8. Fluid pressure on a plane surface

The total pressure on a plane surface inclined at an arbitrary angle a to the horizontal (Fig. 11) can be determined with the help of the hy­drostatic equation (2.2). For this calculate the pressure P of the fluid on any portion of the surface of arbitrary outline of area S. The x axis is directed along the line of intersection of the plane with the free surface, and the у axis is perpendicular to that line in the given plane.

The elementary pressure on a differential area dS is

where p₀ = pressure on the free surface;

h = depth of the area dS.

To determine the total pressure P, integrate over the whole

area S:

where у = distance of the centre of the area dS from the x axis.

The quantity ,it will be recalled from the course in

mechanics, is the moment of the area S about the axis Ox and is equal to the product of the area and the distance y_c of its centre of gravity from that axis. Hence,

whence

where h_c = depth of submersion of the centre of gravity; and finally

(2.6)

i. e., the total pressure of a liquid on a plane area is equal to the prod­uct of the area and the static pressure at its centre of gravity.

If the pressure p₀ is atmospheric, the gauge pressure of the liquid on a plane surface is

(2.6)`

The centre of pressure, which is the point of application of the re­sultant force on the area, is found as follows.

As the external pressure p₀ is transmitted equally to all points of the area £, its resultant is applied at the centre of gravity of the area. The location of the point of application of the resultant gauge pressure (point D) is found from the well-known equation of mechan­ics which states that the moment of a resultant about an axis Ox is equal to the sum of the moments of the component forces about that axis, i.e.,

,

where y_D = distance of the point of application of force P_g from the x axis.

Expressing P_g and dP_g in terms of y_c and у to determine y_Di we have

where moment of inertia of the area S about the axis Ox.

Taking into account that

where -moment of inertia of the area S about its own centre line parallel to Ox, we finally obtain

(2.7)

Thus, the point of application of force P_g is located below the centre of gravity of the area at a distance

If p₀ is equal to atmospheric pressure and if it acts on both sides I of the surface, point D will be the centre of pressure. If p₀ is greater I than atmospheric pressure, the centre of pressure is found according I to the laws of mechanics as the point of application of the resultant p of forces P_g and p₀ S. The greater the latter force as compared with I the former the closer, apparently, is the centre of pressure to the • centre of gravity of the area S.

Fig. 12. Pressure distribution on a rectangular wall

To determine the second coordinate x_D of the centre of pressure, the moment equation should be written with respect to the у axis.

In the special case when the surface is a rectangle with one side in the free surface of the liquid the centre of pressure is located very simply. As the pressure diagram for the forces acting on a submerged surface represents a right-angled triangle (Fig. 12) whose centre of gravity is located at 1/3 the altitude b of the triangle, the centre of pressure will also be located at 1/3 6 from the lower edge of the rec­tangle.

In engineering we often have to deal with the pressure of fluids on plane surfaces, e.g., against pistons in various hydraulic machines and devices (see examples and Chapter XIV). In these cases the pressure p₀ is usually so great that the centre of pressure may be assumed to coincide with the centre of gravity of the area on which it is acting.