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Total Pressure on Plane Areas

The total hydrostatic pressure on a plane surface is equal to the pressure at the centroid times the area of the plane surface.

$$P = \gamma \bar{h} A$$ $$\bar{h} = \text{vertical depth of centroid below liquid surface}$$

For inclined surfaces, pressure still depends on the vertical depth of the centroid, not the sloping distance alone.

$$\bar{h} = \bar{y}\sin\theta$$ $$P = \gamma A\bar{y}\sin\theta$$

Center of Pressure

The resultant hydrostatic force acts below the centroid for vertical and inclined plane areas. The distance from centroid to center of pressure is commonly denoted by $e$.

$$e = \frac{I_G}{A\bar{y}}$$ $$y_{cp} = \bar{y} + e$$ $$h_{cp} = y_{cp}\sin\theta$$
For vertical plates, $\bar{y}$ and $h$ are the same vertical distance. For inclined plates, use distance measured along the plane for $y$ and vertical depth for $h$.

Problem: Rectangular Gate

A rectangular gate is 1.5 m wide and 3 m high, vertically submerged in water with its top edge 2 m below the water surface. Find the total force and the center of pressure depth.

$$\bar{h} = 2 + \frac{3}{2} = 3.5 \text{ m}$$ $$A = 1.5(3) = 4.5 \text{ m}^2$$ $$P = 9.81(3.5)(4.5) = 154.51 \text{ kN}$$ $$I_G = \frac{1.5(3)^3}{12} = 3.375 \text{ m}^4$$ $$e = \frac{I_G}{A\bar{h}} = \frac{3.375}{4.5(3.5)} = 0.214 \text{ m}$$ $$h_{cp} = 3.5 + 0.214 = 3.714 \text{ m}$$

Answer: $P = 154.51 \text{ kN}$ and the center of pressure is 3.714 m below the water surface.

Problem: Square Plate Location

A square plate 3 m on each side is vertical in water. If the center of pressure is 8 cm from the centroid, find how far the upper side is below the water surface and the hydrostatic force.

$$I_G = \frac{3(3)^3}{12} = 6.75 \text{ m}^4$$ $$e = \frac{I_G}{A\bar{h}}$$ $$0.08 = \frac{6.75}{(3)(3)\bar{h}}$$ $$\bar{h} = 9.375 \text{ m}$$ $$y_{top} = 9.375 - 1.5 = 7.875 \text{ m}$$ $$P = 9.81(9.375)(3)(3) = 827.72 \text{ kN}$$

Answer: The upper side is 7.875 m below the surface, and $P = 827.72 \text{ kN}$.

Problem: Circular Plate — Total Force and Center of Pressure

A circular plate with diameter 1.20 m is submerged vertically in fresh water with its center at a depth of 2.50 m below the free surface. Find the total hydrostatic force on one face and the depth of the center of pressure.

$$A = \frac{\pi(1.20)^2}{4} = 1.131 \text{ m}^2, \quad \bar{h} = 2.50 \text{ m}$$ $$P = \gamma \bar{h} A = 9.81(2.50)(1.131) = 27.73 \text{ kN}$$ $$I_G = \frac{\pi d^4}{64} = \frac{\pi(1.20)^4}{64} = 0.1018 \text{ m}^4$$ $$e = \frac{I_G}{A\bar{h}} = \frac{0.1018}{1.131(2.50)} = 0.036 \text{ m}$$ $$h_{cp} = 2.50 + 0.036 = 2.536 \text{ m}$$

Answer: Total force is 27.73 kN. The center of pressure is 2.536 m below the surface — only 36 mm below the centroid because the plate is deep relative to its size.

Problem: Inclined Rectangular Gate — Force and Location

A rectangular gate is 1.0 m wide and 2.0 m long, set at 60° from the horizontal. The upper edge of the gate is 1.5 m (vertical) below the water surface. Find: (a) the total hydrostatic force on the gate, and (b) the center of pressure measured along the inclined gate from the water-surface intersection.

$$\bar{h} = 1.5 + \frac{2.0\sin 60°}{2} = 1.5 + 0.866 = 2.366 \text{ m (vertical depth to centroid)}$$ $$A = 1.0(2.0) = 2.0 \text{ m}^2$$ $$P = 9.81(2.366)(2.0) = 46.43 \text{ kN}$$ $$\bar{y} = \frac{\bar{h}}{\sin 60°} = \frac{2.366}{0.866} = 2.732 \text{ m (along incline)}$$ $$I_G = \frac{1.0(2.0)^3}{12} = 0.667 \text{ m}^4$$ $$e = \frac{I_G}{A\bar{y}} = \frac{0.667}{2.0(2.732)} = 0.122 \text{ m}$$ $$y_{cp} = 2.732 + 0.122 = 2.854 \text{ m (along incline)}$$

Answer: Total force is 46.43 kN. The center of pressure is 2.854 m from the water-surface line measured along the inclined face of the gate.

Problem: Gate Hinged at the Bottom — Reaction at the Top Stop

A rectangular gate is 1.20 m wide and 2.40 m tall, hinged at its bottom edge and held by a horizontal stop (latch) at its top edge. The gate is vertical, and the water surface is 1.00 m above the top of the gate. Determine: (a) the total hydrostatic force on the gate, (b) the depth of the center of pressure, and (c) the horizontal reaction force at the top stop.

$$\bar{h} = 1.00 + \frac{2.40}{2} = 2.20 \text{ m}$$ $$A = 1.20(2.40) = 2.88 \text{ m}^2$$ $$P = 9.81(2.20)(2.88) = 62.11 \text{ kN}$$ $$I_G = \frac{1.20(2.40)^3}{12} = 1.382 \text{ m}^4$$ $$e = \frac{1.382}{2.88(2.20)} = 0.218 \text{ m}$$ $$h_{cp} = 2.20 + 0.218 = 2.418 \text{ m below surface}$$

Distance of resultant from hinge: gate bottom is $(1.00 + 2.40) = 3.40$ m below surface, so $d = 3.40 - 2.418 = 0.982$ m from hinge.

$$\sum M_{hinge} = 0: \quad R(2.40) = P(0.982)$$ $$R = \frac{62.11(0.982)}{2.40} = 25.40 \text{ kN}$$

Answer: Total force is 62.11 kN; center of pressure at 2.418 m depth. The reaction at the top stop is 25.40 kN outward.

Problem: Triangular Plate in Sea Water

An isosceles triangular plate has a base of 1.80 m and height of 2.40 m. It is submerged vertically in sea water (SG = 1.025) with its base (top edge) 0.80 m below the free surface and the apex pointing downward. Find the total hydrostatic force on the plate and the depth of the center of pressure.

The centroid of a triangle is 1/3 of height from the base, so the centroid is 0.80 + (2.40/3) = 0.80 + 0.80 = 1.60 m below the surface.

$$A = \frac{1}{2}(1.80)(2.40) = 2.16 \text{ m}^2$$ $$\gamma_{sw} = 9.81(1.025) = 10.055 \text{ kN/m}^3$$ $$P = 10.055(1.60)(2.16) = 34.75 \text{ kN}$$ $$I_G = \frac{bh^3}{36} = \frac{1.80(2.40)^3}{36} = 1.382 \text{ m}^4$$ $$e = \frac{I_G}{A\bar{h}} = \frac{1.382}{2.16(1.60)} = 0.400 \text{ m}$$ $$h_{cp} = 1.60 + 0.400 = 2.00 \text{ m below surface}$$

Answer: Total force is 34.75 kN. The center of pressure is 2.00 m below the surface (0.40 m below the centroid).

Problem: Plate with Fluid on Both Faces — Net Resultant

A vertical rectangular gate 2.0 m wide and 3.0 m tall is set in a wall between two water reservoirs. On the left (high) side, the water surface is 1.0 m above the top of the gate. On the right (low) side, the water surface is at the same elevation as the top of the gate (i.e., the right water depth equals the gate height). Find the net hydrostatic force on the gate and the location of the net resultant from the bottom of the gate.

Left side — water 1.0 m above gate top, so centroid depth = 1.0 + 1.5 = 2.5 m:

$$F_L = 9.81(2.5)(2.0)(3.0) = 147.15 \text{ kN}$$ $$I_{G,L} = \frac{2.0(3.0)^3}{12} = 4.5 \text{ m}^4$$ $$e_L = \frac{4.5}{6.0(2.5)} = 0.30 \text{ m} \Rightarrow h_{cp,L} = 2.80 \text{ m from surface}$$ $$d_L = (1.0+3.0) - 2.80 = 1.20 \text{ m from gate bottom}$$

Right side — water surface at top of gate, centroid depth = 1.5 m:

$$F_R = 9.81(1.5)(2.0)(3.0) = 88.29 \text{ kN}$$ $$e_R = \frac{4.5}{6.0(1.5)} = 0.50 \text{ m} \Rightarrow h_{cp,R} = 2.00 \text{ m from right surface}$$ $$d_R = 3.0 - 2.0 = 1.00 \text{ m from gate bottom}$$

Net force and resultant location:

$$F_{net} = 147.15 - 88.29 = 58.86 \text{ kN (leftward)}$$ $$\bar{d} = \frac{147.15(1.20) - 88.29(1.00)}{58.86} = \frac{176.58 - 88.29}{58.86} = 1.50 \text{ m from bottom}$$

Answer: Net hydrostatic force is 58.86 kN acting toward the lower reservoir. The resultant acts at 1.50 m from the bottom of the gate.