CHAP. VI.] SUPPLEMENT TO CHAP. XTV. 313 



For 8 = a, this is zero, as should be, since then the load is not upon the 

 span. For 8 = a, V = p r sin a, as should be, for full load over whole 



span. For the semi-circle, a = 90 = , sin a = 1, cos a = 0, and 



2 



TT cos 3 8 

 sin 8 + cos 8 + 8 sin 8 =- 



V pr . 



7T 



If the semi-circle is uniformly loaded over whole span, 8 = a = 90 



= , sin 8 = 1, cos 8 = 0, and V = p r, as should be. The for- 



2 



mula (79) above is precisely the same as that given by Capt. Eads in his 

 Report to the Illinois and St. Louis Bridge Co., May, 1868. 



(ft) Horizontal Thrust. 



In similar manner, from eq. (46), Art. 18, by inserting p d x = p r cos 8 d8 

 in place of P, and integrating between 8 = a and 8. we have similarly to 

 Art. 32 



< 80 > 



pr sin a A t A t B 2 



where H, = i- - =r- , A = , B = =- , and 



12 **! AI JBj 



Ai=3 a 3 sin a cos a 2 a sin 2 a 3 8 9 sin /3 cos /3+12 cos a sin )3 



6 8 sin 2 /3+6 a sin a sin +2 a ^? -, 



Sina 



Aa=3a 3 sinacosa+2a sin* a+6^sin 2 /3 3 /3+3 sin/3 cos)3 



6 a sin a sin 82 a - , 

 sin a 



Bj=a (a+sin a COS a) 2 sin 2 a, B a =a (a+sin a COS a). 

 Formula (80) agrees exactly with that given by Capt. Eads in the Eeport 



above quoted, if terms containing K are neglected. Since K = - -, where I 



Ar 2 



is the moment of inertia and A is the area of cross-section ; r being the 

 radius ; for small central angle r is very large in proportion to , or the 



jA 



square of the half depth. In such case, then, K may be neglected. For 

 |3 = a, we have H = 0, as should be. For 8 = a, we have load over 

 entire span, and 



_ . . 8 a 2 n sin' ? a '3 sin a COS a K (3 a+2asin* a 8 sin a COS a) 



H= p r sin a - - - r - - ; - ^_ . - L 

 (1+K) a (a+sm a COS a) 2 sin* a 



Approximately we have, by series, for full load, from (80) : 



.pa* 

 2h 45I + 4AA 2 



