Plane Surface Gates 1B9D77

### Problem 1: Vertical trapezoidal gate under water pressure 1. **Problem statement:** A trapezoidal gate with upper edge length $b_1 = 5$ ft, vertical sides $h_1 = 6$ ft and $h_2 = 9$ ft, is submerged with the upper edge at the water surface. The water specific weight is $\gamma = 64$ lb/ft$^3$. Find the total hydrostatic force on the gate and the center of pressure. 2. **Formula for total hydrostatic force:** $$F = \gamma \cdot h_c \cdot A$$ where $h_c$ is the depth of the centroid of the submerged surface, and $A$ is the area of the surface. 3. **Calculate the area $A$ of the trapezoid:** The lower edge length $b_2 = b_1 + (h_2 - h_1) \times \text{slope}$. Since the sides are vertical, the width changes linearly from 5 ft at top to some value at bottom. Actually, the problem states two vertical edges of 6 ft and 9 ft, so the trapezoid height is $h = 9$ ft (the longer side). The bottom edge length $b_2$ can be found by similarity: $$b_2 = b_1 + (h_2 - h_1) \times \frac{b_1}{h_1} = 5 + (9 - 6) \times \frac{5}{6} = 5 + 3 \times \frac{5}{6} = 5 + 2.5 = 7.5 \text{ ft}$$ Area of trapezoid: $$A = \frac{(b_1 + b_2)}{2} \times h_2 = \frac{5 + 7.5}{2} \times 9 = 6.25 \times 9 = 56.25 \text{ ft}^2$$ 4. **Find the centroid depth $h_c$ from water surface:** The centroid of a trapezoid from the top is given by: $$\bar{y} = \frac{h}{3} \times \frac{2b_1 + b_2}{b_1 + b_2} = \frac{9}{3} \times \frac{2 \times 5 + 7.5}{5 + 7.5} = 3 \times \frac{10 + 7.5}{12.5} = 3 \times 1.4 = 4.2 \text{ ft}$$ 5. **Calculate total force:** $$F = \gamma \cdot h_c \cdot A = 64 \times 4.2 \times 56.25 = 15120 \text{ lb}$$ 6. **Calculate center of pressure depth $h_p$:** $$h_p = h_c + \frac{I_G}{h_c A}$$ where $I_G$ is the second moment of area about the horizontal axis through the centroid. For trapezoid, $I_G$ about centroidal axis parallel to base: $$I_G = \frac{h^3}{36} \times \frac{b_1^2 + 4 b_1 b_2 + b_2^2}{b_1 + b_2}$$ Calculate numerator: $$b_1^2 + 4 b_1 b_2 + b_2^2 = 5^2 + 4 \times 5 \times 7.5 + 7.5^2 = 25 + 150 + 56.25 = 231.25$$ Then: $$I_G = \frac{9^3}{36} \times \frac{231.25}{12.5} = \frac{729}{36} \times 18.5 = 20.25 \times 18.5 = 374.625 \text{ ft}^4$$ 7. **Calculate $h_p$:** $$h_p = 4.2 + \frac{374.625}{4.2 \times 56.25} = 4.2 + \frac{374.625}{236.25} = 4.2 + 1.585 = 5.785 \text{ ft}$$ --- ### Problem 2: Force to open a hinged rectangular gate 1. **Problem statement:** A vertical rectangular gate 4 ft wide and 6 ft high, hinged at the top, has water on one side. Find the force applied at the bottom of the gate at 45° with vertical to open the gate when water surface is: (a) at the top of the gate, (b) 3 ft above the top, (c) 3 ft below the top. 2. **Hydrostatic force on rectangular gate:** $$F = \gamma \cdot h_c \cdot A$$ where $h_c$ is centroid depth, $A = b \times h = 4 \times 6 = 24$ ft$^2$. 3. **Centroid depth $h_c$:** For rectangle with top at water surface, $h_c = \frac{h}{2}$. 4. **Moment of force about hinge (top):** The force acts at center of pressure depth $h_p$: $$h_p = h_c + \frac{I_G}{h_c A}$$ For rectangle: $$I_G = \frac{b h^3}{12} = \frac{4 \times 6^3}{12} = \frac{4 \times 216}{12} = 72 \text{ ft}^4$$ 5. **Calculate $h_p$ for rectangle:** $$h_c = 3 \text{ ft}$$ $$h_p = 3 + \frac{72}{3 \times 24} = 3 + \frac{72}{72} = 3 + 1 = 4 \text{ ft}$$ 6. **Calculate force $F$ for each case:** (a) Water surface at top: $$F = 64 \times 3 \times 24 = 4608 \text{ lb}$$ (b) Water surface 3 ft above top: New depth to centroid: $$h_c = 3 + 3 = 6 \text{ ft}$$ $$F = 64 \times 6 \times 24 = 9216 \text{ lb}$$ (c) Water surface 3 ft below top: Water depth $h = 6 - 3 = 3$ ft $$h_c = \frac{3}{2} = 1.5 \text{ ft}$$ $$A = 4 \times 3 = 12 \text{ ft}^2$$ $$F = 64 \times 1.5 \times 12 = 1152 \text{ lb}$$ 7. **Calculate moment about hinge for each case:** Moment arm is $h_p$ from hinge. For rectangle: $$I_G = \frac{b h^3}{12}$$ Calculate $h_p$ for each case: (a) $h=6$ ft: $$I_G = \frac{4 \times 6^3}{12} = 72$$ $$h_c = 3$$ $$h_p = 3 + \frac{72}{3 \times 24} = 4$$ Moment: $$M = F \times h_p = 4608 \times 4 = 18432 \text{ lb-ft}$$ (b) $h=9$ ft: $$I_G = \frac{4 \times 9^3}{12} = \frac{4 \times 729}{12} = 243$$ $$h_c = 4.5$$ $$h_p = 4.5 + \frac{243}{4.5 \times 36} = 4.5 + 1.5 = 6$$ $$F = 64 \times 4.5 \times 36 = 10368$$ Moment: $$M = 10368 \times 6 = 62208 \text{ lb-ft}$$ (c) $h=3$ ft: $$I_G = \frac{4 \times 3^3}{12} = 9$$ $$h_c = 1.5$$ $$h_p = 1.5 + \frac{9}{1.5 \times 12} = 1.5 + 0.5 = 2$$ $$F = 64 \times 1.5 \times 12 = 1152$$ Moment: $$M = 1152 \times 2 = 2304 \text{ lb-ft}$$ 8. **Force applied at bottom at 45° to vertical to open gate:** The applied force $P$ must create a moment equal to the hydrostatic moment: $$P \times 6 \times \sin 45^\circ = M$$ $$P = \frac{M}{6 \times \sin 45^\circ} = \frac{M}{6 \times \frac{\sqrt{2}}{2}} = \frac{M}{3 \sqrt{2}}$$ 9. **Calculate $P$ for each case:** (a) $$P = \frac{18432}{3 \sqrt{2}} = \frac{18432}{4.2426} = 4344.5 \text{ lb}$$ (b) $$P = \frac{62208}{4.2426} = 14656.5 \text{ lb}$$ (c) $$P = \frac{2304}{4.2426} = 543.0 \text{ lb}$$ --- ### Problem 3: Gate AB with hinge A and stop B 1. **Problem statement:** Gate AB is 5 ft wide, hinged at A (top), restrained by stop at B (bottom). Water depth $h=9.5$ ft at 20 °C. Find: (a) Force on stop B, (b) Reactions at hinge A. 2. **Hydrostatic force on gate:** $$F = \gamma \cdot h_c \cdot A$$ Water specific weight at 20 °C approximately $\gamma = 62.4$ lb/ft$^3$. 3. **Gate dimensions:** Height $h = 9.5$ ft, width $b = 5$ ft. 4. **Calculate area:** $$A = b \times h = 5 \times 9.5 = 47.5 \text{ ft}^2$$ 5. **Centroid depth:** $$h_c = \frac{9.5}{2} = 4.75 \text{ ft}$$ 6. **Calculate force:** $$F = 62.4 \times 4.75 \times 47.5 = 14070 \text{ lb}$$ 7. **Center of pressure depth:** $$I_G = \frac{b h^3}{12} = \frac{5 \times 9.5^3}{12} = \frac{5 \times 857.375}{12} = 357.24 \text{ ft}^4$$ $$h_p = h_c + \frac{I_G}{h_c A} = 4.75 + \frac{357.24}{4.75 \times 47.5} = 4.75 + 1.58 = 6.33 \text{ ft}$$ 8. **Calculate moment about hinge A:** $$M = F \times h_p = 14070 \times 6.33 = 89063 \text{ lb-ft}$$ 9. **Force on stop B:** The stop B resists the moment, so force at B times gate width equals moment: $$F_B \times 9.5 = M \Rightarrow F_B = \frac{M}{9.5} = \frac{89063}{9.5} = 9375 \text{ lb}$$ 10. **Reactions at hinge A:** Vertical reaction $R_A$ balances vertical forces: $$R_A = F - F_B = 14070 - 9375 = 4695 \text{ lb}$$ Horizontal reaction $H_A$ is zero (no horizontal forces). ---