EXAMPLE 3.10 of betaldua is no Assump 1. Negl 2. Negl Force to Open an Elliptical Gate Problem Statement An elliptical gate covers the end of a pipe 4 m in diameter. If the gate is hinged at the top, what normal force F is required to open the gate when water is 8 m deep above the top of the pipe and the pipe is open to the atmosphere on the other side? Neglect the weight of the gate. (a) Water labicato 8 m Hinge 5 m F Atmospheric pressure 4 m diameter pipe State the F(N) Generat 1. Calcu 2. Find Eq. (3 3. Draw 4. Apply Take Acti 1. Hydro- Define the Situation Water pressure is acting on an elliptical gate. Properties: Water (10°C), Table A.5: y = 9810 N/m³. . P = P = • A = form r. If red the side? Assumptions: 1. Neglect the weight of the gate. 2. Neglect friction between the bottom on the gate and the pipe wall. State the Goal F(N) Force needed to open gate. 3.5 C As engir as tanks Con bution w integrat is easier in Fig. 3 the free Generate Ideas and Make a Plan The lin TE 1. Calculate resultant hydrostatic force using F = PA. 2. Find the location of the center of pressure using Eq. (3.33). where 3. Draw an FBD of the gate. 4. Apply moment equilibrium about the hinge. Take Action (Execute the Plan) 1. Hydrostatic (resultant) force . p=pressure at depth of the centroid =(9810 N/m³)(10 m) = 98.1 kPa P=(Ywater)(centroid) A = area of elliptical panel (using Fig. A.1 to find formula) A = Tab = π(2.5 m)(2 m) = 15.71 m² 1+2 = 12 +Z₁ I T + Z₂ SE S: . Calculate resultant force F, PA (98.1 kPa)(15.71 m²) = 1.54 MN 2. Center of pressure ⚫y = 12.5 m, where y is the slant distance from the water surface to the centroid. • Area moment of inertia I of an elliptical panel using a formula from Fig. A.1 • πα π(2.5 m)³(2 m) = = 4 Finding center of pressure Ī 4 25.54 m² quite dall = 24.54 m² Yep - y = = = 0.125 m YA (12.5 m)(15.71 m²) 3.5 Calculating Forces on Curved Sur SECTION 3.5 CALCULATING FORCES ON CURVED SURFACES 3. FBD of the gate: di gnim Hy -Hinge abdix 2.625 m er Fp- 5 m the wel -F بنا nion Hx 4. Moment equilibrium Mhinge=0 1.541 x 106NX 2.625 m - FX 5 m = 0 F=809 kN Surfaces A