388 NOTE TO CHAP. XIV. [APPENDIX 



Thus, for instance, for P 10 (see Fig. X.), we lay off the reaction 

 at A upwards to scale from C to D, then the horizontal thrust 

 at A from D to 1, then the equal thrust at B from 1 back to D, 

 thru the reaction at B from D to 8, and finally the weight 

 <lo\vn from 8 back to C, thus closing the polygon for the exterior 

 forces. Lines parallel to the pieces then give the strains. 

 Thus the thrust and reaction at A are in equilibrium with E 

 and 1 2. Then 1 2 is in equilibrium with B and 2 3, and so on. 

 Observe that the diagram checks itself. Thus the last diagonal 



7 8 must be in equilibrium with 6 7 and G (flange H being re- 

 moved), and that this is so is shown by the strain in 78 passing 

 exactly through 8, thus making the strain in H zero. We can 

 also check the work by calculating the strain in the last flange 

 D by moments. Thus for P 10 



D x 9.43 = 7.2 x 72.8 - 8.1 x 19.1 - 10 x 25 = 119.45, 

 or D = + 12.6. 



If this agrees with D as found by diagram, and if the diagram 

 also checks, we may have confidence in the accuracy of the 

 work, and at once scale off the strains. Observe that diagonals 

 4 5 and 5 6 are both tension ; also that F and G are tension. 



We have given also the diagram for P n , which the reader 

 can easily follow through for himself. F and G are both ten- 

 sion, 3 4 and 4 5 both compression. The horizontal thrust is 



8 5, and the reaction at A = b 1. 



We thus make a diagram for each separate weight, and then 

 taking the dead load at the live, we can form the following 

 table of strains. Since we wish only the maximum strains on 

 one-half the arch, those on the other half being precisely simi- 

 lar, we can diagram the strains due to all the weights upon the 

 right half at once by taking the sura of their reactions and 

 thrust at A. We have then the following table : 



