APPENDIX.] THE BRACED AECH. 387 



4th. Arch, as in 3d, but pivoted at ends of centre line. 



5th. Arch without hinges, or continuous at crown and fixed 

 at abutments. 



These cases will illustrate all the principles of Chapter XIV., 

 and a comparison of the results obtained in each case ma^ 

 prove instructive. 



2O. Arch hinged at a/pex 8, and at the extremities of the 

 lower flange -flanges H and A being removed. 



From Art. 158 we can easily find the reaction and horizon- 

 tal thrust at left end either by construction or formula for 

 every weight. Thus 





2 A 

 For the first weight P,, then, 



= 0.603, 



For the weight P 10 , 



10(85.3 + 37.5) an( , 



170.6 



10 (85.3 37.5) 

 H = 2x29.5 = 8 ' a tom - 



In similar manner, we find 



P! = 0.603, P 3 = 1.33, P 3 = 2.06, 



H! = 2.74, H 2 = 3.84, H 3 = 5.9, 



P 4 = 2.8, P 6 = 3.53, P 6 = 4.2, 



H 4 = 8.1, H 5 = 10.2, H,, = 12.1, 



P 7 = 5.0, P 8 = 5.7, P 9 = 6.47, P 10 =7.2, 



H, = 14.4, H 8 = 12.1, 1^=10.2, H 10 = 8.1, 



P u = 7.94, P u = 8.6, P u = 9.4, 



H u = 5.9, H 12 = 3.84, H 13 = 1.74. 



It will be at once seen that the reaction of P 8 at A is the 

 game as of P e at B, or equal to 10 P, ; while the horizontal 

 thrust is the same for both. We need them only to find P and 

 H for weights 1 to 7, and can then at once write down the 

 others. We are now ready either to calculate or diagram the 

 strains. 



