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Masonry Walls

Water pressure on masonry walls
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Key Facts

Gyroscopic Couple: The rate of change of angular momentum (\tau) = I\omega\Omega (In the limit).
  • I = Moment of Inertia.
  • \omega = Angular velocity
  • \Omega = Angular velocity of precession.


Blaise Pascal (1623-1662) was a French mathematician, physicist, inventor, writer and Catholic philosopher.

Leonhard Euler (1707-1783) was a pioneering Swiss mathematician and physicist.

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 wH at the bottom as shown in figure. Thus the pressure diagram will be a triangle.

23547/masonry_wall.png
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The total pressure on the wall will be the area of the triangle, i.e.,
P = \frac{wH}{2}\times H = \frac{wH^2}{2}

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

Let, \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 \frac{H}{3} from the base, where H is the height of the triangle. Therefore depth of center of pressure from the water surface,
\bar{h} = H - \frac{H}{3} = \frac{2H}{3}

Thus the pressure of water on a vertical wall will act through a point at a distance \frac{H}{3} from the bottom, where 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
= \frac{wH^2}{2} = \frac{9.81\times 4^2}{2} = 78.48\;KN

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