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Hydrostatic Force on Submerged Gate Example

Medium DifficultyFE Fluid Mechanics

Calculate total hydrostatic force and center of pressure on a vertical rectangular gate.

[Diagram: Vertical rectangular gate submerged in water - to be added]

Concept

Hydrostatic force is the total force exerted by a static fluid on a submerged surface. For a vertical surface: The total force equals the pressure at the centroid multiplied by the area The center of pressure is the point where the resultant force acts For vertical surfaces, the center of pressure is always below the centroid Pressure increases linearly with depth: p = ρ g h Key formulas: F = p_{a v g} \times A y_{c p} = \overset{y}{ˉ} + \frac{I _{c}}{A y ˉ}

A rectangular gate with width b = 4 m and height h = 3 m is installed vertically in a water tank. The top edge of the gate is at the water surface. Find: Total hydrostatic force F acting on the gate (kN) Location of center of pressure y_{c p} below the water surface (m)

b = 4 m h = 3 m Top edge at water surface (depth = 0) ρ = 1000 kg/m^{3} g = 9.81 m/s^{2}

For a rectangle with top edge at the surface, the centroid is at mid-height: \overset{y}{ˉ} = \frac{h}{2} \overset{y}{ˉ} = \frac{3}{2} = 1.5 m The centroid is 1.5 m below the water surface.

Hydrostatic pressure increases linearly with depth: p_{a v g} = ρ g \overset{y}{ˉ} p_{a v g} = 1000 \times 9.81 \times 1.5 p_{a v g} = 14, 715 Pa

The total force is the average pressure times the surface area: F = p_{a v g} \times A A = b \times h = 4 \times 3 = 12 m^{2} F = 14, 715 \times 12 = 176, 580 N F = 176.6 kN

For a vertical rectangular surface with top edge at free surface, use the simplified formula: y_{c p} = \frac{2}{3} h y_{c p} = \frac{2}{3} \times 3 = 2.0 m The center of pressure is located 2.0 m below the water surface (at 2/3 of the gate height).

Verify using the general center of pressure formula: y_{c p} = \overset{y}{ˉ} + \frac{I _{c}}{A y ˉ} For a rectangle about its centroid: I_{c} = b h^{3} /12 I_{c} = \frac{b h ^{3}}{12} = \frac{4 \times 3 ^{3}}{12} = 9 m^{4} y_{c p} = 1.5 + \frac{9}{12 \times 1.5} = 1.5 + 0.5 = 2.0 m Both methods give the same result, confirming our answer.

F = 176.6 kN y_{c p} = 2.0 m below water surface Note: The center of pressure (2.0 m) is below the centroid (1.5 m) because pressure increases with depth.

F = p_{a v g} \times A y_{c p} = \overset{y}{ˉ} + \frac{I _{c}}{A y ˉ} Notation: F = total hydrostatic force, p_{a v g} = pressure at centroid, A = area, \overset{y}{ˉ} = depth to centroid, y_{c p} = depth to center of pressure, I_{c} = second moment of area about centroid.