REINFOKCED CONCRETE TANK 311 



312'5 



the average pressure on the 40 sq. ft. of area = ■ 9 lbs. per sq. ft., hence the 



total horizontal pressure taken by a pier 



= !^X40 = G250lb8. 



The resultant pressure acts at a height of £ of the depth of water, or § feet, 

 above the bottom of the tank, therefore bending moment on buttress at foot 



= M b = (G250 X f) foot-lbs. 

 or M 6 = (6250 X § X 12) inch-lbs. 



Assuming the thickness of the buttress to be 6 inches, i.e. b — 6, then from 

 equation V. — 



M 6 = 1500M — 500/i 2 



substituting for h from IV. we have — 



Mb = 1500 X %'P — 500 X Q£) 2 d* 

 Mb = 479^2 — 50-9^2 

 Mb = 428-1^2 



6250 X 5 X 12 



.\ d 



3 X 428-1 



hence d = 171 inches, and allowing for a depth of imbedment of the steel in 

 the concrete of If inches from the centre of the tension rods to the tension 



surface of the concrete, depth of buttress at bottom = 17 - l + l - 75 



= 1885 inches. 



We next proceed to calculate the vertical reinforcement. 



The total pull in the steel = the total thrust in the concrete 



ft . a =4 • M 



but from IV., h = \^d 



A=if X 17-1 

 .-. /,.«=S|2x6 X^X 17-1 

 _ 500 X 6 X 15 X 1 71 

 2 X 16,000 X 47 

 a = - 512 sq. in. 



Now the cross-sectional area of a f^" round bar is 0'249 sq. in., therefore 

 the cross-sectional area of two such bars = 0*498 sq. in. This seems barely 

 sufficient, but two ~g" dia. bars will be strong enough, when we consider the 

 extra strength gained through turning these two bars by a gentle curve into 

 the base. 



