Engineering Prob?

Question

A vertical retaining wall contains water to a depth of 15 meters. calculate the center pressure and the turning moment about the bottom for a unit width. take the density as 1000 kg/m^3.

Answer 1

pressure at any depth in a liquid = rho * g * h

rho is density you give, g is gravitational constant (9.81 in your SI units) and h is depth.  Zero pressure at the top (h=0), you calculate the pressure at the center (I assume you mean 7.5m deep).

The structure is a cantilever beam with fixed end (at the bottom of the wall).  Load varies linearly along length with zero force (pressure time unit width) at the end (top of wall) and maximum at the fixed base.  Either find equivalent load at the centroid of the triangular varying load, or draw shear and moment diagram and find moment at base.  I'm not going to give the answer since it is obvious that you need to do this.  

Engineers must know how to do these calculations or people die.

Answer 2

The shape of the pressure diagram is a triangle so the centre of pressure is at the C of g of a triangle ie 1/3 of the way up (5m)

The turning moment is thus the total pressure times the distance to the centre of pressure.

The pressure at the base is 15 x 1000 = 15000 kg/m2 so the total force  = 15000 x 15 / 2 kg/m run

The overturning moment = 15000 x 15 / 2 x 5m kgm / m run