160 METHODS OF LOADING [CHAP. XI 



We have first, then, to construct the cross-lines [Art. 100 (4)] 

 as shown in Fig. d. 



Take first the second span. For this span, as shown in Fig. 

 a, the first span is f nlly loaded, as also the last. Make, there- 

 fore, e a = O t PU draw a D, and we thus find the point D, com- 

 mon to all the middle sides of the second equilibrium polygon 

 for different positions of load. So also make on right 54 = 

 O t P! and draw 4 D', and then, since third span is empty, D' F, 

 and we thus have F. Now from D and F lay off the cross- 

 line distances, as D a, D b, D c, equal to e a, e b, e c, etc. Draw 

 lines through these points and D and F respectively, and note 

 their points of intersection abed with the verticals through 

 the supports. [NOTE. Be careful to preserve an orderly nota- 

 tion^ These points give the moments for each position of 

 load in second span. Take the length a a from Fig. e and lay 

 it off from a on the right support vertical, and join the end 

 with a on left. This is the tangent for full load in second 

 span. A parallel to it at distance a from left gives the shear 

 in Fig. b. Then lay off b b taken from Fig. e on right, and join 

 with b on left. This is tangent for load over three-fourths 

 second span from right. A parallel to it at distance a from 

 foot of perpendicular one-fourth of span from left cuts off 

 shear for this position of load. So for tangents c c, d d. We 

 thus obtain the curve for positive shears in second span. The 

 negative shears are obtained by subtracting these from the 

 shear already found for full load. We thus have the lower 

 curve, and the shear diagram for second span is complete. 



For first span, only the third is loaded. We lay off, then, 

 78 equal to the distance between cross-lines corresponding, 

 draw 8 F l5 and thus find F!. Lay off now at left end e d, e c, 

 e J, draw lines from these points through second support, and 

 note intersections with vertical through D. Through each of 

 these intersections draw a line through F 1? and produce to 

 intersections abed with vertical through second support. It 

 is from these last points that the distances a a, bb, etc., taken 

 from Fig. f must be laid off respectively in order to find ' the 

 tangents e a, eb, ec, ed, ee. 



Parallels to these tangents above in Fig. b give, as before, 

 the positive shear for each position of load. The negative 

 shear is, as before, found by subtracting the positive from total 

 load shear. Thus shear diagram for first span is complete. 



