## Total Hydrostatic Force on Plane Surfaces

For horizontal plane surface submerged in liquid, or plane surface inside a gas chamber, or any plane surface under the action of uniform hydrostatic pressure, the total hydrostatic force is given by

$F = pA$

where p is the uniform pressure and A is the area.

In general, the total hydrostatic pressure on any plane surface is equal to the product of the area of the surface and the unit pressure at its center of gravity.

$F = p_{cg}A$

where pcg is the pressure at the center of gravity. For homogeneous free liquid at rest, the equation can be expressed in terms of unit weight γ of the liquid.

$F = \gamma \bar{h} A$

where   $\bar{h}$   is the depth of liquid above the centroid of the submerged area.

Derivation of Formulas
The figure shown below is an inclined plane surface submerged in a liquid. The total area of the plane surface is given by A, cg is the center of gravity, and cp is the center of pressure.

The differential force dF acting on the element dA is
$dF = p \, dA$

$dF = \gamma h \, dA$

From the figure   $h = y \sin \theta$,
$dF = \gamma(y \sin \theta) \, dA$

Integrate both sides and note that γ and θ are constants,
$\displaystyle F = \gamma \sin \theta \int y \, dA$

Recall from Calculus that   $\displaystyle \int y \, dA = A\bar{y}$
$F = (\gamma \sin \theta) A\bar{y}$

$F = \gamma (\bar{y} \sin \theta)A$

From the figure,   $\bar{y} \sin \theta = \bar{h}$,   thus,

$F = \gamma \bar{h} A$

The product   $\gamma \bar{h}$   is a unit pressure at the centroid at the plane area, thus, the formula can be expressed in a more general term below.

$F = p_{cg} A$

Location of Total Hydrostatic Force (Eccentricity)
From the figure above, S is the intersection of the prolongation of the submerged area to the free liquid surface. Taking moment about point S.

$\displaystyle Fy_p = \int y \, dF$

Where
$dF = \gamma(y \sin \theta) \, dA$
$F = \gamma (\bar{y} \sin \theta)A$

$\displaystyle [ \, \gamma (\bar{y} \sin \theta)A \, ]y_p = \int y \, [ \, \gamma(y \sin \theta) \, dA \, ]$

$\displaystyle (\gamma \sin \theta) A \bar{y} \, y_p = (\gamma \sin \theta) \int y^2 \, dA$

$\displaystyle A \bar{y} \, y_p = \int y^2 \, dA$

Again from Calculus,   $\displaystyle \int y^2 \, dA$   is called moment of inertia denoted by I Since our reference point is S,

$A \bar{y} \, y_p = I_S$

Thus,

$y_p = \dfrac{I_S}{A \bar{y}}$

By transfer formula for moment of inertia   $I_S = I_g + A{\bar{y}}^2$,   the formula for yp will become   $y_p = \dfrac{I_g + A{\bar{y}}^2}{A \bar{y}}$   or

$y_p = \bar{y} + \dfrac{I_g}{A \bar{y}}$

From the figure above,   $y_p = \bar{y} + e$,   thus, the distance between cg and cp is

Eccentricity,   $e = \dfrac{I_g}{A \bar{y}}$

## Total Hydrostatic Force on Curved Surfaces

In the case of curved surface submerged in liquid at rest, it is more convenient to deal with the horizontal and vertical components of the total force acting on the surface. Note: the discussion here is also applicable to plane surfaces.

Horizontal Component
The horizontal component of the total hydrostatic force on any surface is equal to the pressure on the vertical projection of that surface.

$F_H = p_{cg}A$

Vertical Component
The vertical component of the total hydrostatic force on any surface is equal to the weight of either real or imaginary liquid above it.

$F_V = \gamma V$

Total Hydrostatic Force

$F = \sqrt{{F_H}^2 + {F_V}^2}$

Direction of   $F$

$\tan \theta_x = \dfrac{F_V}{F_H}$

Case 1: Liquid is above the curve surface
The vertical component of the hydrostatic force is downward and equal to the volume of the real liquid above the submerged surface.

Case 2: Liquid is below the curve surface
The vertical component of the hydrostatic force is going upward and equal to the volume of the imaginary liquid above the surface.

## Stresses on Thin-walled Pressure Tanks

The circumferential stress, also known as tangential stress, in a tank or pipe can be determined by applying the concept of fluid pressure against curved surfaces. The wall of a tank or pipe carrying fluid under pressure is subjected to tensile forces across its longitudinal and transverse sections.

Tangential Stress, σt (Circumferential Stress)
Consider the tank shown being subjected to an internal pressure p. The length of the tank is L perpendicular to the drawing and the wall thickness is t. Isolating the right half of the tank:

$2T = F$

$2(\sigma_t tL) = pDL$

$2t\sigma_t = pD$

$\sigma_t = \dfrac{pD}{2t}$

Longitudinal Stress, σl
At the end of the tank, the total stress PT = σl Aend should equal the total fluid force F at that end. Since the wall thickness t is so small compared to internal diameter D, the area Aend of the wall is close to πDt.

$P_T = F$

$\sigma_lA_{end} = \pi A_i$

$\sigma_l(\pi Dt) = p(\frac{1}{4}\pi D^2)$

$t\sigma_l = \frac{1}{4}p D$

$\sigma_l = \dfrac{pD}{4t}$

Observe that the tangential stress is twice that of the longitudinal stress.

$\sigma_t = 2 \sigma_l$

Spherical Shell
If a spherical tank of diameter D and thickness t contains gas under a pressure of p, the stress at the wall can be expressed as:

$\sigma_l = \dfrac{pD}{4t}$

Spacing of Hoops of Wood Stave Vessels
It is assumed that the wood will not resist tension, only the hoops will resist all the tensile stress caused by the internal pressure p.

$F = 2T$

$pDs = 2 \sigma_t A_h$

$s = \dfrac{2 \sigma_t A_h}{pD}$

where
s = spacing of hoops
σt = allowable tensile stress of the hoop
Ah = cross-sectional area of the hoop
p = internal pressure in the vessel
D = internal diameter of the vessel

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