Calculation of Pressure Distribution:

There are the two forces acting on the dam is its self weight W which is

W = ρ ⋅(a + b)/2 H

and the horizontal pressure of water P as shown. The self weight W does not pass through the middle point M of the base BC, rather it passes through N. It can be easily seen that

NC =   x¯  =a2+ ab + b2/3(a + b)

Thus, the eccentricity 'e' of the weight W is

e=MN =MC-NC = b/2- a²+ab+b²/3(a+b)

Considering 1.0 m length of the dam, a pressures acting at the base of the dam could be divided into the following three types

(a)        Due to the vertical load W, it is W/A (uniformly compressive)

∴          p1 = W/A= W/ b ×1   = W/ b

(b)       Due to the moment W. e caused by eccentricity of W;

P2=± M.y/I=± W.e.(b/2)/( 1 × b³/12)= ± 6W.e/b²                                                          

There would be compressive force at heel C and tensile at toe D, as the point N is nearer to C.

(c)        Due to the moment P ⋅ h/3 caused by the horizontal water pressure acting at a height h/3 from the base

∴ p3   = ± My/I = ± (P.h/3) b/2/ (1 × b³/12) =± 2 Ph/b²

A resultant pressure at any point is the algebraic sum of the pressures p1, p2 and p3 as calculated above.