352 
Theodore Sedgwick Johnson 
or d = \ = 10.1 inches. 
^fsPjb 
A 12 inch wall will then be assumed, and with p = 0.0075, 
there will be required 0.909 square inches of steel per foot width. 
This will require f inch bars spaced 8 inches. 
Vertical reinforcement will be f inch rods, 18 inch centers, 
wired and clamped to horizontal rods. 
The same thickness of wall will be assumed to the top, which is 
on the side of safety. For earth action inwards pressure would 
be 533.3 pounds square feet and M = X 533.3 X 121 X 12^ 
= 64,500 inch-pounds. 
To find d, assuming 6 = 12 inches, d = \ ■ , p and j 
^ fs pjb 
would be as before, d = 7.38 inches and there would be re- 
quired 0.0075 X 7.38 inches X 12 inches or 0.675 inch per 
square foot width, which would require f inch square rods 
spaced 7 inches and with proper reduction. At the 12 foot 
height, the moments are reduced 25 per cent, and d = 76.5 for 
water and 40.7 for earth. The corresponding values of d are 
8.75 inches and 6.4 inches, while the steel required is 
0.0075 X 8.75 X 12 = 0.79 square inch 
0.0075 X 12 X 6.4 inches = 0.58 square inch 
This requires f inch rods spaced 9 inches and f inch rods spaced 
8 inches. At the 8 feet height, 0.67 square inch and 0.50 square 
inch are required, which requires the 
I inch rods to be spaced 10 inches and the 
I inch rods to be spaced 9 inches. 
For the last 4 feet both rods will be spaced 12 inches centers. 
In the case of the circular tank, the stresses will be determined 
in the same conditions. For pressure on the inside, the entire 
tank will be under tension, and would be designed in that way 
with vertical distribution. When empty (i.e., earth pressure 
acting) a uniform compression will be caused in the entire ring. 
In the circular tank, the tensile stress at all points in a circu- 
lar ring 1 foot high will be 
t = 
18 feet X 998.4 pounds 
8985.6 pounds. 
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