124 



STRENGTH OF MATERIALS 



To find the amount of the longitudinal tension, consider a cross 

 section of the cylinder near its center, where the influence of the end 

 restraints can be assumed to be zero (Fig. 91). Then the resultant pres- 

 1^ sure on either end is P = irr l w, and 



the resultant longitudinal stress on t lie 

 cross section is 2 irrhp. Therefore 



rw 



= 2 Trrhp ; whence p = - 



U // 



\B 

 FIG. 91 



This is the same formula as for the 

 sphere, which was to be expected, since 

 the cross section is the same in both cases. 



If p t denotes the longitudinal stress and p k the hoop tension, then 



Till 7"W 



p l -,p h = ' and, consequently, the equivalent stress p e is 



2h> fi 



_! 

 m 



If m = 3^, this becomes 

 (65) 



2m-l 

 2m 



85 - 

 h 



WT 

 T 



Formula (65) is the one to be used in finding 

 the tensile stress in a thin < \ Under subjected to 

 uniform internal pressure, in which the .-inls an- 

 held together by the body of 

 the cylinder and not by inde- 

 pendent stays or fixed sup- 

 ports. 



Problem 129. An elevated water 

 tank is cylindrical in form with 

 a hemispherical bottom (Fig. 92). 

 The diameter of the tank is 20 ft. 

 and its height 62 ft., exclusive of 

 the bottom. If the tank is to be 

 built of wrought iron and the fac- 

 tor of safety is taken to be 6, what 

 should be the thickness of the bot- 

 tom plates, and also of those in the 

 body of the tank near its bottom ? 



NOTE. Formulas (63) and (65) give the required thickness of the plat.-s, prov-- 

 the tank is without joints. The bearing power of the rivets at the joints, however, is. in 

 general, the consideration which determines the thickness of the plates. 



FIG. 92 



I i... '.:; 



