to the 
horizontal component of a normal wind. 

3rd. To tabulate the stresses due to the dead or permanent load and to the wind 
pressure, distinguishing between tensions and pressures ; and, finally, to form a table 
of total stress, by taking the sum or difference of the stresses due to the dead load and 
wind pressure, according as the stresses are of the same or of opposite kinds. 

36. Stress Diagrams for Normal Wind Pressure. — In Fig. 44, PI. III., the 
rafter AD is 1£*4', and the pitch is 30°. Hence, if the trusses are 10' 0" apart, 
and each truss is supposed to support half the adjacent bay, the total normal wind 
pressure supported by AD is 19 • 4 x 10 x 26*4 lbs. If the purlins carrying the roof 
are supposed to be supported at B and C, the wind pressure p x = p A = ■$■ x total 
wind pressure; and p 2 = jt? 3 = £ x total wind pressure; whence/?! = 0*38 tons, and 
p % = 0*76 tons. If the rafter AD is considered as a beam uniformly loaded, and 
supported at A, B, C, B, the distribution of the pressure at the latter points will be 
somewhat different. 

It is necessary first to determine the reactions Q x Q a at A and A 1 . In Fig. 44b, 
PL III., set off the loads j^ = 0-38 tons; p 2 = # 76 tons; jt? s = 0*76 tons; p A = 0'38 
tons, along the load line a b. Take any convenient polo O, and draw the funicular 
polygon of the forces/?!, Pi,Ps 9 pi 9 with respect to O ; then O o drawn parallel to the 
closing side a/3 of the funicular polygon determines the two reactions QiQ 3 . Draw 
o c parallel to the bar 1, and d c parallel to 2. Then the polygon obdeia the reciprocal 
of the joint A, and oc 9 dc represent the stresses due to wind pressure in the bars 1 and 2 
respectively. Complete the stress diagram, and it will be found that the bars 11 and 13 
are unstrained, and consequently the stresses in 10, 12, and 14 are equal. . 

If, as is the case in all large iron roofs, one end of the truss rests on rollers, so as 
to allow of expansion due to variations of temperature, then, since only one abutment 
can furnish the horizontal component of the reaction, it will be evidently necessary to 



STRESS DIAGRAMS. 47 

draw two diagrams for wind pressure, since the stresses in the bars forming the trass 
will be different when the wind acts on different sides of the roof. 

Suppose, in Fig. 44, PL III., that the rollers are placed at the end A 1 , then this 
abutment can furnish only the vertical component of its reaction, and the whole 
horizontal component of both the reactions Q x and Qa must be borne at A. In 
Fig. 44c, PL IV., draw the load line a b> and set off along it the four loads and the 
two reactions. Through o draw a horizontal line, and terminate it on perpendiculars 
dropped from a and b ; then ca is the vertical component of Qa and b d of Q M while cd 
represents the sum of the horizontal components of Q x and Q 2 . Now the forces acting 
at A are the load p l9 the vertical component of Q l9 the horizontal components of Q, and 
Q a , and the stresses in the bars 1 and 2. From c draw ce parallel to the bar 1, and 
from g draw g e parallel to 2. Then the five-sided polygon gbdceia the reciprocal of 
the joint A (Fig. 44, PL III.), and the lines ce, eg represent the stresses in the bars 
1 and 2. Completing the stress diagram, the triangle c af will be the reciprocal of 
the joint A 1 , since at this joint there act only the bar stresses 14, 15, and the vertical 
component (ca) of Q a . It will be seen that the bars 11 and 13 are unstrained, and 
hence the stresses in 10, 12, 14 are equal. 

Now suppose the rollers to be at A, the wind coming from the same side. The 
polygon of forces abdc (Fig. 44d, PL IV.) will be precisely the same as before, but 
the horizontal component cd will now act at A 1 instead of at A Hence at A there is 
only the load jp u the vertical component (bd) of Q u and the stresses 1 and 2. Draw 
de, ge parallel to 1 and 2 respectively, then the four-sided figure g b d e is the reciprocal 
of the joint A (Fig. 44), and the sides de,ge represent the stresses in the bars 1 and 2 
respectively. Completing the stress diagram, the triangle cad iq the reciprocal of the 
joint A 1 , and the bars 8, 9, 11, 13, 15 are unstrained. These bars are, however, brought 
into play as soon as the stress in D A 1 ceases to be a simple pressure, and D A 1 begins 
to bulge, 

37. Diagrams of Horizontal and Vertical Components of Wind Pressure. — 
To draw the horizontal component diagram, obtain first the horizontal components 
Ai» Aj> A 8 > A 4 of the forces pi,p%,ps,pi (Fig. 44, PI. III.), then H (Fig. 45, PI. IV.) 
equal to the sum of the horizontal components h x + h^ + A3 + A 4 will be the magnitude 
of their resultant, and by symmetry H will act at 8, the middle point of the rafter A D. 
Suppose the truss fixed at the support A 1 , and furnished with rollers at A, then in 
order that equilibrium may be maintained, the support A, though unable to furnish a 
horizontal reaction, must be supposed capable of supplying a vertical downwards- 
acting force v sufficient to prevent the truss from rotating about A 1 . Since R, H, and 
this vertical force v are the only forces acting on the frame, their directions must (§ 8) 
pass through a point. Draw a vertical through A, cutting the line of action of H in t. 



48 GRAPHIC STATICS. 

then the line of action of R must pass through t. For the stress diagram (Fig. 45a) 
draw the triangle bdc of the exterior forces, making ba equal to H, then ae, ca 
represent v and R respectively. Set off h ly A a , A3, A 4 successively from a along ab. 
The four forces acting at A are h u v, and the stresses in 1 and 2. Draw ed,cd parallel 
to 2 and 1, then ed, cd represent the stresses 2 and 1 respectively. Proceed to the 
joint B, and complete the diagram ; as before, the struts of the right half-truss are 
unstrained. 

The construction of the stress diagram, when the truss is fixed at A and free to 
move at A 1 , presents no difficulty. In this case the support A must furnish the 
oblique reaction R, while A 1 supplies a vertical wpirard-acting reaction equal 
to v. 

The vertical component diagram can be drawn in the same way as Fig. 39c, PI. I., 
since it is merely the case of unsymmetrical vertical loading. Only one vertical com- 
ponent diagram need be drawn, since the wind blowing from the other side will merely 
cause the stresses in symmetrically situated bars to be interchanged. 

The stresses obtained from the dead load and wind pressure diagrams must now 
be tabulated and added with due regard to their sign. The maximum stress in each 
bar is thus ascertained. 

38. Bowstring Roof with Normal Wind Pressure. Figs. 47, 47a, 47b. — The 
load at any joint due to wind can be taken as equal to the pressure which the wind 
would exert upon an area equal to the portion of roof surface supported by one bay of 
the rafter, and inclined at the same angle to the horizontal as the tangent to the curve 
of the rib at that joint. Supposing the direction of the wind to be horizontal, it will 
be necessary to take from the table, page 45, the normal pressure corresponding to the 
several inclinations of these tangents, and to multiply them by the area of roof surface 
supported by one bay of the rib, or in the case of the end joint by half this area. 
The pressures p l9 p 2f p z , p A , p 6 (Fig. 47) are thus obtained, and their resultant R 
will, if the curve of a rib is a circle, pass through its centre C. If B is the roller 
end of the roof truss, the entire horizontal component of this resultant must be 
balanced by a horizontal reaction at A, while the vertical component of the reaction 
will be borne partly at A and partly at B. Draw the polygon of the forces p t . . . p B 
(Fig. 47a), then the closing line fa of this polygon gives the magnitude and 
direction of R. Through A and B draw lines parallel to the direction of R. Take 
any pole 0, and resolve R by means of the funicular polygon a, fi, y (§ 13) into the 
two parallel components a 0, of (Fig. 47a) acting at B and A respectively ; draw 
a horizontal line g h through 0, terminated by perpendiculars from a and /. Then 
the forces acting at A will be p l9 the reaction H (=^A), V (=/A), and the 
stresses 1 and 2. From g and e draw gk f eh parallel to 1 and 2 respectively ; then 



1 






STfiESS DIAGBAMS. 



49 



Fxo. 47. 




Fig. 47a 



h 






50 



GEAPHIO STATICS. 



the figure efhgk is the reciprocal of the joint A, and g k, ek represent the stresses 
in bars 1 and 2. Complete the stress diagram in the usual way. The diagonals are 
all in tension, the verticals are all in pressure except those nearest the springing on 
each side. 

If A is the roller end of the roof truss, the horizontal component ghof the wind 
pressure must be borne entirely at B. The forces acting at A will bep l9 V, and the 
stresses 1 and 2 ; draw hk } ek (Fig. 47b) parallel to the bars 1 and 2, then the figure 
efhk is the reciprocal of the joint A, and hk, ek represent the stresses in 1 and 2 
respectively. Complete the diagram, and it will be seen that a very considerable 
alteration in the stresses has taken place. The segment of the tie nearest to B is in 
pressure, and the segment of the rib terminating at the same point in tension. If these 
abnormal stresses are not counterbalanced by the stresses due to dead load, the fact of 
their possible occurrence must be taken into consideration in the design of the roof. 

39. Warren Girder. — Fig. 48 represents the frame diagram of a Warren girder 
loaded with equal loads at its upper joints. Fig. 48a represents half the stress 



Fig. 48. 




diagram ; the stresses in symmetrically situated bars are equal. Fig. 48b is the stress 
diagram for the same girder when loaded with a single load/? at one of the lower joints t. 
The magnitudes Bx and B a of the reactions of the left and right support are obtained 
by a funicular polygon in the usual way. It should be noticed that with this loading 
the diagonals 7 and 9 are both in tension, while the diagonals on either side are 
alternately in pressure and tension. Moreover, the stresses in all the diagonals to 
the right and left of the joint t are respectively equal, and these stresses are directly 






STRESS DIAGRAMS. 



51 



proportional to the reactions at B and A, and inversely proportional to B t 9 A t. The 
stresses in the segments of the boom increase in arithmetic series from the abutments 
to the loaded joint. If p is applied on the left half girder, 13 is a tension and 15 a 
pressure : the reverse is the case if p acts on the right half girder. 

40. Bollman Girder. — In this girder (Fig. 49) the span is divided into equal 
bays by equally long verticals, the lower ends of each of which are tied back to the 
extremities of the boom. The stress diagram (Fig. 49a) of this girder is drawn for 
equal loads w l9 to %9 w z applied at the upper ends of the verticals. The drawing of 

Fig. 49 



• 
















#* > 


' 




r 


a >r 


* 






II 




9 




• 




^ \ 



Fig. 49a. 




this diagram brings out one peculiarity worthy of notice. For the purposes of the 
reciprocal the loads may be considered to be applied at the lower sides of the verticals, 
and the pressures 15, 9, 3 in the latter will be respectively equal to w l9 w 29 w z . Now 
it will be found impossible to construct a stress diagram which closes properly unless 
the points a, b 9 c (Fig. 49), at which the ties cross, are considered as joints. Thus, for 
the purposes of the stress diagram, the tie B must be considered as made up in three 
segments, figured 14, 18, 11, and A D in two lengths 13 and 10. In the stress diagram 
the reciprocals of the joints a, b 9 c form parallelograms, as shown in Fig. 49a. In the 

H 2 



52 



GBAPHIC STATICS. 



present case it would not, of course, be necessary to draw the complete stress diagram 
in order to obtain the stresses in all the bars of the frame diagram, Fig. 49. The 
reciprocal triangles of the joints C, D, E give the stresses in all the ties, and the stress 
in the boom is equal to the horizontal component of the resultant of the tensions of all 
ties meeting at either A or B. If, however, in such case as this it is desired to draw 
the complete stress diagram, it must be remembered that such intersections as a, 6, c 
(Fig. 49) must be treated as joints, otherwise a diagram which will close up properly 
cannot be obtained. 

The method of reciprocal diagrams is not in general adapted to girders subject to 
travelling load, as the construction of the number of diagrams required in investigating 
the stresses for different positions of the load would be a tedious operation. The 
method, however, is sometimes useful for obtaining the stresses in the booms, since the 
latter suffer their maximum stress when the girder is fully loaded, and it may often be 
employed with advantage in ascertaining the stresses due to dead load ; i.e to the 
weigHt of the structure itself and of the roadway. 



Fig. 50. 



CHAPTER V. 

ACTION OF STATIONARY LOADS. 

Beams fixed at one end. 

41. Single Concentrated Load. — By a beam is understood a single continuous 
bar, such that the axis, i. e. the line joining the centres of all transverse sections, is 
straight ; and further, that there is a plane of symmetry containing the axis. The 

loads are supposed to act in this plane of symmetry with 
directions normal to the axis. Such a bar A B (Fig. 50) 
is fixed at A, and supports a load W at the free end B. 
Considering any section X, suppose two opposite forces S' 
(acting up) and S (acting down) equal and parallel to W 
to be applied at X. Of these forces W and S' form a couple 
whose arm is x 9 and S is unbalanced. The moment W . x of 
the couple is termed the bending moment at the section X, 
and is resisted by the moment of an opposite stress couple, 
termed the moment of resistance, acting in the interior of the 
beam. The force S is termed the shearing force at the section X, and is met by the 
equal and opposite shearing stress, or resistance of the section to shearing. 




ACTION OP STATIONARY LOADS. 



53 



Shearing Stress. — The shearing stress at any section of a loaded beam is therefore 
equal and opposite to the resultant of all the exterior forces acting on one side of that 
section. The shearing stresses at all sections of the beam A B (Fig. 50) are equal to 
W. Hence the distribution of shearing stress will be graphically represented by a 
rectangle A' B' h s y of which the length is equal to that of the beam and the height is 
equal to W, set off from any convenient scale. The action of S' ( = W) at any section X 
tends to cause the part X B to the right of that section to move down relatively to the 
remaining portion X A of the beam, the shearing stress S, or molecular resistance to 
shearing at X, must therefore act upwards. Shearing stress at any section which resists 
a tendency of the portion of the beam to the right of that section to move down under the 
action of the exterior forces will be termed negative. Thus if the load W were exchanged 
for an upward thrust, the shearing stress at any section of the beam A B would be 
termed positive. Figs. 51 and 51a show these two cases. Shearing is supposed to 




Fig. 51. 



Fig. 52. 



Shearing stress 





Fig. 51a 




>hendbuj numunt 



Fig. 52a. 




Shearing stress 





-henduuj mmnjsnt 




have begun to take place at the section X. In Fig. 51 the shearing stress from A to X 
is negative : in Fig. 51a it is positive. If W acts at any point between A and B (Fig. 50), 
the shearing stress at any section to the right of that point is nil if the weight of the 
beam itself is neglected ; while for sections to the left it is negative, and equal to W. 
In accordance with the above convention with regard to the signs of shearing stresses, 
loads will be considered as positive forces, upward-acting reactions as negative. 

Bending Moments. — The bending moment at any section of a beam is equal to the 
moment of all the exterior forces acting on one side of that section. The bending 
moment at X (Fig. 50) is equal to W. x and becomes zero, and W. I when x = o and I 
respectively. Draw the right-angled triangle abm, making a b equal to I, and a m 
equal to W. Z, then the ordinate {yy) of this triangle under any section X gives the 
bending moment at X on the same scale as that adopted in setting off a m. The 
triangle has been drawn right-angled, but this is obviously non-essential, any triangle 
which has its base equal to a m and height equal to I will equally represent the bending 



54 



GRAPHIC STATICS, 



Fig. 53. 



moments. In accordance with the convention adopted in § 17, the bending moment for 
the case shown in Fig. 52 is termed positive, and for that shown in Fig. 52a negative. 

42. Several Concentrated Loads. — Shearing Stress. — The beam A B (Fig. 53) 
supports those loads W u W„ W 8 . The shearing stress for any section lying between 
A and the line of action of W is W x + W % + W 8 . Between W x and W a the shearing 

stress is W a + W 8 , and bet ween W a and W 8 it is W 8 . Hence 
the distribution of shearing stress may be represented by a 
figure made up of rectangles whose breadths are equal to 
W x + W a + W 8 ; W, + W 8 ; and W 8 ; and whose lengths 
are equal to that of the segments into which the beam is 
divided by the points of application of the loads. The 
shearing stress is all negative. If the end of the beam 
projects beyond W 8 , the shearing stress in the projecting 
portion will be zero, if the weight of the beam is left out of account. 

Bending Moment. — The bending moment diagram for each separate load can be 
drawn as in the preceding paragraph, then the sum of the three ordinates under any 
section is the ordinate of the figure required. Another construction will be given later. 

43. Uniformly Distributed Load.— Shearing Stress.— The beam A B (Fig. 54) 
supports a load w per unit of length. The shearing stress at any section X is negative, 
and equal to w.x, becoming zero at B, and w.l at A. The distribution of shearing 
stress will therefore be represented by a triangle she 9 whose height s h is equal to J, 
and base s e is equal to to J set off from any convenient scale. 




Fig. 55. 



Fio. 51 





X 



Bending Moment. — The bending moment at X is w.x. 5, and becomes zero at B, 

in fl 

an( j _j_. a t A. The bending moment at any section is therefore given by the ordinate 



ACTION OP STATIONARY LOADS. 



55 



of a parabola b d (Fig. 54) whose vertex is at b, and whose ordinate a d at a is equal to 

to. I 2 

— jr- , and which touches ab sib. If the scales are so arranged that a d is not greater 

than j, a circular arc will sufficiently approximate to the curve. 

If the beam A B (Fig. 55) is loaded with a uniformly distributed load from B to 
C only, the distribution of shearing stress is evidently represented by a rectangle 
under the segment A C, and a triangle under C B, the rectangle and triangle having a 
common side s h equal to w . x. The bending moments under this condition of loading 
are given by the ordinates of a straight line under A C and a parabola under C B, the 



parabola and straight line having a common ordinate tp equal to 

w.x 



w.a? 



, while the 



extreme ordinate ar of the straight line is equal to -y- (2 /— x). The representation 

of the distribution of shearing stress and bending moment under other conditions of 
loading can be obtained by combining the results above obtained. 

Beams supported at both ends. 

44. Single Concentrated Load. — Shearing Stress. — The beam A B (Fig. 56) 
supports a single concentrated load W acting at G. The only exterior force acting to 

Fig. 56 





i 
i 


+ 


' 





m» 



the right of the section C is — Rj, the reaction of the abutment B. Hence the shearing 
stress at D, or any other section between C and B, is positive and equal to B I# 
Similarly the shearing stress from A to C is negative and equal to - (W - 1^). (Taking 



56 GRAPHIC STATICS. 

moments about A ; Rx . A B = W . A C, hence the shearing stress from C to B is equal 
to W,— r-, and from A to B is— W.— ^-.) 

Draw the load line a b representing W on any convenient scale, and draw the 
funicular polygon a/3y with respect to any pole ; thus (§ 14) s parallel to y a , the 
closing line of this polygon, divides a b into two parts as 9 sb 9 representing R x and R 
respectively. 

The graphic representation of shearing stress will be two rectangles, one of which 
is vertically under the segment A C and the other under C B, the heights of these 
rectangles being respectively equal to s b and a s. 

It will be noticed that in the figure as drawn, a horizontal line pm drawn 
through s separates the positive and negative portions of the shearing stress diagram. 
The positive shearing stress is shown above the line p ra, and the negative below. 

Bending Moment. — In Fig. 56 the only exterior force acting to the right of the 
section D is R x . Hence the bending moment at D is R t . D B ; or, Rx . x. Let fall a 
vertical from D, cutting the two sides a y , /3 y of the funicular polygon in e and /. 
Since the triangle efyia similar to the triangle a 8, 

ef x 

at H ' 

where H is the polar distance (§ 15). Similarly the bending moment at any section is 
equal to the product of the ordinate (measured parallel to the load) of the funicular 
polygon under that section into the polar distance. The bending moment at any 
section of the beam is, therefore, proportional to the ordinate of the funicular polygon 
vertically below it ; and, if the polar distance is taken as unity on the linear scale, 
these ordinates read off from the scale of loads will give the bending moments at all 
sections. 

45. Any Number of Concentrated Loads. — Shearing Stress. — The beam A B 
(Fig. 57) supports four concentrated loads W x . .W 4 acting at P, E, D, 0, respectively. 
On a load line a b set off a c = W 2 ; c d = W a , and so on. Draw the funicular polygon 
1 .... 6 of the loads with respect to a pole O ; then (§ 14) S parallel to the closing 
line 16 of this polygon divides ab into aS, S6, representing Rj and R, respectively. 

The shearing stress at X is equal to R x '— W x = aS — ac = Sc. Similarly the 
shearing stress at any section between E and D is R! — ( W 2 + W a ) = S tf, and generally 
the shearing stress for the several segments of the beam is given by the distances from 
S of the divisions on the load line. 

The whole figure representing the distribution of shearing stress is made up of 
rectangles obtained by drawing vertical lines through the points of application of the 
loads, and horizontal lines through the divisions on the load line. The horizontal 



ACTION OP STATIONARY LOADS. 



67 



through S separates positive and negative shearing stress, and the change of sign takes 
place at the section D. 

It follows from the above that, if the loading is symmetrical and symmetrically 
applied with respect to the centre of the beam, no load being applied at the centre, 
there will be a segment at the centre, in which the shearing stress is nil. 




Bending Moments. — The bending moment at X is equal to the moment 
of the resultant of B t and W x about X. The magnitude of this resultant is 
— Ri + W x = — Sa + ac = — 8c: it is therefore negative, and its line of action 
passes through i, the intersection of the sides 16, 45 of the funicular polygon pro- 
duced (§ 12). The bending moment atX is therefore equal to — Sc.fr, when i r is 

I 



58 GEAPHIO STATICS. 

the perpendicular from i on the vertical through X. The triangle mni is similar to 
S c 0. Hence 

g — =gS orron. H = Se.tr, 

where H is the polar distance (§ 15). 

Hence, as in the preceding section, the bending moment at X is equal to the 
product of the ordinate m n of the funicular polygon under X, into the polar distance. 
If H has been taken as unity on the linear scale, then m n read off from the scale of 
loads gives the bending moment at X, and similarly for all other sections of the beam. 
In Figs. 56 and 57, H, if not unity, should be taken at some convenient whole number 
from the linear scale, then the ordinates of the funicular polygon read off from a 
properly constructed scale (§ 17) will give the bending moments. 

If the equal similarly loaded beam A B (Fig. 58) is fixed at A, the reaction Ej at 
A will be equal to W x + W a + W 8 + W 4 . Using the same pole and polygon of forces, 
the funicular polygon takes the form shown, and its two last sides intersect in t, a 
point on the line of action of the resultant P of the four loads (§ 12). Produce 5 i to 
meet the vertical through A in r, then the moment of P about A is P . m t, and, since 
the triangles lir and aOb are similar, P.rai=lr.H. Hence lr represents the 
bending moment at A, and similarly the ordinate of the funicular polygon under any 
other section gives the value of the bending moment at that section. 

46. Uniformly Distributed Load. — Shearing Stress. — The beam A B (Fig. 59) 
of length I is subjected to a load w per foot run, the whole load is therefore equal to 

to . l 9 and the reactions are each equal to ^- . The exterior forces acting to the 

right of a section X, distant x from A are — R i9 and a load equal to w(l—x). 
Hence the shearing force at X is 

-^ + »(J -«)=£(! -2*). 
and the shearing stress is — ^ (Z — 2 x). 

This equation shows that the ordinates which represent the shearing stress at the 
several sections of the beam are those of a straight line. Putting x = o ; or x = l 9 the 

shearing stresses at the abutments A and B are ^- and + -s- respectively. 

Putting x =s - , the shearing stress at the centre of the beam is 0. 

to I 
Make r s 9 r' s' each equal to —- , and join s s' ; then the whole figure srr' s 9 

represents the distribution of shearing stress, and the ordinate of this figure under any 



ACTION OP STATIONABY LOADS. 



59 



section of the beam gives the shearing stress at that section. The shearing stress is 
positive for the right half of the beam, and negative for the left half. 

Bending Moments. — The bending moment at X is equal to — TO* . x + w .x . ■$ 

The bending moments at the several sections of the beam A B are therefore 
represented by the ordinates of a parabola. The uniformly distributed load may in 



Fig. 59. 




—.—.».... 0.. 




fact be regarded as an infinite number of equal concentrated loads infinitely near 
together, and the parabola is the curve which the funicular polygon of such a system 

of loads becomes. Putting x = 0, x = Z, and x = -5-, it is evident that the bending 

moment at either abutment is 0, and at the centre of the beam — — ^- . The parabola 

in question is of the form A'v B' (Fig. 59a); the vertex is under the centre of the 
beam, and the axis is at right angles to the beam. 

Draw the load line a b representing w. I (the total load), bisect a b in S, and take 
any pole on a line from S at right angles to a b. Draw A' t! parallel to b and t f B' 
parallel to a, then A' 1 B' is the funicular polygon of the total load, and if w . I acts 

1 2 



60 GEAPHIO STATICS. 

as a concentrated load at the centre of the beam, the bending moment at the centre of 
the beam is (§ 44) f *\ H' 

The triangle A' 1! t 9 b S are similar. 
Hence *_' 

At " &o ; or »* f = 2 'H~4.H 

The bending moment at the centre of the beam, supposing to. I to act there as a 

concentrated load, and H to be taken as unity, is therefore — ^-, or half the bending 

moment at the centre, when w . I is uniformly distributed. Hence a, the vertex of the 
parabola, bisects 1 1\ 

Any number of tangents to the parabolic funicular curve can readily be drawn : 
thus, to obtain a tangent at a point vertically below the section X (Fig. 59), set off b c 
on the load line equal to w . x. (This will amount to dividing a b in the same propor- 
tion as B A is divided at X, since the load distribution is uniform.) 

The bending moment at X will not be altered by considering two concentrated 
loads w.x and w(l—x) to act at the middle points of AX and XB respectively. 
Join c 0. Thus yy', the ordinate below X of the funicular polygon 1 2 3 4 of these 
concentrated loads, gives (§ 45) the bending moment due to them at X : y y f is also 
an ordinate of the parabolic funicular curve at y 7 , and 2 3 is a tangent to this curve 
at y'. Any number of tangents and their points of contact can similarly be obtained ; 
and following out the above principles, the following simple construction is arrived at. 

Bisect A' B' (Fig. 59a) equal and parallel to the beam by a perpendicular, and 
draw A' **, B' i parallel to the rays 6, a. Divide A' i and B' f into any number of 
equal parts, 1 2 3 . . . . , 1' 2 ; 3' . . . Join 1 1' 2 2', &c, as shown in the figure ; the lines 
obtained are tangents to the parabola ; and moreover, the point of contact of any 
tangent bisects the distance between the points in which adjacent tangents cut that 
tangent. Thus r, obtained by drawing a vertical line up through 4, is the point of 
contact of 2 2'. 

A parabola of which the vertex, axis, and one point are known, can also be drawn 
as follows. From V, the vertex (Fig. 60), draw a line V T perpendicular to the axis 

V F. From B, the given point on the curve, draw B T parallel to the axis. Divide 

V T and B T into the same number of equal parts. Join each of the divisions of B T 
with the vertex, and cut the lines obtained by lines drawn parallel to the axis through 
the divisions of V T, as shown in the figure. To determine the focus, draw B X 
perpendicular to the axis ; make V t = V X. Join B t, then B t is the tangent at B, 
and a line bisecting B t at right angles determines the focus F. 

The following is a very convenient way of drawing the parabola approximately 
by arcs of circles if the vertex and focus are known. F a and V (Fig. 61) are the focus 



ACTION OF STATIONARY LOADS. 



61 



and vertex respectively of the required parabola. Draw any lines 1 2 3 ... at right 
angles to the axis. It will be more convenient to divide V F a into equal parts, and 
draw the lines 12 3... through the parts of division, continuing these lines at the 
same intervals. Make OiY= 2 . VF a , and with o x as centre and ^V as radius, describe 
a circular arc as far as a, a point approximately on a line bisecting V 1 and perpen- 
dicular to the axis. With 2 as centre, describe an arc with radius 2 . V F 2 cutting the 
axis in o 2 ; with o 2 as centre and radius to a describe the arc a b as far as b 9 about half- 
way between the lines 1 and 2. With 3 as centre, describe an arc of radius 2 . V F a 
cutting the axis in o 8 ; with o z as centre and radius to b, describe the arc be, and 
continue the process till the curve has been drawn as far as required. The symmetrical 



Fig. 61. 



Fig. GO. 








portion of the parabola on the other side of the axis is, of course, drawn at the same 
time. By drawing a line g h at starting parallel to and at a distance from the axis 
equal to 2 .V F a , the construction is shortened, as the constant radius 2 . V F a employed 
in obtaining the centres o % , 8 , Ac, is then always ready to hand. If (as in the figure) 
the distance V F a is in the first instance divided into equal parts, and all the other 
ordinates drawn at equal intervals, the centres o v o 2 o S9 &c, require no further deter- 
mination. If the lines 12 3.. are not taken too far apart, the curve will approximate 
very closely indeed to the parabola. In the figure a second parabola with the same 
vertex V and ¥ x as focus has been similarly drawn.* 

In practice the parabola may be represented by a circular arc if t v, the ordinate 
of the vertex (Fig. 59 a), is not greater than |th of the span ; but it is frequently necessary 

* This construction is due to Prole. 



62 



GRAPHIC STATICS. 



to draw parabolas of greater eccentricity, in which case one of the two constructions 
given above should be employed. The parabola having been drawn, the ordinate 
under any section gives the bending moment at that section. 

The case of a uniformly distributed load partially covering a beam will be treated 
under the head of travelling loads. 

47. Load with any Fixed Distribution. — If the load has any distribution other 
than uniform, it will be necessary to represent it by an area (termed the loading area) 
bounded by a curve, the ordinate at any point of whicb represents the load per unit of 
length at that particular point. This area is then cut up into vertical strips, and the 
load represented by each strip is dealt with as if it were replaced by an equal concen- 
trated load acting through the centre of gravity of the strip. Thus if the ordinates 
A A', cc' 9 dd/j &c. (Fig. 62), represent the respective loads per unit of length at the 

Fig. 62, 





equidistant sections A, c, d . • . of the beam A B, the whole area ABB'A' will represent 
the load distribution. The strips are approximately trapeziums whose centres of gravity 
g lf g 2 ... may be taken to lie in the* centre lines, aTd whose weights t*, + „. *re 
approximately proportional to their mean heights. 

Draw a b representing the total load on the beam, and divide a ft in the proportion 
of the mean heights of the six trapeziums. Join the points so obtained to any pole 0, 

and draw the funicular polygon 12 8. The sides of this polygon are tangents to 

the funicular curve at the points 1, c", d" . . . g'\ 8, the funicular curve can therefore be 
drawn, and its ordinates give the bending moment at all sections. 

The shearing stresses, if the beam is considered as loaded by the concentrated 



ACTION OF STATIONARY LOADS. 



63 



loads w u w 2 . . . acting through ffi,ff i9 ... will be given (§ 45) by the ordinates of a 
figure made up of rectangles, and these stresses will for the sections c, d, e ... be 
evidently unaffected by the hypothesis of concentrated loads in place of continuous 
loading. Hence the ordinates of a curve v tf drawn through points on the sides of the 
respective rectangles vertically under c, d, e . . . (i. e. the middle points of these sides) 
will give the shearing stress at any section. 

As in the previous figures, the horizontal through S separates positive and 
negative shearing stress, and shows when the change of sign takes place. 

By cutting the loading area up into a larger number of strips a closer approxima- 
tion to the actual conditions of loading is of course obtained. The curve A' B is termed 
the extradoa of the load. 



Fro. 63. 




48. Forces in one Plane but not Parallel. — Consider first the effect on a 
section X (Fig. 63) of a single force P acting in the plane of symmetry of a beam A B, 
and at an angle to the axis. The line of action of P cuts the axis in s, and the plane 
of the section in t. Suppose P to act at s 9 and to be replaced by its vertical and 
horizontal components. The former P sin. gives rise 
(§ 41) to a couple whose moment is a . P sin. 0, and 
to an unbalanced force P sin. acting in the plane 
of the section, and causing an equal and opposite 
shearing stress on that section. The horizontal com* 
ponent P cos. acts as a direct stress, or pressure on 
the section. It is the same if P is supposed to act 
at t ; the horizontal component gives rise to a couple 
P whose moment is d . P cos. 0, and an unbalanced force P cos. causing direct stress on 
the section, while the vertical component P sin. causes an equal and opposite shearing 
stress on the section X. Since d = a tan. ; a P sin. = d . P cos. 0. 

We may, therefore, deal only with the vertical components of the forces p x . . p A 
(Fig. 64), and determine the bending moments due to these components acting as 
vertical loads at C . . F, leaving the direct stresses in the several segments A C, C D . . . . 
to be separately obtained. 

Draw a vertical a r through a, and horizontals through 6, c 9 d, e, cutting it in 
1 2 3 r. Thus a 1 , 12, 2 3, 3 r are the vertical components of the forces, and a r is that 
of their resultant. Draw a funicular polygon of the forces a 1, 12, 2 3, 3 r, with 
respect to any pole ; the shearing stresses and bending moments are then obtained 
as in § 45. 

Draw the funicular polygon of the vertical components, then s parallel to a /3 
determines the magnitudes as, sr of the vertical components of B 19 B, respectively. 
Drop verticals b V, c2', dr from 6, c, d, on er 9 then rl' f V 2', 2' 3', &e are the 



64 



GRAPHIC STATICS. 



horizontal components of the four forces pup 29 p*,Pi% and these components together 
with the components of the reactions form a system in equilibrium. The direct stress 
on any section is equal to the resultant of all the horizontal components acting on 
either side of that section, and is a pressure, or a tension, according as this resultant 
acts towards, or away from the section. Thus the stress in the segment AG is 
a pressure, and equal to eg (the horizontal component of the reaction at A); in CD 

FiQ. 64. 




it is a tension, and equal to e 3'— eg = g 3' ; in D E it is a pressure, and equal to 
2' 3'-^ 3'= 2'g ; in E F a tension, and equal to 2' 1'- 2'g = gV; and finally in F B 
it is a pressure, and equal to ^ 1' — r 1' = <j r = sd, the horizontal component of the 
reaction at B. The distribution of direct stress in the beam can therefore be repre- 
sented by a figure made up of rectangles as shown, the depth of the figure under any 
section representing the stress on that section. Pressures and tensions are shown on 
opposite sides of the line m n : the sum of the former is of course equal to the sum of 
the latter. 

49. Beam resting on three Supports forming two equal Spans, and 
Uniformly Loaded.— Shearing Stress. — The beam A B of length 2 I (Fig. 65) rests 
on three supports at the same level, and sustains a uniformly distributed load w per 
foot run. By means of the " Theorem of three moments " it can be shown that the reac- 

O ft Q Q 

tions of the supports are ^wJ; -;U>.1\ q*0.J; hence making al = c3 = ^w.l and 

o 4 o 8 

b2 = b2' = -w.l, and joining 1 2, 2' 3, the diagram of shearing stress is obtained. 

o 



SIMPLE BEAK— TRAVELLING LOAD. 



65 



Bending Moment. — For any section distant x from A the bending moment is 

3 x /3 I x\ 3 

^ mj . / — 10 . a? a = w . # ( -£ ■" s )• Putting x = 0, or x = - 1, the bending moment 

vanishes ; putting x = I, the bending moment becomes ^— . The curve of bending 

o 
Fig. 65. 




moments for the span A B is a parabola passing through d % d and e, where 

3 w P 

a'd = -l and V e = —5— , and the vertex of this parabola is at a horizontal distance 

- / from at* For the span B the curve is of course an equal symmetrically situated 

o 

parabola passing through e and d. 



CHAPTER VI. 

SIMPLE BEAM.— TBAVELLING LOAD. 

50, Single Concentrated Load. — Shearing Stress. — A beam must be sufficiently 
strong to resist the maximum shearing stresses and bending moments which can arise 
under any possible disposition of the given load. Hence, in dealing with a travelling 
load it is the maximum stresses and moments which have to be ascertained. 

The beam AB (Fig. 66) sustains a load W acting at a point D to the right of a 

DB 



section X. The shearing stress at X is therefore (§ 44) equal to— W . 



/ 



Hence as 



66 



GRAPHIC STATICS. 



D B increases, or as W moves towards X, the shearing stress increases and is negative. 

AD 

If D is between A and X, the shearing stress at X is equal to + W. — p . Hence as 

W approaches X from the other abutment A, the shearing stress increases and is 
positive. The shearing stress at X is therefore a maximum when W is at X, and is 
positive or negative according as W approaches X from the right or left abutment. 

Hence, if W moves from B to A, the successive maximum shearing stresses at the 
several sections of the beam are represented by the ordinates of a triangle rr f h, in 
which r* h is equal to W, the shearing stress in this case being all negative. If W 




moves from A to B, the maximum shearing stresses are all positive, and are repre- 
sented by the ordinates of an equal triangle r r 9 h 9 . 

Bending Moment. — Draw the funicular polygon 12 3 (Fig. 66) of the load W with 
respect to any pole 0. Thus (§ 44) the bending moment at X, when W is at D, is 
y' y . H. If W moves to the position D', the funicular polygon takes the form 1 2' 3', 
and the bending moment at X becomes y" y. H. Hence evidently the bending 
moment at X increases as W moves towards X, and is a maximum when W arrives 
at X. 

The maximum bending moment (M) at X is equal to Rx . x y and 



B^ to. 



I — m 

I 



SIMPLE BEAM,— TEAVELLING LOAD. 67 

W 
Hence M = -y . x (J — x ). The curve whose ordinates would represent the 

V 

maximum moments at all sections of the beam is therefore a parabola A' v B', and 

I W / 

putting x = ~ , it is evident that the vertex t; of this parabola is distant -5- H (or 

W I 

— 7^- if H is taken as unity) from A' B'. 

51. Any Number of Concentrated Loads. — Shearing Stress. — If any load 
additional to W (Pig. 66) is applied between B and X, the shearing stress at X is 
evidently increased. If on the other hand the new load (P) is applied at a point D' 
between A and X, the shearing stress is diminished, provided that P. AD' is not 
greater than 2.W.DR If P . A D' = W . A B, the shearing stress at any section 
between D' and D is zero. As the product P. AD' increases, the shearing stress at 
X increases, but changes sign. If P . A D' = 2W.DB, the shearing stress at X 

has the same value ( W • —r~ ) as before the addition of the new load P, but its sign 

has changed. 

Hence, generally, if two or more equal or nearly equal loads preserving constant 
intervals move over a beam A B, the maximum shearing stress at any section X will 
arise when the leading load reaches X moving from the support farthest from X. In 
the case of a railway bridge, therefore, the maximum shearing stress at any section 
will usually occur when the longer segment is covered by the train, and the leading 
axle has reached that section. 

A beam A B (Fig. 67, PL V.) of 45' 0" span is traversed by a train of passenger 
engines and tenders. The loads on the three engine-axles are 9, 15, and 7 tons; 
those on the three tender-axles are each 7 tons. The axle intervals are indicated. 
Supposing the train to move from right to left, the shearing stress at any section will 
be a maximum when either the leading or driving axle of the leading engine reaches 
the section. From a set off the loads W u W a . . . successively along the load line, 
making a b equal to W u b c equal to W a , and so on. Draw the funicular polygon of 
the loads with respect to the pole (Fig. 67a, PL VI.). (The polar distance H is 
made equal to 45 tons taken from the scale of loads.) It will be necessary for the 

funicular polygon to embrace the loads W x W 8 , as some of the maxima shearing 

stresses (those for a length of 8' 0" from A) will evidently arise after W x has moved 
off the beam. To obtain the maximum shearing stress when W x arrives at a section X 
distant 15' 0" from A, take any point r on the line of action of W lf draw r I horizontal 
and equal to 15' 0", produce r I to k, making Ik equal to AB. Drop verticals from 
I and kj cutting the funicular polygon in v and t respectively. Join v t 9 and draw the 

k 4 2 



68 GEAPHIC STATICS. 

vector 8 (Fig* 67a) parallel to t v, then (§ 44) the shearing stress at the section X 
is a 8 when W t arrives at that section. To obtain the shearing stress when W, reaches 
X, take r* on the line of action of W a , make i* V equal to 15' 0", and V if equal to 
45' 0". Drop verticals from V and if cutting the funicular polygon in tf and i, and 
draw the vector ^ (Fig. 67a) parallel to if vf. Then b ef (Fig. 67a) represents (§ 44) 
the shearing stress when W, arrives at X. This shearing stress is less than that 
arising when W 1 reaches X. Proceeding in this way, the maximum shearing stress 
at any section can he obtained. In Fig 67b, PI. VI., the maxima shearing stresses 
for sections 1' 6" apart have been plotted, and the curve of maximum shearing stress 
has been drawn. This curve has cusps, which occur when each new load comes on 
the bridge at B, and also when the leading load W l goes off the bridge at A. The 
position of these cusps is indicated in Fig. 67b by dotted ordinates. The maxima 
shearing stresses will usually occur when W 1 arrives at any section ; but for a portion 
of the beam A' C (Fig. 67b) equal to the interval between W t and W, (8' 0") measured 
from A, the maxima stresses will be obtained by treating W 2 as the leading load. 
For some small distance to the right of C, W a arriving at a section will also be found 
to produce the maximum shearing stress. 

For every negative maximum shearing stress there will of course be a corre- 
sponding positive maximum shearing stress at a symmetrically situated section when 
the same train traverses the beam from left to right. 

Bending Moments. — Referring to Fig. 66, it will be evident that any additional 
load applied on either side of the section X will increase the ordinate of the funicular 
polygon under that section, and therefore the bending moment. Moreover the 
maximum ordinate of any funicular polygon must evidently be at one of its angles. 
Hence, generally, the maximum moment at any section will occur when the heaviest 
load reaches that section, and the beam on both sides is as fully loaded as possible. 
It has been shown (§ 50) that when a single concentrated load traverses a beam, the 
curve of maximum moments is a parabola. If two or more loads preserving a fixed 
interval traverse the beam, the curve becomes much more complex ; its ordinates may, 
however, be obtained by adding the ordinates representing the maximum moments due 
to the separate loads.* Thus, if A t; B (Fig. 68) is the curve of maximum moments 
due to W lf and A tf B that due to W„ then when W x arrives at X, yyf will (§ 50) 
represent the bending moment at X due to W x ; but when Wi is at X, W, will be at X', 
and the bending moment at X due to W a at X' will be yy M . Hence the total bending 
moment at X due to W x at X and W a at X' will be represented by y a = yy 1 + yy". 
In 'this way ordinates representing the maximum moment at any section of the beam 
can be obtained, and a curve of maximum bending moment can be drawn. 



• « 



Wrought Iron Bridges and Roofs. 9 — Unwih, 



SIMPLE BEAM.— TRAVELLING LOAD. 



69 



Fin. 68. 



This curve can also be obtained by a method analogous to that above described for 
obtaining the curve of maximum shearing stress. Referring to Pig. 67, PI. V., the 
maximum bending moments at any section X will occur when either W x or W a is at 
X, and the beam on both sides of X is as fully loaded as possible. In order, therefore, 
to obtain all possible conditions of loading, it will be necessary to extend the train by 
taking in a new load W I0 corresponding to the trailing axle of the tender of an engine 
in front of W,. The funicular polygon as altered to embrace this new load will take 
the form 1' 2' 2 3 4 ... . Similarly new loads W 9 and W u may have to be introduced 
at the other end of the beam, and the funicular polygon correspondingly extended. 

To ascertain the maximum bending moment at a section X distant 15 ; 0" from A, 
draw r / equal to A X and I k equal to A B. Let fall verticals from I and k cutting the 
funicular polygon in m and t. Join m t ; then y 2, the intercepted ordinate under r, 
represents the bending moment at X when W x is at X. To ascertain whether the 
bending moment at X is greater 
when W a is at X, draw / 1', C k J 
as before. Let fall verticals /' i/, 
V i from V and k! 9 and join v f t'. 
Then %f 3 is the bending mo- 
ment at X when W a is at X ; 
and since y 1 3 is greater than 
y 2, the maximum moment at 
X arises when W a is at X. By 
precisely similar construction y" 
8 is the bending moment when 
W T is at X', and y"' 9 when W a 
is at X'. Hence y ,n 9, which 
is greater than j/' 8, is the maximum bending moment at X'. In this last construc- 
tion W t and W 8 have been used in place of W 1 and W„ as otherwise the condition of 
maximum loading on each side of X' would not have been satisfied. 

The maxima moments for sections V 6" apart have been thus obtained and plotted 
in Fig. 67c. The resulting curve is the curve of maximum bending moment when the 
given train traverses the bridge from right to left. This curve is not necessarily 
symmetrical ; hence, for trains running both ways, the maximum bending moment will be 
represented by the longer of the ordinates under any two symmetrically situated sections. 

It should be noticed that the principle involved in the above construction is tho 
moving of the beam under the loads. The funicular polygon having been once drawn 
for all the loads, this movement of the beam gives a new position of its closing line, 
and on the position of the closing line of the funicular polygons, both shearing stresses 
and bending moments depend. 




i 



70 



GRAPHIC STATICS. 



52. Uniformly Distributed Ijoa.&.~-Shearing Stress. — The beam A B (Fig. 69) 
of length I supports a uniformly distributed travelling load of to per foot run. If the 
load comes on the beam at B, the shearing stress at a section X distant x from A will 
be a maximum when the length X B is covered by the load, and will not be altered if 

a concentrated load equal to w.x acting at 
C, the middle point of X B, is supposed to 
replace the uniformly distributed load. 
The shearing stress at X is therefore 

w.x — Kj = R| 

and Bt_AC2Z-» 

Hi 



Fig. 69. 






A* 




Hence 



CB 



B * = 21 * * 



The curve of maximum shearing stress is therefore a parabola. Putting x = o 

and x = I the maximum shearing stress at B and A are o and — ^— respectively. If 

to I I 

the scale of loads is such that — ^- is represented by a length a d not greater than -» 

the circular arc through b and d having its centre vertically below b will approximate 
sufficiently closely to the parabola. If otherwise one of the methods given in § 46 
may be employed. The shearing stress when the load moves from right to left is all 
negative. If the load moves from left to right, the shearing stress is all positive and 
may be represented by an equal parabola a d', having its vertex at a and axis on d a 
produced. 

Bending Moments. — The maximum bending moment at any section will occur 
when the whole beam is fully loaded. The curve of maximum bending moment is 
therefore a parabola (§ 46). 

Shearing Stress^ obtained from curve of Bending Moments. — It is sometimes 
convenient to obtain the maximum shearing stress at any section without drawing the 

Fig. 70. 





curve of maximum shearing stress. The beam A B (Pig. 70) is gradually loaded by a 
travelling load of w per foot rim, which moves from B to A. Draw the load line 



SIMPLE BEAM.— TRAVELLING LOAD. 71 

ab and bisect ao by a perpendicular H, making H equal to any convenient whole 
number taken from the scale of loads. Draw At, Jit parallel to a 0, b, then At, Bt 
are tangents to the funicular curve of the whole load at A and B, and v the middle 
point of m t is its vertex (§ 46). The maximum shearing stress at a section X will 
occur when the load covers the longer segment B X. Bisect A X, X B by verticals 
cutting A t and B i in p and g. Then p q is a tangent to the funicular curve of the 
whole load at x vertically under X. In order to cut away the portion of the total load 
covering the segment A X, produce qp to cut the vertical through A in 1 ; join 1 B 
and draw 1' parallel to 1 B. Set off from a along the load line, a c equal to the load 

on the segment AX I— r = y^I- Then e 1' is the maximum shearing stress at X. 

In the same way the maximum shearing stress at any other section can be obtained. 
The bending moment at X when the load reaches that section is y x . H. 

53. Combined Stationary and Travelling: Load. — A railway bridge girder is 
Bubject to stress due to its own weight together with that of the platform and rails 
which constitutes its stationary or permanent load, and is sometimes assumed to 
be uniformly distributed. It is further subjected to stress due to passing trains which 
constitute its travelling load, and this load is almost always assumed to be uniformly 
distributed, an hypothesis which in bridges of considerable span introduces very 
Blight error. 

Sharing Stress. — By combining the diagrams of shearing stress for a stationary 
and travelling load, the actual total shearing stress is obtained. 

TbuB, in the case of a uniformly distributed load moving from right to left over a 
girder which also sustains a uniformly distributed dead load, the diagrams of shearing 
stress are (§52 and § 46) a parabola Fia 71 

bp and two triangles a a c and b d' c 
(Fig. 71). The triangle asc represents 
negative shearing stress, thus in order 
that the ordinates of the triangles and 
parabola may be conveniently summed, 
the position of the negative triangle 
should be reversed with respect to ab. 
The ordinates of the hatched figure 

spobd then give the total shearing stress. The shearing stress changes sign at X 
and is negative from a to X, and positive from X to o. If the figure is drawn for a 
uniformly distributed load moving from left to right, the change of Bign will take place 
at a section X' symmetrically situated with respect to the centre of the girder. Hence 
in a railway girder there is a portion X X' at the centre in which the shearing 



72 GRAPHIC STATICS. 

stress is positive or negative according as the load comes on the bridge from the left or 
right abutment. It is this peculiarity in the distribution of shearing stress which 
necessitates counter-bracing near the centre of a girder, inasmuch as the bracing bars 
may here be called upon to act as struts and ties alternately. 

Similarly, by combining the shearing stress diagrams for a system of concentrated 
stationary loads (§ 45) with that for a system of concentrated travelling loads (§ 51), 
or with that for a uniformly distributed travelling load (§ 52), the maximum total 
shearing stress at any section due to these respective conditions of loading can be 
obtained. 

Bending Moment. — The maximum total bending moment at any section will be 
simply the sum of the ordinates of the funicular polygons or curves of the stationary 
and travelling loads, provided of course that these polygons or curves have been 
drawn with the same assumed polar distance. By summing these ordinates a curve of 
maximum total bending moment can be drawn if required. 

54. Curve of Total Stress in Booms. — It is sometimes desirable in dealing with 
girders to make use of a curve the ordinates of which represent the total stress at any 
section of either boom. This curve of total stress will, for a girder of uniform depth, 
evidently be of the same form as that of total bending moment, since the total stress in 
the booms at any section is equal to the total bending moment at that section divided 
by the depth of the girder. The one curve can therefore be used in place of the other 
by a mere adjustment of scales. If the depth varies — as in the case, for example, 
of a " hog-backed " girder — the curve of total stress must be deduced from the curve 
of bending moment. 



CHAPTER VII. 

BBAOED GIRDEBS. 

55. General Considerations. — In a perfect braced stracture each member is 
subject only to direct tension or pressure, so that the total stress is the same for 
any cross section of a given member, and its intensity is uniform over that section. 
In dealing with braced structures, therefore, it is usual to assume that there is a joint- 
pin at all the meeting points of the bars, and that the axis of this pin passes through 
the intersection of the axes of the bars. The travelling load and the weight of 
the platform are usually carried by cross girders, attached at the joints of one of the 
booms ; the girder is therefore supposed to be loaded only at these joints. The weight 
of the girder itself is also supposed to act at the joints of the booms, and if this weight 
is considerable, it should be supposed to be equally divided among the joints of both 



BRACED GIKDEBS. 73 

booms. In small girders the whole load may be supposed to be borne on the same set 
of joints. The proportion of load to be borne on each joint will therefore depend on 
the distance apart of the cross girders. 

56. Nature of the Loads on a Railway Girder. — It is usual to reduce the 
concentrated loads to an equivalent uniformly distributed load. On short spans this 
average load will be greater than on large spans, and the stresses in the former case 
will be greater in consequence of the concentration of the actual loads at points whose 
distances apart are small, relatively to the span. A higher average should therefore be 
taken for small span girders. The following may be adopted, viz. 2 tons per foot run 
of each line of rail for 25' 0" span ; If ton for 30' 0" ; 1£ ton for 40' 0" ; and 1£ 
for 60' 0" and upwards. The platform* is either carried on cross girders at short 
intervals which receive the load from timber sleepers, or the cross girders receive their 
load from longitudinal girders under each railway bar, whose length is a multiple of 
the length of the bay of the loaded boom of the main girders. If the former, the load 
on the cross girder should be taken to be that on the most heavily loaded axle of the 
heaviest locomotive (about 15 tons), multiplied by the number of lines of rail supported 
by a pair of main girders. This load will be concentrated at points whose distances 
apart will depend upon the gauge. If the latter, the load on the longitudinal girders 
may be taken as 2£ tons per foot run of each line of rail, the loads on the cross girders 
will then be equal to the total load on the longitudinal girders, and will be con- 
centrated at the points of attachment of the latter. The curves of shearing stress and 
bending moment for longitudinal girders may therefore be drawn by § 52. No general 
method can be given for the determination of the stress in girders subject to travelling 
loads, but it may be noticed that the stress diagram can usually be employed to 
determine the stresses in the booms which attain a maximum when the girder is fully 
loaded. The stress diagram can also be employed to determine the stresses due to the 
dead load, these stresses being afterwards added to the stresses due to the travelling 
loads. When only two bars meet at the end joints of the girder, the stresses in these 
bars can be obtained by a direct resolution along their directions of the reaction of the 
abutment when the girder is fully loaded. 

57. Warren Girder. — Load on one set of Joints. — Fig. 72 represents the frame 
diagram of a Warren girder of 50' 0" span, made up of five 10' 0" bays. The 
travelling load is taken at 1£ ton per foot run, the weight of the girder, platform, &c, 
at 1 ton per foot run. The whole of the load is supposed to be borne on the joints of 
the lower boom, and there is a cross girder at each joint. The bridge is supposed to 
consist of two girders carrying a double line of rail. Thus, supposing each joint to 
sustain a portion of the load extending as far as the centres of the adjacent bays, 

* For weight of platform, <&o., see Appendix. 

L 



74 GBAPHIO STATICS. 

there will be 10 tons of dead load and 15 tons of travelling load at each of the four 

bottom joints. 

Stresses in Bracing Bars. — It can easily be shown that the stress in any 

bracing bar is a maximum when the shearing stress in the bay to which the bar 

belongs is a maximum, or (§ 51) when the bridge is fully loaded on one side of that 

bay. Since, in the present case, only the bottom joints are loaded, the shearing 

stress will be uniform over each of the several bays A 2, 2 4, &c., of the bottom 

boom. Thus, any pair of bracing bars 2 3, 3 4 will be in a condition of maximum 

stress when the joints 4, 6, and 8 are fully loaded.* Similarly the bars A 1, 1 2 

will be in a condition of maximum stress when the joints 2, 4, 6, 8 are fully 

loaded. In thus considering each joint to be suddenly loaded as the train moves 

over the bridge, a supposition is made which, if not in accordance with the actual 

condition of things, is in favour of the bridge, since obviously as each point 

becomes loaded, the joint immediately in front of it will take up some of its load. 

Since the maximum shearing stress is uniform over each bay, it is evident that the 

distribution of maximum shearing stress during the passage of the travelling load from 

B to A will not be represented by a curve as in § 52, but a stepped polygon, the 

depth of which will vary for each bay. To draw this polygon make a v equal to the 

travelling load on one joint (15 tons), join vb and let fall verticals through all the 

joints 2 4 » • of the lower boom, the vertical through 8 cutting a b in c and v b in d. 

cb 
Then cc* = av . —.or ccf represents the shearing stress from 8 to A due to 15 tons 

at 8. Draw a horizontal through c' to cut the vertical through 6 in d, the same vertical 
cutting ab in e and vb in e*. Now e d is the shearing stress from 6 to A due to 15 tons 
at 6, make dd! equal to ed \ then e d! represents the maximum shearing stress in the 
bay 6 4. Draw a horizontal through d 1 to cut the next vertical and proceed in the 
same way till the stepped outline is complete. 

This outline can also be obtained from the parabolic curve of maximum 
shearing stress due to a uniformly distributed travelling load (§ 52). Make ap equal 

to the half-total travelling load over the whole bridge (-_- = 37£ tons], and by mean 

of the method of § 46 draw the parabola bp. If then verticals are dropped from the 
middle points of the bays of the lower boom to cut the parabola in s x s % s^ then 
horizontals through s x s 2 s 3j <fec, terminated by verticals through A 2 4, &c, give the 
stepped figure as before. 

In order now to combine the shearing stress due to dead load with that due to 

* While the load passes from 4 to 2 the stresses in the bars 2 3, 3 4 are increased by the loading up 
of 4 and decreased by the loading up of 2. These stresses, therefore, will never ezoeed those obtained by 
assuming 4 fully loaded and 2 without any load. 



BRACED GIRDERS. 



75 



travelling load, make ar^br* equal to half the total dead load on the girder, and join 
r r*. Then (§ 46 and § 53) the ordinates of the figure contained between the line r r' 
and the stepped outline beef dd\ &c, represent the maximum total shearing stresses at 
all sections of the girder : thus the ordinate s 2 1 2 represents the maximum total shearing 
stress at a vertical section through 3. Draw # a Aa, 8 2 g 2 respectively parallel to 32, 34, and 



Fig. 72. 




Fig. 72a. 



Linear ScaU>*f- 10' 0* 



9 • # 

IHHriHK 



toads flfcVfP Vmu 



termmated by a horizontal through * 2 , then 8 2 A a , s 2 g 2 represent the maximum stresses 
in the bars 23, 34 ; the former a pressure, the latter a tension. Similarly the stresses 
in all the bars can be obtained, but it will not be necessary to carry on the construction 
beyond 6, the point at which (§ 53) the shearing stress changes sign. Starting from 
the other abutment B, the stress determined for the bar A 1 must be written against 
the corresponding bar B 9 as its maximum stress when the travelling load comes on 

l 2 



76 GRAPHIC STATIOa 

the girder at A, and similarly for the other corresponding bars of the right half 
girder. It may therefore happen that at the centre of the girder there are bars which 
have both a tension and a pressure written against them, showing that in these bars 
the nature of the maximum stress depends upon the direction in which the load moves. 
Such bars must be designed to resist both kinds of stress. 

Stresses in Booms.— Since the stress in both booms will be a maximum when the 
whole girder is fully loaded, a stress diagram can be drawn by the methods of 
Chapter IV., the sides of which, in accordance with the principle of reciprocal figures, 
will give the maximum stresses in the several segments of the booms. Such a diagram 
is shown in Fig. 48a, p. 48, the only difference being that the upper joints only a^ 
supposed to be loaded, instead of the lower joints as in the present case. This method^ 
will in some cases be found to be the quickest, while in others it is more convenient to 
obtain the maximum stresses in the booms from the curve of maximum bending moment. 
In a girder loaded at the joints only, the distribution of maximum bending moment 
will be represented by a polygon instead of a curve, as in § 52. This polygon can 
be obtained from the curve of maximum moments due to a uniformly distributed load 
(§ 52), but it will be quicker to draw the funicular polygon at once. Thus, on a load J 
line /n, (Fig. 72a) set off the loads* on half the girder, draw n at right angles to Zn, ! 
and make n O equal to any convenient whole number taken from the scale of loads ; \ 
draw the funicular polygon a' afiy. . (Fig. 72b) of the loads with respect to the pole ; 
0. Now the bending moment on a vertical section of the girder through the joint 4 
is equal to m fi . n (§ 45) : taking moments about 4, it is evident that the stress in 
the segment 35 of the upper boom multiplied by A, the depth of the girder must be 
equal to the bending moment at the joint 4. Hence the stress in 35 is equal to 

-t— . n . , and can be readily calculated or obtained by construction as follows : — 

Make n q equal to A, join q 0, make nf equal to m /J, and draw/t parallel to q ; then 

n t= -j- . nO = maximum stress in the segment 35 of the upper boom. Make mf?\ 

equal to n t ; then, since h and n are constant, it is only necessary to make — , = -^3 . 

in order to obtain k a', the stress in the segment 13 of the upper boom. By thus, 
multiplying all the ordinates of the angles of the funicular polygon by the constant; . 

-T- , a new polygon aa'fif*/ is obtained. This is the polygon of total stress (§ 54) for . 

both booms, and the ordinate of this polygon under the middle point of any segment: 
of either boom gives the stress in that segment, all the stresses in the upper boom 
being pressures and in the lower boom tensions. The above construction need only 
be carried out for half the girder, as symmetrically situated segments of the booms will \ 

• The loads are plotted to half the given scale in order to keep down the size of the funicular polygon. 



BRACED GIRDERS. 77 

undergo similar stresses. It may be noticed that in the present cases the stresses in 
35, 67, and 46 are equal. 

The stresses in the booms may also be determined from the diagram of total 
shearing stress. Join/? to f„ the middle point of a b (Fig. 72), and produce p t z top' ; 
then the ordinates of the figure made up of the two triangles rp t z and r 1 // t z will 
represent the shearing stresses at all sections of the girder when the latter is completely 
loaded with 1 ton per foot dead load and li ton per foot travelling load. Dealing 
with the ordinate of this new figure under the joint 3 precisely as before, fresh values 
will be obtained for the stresses in the bars 32 and 34 The horizontal components of 
these stresses will be given directly by the figure, just as g 2 t % A, are the horizontal 
components of the stresses in 34, 32 when only the joints 46 and 8 are loaded with 
the travelling load. Similarly the horizontal components of the new stresses in all the 
bracing bars of the left half-girder can be obtained. Now the stress in the segment of 
the boom A 2 is equal to the horizontal component of the stress in A 1, and the stress 
in 12 is equal to the sum of the horizontal components of the stresses in 12 and A 1, 
and generally the stress in any segment of either boom is equal to the sum of the 
horizontal components of the stresses in all the bracing bars betweeu that segment and 
the nearest abutment. 

The total maxima stresses in all the bars having been obtained and carefully 
tabulated, the dimensions of the bars must be calculated. The weights of the latter can 
now be ascertained, and the stresses should be redetermined, taking the weight of the 
girder itself into consideration. 

58. Warren Girders Loaded at all Joints. — Taking the weight of the girder 
(Fig. 73) at 0*4 ton, the weight of the platform, platform girders, rails. &c, at 
• 6 ton per foot, and supposing one-half the weight of the former to be borne on the 
upper joints 13 5 7 9, there will be — 

Tons. 
Dead load on eaoh upper joint •• •• •• 2 

„ „ lower ff 8 

the travelling load remaining as before, 15 tons on each lower joint 

Stresses in Bracing Bars. — The effect of this transference of a portion of the load 
to the upper joints will be that the shearing stress will no longer be uniform over each 
of the bays A 2, 24, &a, but will now be uniform only under each half bay or under 
each bracing bar. Draw the stepped outline s L s 2 s 3 . . .b representing the distribution 
of maximum shearing stress due to live load. This outline will be, of course, precisely 
the same as that in Fig. 72. Make a d (Fig. 73) equal to half the total dead load, and 
set down successively from d along d a the loads at 1, 2, 3, 4 . * . . Then by drawing 
horizontals through the divisions on d a thus obtained, a new stepped outline d e ef d! is 
arrived at, the ordinates of which give the shearing stress due to dead load (§ 45). 



78 



GRAPHIC STATICS. 



Thus the ordinates of the figure comprised between the two stepped outlines give the 
maximum total shearing stresses at all seqtions of the girder, and d 8 X is the maximum 



Fig, 73. 




Loads jfc* fbAn* 

9 w » » yw y y y » w « 



total shearing stress in the left half of the bay A 2 (Fig. 73;. Draw s x t x parallel to 
the bar A 1 terminated by the horizontal d t u then s x t x is the maximum stress in A 1. 
Similarly *, t % and s s t z are the stresses in the bars 12, 23 respectively* 



j 



BRACED GIBDEBS. 79 

Strtss in Booms. — On the load line In (Fig. 73b) setoff successively the loads on 
the lower joints of half the girder, and from n on In produced set off the loads on the 
upper joints of half the girder ; thus the first load set off from n will be half the load 
at 5, or 1 ton. Take a pole 0, such that n is at right angles to I n, and equal to a 
convenient whole number taken from the scale of loads. Draw the two funicular 
polygons a 2'4' . . . and a V 3' ... as shown in Fig. 73a ; the polygon a' 3' 5' repre- 
senting the distribution of bending moment on the upper and a 2' 4' 6' on the lower 
boom. The ordinates of the figure comprised between these two polygons will 
represent the bending moments over the whole girder. By multiplying the ordinates 

n 

of this figure under each joint by -7- as before, and setting down from a! V (Fig. 73b) 

the lengths obtained, the points 1", 2", 3" . . . are arrived at. By joining a' 2", 2" 4", 
4" 6" the polygon of total stress of the lower boom would be obtained, and similarly 
1", 3", 6" is the polygon of total stress for the upper boom. Then the maximum stress 
in any segment of either boom is given by the ordinate under the middle point of that 
segment of the polygon of total stress of the boom to which it belongs. 

Substitute in Fig. 73 for the outline s l s 2 s 3 . . ., &c, a new outline (shown in 
dotted lines) representing the distribution of shearing stress when the whole of the 
bridge is covered by the travelling load ; then new values for the stresses in the bracing 
bars will be obtained, and by summing the horizontal components of these stresses as 
described above the stress in any segment of either boom can be arrived at. 

The stresses in the booms can also be obtained by an ordinary stress diagram. 

If in a girder of this kind the travelling load is transferred from longitudinal 
girders to cross girders attached say at alternate joints of the lower boom, this will 
modify the outline of the figure representing the shearing stress due to travelling load, 
as the shearing stress due to travelling load will be uniform between the points of 
application of the latter, i. e. between alternate joints instead of adjacent joints, as in 
the case dealt with above. 

59. Lattice, or Trellis Girder. — Fig. 74 represents a lattice girder of 50' 0" 
span. The bracing consists of four systems of triangles. The angle of inclination of 
the bars is 45°. The travelling load is taken at 1 ton per foot, dead load due to 
platform, rails, &c., 0*5 ton per foot, dead load of girder # 4 ton per foot. The 
latter is supposed to be carried half on the lower and half on the upper joints. The 
booms are divided at the joints into 12 bays of 5' 0". The loads are therefore as 

follows : — Tons. 

Dead load on upper joints 1 

„ lower „ 3*5 

Travelling load on lower joints .. . 5'0 

Total « - .... 8-5 



80 GRAPHIC STATICS. 

Stresses in Bracing Bars. — It will be necessary to deal with each system of 
triangulation separately. Selecting the system 1 2 3 4 5 6 7 8, set off the dead loads on 

the joints 2 7 successively on a load line, and by drawing horizontal lines through 

the divisione on this line to meet verticals through the loaded joints, the shearing 
stress diagram due to dead load is obtained (§ 45). On account of the unsymmetrical 
distribution of the loads, the horizontal line separating positive and negative shearing 
stress will not bisect the load line, but must be found by drawing the funicular polygon 
of the dead loads with respect to any pole, as in § 47. Now draw the shearing stress 
diagram due to the travelling loads at 2 4 and 6, supposing them to come on the 
bridge from the right abutment. The construction is precisely similar to that of § 57. 
The resulting figure, combined with the previously drawn shearing stress diagram for 
dead load, represents the distribution of total shearing stress along the girder, when 
only the numbered joints are loaded, and the travelling load comes on the bridge from 
the right. Then lines drawn parallel to each bracing bar and terminated by the sides, 
or produced sides, of the diagram of total shearing stress under that bar, give the 
maximum stresses in those bars. These tresses can now be tabulated as far as the 
bar 45, after which the shearing stress changes sign, the corresponding bars beginning 
at the right abutment will have the same respective maxima stresses when the load 
comes on the bridge from the left 

The maxima stresses in the bracing bars forming the remaining systems of 
triangulation can similarly be obtained. 

stresses in Booms.— The maxima stresses in the booms arise when the girder is 
fully loaded. Draw therefore the diagram of shearing stress due to travelling load 
acting at all three of the lower joints of the system of triangulation under considera- 
tion (§ 57). From this diagram the stress in any bracing bar, when the travelling 
load acts at all the lower joints, is obtained, and it is only necessary to add the stress 
due to dead load only, taken from the other diagram. The horizontal component of 
this stress should be obtained and written down. The horizontal components of the 
stresses in all the bracing bars having been written down, the maximum stress in any 
segment of either boom will be equal to the sum of the horizontal components of the 
stresses in all the bracing bars between that segment and the nearest abutment. 

If the bracing bars are inclined at 45°, as in the present case, the horizontal 
component of the stress in any bracing bar will be equal to the shearing stress in the 
bay under that bar. 

The maxima stresses in the booms may be approximately obtained from a curve 
of total stress drawn for a uniformly distributed load covering the whole girder. 
This uniformly distributed load amounts to 1 • 9 tons per foot run, the total load on 
the girder is therefore 1*9 x 60 = 114 tons, this curve of total stress is a parabola 
(§ 54). 



BRACED GIBDEBS. 



81 



Set off along a load line a (2 a length representing the total load (114 tons), 

bisect a d by a perpendicular S making S equal to h the height of the girder 

(Fig. 74). Draw at, bt parallel to aO, Od (§ 46), and bisect tc in v. Draw a 

parabola with v as vertex and t 6, t a tangents at b and a respectively (§ 46) : 

then the bending moment N at any section X is equal to y • S 0, but the total stress 

M 
in either boom at a vertical section through X is -r- or since S = A, the total stress 

in either boom at X is equal to y. 

The ordinates of the parabola avb (Fig. 74) read off from the scale of loads 
therefore give the t )tal stress in both booms at all sections along the girder, and the 

Fig. 47. 




Scale for Stress in. booms, £2x 

so » 



maximum total stress in any segment of either boom is given by the ordinates of the 
parabola vertically under the middle point of that segment. The stresses obtained 
will be very approximately those arising under the actual conditions of loading, and 
the approximation will be closer as the reticulations into which the web of the girder 
is cut up become smaller, or as the ratio between the length of a bay of the boom to 
the total span becomes larger. 

The above is given merely as an example of graphic treatment, which is, never- 
theless, in this case too tedious for actual practice. In the case of the lattice girder, 
arithmetical methods are simpler and more expeditious. 

60. Bowstring Suspension Girder. — Fig. 75, PL VIL, represents the frame 
diagram of one of the girders of Sarpsfos Bridge, Norway. The span is 60' 9", made 

M 



82 GRAPHIC STATICS. 

up of nine bays of 6' 9". The travelling load is taken at 1 ton per foot ; weight of 
girder, • 2 ton per foot ; and of platform, • 4 ton per foot. Supposing the upper 
joints to carry half the weight of the girder only, there will be — 

Tom. 
Dead load on upper joints 0*675 



„ lower „ 8-375 

Travelling load on lower joints •• .. 6*750 



Total „ „ .. .. 10-125 

The diagonals are supposed capable of taking up tensile stress only. In the 
preceding paragraphs no allowance was made at starting for the weight of the girders 
themselves. In the present instance it is proposed to make such an allowance, but 
the remark at the end of § 57 will still apply. The stresses having been obtained, the 
corrected weight of the girder is arrived at, and the stresses should be redetermined for 
this corrected weight, unless it proves to be very near the original estimate. 

Stress in Diagonals. — Make a s, b s f (Fig. 75a), each equal to half the total dead 
load, join sd. Draw below a b the stepped outline (§ 57), which forms the shearing 
stress diagram for the travelling load coming on the bridge at B, and loading each 
joint successively. Then the ordinates of the complete figure, under the middle of any 
bay, represent the maximum shear in that bay ; and also (§ 41) the reaction at the 
abutment A when the portion of the bridge to the right of that bay is loaded with the 
travelling load. Suppose the girder cut across by a section plane xy and the right 
portion removed. The left portion would then be in equilibrium under the action of 
the reaction of the abutment A, and of three forces acting at the points of section of 
the three bars 4 6, 4 5, 3 5, and equal and opposite to the stresses in those bars. It is 
required therefore to resolve a force whose magnitude is represented by c d (Fig. 75a) 
and which acts at A, along three given directions. 

Produce 6 4 to cut the vertical at A in e : join e 5. For convenience, c d is 
transferred to d d'. Draw d 4! parallel to 6 4, and d' 4' parallel to e 5, 4' 5' parallel to 
4 5, and d! 5' parallel to 5 3. Then (§ 11) 4' 5' is the component of the reaction acting 
along 4 5, and 4' 5' is therefore the maximum stress in 4 5. The determination of the 
stresses in the other diagonals 2 3, 6 7, 8 9 is similarly carried out ; these stresses are 
all tensile. The stresses in the dotted bars will, by symmetry, be respectively similar. 
Thus, the maximum in 1 4 is equal to that in 2 3, and so on. 

These stresses may also be determined by taking moments about the intersection 
of the segments of the boom produced. Thus, if p is the stress in 4 5, and o the 
intersection of the bars 6 4, 5 3 produced, 

jp . L = cd.r ; 

p = en • — • 
Li 



»< 



BBACED GIRDERS. 83 

The lengths L and r can be taken from the drawing, and the value of p thenee 
deduced. The great distance of the point o will, however, often render this method 
impracticable. 

Stresses in Verticals. — Suppose the girder cut by a plane zz 9 and resolve the 
maximum shearing stress c d in the bay 3 5, acting as a reaction at A along the three 
bars 2 4, 4 3, 3 5. The construction is carried out as before, cd being transferred to 
<? d" (Fig. 75b), 4" 3" is obtained as the maximum stress in the vertical 4 3. This 
stress is a pressure. This construction is carried out in Fig. 75b for the verticals 4 3, 
6 5, 8 7. 

The principle of moments might be employed here also, the moments being 
taken about the intersection of the bars 4 2, 5 3 produced. 

The stress in the end vertical 2 1 cannot be obtained by either of these methods. 
By resolving the maximum reaction at A along the bars A 2, A 1 the stresses in those 
bars are obtained. Compounding the stress in A 1 with the load at 1 (10* 125 tons), 
and with the stress in 1 4, the stresses in 2 1 and 1 3 are determined. 

Stress in Booms. — By combining the stress diagrams for the dead load with that of 
the travelling load covering the whole girder, the stress diagram for the girder when 
fully loaded is obtained. The ordinates of this diagram under the centre of each bay 
can then be severally resolved along the bars in that bay as above, and the maximum 
stresses in the booms are arrived at. 

The stresses in the booms can also be derived from the funicular polygon of 
total load, or from the curve of maximum bending moment. 

On a load line I F (Fig. 75d) set off successively the total loads on the pairs of joints 
2 1, 4 3, &c. (these total loads are equal to 10 • 8 tons). Take as pole, making V equal 
to a convenient distance. Draw half of the funicular polygon of these loads as shown 
(Fig. 75c). Then the bending moment at 5 (Fig. 75) is equal tomn. H, where H = 
V. From 5 draw a perpendicular 5 1 on 4 6 (Fig. 74), then, if Q is the maximum 
stress in 4 6, Q . 6* = mn . H; 

or, Q = m».g < ; 

whence Q can be obtained by calculation or construction. "By drawing a perpendicular 
from 6 on 3 5 the maximum stress in 3 5 can be determined ; in this case, however, by 
symmetry the compressive stress in 4 6 is equal to the tensile stress in 3 5. 

As has been stated, the stresses in the end segments of the booms can be obtained 
by a resolution of the maximum reaction at the abutment along the directions of those 
segments. 

The maximum stresses in the booms may also be obtained by drawing a 
reciprocal diagram of the whole girder, according to the method of Chapter IV. 

If, as is occasionally the case, the cross girders bearing the platform are attached 

m 2 



84 GBAPHIO STATICS. 

to the verticals at points intermediate between the upper and lower ends of the latter, 
the load may be distributed at the upper and lower joints in the ratio of the upper 
and lower portions into which the verticals are divided by the points of attachment. 

61. Bowstring Girders.— Pig. 76, PI, VIII., represents the frame diagram of a 
bowstring girder of 192' 0" span, divided into sixteen bays of 12' 0". The whole of 
the loading is taken to be uniformly distributed, the actual concentrated loads being 
supposed replaced by an equivalent distributed load. The travelling load on the 
girder has been taken at 1 • ton, and the dead load at • 65 ton per foot, the latter 
being assumed to act half at the upper and half at the lower joints. 

The loading will therefore be— 

Tons. 
3-9 

3-9 
12-0 

3-9 
15-9 

9-9 



Dead load on upper joints •• 

„ lower „ 

Travelling load on lower joints 
Total „ upper „ 

„ „ lower „ 

it »» en d „ 



The verticals are intended to act both as struts and as ties, the crossed diagonals 
as ties only. 

The stresses due to dead load, and also the stresses in the booms due to both dead 
and travelling loads, can be obtained by means of reciprocal diagrams. The latter will 
somewhat resemble that given in Fig. 46a, the difference being, that in the present 
instance the lower boom is straight instead of polygonal. In drawing these reciprocal 
diagrams it will be necessary, as in § 32, to deal with that system of triangles only in 
which the diagonals are in tension. 

Diagonals. Travelling Load. — The load is supposed to come on the girder at B, 
Fig. 76, PL VIII., the diagonals shown in continuous lines will then be in tension, and 
the maximum stress in any diagonal ef will arise when the travelling load reaches *, 
the middle point of the bay under ef; i. e. when the shearing stress at * T is a maximum. 
Suppose the girder cut by a section plane a /J, and, considering the maximum shearing 
stress at 6 T to act as a reaction at A, resolve this shearing stress along the directions of 
the three bars cut by a /J. The component along ef is the maximum stress in ef due to 
live load, and similarly for the remaining diagonals. 

The maxima shearing stresses at the middle points of the several bays of the 
girder are found by the construction of § 52 (Fig. 70). The load line ba' (Fig. 76a) is 
made equal to the travelling load (192 tons), and is bisected at right angles by c ; Oc 
is made equal to 100 tons on the scale of loads : at is drawn parallel to O a', b O being 
produced to cut at in t, A « f , 8 n B are bisected, and perpendiculars are dropped from 
their points of bisection to cut at and bt in p and r, then (§ 52) rt is a tangent to the 
bending moment curve at a point on the latter vertically below s T . Produce rp to cut 



CENTRE OF PARALLEL FORCES.— CENTRE OF GRAVITY OF PLANE FIGURES. 85 

the vertical through A in 7 : join 7 6, and draw 7' parallel to 7 b. Then if b 7" 
the total load covering the length Bs 1 of the girder, 7' 7" is the maximum shearing 

stress at 6, (§52). The maximum shearing stresses at the points 6 S s 9 are 

similarly obtained. 

A stress represented by V 7" (Fig. 76a) is now supposed to act as a reaction at A 
(Fig. 76), and must be resolved along the bars cut by the section plane a fi. Produce 
the cut segment of the upper boom to meet the vertical through <A. in g. Draw I 7 
(Fig. 76b) vertical and equal to 7' 7" (Fig. 76b) and Id horizontal: draw 7n, In 
parallel to hg, gf respectively, and nd parallel to ef\ then (§ 11) nd represents the 
component of 7' 7" resolved along ef 9 and n d is therefore the maximum stress in ef 
due to travelling load. This construction is carried out in Fig. 76b as far as the eighth 
diagonal from the left abutment. 

Verticals. — The mode of determining the stresses in the verticals is very similar. 
Suppose the girder to be cut by a plane y 8 (Fig. 76), and resolve the maximum 
shearing stress at s 7 along the bars cut. In Fig. 76c make 1 7 equal to V 7", and draw 
7v 9 lv parallel to eg', g f f (Fig. 76), then the vertical v k gives the component of 7' 7" 
alone e i, and therefore the maximum stress in e i. The construction is carried out in 
Fig. 76c for all verticals from the second to the seventh inclusive. The first vertical does 
not, properly speaking, come into the system of triangulation under consideration ; it 
will be sufficient to make this vertical able to sustain the load on one bay of the girders 
or the heaviest concentrated axle load if th$ latter is greater. The dotted diagonals will 
be similarly strained when the load comes on the girder at A. The above construction 
can be employed to determine the stresses due to dead load if a shearing stress diagram 
(§ 46) for dead load is first drawn, and also the stresses in the booms due to travelling 
load if a shearing stress diagram of travelling load covering the girder (§ 52) is drawn. 
The stresses in the booms may also be obtained from the curve of bending moments. 

The stresses due to dead and travelling load should in all cases be carefully 
tabulated and then added , having due regard to their sign. 



CHAPTER VIII. 

CENTBE OF PABALLEL F0BCE8.— CENTEE OP GBAVITY OF PLANE FIGUBEa 

62. Parallel Force© acting at Points in one Plane. — If a system of parallel 
forces p x ; p 2 . . .acting at any points a x ; a,. . .(Fig. 77) in the plane of the paper are 
supposed to turn about those points still retaining their parallelism, then the resultant 



86 GRAPHIC BTATIO& 

of the whole system will always pass through a fixed point, in the same plane with 
<h ; <h • • • * termed the centre of the parallel forces. 

This centre can he found as follows : join a x a a (Fig. 77), and divide a x a* in c l9 such 

that — = P * : then (§ 13) c x is the centre oip x and/?* Suppose j^ + p 2 to act at c x and join 

£l a 2 Pi 

c x a z . Divide c l a z so that — = P * . .. Then c t is the centre of p x , p %y and p» and 

c 2 <h Pi+P 

by carrying on the process till the last force has been dealt with, the centre of the 




system is obtained. If any one of the forces ; eg 9 p b is of opposite sense to the rest, then 

C C 2?k 

c A will lie on c z a^ produced, so that — = ^ . The production must 

a 6 c A p x +p 2 +ps+p A 

always be made from the point of application of the less in the direction of the point of 
application of the greater of the two forces. 

The above method would be tedious if the number of forces were great, and in 
this case the centre can be more readily found by the employment of the funicular 
polygon. Since the parallel forces may be supposed to act in any direction, provided 
that their parallelism is preserved, they may be supposed to act in the plane of the 
paper. Through 0|. . .Oj (Fig. 78) draw the parallel lines of action in any direction. 
Parallel to this direction draw a load line o 5, and set off successively on o 5 lengths 
proportional to the magnitudes of the forces p x . . ,p 6 acting at c^ . . .a,. Take any pole 
0, and draw a. . .€, the funicular polygon of the forces. The last sides of this polygon 
intersect in i, then (§ 12) the resultant of the forces passes through i, and consequently 
a line through i parallel to the assumed direction of the forces will contain the 
" centre " of the forces. By drawing the lines of action of the forces through a x . . .a $ 
in any other assumed direction, and repeating the above construction, a second line 
will be found containing the centre. It will be simpler to draw these new lines of 
action at right angles to the previously assumed direction, and then it is not necessary 
to draw a new polygon of forces, but merely to draw a new funicular polygon, the 
sides of which are respectively at right angles to those of the first. The funicular 
polygon eh . . . € x is thus drawn, and its two last sides intersect in %\. Hence c is the 
centre of the whole system of forces. 

It will thus in general be necessary to draw one polygon of forces and two 



CENTRE OF PAEALLEL FORCES.— CENTRE OF GRAVITY OF PLANE FIGURES. 87 



funicular polygons in order to obtain the centre of a system of parallel forces whose 
points of application lie in one plane. If, however, the forces were symmetrical, and 
their points of application symmetrically situated with respect to any line, or if all the 



Fig. 78. 





points of application were on a line, this line must contain their centre, and only one 
polygon of forces and funicular polygon are needed. 

63. Parallel Forces in Space. — If the points of application of a system of 
parallel forces are not in one plane, the forces must be projected on to two co-ordinate 
planes which it is convenient to take at right angles to each other. In Fig. 79 
a \ a/ ; <h <h! • • • are the orthographic projections of the points 'of application of a system 
of parallel forces p x , p 2 , . . . . ; x y is the ground line of the planes on which the forces 
are projected. Dealing with the plans Oj, a,, ... of the points of application of the 
forces precisely as in the preceding section y by means of one polygon of forces and 
two funicular polygons, a point c is obtained, which is the plan of the required centre. 
Dealing with a/, a/ . . . , the elevations of the points, by means of a funicular polygon, 
a line is obtained containing the elevation of the centre. By projecting up from the 
plan e on to this line the elevation d of the centre is determined. 

Thus, set off the forces p u p 2j * . . in succession along a load line 5, which load 
line may conveniently be taken on x y, the forces being supposed temporarily to act in 
directions parallel to the vertical plane, so that the plans of their lines of action drawn 
through <*! Os . . . will be parallel to x y. Draw the funicular polygon a ... c of the 
forces, then a line parallel to x y drawn through the intersection i of the first and last 
sides of this funicular polygon will contain the plan of the centre. Suppose the parallel 



88 



GRAPHIC STATICS. 



x- 



forces to act at right angles to the vertical plane, so that their plans are at right 
angles to xy 9 and draw a second funicular polygon a x . . .€ 2 whose sides are respectively 
perpendicular to those of the first. 

Through i u the intersection of the first and last sides of this second funicular 
polygon, draw a line perpendicular to x y and cutting the line through % in c. Then 
c is the plan of the required centre of the system. 

Dealing with the elevations a/, a^, . . . of the points of application of the forces, 



Pig. 79 




suppose the later to act in directions parallel to the horizontal plane and draw their 
elevations parallel to x y. Draw a third funicular polygon a, . . . c, whose sides are 
respectively parallel to those of the first funicular polygon a . . . c, then a line parallel 
to xy through i,, the intersection of the first and last sides of this funicular polygon 
will contain the elevation of the centre of the system ; projecting from e on to this line, 
we obtain c f 9 and the required position of the centre of the parallel forces is therefore 
determined. 

The magnitude of the resultant of the system is equal to tp 9 the algebraic sum of 
the forces. 



CENTRE OP PARALLEL FORCES.— CENTRE OP GRAVITY OP PLANE FIGURE& 89 

In the two preceding sections the forces have been all supposed to have the same 
sense, if any of them are of opposite sense, the fact must be remembered when the 
forces are being set off along the load line. The forces forming a system may be split 
up into groups, if then the centre of each group is found, and the resultant of the 
group is supposed to act as a single force at that centre, the centre of the whole system 
can be found by dealing with these resultants acting as parallel forces at the centres of 
groups by means of the preceding section. 

64. Centre of Gravity of Lines and Curves. — If the forces in the preceding 
sections are supposed to be replaced by the weights of a system of bodies, or of the 
several portions of the same body — which weights form a system of parallel, because all 
vertical, forces — then the centre of the parallel forces is termed the centre of gravity of 
the system of weights, or of the body. The centre of gravity of a body may therefore 
be defined as the point through which the resultant of the weights of the particles of 
the body acts in whatever position the body is placed, or as u the point at which the 
whole weight of the body may be supposed to be concentrated without altering its 
statical effect." In applying the term to lines or plane figures, the former must be 
conceived as made up of heavy points, and the latter of heavy lines. 

Broken Line.— -The centre of gravity of the several segments of the broken line 
will be at their respective middle points. Suppose parallel forces proportional to the 
lengths of the segments to act at their middle points, and find their centre as in § 62 if 
the line lies in a plane, or § 63 if otherwise ; this is the centre of gravity of the line. 
If the segments are all equal and make equal angles with each other so that the broken 
line is part of a regular polygon, as in Fig. 80, the centre of gravity can be found in a 
much shorter way as follows : Join a/, and bisect af by a perpendicular : this per- 
pendicular is evidently a line of symmetry, and contains the centre of gravity of the 
line, as well' as the centre of the circle passing through a . . . . /. Through 
draw any line xy. Then if I is the length of a 6, be . . . . ; x x x 2 . . . the perpendicular 
distances of their respective middle points from xy ; L the whole length of the line, 
and X the perpendicular distance of its centre of gravity from x y — 

3 (i . w) = L . X. 

But if d d f is the projection on x y of any segment c d — 

cd r , , - 

-ttj = — ; or, ca.a?i = c a.r 
o a &s 

where r is the radius of the circle through a . . . e. 
Hence — L . X = r (<f d') 

and — L . X = r . a'f. 

Construct a right-angled triangle fn'n whose hypothenuse f n is equal to L 

s 



90 



GEAPHIO STATICa 



and perpendicular rl n equal to r. With / as centre, and / a' as radius, describe 

r • a! f 
a circle cutting f n in s ; draw 8 d perpendicular to x y ; then, s s' = — 1 y- — = X ; 

draw s G parallel to xy, cutting the line of symmetry h in G. Then G is the centre 
of gravity of the whole line a . . . e. 





Circular Arc. — The construction of the preceding section can he applied to the 
circular arc ab, Fig. 81, since the latter can be regarded as a polygon whose sides are 
infinitely small. 

The centre of gravity of the arc lies on O c the line of symmetry, being the 
centre of the arc, and c its middle point ; draw the tangent at c and make c d equal 
to the arc a c ; join d, draw a e parallel to c, and e G parallel to a b 9 then G is the 
required centre of gravity. 

In order to set off the length of the arc a c approximately along the tangent at c ; 
join a c and produce a c to /, making cf equal to half a c. With / as centre, and fa as 
radius, describe an arc cutting the tangent in d. Then d c is approximately equal to 
the arc a c. 

The centre of gravity of any curved line can be found by cutting it up into small 
segments and treating these segments as straight lines. 

65. Centre of Gravity of Plane Figures. — The centre of gravity of a triangle 
is found at the point of intersection of lines drawn from any two angles bisecting the 
opposite sides ; this point is at a distance of £ of either of these lines, measured along 
them from the point of bisection. 

Parallelogram. — The centre of gravity is at the intersection of the diagonals. 

Trapezium. — The centre of gravity must lie in the line of symmetry ef (Fig. 82) 
bisecting the parallel sides be, ad. Draw the diagonal bd dividing the trapezium 
into two triangles, and obtain g x g 2 the centres of gravity of these triangles. Join 
9 \ ffi} cutting efin G, then G is the centre of gravity of the trapezium. 



CENTEE OP PARALLEL FORCES.— CENTBE OP GRAVITY OP PLAKE FIGURES. 91 

If b c= I and a d = L, then it can be shown that Qt divides ef in the proportion 

l + 2L:2l+L; 
i +!■■$ + * 

Produce cb, ad in opposite directions to n and m, making dm = be = U and 
b n = a d = L. Join to n, then m n cuts 0/ in G. 



Fio. 82- 




Fig. 8a 



Four-sided Figure. — Draw the diagonals ac 9 bd of the four-sided figure abed 
(Fig. 83), fiud g l9 g % the centres of gravity of the two triangles into which the figure 
is divided by the diagonal b d : join g x g 29 then the centre of gravity of the whole 
figure lies on g x g %9 and if the centres of gravity of the two other triangles in which 
the figure is divided by the diagonal a c are obtained and joined, the intersection of 
the lines joining these pairs of centres of gravity will be the centre of gravity G of 
the whole figure. The point G divides the 
distance g x g % in the inverse proportion of 
the areas of the triangles bad and bed. 
If i is the point where g x g 2 cuts b d 9 it can 
easily be shown that g 2 i is equal to g x G. 
It can further be proved that if o is the 
intersection of the diagonals and a o l9 d o 2 
are equal to c o 9 b o respectively, then the 
centre of gravity of the whole figure coin- 
cides with that of the triangle o o x o 2 . 

Polygons. — The centre of gravity of any polygon can be found by dividing it 
into four-sided figures or triangles, and finding the centre of gravity of each. At the 
centre of gravity of each portion a force proportional to the area of that portion must 
be supposed to act ; then, by means of a polygon of forces and two funicular polygons, 
the centre of the forces or the centre of gravity of the whole polygon can be found ae 
in § 62. To obtain lines whose lengths are proportional to the areas of the triangles 
into which the polygon is split up, the construction of § 6 may be employed. The 
following simple construction can also be adopted. Draw two lines Ox Oy (Fig. 84) at 
any convenient angle. Suppose H, B; H n B t ; H a , B a . . .to be the heights and bases 

N 2 




92 



GRAPHIC STATICS. 



of the triangles to be dealt with. It is merely necessary to reduce these triangles to 
any common base. Taking B as this base, H represents the areas of the 1st triangle. 
SetoflF06( = B)and0 1 ( = B l ) along y, and h ( = H,) along x. Join&A,, 
and draw b x x x parallel to h h x . Then 

25 .9*!; or, 

06, 06 

0«i,BaH„B,. 

Hence x x represents the area of the 2nd triangle at the base B. Any number of 
triangles, rectangles, or parallelograms can be similarly reduced to a common base. 



Fig. 84. 



Fig. 85. 





Circular Sector. — The sector a c b (Fig, 85) may be supposed made up of a 
system of triangles whose common vertex is at 0, and whose bases, infinitely small, are 
on the arc acb. The arc a x c x b l9 struck with a radius equal to £ of a 0, contains the 
centres of gravity of all these triangles, the weights of the latter can therefore be 
supposed to act in the arc a x c x b x , and the centre of gravity of this arc is the centre of 
gravity of the sector. The construction above described can be employed : draw c t 
tangent to the arc a c b at c, and equal to the arc c b : join 1 0, cutting a line through 
b x at right angles to a^ b x in e, draw eg x through e parallel to b x a x , and cutting the line 
of symmetry c in g l9 then g x is the centre of gravity of the arc a^ c x b X9 and of the 
sector acbO. 

Circular Segment. — The segment abc (Fig. 85) is equal to the difference between 
the sector acbO and the triangle a b 0. Join x b x , cutting c in g 2 , then g % is the 
centre of gravity of the triangle. Supposing then that weights proportional to the 
areas of the sector and triangle act at g x and g % respectively, it is necessary to find a 
point y,onOc such that 



g x g % area of triangle if hf 

g t g 9 area of sector arc cb ct 9 



where hf is the perpendicular from £, the middle point of a b, on b. 



CENTRE OP PABALLEL FOBCES.— CENTBE OP GRAVITY OP PLANE FIGURES. 93 



Produce g 2 b x to d 9 making g 2 d equal to c t. Draw g x a parallel to a b and equal to 
kf. Join d s 9 and produce d 8 to cut Oc in g z ; then g 3 is the centre of gravity of 
the segment acb. 

Annular Segment. — The annular segment acbefd (Pig. 86) is the difference 
between the two circular sectors acbO and dfe0 9 determine first the centres of 
gravity g x g % of these sectors respectively, and then obtain a point G-, such that 

g x G area of a e b O (a 0)* 
frG " area of dfe = (Ol/ # 

Join af 9 and draw df parallel to af. 
Then OfOd 



0/ 



.oror- ( °*>. 

0a ,or,u/ - Qa 



Draw g x % and g 2 h at right angles to c 0, and respectively equal to Of and a. 
Join k i 9 and produce it to cut c in Qt. Then Q- is required centre of gravity. 



Pig. 86. 



Fio. 87. 




.&• 




Parabolic Segment. — The centre of gravity g of the parabolic segment bed (Fig. 
87) is on d c 9 the line which bisects b a and all other parallel chords, and gd = f . dc. 

The centres of gravity of the half segments bcd 9 acd are situated at points g x g % 
on a line through g parallel to a b 9 and 

The area included between the arc acb 9 and the tangents a t 9 b t at a and b 
respectively, is the difference between that of the triangle atb and the segment acb. 
The area of this segment is -| cp . ab where c p is the perpendicular from c on a b. 
The centre of gravity of the area at be can therefore be found by determining the 
centres of gravity of the triangle and of the area at be, and proceeding as in previous 
cases. The triangle and the rectangle % cp .ab must be reduced to a common base 
(§ 6), hx.order to obtain the lengths of lines proportional to their areas. 



94 GEAPHIO STATICS. 

Irregular Figures. — The centre of gravity of such a figure as the section of a 
railway bar can be found by cutting up the section into strips by lines perpendicular to 
its axis of symmetry. These strips will be very approximately rectangles or trapeziums 
whose areas, if their breadths are the same, will be proportional to their mean heights, 
or if the breadths vary they must be reduced (Fig. 84) to any convenient common 
base. Forces proportionate to these reduced areas must be supposed to act at their 
respective centres of gravity in a direction perpendicular to the axis of symmetry, 
then, by means of the polygon of forces and funicular polygon, a line is obtained (§ 62) 
which will intersect the axis of symmetry in the required centre of gravity. 

If the section, the centre of gravity of which is required, has no axis of symmetry, 
it must be divided up into simple areas in the most convenient way, forces proportional 
to the reduced areas of the several portions must be supposed to act at their respective 
centres of gravity, the centre of these forces can then be found by the construction of § 62. 

66. Carved Surfaces and Solids. — It is not proposed to give any constructions 
for determining the centres of gravity of curved surfaces and solids, but merely to 
state their positions for simple forms. If any surface or solid can be split up into 
portions such that the area, or volume and the position of the centre of gravity of each 
portion is known, then the centre of gravity of the whole surface or solid can be 
obtained by supposing parallel forces proportional to the areas or volumes of the 
several portions to act at their respective centres of gravity, and determining the 
centre of these parallel forces as in § 63. 

Pyramidal or Conical Surface. — The centre of gravity is on the line joining the 
centre of gravity of the base to the vertex at a distance of £ of this line from the base. 

Hemispherical Surface. — The centre of gravity bisects the axis of symmetry. 

Surface of Hemispherical Segment. — The centre of gravity bisects the axis of 
symmetry. 

Surface of Frustum of Pyramid. — If the frustum is contained between two 
parallel planes, the centre of gravity divides the line joining the centres of gravity of 
the two ends in the proportion a + 26:2a + 6 

when a and b are the two parallel sides of any one of the trapeziums which form the 
faces; or, P + a^P + p 

when P and p are the perimeters of the polygons which form the ends. In either case 
the smallest portion of the divided line must of course be that nearest to the largest end 
of I he frustum. 

Surface of Frustum of Cone. —The centre of gravity divides the line joining the 
centres of gravity of the two ends in the proportion 

D + 2d:2T) + d 

when D and d are the diameters of the two ends, if circular. 



CENTEE OP PARALLEL FORCES.— CENTRE OF GRAVITY. 



95 



Pyramid. — The centre of gravity is on the line joining the centre of gravity of 
the base with the vertex, and is at a distance of \ of the whole line from the base. 

The centre of gravity of a triangular pyramid coincides with that of four equal 
weights placed at its angular points. 

Cone. — Same as the pyramid. 

Hemisphere. — The centre of gravity is at a distance of f of the radius, measured 
from the centre of the base up the axis of symmetry. 

Frustum of Pyramid. — The centre of gravity is on the line joining the centres of 
gravity of the ends, and if Gr is the centre of gravity of the frustum, g x that of the 
whole pyramid, and g 2 that of the pyramidal portion cut away, then — 

^G_ V a 3 
Ggx" v' 0T b> 

where V and v are the volumes of the whole pyramid and of the portion cut away, and 
a and b are any homologous sides of those pyramids respectively. 
Frustum of a Cone. — Using the same lettering as befor< 



g*Q V L 8 
a- ■*-» or »» or 7» 



R" 
r 1 



where L and I are the respective slant heights, and R and r the respective radii of the 
bases of the whole cone and of the conical portion cut off. 





Conical Sector. — The centre of gravity of the conical sector generated by the 
revolution of the sector A C (Fig. 88) about C may be found by describing the are 
acb with radius equal to £ A, joining a b and bisecting c d in g : g is the centre of 
gravity, and the distance g is equal tofOC — fCD. 

Segment of a Sphere. — The centre of gravity of the spherical segment generated 
by the segment A C A' (Fig. 89) revolving about B is situated at a point Q- on B C, 
such that 






a 

r 



96 



GRAPHIC STATICS. 



where g x is the centre of gravity of the cone A A', g 2 the centre of gravity of the 
conical sector A A', r the radius A, aud 



a = 



_ (2r-h)(r-h) 
~ 2r 



where h is the height B C of the segment : a is readily constructed as a fourth propor- 
tional to 2 r ; r — h ; and 2 r — h. 

Solids bounded by Irregular Surfaces. — The centres of gravity of such solids may 
be found by supposing them cut up by parallel planes into a large number of slices. 
If the thickness of these slices is small, the centre of gravity of each will be very 
nearly the same as that of its mean section, i.e. its section by a plane parallel to and 
bisecting the distance between the two planes by which it is bounded. If the thickness 
of the slices is the same, their .volumes will be approximately as the areas of their 
mean sections. Supposing parallel forces proportional to the volumes of the slices to 
act at their respective centres of gravity, the centre of these parallel forces can be 
determined as in § 62, or § 63. 



Fro. 89a. 



CHAPTER IX. 

MOMENT OP INEBTIA— CENTBAL ELLIPSE, Ac. 

67. Moments of Parallel Forces. — The moment of a system of co-planar forces 
about a point in their plane has been explained in § 16, as the sum of the products 
of each force into its perpendicular distance from the point, and is equal to the 

product of the resultant of the forces into its perpendicular 
distance from the point The physical effect of the action 
of such forces is a tendency to produce rotation about an 
axis perpendicular to the plane of the forces and meeting it 
in the given point. 

The moment of a force P about an axis A B (Fig. 89 a), 
when the force does not act in a plane perpendicular to the 
axis, may be defined as the product of the projection, P lf 
of the force on a plane M M t perpendicular to the axis A B, 
into the distance A S of the point in which A B intersects 
the plane, from the projection P^ in other words, the 
moment of a force with respect to an axis is the moment of the projection of the 
force on a plane normal to the axis with respect to the point in which the axis 
intersects the plane of projection. 




j 



MOMENT OF INEBTIA.— CENTEAL ELLIPSE, &a 



97 



The distance A S of the foot of the axis from the projection of the force is, 
moreover, equal to the length of the line C D, which is perpendicular to P and the 
given axis, i.e. is the shortest distance between them. 

A definite physical meaning can be attached to the foregoing, viz. : that the force 
being resolved into 2 components, one P x in a plane perpendicular to A B, and the 
other parallel to A B, i.e. into P. cos. 6 and P. sin. 0, where 6 is the angle between P 
and the plane perpendicular to A B. The former tends to produce a rotation measured 
by the product Y x x A S about A B, and the latter tends to produce a translation 
along A B. If we have a second force Q parallel to P, its moment about A B is 
similarly Q x or Q cos. multiplied by A Si, A S and A 8 X being in the same straight line. 
The algebraical sum of the moments of the components P x and Q x about A B is equal 
to the moment of their resultant, i.e. Pi x A S — Qx x A S x = moment of Rx about 
A B, where Rx is the resultant of Pi and Qx ; but R x = R cos. 0, where R is the resultant 
of P and Q. Hence, the moment of two parallel fig. 89b. 

forces about any axis is proportional to the 
product of their resultant into the shortest 
distance between it and the given axis. This 
same is evidently true for any number of parallel 
forces. 

Draw any plane as N N x (Fig. 89 b) through 
AB and let the line of action of P meet it in a^ 
and that of Q meet it in a,, and let the line of intersec- 
tion of the two planes M M lf N Nj meet Pi in b x 
and Qx in b 2 . 

Then the algebraical sum of the moments 
of P and Q about A B is proportional to 
Pi x A b x ± Qx x A b % and also to the projection of 

their resultant on the plane M Mi multiplied by the distance between A and the point 
in which such projection intersects the line b x b 2 . (= x suppose). The equation 
P x A b x ± Q x A b 2 = R . x is, therefore, evidently true. 

Lastly, if from a x and a 2 any parallel lines are drawn in the plane N N x meeting the 
line A B in x x and x 2 respectively ; then, since a x x x = A b x multiplied by some constant, 
and Oa x 2 = A b 2 multiplied by the same constant, it follows that the moments of 
P and Q about AB are proportional to P x a x x x and to Q x a,^, and that 
P xa^j + Qxflj^sEx the line drawn from the point in which R intersects the 
plane N Nx parallel to a x x x (or a^ x 2 ) to meet A B. 

No direct physical meaning can be attached to such an expression as 
P x a x x x ± Q x a 2 x 2 ± .... but it nevertheless leads to useful results, and therefore 
the moment of a system of parallel forces about any axis is sometimes defined as the 






98 GBAPHIO STATICS- 

algebraical sum of tlte products of the forces into the distances, measured in any parallel 
direction, of the points in which their directions meet any plane through the axis, from that 
axis. 

Thus if the forces p x . . . p 6 act at a x . . . a 6 (Fig. 90) respectively, their moments 
about an axis X X lying in the plane containing a x . . . a 6 are equal to p l ,y l9 p 2 .y%, . . . . , 
where y\ % y%... are the distances of a x a* . . . from X X measured in directions making 
any angle with X X. 

The total moment of the systems of forces is 2 (p . y) , and 

3(p.y) = 2i>.Y, 

where Y is the distance of the centre of the system from X X measured in a direction 
parallel to the distances y l9 y 2 . . . If the points of application a^ Oj . . . move on 
parallel lines, the centre of the system will move on a parallel line. Further, the sum 
of the moments of a system of parallel forces about any axis passing through their 
centre is nil, and conversely if the sum of the moments about any axis is nil, the axis 
must pass through the centre of the system. 

If the sum of a system of parallel forces is zero, then any force must be equal and 
opposite to the resultant of the remainder, and will form with that resultant a couple. 
Such a system can be formed into as many couples as there are groups into which it 
can be split up. If the centres of a pair of groups coincide, the couple reduces itself 
to two equal and opposite forces acting at a point. 

The moment of a force with respect to a plane has no physical meaning unless 
the plane is parallel to the direction of the force. In that case the moment of the force 
is the product of the force into the perpendicular distance between it and the plane. 
Analogously, however, to the above convention as to the moment of a system of parallel 

forces about an axis, if a plane cuts a system of parallel forces in Oy , a* and 

parallel lines are drawn in any direction through a a% . . . , &c, meeting a second 
plane in b l9 b 2 . . . , then the sum of the products of each force by the corresponding 
distance, i.e. 2 (p . a b), is called the moment of the system about such second plane, 
and is equal to R . Y where R is the resultant of the forces and Y is the distance 
measured parallel to aj? l9 &c, between the point in which R intersects the plane 
#i? #29 <h> & c *9 an d the plane with respect to which the moments are taken, i.e. the 
plane b l9 b 2 , b 3 . . . , &c. 

If the lines oA, &c, are all drawn in the plane a l9 a*, a s , the moment about the 
plane is evidently identical with the moment about the line of intersection of the two 
planes, i.e. with the moment about an axis. 

It should be noticed that the expression thus defined as the moment of a system 
of parallel forces about a plane is of an entirely different character to the moment of 
a system of co-planar forces about a point in their plane. The latter depends essen- 



MOMENT OF INBETIA--CENTBAL ELLIPSE, <&c. 99 

tially on the directions of the lines of action of the forces, each of which may be 
supposed to act at any point of its line of action ; the former, on the contrary, is entirely 
independent of the direction of the lines of action of the forces, and varies with the 
co-planar points which are selected as the points of application of the forces. 

68. Reduction of Moments to a Common Base. — The reduction to a common 
base of the moments of a system of parallel forces about a, point in their plane has already 
been treated in § 20 ; a similar construction serves for the reduction of moments 
about an axis parallel to the direction of the forces. In Fig. 90 set off the forces 
PuP% • • • acting at a M 03 . . . along a load line Id parallel to X X, the axis about which 
moments are to be taken. Through a lf a >29 . . . draw the lines of action of the forces 
parallel to X X. Draw a funicular polygon of the forces with respect to any pole O, 
and if necessary produce the sides of this funicular polygon to cut X X in b 09 b 19 . . b s 
successively. From 0, draw S parallel to the directions of y 19 y 2 . . . ; then b b x 
represents the moment of p x about X X, b x b 2 that of p 2 , and so on, all these moments 
being reduced to a common base S. Moreover, b J 6 represents the reduced moment 
of the resultant, and b b 6 . S is equal to the sum of the moments of all the forces, 
or— &A.OS = 3(p.y). 

If the lever arms are supposed to be perpendicular to the axis, then = 90° and 
the moment base is H drawn from perpendicular to the load line I d 9 so that in 
this case — 6 6 5 .OH » 2(py). 

In future the lever arms will always be supposed perpendicular to the axis. 

69. Moment of Inertia of a System of Parallel Forces. — If jt?i, p 2 . . . are 

parallel forces acting at points a l9 a* . . . distant y X9 y 2 . . . from an axis X X in the same 
plane, then 2 (py 2 ) is the moment of inertia of the system about XX. The sign of 
the moment of inertia is therefore independent of that of the distances y X9 y 2 . . . and 
dependent only on the sense of the forces. 

The moment of inertia can be obtained graphically as follows : — Reduce the 
moments of the forces to any common base OH (Fig. 90), as in the preceding 
section, and suppose that parallel forces whose magnitudes are represented by the 
reduced moments b b X9 b x b 2 . . . . act parallel to X X at the given points of application ; 
b b x at a u b x b 2 at Oj, and so on. By means of a new funicular polygon, reduce the new 
system of forces to a common base O'H' (Fig. 90), setting off the distances b b x , 
b x b 2 . . . on a new load line also parallel to X X. The sides of the new funicular 
polygon (produced if necessary) successively cut X X in c 09 c l9 .... 

Then— 

CoCi . 0'II'= bobi . jf|. 

But— 

o 2 



100 



GRAPHIC STATICS. 



Hence— frVx = w . OH . 0' W. 

Similarly — # ¥%y* = c, c> . H . 0' H'. 

And— S(pjf) = (€0^ + ^^+ ...)OH.O'H', 

the sign of the intercepts c^, c x c 2 , ... depending on the direction in which the force 
tends to produce rotation. 
. Thus— J(p^ = ^.OH.O'H', 

or if 0' H' has been taken equal to H — 

S(ptf)«*«b.olil 
By repeating the above process moments of higher order, such as 2 (p y 8 ), could be 
dealt with graphically. 

In Fig. 90 the intercepts b b l9 ^^...have all been transferred to a load line 

Feo.90. 



X 






I 

$ 



/ 6\ 







* 

I 



*i 



V  >  A b / fc— / / K 

7 A v ^ *• / / A ». \ 

— • '^ i\v. &+. ' ' — -\>K \ i $\ k' — )% — 

/ i / \ \ \ H i\ / 




\» » • : • • i *■ 



«* 



O** 



fo *♦ Qi ^<" Q 




parallel to XX; it will, however, save trouble to use the intercepts themselves as 
forming the load line : this is shown by the dotted vectors. Similarly it will usually 
be convenient to take the load line on the axis about which moments are to be taken. 
It will be noticed that two of the sides of the second funicular polygon accidentally 
cross on the line of action of jp 3 . 



MOMENT OF INERTIA— CENTRAL ELLIPSE, &o. 101 

» 

It is evident that the second system of forces — i. e. the reduced moments of the 
forces supposed to act at their respective given points of application— will have a 
u centre," which could be found by § 62. This centre is termed the centre of the reduced 
moments. 

The moment of inertia of a system of parallel forces is of two dimensions in 
space and one in mass, and is said to be in square foot pounds, or square inch 
pounds, according as the foot or inch is the linear unit. 

70. Radius of Gyration.— If SQpy 8 ) is the moment of inertia of a system of 
parallel forces p l9 p 2 . . . about an axis X X, and k is a linear magnitude such that 
i?. 2 (p) = 2 Qfly 3 ), then k is the radius of gyration of the system about the axis X X. 

It has been shown in the preceding section that, for the system of parallel forces 

indicated in Fig. 90, 

2(py*) = *<vOH.O'H\ 

Hence — 

p q,c a . OH . O'H' ^ cc, . OH . Q , H > 
*(P) Id 

Make Ih = O'H'. Draw ht parallel to dO and ts parallel to H. Then by 

. ., ± . . ts Ih OH.O'H' 
similar triangles tt-^ = ,-^; or ts — j-j . 



Hence — *=Vc w c 4 .<«, 

whence k can be obtained by const ruction. Or, the second base 0' H f might have been 
taken equal to Id [ = 2(/?)], in which case — 

If K is the radius of gyration of a system of parallel forces about an axis passing 
through their centre, and k the radius of gyration of the system about a second axis 
parallel to and distant d from the first, then — 

PbK' + A 

Hence, if the radius of gyration about an axis through the centre of the system 
has been obtained, the radius of gyration about any parallel axis at a known distance 
from the first can be determined, and vice versa. 

The radius of gyration determined as above must be regarded as a line making 
the same angle with the axis as that made by the lever arms of the moments of the 
forces. Like these lever arms, therefore, it will here be supposed invariably perpen- 
dicular to the axis. 

71. Curve of Inertia and Central Curve.— Let k be the radius of gyration of 
a system of parallel forces about any axis X X, and suppose two lines to be drawn 
on either side of X X parallel to and distant k from X X. Then, if the axis X X is 



«. i 



102 GEAPHIC STATICS. 

supposed to turn about any point on its direction, lines drawn parallel to, and at 
distances from the new positions of the axis respectively equal to the radii of gyration 
about those new positions, will be tangents to a curve of the second order having as 
centre. This curve, which is either an ellipse or an hyperbola, is termed the Ellipse, or 
Hyperbola of Inertia of the system of parallel forces. In general, the curve is called 
the Curve of Inertia. This curve of inertia is fixed if its centre is given or 
assumed. 

If the sign of the moment of inertia 2 (py 2 ) is the same as that of tp, If is 
positive, and the curve of inertia is an ellipse. If all the forces have the same sense, 
k* is positive and the curve is always an ellipse, and as this is commonly the case in 
practice, the ellipse of inertia alone will here be dealt with. If there are positions of 
the axis for which k 2 is negative (i. e. the sign of 2 (p y a ) differs from that of 2 (/>), the 
curve is an hyperbola. The same constructions will, however, apply in the latter 
case. 

If the point (the centre of the curve of inertia) is the centre of the system of 
forces, the curve is termed the Central Curve ; or, in the general case which arises in 
practice, the Central Ellipse. A system of parallel forces can, therefore, have an infinite 
number of curves of inertia, but only one central curve. 

If the points of application of the forces are not in one plane, the curve of inertia 
becomes a surface, termed the Ellipsoid, or Hyperboloid of Inertia, and the central curve 
becomes the Central Ellipsoid, or Hyperboloid. 

If the ellipse of inertia, or the central ellipse, of a system of forces can be drawn, 
then the radius of gyration about any axis passing through its centre is readily obtained. 
Suppose aV a! b (Fig. 91) to be the ellipse of inertia, or the central ellipse of a system 
of forces, and that the radius of gyration about any axis Y T passing through its 
centre, is required. Draw tangents to the ellipse parallel to Y Y ; then, by the 
property above enunciated, the distance k of these tangents from Y Y is the radius of 
gyration about Y Y. Moreover, if the ellipse in question is the central ellipse of a 
system of forces, the radius of gyration (k) about any cuds whatever can be obtained. 
For, from the ellipse itself, the radius of gyration (K) about a parallel axis passing 
through its centre can be obtained, and k?=K 2 +cP y where d is the distance apart of 
the parallel axes. 

The ellipse of inertia and the central ellipse will in general have to be drawn from 
two conjugate diameters, or from the principal axes. (Two diameters of an ellipse 
are conjugate when either bisects all chords parallel to the other.) Thus, determine the 
radius of gyration k x about any assumed axis a a! (Fig. 91), and draw eh,fg parallel 
to and distant k x from a a!. Then if b b 1 were the direction of the axis conjugate to a a 9 , 
determine k 2 the radius of gyration about b V, and draw ef, h g parallel to and distant 
k 2 from bb 1 . Then eh, hg,gf,fe are all tangents to the ellipse, and ehgf is its 



MOMENT OF INERTIA.— CENTRAL ELLIPSE, <fcc. 



103 



circumscribing parallelogram. To draw the ellipse, divide a e and a into any the 
same number of equal parts, and draw lines from b through the divisions of a 0, and 
from V through those of a e. The lines obtained severally intersect on the curve as 
shown. The ellipse can be completed by similar procedure. Or, if the lengths of the 



Fig. 91. 




Fiq. 92. 



£*.«*■- 




conjugate axes a a', b b' are first determined as above, the principal axes can be obtained 
directly as follows. Through b (Fig. 92) draw a perpendicular to aa\ and make b d' 
and b d on this perpendicular both equal to a 0. The line bisecting the angle d'Od 

is the major axis, and its half-length MO is equal to ^ . The minor axis 

has a length N equal to - * The ellipse can then be drawn by the paper 

trammel in the usual way. 

72. Properties of the Central Ellipse and Ellipse of Inertia.* — The following 
properties of the central ellipse and ellipse of inertia are important. 

1. If the reduced moments of a system of parallel forces about an axis XX are 
obtained, and these reduced moments are supposed to act as parallel forces at the 
original points of application ; then the centre of reduced moments, if joined to any 
point on X X, gives the direction of the diameter conjugate to X X of that ellipse of 
inertia which has as centre. 



* In order to avoid somewhat lengthy mathematical proofs, it has been thought best simply to state 
the most important properties of the Central Ellipse and Ellipse of Inertia. A much fuller treatment of 
this branch of the subject will be found in ( Element© der Graphischen Statik, 9 Bauschinger, X. Abechnitt, 
and in ' Die Graphisohe Statik,' Culmann, II. Absohnitt, Kapitel 8. 



104 



GBAPHIC STATICS. 



2. If the centre o of any curve of inertia of a system of parallel forces is joined 
to the centre of the central ellipse, the diameters of the ellipse of inertia and of the 
central ellipse conjugate to the line o are parallel and equal. The lengths of the 
other diameters (those which lie on o 0) bear the following relation to each other. If 
A is the semi-diameter of the central curve, a that of the ellipse of inertia, and d the 
distance between their centres— 

a* = A 9 + <P 

3. If any line X X is assumed as the direction of a diameter of the central ellipse 
of a system of parallel forces, and the latter are divided into two groups whose reduced 
moments about X X are determined, then if the centres of the reduced moments 
(§ 68) are separately determined for each group, the line joining these centres is 
parallel to that diameter of the central ellipse which is conjugate to X X. 

4. The centre of the reduced moments about an axis X X of a system of parallel 
forces is in the central ellipse the pole of a line drawn parallel to X X at the same 
distance from the centre of forces as X X, but on the opposite side. (In any conic 
section, an external point is termed the " pole " of the chord of contact of the tangents 
drawn from it. An internal point is the " pole " of the locus of the intersection of 
pairs of tangents drawn from the extremities of all chords drawn through it.) The 
centre of reduced moments is therefore a point on that diameter of the central ellipse 
(produced if necessary) which is conjugate to the direction of the axis X X about which 

the moments are taken. Thus, if A B A' B' (Fig. 93) is 
the central ellipse of a system of parallel forces ; then, C 
the centre of reduced moments of these forces about 
an axis X X, lies on the diameter A' A produced, this 
diameter being conjugate to BB', which is parallel to XX. 
Moreover, ifOA = a; OE = d; and 00 = x\ — 

a 1 

from which the position of C can be determined if the 
central ellipse of the system, or merely its two con- 
jugate diameters A' A, B' B are known, or have been 
obtained. The construction for determining x is shown. 

5. If a system of forces can be split up into pairs such that the lines a^ a*, b x b 2 . . . . 
joining the points of application of the forces forming each pair are all parallel, and 
if the centres of all pairs lie on a straight line X X, then X X is conjugate to the 

direction of aid*, b x 6 a , in the central curve of the system, and in all curves of 

inertia whose centres lie on X X. Similarly, if the forces forming a system can be 
split up into pairs of groups, A x A a being the centres of one pair of groups, B % Bj the 



Fia. 98. 




MOMENT OF INERTIA.— CENTRAL ELLIPSE, &o. 



105 



centres of another pair, and so on ; and if the centres of all pairs of groups lie on a 

straight line XX, then XX is conjugate to the directions A x A 2 , B x Bj in the 

central curve of the system, and in all curves of inertia whose centres lie on X X. In 
either case the line X X obviously passes through the centre of the whole system. 
If the directions of a pair of conjugate diameters are at right angles to each other, 
these directions are those of the principal axes of the central curve. 

6. If x x y X9 x 2 y 29 are the respective co-ordinates of the points of application 

a l9 a* . . . . of a system of parallel forces referred to two axes XX, Y Y, and if 
X (x y) = o, then X X and Y Y are conjugate diameters of the central curve of the 
system. 



Fig. 94. 



73. Moment of Inertia of a System of Forces determined by means of the 
Central Ellipses of its Groups. — In Fig. 94, E x is the central ellipse of a group of 
forces p'i + p f 2+ • = 2p', and B 2 the central ellipse of a group p x "+ p" + . = tp", the 
two groups being portions of a system. 
It is required to determine the moment of 
inertia of the system about an axis X x X x . 
The centre of the system is supposed to 
be known, or to have been already obtained, 
as also YY the direction of the diameter 
conjugate to X X of the central ellipse of 
the whole system : X X being parallel to 
X x Xx . Dealing first with the group p\ , 
p\ . . . , draw B x B/ the diameter of its central 
ellipse Ex parallel to XX, and determine 
A x A/ the diameter conjugate to B L B/. 
Produce A/ A x to cut X! X! in G x . Then 
the ellipse of inertia of the group jd/, p 2 ', . . . . 
will (§ 72. 2) have one diameter parallel 
and equal to Bx B/, while the length of 
the other which lies on A/ G x is the hypo- 
thenuse of a right-angled triangle whose 
sides are equal to Ci O x and A x O x . Draw O x a x equal and at right angles to A x O x : 
make d Di equal to G x a x . Then D x is the extremity of the diameter conjugate to 
X! X! of that ellipse of inertia of the forces p X9 pi . . . which has G x as its centre, 
and the perpendicular from D x on X! X x is (§ 71) the radius of gyration (k x ) of the 
group p x > pi ... about the axis Xx X x . Hence (§ 70) h x . 2 p' is the moment of 
inertia of this group about X X. Repeating the above construction for the central 

p 




106 GEAPHIC STATICS. 

ellipse E a of the group of forces pi\pi\ . . . , with similar lettering kf. %p" is the 
moment of inertia of this second group about Xx X^ The moment of inertia of the 
whole system about X x X x is therefore k*. %p* + k 2 2 . tp"; or, if there are more than two 
groups, v. 2 j/+ V. V+ V. *f +.... 

Having determined the points D lf D 2 , D s . . . for all the groups of a system it is only 
necessary to suppose the sums of the forces %p', 2jp"... to act as parallel forces 
respectively at D 1? D 3 ,. . .(tp f at D x ; 2jt/' at D a and so on), and then to treat them 
precisely as single forces by the method of § 69. Since in this construction two 
funicular polygons will have to be drawn between lines passing through D n D a , . . . 
parallel to X! X x , it is not absolutely necessary to find the positions of D x , D 2 . . . 
Thus, if a tangent to the ellipse E x is drawn parallel to X. x X x and a perpendicular is 
let fall on this tangent from O x cutting the tangent in t x and the axis X x Xi in r x : then 

h Xf determined by making h x r x = O x r x + O x t x , is the point through which to draw 
the parallel to X x X x and h x r x = k X9 the radius of gyration of the group p X9 p % \.. 
about X! Xx . 

Instead of being supposed to act at D x the sum of forces 2,p' may bo taken to 
act at D, the other extremity of the diameter conjugate to the direction of X x Xj of 
the ellipse of inertia which has O x as centre, 

74. Ellipse of Inertia and Central Ellipse of a System of Five Parallel Forces. 
— An A a . .A 6 (Fig. 94, PI. IX.) are the points of application of five parallel forces 
p x = 40 lb. ; jt? a = 60 lb. ; jo 3 = 35 lb. ; p € =z 80 lb. ; p 6 = 50 lb. of similar sense ; it is 
required to draw that ellipse of inertia of the system which has a given point Q as 
centre. 

Draw any axis X X through Q ; then (§ 72) the diameter of the ellipse of inertia 
conjugate to X X will pass through the centre of reduced moments of the forces about 
X X. Set off the forces successively on a load line 5, which can conveniently be 
taken on XX. Assuming the forces to act parallel to XX, draw the funicular 
polygon I ... V with respect to the pole 0. The sides of this polygon severally cut 
XX in ; , V, . . . 5', thus determining 0' 1', 1' 2\ . .the reduced moments of p x , /? a , . . . 
Suppose those reduced moments to act as forces parallel to X X and draw the funicular 
polygon I L . . . Y x with respect to the pole O', which should be taken at a distance from 
X X equal to 5 ; ie,%p. The extreme sides of this funicular polygon intersect in 
m 2 . Now suppose the reduced moments to act as forces at right angles to X X, and 
draw the funicular polygon I/. . .V/ with respect to the same pole O'. The sides of 
this polygon must of course be drawn at right angles to the vectors OV, O' 1'. . . The 
first and last sides of this polygon intersect in m X9 then lines from m a and m x respec- 
tively, parallel and perpendicular to X X intersect (§ 69) in M the centre of reduced 



MOMENTS OP BESISTANCE.— CENTRAL ELLIPSE AND KEEN OP A SECTION. 107 

moments about XX. Thus MQ is the direction of the diameter of the ellipse of 
inertia conjugate to X X. 

To find the length of this diameter it is necessary (§ 72) to obtain the radius of 

gyration (k) of the system about XX. By § 70, k = y/ ef . H ; where e and / are 

the intersections of the first and last sides of the funicular polygon I x . . . V! with X X, 
and H is the polar distance in the original polygon of forces. A line parallel to 
and at a perpendicular distance k from XX therefore (§71) cuts YYj i.e. MQ 
produced, in the extremity of the required diameter of the ellipse of inertia. 

In order to find the length of the other diameter it is necessary to obtain k x the 
radius of gyration about Y Y. The construction, which is precisely similar to that 
described above, is carried out by means of the two funicular polygons I a . . . V a and 

I a r . . . V a f , and the last sides of the latter cut Y Y in e lf f x . Then k x = ^/ e x f x . O a H 9 
where O a H a , the polar distance of the new polygon of forces, has been taken equal to 
OH. A line parallel to and at a perpendicular distance k x from Y Y cuts X X in the 
extremity of the other diameter, and the ellipse of inertia can now be drawn (§ 71). 

The centre of the system is determined by means of the two funicular polygons 
I . . . V and I'. . • V (§ 67). Joining M the centre of reduced moments to C, the 
direction of the diameter of the central ellipse conjugate to a line through G parallel to 
X X is obtained. The radius of gyration about the axis through M and 0, obtained 
precisely as above, determines the length of one semi-diameter of the central ellipse. 
The radius of gyration K about the conjugate diameter through C parallel to X X can 
be deduced from the already obtained radius of gyration (k) about X X, since I? = K a 
+ d 2 , where d is the perpendicular distance from C to X X. 

To avoid confusion of lines the construction for the central ellipse is not carried 
out in Fig. 94, PI. IX. This figure appears complicated, but its construction will be 
found extremely simple and it affords an excellent exercise in dealing with funicular 
polygons. 



CHAPTER X. 

MOMENT OF RESISTANCE.— CENTRAL ELLIPSE AND KEEN OF A SECTION. 

75. Bending Stress. — A beam defined as in § 41 is subject to simple bending 
stress only, when the loading can be reduced to a couple acting in the plane of 
symmetry containing the axis. If all, or any of the external loads act out of this plane, 
a stress due to twisting arises. If they are not normal to the axis, the beam has to 

p 2 



108 



GRAPHIC STATIC3. 



sustain direct in addition to bending stress (§ 48). If the loading can be reduced to a 
couple and a vertical force, then (§ 41) the beam is subject to bending and shearing 
stress. This is the case of a horizontal beam supported at both ends, and sustaining 
vertical loads acting in its plane of symmetry. 

Considering the section C C of a beam A B (Fig. 96) supported at A and B, and 
sustaining a load P. The action of P gives rise at C to a couple whose moment 
M ( = Ra . BO, the bending moment at C) tends to produce rotation contrary to 
the direction of the hands of a watch, and also to a vertical tangential force F (= Ra) 
acting in the plane of the section C — the shearing force at — tending to cause the 
right segment C B of the beam to move upwards relatively to the left segment 

It has been shown in Chapters V. and VI. how the bending moment and shearing 
forces at any section of beams variously loaded can be obtained. It remains to deal 



ft 



wkimA 



i 



Fig. 96, 



? 



P*- 



*--- » p 



neutral/ 



 



CUCLS 




with the resisting forces, or stresses, so as to be able to ascertain whether a beam is 
sufficiently strong at a given section. 

In the case of Fig. 96, for example, the bending moment M must be resisted by a 
stress couple whose moment is equal and opposite : this couple is the resultant of the 
molecular stresses in all the fibres cut by the section plane, and its moment is termed 
the Moment of Resistance of the section. 

Again, the shearing force F must be met by an equal and opposite shearing stress, 
or molecular resistance of the section to shearing. 

If the beam is bent under the loading, as shown in Fig. 97, its lower surface is 
evidently extended and its upper surface compressed. Between the upper and lower 
surface there is a layer of fibres whose lengths are unchanged, and which is termed the 
Neutral Surface. This layer, dividing fibres in tension from those in pressure, cuts the 
plane of symmetry of the beam in the Neutral Axis of the Beam, and the plane of the 
section C C in the Neutral Axis of the Section. 



MOMENT OF RESISTANCE.— CENTRAL ELLIPSE AND KEEN OP A SECTION. 109 

Id Fig. 97 CC, GCf are supposed to be two sections originally parallel and 
indefinitely near to each other. Through g, the point where C C cuts the neutral axis 
of the beam, y y is drawn parallel to C C. Then evidently, if the curve taken by the 
neutral axis is considered to be a circle with O as centre, the contraction, or loss of 
length in the upper fibres is proportional to their distances from the neutral axis, and 
similarly for the elongation, or gain of length in the 
lower fibres. The actual bending is supposed to be l '" 97 

very small, so that the straining forces remain sensibly 
parallel, and the resultant of the pressures above and the 
tensions below the neutral axis are equal and opposite. 
These resultants form a couple, whose moment p . D 
(Fig. 96) is the moment of resistance of the section. 

But, provided that the limit of elasticity is not 
exceeded, the elongations and contractions are propor- 
tional to the tensions and pressures producing them. If, 
therefore, t and p are the tension and pressure respec- 
tively in fibres distant y' and y from the neutral axis — 

t <x y' and p <x y. 

Hence if A B (Fig. 98) represents the section C C 
(Figs. 96 and 97), and XX its neutral axis, the 
intensity of stress in a layer of fibres a a distant y 
from X X is c . y, where o is a constant. 

If a is the area of the layer, its breadth being supposed very small, the total stress 
(a) in the layer is c . a . y ; the moment of this stress about X X is c . a . y\ and the  
total moment about X X of the stresses in all the fibres cut by the section plane is 
2 (c. a. y*) t or c . % {ay'). If o = 1 and y =1, then $ = c : hence the constant c is the 




o ..- , 

O- * 



stress per unit of area at a unit distance from the neutral axis. (Since no other horizontal 
forces come into play, the sum of the tensions is equal to the sum of the pressures, or 
s.% {ay) = o: hence the neutral axis in the case of simple bending passes through the 
centre of gravity of the section.) 

The total moment of resistance of the section is s . S {ay'). If d p and d, are the 



110 GRAPHIC STATICS. 

distarices of the extreme fibres in pressure and tension respectively, and£,/, the safe 
working stress per unit of area of the material in pressure and tension— 

1 d p ' 1 d t 

Hence, M _ ft nT /« 

*-d; 0T d t > 

and the moment of resistance of the section is 

Thus, in order that the section may have sufficient resistance, M, the bending 
moment at the section, must be equal to or less than the lesser of the two values 

f f 

-2 . 2 (ay 2 ) or*^-. t (ay 2 )- ^ *^ e material is equally strong to resist pressure and 

tension, as in the case of wrought iron and steel, f 9 =/*• If the section is symmetrical 
about its neutral axis, d p = d t . 

76. Resistance Area. — If the position of the neutral axis of a section is 
known, and if the stresses are proportional to the distances from this axis, then 
evidently a figure can be drawn which represents in magnitude and distribution 
the stresses over the section at the moment when the extreme fibres on one side are 
strained to the extent of their safe load. Such a figure is termed the Safe Resistance 
Area of the section. 

In the case of the rectangular section (Fig. 98), since the breadth is everywhere 
the same, the stress in any layer of fibres a a is simply proportional to the distance y 
of that layer from X X. Draw r q equal to the depth of the section and perpendicular 
to the neutral axis. Then if f v is the safe load per unit of area of the material in 
pressure and tension, and b the breadth of the section,^ . b is the stress in the upper 
layer of fibres. Set offrr' and qq 1 at right angles to r q and equal tof p .b 9 read off 
from any convenient scale which may be termed the scale of resistances : join r' q[. 
Then the whole area rr 9 q' q represents the stress over the section A B in magnitude 
and distribution. The area of the triangle orr* represents the sum of the pressures, 
and the latter have a resultant passing through its centre of gravity. Similarly oqq* 
represents the sum of the tensions. The ordinate of the figure rr* q? q opposite any 
horizontal layer a a gives the stress in that layer. 

If now the stress f p is supposed to act uniformly over the whole area rr* qfq 9 a&& 
pressure over orr f and a tension over oqqf : then the moment about X X of resistance 
of the whole figure rr* q 1 q with its varying stress is the same as that of the section 
A B with its varying stress. In the present case, as the section has a horizontal and 
vertical axis of symmetry, the pressure and tension areas are similar as well as equal. 



MOMENT OP BESISTANOE.— CENTKAL ELLIPSE AND KEKN OF A SECTION. HI 

 

They will of course be dissimilar in the case of sections which are not thus symmetrical, 
but they must always be equal, since the resultant pressures and tensions form a couple. 
In Fig. 98 the centres of gravities of the two triangles give the distances* of the 
points of application of the resultant pressures and tensions from X X ; and if D is 
the vertical distance between these centres of gravity, and Fl0 99 

A the area of one of the equal triangles, the moment of r r- 

resistance is D . A. ^ ^^^^w^ ? 

If rr* is taken as equal to 6, or if f p is taken as unity ^ ;lr i Q ^ 

on the scale of resistance, then the resistance area becomes >^lfll^ » 

an equivalent area to the section ; i. e. the area which, with j^M£MMfofo>^ 

the same material and uniform in place of a varying stress, * '* 

would have the same moment of resistance. The moment of resistance in this case 
would be D . A ,f p . It may sometimes be convenient to follow this plan if it is 
desired to form a comparison of the resistance areas of different sections of the same 
material. 

The beam tends to fail by pressure or tension, according as ^ is less or greater 

d . * 

than --• If, therefore, the resistance area is drawn (1) for safe stress in pressure; 

(2) for safe stress in tension : then that area for which neit/ier of the limiting stresses 
exceeds the safe working load of the material, will be the one corresponding to the 
actual load which the section can safely bear, or the safe resistance area of the section. 

77. Examples of Resistance Areas. — To apply the preceding section prac- 
tically, suppose that the moment of resistance of the I section given in Fig. 100 is 
to be determined by means of a resistance area. The latter to be drawn for a safe 
working stress in pressure and tension of 5 tons per square inch. Draw r o as before 
at right angles to XX, and rr* at right angles to ro, making rr'=6x5 = 22| 
tons. Join i* 0, and draw a horizontal line through the lower edge of the upper flange 
cutting r in c and ¥ in d ; then rcdr* is the resistance area of the upper flange. 

c h CO 

Make r e equal to V x 5 ( = 5 tons), and join e 0, cutting d c in h. Then — = — 

and c h represents (on the same scale as r r 7 ) the resistance of the top layer of fibres of 
the web of the section. Thus the figure orr* dh is the resistance area of the upper 
half of the I section. The resistance area of the lower half is precisely similar, and 

* As drawn, the tension area is shown to the left, and the pressure area to the right, of a vertical 
rq; this plan possesses some advantages, but has the effect of equally displacing the centres of gravity 
of these areas left and right of the vertioal. As long as the distances of these centres of gravity from 
rq are equal, therefore, the stresses have no moment about rq. The resistance area might of course be 
drawn as shown in Fig. 99, where it is symmetrical about a vertioal line. 



112 



GEAPH1G STATICS. 



need not be drawn. Find g v and g a (§ 65), the centres of gravity of the trapezium 
red r', and the triangle coh* respectively, and divide g t g a in G such that 

g!<? _ area of coh p 
g t Q area of r c d r 1 ' 

V-t'-J Fid. 100. 



then G ia the centre of gravity of the whole resistance area ; and if y is the vertical 
distance of G- from X X, the moment of resistance (M) of the I section is— 

2y. area o r r 1 d h = 2y. ^(rr'.ro — hd .eo). 




It is only necessary now to read off rr 1 and h d from the scale of resistances, and 
y, ro, and co from the linear scale, and we obtain — 
M = 97-87 inch tons, 
which must not be less than the bending moment at the section under consideration 
expressed in the same units. 

If an equivalent area is to be drawn, rr 1 must be made equal to b and re to b'. 
* The oontre of gravity of or r' d A may also be found by considering it to be the difference of the 
triangles otV and ohd. 



MOMENT OF RESISTANCE.-CENTKAL ELLIPSE AND KEEN OP A SECTION. 113 

Or the equivalent area may be drawn symmetrical about the vertical axis of the I 
section, and superposed on the section itself, as shown in Fig. 100. The construction 
needs no explanation. 

In Figs. 101, 102, 103, and 104, equivalent areas are shown for sections of given 
forms. It will be noticed that wherever the outline of the section is perpendicular to 

Fig. 102. 







r 






/ 


«^m * ^^m* ^^m» •«■» * mm 


.•^ps«^mm»^^^»< 












t 


• «l«Mfl«^l«li«B 


■1 





XX, the corresponding outline in the resistance area is a line passing through o. 
When the outline is parallel to XX in the section, it is parallel to XX in the 
resistance area. Any other straight line in the section gives an arc of a parabola in 
the resistance area. In other cases the outline of the latter is a curve, any number of 
points on which can be obtained by dealing with separate horizontal layers of fibres in 
the section. Thus, taking the layer a a (Fig. 101), make re equal to a a : join eo 9 

Fir. 103. 




cutting a a produced in h. Then A is a point on the curve outline, and any number 
of points can similarly be obtained. In the case of the hollow circular section 
(Fig. 103), points on the outline of the equivalent area can be obtained by treating 
the section as the difference of the two circles, setting a b and a 1 V from r along r r', 
and drawing lines to o cutting a b produced. Or a a' + V b ( = 2 . a a 1 ) can be set off 
from r 1 to e f , and e'o' joined. A different outline is of course obtained, but the same 
equivalent area. Both outlines are shown in Fig. 103. 

If the section has only a vertical axis of symmetry, as in the case of the trapezium 
(Fig. 105), it will be necessary first to determine its centre of gravity (§ 65), and then 

Q 



114 



GEAPHIC STATICS. 



to draw the neutral axis X X. Two resistance areas can now be drawn ; one for safe 
stress in tension, the other for safe stress in pressure (§ 76). The lower half, or tension 
area, of tbe former will be obtained precisely as before. In the present instance an 



Fig. 104. 




equivalent area has been dra^rn, and q <l is therefore made equal to b V. Points on the 
curve o q are then obtained as before. For the upper half, or pressure area, draw x x 
parallel to XX; and at a distance from it, equal to d tJ the depth of the extreme 
extended layer of fibres from the neutral axis X X. Then make the reduction of 



Of 



Fio. 105. 

of r 




b q,' 7 q 

resistances on the line xx. Thus, for the upper layer of fibres aa\ set off ce on x x 
equal to a a 1 : join o e> and produce o e to cut a a' produced in r'. Then r r' represents 
the pressure in the upper layer of fibres a a'. If this pressure is found to be greater 
than the assumed safe stress of the material in pressure, the section tends to fail by 
pressure. The construction of an equivalent area for safe stress in pressure is precisely 
similar, A line must be drawn parallel to X X, and a distance from it equal to d p : on 
this line, the reductions of resistances for the lower half of the section must be made. 
If the areas have been drawn as resistance areas, and in one of them neither of the 
limiting stresses is found to exceed the safe working load of the material, this will be 
the resistance area corresponding to the safe load of the section. 



MOMENT OF RESISTANCE.— CENTRAL ELLIPSE AND KERN OF A SECTION. 115 

It will be evident that by reversing the process above described a section can be 
drawn to correspond to any given or assigned distribution of stress. 

In Fig. 106 a resistance area for a safe stress in tension of l a 5 ton per square 
inch has been drawn, in the case of the section of a cast-iron beam. The centre of 
gravity of the section is first found by means of a funicular polygon. The section is 
cut up into five parts, as shown, and the number of square inches in each part, 
obtained arithmetically, is read from any convenient scale, and set off along a load line 
in the usual way. The reductions for the pressure area are made to a line x x parallel 

Scale %' - 70 SqVP w*. FlG ' 106 « 




S c cblf e s. 



4 .9- 9 ^ ^" • 

Zvnear yo — 1 Resistance* ¥$> - 10 Tans 

to X X, and a distance d t from it. The pressure in the upper layer of fibres at the 

moment when the lower layer is subject to a tension of 1 • 5 ton per square inch is 

rr' 12 
represented by r /, and the stress per unit of area in the upper is -g- = -g- = 2 • 4 

tons, r r 7 being read off from the resistance scale. 

* 

78. Moment of Inertia of a Section. — An area considered as made up of heavy 
elements can be said to have a moment of inertia in the same sense as a centre of 
gravity. If c^ , 03 . . . are the areas of indefinitely small elements, and y Y , y 2 . . . their 
respective perpendicular distances from an axis X X, then X (ay 3 ) (= I) is the moment 
of inertia of the whole area about XX. If 2 a (the whole area of the figure) = A, then 

V ( a y*\ 

^ A y ' = P, where k is the radius of gyration of the area about X X. Thus k is the 

distance from X X of a point such that the whole area, if considered there concentrated, 
would have the same moment of inertia as the actual distributed area. 

Q2 



116 



GRAPHIC STATICS. 



Fig. 107. 



If a figure is cut up into several portions, the moment of inertia of the whole 
figure about any axis is equal to the sum of the moments of inertia of the several 
portions about the same axis. If a section .is conceived as the difference of two figures, 
the moment of inertia of the section about any axis is equal to the difference of the 
moments of inertia of the two figures about the same axis. If I is the moment of 
inertia of an area (A) about an axis through its centre of gravity, and I' its moment of 
inertia about a parallel axis at a distance d from the first, then — 

r = i + a . #. 

Thus, if the moment of inertia of a few simple figures is known, that of a more 
complex figure can be obtained, and from the moment of inertia about one axis that 
about any parallel axis can be deduced. 

79. Moment of Inertia of a Section determined by the Funicular Polygon.— 
If a section is cut up into strips parallel to any axis, and the areas # of the strips are 

supposed to act as parallel forces at their respective 
centres of gravity, then the moment of inertia of 
the system obtained as in § 69 will be approximately 
the moment of inertia of the area, the approximation 
being closer the greater the number of strips thus 
dealt with. 

Again, suppose the section (Fig. 107) to be cut 

parallel lines into a number of strips, as 

i shown. Set off the areas of these strips 

<*i, a,, 03. . . .along a load line ll\ and 

supposing a lf a,,, 03. . .to act as parallel 

forces at the centres of gravity of the 

respective strips, draw the funicular 

polygon I II III . . . with a polar distance 

2 al IV\ 

equal to —I = — L The first and last 

sides of the funicular polygon intersect 
on X X, the neutral axis of the section. 
Produce any pair of adjacent sides VII 
VI, VIII VII of the funicular polygon to 
cut XX in d lt d 2 . Let y l9 y a . . . be the 

distances of the centres of gravity of the strips a l9 03, . . . from X X respectively. 

Then since the triangles be 0, d, d 2 VII are similar — 

&l ^t _ OC _ <*7 




y: 



or 



MOMENT OF BESISTANCE.— OENTBAL ELLIPSE AND KEBN OF A SECTION. 117 

Hence— *4.i«^lL. 

But d x d 2 . & is the area of the triangle VII d x d 39 and the whole figure contained by the 

sides of the funicular polygon is made up of all the triangles corresponding to VII d x d*. 
Hence if A 2 is the area of this figure — 

A _*(«lD 

or I = Aj . A, 

when A is the area of the section, and I its moment of inertia about X X.* 

80. Moment of Inertia of a Section obtained from the Resistance Area. — 
It was shown (§ 75) that the moment of resistance (M) of a section about its neutral 

f f f f 

axis is ^ 2 (ay*) or =y 2 (ay % )\ i.e. ^ . I or J y- . I, where I is the moment of inertia of 
a p a d p d t 

the section aboilt its neutral axis. By § 76, M = D . A .f p , where A is the equivalent 

.tension or pressure area, and D the distance apart of their centres of gravity. 

Hence— 

D.A./, = £.I, 

and 

I = d,.D.A. 

Thus, the moment of inertia of any section can be determined from its equivalent 
area. In the case' of the rectangular section (Fig. 99), for instance — 

where b is the breadth and h the height of the section : if ft = A the section becomes 
a square, and 

Is= f2' 

(The moment of inertia of figures bounded by regular curves is most readily obtained 
by means of the integral calculus.) The following table gives the values of I for 
sections of various forms : — 

* This construction must be considered as merely approximate. An error is introduced in supposing 
the reduced moments to act as parallel forces at the centres of gravity of the laminae. These moments 
should be taken to act at the centres of moments about X X of the elementary areas which make up each 
lamina. For laminae at a distance from X X, and for laminae whose breadths are very small, these 
centres of moments and the centres of gravity nearly coincide. The approximation is more olose 
therefore, the smaller the laminas are taken. 



118 



GRAPHIC STATICS. 



No. 



IL 



III. 



IV. 



V. 



VI. 



VII. 



VDX 



IX. 



Section. 



»._.fr — » 






«n 



fa^MII» 



.£._ 



. A 





>_.& -4 





IPS*"* - 

4 







...«. 






■*!•> M 



4 .. 






T 



I 






ft* 



J?" 



6i 



J 



Moment of Inertia 



6A» 
12 



6 (y - ^t 8 ) 

12 



12 



5^3 
16 



.6* = 0-6413 6 4 



5^3 
16 



. V 



12 



12 



W-V) + W + *' 3 )] 



*[*.(*-V) + 
W + V) + M**-V)] 



Area. 



bh 



bik-hj 



3 V'u _ 



V = 2-598 &• 



3^3 



V 



bA-(6-6,)Ai 



&* + &,*, 



6. *. + &*» 



(* - A,) + 6, (*, + *,) 



Distance x. 



h 

2 



2 



V2 



6^1 = 0-866 6 



h 
2 



2 



bh +ln*i(k + K) 



2 [6 A -(6-6^ A,] 



Best found by 
construction. 



MOMENT OF RESISTANCE.— CENTRAL ELLIPSE AND KERN OF A SECTION. 



119 



No. 



X. 



XL 



XII. 



XII L 



XIV. 



XV. 



XVI. 



Section. 







« bj * 




« *■;■* 



...., 





XVII. : £ 




XVIII. 




Moment of Inertia. 



Area. 



Distance as. 



6ft? 
36 



6ft 
2 



&»_}_466 1 + 6 1 a 
36 (6 + 60 



.ft 3 






6 + ti 



.ft 



3 



6 + 26, ft 
6 + 6 2 '3 



*t 



bi — bt 



(V-VH^-V) 






61 



Cii cle. 
«P = 0-0491<f 



K—*0+^(*rHO 



Beet found by 
construction. 






4 



Ellipse. 



Semicircle. 
0-110r 4 



Parabolic segment. 
T fy6ft a =:0•04o7 6ft , 



Parabolic segment. 



6ft 



Rest found by 
construction. 



2 



ft 
2 



~2 



|6* 



3&* 



0-4244r 



I* 



«» 



120 



GRAPHIC STATiCS. 



The following construction for determining the moment of inertia may sometimes 
be found convenient in dealing with figures of irregular outline. The case taken is 
the rail section (Fig. 107a), the moment of inertia I of which about the axis X X is to 
be determined. 

Draw the axis r o i* at right angles to X X' and construct the curve r V for the 
upper portion of the section precisely as an equivalent area : complete the curve for 
the lower portion, making the reductions to a layer of fibres at a distance h 1 from the 

Fio. 107a. 



.4 — .._ . — .C.___...^ fy 




axis X X equal to the distance of the top of the section from X X. In the figure the 
curves are drawn on the same side of r r\ Let x be any ordinate of the section parallel 
to X X ; 1 1' ( = a/) the corresponding ordinate of the equivalent area ; y the distance 
of x from X X. From r draw any length r t ( = C) parallel to X X, and from t draw 
a line perpendicular to X X, cutting 1 1' produced in 1". Join 1" 0, and from 1' draw 
a line parallel to 1" cutting rr* me. From a new axis r x r/ set off an ordinate ef ef 1 
equal to 1 e 9 on 1 1' produced. Let d eP = #". 



Now 



Hence 



X* x , x ' of 
— = r- f and — = =. 

y h y 



h'Q 



(a). 



Repeat the above construction for a sufficient number of ordinates of the curved outline 
r V o s, and draw the new outline r x e! 1 o 1 d through the points obtained. Let A be the 
whole area contained by this second outline and the axis r x r t f . 
By (a) 



but, 
Hence 






I = A.*\C. 



MOMENT OF KESISTANCE.— CENTRAL ELLIPSE AND KEEN OF A SECTION- 121 

If, therefore, A is obtained by a planimetcr, or by cutting the figure up into strips 
parallel to X X, and considering the strips as trapeziums,' the moment of inertia of the 
rail section about X X is determined. The construction is very simple, and when 
carried out it remains to determine the area of a figure merely, and not an area and a 
centre of gravity as in the previous construction. 

If XX is the neutral axis of the section, the moment of resistance is 

V • I = ^i • A . C. 

The figure A also gives the moment of inertia about X X of any portion of the 
section cut off by a line parallel to X X, for evidently the moment of inertia of the 
portion above x is equal to h' . C x area r x d d 1 . 

81. Central Ellipse and Kern of a Section. — In the sense in which an area is 
said to have a centre of gravity and a moment of inertia, it will also have a central 
ellipse conforming to the definition of § 71, and possessing all the properties of the 
central ellipse of a system of parallel forces stated in § 72. 

The central ellipse for any section having been once drawn, the moment of inertia 
or radius of gyration of the section about any axis whatever lying in its plane can be at 
once obtained (§ 71). To draw the central ellipse of an area, it will in general be 
necessary to know the radius of gyration about two conjugate axes passing the centre 
of gravity of the area. If the area — the section of an I or T shaped beam, for 
example — has a line of symmetry, this line will be one of the principal axes of the 
central ellipse. In this case it will only be necessary to know the radius of gyration 
about the line of symmetry, and about a second line through the centre of gravity of 
the area, and at right angles to the first : these radii of gyration are the semi-major 
and semi-minor axes of the central ellipse. In other cases, such as a trapezium or 
triangle, the directions of two conjugate diameters can be obtained by inspection 
(§ 72-5). The radii of gyration about these diameters as axes give respectively the 
lengths of the diameters, provided that the lever arms (§ 68) are taken parallel to the 
direction of those diameters, whence the ellipse can be drawn (§ 71). If the lever 
arms are taken perpendicular to the axes, the radii of gyration give the perpendicular 
distances from the extremity of each diameter on the direction of the other. 

Suppose that a system of equal parallel forces p 19 p 2 . . . . act perpendicular to a 

section at points infinitely near to each other ; in other words, suppose a uniformly 

distributed stress to act over the section, then the resultant stress passes through the 

centre of gravity (G) of the section ; and if X X is any axis in the plane of the section 

!/i 9 y* • • M *^ e distances from X X of the points of application of p u p 2 . . . ., and Y the 

distance of G from X X — 

3(py)=Y.2jp (a> 

Suppose now that the original forces p l9 p% .... are replaced by their moments 

R 



122 GEAPHIC STATICS. 

PiJ/u P*y*y • • • • about XX; i. e. suppose the section to be acted upon by a uniformly 
varying stress — a stress which varies directly as the distance from X X. The new 
system of forces will have a centre (x , and if Y' is the distance of G' from X X — 

a(ptf)=Y\S(pf> 
Substituting from (a) — 

2(py«)=Y',Y. 2j>. 

The point Y' at which the resultant of the uniformly varying stress acts may be termed 
the Stress Centre. This stress centre corresponds exactly to the centre of the reduced 
moments (§ 68) of a system of isolated forces. 

If the axis about which the moment of inertia of a section is taken is supposed to 
turn about a fixed point P, the locus of the corresponding positions of the stress centre 
is a straight line, which in the central ellipse is the polar of a point P' ; P' and P being 
symmetrically situated with respect to the centre of gravity of the section. 

Suppose now that the axis, about which the moment of inertia of a section is 
taken, moves round the section, always touching it, but never cutting it ; taking, for 

example, the successive positions t x t l9 1 2 t 2y t 3 t 3 . . . . (Fig. 108). 
Then the locus of the corresponding positions of the stress 
centre is a closed figure, termed the u Kern" of the section. 
It follows from the above that the kern of any rectilinear 
figure is itself a rectilinear figure having the same number of 
sides. The Central Ellipse and the Kern being geometrically 
related, either can be obtained from the other. Suppose that 
a plane figure is suspended vertically in a liquid so as to be 
always wholly immersed, but always touching the surface: 
then the Kern is the locus of the Centre of Pressure for all positions fulfilling the 
above conditions. 




82. Examples. Central Ellipse and Kern of Simple Figures. — Paral- 
lelogram. — The lines X X, Y Y bisecting pairs of parallel sides (Fig. 109) are the 
directions of two conjugate axes of the central ellipse. The moments of inertia of the 
parallelogram about X X and Y Y — if the lever arms are taken parallel to Y Y, X X — 

ba* . ab z 

are ^r. sin <f> and — . sin <f> respectively : the area of the figure is a b . sin <£, where 

AB'=a and AB = b. Hence if i^ and h* are the radii of gyration about X X and 
Y Y, and parallel to Y Y, X X respectively, 

Thus 

*» = </ia.±a, and ^ = ^/JbTJb. 

Make x d equal to i A! x and describe a semicircle on A' df, cutting the perpendicular 



MOMENT OF EESISTANCE.— CENTRAL ELLIPSE AND KERN OP A SECTION. 123 



at x to A' d in /: then x f = h x . Make m equal to k x and n equal to i 2 , deter- 
mining k % in precisely the same way. The ellipse can then he drawn (§ 71). 

The kern will be a four-sided rectilinear figure, for the stress centre describes a line 
while the axis turns about the angles of the parallelogram, and is a point when the 



Fig. 109. 




axis coincides with the sides of the parallelogram. When the axis turns about A the 
locus of the stress centre is the polar of A'. Draw tangents to the central ellipse from 
A' then a fS the portion of the chord of contact intercepted by Y Y, X X is one side 
of the kern. The opposite side a! j3 is traced when A' is the fixed point about which 
the axis turns : it is the polar, therefore, of A, and obviously from the symmetry of the 
figure ft = f$ and Oa' = Oo. Join a ft ft a! and the kern is complete. 



Again (§ 72-4) — 
Similarly — 



Oa=^- = 



**' A«* 



= ta. 



Oy~ ±a 

• It is easy to show that this is the oase. By the preceding §, if K is the radius of gyration about 
an axis A B, Y and Y' being the respective distance of the centre of gravity and the centre of stress of 
the section from AB, then— K" . Sp = Y . Y'Sp; or K 9 = Y . Y\ 

But, if i t is the radius of gyration about an axis through parallel to A B— 

k^v + o?: 

Hence— If + 0? = Oy(Oa + Oy) 

and hi = a • y. 

B 2 



124 



GEAPHIC STATICS. 



The kern can therefore be drawn at once without the aid of the central ellipse. 
From the kern any number of points on the latter can be obtained. Thus to 9 = 
Oct . Oy, and On 2 = 0£ . Ox. The diagonals A' A, B' B are the directions of 
conjugate axes of the ellipse, and if A' A cuts the side a ft of the kern in c, then 
? = c . A'. Eight points on the ellipse or two pairs of conjugate diameters are 
thus obtained. The curve could be drawn from either pair of diameters, or other 
points could be obtained by means of its geometrical properties. 

Rectangle. — For the rectangle sin <f> = 1 and k l9 k 2 , determined in precisely the 
same way, are the semi-major and minor axes of the central ellipse. The kern is in 
this case a rhombus whose diagonals are respectively one-third of the sides of the 
rectangle parallel to them. 

Square. — For the square sin <j> = 1 and a = b. The central ellipse becomes a 
circle whose radius is a y/^ ^here a is the side of the square. The kern is a square 
whose diagonals are parallel to and one-third of the sides of the square. 

Trapezium. — In Fig. 110 — X X and Y Y drawn through 0, the centre of gravity 
of the figure, the former parallel to, the latter bisecting its parallel sides A A', B B', are 
the directions of two conjugate diameters of the central ellipse. 

Fig. 110. 




The moment of inertia about X X (if the lever arms are taken -parallel to Y Y) 



is (see Table, p. 119.) 

The area is — 
where 



£ y + ib^ + b* 
8 6" b + b x 
h 



. sin 0. 



g(& + &i) sin*, 

of gyration about X X — 



^ V + ibb + bf _ r bb t 1 

151 -18- (b+btf L^^sKM-fri)'] 



MOMENT OF BESISTANOE.— CENTRAL ELLIPSE AND KEEN OF A SECTION. 125 

To obtain h by construction, since 

describe a semicircle on yy' cutting a perpendicular to yy drawn from c its middle 
point in d, and a second perpendicular drawn from /, the point where the diagonal 
A' B cuts y y, in g. Then — 

Make d I equal to fg. Then — 



*!. V /*IP+p!*y.»..81 k . 



Set off m, m' from along Y Y equal to k^ then mm! is the diameter of the ellipse 
conjugate to the direction of X X. 

The moment of inertia of the trapezium about Y Y is 

Hence V = A(* + y), 

which can easily be obtained by construction. 

The kern can be determined by means of the central ellipse by drawing in the 
latter the polars of points symmetrical to A, A', B, B', with respect to : it can also 
be obtained directly from the radii of gyration about XX and YY, as follows. 
Suppose the axis to coincide with BB', then a the stress centre lies on YY the 
direction conjugate to B B' in the central ellipse, and 

Oa = ^. Similarly O a' =^. 

When the axis turns about B' into a position B' Y' parallel to Y Y a , the centre of 
moments is on X X at a point y such that 

' ox 

Suppose that the axis comes into the position A' C parallel to Y Y cutting X X in a!. 

Then if 

V 



08 = 



Oa" 



8 is a point on the produced locus of the stress centre when the axis turns about A' 
The corresponding points y 8' are obviously symmetrical to y and 8 with respect to 0. 
Join a y, a -/ and produce them to cut a' 8, a' 8' in /3 and ft : Then a £ a' ft is the kern 
of the trapezium. Points on the central ellipse can now be obtained from the kern. 
Triangle. — In Fig. 113. AD bisecting B C and X X through (the centre of 



1 



126 



GEAPHIC STATICS. 



gravity of the triangle) parallel to B C are conjugate axes of the central ellipse. The 

bh* 
moment of inertia about X X (the lever arms being taken parallel to Y Y) is -^- sin $, 



where A = A D ; J = BC. Hence 



36 



*.* = 



b h* sin </> 

36 
b h sin <f> 






make m = m' = i t . 



1 "" V 6 3 



A /ft' 



The moment of inertia of the triangle ABD about TY is ^(h)- sin ^, where 



Fro. 118. 




A D = h and B C = ft. The moment of inertia of the whole triangle about Y T is 



therefore 



tit)™ * 



and- 



hth\t 



V-TT 



b (aT 8in * v 



* 



24 



MOMENT OF BESISTANCE.— CENTRAL ELLIPSE AND KEEN OF A SECTION. 127 



'BD BD 



Thus K-y/t±*.\-</ 

make n = n' = i* A pair of conjugate diameters m m', n n 1 are thus obtained. 
Drawing axes parallel to AC, AB, and proceeding as before, two new pairs of 
conjugate diameters can be determined, thus giving twelve points on the ellipse. Or, 
the latter can be drawn at once by means of the conjugate axes m m\ n n'. 

For the kern, suppose the axis to coincide with B C ; then the stress centre a lies 
on Y T and is in the central ellipse the pole of a line symmetrical to B C with respect 
to 0. Hence — y 

Ua== OD""T"** A ' 
8 

Similarly O0 = iBD'; Oy = i.CD". 

Join a /?, /J y, y a ; then a fi y is evidently a similar, similarly situated triangle to A B C, 
whose sides are respectively one-fourth of the corresponding sides of A B C. 

Again, suppose the axis to turn about A then fly is the polar of a point A' on 
Y Y symmetrical to A with respect to 0, and 

But 08=AandOA' = $*. 



12 



Hence as before k x = \/ ^ • «• Thus, if the kern is first drawn, the ellipse can be 

obtained. 

Isosceles Triangle. — If A B = A the diameters m m', n n' become the principal 
axes of the central ellipse. 

Equilateral Triangle.— -If AB = BC = AC, then h = ^ ^ b. Thus 

The central ellipse therefore becomes a circle whose radius is b y/ £ where b is the 
side of the triangle. The kern is a similarly situated equilateral triangle whose sides 

are equal to 7. 

Circle. — The moment of inertia about an axis through the centre is tti d\ where 
d is the diameter. The radius of gyration about any axis is therefore — 




The central ellipse is therefore a concentric circle whose diameter is 5 . 



128 GEAPHIO STATICS. 

¥ 

The kern is evidently another concentric circle, and its radius 



r = 



d ~S* 
2 



Ellipse. — Draw two conjugate diameters A A', B B'. Let A A' = 2 a ; B B' = 2 b 
and = the included angle. The moment of inertia about A A' is 

2 a 6*. 8in0, 

the area is tr a b sin 0. Hence if h x is the radius of gyration about A A' 

similarly for the other axis BB' K = g. 

The central ellipse is therefore a similar concentric ellipse whose axes are half 
the length of those of the ellipse itself. 

When the axis touches the ellipse at A, the position of a the stress centre is on 
A A', and if is the centre of the ellipse 

The kern is therefore another similar concentric ellipse whose axes are one-fourth 
of those of the original figure. 

83. Central Ellipse and Kern of an I Section. — The section (Fig. Ill, PL X.) 
is symmetrical about XX and YY. These lines of symmetry are therefore the 
directions of the principal axes of the central ellipse, and their intersection G- is its 
centre. 

To determine the moment of inertia about X X, the section is cut up into 10 
laminas by lines parallel to XX. Since the section is symmetrical about X X, it will 
only be necessary to deal with the 5 laminas A x ... A+ forming its upper half. Of these 
A! and A+ are rectangles; A a , A 8 , A 4 approximately trapeziums. Set off the areas 
Ai . . . A 5 (reduced to any common base) along a load line 5. In the present instance 
the areas were found by scaling off the dimensions and multiplying out. Through 
ffi > 02 • • • » the centres of gravity of the laminas, draw lines parallel to X X. Consider 
the areas A lf A,. . .to act as parallel forces along these lines, and draw the funicular 
polygon I ... V of these forces with respect to a pole at a distance H from the load 
line 5. Produce the sides of this polygon to cut X X in 0' 1'. . 5' respectively. Then 
(§ 68) 0' 1', 1'2'.. .are the reduced moments about XX of A x , A a . . . These reduced 
moments must now be supposed to act as parallel forces at the stress centres with 



MOMENT OP RESISTANCE.— CENTBAL ELLIPSE AND KEEN OP A SECTION. 129 

respect to X X of each lamina ; e. g. 4' 5' must be taken to act at the stress centre with 
respect to X X of all the elementary areas which go to make up A 8 . Now g 6 is the 
centre of the central ellipse of the area A 6 , and if m 6 (obtained as in § 82) is its semi- 
axis on Y Y, then the distance of the required stress centre (a $ ) from g s is equal to 

— ^-(§72-4). The position of a 5 can, therefore, be obtained by construction. Draw a line 

through a* parallel to X X. In the case of the remaining areas, A x . . A 4 , the corre- 
sponding semi-axes m x . . m A are very small relatively to the distances g x G-, g 2 Q-, : 

the distances a x g ly a 2 g 2 . . are very nearly nil, and the reduced moments 0' 1' . . . 3' 4' 
may be taken to act at g x . . g^. 

Using 0' 5' as a load line, draw the funicular polygon I' ... V with a polar distance 
H'. The first and last sides of this polygon cut XX in 0", 5". Then (§ 68) the 
moment of inertia of the upper half of the section about X X is— 



H.H\ 0"5", 
and the moment of inertia of the whole section is- 



2H.H'. 0"6". 

The radius of gyration (jfc x ) about X X is — 

B.ff.VP 



/ 2H.H'.0"5" / H.H r 

V2(A 1 +A l .„+A,rV 



But H' was taken equal to 5. 

Hence — h = Vh.<T5* 

Make Gtm and Qtmf equal to k t . Then mm! is the principal axis of the centra] ellipse 
lying on Y Y. 

For the moment of inertia about Y Y, the same process might be followed, the 
section being cut up into laminas by lines parallel to TT. It will be simpler 
however, to make use of the original laminas, and to adopt the construction indicated 
in § 73. Obtain b l9 6 a . . . , the extremities of the axes parallel to XX of the central 
ellipses of the several laminas, and suppose the areas A 19 A a . . . to act as forces parallel 
to Y Y at b l9 b 2 . . . respectively. Between parallels to Y Y from 6 X , b 2f . . . draw the 
funicular polygon I x . . . Y t of the forces A M A, . . . * It is not necessary to draw a new 
load line parallel to TY, the sides of the funicular polygon are merely drawn 

respectively at right angles to the vectors 0", 10" A shorter polar distance 

H" has been taken, in order to spread out the funicular polygon somewhat. The 
intercepts OTIi , I72i . . . on Y Y, given by the produced sides of the polygons, are now 
treated as parallel forces acting also at Jj, 6 2 . . . . A second funicular polygon I a . . . V, 

* In the figure the parallel to T Y from b 4 coincides with the outline of the section. 

S 



130 GRAPHIC STATICS. 



with a pole 0'" at a distance H"' from the load line 0! 5, is now drawn. Its first and 
last sides cut Y Y in a , 5,. Then &, the radius of gyration of the whole section about 
Y Y is equal to— /h». H"\ 6^ 

_ V 5T 

In the figure H"= H'"= i . 6. 



'%°a 



Hence — k = . /?" 



^"Vy '°a 5 «' 



Set off G n, G n! each equal to k* from G along X X ; then n v! is the other axis of the 

central ellipse. 

The kern is evidently a rhombus whose angles lie on X X and Y Y. When the 

axis coincides with the lowest layer of fibres D x D 3 , the centre of moments a lies on 

Y Y at a distance from G equal to 

GTS? if 



QD GD 

This determines the position of a. 

When the axis is in the position E x E l9 the stress centre ft lies on X X at a distance 

k 2 
from G equal to ^pp ; a! and ft! are respectively symmetrical to a and ft. The line a ft 

is the locus of the stress centre when the axis turns about the point D 2 . 

The central ellipse of such a section having been drawn once for all, the moment 
of resistance (M) of the section about any axis whatever can be immediately deduced. 

For (§ 75) M = J -j . I; or^ . I, and I = A K a , where A is the area of the section and 

K the radius of gyration. From the central ellipse the value of K for any axis passing 
through G (Fig. Ill, PL X.) can be obtained (§ 71). The radius of gyration (k) 
about any parallel axis in the same plane distant D from the first can thence be 
deduced, for tf = K a + D a (§ 70). 

84 Central Ellipse and Kern of an Angle Iron. — Fig. 112, PI. XI. The 
angle iron is dealt with merely as two rectangles, the rounding off of the angles being 
neglected. This rounding off can be approximately taken into account by considering 
the ends of the section as trapeziums, and the filling up of the angle as a triangle ; 
very little difference is, however, occasioned by neglecting it altogether, and the 
construction is simplified. 

The centre of gravity G of the whole section divides g x g 2 , the line joining the 
centres of gravity of its two component rectangles, inversely as the areas of the latter. 
Through G draw an axis X X parallel to one of the arms of the section. Set off 1, 1 2, 
representing the areas of the rectangles, along a load line parallel to X X. Take a 
pole at a distance H from 2. Draw lines through g X9 g 2 parallel to XX, and 
place the funicular polygon 0' I II between them. The first and last sides of this 



MOMENT OP RESISTANCE.— CENTBAL ELLIPSE AND KERN OP A SECTION. 131 

polygon intersect on X X, since X X passes through the centre of gravity : the reduced 
moments of the two rectangles about X X are equal, and equal to the intercept 0' 1'. 
It is necessary now to find the stress centres of the rectangles with respect to X X, and 
to suppose the reduced moments to act as parallel forces at these stress centres. The 
principal axes of the central ellipses of the rectangles are obtained (§ 82) ; then, if a is 
the semi-major axis of the central ellipse of the larger rectangle and d the distance of 

g x from XX — ^ = J 

which determines the position of <f> the stress centre of the first rectangle. The 
required stress centre of the other rectangle is similarly obtained ; it lies at a very 
short distance from g 2 . From these stress centres draw lines parallel to XX, and 
taking a pole 0' at a distance H' from the new load line 1' 0', draw the funicular 
polygon 0" F II' 2". Then (§ 70) k x the radius of gyration of the whole figure about 



A 



X X is equal to / ; and since H' has been taken equal to half 2 — 

V 02 

We must now either draw any two axes through G, and obtain the radii of 
gyration about each separately, or determine the direction of Y Y, the axis conjugate 
to X X, and obtain the radius of gyration about Y Y : the latter course is here followed. 
The line joining the two stress centres with respect to X X is (§ 72-3) the direction 
conjugate to X X. Draw Y Y through G- parallel to this line. To find the radius of 
gyration about Y Y, the method of § 73 may be employed. Thus, draw tangents 
to the central ellipses of the rectangles parallel to Y Y. Then if r x and r a are the 
distances of g 19 g % from the respective tangents and d X9 d 2 , the distances of g x , g 2 from 
YY, draw lines parallel to Y Y at distances from it equal to y/ d x + r x and y/d? + r a 2 , 
and on the same side of Y Y as g x and g 2 respectively : the construction is shown. 
Between these lines draw the funicular polygon 0/ Ii U x 2/, its sides being respectively 
parallel to the vectors t 19 1, 19 2 & 19 of a new load line X 2 M parallel to Y Y. The 
intercepts 0/ 1/, 1/ 2/, of the sides of this funicular polygon on Y Y are the reduced 
moments about Y Y of the two rectangles. Suppose these reduced moments to^act as 
parallel forces along the same lines, and draw the funicular polygon a I, II a 2 a with 

respect to the pole O a . Then £ a , the radius of gyration of the whole section about 

i = 

Y Y, is equal to / H t . H a a 2 a . or 8 i nce H a has been taken equal to half t 2 X — 

V 0,2, 

whence ^ is obtained. Lines parallel to and distant £, from Y Y cut X X in the 

8 2 



132 GRAPHIC STATICS. 

extremities of one axis of the required central ellipse of the section. Lines parallel to 
and distant k x from X X cut Y Y in the extremities of the conjugate axis. From these 
two conjugate axes the central ellipse is drawn. 

The kern is a five-sided figure, the construction of which presents no difficulty. 
When the axis coincides with A B, the stress centre lines on Y Y, at a distance G a 

from G equal to ~-j: • When the axis coincides with C D, the stress centre is at a point 

fi on Y Y at a distance from G equal to -=—; • When the axis takes the position R S, the 

stress centre is at y, a point on the axis of the ellipse conjugate to the direction of R S, 

such that G y = ^— , • The point 8 on the same axis of the ellipse, obtained in the 

same way, is the position of the stress centre for an axis parallel to R 8, and passing 
through A. The points € and 6, found in the same way, are the positions of the stress 
, centre for vertical axes through B and A respectively. 

85. Resistance to Shearing. — In order that a section may be able to resist the 
action of shearing force, it is necessary that the maximum intensity of shearing 
stress on any layer of fibres of the section should not exceed the safe resistance to 
shearing of the material. Shearing stress is not uniformly distributed over a section, 
but may be taken as uniform along lines parallel to the neutral axis of the section. 

If Y is the resultant shearing stress which a section of area A can safely resist ; 

then — V = i A .p; or, V = A . «, 

where p is the less of the safe stresses (per unit of area) of the material in tension and 
pressure, and s is the safe load in shear.* 

86. Intensity of Stress at the Neutral Axis. — If a a, fi/3 (Fig. 114) are two 
cross sections of a beam distant z apart, and P lf P, are the resultant stresses in all the 
fibres above the neutral surface of the beam at a a and ££ respectively, then the 
mean shearing stress over the layer of fibres a a 1 at the neutral surface is equal to 

P — P 

Pi — P a , and the intensity of stress is * * i — • This shearing stress approximates 

more nearly to the tangential, or vertical shearing stress at the neutral axis of the cross 
section at a a, as the distance z diminishes. 

For beams of simple section, the magnitude of this shearing stress may be 
ascertained from a consideration of the resistance areas at two adjacent cross sections. 
Thus, suppose the beam (Fig. 114) to have a uniform rectangular section — height A, 

* See Appendix. 



MOMENT OF BESISTANCE.— CENTBAL ELLIPSE AND KERN OF A SECTION. 133 

breadth b. Let R be the reaction at B, I the distance from B to the cross section a a ; 
M x and M a the bending moments at a a, /J ft respectively, and A! and A a the areas of 
the corresponding resistance areas. Then (§ 76) — 

Mtsp. Aj and M, = § h . A* 



but — 
Hence— 

Now- 



M, = R . I and M 8 = E (I - *). 

AB.Z , A «B . J . B . » 



P l -P 2 = A 1 -A, = |-^ = V 



the total shearing stress on a a\ The intensity of shearing stress (v) on a a' is equal to 
V 
b.z* 



Thu 



3 B 



The total shearing stress on the section a a is equal to R, and the mean intensity of 



Fio. Hi. 




R 
shearing stress is t-t. Hence the ratio of the intensity of shearing stress at the 

neutral axis of the section a a to the mean intensity of shearing stress over the whole 
section is f : 1. For a rectangular section the intensity of shearing stress is greatest 
at the neutral axis. 

87. Intensity of Stress at any part of a Section. — The shearing stress on any 
layer of fibre cd (Fig. 114) parallel to the neutral surface of the beam is equal to the 
difference between the resultant pressures at a a and ft fi of all the fibres above c d 



134 GEAPHIC STATICS. 

Let Y u Y a be the vertical ordinates under a a, flfi of the curve of bending moments; 
then, if H is the polar distance — 

M l= : Y 1 .H(§44). 



Hence — 



and similarly- 



Al = 2h ' Yl ' Hf 



A>a= 2A Ys,,H ' 



4A 4A 

Thus, if t t is made equal to — ^- 1 and r t f equal to —7-^, the triangles rto,rti are the 

resistance areas corresponding to M x and M a for the sections aa,j3j8. Produce c </ to 
cut ^ 0, f in /, $ : then the cross-lined figure 1 1! ef e represents the total shearing stress 
on e c'. 

From the above considerations a general expression for the intensity of shearing 
stress on any layer of fibres parallel to the neutral axis in a section of any form can 
be directly deduced. Let /i be the intensity of direct stress (tension, or pressure) on a 
horizontal layer of fibres of length x distant y from the neutral axis of any section. 
Let Y be the shearing force, M x the bending moment at this section, and I the distance 
of the section from the support of the beam. Let f 2 and M 2 be the corresponding stress 
jand bending moment for a second section distant z from the first : then — 

M, = V.l 

M, =V(Z-«). 
Now (§ 75) 

/1 = — -J— and/ t a — j— , 

where I is the moment of inertia of the whole section about the neutral axis. Since 
the two sections are supposed very near, the value of I is the same for both. The 
resultant of all the direct stresses in the fibres above the layer x in the first section is 

where h! is the distance of the uppermost layer of fibres from the neutral axis. The 
corresponding resultant for the second section is similarly 

Hence the total shearing stress for the layer x of the first section is, if z is very small, 

^2{'«y(M t -M,). 

But 

If, - M, = Vt. 



MOMENT OF RESISTANCE.— CENTRAL ELLIPSE AND KERN OP A SECTION. 135 

Hence the required shearing stress is equal to 

and its intensity (v) is 

or, expressed in the symbols of the Integral Calculus, 






The intensity of shearing stress is a maximum for a layer of fibres at a distance y from 
the neutral axis such that 

\r,* ydy 

is a maximum. 

The following* are the intensities of shearing stress for a few simple sections 
calculated from equation (a). 

Rectangle. — Height h, breadth 6. (No. I., p. 118.) 



v 

v max. = J r-rz- for y =s 0. 
o a 

Circle.— Radius r. (No. XIV., p. 119.) 

v max. = - — -j for y = 0. 

oirr 

Square on edge. — Side b. (No. III., p. 118.) 

vma ' m Tb* {ot y " V 

I Section. (No. VI., p. 118.) 



rmax. 



88. Curve of Shearing Stress. — A curve representing the distribution of 
shearing stress over any section can be .drawn by the following method. Taking first 
the rectangular section (Fig. 115), construct the equivalent area r t o in the usual way 

* ' Tasohenbuoh der Hfitte.' 



136 



GRAPHIC STATICS. 



(§ 76). Draw the equidistant horizontal ordinates 1 1', 2 2',... cutting the area up 
into strips of equal depth 8. Let y lt y t . . . be the distances of 1 1', 2 2'. . . from the 
neutral axis : then 



A 
2 



h 



Now the shearing stress for the layer of fibres a x a x is proportional to S* b y, and the 

2 * 
area r t V 1 above 1 1' is equal to j 2* by ; similarly the area r t 2' 2 represents the 



necessary 



Fia. 119. 



—ft> 




curve whose ordinates are respectively proportional to the areas of the portions of the 

equivalent areas above them. Let x X9 x% . . . be the mean breadths of the trapeziums 

rV 9 12'.... From any vertical axis so' set off x x on 1 1' produced, x x + ^on2 2' 

produced, and so on ; finally, set off 2 (x x ) on the neutral axis. The curve s s f thus 

obtained gives the distribution of shearing stress on the section. Thus the ordinate of 

V h 
this curve multiplied by -y . 8 . ^ gives the magnitude of the shearing stress on any 

1 L 

V 8 A 
horizontal layer of fibres, and multiplied by -= . j- . ^ gives the intensity of shearing 

X O Ik 

stress on the same layer. Moreover I = % h . <71? . 8 . «r. Hence the intensity on 
any layer is given by the corresponding ordinate of ss 9 multiplied by ^ — j — j-j. 

O • lb . & 

For a section symmetrical about the neutral axis, the curve of shearing stress 
distribution is, of course, symmetrical also. If the ordinate o V ( = 2 #i) is incon- 
veniently large, the distances a x , x x + # a , &c, should be plotted to a scale of \ or \. 
It follows from the above that the total shearing stress is always a maximum at the 
neutral axis, although the maximum intensity may occur at quite another position. 

The curve ssf (Fig. 116), obtained from the parabolic equivalent area in precisely 
the same way, gives the distribution of shearing stress on the square section shown : 
the intensity of shearing stress (v) on any layer of fibres x of this section is equal to 



MOMENT OF EE8I3TANOE.— CENTRAL ELLIPSE AND KEEN OF A SECTION. 137 

~ ' T 9 • ^' w ^ ere d is the ordinate of the curve ss' corresponding to a. In this case 

I=* oV.S * Hence * ". 

2 2 •--••i777 

A carve can easily be drawn representing the intensity of stress at all parts of the 

Fig. ne: 




section. If of' is the ordinate of the new curve corresponding to a*, then af'= — . 

v_ a" d . x 

Thus by constructing ^ = — t where C is a constant, any number of points on the 

intensity curve ss" are obtained. The constant C should be chosen bo as to bring the 
ordinates of ss" within convenient limits. In drawing the curve ss", the constant C 

was taken equal to «, and the two curves have a common ordinate at a distance -. 

from the neutral axis. 

For the T section (Fig. 117) the equivalent area is drawn in the usual way, the 

Pw. 117" 



, I _.. 




reduction for its lower portion being made to a layer of fibres at the same distance (A*) 
from the neutral axis as the uppermost layer (§ 77). In dividing np the equivalent 



138 



GEAPHIC STATICS. 



area into strips, it will be most convenient to make 8 the height of a strip a simple 
fraction — in the figure i — of the thickness of the flange of the section, setting this 
distance off on rr', from r down and from r' up, as far as the neutral axis. The 
distribution curve s a 1 s 8 is drawn as before. In deducing the intensity curve, the 
constant has been taken equal to the breadth of the web of the T section : the 
portion s' s s of this curve, therefore, coincides with the distribution curve, while the 
upper portion takes the form 8 s" s' as shown by the dotted line. In a section of this 
form the maximum intensity of shearing stress obviously occurs at the neutral axis. 



1 



89. Resistance of Long Struts. — A pressure bar or strut, if the ratio of its 
length to its least cross dimensions exceeds a fixed value depending both on the 
material and on the form of the section, tends to fail by cross-breaking. Cross-breaking 
involves a combination of bending and direct stress in the bar, which may cause the 
latter to fail before the thrust it has to sustain reaches the safe load in simple pressure. 
Various empirical formulas have been proposed for the resistance of long struts, but the 
experiments on which they are based are few in number, and the formulae themselves 
are not free from objection. The rational formula alone is therefore given, which, 
moreover, has the advantage of being easy to apply. 

The following four cases arise (Fig. 118) : — 

L Strut fixed at one end and free at the other. 
II. Both ends free, bat guided in the direction of the thrust. 
HL One end fixed, the other free, but guided in the direction of the thrust. 
IT. Both ends fixed. 

Fig. 118. 













m 



IF 



The corresponding formulae are — 



LTV IT E • I • 



U. P = n . *» 



EI 



. . (a) 



MOMENT OP EESISTANCR— CENTRAL ELLIPSE AND KERN OF A SECTION. 139 



HI. P = 2nir>-, r 

r 

IV. P = 4^^ 



(r) 



Where — 



P is the safe load. 
n the coefficient of safety. 
I the length of the strut. 
E the modulus of elasticity.* 

I the least moment of inertia of the section about an axis through the centre of gravity. 
(If A is the area of the section, then I = A • P, where k is the semi-minor axis of 
the central ellipse of the section.) 



The values assigned to n may be- 



For wrought iron 
» cast 



» 



*» 



wood 






I I 



The formulae (a), (/J), (y), (8) are to be applied only when r, or -i exceed the 

following values, h being the shortest side of a rectangular section, d the diameter of a 
circular section. 



Material. 


1 
h 


I 
d 


Formula. 


"Wrought iron .. 

V^aSu }} .. .. 

Wood 


14 

H 

8 


12 
5 
6 


(«) 


Wrought iron .. 

V^&O V f 9 .a . . 

Wood 


28 

11* 
13£ 


24 
10 

11* 


(0) 


Wrought iron .. 
v^ast* )) •• •• 
Wood 


38 
16 
19 


33 
14 
16 


(y) 


Wrought iron .. 
Wood 


56 
23 
27 


48 
20 
23 


(8) 



• See Appendix. 



APPENDIX. 



1 



WEIGHT OP A CUBIC FOOT OP VARIOUS SUBSTANCES. 



Substance. 


Lb. 


Tons. 


Wrought Iron .. 


480 


•2143 


Cast Iron 


450 


•2009 


Steel 


487 


•2174 


Oak 


52 


•0232 


Elm 


35 


•0156 


Bed Pine 


37-5 


•0167 . 


Masonry 


130 


•0580 


Brickwork 


112 


•0500 


Concrete 


120 


•0536 


Earth 


100 


•0446 


Water 


62-5 


•0279 



EOOFS. 



WEIGHT OF VARIOUS ROOF COVERINGS IN LB. PER SQUARE FOOT. 



Sheet lead .. 


•• 


• • 


• • 


• • 


• . 


7 


Sheet zinc .. 


•• 


• • 


• • 


• • 


• • 


l-25tol-63 


Sheet iron Qfo" thick) 


•• 


• • 


• t 


• • 


• • 


3 


„ corrugated 


•• 


• • 


. • 


• • 


• • 


3-4 


Slating 


•• 


• • 


• • 


• • 


• • 


5 to 11 


Tiles 


• a 


• • 


• • 


• • 


• • 


7 „ 20 


Pantiles 


• • 


m m 


• 9 


• • 


• • 


10 


Cast-iron plates (f " thick) 


• • 


• • 


• • 


• • 


• • 


15 


Boarding (£" thick) 


• • 


• • 


• • 


• • 


• • 


2-50 


Thatch 


• • 


• • 


• • 


• • 


 • 


6-50 


Slates and iron laths 


• • 


• • 


 • 


• • 


• • 


10 


Sheet iron (16 W.G.) and laths 


• • 


• • 


• • 


• • 


• • 


5 


Corrugated iron and laths .. 


• • 


• • 


• • 


• • 


• • 


5-5 


Boarding and sheet iron (20 W. 


G) 


• • 


• • 


• • 


• • 


6-5 


Timbering of tiled and slated roofs, additional .. 


• • 


• • 


5-5to6-5 



I 



APPENDIX. 



141 



WEIGHT OP ROOF FRAMING. 



Nature of Roof. 



'Pent .. .. 
Common Truss 



E 



efc 



n 
» 

99 
99 
99 
99 



I* 

o 



■8 

s 



r Manchester 

J Lime Street 

[ Birmingham 
Strasburg Railway 
Paris Exchange 
Dublin .. .. 
Derby 
Sydenham .. 



99 



St. Pancras 
Gremorne . 







Weigh! 


fc per square 


foot of 


Clear 


Diatanoe 

apart of 

Principal. 


covered area in 


lb. 


Span. 


Purlins, 
Ac 


Principal, j 

1 


Total 
Ironwork. 


feet. 


feet 








15 


•  


• • 


• • 


3-5 


87 


5 


11 


3-5 


4-6 


40 


12 


20 


3-5 


6-5 


50 


10 


•• 


•  


3-0 


54 


14 


6-5 


3-0 


9-5 


55 


6-5 


4-6 


7-0 


11-6 


72 


20 


4-2 


2-8 


7-0 


84 


9 


2-6 


5-9 


8-5 


100 


14 


• • 


• • 


7-0 


180 


26 


0-8 


6-6 


G-4 


50 


11 


• a 




9-6 


154 


26 


• • 


4-9 


• • 


211 


24 


• • 


7-3 


11-0 


97 


18 


•  


• • 


12-0 


153 


26 


9-5 


5-5 


j 150 


41 


16 


8-4 


7-3 


10-7 


81-5 


24 


10-8 


6-0 


16-8 


120 


. • 


7-9 


3-9 


11-8 


72 


•• 


8-4 


2-9 


11-3 


240 


29-33 


7-4 


17-1 


24-5 


45 


14-5 


6-2 


5 3 


11-5 



From * Wrought-iron Bridges and Roofs.' — Unwin. 



ROAD BRIDGES. 

WEIGHT OF PLATFORM, &o. 



Planking and joists .. 
Broken stone, or gravel, roadway 
Densely packed crowd 
Heavy draught horse .. 



• a 



a* 



• • 



30 lb. per square foot. 
100 „ 
120 „ 
1400 lb. 



»» 



• • 



Bails 
Ballast .. 
Timbering 
Platform girder 



•• 



RAILWAY BRIDGES. 

WEIGHT OF PLATFORM, &a 

•03 tons per foot ran of each. 
•15 to -21 line of rail. 
•07 to -17. 
•10 to -25. 



•• 



a. 



• • 



142 



APPENDIX. 



Weights 
Distances 



Weights 
Distances 



Weights 
Distances 



Weights 
Distances 



WEIGHTS OF LOCOMOTIVES AND TENDERS, G.N.tt. 

Tank Engine. 

Tender. Engine. 

13 tons 10 cwt. 14 tons 10 cwt. 12 tons cwt. 

12 ft. 9 in. 7 ft. 6 in. 



•• 



•• 



•• 



•• 



•• 



•• 



Express Passenger Engine and Tender. 
Tender. 



Engine. 



.. 10 t. 8 c. | 9 t. 18 c. | 7 t. 16 c. || 8 t. 18 c. | 15 t. 3 c. | 7 *. 17 c. | 7 /. 7 c. 
6' 6" 6' 6" 7' 1" 8' 8" 7' 9" 6' 6" 



Express Passenger Engine (four coupled) and Tender. 

Tender. Engine. 



•• 



.. fO t. 8 c. | 9 t. 18 c. | 7 L 16 c. [| 12 *. c. | 14 t. 10 c. | 9 t. 12 c 
6' 6" 6' 6" 8' 6" 8' 3" 9' 6" 

Goods Engine (six coupled) and Tender. 

Te nder. Engine. 

.. 10 U 8 c. | 9 t. 18 c. | 7 t. 16cT|| 11 t. 10 c. \ lit. c. | 9 U 2 & 
6' 6" 6' 6" 7' 9" 8' 3" T 3" 



WORKING STRENGTH OF VARIOUS MATERIALS. 



Materials. 



Wrought-iron Bars . 
Wrought-iron Plates 
Drawn Iron Wire 
Cast Iron 



Soft Steel, unhardened .. 

„ hardened and tempered 
Steel Wire 



Phosphor Bronze 



„ Beech \„ 



Wood Ash 
- Oak 



» 



Pine 



{} 







Good Brickwork 
Ordinary „ 
Stone 



Safe load in lb. per square inch. 



Tension. 

10,400 

10,000 

13,200 

3,600 



17,700 
35,400 
27,300 



9,900 



1,700 
1,560 
1 ,700 
1,000 



Pressure. 



10,400 
10,000 

10,000 



17,700 
85,400 



940 
510 
940 
510 
940 
510 
620 
310 



140 

85 

200 



Shearing. 



7,800 
7,500 

2,700 



13,200 
26,600 



7,400 



100 

• • 

85 

. . 
55 






Modulus of Elas- 
ticity EL 



28,500,000 



17,000,000 



30,000,000 



14,000,000 



| 1,500, 
} 1,400, 



000 



000 



* fa, stress parallel to fibres. 
\/3, stress perpendicular to fibres. 

Note. — The above values of the safe load may be taken for structures subject to travelling load. For structures subject to dead 
load only, these values may, in the case of Iron and Steel, be multiplied by £. 



LONDON : PRINTED BY WILLIAM CLOWE8 AND SONS, LIMITED, STAMFORD STREET AND CHARING CROSS. 



ir 



PLATE 1 



Kio. -,i). 



Fi£. 39a 



**•» 



H 




Fig. 40. 



Fig. 4.0a. 




:. nicN* v.- \' :v. 



PLATE 11. 



Fig. 41. 




Fig. 43. 




Fig. 43a. 



w. i 




; .v .• x 



•n.i.cn : r. v N* j \.v V', rk 



>> 



PLATE 111 



Fig. 44. 




Fig. 44a. 



Fi g . 44«b. 




Scales. 

i)S — 1 Tan 



1'- 10 Feet 



J C £ 

1 ■■""' r-r-T-r-r- 



-I 



Httt 



h * I.X.^]':^...' ni r.JcSf-wYork 




7 <*< YS r:>: ::." oTii-iLAcliewYorK . 









ij_ 






| 


j! 


V 


4i_ 


! 

L 


j! 

A 



-U=± .1 1 i 



4 



1 



rr - 



PLATE M. 



1 



L._ 






■00 




1 



* 



*l 



s* 



» 



I 

In 



8 



I 



3 



I 



3 
1 

4 






-co 





// / / 
/ / / / 



c=^L=^d 



,-V -r 



'* t 




/ / / , 



A -.\ . i i'L„ rr..;: % r. '/Kew Yrr*c 



1 



PLATE VII. 



Fig. 75 



0^ 




1 



Fig. 75 d. 



10 6 Q 



Loads & — W Tons. 

— y * >- 



—r— 



Jfi Twit 



Linear Scale f£ - H) Feet. 

r , , , . f . ,  , v y y — 



jpFe** 



it 



r fr r.K /t^h.!  ndin&KewY^rk. 



TIATE IX 



Fig. 94 



^ 




Scales. 
T~2Q ft 



v '..-wY • /. 



Fig. 111. 



Scale, 1'— 1 &/T JruHv. 

1,', J 



7FPPF7 



4**' 



/• k j> Ik J t 



yg&Z 




I 





i s 


\ 


j / 


\ 


i ' / 


\ 


1 / 

1 / 




I 

-Vu& Size 



4 



AlOAOttlASO 




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