APPENDIX.] NOTE TO CHAP. L 359 



zero, and we have the moment of the strain in b c balanced 

 by the moment of the reaction and the moment of X b ; the 

 first causing compression, the second tension, and the difference 

 then giving the resultant moment strain, which, divided by the 

 lever arm of b c, gives the strain itself. 



The method is easy of application, but, as we have already 

 remarked, the determination of the lever arms for each case is 

 frequently tedious. These may, however, be scaled off from 

 the frame diagram with sufficient accuracy in practice. 



As before, we need only the strains due to the first and last 

 weight, and can then form our tabulation as above. This 

 tabulation we can also check by finding the strains due to uni- 

 form load independently, and seeing whether it agrees with 

 the sum of the separate apex weight strains. 



Thus, for all the weights acting, suppose we wish the strain 

 in Y g. The lever arm of Y g is 9.85 feet. We have then re- 

 action = 35 tons, multiplied by 35 feet = 1225. This must be 

 diminished by P 7 x 25 = 250, P 6 x 15 = 150, and P 5 x 5 = 

 50. We have then 1225 - 450 = 775, which, divided by 9.85, 

 gives 78.7 tons tension in Y g, agreeing with our tabulation 

 above. 



For the method of calculation by resolution of forces, see 

 Art. 16 of this Appendix. 



1O. Girder with Straight Flanges. In such a case, as the 

 lever arms are at once known and are constant, the above 

 method is of very easy application. In this case the strains in 

 the diagonals are best found by multiplying the shear by the 

 secant of the inclination of the diagonal with the vertical.* 

 Thus, if this angle is 45, we have simply to multiply the shear 

 at any point by 1.4142, and we have at once the strain in the 



* This is but a particular result of the general method of moments. Thus, 

 for any diagonal, as ab (Fig. VII.), according to our rule, we take the centre of 

 moments at the intersection of the two other, sides cut by a section through 

 the truss, viz. , the flanges. But these two sides are here parattd, hence their 

 intersection is at an infinite distance. The lever arm of a b is then oo x cos <, 

 <p being the angle with the vertical. If the weight PI acts, we have then, 

 calling the reaction R, Rxoo P oo = S cos </> x oo, where S is the strain in 



a b. This can be put (R P) oo = S cos x oo, hence S = - . But 



COS y> X 00 

 00 1 



R P is the shear at b. = - = see & : hence we have only to 



' COS 1/1 X 00 COS <p 



multiply the shear by the secant. 



