CHAP. XII.] METHODS COMBINED. 191 



and the support. The known moment gives, then, the charac- 

 ter of the strain only for that flange which does not meet a 

 diagonal at the support. The moment at B, therefore, being 

 positive, gives us compression here in lower flange, because, for 

 the system of triangulation corresponding to the weight, that 

 flange does not meet a diagonal at B. For a weight upon the 

 other system of triangulation (dotted in Fig.), the same moment 

 would give us the tension in E 7. The construction assumes 

 equilibrium between F 7, 7 8, and E 8, and the shear at B ; 

 that is, between the pieces cut by an ideal section to the right 

 of B through the truss and the shear at that section. That this 

 is so is shown by the strain diagram, since there we see that the 

 strains in these pieces form a closed polygon with the shear at 

 B = E! Fj. This must evidently be so if these are the only 

 pieces cut by such a section, since then the horizontal com- 

 ponents of the strains in these pieces must balance, and the 

 resultant vertical component must be equal and opposite to the 

 shear. 



It is important to know which side of E x F x to lay off F t 7, 

 since, if we had laid it off in this case to the right, we would 

 have obtained a very different value for E : 8. For this pur- 

 pose we have only to suppose the strain in the flange (either 

 upper or lower, as the case may be) to be applied at the point 

 of junction or apex of the other two pieces, and then lay it off 

 in the direction with reference to that apex corresponding to 

 the known character of its strain. The direction of the shear 

 is always known from the reactions. 



Thus in our Fig. the shear between B and C acts down from 

 E! to F!, because P 4 , which also acts down, is greater than the 

 sum of the upward reactions at A and B. The strain in F 7 

 is also known to be compressive, and therefore, in following 

 round the strain polygon commencing from E x to Fj, it must 

 act towards apex at 7. We must, therefore, lay it off to the 

 left of E! F lt In similar manner, for the other triangulation, 

 the strain in flange E 7 is, in span B C, in equilibrium with 7 8 

 (dotted diagonal) and F 8 and shear E t F l5 and is, moreover, 

 known to be tension. Consider it acting then at B ; and then, 

 since it is tension, we go round the polygon from E x to F 1} and 

 then to the right of E x F t , or away from B, the point at which 

 it is supposed to act. 



Now for the case of the " tipper ; " the reaction at D, and 



