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# Masonry Walls

Water pressure on masonry walls

## Overview

Consider a vertical masonry wall having water on one of its sides as shown in figure. Now consider a unit length of the wall. We know that the water pressure will act perpendicular to the wall. A little consideration will show, that the intensity of pressure, at the water level, will be zero, and will increase by a straight line law to $\inline&space;wH$ at the bottom as shown in figure. Thus the pressure diagram will be a triangle.

The total pressure on the wall will be the area of the triangle, i.e.,
$P&space;=&space;\frac{wH}{2}\times&space;H&space;=&space;\frac{wH^2}{2}$

This pressure will act through the center of gravity of the pressure diagram.

Let, $\inline&space;\bar{h}$ = Depth of the center of pressure from the water surface.

We know that the c.g. of triangle is at a height of $\inline&space;\frac{H}{3}$ from the base, where $\inline&space;H$ is the height of the triangle. Therefore depth of center of pressure from the water surface,
$\bar{h}&space;=&space;H&space;-&space;\frac{H}{3}&space;=&space;\frac{2H}{3}$

Thus the pressure of water on a vertical wall will act through a point at a distance $\inline&space;\frac{H}{3}$ from the bottom, where $\inline&space;H$ is the depth of water.

Example:
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##### Example - Water pressure on masonry walls
Problem
One of the walls of a swimming pool contains 4m deep water. Determine the total pressure on the wall, if it is 10m wide.
Workings
Given,

• Depth of water, H = 4m
• Width of wall = 10m

We know that pressure on the wall per meter length
$=&space;\frac{wH^2}{2}&space;=&space;\frac{9.81\times&space;4^2}{2}&space;=&space;78.48\;KN$

and total pressure on the wall,
$P&space;=&space;10\times&space;78.48&space;=&space;784.8\;KN$
Solution
Total pressure on the wall = 784.8 KN