APPENDIX.] 



PIVOT SPAN. 



373 



or in tension, as indicated by the sign, since the momlfat of P, 

 overbalances that of the reaction. 



\Note. The different lever arms are easily obtained from 

 the known dimensions of the truss. We have considered it 

 unnecessary to detail how they are to be found. They may 

 either be measured to scale from the frame or computed trigo- 

 nometrically.] 



The lower flanges are found in similar manner. 



Thus, strain in F is zero, since it passes through the point of 

 moments. 



For G and H, we have 



G- x 8 = 7.415 x 20 + 10 x 10, or G = 6.04 tension. 

 In like manner, for I, 



I x 10 = - 7.415' x 40 + 10 x 30 = + 0.34. 



For K, for similar reasons as above for E, we have centre at 

 11, and therefore the reaction at B also enters into the equa- 

 tion of moments, and 



K x 10 = - 7.415 x 50 + 10 x 40 - 1.58 x 10, or K = + 1.34. 

 We have then, finally, for the strains in the flanges due to P t 



In a precisely similar manner we Ad the strains due to P 2 , 

 P 8 and P 4 . 



We have only to observe that for P 7 , the first weight to the 

 right of C in the other span, the reaction at A is negative and 

 equal to the reaction of P 8 at D, already found, or 1.22. 



Now as we suppose the end A bolted down, this reaction acts 

 as a weight of 1.22 tons suspended from the end. So for the 

 reactions of P 8 and P 9 , viz., 1.035 and 0.584. These reac- 

 tions, moreover, must all take effect through diagonal 1 2 and 

 flange F, as the end vertical cannot take tension. 



Finding then the strains due to each of the other weights, 

 we can, finally, tabulate our results as on next page : 



