Question: Problems solving (Plane surface):
1. A vertical trapezoidal gate in the face of a dyke is subjected to sea water pressure ($\gamma=64$ lb/ft$^3$) on one side. The upper edge is in the water surface and is 5 ft long. Two edges are vertical and measure 6 ft and 9 ft each. Determine the total pressure force on the gate and the location of the centre of pressure. [King, p-47, Exc-7]
2. A vertical rectangular gate 4 ft wide and 6 ft high, hinged at the top, has water on one side. What force applied at the bottom of the gate, at an angle of 45 with the vertical, is required to open the gate when the water surface is (a) at the top of the gate, (b) 3 ft above the top of the gate, (c) 3 ft below the top of the gate? [King, p-47, Exc-9]
3. Gate AB in Fig. P2.55 is 5 ft wide into the paper, hinged at A, and restrained by a stop at B. The water is at 20 °C. Compute (a) the force on stop B and (b) the reactions at A if the water depth $h=9.5$ ft. [Frank, p-110, Exc-2.55]
### Problem 1: Vertical trapezoidal gate under water pressure
1. **Problem statement:**
A trapezoidal gate with upper edge length $b_1=5$ ft at water surface, vertical sides $h_1=6$ ft and $h_2=9$ ft, is subjected to water pressure with specific weight $\gamma=64$ lb/ft$^3$. Find total pressure force $F$ and center of pressure $y_{cp}$.
2. **Formula for pressure force on vertical surface:**
$$F = \gamma \times A \times \bar{h}$$
where $A$ is area of gate, $\bar{h}$ is depth of centroid from water surface.
3. **Formula for center of pressure:**
$$y_{cp} = \bar{h} + \frac{I_G}{\bar{h} A}$$
where $I_G$ is second moment of area about horizontal axis through centroid.
4. **Calculate area $A$ of trapezoid:**
$$b_2 = b_1 \times \frac{h_2}{h_1} = 5 \times \frac{9}{6} = 7.5 \text{ ft}$$
$$A = \frac{(b_1 + b_2)}{2} \times h_2 = \frac{5 + 7.5}{2} \times 9 = 56.25 \text{ ft}^2$$
5. **Calculate centroid depth $\bar{h}$:**
Centroid from base:
$$\bar{y} = \frac{h_2}{3} \times \frac{2b_2 + b_1}{b_1 + b_2} = \frac{9}{3} \times \frac{2 \times 7.5 + 5}{5 + 7.5} = 3 \times \frac{20}{12.5} = 4.8 \text{ ft}$$
Depth from water surface:
$$\bar{h} = h_2 - \bar{y} = 9 - 4.8 = 4.2 \text{ ft}$$
6. **Calculate moment of inertia $I_G$ about centroidal axis:**
$$I_G = \frac{h_2^3}{36} \times \frac{b_1^2 + 4 b_1 b_2 + b_2^2}{b_1 + b_2}$$
Substitute:
$$I_G = \frac{9^3}{36} \times \frac{5^2 + 4 \times 5 \times 7.5 + 7.5^2}{5 + 7.5} = \frac{729}{36} \times \frac{25 + 150 + 56.25}{12.5} = 20.25 \times \frac{231.25}{12.5} = 20.25 \times 18.5 = 374.625 \text{ ft}^4$$
7. **Calculate total force $F$:**
$$F = \gamma A \bar{h} = 64 \times 56.25 \times 4.2 = 15120 \text{ lb}$$
8. **Calculate center of pressure $y_{cp}$:**
$$y_{cp} = \bar{h} + \frac{I_G}{\bar{h} A} = 4.2 + \frac{374.625}{4.2 \times 56.25} = 4.2 + \frac{374.625}{236.25} = 4.2 + 1.586 = 5.786 \text{ ft}$$
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### Problem 2: Force to open rectangular gate hinged at top
1. **Problem statement:**
Rectangular gate width $b=4$ ft, height $h=6$ ft, hinged at top. Water on one side. Find force $F$ applied at bottom at 45° to vertical to open gate for:
(a) water surface at top,
(b) water surface 3 ft above top,
(c) water surface 3 ft below top.
2. **Water pressure force on gate:**
$$F_w = \gamma b \frac{h^2}{2}$$
Depth to centroid:
$$\bar{h} = \frac{h}{2}$$
3. **Moment of water force about hinge (top):**
$$M_w = F_w \times \bar{h} = \gamma b \frac{h^2}{2} \times \frac{h}{2} = \gamma b \frac{h^3}{4}$$
4. **Force $F$ applied at bottom at 45° to vertical produces moment:**
Lever arm = $h$, force component perpendicular to gate:
$$F_{perp} = F \sin 45^\circ = \frac{F}{\sqrt{2}}$$
Moment:
$$M_F = F_{perp} \times h = \frac{F}{\sqrt{2}} h$$
5. **Equate moments to open gate:**
$$M_F = M_w \Rightarrow \frac{F}{\sqrt{2}} h = \gamma b \frac{h^3}{4}$$
$$F = \frac{\gamma b h^3}{4} \times \frac{\sqrt{2}}{h} = \frac{\gamma b h^2 \sqrt{2}}{4}$$
6. **Calculate for each case:**
(a) Water surface at top:
$$\gamma = 62.4 \text{ lb/ft}^3$$ (fresh water assumed)
$$F = \frac{62.4 \times 4 \times 6^2 \times \sqrt{2}}{4} = 62.4 \times 4 \times 36 \times \frac{\sqrt{2}}{4} = 62.4 \times 36 \times \sqrt{2} = 62.4 \times 36 \times 1.414 = 3177.5 \text{ lb}$$
(b) Water surface 3 ft above top:
New height $h' = 6 + 3 = 9$ ft
$$F = \frac{62.4 \times 4 \times 9^2 \times \sqrt{2}}{4} = 62.4 \times 81 \times \sqrt{2} = 62.4 \times 81 \times 1.414 = 7149.5 \text{ lb}$$
(c) Water surface 3 ft below top:
New height $h' = 6 - 3 = 3$ ft
$$F = \frac{62.4 \times 4 \times 3^2 \times \sqrt{2}}{4} = 62.4 \times 9 \times \sqrt{2} = 62.4 \times 9 \times 1.414 = 794.5 \text{ lb}$$
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### Problem 3: Gate AB with hinge A and stop B
1. **Problem statement:**
Gate AB width $b=5$ ft, water depth $h=9.5$ ft, hinged at A, restrained by stop at B. Find:
(a) force on stop B,
(b) reactions at A.
2. **Water force on gate:**
$$F_w = \gamma b \frac{h^2}{2} = 62.4 \times 5 \times \frac{9.5^2}{2} = 62.4 \times 5 \times \frac{90.25}{2} = 62.4 \times 5 \times 45.125 = 14070 \text{ lb}$$
3. **Depth to centroid:**
$$\bar{h} = \frac{h}{2} = 4.75 \text{ ft}$$
4. **Moment of water force about hinge A:**
$$M_w = F_w \times \bar{h} = 14070 \times 4.75 = 66832.5 \text{ lb-ft}$$
5. **Force on stop B:**
Stop B prevents rotation, so force at B balances moment:
$$F_B \times L = M_w$$
Assuming length $L = 9.5$ ft (height of gate),
$$F_B = \frac{M_w}{L} = \frac{66832.5}{9.5} = 7035.5 \text{ lb}$$
6. **Reactions at hinge A:**
Vertical reaction $R_A$ balances vertical forces:
$$R_A = F_w = 14070 \text{ lb}$$
Horizontal reaction $H_A$ balances horizontal forces (if any), here assumed zero.
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