CHAP. XII.] METHODS COMBINED. 173 



find A I tension and I m tension. [The strains in the upper 

 flanges must always be tension, since the moments at the sup- 

 ports for loaded span are always positive.] Moreover, the Fig. 

 thus far shows that A.l,lm and m B are in equilibrium with 

 the shear B A = H M, as evidently should be the case ; hence 

 the strain in B in is compressive. 



We have thus the strains in the three pieces at the right, and 

 can proceed from these to find all the others. Thus the strains 

 in B k and k I are in equilibrium with m I and B m. Lines 

 parallel to B k and k I, therefore, which close the polygon com- 

 menced by B m and m I, give us the strains in B k and k I, 

 Observe that the line I k crosses A B, thereby making B k op 

 posite in direction, consequently in strain, from B m. This ma}- 

 also be seen by following round the triangle m I k B, remem 

 bering that, as m I is always found to be in tension, it must act 

 away from the new apex, that is, from m to I. "We thus find 

 k I in compression, and k B acting away from this apex, or in 

 tension, therefore of opposite strain from the preceding flange 

 B m, which, as we have seen, is in compression. The reason is 

 obvious. The inflection point * 2 falls in the flange B k. If the 

 beam were solid, the strain at 4 would be zero ; to the right of 

 i,j we should have compression, to the left, tension. In the 

 framed structure the strains can only change at the vertices. 

 The crossing of A B by Ik indicates such change, and B k gives 

 its amount by scale. 



Now taking the upper apex, we have here A I and I k in 

 equilibrium with k h and A h. As we already know, k I is in 

 compression. We must, therefore, now take it acting from k to 

 I, and following round the triangle we find A A compression, 

 and h k tension. From h on, the traverses between A h and 

 B k produced towards the right [Fig. 80 (#)] will give the di- 

 agonals, while the upper and lower flanges will be given by the 

 distances to them from A and B respectively, until we arrive 

 at the weight P. Observe the influence of the weight. We 

 have k h and B k in equilibrium with h g and g B, and also the 

 weight P = B' B. We must take, therefore, A B' = P H, and 

 then draw h g and B' g. Distances to the right of B' along 

 B' g are compressive lower flanges, to the left, tensile ; while to 

 the right of A we have compressive upper, and to the left of A 

 tensile upper flanges. The two diagonals at the weight k h and 

 h g are in tension. From h g on, the diagonals are alternately 



