Chapter 2 : Fluid Statics
Lecture 5 :

Hydrostatic Thrusts on Submerged Plane Surface

Due to the existence of hydrostatic pressure in a fluid mass, a normal force is exerted on any part of a solid surface which is in contact with a fluid. The individual forces distributed over an area give rise to a resultant force.

Plane Surfaces

Consider a plane surface of arbitrary shape wholly submerged in a liquid so that the plane of the surface makes an angle θ with the free surface of the liquid. We will assume the case where the surface shown in the figure below is subjected to hydrostatic pressure on one side and atmospheric pressure on the other side.

Fig 5.1   Hydrostatic Thrust on Submerged Inclined Plane Surface

Let p denotes the gauge pressure on an elemental area dA. The resultant force F on the area A is therefore

 (5.1)

According to Eq (3.16a) Eq (5.1) reduces to

 (5.2)

Where h is the vertical depth of the elemental area dA from the free surface and the distance y is measured from the x-axis, the line of intersection between the extension of the inclined plane and the free surface (Fig. 5.1). The ordinate of the centre of area of the plane surface A is defined as

 (5.3)

Hence from Eqs (5.2) and (5.3), we get

 (5.4)

where is the vertical depth (from free surface) of centre c of area .

Equation (5.4) implies that the hydrostatic thrust on an inclined plane is equal to the pressure at its centroid times the total area of the surface, i.e., the force that would have been experienced by the surface if placed horizontally at a depth hc from the free surface (Fig. 5.2).

Fig 5.2   Hydrostatic Thrust on Submerged Horizontal Plane Surface

The point of action of the resultant force on the plane surface is called the centre of pressure . Let and be the distances of the centre of pressure from the y and x axes respectively. Equating the moment of the resultant force about the x axis to the summation of the moments of the component forces, we have

 (5.5)

Solving for yp from Eq. (5.5) and replacing F from Eq. (5.2), we can write

 (5.6)

In the same manner, the x coordinate of the centre of pressure can be obtained by taking moment about the y-axis. Therefore,

From which,

 (5.7)

The two double integrals in the numerators of Eqs (5.6) and (5.7) are the moment of inertia about the x-axis Ixxand the product of inertia  Ixy about x and y axis of the plane area respectively. By applying the theorem of parallel axis

 (5.8)

 (5.9)

where, and are the moment of inertia and the product of inertia of the surface about the centroidal axes , and are the coordinates of the center c of the area with respect to x-y axes.

With the help of Eqs (5.8), (5.9) and (5.3), Eqs (5.6) and (5.7) can be written as

 (5.10a) (5.10b)

The first term on the right hand side of the Eq. (5.10a) is always positive. Hence, the centre of pressure is always at a higher depth from the free surface than that at which the centre of area lies. This is obvious because of the typical variation of hydrostatic pressure with the depth from the free surface. When the plane area is symmetrical about the y' axis, , and .