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STRENGTH OF BEAMS 
 
 TRANSVERSE LOADS. 
 
 WILLIAM ALLAN, A.M., L.L.D., 
 
 i\ 
 
 LATE PRINCIPAL OF McDoNOCte SCHOOL, MARYLAND. 
 
 AUTHOR OF "CHANCELLORSVILLE," "JACKSON'S 
 
 VALLEY CAMPAIGN," " THE ARMY OF NORTHERN 
 
 VIRGINIA IN 1862," ETC., ETC. 
 
 SECOND EDITION, REVISED. 
 
 NEW YORK: 
 
 D. VAN NOSTRAND COMPANY, 
 
 23 MURRAY AND 27 WARREN STREET. 
 
 1893. 
 
PREFACE. 
 
 The usual discussion of stress in beams loaded 
 transversely, involving as it does the calculus, is 
 unintelligible to that large class of builders, and 
 others, whose mathematical training has not exten- 
 ded beyond the elements of conic-sections. Yet, the 
 most important cases may be explained without 
 using the higher mathematics. 
 
 The aim of the following pages is to put into 
 brief and convenient shape, without resorting to 
 the higher mathematics, the discussion of the most 
 important and common cases of horizontal beams 
 under vertical loads. In beams of rectangular 
 cross-section the formulae given are exact under the 
 assumptions generally made in regard to the laws 
 of elasticity. In flanged beams the approximations 
 made use of are those adopted in ordinary practice. 
 
 A graphic method of determining the amount 
 of resistance is added, which is applicable to all 
 cross sections, and which, if used with ordinary 
 skill and care, will give results exact enough for 
 every purpose. This method is so simple, and of 
 such general application, that it commends itself to 
 
6 
 
 practical builders. It is easily understood, and re- 
 quires no other appliances but those in the hands 
 of every draughtsman. In the case of unsymmetrical 
 cross-sections, (as for instance, iron rails) it affords 
 an easy means of obtaining sufficiently accurate 
 results, without resorting to the tedious calculations 
 otherwise necessary. 
 
 This discussion (in which there is no attempt at 
 originality) was prepared for the use of the inter- 
 mediate classes in Engineering, at Washing- 
 ton and Lee University. Having been found 
 serviceable to them, both during their college 
 career and in subsequent practice, I am induced 
 to hope that it may save time and labor to those 
 engaged in the same pursuits. 
 
 In prepanng the manuscript for the press, and 
 correcting the proofs, I have been much indebted 
 to one of my former pupils, Mr. Julius Krutt- 
 schnitt, C. E., now Instructor in the McDonogh 
 School. 
 
 AUGUST, 1875. 
 
STRENGTH OF BEAMS 
 UNDER TRANSVERSE LOADS. 
 
 IN the following discussion the ordi- 
 nary cases of loaded beams are treated 
 without resorting to the higher mathema- 
 tics. It is an attempt to compile from 
 various sources the simplest methods of 
 treating such cases as arise most frequently 
 in practice. 
 
 Transverse stress is produced by a load 
 applied to a beam in a direction perpen- 
 dicular, or inclined, to its length. A D 
 (Fig. 1) is a beam subjected to such a 
 stress. 
 
 In this kind of stress a compression of 
 the particles or fibres on one side of the 
 
A 
 
 
 B 
 
 FIG. 1. 
 
 beam and an extension of those on the 
 other, are produced. In consequence of this 
 the beam bends. Experiment shows that 
 the amount of compression on the one side, 
 and of extension on the other, diminishes 
 as we go inwards from the top or bottom 
 towards the centre, and at some intermedi- 
 ate plane, O O', becomes zero. The fibres 
 at this plane being neither lengthened nor 
 shortened, it is called the neutral plane, and 
 its intersection by the plane of vertical 
 section is called the neutral line or axis. 
 Experiment also shows that, from this 
 plane towards the top and bottom, the 
 amount of extension and compression may, 
 for the stresses that occur in ordinary prac- 
 
tice, be considered as varying directly with 
 the distance from the neutral plane. 
 
 The extreme top and bottom fibres suffer 
 the greatest compression and extension, 
 and in case of rupture, the rupture begins 
 with them. Some question exists as to the 
 exact location of the neutral line or plane, 
 but for slight deflections it passes through 
 the centre of gravity of the cross section 
 of the beam, and it is very probable that it 
 never deviates from this position. 
 
 In discussing transverse stress, the as- 
 sumptions based upon experiment may be 
 stated as follows : 
 
 1. The forces on the fibres are directly 
 as the amount of extension or compression 
 they produce; ( lit tensio sic vis,) and since 
 the extension and compression increase as 
 the distance from the neutral axis, the 
 forces vary in the same proportion. 
 
 2. Within elastic limits the extension 
 and compression at equal distances from 
 
10 
 
 the neutral axis arc equal, and the foices 
 producing them are equal. 
 
 3. The neutral axis passes through the 
 centre of gravity of the cross section. 
 
 RECTANGULAR BEAMS. 
 
 Let us now discuss the relations existing 
 between the forces in, and on, transversely 
 loaded rectangular beams, the load being 
 supposed to be vertical in direction and 
 the beam horizontal. 
 
 Case I. 
 
 Let A D (Fig. 2) be a beam so thin that 
 it may be considered as composed of but 
 one layer of fibres or particles. Let it be 
 fastened in a wall at AB, and be loaded 
 with W at the other end. Neglect for 
 the time the weight of the thin beam it- 
 self, which is small. Imagine it to be cut 
 by a vertical plane E F at any point, and 
 let us see under what forces the part E D 
 is held in equilibrium. 
 
 The only external force on E D is W 
 acting at D downwards, and E D is pre- 
 
11 
 
 G<* 
 
 d 
 
 vented from falling under this weight by 
 the resistance of the fibres at EF. To 
 analize these forces, let us take O, as an 
 origin of co-ordinates, and O l O f as the 
 axis of x, and O l E as the axis of y, and 
 as the forces are all in one plane, find 
 their components along these axes. The 
 internal forces, or resistance of the fibres 
 at E F, are : 
 
12 
 
 1. The horizontal forces which are ten- 
 sile above and compressive below, and 
 which increase from zero at Oj just in pro- 
 portion as we go from that point towards 
 the upper or lower edge of the beam. (Fig. 
 3.) 
 
 w 
 
 (FiG. 3.) 
 
 2. The vertical force. This is called the 
 shearing force or transverse shearing force. 
 It resists the tendency of the part of the 
 beam E D to slide down on the surface E 
 F. The existence of this force may be 
 realized if we conceive the beam to be 
 divided into two parts by the vertical plane 
 E F, and those parts to be united by some 
 very elastic substance, as india-rubber. 
 
13 
 
 Then the beam would take the form shown 
 in Fig. 4, the part F C sliding down on the 
 
 I 
 
 other. The force in the beam that resists 
 this sliding is represented in Fig. 3, by the 
 vertical arrow at E. Let it be called T. In 
 Fig. 3 are represented all the forces we 
 have to deal with. Since this system of 
 
14 
 
 forces is balanced, the following equations 
 must be fulfilled : 
 
 2X=0 2Y=Q J2M=0 (1) 
 
 That is : the sum of all the horizontal forces 
 (^X), and the sum of all the vertical 
 forces (2 Y) must each be equal to zero, 
 and the sum of the moments about any 
 point as O l must also equal zero. 
 
 The only horizontal forces in the system 
 are the two triangular groups of forces E O, 
 H and F. Oj L, representing the sum of 
 the tensile and compressive stresses on the 
 fibres. As the group EO, II acts in a di- 
 rection opposite to that of the group FO, 
 L, and as the algebraic sum of the two 
 groups is zero (^X=z:0), the groups must 
 be equal to each other. This is indicated 
 in the figure by the equality of the triangles 
 EOjHandFC^L. 
 
 The vertical forces are W and the 
 shearing force T at E F, and since 
 
 J2Y=T W' = 0. We have 
 
 T=W (2) 
 
15 
 
 Next obtain the moments of all the 
 forces about O 1 and place the sum of these 
 moments = zero. Replace the tensile and 
 compressivc forces by their resultants. The 
 resultant or sum of all the tensile forces 
 represented by the triangle E O t II (Fig. 3) 
 may evidently be represented by the area 
 of the triangle of which the base EOjis 
 the distance over which the forces are dis- 
 tributed, and the altitude E II is the stress 
 in the outside fibre. Let 
 
 S = this stress E H 
 and d = E F = depth of beam 
 
 Then area E O : H^=i S c?=N= resultant 
 of tensile forces. Similarly 
 
 Area F O, L ^Sd T$'= resultant of 
 compressive forces, 
 
 These resultants will pass through the 
 centres of gravity of the triangles E O l H 
 and O l FL, since the little forces of which 
 they are composed are represented by these 
 triangles. Hence the direction of JVwill 
 intersect O 14 E at a point G (Fig. 5), whose 
 
1G 
 
 M 
 
 c 
 
 distance from O 1 is 2 / 3 E O 1 =f-. - = . 
 
 \$ o 
 
 This is the lever arm of JV about O x . That 
 
 of N' is G' Oj also = Hence the sum of 
 
 3* 
 
 the moments of these two forces about O l 
 (since they both tend to produce left- 
 handed rotation) is 
 
17 
 
 The force T since its direction passes 
 through Oj has no lever-arm, and hence 
 its moment is zero. 
 
 If the distance from O' to O l be called 
 x, the moment of the weight W is -f-W 
 x (since it tends to produce right-handed 
 rotation), 
 
 ; -Srf f = ^M = 
 
 6 
 
 (3) 
 
 So far we have considered a beam whose 
 breadth is that of only one row of fibres, 
 but a beam of any breadth may be made 
 up of a number of such slices placed side 
 by side, and if b the number of slices, or 
 breadth of the beam, arid W=the weight 
 hung at the end of it, then eq. (3) becomes 
 
 Wx^^Sbd 2 (4) 
 
 This discussion is general and will apply 
 to any section as well as to E F. S and x 
 are the variables in eq. (4), and these 
 quantities will have different values at 
 the different sections, which values in- 
 
18 
 
 crease as we go towards AB (Fig. 2), but 
 the form of the equation will evidently be 
 unchanged. If A C (Fig. 2) be = /, we 
 have for the section at A B (Fig. 2) 
 
 -Sbd* (5) 
 
 6 
 
 AB is the section of greatest stress, and 
 the beam if overloaded will break there. 
 
 The quantity S b d 2 is called the mo- 
 
 ment of resistance of the fibres, or moment 
 of the internal forces, and is often written 
 Jl/for brevity. Wx is called the moment 
 of the weight, or moment of the external 
 forces. 
 
 Let the maximum value of S b d 9 (eq. 5) 
 be called M . 
 
 We may illustrate geometrically the 
 variation of the moments M TP#, and 
 consequently of the stresses produced on 
 the outside fibres from A to C. 
 
 In Fig. 6 let A be the beam. Take 
 a line on some scale to represent the value 
 
19 
 
 FIG. 6. 
 
 of M =W, and lay it off from A per- 
 pendicular to A C. Let AL be this line. 
 Draw L C. Then the dotted perpendiculars 
 in the triangle LAC will represent the 
 moments of resistance in the beam at the 
 several points at which they are drawn. 
 
 From eq. (2) it is seen that the shearing 
 force is constant at every section of the 
 beam. This force we may assume with 
 sufficient accuracy, for our present purpose, 
 to be uniformly distributed over the cross 
 section of the beam on which it acts. 
 Hence if A = area of cross section and, 
 
20 
 shearing force on a unit of the surface, 
 
 T = W = A* (6) 
 
 Lay off A C (Fig. 7) = I and CP = W. 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 1 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 I 
 
 I 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 FIG. 7. 
 
 Then the rectangle AP represents geo- 
 metrically the shearing stress at every 
 point of the beam. 
 
 Corollary. When several weights as 
 W WjW Fig. 8) are suspended from the 
 beam at different points, the moment of 
 resistance at any point is equal to the joint 
 moments of the weights at that point. 
 Thus, calling distances measured from C, 
 G, and E towards A, x, x\ and x 2 respect- 
 
21 
 
 f a^ 
 
 o> 
 
 I 
 
 ively, we have for the equation of moments 
 for points between C and G 
 
 Wx = ~ S b d' 2 Between G and E 
 
 D 
 
 Waj+W.a;, =-Sbd* While at K, 
 6 
 
 for instance, it is 
 
22 
 
 i^i 4 ,^ 2 ~ 
 
 The shearing force at K is 
 
 T W-hWj-f-Wa (8) 
 
 Geometrically. Let A C (Fig. 8) = I A G 
 
 = 1, and AE = J 2 . Lay off 
 L H = W 4 1, and A I = W 2 1 2 . Draw the 
 triangles as in Fig. 8. Then N P = total 
 moment at K, for instance. 
 
 The shearing force is represented by the 
 rectangles A P, N O, and S T (Fig. 9), and 
 at any point in the beam is equal to the 
 sum of the weights between that point 
 andC. 
 
 EXAMPLES. 
 
 (1.) Suppose the safe stress per square 
 inch to be 1,000 Ibs. (=S), and Z=10 ft., 
 b 3 inches, and d = 12 inches, what 
 weight will the beam support ? 
 
 (2.) Suppos3 W= 8 / 4 ton, 1=12 ft., b=2 
 inches, what must be the depth (d) of a 
 rectangular cast iron beam, so that S shall 
 not exceed 4 tons ? 
 
23 
 
 p A* 
 
 I 
 
 Let the beam be as in the last case, but 
 with the load distributed uniformly over it 
 (Fig 10). Let w weight on a unit of 
 length, W = total weight on A C, 1= A C 
 = length, d= A B =: depth, x= E C as be- 
 fore. Then the forces to be considered are 
 represented in Fig. 11, the little arrows 
 
24 
 
 along E C representing the weight distrib- 
 uted along the beam. 
 
 I 
 
 1 
 
 
 
 B 
 
 FIG. 10. 
 
 Replace the weights along E C by their 
 resultant, which is = w x, and which should 
 be applied at the middle point of E C, since 
 the little weights on the beam are uni- 
 formly distributed. Then putting the 
 resultants N and N' in place of the tensile 
 and compressive force, and proceeding as 
 before, we have 
 
 T wx=0 
 
 Tw x (9) 
 
25 
 
 Also wx.-~ 
 
 ~M (10) 
 At AB (Fig. 10) these equations become 
 
 <") 
 
 1 
 
 ) 
 
 By comparing the last equation with eq. 
 (5), we see that if the weight and beam be 
 the same the stress on the fibres in this last 
 case is only one-half what it was in the 
 former, or, what amounts to the same, the 
 beam will bear twice as much distributed 
 over it, as it will when the weight is con- 
 centrated at the extremity. 
 
 From eq. (9) we see that the shearing 
 force is not constant as in the last case, but 
 varies as x. It is greatest at A. 
 
 Geometrically. The equation M = % w x 2 
 corresponds with that of a parabola with 
 vertex at C and axis vertical. Lay off A L 
 (Fig. 12) =i?of, and through L and C 
 
o 
 
 T 
 
 LLLLUUUiiU, 
 
 -is 
 
 FIG. 11. 
 
27 
 
 draw a parabola. The ordinates of this 
 parabola (dotted in the figure) will repre- 
 sent the moments at the several points. 
 
 The equation T =.w x is represented by 
 the triangle A P C (Fig. 13), which there- 
 fore gives the shearing stress at every point 
 of the beam when A P is taken -=.wl. 
 
 Corollary 1. When the load is distrib- 
 uted over only a part of the beam as in 
 Fig. (14), let R C=m=the loaded part, 
 and take the other letters as before. Then 
 the equations for any section in the loaded 
 part are evidently the same as those just 
 obtained, viz. : 
 
28 
 
 FIG. 13. 
 
 O 
 
 3 
 O 
 
 ffl 
 
29 
 
 (12) 
 
 At R V 
 
 And T=wx 
 
 But at any section E F between A and 
 R the moment of the load is = wm 
 (x Vs m )> the latter factor being the dis- 
 tance from the centre of gravity of the 
 load to the section E F. The moment of 
 resistance having the same form as before, 
 we have for the equation of moments for 
 any section in R A 
 
 wm (x l / 2 m) -Sbd* =M (13) 
 
 At A this becomes 
 
 w m (I \ m)=M the greatest moment. 
 
 The shearing force at E F being equal 
 in amount and opposite in direction to the 
 whole load between E F and C will be 
 
 T=w m (14) 
 
 Geometrically. For the moments: lay 
 off A C = I and C R m (Fig. 15). At R 
 
 erect PR and at A make AL=:M . 
 
 /v 
 
30 
 
 Through C and D draw a parabola as in 
 the last case, and (since eq. [13] is of the 
 first degree) through D and L draw a 
 straight line. Then from C to R the ordi- 
 nates of the parabola represent the mo- 
 ments, and from R to A they are represented 
 by the ordinates of the trapezoid R L. For 
 the shearing stress (Fig. 16), lay off from 
 
31 
 
 <q fc, 
 
 R, R N=w wi. Draw C N and N P. Then 
 the triangle CRN (corresponding to the 
 equation T=wx) gives the shearing force 
 at each point in C R while the rectangle R 
 P (corresponding to the equation Tw m) 
 gives the force in the remaining segment 
 of the beam. 
 
 Corollary 2. When there is a load W 
 at the extremity C, in addition to the load 
 
uniformly distributed over the beam (Fig. 
 17), we have a combination of Cases I. and 
 
 B 
 
 E 
 
 a 
 
 E 
 
 FIG. 17. 
 
 II. and the moment of the external forces 
 at any section, E F, is 
 
 Hence the equation of moments is 
 
 -f-W x (15) 
 
 M=-S5<f= 
 6 
 
 This is greatest at A, or 
 
 M =-S b d =i w l*+ W I 
 6 
 
 The shearing force 
 At A 
 
 (16) 
 
 (17) 
 (18 
 
33 
 
 Geometrically. The simplest way of 
 representing the moments is to construct 
 those due to each kind of weight, and 
 
 qn /v* 
 
 then combine them. Thus, let M' = 
 
 and M" =W x. Construct M' as in Case 
 II., it being represented by a parabola 
 with vertex at C and axis vertical, and M" 
 as in Case 7"., it being represented by a 
 triangle (placed under A C for conven- 
 ience). Since M = M' -f- M" from eq. 
 (15) we have the total moment at any point 
 E (Fig. 18) represented by the sum of the 
 ordinates of these two figures=N P Fig. 
 (18.) 
 
 The moments may also be represented 
 by the parabola corresponding to eq. (15) 
 as in Fig. (19.) This parabola has its ver- 
 tex at C' and not at C. Of course, only 
 that part of the curve between C and A is 
 applicable to our purpose. 
 
 The shearing force is represented by 
 adding the triangle N PP' (Fig. 20) repre- 
 senting the variable part w x of T to the 
 rectangle C P which represents the constant 
 part W of T. 
 
34 
 
 EXAMPLE. 
 
 Discuss the forces when the load is distrib- 
 uted as shown in Fig. (21) assuming 
 various values for LN and N C as well as 
 for w and W. 
 
 Case IIL 
 Let the beam whose length is I rest upon 
 
35 
 
 supports at B and D (Fig. 22) and let it be 
 loaded at some point G with a single 
 weight W. Let m and n be the segments 
 into which the beam is divided at the point 
 G of the application of the weight. 
 .First find the proportions of the weight 
 
36 
 
 2 
 fa 
 
 supported at B and D, or in other words, 
 the reactions of the supports. By the prin- 
 ciple of the lever the respective portions of 
 - the weight supported at B and D are in- 
 versely proportional to the distances of these 
 points from G. Thus, let W' reaction at 
 D and W" = reaction at B and then 
 
 W : W": :m:n 
 \y"-f W : W' : : m+n : m 
 
o tv 
 
 ot 
 
 g g 
 
 a a o 
 
 FIG. 21. 
 
 1 
 
 I 
 
 FlG. 
 
38 
 But W'+W"=Wand m + n=l 
 
 And so W"=4w 
 
 I 
 
 Now apply the conditions of equilibrium 
 to any part A E of the beam counting from 
 .A. (Fig. 23), 
 
 H 
 
 2 
 fe 
 
 S 
 
39 
 
 1st. Between A and G. ^'X~0 merely 
 indicates the equality of X and 1ST', as these 
 are the only horizontal forces. 2 Y~0 
 shows that the shearing force at the section 
 E F is downwards 
 
 and=W" . . T= W" = ^W (20) 
 
 2 M= 0. The joint moment of N and 
 
 N' is as before=-S b d\ That of T is 
 6 
 
 zero. The only other force acting on A 
 E is W", the reaction of the abutment. 
 Let O O l x. Then the moment of W" 
 is 
 
 W"X= .Wa? 
 
 11 1 
 
 Hence j.Wx 
 
 .~Wx=-Sbd 2 = M (21) 
 
 i 6 
 
 2d. Between G and C (Fig. 24). Here, 
 between A and E are the two external 
 forces W" at B and W at G. Hence the 
 shearing force at E F is upwards and 
 
40 
 
 P S3 
 
 Sz; 
 
 T= W W= W' = yW. (22) 
 
 For >' M =rr we have 
 
 - - S b c7 2 + W 'xW (xm) = 
 . . y Wx W (xm) = l - S b d 2 M (23) 
 
 fr U 
 
41 
 
 The greatest value of eq. (21) is at G, 
 where* it becomes identical with eq. (23) 
 for the same point. This value is 
 
 At A and C the moments are zero. 
 
 Geometrically. From eq. (21), which is of 
 the first degree, it is seen that the moments 
 
42 
 
 vary in A G as they did in Case I. Hence 
 they may be represented by the ordinates 
 of the triangle A L G. (Fig. 25). 
 
 Eq. (23) is also that of a straight line 
 cutting the axis of X at C. Hence the 
 moments in G C are represented by the 
 triangle G L C (Fig. 25). The shearing 
 
 I 
 
43 
 
 force in A G is represented by the rectangle 
 A P and that in G C by the rectangle CP'. 
 (Fig. 26.) 
 
 Note. That the maximum moment (at G) 
 corresponds to the point where the shear- 
 ing force passes through zero. 
 
 Corollary 1. When the weight W is at 
 the middle of the beam, we have 
 
 Then eq. (21) becomes 
 
 V, 
 
 and (23) is 
 
 - x ,=\ 
 
 (25) 
 
 At the centre M = Y 2 W. V, I = l /,W I (26) 
 The triangles A L G and G L C (Fig. 27) 
 
 represent the moments in this case. G L 
 
 having been laid off ==. */4 W L 
 
 The shearing force throughout the beam 
 
 is then T= l / 2 W (27) as is shown in the 
 
 rectangles (Fig. 28). 
 
 Comparing the value of M given in eq. 
 
44 
 
 (5) Case I. with that of M in eq. (26) we 
 see that the load and the length being the 
 same, a beam will bear four times as much 
 with both ends supported, and the load 
 placed in the middle as it will do with one 
 end fixed and the other loaded. 
 
 Corollary 2. When there are several 
 weights, as in Fig (29), the moment of the 
 
external forces at any section is that due 
 to the action of all the weights. Let the 
 segments into which the weights W, W, 
 and W 8 divide the beam be m and n for 
 W, m l and n l for W l and m 2 and n^ for 
 
 w f . 
 
 Let R a reaction of abutment at B and 
 R 2 = reaction of abutment at D and 
 = length and let x be counted from 
 A as before. 
 
46 
 
 
 
 
 >.S 
 
 ^ ^ 
 
 v 
 
 
 
 
 
 s 
 
 
 ., 
 
 
 
 a 
 
 1 
 
 t-/t% 
 
 %h 
 
 
 
 - 
 
 X 
 
 ; y| 
 /^% 
 
 
 
 , 
 
 a 
 
 ' 
 
 <J 
 i__/^ifc 
 
 
 
 y 
 
 s 
 
 > 
 
 - V 
 
 tt 
 
 
 
 , > 
 
 9 
 
 <i PQ 
 
 |f^^W 
 
 The reaction of the abutment at B due 
 to these weights is 
 
 ???__.. m ^ , m 9 ^ 
 So at D R 2 =z W+ 7 -'Wj+^Wg 
 
 It i 
 
 t 
 
 For any section between A and G, Rj 
 
'47 
 
 is the only external force and hence the 
 equation of moments is 
 
 R^ifi b cf=M (29) 
 
 The shearing force is T=R, l (30) 
 
 For every section between G and G' 
 the weight W is to be taken into consider- 
 ation and the equations are : 
 
 R^ W(x m)=^S b <F=M (31) 
 
 TznR, W (32) 
 
 For any section between G' and G") 
 we have : 
 
 (33) 
 
 T=R 1 _W W, 
 
 For any section between G" and C 
 R i xW(xm)W 1 (xm } )W 2 \ 
 
 (xm^)=^Sbd 2 = 'M. I (34) 
 
 T=R l W W, W, ) 
 
 The location of the greatest moment is 
 most readily determined by geometrical 
 construction. 
 
48 
 
 Geometrically. The moments are rep- 
 resented in Fig. (30) by constructing sep- 
 
 o 
 
 CO 
 
 arately those due to each weight and then 
 combining them. Thus, the moments pro- 
 duced at every point in the beam by the 
 weight W are represented by the triangle 
 
> 49 
 
 A L C (Fig. 30), in which G L equals the 
 greatest moment due to W= W m. Sim- 
 ilarly A L' C represents those due to W,, 
 
 n 
 G' I/ being = 1 W 1 m 1 and A L" C gives 
 
 ?i 
 those due to W 2 , G" L" being=i- 2 W 2 m 2 . 
 
 Now, if at every point we add together 
 the ordinates of these three triangles for 
 that point, and lay them off above A C we 
 shall get a polygon A H H' H" C, which 
 represents eqs. (29) (31) (33) and (34) and 
 gives the total moment at any section. The 
 greatest ordinate of this polygon will, of 
 course, show the location of the maximum 
 moment. This will be at G or G' or G"> 
 according to the relative amounts and pos- 
 itions of the weights W, W,, and W 2 . In 
 the Fig. it is at G'. Hence from eq. (31) 
 
 R m l W(m l m)=-. S b d 2 = M (35) 
 The shearing force may be represented 
 
50 
 
 3 
 
 
 
 ..*.... 
 
 ~" 
 
 as in Fig. (31). It is greatest in that one 
 of the two end segments which corresponds 
 to the greater of the two quantities R a and 
 
 Note. That in this case the simplest way 
 of finding the point of maximum moment 
 
is to construct the figure representing the 
 shearing force, and the point when the 
 shearing force passes through zero, (G' 
 Fig. 31) is the point sought. 
 
 EXAMPLES. 
 
 1. Let fc20 ft., m 6 ft., ^ = 10 ft., w f = 
 15 ft., W=l ton, W, = 2 tons, W 2 =3 tons. 
 
 Find the maximum moment. 
 
 2. Find the size of a rectangular wooden 
 
 beam where 
 
 /=15 ft., m=3 ft., m 1 =. 6 ft., wi 2 = 14 ft. 
 W=l ton, W,=i ton, W k =2 tons 
 8=1,000 Ibs. and d^ b. 
 
 Case IV. 
 
 Let the load be equally distributed over 
 the beam (Fig. 32). In this case the reac- 
 tion of each abutment =% the load, or 
 
 R,=^=K, (36) 
 
 rj 
 
 Take any section E F whose distance 
 from A=&. Then the external forces 
 
CC 
 
 6 
 
 acting between A and E F are, R 1 and the 
 resultant of all the little weights from A 
 to E (w x). This last force acts at its 
 centre of gravity (Fig. 33), which is half 
 way from A to E. Its lever arm is there- 
 
 oc 
 
 fore = Hence the equation of moments 
 & 
 
 will be 
 
53 
 
 CO 
 CO 
 
 or 
 
 (37) 
 
 This is a maximum at the centre where 
 M = /. to P (38) 
 
 The shearing force at E F is 
 
T=- wx (39) 
 
 This is greatest at the abutments where x 
 I, or 
 
 . T _ 
 ~ 2 
 
 (40) 
 
 At the centre T=0 
 
 Geometrically. The values of M in eq. 
 (37) may be represented by a parabola 
 
^ 55 
 
 with vertex at L, the ordinates G L 
 (Fig. 34) being taken = M . The shear- 
 ing force is represented by the two triangles 
 A P G and G Q C (Fig. 35). 
 
 
 - 
 
 The maximum moment exists at the point 
 (G) when the shearing force is zero. 
 
56 
 
 Corollary. When the uniform load ex- 
 tends only over a certain distance from one 
 of the supports, as in (Fig. 36), let A T>= 
 
 CO 
 
 CO 
 
 A 
 
 loaded segment m. The reactions of the 
 abutments are: 
 
 At A, R^w ml - Y~ ) ) 
 
 V l ' 
 
 At C, l&i= 
 
 (41) 
 
Then for any section in A D the moments 
 of the external forces will be as in the case 
 just discussed. 
 
 l 
 
 \ 
 
 w 
 
 And the equation of moments in A D will 
 be: 
 
 w m 
 
 tl% m\ wx* 1 
 
 ( - x =- S o a =M (42) 
 \ I / 2 b 
 
 For any section in D C the whole load 
 (w m) is to be considered as acting through 
 its centre of gravity (G), and the equation 
 of moments is: 
 
 fl - m\ , . 1 C1 7 72 T/T 
 
 l \x wm(x> %m)= S o a =M 
 
 Reducing 
 
 '^f(l x ) = I S b c7 2 ==M (43) 
 
 /v 6 O 
 
 The shearing force in A D is 
 
 /l=$ m \ ^ 
 
 L w m.l - - J w x I 
 
 (44) 
 In D C, T=w m. 
 
58 
 T is a maximum at A, or 
 
 Geometrically. Eq. (42) corresponds to 
 the parabola A L K (Fig. 37), which cuts 
 
 A C at A and K (whose distance from A= 
 y- [I i m] ) and whose axis is vertical 
 
59 
 
 Eq. (43) corresponds to the straight line H 
 C. We can only use the part A L H of 
 the parabola, the moments in D C being 
 represented by the triangle D H C. The 
 maximum moment is at N corresponding 
 to the vortex L of the parabola. The value 
 of this moment is: 
 
 (45) 
 
 which is obtained from eq. (42) by substi- 
 tuting for x the value A N" (=4 A K)= 
 
 tn 
 
 *(l 4 m). This M is always less than 
 
 the maximum moment that exists when the 
 load extends all over the beam as will ap- 
 pear by making m to vary in eq. (45) and 
 applying the tests for a maximum to it. 
 
 The shearing stress for the loaded seg- 
 ment is represented by the triangles A P K 
 and K P' D (Fig. 38), and for the other 
 Segment by the rectangle D H. The point 
 K, where the stress is zero, is found by 
 making in eq. (44) 
 
60 
 
 rp il\m\ 
 
 I =10 m f * w x=0 
 
 and finding the value of x. 
 
 This point corresponds to the maximum 
 moment. It may also be found graphi- 
 cally by constructing Fig. 38, and, as be- 
 fore, affords the easiest method of deter- 
 
mining the point of the beam where the 
 maximum moment exists and where conse- 
 quently there is greatest danger of rupture. 
 
 EXAMPLES. 
 
 1. Let 1=20 ft. -20=500 Ibs. per ft. m= 
 15 ft. and let there be a weight in addition. 
 
 G 
 A 
 
 ) @ Q Q Q 
 
 c 
 
 
 1 
 
 i 
 
 f 
 
 ! 
 
 TV 
 
 FIG. 39. 
 
 W=5 tons at a point 18 ft. distant from 
 A. Required the maximum moment. 
 
 AT 
 
 FIG. 40. 
 
62 
 
 2. Let one-half of the above beam be 
 loaded with a uniform load, w=~L ton per 
 foot, and the other half with a uniform 
 load of w =i ton per foot. Required 
 the moments. 
 
 Case V. 
 
 A single moving load. When a single 
 moving load passes over a beam, as in 
 (Fig. 41), the maximum moment at each 
 instant (as appears from Case III.) takes 
 Affect at the section just under the weight. 
 To determine the law of variation of these 
 maxima as the weight travels over the 
 beam: Let rc=the distance at any instant 
 from A, and then the reaction of A at that 
 instant (= the part of W transmitted to it) 
 
 I x 
 
 W Multiplying this by the lever 
 
 V 
 
 arm x we have for the moment under the 
 
 W 7* 1 
 
 weight : (l-x) = ~ S b d a =M (46) 
 
 This is a maximum at the centre, where 
 
Eq. (46) corresponds to a parabola (Fig 41) 
 with vertex at L, the ordinate G L being= 
 V 4 W L The shearing force for each seg- 
 ment into which W (Fig. 42) at any instant 
 divides the beam is equal to the reaction of 
 
61 
 
 the abutment corresponding to that se# 
 ment. Thus, if W is at a distance x from 
 
 A the reaction of A is =W~ ^L a nd of C 
 
 T x 
 
 r 
 
 If W has the position marked 2 in (Fig. 
 42), then the shearing stress in the left 
 
 I 
 
 ^ <9 P4 fe 
 
segment is shown by the rectangle 
 W and in the right segment by the rec- 
 tangle WN" N"' C. The diagram shows 
 in a similar way the stress at other points. 
 If the third position of W in the figure is 
 at the centre of the beam then evidently 
 the greatest shearing stress to be provided 
 for in the left half of the beam will be 
 represented by the locus of the points like 
 L, N', P', and for the right half it will be 
 the locus of the points P", Q", L', etc. 
 
 These loci are given by the equations : 
 
 W 
 
 T= *)= eq. of L C ) 
 
 ^ (48) 
 
 T= x =eq. ofALM 
 
 L * ' 
 
 Turn the triangle A L' C down for con- 
 venience, as in (Fig. 43), and then the 
 shearing stress to be provided for is given 
 b.y the figure A L P I/ C. In this figure 
 
 AL = CL' = W and D P==^. 
 
Case VI. 
 
 A distributed moving load. When a 
 moving load gradually covers a beam, 
 (Fig. 44), moving on from one end as a long 
 train of cars, the maximum moments pro- 
 duced is that due to the load when it covers 
 the entire length of the beam, and conse- 
 
67 
 
 o 
 
 
 
 queritly this case is provided for in Case 
 IV. 
 
 But with the shearing stress it is dif- 
 ferent. Here, as in Case V., we need the 
 locus of the greatest shearing stresses that 
 be brought upon the beam. This maximum 
 at any section D occurs when the longer 
 segment into which D divides the beam is 
 
68 
 
 loaded, and the other is not. In that case 
 the shearing force at D (=the reaction of 
 the abutment C) is 
 
 T- (49) 
 
 ~ 2 I 
 
 This equation gives the parabola A 
 N P' (Fig. 45) with vertex at A, where 
 
' 69 
 
 nn 7 
 
 CP' = - . When the load comes from 
 
 the other end of the beam we get the 
 parabola C N P. Hence the figure A P 
 N P' C gives the maximum shearing stress 
 to be provided for. 
 
 It is easy to see that the shearing 
 stresses thus obtained are greater than 
 those which exist whep the load covers 
 the entire beam. In the latter case the 
 forces are represented by the triangles 
 A P G and G P' C (Fig. 45), the shearing 
 ' stress at D being given by eq. (39J 
 
 In the case of the passing load we have 
 just seen that 
 
 T WX * TK 
 ~l 
 
 /C i 
 
 The value of D H is always less than that 
 
 of D K when x >_-^ ; for if 2 x be a certain 
 
 
 quantity, then the product of the halves 
 of that quantity (=x 2 ) is greater than the 
 
product of any other two parts ('such 
 as I and [2 xl\ ) into which it can be 
 separated. 
 
 That is 
 
 w x 2 w I (2 xl 
 TT > 2 I 
 
 W X 
 
 TT> * 
 
 (50) 
 
 In the expressions of the moment of 
 resistance M= S b d~ the quantity de- 
 
 noted by S ( = the stress on the outside 
 fibres) varies directly as M. Hence, all 
 the geometrical illustrations we have 
 given of the moments may apply equally 
 well to the values of S. The maximum 
 moments give the maximum stress on 
 the fibres, and indicate the points of rup- 
 ture when the beam is loaded with its 
 breaking weight. 
 
 ULTIMATE VALUES OF S. 
 
 If beams are loaded transversely until 
 fracture takes place, the value of S or the 
 stress on the outside fibre which exists at 
 
71 
 
 the moment of fracture, gives us a value for 
 the tensile or compressive strength of 
 the material according to the manner of rup- 
 ture. If the beam yields by tearing, S 
 gives us the tensile strength, if by crushing 
 S gives the compressive strength. We 
 readily obtain the value of S answering 
 to the ultimate strength from any of the 
 formulas under " Transverse Stress," by 
 substituting given values for /, #, and 6?, 
 
 and the actual breaking weight for W. 
 
 - 
 
 But the tensile and compressive strengths 
 of materials are also obtained by direct 
 tension and compression, the force being 
 applied in the direction of the length 
 of the .bars until rupture takes place. 
 
 If our theory were perfect the values 
 of tensile and compressive strength thus 
 deduced would agree with the ultimate 
 values of S found in tranverse stress; but 
 they do not. 
 
 The difference is very wide sometimes. 
 Thus in cast iron, S (in this case it repre- 
 sents the tensile strength) derived from 
 
breaking rectangular beams by a trans- 
 verse load is nearly 20 tons per square 
 inch, while the tensile strength obtained 
 directly is only about 8 tons. This dis- 
 crepancy has been accounted for in two 
 ways. 
 
 1. That the neutral axis moves towards 
 the compressed side, and that therefore 
 a larger portion of the beam is subjected 
 to tension than the formula supposes. 
 
 2. That the neutral axis always remain- 
 ing at the centre of gravity of the beam, 
 the additional strength is due to the 
 adhesion of the fibres which is developed 
 by the unequal lengthening, and short- 
 ening of them as we go from the neutral 
 axis towards the surfaces. In favor of this 
 view is the fact that we know such ad- 
 hesion to be an element of strength; for 
 the compression or extension due to a 
 given force is not so great in a trans- 
 versely loaded beam as in one directly 
 compressed or extended. 
 
 The action of this adhesive force may be 
 illustrated as follows: 
 

 FIG. 46. 
 
 In the beam A B (Fig. 46), strained by 
 the weight W, all the fibres are equally 
 elongated, and they only resist by their 
 direct tenacity. But in the beam A'C 
 to the one-half of which is appended the 
 weight, while the other half, E 0, is less 
 strained or altogether prevented from 
 extending, evidently W will have to 
 overcome not merely the tenacity of the 
 fibres in A' B' but the adhesive force of 
 
74 
 
 the fibres along the plane E F, where the 
 two parts of the beam join ; for this force 
 will tend to prevent the stretching of the 
 fibres in A' 13', and consequently increases 
 the strength of A' B'. This kind of force 
 exists between every two layers of hori- 
 zontal fibres in a beam under transverse 
 loading, and is called the longitudinal 
 shearing stress. It is neglected in the 
 formulae we have given. 
 
 From the variation between the ulti- 
 mate values of S (called moduli of rupture) 
 and the values for strength obtained by 
 direct tension and compression, it results 
 that the values should be determined in 
 both ways, and that the values gotten by 
 one method should not be used in 
 calculations involving the other kind of 
 stress. 
 
 BEAMS OF UNIFORM STRENGTH. 
 
 As already stated, in solid rectangular 
 beams, S has different values for the va- 
 rious points in the length of the beam. 
 There is always a point of maximum 
 
. 75 
 
 stress where the beam, if loaded suffi- 
 ciently, will break. Now at all other 
 points there is an excess of material 
 which is useless and injurious from its 
 own weight. To secure the requisite 
 strength with the least material is an ob- 
 ject usually desirable, and this can be 
 readily accomplished in certain materials 
 (as cast iron), by giving the beam such 
 a shape as will make S, the stress on the 
 outside fibre, constant throughout its 
 length. In wood the injury resulting 
 from the cross cutting of the fibres fre- 
 quently prevents the putting of the theory 
 into practice. 
 
 The application of the theory of uni- 
 form strength to beams of rectangular 
 cross section may be most simply explained 
 by taking up the cases we have discussed 
 in detail. 
 
 In Case I. from eq. (5) the maximum 
 stress in the outside fibre is, 
 
 (51) 
 This stress only occurs at A, where the 
 
76 
 
 beam will ultimately break, and it is 
 evidently possible to take away some 
 of the material between that point and 
 
 FIG. 47. 
 
 C without diminishing the strength. If 
 this be so done that at every point be- 
 tween A and C there shall exist on the 
 outside fibre a stress equal to that at A, 
 the beam will be one of uniform strength, 
 and we shall have attained the greatest 
 economy of material. Let us suppose, 
 the use we have for the beam requires 
 the depth to be uniform. What must be 
 its plan in order that S shall be constant 
 in value, or the beam be as liable to 
 break at any other point as at A 6 . 
 
 In eq. (4) S=-^- - , if we assume S to 
 
77 
 
 be constant, the other side of the equation 
 must be constant also, and since 6 Wis 
 constant, and we have made d constant 
 by assuming the depth to be uniform, the 
 whole expression can be constant only 
 when b varies as x. 
 
 byox whence b=c x (where c = 
 some constant factor). This equation 
 which is that of a straight line shows that 
 the breadth must vary directly as the 
 length. Hence the plan should be a tri- 
 angle with vertex at C (Fig. 48). 
 
 FIG. 48. 
 
 On the other hand, if we suppose the 
 breadth to be uniform and wish to have 
 
 6 Wx 
 
 S constant, in the eq. S= v~7i~> x 
 
 u d 
 
 vary as d, or 2 
 
 d 2 c x 
 
78 
 
 This corresponds to a parabola, and the 
 beam, if the top be straight will have the 
 elevation shown in (Fig 49). 
 
 Suppose that b varies as d, then d =n b 
 (?i being a constant) and 
 
 S= 
 
 6 W x 
 
79 
 
 To render S constant we must have, 5 8 x> 
 x . . b*=cx and d*=n B ex. 
 
 These are the equations of a cubic para- 
 bola. Hence the horizontal section (Fig. 
 
 FIG. 50. 
 
 50), and the vertical section (Fig. 51),, 
 should be curves of that kind. The cross 
 
 FIG. 51. 
 
80 
 
 section is rectangular as in (Fig. 52). 
 In Case II. we have 
 
 FIG. 52. 
 
 Hence, if we make suppositions similar 
 to those above we shall have, when the 
 depth is uniform (or d=& constant) cc 2 
 varying as 6, or 
 
 b=cx* 
 
 Hence the plan (Fig. 53) should consist 
 r of parabolas with vertices at C. If b feS 
 " constant then 
 
 d^ ex 2 or d=<\/ ex. 
 This is the equation of a straight line, and 
 
81 
 
 FIG. 53. 
 gives for the elevation the triangle (Fig. 54). 
 
 FIG. 54. 
 
 In Case III. the analysis gives results 
 > i milar to those in Case I. 
 
 6 " 
 Thus from equation (21) S= 
 
 x 
 -, 7a - 
 
 0Cf . 
 
 - ? - . 
 i 
 
 TT 6W n . 
 
 Here - is constant, and if a be 
 
 constant also, b must vary as x. 
 . . b = ex. 
 
This gives the triangle A G H (Fig. 55.) 
 We obtain similarly the triangle G C H for 
 the other end of the beam. 
 
 If the breadth be constant we have, 
 
 = ex- 
 
 which gives a parabola A K (Fig. 56) with 
 vertex at A. So, for the right hand end 
 of the beam the proper elevation is the 
 
83 
 
 parabola C K (Fig. 56). The elevation 
 (Fig. 56) assumes that the top of the beam 
 needs to be horizontal. 
 
 In Case IV., equation (37) gives 
 
 S= 7 r ' Here, if d be constant in 
 b d 
 
 order to render S constant we have 
 b = cx(l x). 
 
84 
 
 This may be represented by parabolas 
 with vertices at G and H (Fig. 57) oppo- 
 site the middle of the beam. If b is con- 
 stant, then, 
 
 FIG. 57. 
 
 which is the equation of an ellipse, and 
 the beam (if it is required to be horizontal 
 on top) may be made as in (Fig. 58). 
 
 In these cases of beams of uniform 
 strength, we have so far only considered 
 the moments of the weights or the bending 
 moments as they are called. But the re- 
 sults are to be modified by the transverse 
 shearing stress. In ordinary rectangular 
 
B: 
 
 FIG. 58. 
 
 beams this shearing stress is so small 
 compared with the bending moment, that 
 it may be left out of consideration. But 
 in beams of uniform strength the ends must 
 not taper to a point, but must always be 
 left large enough to bear the shearing 
 stress. In the case represented in (Fig. 57) 
 the beam should have, near the ends, the 
 shape shown in (Fig. 59). 
 
86 
 
 DOUBLED FLANGED BEAMS. 
 
 So far we have considered beams with 
 rectangular cross sections, and beams of 
 uniform strength deduced from these. We 
 will not consider beams of x shape. 
 
 It is evident from the investigation al- 
 ready given of the condition of stress, in 
 transversely loaded beams, that those por- 
 tions of the beam nearest the centre bear 
 but a small proportion of the stress, while 
 the contrary is the case with the outside 
 fibres. Hence we would gain strength by 
 moving a considerable portion of that 
 about the neutral axis and placing it on 
 the top and bottom. 
 
 The first form in which the idea was 
 applied was in the T or _[_ cast iron beam. 
 The fact that rectangular cast-iron beam 
 always broke by the tearing of the fibres 
 on the side subjected to tension, suggested 
 the idea of reinforcing that side of the 
 beam with a flange. The results of this is, 
 that the neutral axis still passing through 
 the centre of gravity of the cross section, 
 
8? 
 
 the extreme fibres subjected to compression 
 are farther off than those subjected to ten- 
 sion, and consequently are strained more 
 nearly to their full strength before fracture. 
 This form of beam gives a large increase 
 of strength for the same amount of iron. 
 
 It was still plain that the fibres in that 
 part of the web about the neutral axis were 
 but little strained as compared with the 
 fibres on the outside, and it was proposed 
 to leave as little material there as possible, 
 and to place the mass of it in two flanges 
 ( i ), one above and the other below, giv- 
 ing to these flanges sizes inversely propor- 
 tional to the tensile and compressive 
 strength of the material. The question 
 then was, how much of the material should 
 be left in the web, for plainly all could net 
 be taken. The amount to be left is deter- 
 mined by experiment. If the web is left too 
 thin, the beam will twist and break under 
 the shearing force, and in some cases, from 
 the want of stiffness in the compressed 
 flange. 
 
 To simplify the calculations, the web is 
 consid jred as bearing all the shearing stress, 
 
and no other, and the flanges as bearing 
 all the extension and compression due to 
 the bending moment ; and these parts 
 should be proportioned accordingly with 
 due reference to the practical difficulties 
 that sometimes occur. The ordinary for- 
 mulas for the strength of such beams are 
 gotten by the following approximation : 
 We first neglect the compressive and tensile 
 forces of the web, which are small com- 
 pared with those of the flanges, and con- 
 sider it as bearing only the shearing stress- 
 Then as the depth of the flanges is 
 generally small as compared with the 
 depth of the beam, we consider all the 
 fibres in each flange as strained alike, and 
 as bearing the average stress that is brought 
 on that flange. (Fig. 60.) 
 
 The resultant of the force on each flange, 
 then, is equal to the stress on a unit of sur- 
 face (S ) multiplied by the flange area (A) : 
 that is =S A. 
 
 The point of application of the force will 
 be at the middle of the depth of theflangea 
 
A. 
 B 
 
 89 
 
 FIG. 60. 
 (at O and O', Fig. 61). Fig. 61 shows the 
 
 FIG. 61. 
 
 forces we have to deal with in the Case 
 corresponding to Case I. under rectangular 
 beams. 
 
90 
 
 Let O' O = <7. 
 
 S' = stress on upper flange per unit 
 
 of surface. 
 S"=stress on lower flange per unit 
 
 of surface. 
 
 A" = area of lower flange. 
 A' =area of upper flange. 
 
 Then if we take O (Fig. 61) as a centre 
 of moments we have : 
 
 N d N'.O.+W x = (ButN=S' A') 
 . . S' A' d= W x (57) 
 
 If we take O' as the centre of moments 
 
 we will get 
 
 S" A" d = W x (58) 
 
 The formula for shearing force is iden- 
 tical with that under Case I. of rectangular 
 beams; that is: 
 
 T= W x (59) 
 
 If A' = A", then plainly S' = S" (from 
 equations 57 and 58), or, the forces of 
 tension and compression are equal (as in 
 rectangular beams); but if A' and A" are 
 n ot equal, we have: 
 
91 
 
 That is, the unit stresses in the flanges are 
 inversely as the areas. Now, to have the 
 material distributed between the flanges 
 most efficiently for strength the unit stress 
 should be in proportion to the ultimate 
 strength of the material against tension 
 and compression, and hence the areas of 
 the cross sections of the flanges should be 
 inversely as the ultimate strength. 
 
 Thus, if A D (Fig. 62} be of cast-iron, 
 which is six times as strong against com- 
 pression as against tension, the unit stress 
 in the lower flange should be made six 
 times as great as in the upper, and to effect 
 this the area of the lower flange should be 
 one-sixth that of the upper. 
 
 The Casexu nder x beams are similar to 
 those under rectangular beams. 
 
 Case I. Beams fixed at one end and 
 loaded at the other. ^ 
 
 S' A' d = W *, and S" A" d = W a; (60) 
 
I 
 
 - 
 
 Case II. Beams fixed at one end and 
 loaded uniformly. 
 
 , and S" A" ct = l / 9 
 
 (61) 
 
 Case III. Beams supported at both ends 
 and loaded at some intermediate point. 
 
93 
 
 (62) 
 
 W.-. W(aj m) 6' A' d, or, =S" A" d 
 I 
 
 Case IV. Beams supparted at both ends 
 and loaded uniformly. 
 
 S' A'rf=Vs wx(l a;) = S" A" d 
 
 Case V. A single moving load over a 
 beam supported at both ends. 
 
 W r 
 
 8' A' d= . - (lx) S" A" d (64 
 
 i 
 
 Case VI. A distributed moving load 
 may be considered as included in Case 
 IV 
 
 The formulae for shearing stress are 
 identical with those in rectangular beams. 
 
 The principles of the uniform strength 
 of beams may be applied to flanged beams 
 as they were to rectangular beams. The 
 discussion is analogous to that already 
 -iven. 
 
MOMENT OF RESISTANCE OF BEAMS DETER- 
 MINED GEOMETRICALLY. 
 
 The following method of obtaining the 
 moment of resistance of beams is of easy 
 application, and in many cases of unsym- 
 metrical cross section is the simplest that 
 can be used : 
 
 1. For illustration, take a beam of rec- 
 tangular cross section. Let G P (Fig. 63) 
 be the cross section at some point of this 
 beam. The stresses on the fibres, as we 
 have already seen, increase just in propor- 
 tion as we go from the neutral axis 
 towards the upper or lower surface of the 
 beam, and may for any vertical slice (as 
 that at E F) be represented by the ordi- 
 nates of two triangles, as shown in Fig. G4, 
 where E I (=F J) represents the stress on 
 the outside fibre. For the cross section G 
 P (Fig. 03) the stresses will be represented 
 by two wedges, the bases of which are G 
 M and M R, and the elevations of which 
 are the triangles shown in Fig. 64. The 
 volumes of these wedges give the amount 
 
95 
 
 CO 
 
 CO 
 
 of compressive and tensile force exerted p.t 
 the cross section in question, arid the 
 points in G P under the centre of gravity 
 of the wedges give the " centres of resist- 
 ance," or the points of application of the 
 resultants of these forces. 
 
96 
 
 As a geometrical representation of the 
 stresses on the fibres, these wedges are 
 perfect, for the perpendicular ordinate of 
 the wedge gives in every case the stress 
 which exists in the fibre over which it 
 stands. Thus the line T' V (Fig. 64) 
 represents the stress on each fibre in the 
 row T V (Fig. 63). 
 
 But it is often difficult to find the centre 
 of gravity of these wedges in the case of 
 curved and irregular cross sections, and 
 yet this must be done before we can know 
 the lever-arms of the stresses. To render 
 this easier to do we may represent the 
 stresses, not by wedges, but by prisms, 
 the centres of gravity of which are over 
 the centres of gravity of their bases. 
 
 Thus, if in Fig. 65 we draw the two 
 shaded tringles, and conceive prisms of a 
 height = E I (the stress on the outside 
 fibre, Fig. 64), to be constructed on them 
 as bases, we shall have a geometrical 
 representation of the stress on the sec- 
 tion G P, less perfect in some respects 
 
97 
 
 FIG. 65. 
 
 than that given by the wedges but better 
 suited to our purpose. 
 
 For, note that, 
 
 1. The volume of the prism G O 1J 
 (Fig. 65) is equal to that of the wedge 
 G M (Fig. 63), and the volume of any 
 part of the prism cut off by a plane 
 parallel to the neutral axis, as that whose 
 base t' O v f is equal in volume to the 
 corresponding part of the wedge T M, 
 or since the height of the prism is con- 
 stant, the stress on the surface G M as 
 we go out from the neutral axis varies 
 
98 
 
 as the area of the triangle which forms 
 the base of the prism. 
 
 2. The vertical slice of the prism stand- 
 ing on any line t v' represents in amount 
 the stress on the line of fibres t r, for this 
 stress is equal to the corresponding one 
 in the wedge, the slice of the prism being 
 as much higher than that of the wedge 
 as t v exceeds t' v'. Of course (except in 
 the case of the outside fibres in the row 
 G H) each ordinate in the slice of the 
 prism no longer represents the stress on 
 the fibre over which it stands, as was the 
 case in the wedge. 
 
 3. The moment of the tensile forces, 
 for instance, will equal the area of the 
 prism G O H multiplied by its height, 
 (E I = stress on outside fibre = S). The 
 centre of resistance of these forces, or 
 the centre of gravity of the prism is at C 
 (Fig. 65), the centre of gravity of the 
 base G O H. The triangle G O H is 
 sometimes called the ^effective area" of 
 the surface G M, because a uniform stress 
 on it of an intensity = the unit stress 
 at G H gives the same amount of resist- 
 
99 
 
 FIG. 66. 
 
 ance, as that on the whole area G M, acted 
 on as the latter is by a varying stress. 
 
 Considering the stresses represented by 
 the two prisms whose bases are G O II 
 and R O P (Fig. 65), as concentrated at 
 the centres of gravity C and C' (Fig. 67) 
 of these bases, and taking one of these 
 points as C', (Fig. 67) as the centre of 
 moments, we have in the case represented 
 in the figure: 
 
 (Vol. of prism G O H)xC C' =W x 
 or if b breadth and d = depth of beam 
 
100 
 
 FIG. 67. 
 
 t 
 
 (65) 
 
 as before. 
 
 Corollary. If the beam be square, 
 b d y and 
 
 M = |s& 3 (66) 
 
 II. As a second example, take a square 
 beam so placed that its diagonal will be 
 vertical. Fig. 68 is the cross section. 
 Here we find the base of the prism of 
 stress by points. To find the line in the 
 base of the prism corresponding to th e 
 
FIG. 68. 
 
 stress in any row of fibres, such as A B 
 whose distance from the neutral axis is O 
 X, proceed as follows: 
 
 We see that if the cross section were 
 the square of which H L M K is the 
 half, then a' b' would be the line required, 
 since this is the breadth at that point of 
 the triangle H O K, which would in that 
 
102 
 
 case represent the base of the prism of 
 stress. Project the points A and B upon 
 H K. Then the actual row (A B) of 
 fibres is as much shorter than the corres- 
 ponding row in the supposed section H 
 M, as R T is less than H K, and conse- 
 quently to obtain the proper line in the 
 base of the true prism of stress, a! b' must 
 be shortened in this proportion. Draw 
 lines from R and T to O. These lines 
 intersect the row of fibres at a and b. 
 Then 
 
 HK: RT( = AB)::a'5': ab (67) 
 
 Hence a b is the line required, and a and 
 b are two points in the outline of the base 
 of the prism of stress. Any number of 
 lines as n v, <fcc., may be gotten similarly 
 and the curve drawn through the points 
 a .. n .. b .. v, &c., will give the form of the 
 base of the prism of stress. This base is 
 shaded in the diagram. 
 
 For any ordinate of the curve O n G, as 
 a X, we have 
 
 OX: aX:: O G: RG = AX 
 
103 
 
 jt 
 
 But A X = X G and making G O = 
 
 and putting O X = x and a X =y, we 
 have 
 
 _ __ .... (68) 
 
 This is the equation of a parabola with 
 vertex at n, half way between II K and 
 the neutral axis. Hence the base of each 
 prism is composed of parts of two sym- 
 metrical parabolas. 
 
 Areas of the bases. Since the area of a, 
 parabola is two-thirds of the circumscrib- 
 ing rectangle, the area of each base 
 
 But n v = y,R' T' and R' T =V a H K 
 .-. nv = l / 4 H K = V 4 <*' 
 
 d' d' 1 , 
 /. Area =: V,^.^= r / 2 
 
 The centres of gravity of these bases 
 (and consequently of the prisms) are at C 
 and C', and the distance 
 
104 
 
 Hence the moment of resistance of the 
 fibres about C or C' is 
 
 (69) 
 
 (S=height of prism or stress on external 
 fibres at G and P.) 
 
 Corollary. To compare the resistance 
 of the beam in this position with its resist- 
 ance when lying flat: 
 
 Let d=side of the square as L G. 
 Then d' d fa and eq. (69) becomes 
 
 M^S*V*=^S#' (70) 
 
 Comparing this with eq. (66), we see 
 that the beam offers greater resistance 
 when flat in the proportion of 
 
 1 1 
 
 In solving problems with diagonally 
 placed beams, place the above value of M 
 equal to the moment of the weight as be- 
 fore. 
 
105 
 
 III. Let us apply this method to a T 
 beam. Take for example the cast-iron j_ 
 beam, calculated in part on p. .257 Ran- 
 kine's Civil Engineering, in which the area 
 of the flange = 2 / 3 that of the web. As- 
 sume the flange to be 6 inches by 1 inch, 
 and the web to be 5 inches by .8 of an inch, 
 and draw a figure of the cross section to 
 scale (Fig. 69). 
 
 1. As the top and bottom of this section 
 is not symmetrical, it. is necessary to find 
 the position of the neutral axis, which is 
 no longer at the half-depth. This may be 
 done by calculation, or by a simple me- 
 chanical process as follows: 
 
 The centre of gravity of the cross section, 
 since this last is symmetrical with regard 
 to the vertical line through the middle of 
 the web, must lie on this line. Cut accur- 
 ately the figure of the cross section out of 
 card board, or tin, or good paper, and 
 suspend it freely by one end of the flange, 
 suspending also from the same point a 
 plummet, Mark the line of the plummet 
 on the card board, and the centre of grav- 
 
<r 
 
 ity being on this line and also on the 
 middle line of the web, will be at their 
 intersection O. By measurement this 
 point was found to be distant from A B 
 four and three-tenths inches (4.3"), which 
 is also the value by calculations. The 
 line L M drawn through this point is the 
 neutral axis. 
 
*, 107 
 
 2. To determine the bases of the prisms 
 of stress. On the upper side the base is 
 the triangle O A B, if the altitude be taken 
 equal to the stress on the fibres along A B. 
 For if L' M' H G were the upper half sec- 
 tion, O G H would be the base of the- 
 prism and O A B is less than O G H, in 
 the same proportion that L M A B, the 
 real half section, is less than L' M' II G. 
 Hence (if the upper be the compressed side) 
 the total compressive force is equal to the 
 prism erected on A O B, with the height 
 equal to the unit stress at A B. 
 
 Below the axis L M. For convenience 
 we should have the height of the tension 
 prism equal to that of the compression one, 
 and the base must be determined under 
 this condition. Complete the large rec- 
 tangle G H Q Y, making the distance of 
 Q Y below O=4.3 inches. Draw O Q and 
 O Y and the shaded trapezoid J N" R Z cut 
 out on the flange by these lines, will evi- 
 dently be the portion of the base due to 
 the flflnjrp. Having prolonged the lines of 
 the web to T and V, draw O T and O V, 
 
108 
 
 and then the shaded triangle OKI will be 
 that part of the base due to the portion (L 
 Z) of the web below the neutral axis. 
 
 The total tensile and compressive forces 
 being always equal, and the height of the 
 prisms having been assumed, each =S'= 
 stress on fibres at the distance of A B from 
 O, the bases of these prisms must be equal 
 also. This necessary equality between the 
 area of O AB and that of OKI + JNRZ, 
 affords a means of testing the accuracy of 
 our work in finding the position of O. 
 
 3. Area of base GAB. This is, 
 
 AB) = y f (4.3x.8) = 
 inches. 
 
 4. To determine the distance C C' (Fig. 
 6D) between the centres of gravity of the 
 prisms, which distance is the lever-arm to 
 to be used when one of these points is 
 taken as the centre of moments. These 
 centres of gravity (C and C') can be readily 
 <U j U i rmined by means similar to those em- 
 ployed in finding the centre of gravity 
 
t, 109 
 
 of the cross-section itself. Thus, cut the 
 shaded areas (Fig. 69) out of card board of 
 paper, and suspending each of them from 
 two points in sucession, draw vertical lines 
 through the points of suspension. The in- 
 tersection of these two lines gives the cen- 
 tre of gravity. In the present case they 
 may be so simply obtained by calculation 
 that we adopt that method. The centre 
 of gravity of A O B is =% the distance 
 from O to A B or O C =% (4.3)= 2.87 in- 
 ches. As to the shaded part below O, by 
 using the ordinary formula for the centre 
 of gravity and taking moments around O, 
 we find the distance 
 
 OC'= 
 
 j N R x % (1.7) (O J K+Q I Z)%(.7) ) 
 | ON R 2 (O J Kj 
 
 2.2764 .01388 
 
 3.0145 .2975 
 Hence the distance 
 
 = 1.318 inch. 
 
 CC'=O C+O C' = 2.87 + 1.318=4.18 ins. 
 Hence, since S' is the height of the 
 
110 
 
 prisms, the moment of resistance of the fi- 
 bres is 
 
 M=4.18xl.72S' = 7.19S' 
 
 If it be desired to have M, not in terms 
 of S', the stress along A B, but of S" the 
 stress on the lowermost fibres (at F P), 
 we have since the stresses increase directly 
 with the distance from O, 
 
 S' :S":: I/ G : L' F :: 4.3" : 1.7" 
 . . S'=2.53. S" and M=18.19.S" 
 
 IV. As an illustration of the great sav- 
 ing of labor sometimes effected by this 
 process, take the steel rail now widely used 
 in England, the cross- section of which is 
 given to scale in (Fig. 70). The determin- 
 ation of the moment here by calculation 
 would be long and tedious. The dimen- 
 sions of the cross section are given on the 
 figure. 
 
 1. The centre of gravity O of the cross 
 section is found by making a template as 
 in the last case, and suspending it freely by 
 a corner. The vertical through the point 
 
FIG. 70. 
 
 of suspension intersects A X at O, which is 
 2.55 inches below A. Through this point 
 draw L M, the neutral axis. 
 
 2. The bases of the prisms of stress are 
 
determined by points as in example 
 II. 
 
 Lay off OX=2.55 inches. Draw the 
 rectangle G H Q Y. Assume the height 
 of the prisms to be the stress in the fibre 
 at A. Then proceeding as in example II., 
 the line of the base correspending to any 
 row of fibres, as B D, is b d. Obtain any 
 number of points in the same way as b and 
 d, and through these points draw a curve 
 bounding the shaded figure A b Od. Simi- 
 larly below the neutral axis, t v is the line 
 in the base of the stress prism correspond- 
 ing to T Y, and the shaded figure O N R, 
 is 'that base where the height is taken 
 equal to the unit stress at A. The equality 
 of the bases in area is the test of accuracy. 
 
 3. To determine these areas. The simplest 
 plan in the present case is first to find the 
 area of the cross section itself. This is done 
 as follows: The rail in question weighed 
 84 Ibs. per yard, and the steel, of which it 
 was made weighed, .277 Ib. per cubic inch. 
 
113 
 
 Hence if A = area of cross section in square 
 inches 
 
 (36 . A).277=84. . . A=8.4 sq. inches. 
 
 Now cut out of the same card board, or 
 paper, templates of the two shaded parts 
 in the diagram, and also of the cross section 
 itself, and weigh them. The ratio of the 
 weights will equal that of the areas. 
 
 The comparison of weights may be 
 readily made by the means of a suspended 
 wire, which may serve as a temporary bal- 
 ance, the templates to be compared being 
 stuck on the opposite ends, and one or both 
 moved until the wire is evenly balanced. 
 The weights of the templates being in- 
 vesely as their distances from the point of 
 suspension of the wire, their areas will be 
 in the same proportion. The areas of the 
 prisms in the case before us were found to 
 be equal each to 2.49 square inches. 
 
 4. The centres of gravity of these bases 
 are found in the same way as that of the 
 cross section itself. The point C was thus 
 
114 
 
 found to be 1.84 inches above. O and C r 
 to be 1.66 inches below it. Hence the dis- 
 tance 
 
 C C'=3.5 inches. 
 
 Therefore, finally, if S'=unit stress at 
 A, the moment of resistance is, 
 
 If we desire M in terms of S" (stress at 
 F P, we have 
 
 S' : S" : : 1^84 : 1,66 
 . * . S' = 1.11S" . .M=9.67S" 
 
 V. A circular cross section (Fig. 71). 
 Here the neutral axis of course = L M, pass- 
 ing through the centre. Draw the circum- 
 scribing rectangle G Y and obtain the 
 points t, by v, d, &c., in the curve bound- 
 ing the base, as heretofore. Through the 
 points so found draw the curves. 
 
 Determine the areas of the bases of the 
 stress prisms by comparing them with the 
 half-square G L M H. Thus, if a template is 
 cut to the surface of one of these bases it 
 will just equal in weight the template cut 
 
115 
 
 *' x^x-x ^- 
 
 FIG. 71. 
 
 to the surface A G L O tb A, or, in other 
 words, the shaded surface AbOdA is just 
 
 (f 
 
 one-third of the rectangle G M, or l / s .-r = 
 
 </ 
 
 * '"'', ': 
 
 The centres of gravity C and C' are 
 
116 
 
 found as before by means of templates. 
 In this way it was found that 
 
 OC=OC'=.587 (O A) =.587.^ 
 
 & 
 
 The height of the prisms being S ( 
 stress at A or P) the moment of resistance 
 is, 
 
 M=S.-^(-.587 d\ =.0978 d* S 
 o 
 
 (The accurate value by calculation is 
 M=.0982 d* S.) 
 
 Note. The curve of the base of the 
 prisms is a lemniscate. To find its equation 
 we have in the triangles O b X' and O B'A, 
 
 bX f : AB' (=BX') :: OX': O A 
 
 And taking the vertical axis as that of X, 
 and the horizontal one, as that of Y, the 
 origin being at O, and calling the co- 
 ordinates of the circle x' and y\ and those 
 of the lemniscate x and y we have : 
 
 y: y'::a:R, or y: ^/E, 2 x 2 : :x : R 
 . . R 2 y 2 = x (R 2 -a 2 ). 
 
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THE VAN NOSTRAND SCIENCE SERIES. 
 
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THE VAN NOSTRAND SCIENCE SERIES. 
 
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 adjustment and use. By F. R. Brainard, U. 8. 
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 No. 103. THE MICROSCOPICAL EXAMINATION OF 
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 numerous illustrations. 
 
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 advantages of Gas for Cooking and Heating, and 
 Useful Hints to Gas Consumers. Second edition, 
 rewritten and enlarged. By Wm. Paul Gerhard, 
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 Gould, M. Am. Soc. C. E. 
 
 J 
 
THE VAX XOSTRAXD SCIENCE SERIES. 
 
 No. 113. PHYSICAL PROBLEMS and their Solution. By A. 
 Bourgougnon, formerly Assistant at Bellevue Hos- 
 pital. 
 
 No. 114. MANUAL OF THE SLIDE RULE. By F. A. Halsey, 
 of the American Machinist. 
 
 No. 115. TRAVERSE TABLES showing the difference of Lati- 
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 and for Angles to Quarter Degrees between 1 degree 
 and 90 deTrees. (Reprinted irom Scribner's Pocket 
 Table Book.) 
 
 No. 116. WORM AND SPIRAL GEARING. Reprinted from 
 " American Machinist." By F. A. Halsey. 
 
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