286 PRACTICAL STRUCTURAL DESIGN 



D = diameter in feet, 



W = the weight of one cubic foot of a fluid. 



It is customary to take one-half the weight of the fluid, or 

 equivalent fluid, and multiply it by the diameter. This is the 

 constant m for each material shown in Fig. 182. 



The circumferential tension divided by the allowable fiber 

 stress in the material gives the number of square inches required. 

 If the tank is to be of steel the thickness of the plate is found by 

 dividing the area by 12, the depth of the strip. The proper allow- 

 ance must then be made for rivet holes. 



If the tank is to be of reinforced concrete the steel area must 

 be sufficient to carry all the tension. Sometimes cracks will 

 open in concrete walls, and if the concrete is relied on to carry 

 part of the stress, the tensional strength of the concrete is lost 

 with the first crack and the steel immediately carries this addi- 

 tional stress. 



When there are no cracks in the concrete it does carry part of 

 the load, so the thickness of the shell is fixed by assuming that 

 the strength of the concrete in tension is 150 Ibs. per sq. in. Di- 

 viding the total stress by 150 the concrete area is found and 

 dividing this by 12 the thickness of the shell is fixed. It should 

 never be less than four inches when first class experienced work- 

 men are employed and six inches is a safe enough minimum to 

 use for all tanks. 



The shell of the tank as thus designed will carry double the 

 tension, part of which is carried by the steel and part by the 

 concrete. 



Let A = total area = A c + A s = A c + nA c , 

 in which A c = area of concrete in square inches, 

 A 8 = area of steel in square inches, 

 n = ratio of deformation between steel and concrete. 



The thickness of the concrete multiplied by 12 is the total area 

 from which must be subtracted the area of the steel, leaving A c , 

 the net concrete area. The area of the steel is multiplied by n 

 and added to the net concrete area, this giving the area, A, in the 

 formula. Dividing the total stress by A the average unit stress 

 in tension is obtained and this is the stress on the concrete. Mul- 

 tiplied by n it gives the actual unit tensile stress in the steel. This 

 stress is very low but the instant a crack appears in the concrete, 

 thus reducing the section, the steel stress is increased by an 



