Adopt M$_{20}$ concrete and Fe$_{415}$ steel. The height of tank in limited to 3 m. draw the reinforcement details. Use tables for IS coefficients.

$V=400 KL=\cfrac{ (400×10^{3})}{10}^{3} =400 m^{3}$ $H_{t}$=3 m (Excluding free board assumed) $M_{20}, M_{415}$ $V=\cfrac{(πD^{2})}{4}×H=400= \cfrac{(πD^{2})}{4}×3$ D=13 Assume; $t_{1}$ = 150 mm $t_{2}$=30 H+50=30×3+50=140 mm Whichever greater should be used as thickness $\cfrac{H^{2}}{Dt}=\cfrac{3^{2}}{(13×150)}=4.61

Form IS code: 3370: 1967 :- page no.-35 table no 9

T= Coefficient × WHR

T= Hoop tension …