Centre of Pressure

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The key to defining the forces on a submerged surface is to calculate the centre of pressure.

The centre of pressure is a point force acting at right angles to the surface from which the pressure can be represented.

Another way of displaying the pressure as a prism, in this representation the centre of pressure is at the centre of 'mass' of the prism.

Calculating the centre of pressure:

Here we an inclined plane, here we have to take account the pressure at the surface, the distance from the surface, the incline and finally the shape of the plane.

Luckily it can all be simplified into this equation as shown

This can be simplified when Po is zero or when using gauge pressure to: 

An Example!

Solution

Firstly to save us time later we can work out second moment of area (Ixx ) and area (A) as we have all the information we need.

Ixx = (0.5 * 13)  / 4

      = 0.3125 m4

A = (0.5 * 1) / 2

    = 0.25 m2

Now if we look back to the problem:

We can calculate yc easily from the cross-section:

yc   = 2/3b = 2/3 * 1= 0.667m

The only value left from our centre of pressure equation is Po

This is the pressure at the top of the object which we are using as our origin

To find this we can work out the pressure due to the water,  then add atmospheric pressure.

Po = 101*103 + 1000 * 9.81 * 3 = 130kPa

We now have everything required to work out the centre of pressure using

yp = 0.667 +  0.3125 / ((0.667 + (130 * 103 )/ (1000*9.81*sin(60)) * 0.25)

     = 0.745m

All that is left is to calculate the force :

Firstly Pc ,the distance below the origin, can be calculated using trigonometry.

zc = 0.745* sin(60) = 0.645m

Then Pc  = 130 * 103 (Pc) + 1000 * 9.81 * 0.645

              = 136kPa

Fc = 136*103  * 0.25 

      = 34kN