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Curved Surface Force Components

For a curved submerged surface, resolve the total hydrostatic force into horizontal and vertical components.

$$P_H = \gamma \bar{h} A_v$$ $$P_V = \gamma V_i$$ $$R = \sqrt{P_H^2 + P_V^2}$$ $$\tan\theta = \frac{P_V}{P_H}$$

$A_v$ is the vertical projection of the curved surface. $V_i$ is the imaginary volume of liquid directly above or displaced by the curved surface, depending on the geometry.

Location of Components

The horizontal component acts through the center of pressure of the vertical projection. The vertical component acts through the centroid of the imaginary volume of liquid.

$$e = \frac{I_G}{A\bar{h}}$$ $$h_{cp} = \bar{h} + e$$ $$\sum M = 0$$
For curved gates, many board problems ask only for $P_H$, $P_V$, and the resultant. If a hinge or bolt is present, use moment equilibrium after computing the components.

Problem: Quadrant of Circular Cylinder

A curved surface is a quadrant of a circular cylinder 3 m long. For $r = 3 \text{ m}$ and $z = 0$, determine the horizontal and vertical force components.

$$P_H = \gamma \bar{h}A_v$$ $$P_H = 9.81(1.5)(3)(3) = 132.44 \text{ kN}$$ $$P_V = \gamma V_i$$ $$P_V = 9.81\left(\frac{\pi(3)^2}{4}\right)(3) = 208 \text{ kN}$$

Answer: $P_H = 132.44 \text{ kN}$ and $P_V = 208 \text{ kN}$.

Problem: Crest Gate Resultant

For a crest gate with $P_H = 66.22 \text{ kN}$ and $P_V = 16.981 \text{ kN}$, determine the resultant and its direction.

$$R = \sqrt{P_H^2 + P_V^2}$$ $$R = \sqrt{(66.22)^2 + (16.981)^2} = 68.36 \text{ kN}$$ $$\tan\theta = \frac{P_V}{P_H}$$ $$\theta = \tan^{-1}\left(\frac{16.981}{66.22}\right) = 14.4^\circ$$

Answer: $R = 68.36 \text{ kN}$ at $14.4^\circ$ above the horizontal component.

Problem: Semicircular Gate — Full Component Analysis

A semicircular gate has its flat diameter (2.0 m) at the water surface and the curved surface bowing downward. The gate is 1.50 m long perpendicular to the cross-section. Water is on the concave (upper) side of the curved gate. The radius of the semicircle is 1.0 m. Determine the horizontal and vertical hydrostatic force components on the gate and the resultant.

Horizontal component: acts on the vertical projection, a 2.0 m × 1.0 m rectangle with centroid at 0.50 m depth.

$$P_H = \gamma \bar{h} A_v = 9.81(0.50)(2.0 \times 1.0)(1.50) = 14.72 \text{ kN}$$

Vertical component: weight of water in the semicylindrical volume above the curved surface.

$$V_{semi} = \frac{\pi r^2}{2} \times L = \frac{\pi(1.0)^2}{2}(1.50) = 2.356 \text{ m}^3$$ $$P_V = \gamma V = 9.81(2.356) = 23.11 \text{ kN (downward)}$$ $$R = \sqrt{14.72^2 + 23.11^2} = 27.46 \text{ kN}$$ $$\theta = \tan^{-1}\left(\frac{23.11}{14.72}\right) = 57.5°\text{ below horizontal}$$

Answer: $P_H = 14.72 \text{ kN}$, $P_V = 23.11 \text{ kN}$ (downward), resultant = 27.46 kN at 57.5° from horizontal.

Problem: Upward Force on a Conical Tank Bottom

A cylindrical water tank has a conical bottom. The circular edge at the base of the cylindrical wall has a diameter of 3.0 m. The water depth above this rim is 4.0 m. The cone narrows to a point 1.5 m below the rim. Determine the total upward hydrostatic force exerted by water pressure on the inner surface of the conical bottom.

The upward vertical force equals the weight of the imaginary water volume above the curved surface up to the free surface level.

$$V_{cyl} = \frac{\pi(3.0)^2}{4}(4.0) = 28.27 \text{ m}^3$$ $$V_{cone} = \frac{1}{3}\cdot\frac{\pi(3.0)^2}{4}(1.5) = 3.534 \text{ m}^3$$ $$V_{total} = 28.27 + 3.534 = 31.80 \text{ m}^3$$ $$P_V = \gamma V = 9.81(31.80) = 311.97 \text{ kN (upward)}$$

Answer: The upward hydrostatic force on the conical bottom is 311.97 kN. This force tends to push the cone off and must be balanced by the structural connection to the cylinder.

Problem: Radial Tainter Gate — Resultant Through Pivot

A radial (Tainter) gate has radius 3.0 m and width 2.0 m into the page. The top of the gate arc is at the water surface and the gate subtends a vertical depth of 2.4 m. Calculate the horizontal and vertical hydrostatic force components on the gate, and explain why the resultant passes through the center of the arc (the pivot).

$$P_H = \gamma \bar{h} A_v = 9.81(1.20)(2.4 \times 2.0) = 56.51 \text{ kN}$$

For the vertical component, find the submerged arc area (waterway cross-section wedge between vertical line and arc):

$$\sin\alpha = \frac{2.4}{3.0} = 0.8 \Rightarrow \alpha = 53.13°$$ $$A_{sector} = \frac{\alpha}{360°}\pi r^2 = \frac{53.13}{360}\pi(3.0)^2 = 4.189 \text{ m}^2$$ $$A_{triangle} = \frac{1}{2}(3.0\sin\alpha)(3.0\cos\alpha) = \frac{1}{2}(2.4)(1.8) = 2.16 \text{ m}^2$$ $$A_{wedge} = 4.189 - 2.16 = 2.029 \text{ m}^2$$ $$V_{wedge} = A_{wedge} \times 2.0 = 4.058 \text{ m}^3$$ $$P_V = 9.81(4.058) = 39.81 \text{ kN (downward)}$$ $$R = \sqrt{56.51^2 + 39.81^2} = 69.07 \text{ kN}$$

Every pressure element is normal to the curved surface and thus directed toward the center of curvature. The resultant of all these normal forces must also pass through the center — the pivot point of the gate.

Answer: $P_H = 56.51$ kN, $P_V = 39.81$ kN downward, resultant = 69.07 kN. It passes through the pivot because all pressure forces are radially directed toward the center of the arc.

Problem: Log Gate with Water on Both Sides

A cylindrical log gate has diameter 1.20 m and length 3.0 m. It sits in a channel slot. The upstream water depth is 1.20 m (equal to the gate diameter) and the downstream water depth is 0.60 m. Find the net horizontal hydrostatic force acting on the log gate and state its direction.

Net horizontal force = hydrostatic force on the upstream projected area minus that on the downstream projected area.

$$F_{up} = \frac{\gamma h_1^2}{2} \cdot L = \frac{9.81(1.20)^2}{2}(3.0) = 21.19 \text{ kN}$$ $$F_{down} = \frac{\gamma h_2^2}{2} \cdot L = \frac{9.81(0.60)^2}{2}(3.0) = 5.30 \text{ kN}$$ $$F_{net} = 21.19 - 5.30 = 15.89 \text{ kN}$$

Answer: The net horizontal force on the log gate is 15.89 kN acting in the downstream direction. The vertical (buoyant) forces cancel by symmetry for a fully submerged cylinder.

Problem: Curved Gate Held by a Bolt at the Top

A quarter-circle curved gate of radius 2.0 m and width 1.0 m (into the page) is hinged at its bottom corner and held by a single horizontal bolt at its top edge. The flat top of the gate is at the water surface, and the curved face is concave toward the water (water is on the outside of the curve). Determine the bolt force required to hold the gate closed. Take the hinge at the curved bottom corner.

With the gate curving from the water surface down to the hinge at the base, water presses on the concave face.

$$P_H = \gamma \bar{h} A_v = 9.81(1.0)(2.0 \times 1.0) = 19.62 \text{ kN (at }h_{cp}=2r/3=1.333\text{ m below surface)}$$ $$P_V = \gamma V = 9.81\left(\frac{\pi(2.0)^2}{4}\right)(1.0) = 30.82 \text{ kN (upward, quarter-circle displaced volume)}$$

Taking moments about the hinge (bottom corner at 2.0 m depth, 2.0 m horizontally from bolt):

$$\text{Arm of }P_H = 2.0 - 1.333 = 0.667 \text{ m (horizontal dist. from hinge to }P_H\text{)}$$ $$\text{Arm of }P_V = 2.0 - \frac{4r}{3\pi} = 2.0 - 0.849 = 1.151 \text{ m (vertical dist. to centroid of quarter-circle)}$$ $$F_{bolt}(2.0) = P_H(0.667) + P_V(1.151)$$ $$F_{bolt}(2.0) = 19.62(0.667) + 30.82(1.151) = 13.09 + 35.47 = 48.56 \text{ kN·m}$$ $$F_{bolt} = \frac{48.56}{2.0} = 24.28 \text{ kN}$$

Answer: The bolt at the top must resist a force of approximately 24.28 kN to keep the curved gate closed.