CHAP. XH.] METHODS COMBINED. 195 



For 5th span loaded ........ moment at D = 181.88. 



shear at D = 10.67 tons. 

 For 6th span loaded ........ moment at D + 49.46 ft. tons. 



shear at D = + 2.91 tons. 

 For 7th span loaded ........ moment at D = 16.48 ft. tons. 



shear at D == 0.97 tons. 

 Also for the loads in the span D E itself : 

 For the 1st load P t ........ moment at D = -f- 158.92 ft. tons. 



shear at D = + 36.17 tons. 

 For* the 2d load P 2 . . . . ____ moment at D = + 271.96 ft. tons. 



shear at D = +25.88 tons. 

 For the 3d load P 3 . . . ..... moment at D = + 202.36 ft. tons. 



shear at D = + 14.1 6 tons. 

 For the 4th load P 4 ........ moment at D == + 62.88 ft. tons. 



shear at D = + 3.82 tons. 



In Fig. 88 we have found by diagram the strains due to P 3 . 

 [For notation, see Art. 114.] We lay off to scale the shear 

 14.16 upwards, since it is positive, and then, since the moment 

 203.36 at D is positive, and hence the strain in A a must be 



nno no 



tension, we lay off A a = -j- = 20.3 tons to the right of 



B A (Art. 125). "With B A and A a thus given, we can rapidly 

 and accurately find all the other strains. Thus from our dia- 

 gram we have, representing tension by minus and compression 

 by plus : 



A0=-20.4 Ac =+8.0 



A k = - 27.2 

 BJ= + 6.0 Brf=--22 B/=-50.8 BA= 4- 1.2 tons; 



and for. the diagonals : 



a& = + 19.6 Jc= 19.6 c^=+19.6 de= -19.6, etc. 



Heavy lines in the diagram represent compression. In a 

 manner precisely similar, we can find the strains due to each of 

 the other "interior" weights, as also to the "exterior" loads 

 upon the other spans. Suppose all these strains thus found. 

 Then the method of tabulation is as follows : 



