Assume that each tank is an unstiffened shell fabricated from 

 1/4-inch plate. 



Then the total weight of each tank is 



wt = h7rdt7 + 47rr^t7 



where h is the length of the cylinder minus end caps, d is the diameter of the 

 cylinder, r is the radius of the cylinder, t is the thickness of the cylindrical 

 shell, and 7 is the specific weight of the steel. 



(1 9.6) (n) (4.0) (0.25) (480) (4) (tt) (4.0) (0.25) (480) 

 wt = ^ + ^ 



wt = 2,460 + 503 = 2,963 lb (dry), 2,570 lb (wet) 



The total volume displaced by the two tanks is 625 ft^. 



2. Personnel Sphere. The personnel sphere would be designed for 

 a collapse pressure of 10,000 feet. The sphere will have an outside diameter 

 of 6.0 feet and will be constructed from HY-80 steel. 



For thin-walled spherical pressure vessels 



-2^ 



where t is the required thickness, p is the design pressure, r is the radius and 

 a is the yield strength of the material. 



(4.45 X 10^) (3.0) (12.0) ^ ^ ^^ . 

 (2) (8.00 X lO'*) ^^^'"■ 



140 



