114 TOWERS AND. TANKS FOR WATER-WORKS. 



By mechanics* it can be shown that T m = T n = \lX com- 

 ponents of the radial pressures = pr. 



Since the diametral plane can be in any position, the tension 

 is the same at all points on the circumference of the ring. 



Taking n as a centre of moments, 



hence there is no bending moment in the shell. 

 Consider now a quarter-ring ml. 



Ti-T m , 



TI IY components of ^ = o, 

 T m 2X components of p = o\ 



therefore there is no horizontal shear at m or at /. 



Thus it is shown that the stress on a vertical joint is tension 

 only, and the amount per lineal inch of joint is T = pr = o.^^Hr 9 

 where T is the tension in pounds per lineal inch of the joint. 

 Note that if p acts inward instead of outward we have compres- 

 sion instead of tension. 



Stresses in the Cone. To find the stress on a circumferential 

 joint or on the section cut by a horizontal plane (Fig. 21), let 

 ee be such a section. The load, W, on the cone eej is the weight 

 of the cylinder of water whose base is ee and whose height is 

 the distance from ee to the top of the tank, plus the weight of 

 water in the cone eej, plus the weight of steel in the cone ccj. 

 This load must be supported by the forces T acting along the 

 elements of the cone and around the perimeter of ee. Then 



The forces T equal the tension on the joint. This tension 

 is uniformly distributed on the perimeter of ee. Hence the 



* For proof see page ; also* "Stresses in Tank Bottom," by Prof. A. N* 

 Talbot, in The Technogrnph, No. 16, page 137. 



