396 NOTE TO CHAP. XIV. [APPENDIX. 



Hence we have, for the total maximum strains for the case of 

 the preceding article, 



B =+ 57.4-63.4 C=+ 81.1 - 77.1 D= + 64.2-91.4 

 E =+279.4 P =+286.3 G= + 286.5 



H =+285.3 12 = + 53.4-39.723=+ 43.7-28.8 



34=+ 37.3-22.445=+ 40.428.456=+ 23.4-12.5 

 67=+ 36.4-22.778=+ 42.5-23.8 



24. Arch continuous at the Crown, pivoted at the 

 extremities of the Centre One. The method of solution is 

 precisely similar to the preceding. We have only to take the 

 rise of the centre line, or h = 20, instead of h =19.5, as before, 

 and the radius and span of centre line, instead of the radius and 

 span of the lower rib. 



One point only needs to be noticed. Having found the reac- 

 tion and thrust for any weight, these forces now act at the ex- 

 tremity of the centre line. We can therefore form the strain 

 diagram as follows : 



First calculate by moments the strain in A and E. Then, in 

 diagram (c), Fig. X., having laid off the thrust o C and the reac- 

 tion C 5, draw from o and b lines parallel to A and E, and lay 

 off o A equal to the strain in A, and 1) d to the strain in E. 

 Then, if these strains have been correctly found, the line A d 

 must be parallel to and give the strain in diagonal 1 2. The 

 diagram thus commenced, can then be continued as shown, and 

 the strains in all the pieces determined. 



We may also form the strain diagram without calculating A 

 and E. Thus o b is the resultant acting at the end of the 

 centre line. Since it acts then half way between A and E, 

 bisect it in <z, and draw a A perpendicular to flange A. Then 

 o A is the strain in A, and drawing A d parallel to diagonal 1 2, 

 we have at once the strain in E and 1 2. 



Performing the operations indicated, we obtain the following 

 table of strains : 



