Hydrostatic Thrusts on Submerged Plane Surface - 1

Hydrostatic Thrusts on Submerged Plane Surface

Due to the existence of hydrostatic pressure in a fluid mass, a normal force acts on any portion of a solid surface in contact with the fluid. The distributed pressure on an area produces a resultant force whose magnitude and line of action depend on the pressure distribution. This chapter develops expressions for the resultant hydrostatic thrust on plane and curved submerged surfaces and locates the point of action (centre of pressure) of the resultant.

Plane Surfaces

Consider a plane surface of arbitrary shape, wholly submerged in a liquid, whose plane makes an angle θ with the free surface of the liquid. Assume hydrostatic pressure acts on one side while atmospheric pressure acts on the other side (so gauge pressure may be used). The geometry is shown in the figure below.

Fig 5.1   Hydrostatic Thrust on Submerged Inclined Plane Surface

Let p denote the hydrostatic (gauge) pressure at an elemental area dA. The elemental normal force on dA is p dA and the resultant force F on the total area A is the integral of these elemental forces over A.

(5.1)

Using the hydrostatic pressure variation with vertical depth (pressure = γ h, where γ is the specific weight of the liquid and h is the vertical depth measured from the free surface), Eq. (5.1) reduces to the integral form:

(5.2)

In the figure, h is the vertical depth of the elemental area dA from the free surface and the distance y is measured from the x-axis, which is the line of intersection between the extension of the inclined plane and the free surface. The ordinate of the centroid (centre of area) of the plane surface A measured from the x-axis is defined by

(5.3)

Substituting the centroid definition into Eq. (5.2) gives

(5.4)

Here h_c = y_c sin θ is the vertical depth of the centroid c of area A measured from the free surface.

Equation (5.4) implies the following important result: the hydrostatic thrust on an inclined plane is equal to the pressure at its centroid multiplied by the total area. In other words, the resultant is the same as that which would act on an identical plane placed horizontally at depth h_c from the free surface.

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, denoted c_p. Let x_p and y_p be the distances of the centre of pressure from the y and x axes respectively. Equating the moment of the resultant about the x axis to the sum of moments of elemental forces about the x axis gives

(5.5)

Solving Eq. (5.5) for y_p and substituting the expression for F from Eq. (5.2) yields

(5.6)

Similarly, by taking moments about the y axis the x coordinate of the centre of pressure is obtained:

(5.7)

The double integrals appearing in the numerators of Eqs. (5.6) and (5.7) are the second moments of area (moment of inertia) and product of inertia of the plane area about the x and y axes. Using the parallel-axis theorem to relate moments about the x-y axes to centroidal axes x'-y' gives

(5.8)

(5.9)

where I_x'x' and I_x'y' are the moment of inertia and the product of inertia of the surface about the centroidal axes (x'-y'), and x_c, y_c are the coordinates of the centroid c with respect to the x-y axes.

Substituting Eqs. (5.8)-(5.9) and the centroid expression (5.3) into Eqs. (5.6) and (5.7) leads to the standard expressions for the co-ordinates of the centre of pressure:

(5.10a)

(5.10b)

The first term on the right-hand side of Eq. (5.10a) is always positive because it contains I_x'x' (which is positive) divided by the area and multiplied by γ/(γ A). Hence the centre of pressure lies deeper than the centroid; that is, the centre of pressure is always at a greater vertical depth from the free surface than the centre of area. This follows from the increasing pressure with depth.

When the plane area is symmetric about the centroidal y'-axis, the product of inertia I_x'y' = 0 and therefore x_p = x_c, i.e., the horizontal coordinate of the centre of pressure coincides with that of the centroid.

Hydrostatic Thrusts on Submerged Curved Surfaces

On a curved surface, the normal direction varies from point to point; therefore pressure forces on elemental surface patches differ in direction and cannot be summed as scalars. Instead, resolve the elemental force into components and integrate each component to obtain the resultant vector.

Introduce a Cartesian co-ordinate system with the xy plane coinciding with the free surface of the liquid and the z axis directed downward from the free surface. A general submerged curved surface is shown in the figure.

Fig 5.3  Hydrostatic thrust on a Submerged Curved Surface

Consider an elemental area dA on the curved surface at depth z below the free surface. The hydrostatic pressure at this element is p = γ z. The magnitude of the elemental force dF acting normal to dA is

Each equality on a new line for clarity:

dF = p dA

dF = γ z dA

(5.11)

The force dF acts normal to the surface. Let l, m and n be the direction cosines of the outward normal to dA relative to the x, y and z axes respectively. The components of dF are therefore

The components of the surface element dA projected on the coordinate planes are

where

• dA_x = l dA is the projection on the yz plane;
• dA_y = m dA is the projection on the xz plane;
• dA_z = n dA is the projection on the xy plane.

Substituting these relations into the component expressions gives the elemental component forces in terms of projected area elements. Integrating over the entire curved surface produces the total components of the hydrostatic force along the coordinate axes:

In the above, z_c is the z-coordinate of the centroid of the projected area A_x (projection of the curved surface on the yz plane) and similarly for other projected areas. If z_p and y_p denote the coordinates of the point of action of F_x on the projected area A_x (on the yz plane), then equating moments about the appropriate axes yields

Here I_yy is the moment of inertia of the projected area A_x about the y axis and I_yz is the product of inertia of A_x with respect to y and z axes. In similar fashion, if z'_p and x'_p denote the coordinates of the point of action of F_y on the projected area A_y, then

where I_xx is the moment of inertia of A_y about the x axis and I_xz is the product of inertia of A_y about x and z axes.

From these relations we conclude the following important practical results:

• The horizontal component of the hydrostatic force on a curved surface in a given direction equals the hydrostatic force on the plane projection of the curved surface perpendicular to that direction, and this component acts through the centre of pressure of the projected plane area.
• The vertical component of the hydrostatic force on a curved surface equals the weight of the liquid volume vertically above the curved surface up to the free surface, and it acts through the centre of gravity of that liquid volume.

Mathematically, the vertical component F_z can be expressed as

(5.18)

where

is the volume of the liquid contained in the vertical prism (or region) extending from the curved surface up to the free surface. Hence F_z = γ × (volume of liquid above the surface) and acts through the centroid of that liquid volume.

Remarks and Practical Implications

• For design of gates, retaining walls, dams and submerged plates, the resultant hydrostatic force and the centre of pressure must be known to determine overturning moments and stabilising reactions.
• For plane surfaces the simple result F = p_c A (pressure at centroid times area) gives the magnitude, while Eqs. (5.10a)-(5.10b) locate the centre of pressure relative to the centroid. The centre of pressure is always deeper than the centroid for a vertical coordinate measured downwards.
• For curved surfaces, resolve forces into components; horizontal components use projected areas, vertical component equals the weight of the liquid above the surface.
• When symmetry exists, products of inertia vanish and centre of pressure coordinates simplify to the centroidal coordinates in the symmetric directions.

Worked Example (illustrative)

Problem: A rectangular plane surface of width b and height h is submerged with its plane inclined at an angle θ to the free surface so that its top edge just meets the free surface. Determine the hydrostatic thrust on the surface and the depth of the centre of pressure measured from the free surface.

Solution:

Write pressure at a depth z as p = γ z.

Express elemental force on a strip of width b at depth z and thickness dz as dF = γ z (b dz).

Integrate over the depth of the rectangle from z = 0 to z = h to obtain the resultant force F.

Compute the centroid depth h_c of the rectangular area measured from the free surface: h_c = h/2.

Use F = γ h_c A to verify the integrated value.

Locate the centre of pressure using the moment equation about the top edge and employ the second moment of area I_top for a rectangle about its top edge (I_top = b h³/3) and the parallel axis relation to obtain y_p = h_c + I/(h A).

These steps give the magnitude and position of the hydrostatic thrust; substitute numerical values where b, h and γ are known to compute the final numbers.

Summary

The hydrostatic force on a submerged plane surface equals the pressure at the centroid times the area. The centre of pressure lies below the centroid because pressure increases with depth; its co-ordinates are obtained by equating moments and using second moments (moments of inertia) of the area. For curved surfaces, resolve forces into orthogonal components: horizontal components equal the forces on the corresponding projected plane areas (acting at their centres of pressure), and the vertical component equals the weight of fluid above the surface acting through the centroid of that fluid volume.