174 GRAPHIC AND ANALYTIC [CHAT XII. 



tension and compression. Moreover, the diagonal e d passes 

 through A, that is A d is zero. The weight P causes no strain 

 in A d, and for this one position of P, A d might be omitted 

 from the structure. The reason is again obvious. The point 

 of inflection ^ coincides with the apex under A d. Since at 

 *! the moment of rupture is zero, if the flange A. d were cut 

 there would be no tendency to motion. We have at i v the 

 shearing force only, A B' giving the strains in the diagonal 

 e d and d c. The upper flange A b', we see again, is in tension, 

 which is also shown by its lying to the left of A. 



Thus we have the strains in every piece by a very simple 

 construction for any position of P, without any calculation what- 

 ever. The method in this case is purely graphical. We have 

 only to find the points of inflection and then proceed as above. 



Fig. 80 (b) gives the strains for the same girder and position 

 of weight P, merely supported at the ends. For this case P D 

 in Fig. 80 not only passes through D, but P O also passes through 

 the upper left-hand corner at C. Hence A B will be less than 

 H M, and A B' greater than P JJ. Moreover, the end lower 

 flanges B a and B m no longer act, and must be removed. 

 Starting now with the reaction B' A [Fig. 80 ()], we go along 

 to the weight, from which point at h k we go back towards the 

 force line, and the reactions are such that the last diagonal 

 must pass exactly through B, just as in Fig. (a] e d passed 

 through A, because the points of inflection or zero moments are 

 now at the ends C and D. A careful comparison and study of 

 the two cases and their points of difference will be advanta- 

 geous to the reader. 



115. Counterbracing. The objection may arise that the 

 above method applies only to a system of bracing such as rep- 

 resented in the Fig., where the diagonals take both compres- 

 sive and tensile strains. In case, as in the Howe or Pratt Truss, 

 for instance, we had vertical pieces as also two diagonals in 

 each panel, then the strain in any diagonal as m I and flange as 

 B m, even if found, are apparently in equilibrium with three 

 pieces, viz., k I, B k and a vertical strut or tie at the intersec- 

 tion of these pieces. Hence, having only two known strains 

 and three to be determined, the method would seem to fail, as 

 any number of polygons may be constructed with sides par- 

 allel to the forces, and hence the problem is indeterminate 

 (Art. 9k 



