CHAP. I.] COMMON POINT OF APPLICATION. 11 



BRIDGES. 



12. For bridges the strains due to a uniform load are of 

 course easily found. In most cases a rolling load can be man- 

 aged also, without making a separate diagram for each position 

 of the load. Thus, if we diagram the strains for the load at 

 the first and last apex, the strains due to intermediate loads 

 will be multiples or submultiples of these, provided all the 

 bays are equal. A calculation for a simple Warren girder of 

 small span, and a consideration of the reaction for each position 

 of the load, will at once illustrate what is meant. [Compare 

 Stoney, Theory of Strains. Pp. 99-111, Yol. I.] 



Thus Stoney, in his Theory of Strains, Yol. I., p. 99, gives 

 the girder represented in Fig. 7, PI. 2, span 80 ft., depth of 

 truss, 5 ft., 8 equal panels in upper flange, 7 in lower. 



For the first weight of 10 tons, P l5 the strains are given by 

 Fig. 7 (a) to a scale of 10 tons to an inch. We form first the 

 force polygon by laying off from o, 10 tons, to P x . From the 

 end of this line we lay off upwards the reaction at right abut- 

 ment % of 10 tons, or 1.25 tons ; and then the reaction at the 

 left abutment = -| of 30 tons, back to o, thus closing the force 

 polygon. [Note. In any structure which holds in equilibrium 

 outer forces, the force polygon must close. If it does not, there 

 is no equilibrium, and 'motion ensues (see Art. 20).] Com- 

 mence now with the reaction at a in the frame diagram, Fig. 7, 

 because here we have a known reaction, a o (force polygon), 

 and only two unknown strains to be determined. Drawing 

 lines parallel to A and 1, we obtain the strains in A and 1. 

 Then pass on to apex b. With the now known strain in 1, we 

 can determine 2 and E. 



Passing now to the next apex, we have A and 2 known, and 

 also the weight P t . Join therefore P x and E [Fig. 7 ()] by 

 lines parallel to B and 3. B and 3 are both in compression. 

 We find diagonal 2 also in compression, and 1 in tension. That 

 is, both the diagonals under the weight are compressed, as evi- 

 dently should be the case. From 4 on we have tension and 

 compression alternately. 



Fig. 7 () gives the strains due to the last position of the load 

 P 7 . The strains in the diagonals are evidently all equal, and 

 alternately tension and compression. 



Now it is not necessary to construct more than these two dia- 

 grams. From these two alone we can determine the strains for 



