Question 6.3.4: Figure1 shows a cross section through a tank with a sloped r...
HYDROSTATIC PRESSURE ON SLOPED GATE
Figure 1 shows a cross section through a tank with a sloped rectangular gate that is 10 m long and 3 m wide (width is measured into the page). The gate is hinged along its top edge and held closed by a force acting on its bottom edge at A. Friction in the hinge and the weight of the gate can be neglected. If the tank is holding water, find the total load on the gate and the magnitude of the force acting on the bottom edge of the gate.
Goal Find the total load on the gate due to the liquid and the value of F_{A} needed to keep the gate closed.
Given The dimensions of the rectangular gate and the depth of water. The gate is held closed by a force at A. We can ignore the weight of the gate and any friction at the hinge.
Assume Since no water acts on the sides of the gate, the system can be treated as planar. The surface the gate rests against at A is frictionless, requiring F_{A} to be perpendicular to the gate. Also, we assume that the liquid is static and the system is in equilibrium.
Draw We draw a free-body diagram of the sloped gate with the pressure load (represented ω(x) in units of force/length) and include the loads at the supports (Figure 2a). Because we will work with gage pressure (6.22), we do not need to include the atmospheric pressure that acts on the back side of the gate. We then represent the distributed load by an equivalent point force R at a distance d_{R} from A (Figure 2b).
P_{gage} = \rho_{liq} gh (6.22A)
P = \gamma _{liq} h (6.22B)
Formulate Equations and Solve We recognize the load distribution as a line load made up of standard line load distributions, so we use Appendix C to calculate the total load due to the liquid pressure and its location. Since the pressure distribution is a trapezoid, we will break it into a rectangle and a triangle. The magnitude of the force/unit length at point B is the hydrostatic gage pressure at B multiplied by the width of the gate:
\omega _{min} =\rho_{liq} g h_{B} w = (1000 kg/m³ )(9.81 m/s² )(9.0 m)(3 m)
\omega _{min}=264.9 kN/m
The magnitude of the force/unit length at point A is the hydrostatic gage pressure at A multiplied by the width of the gate:
\omega _{min} =\rho_{liq} g h_{A} w =(1000 kg/m³ )(9.81 m/s² )(15 m)(3 m)
\omega _{min} = 441.5 kN/m
Figure 3 shows how we divide the load into two standard line load distributions: a rectangle (A_{1}) of height 264.9 kN/m and a triangle (A_{2}) of height 441.5 – 264.9 = 176.6 kN/m. We then calculate the total force for each standard line load and locate them at their centers of pressure as shown in Figure 4.
Using Appendix C we find that
F_{1} = (10 m)(264.9 kN/m) = 2649 kN , and d_{1} = 10/2 = 5.0 m
F_{2} = (10 m)(176.6 kN/m) 2 = 883 kN , and d_{2} = 10/3 = 3.33 m
Forces F_{1} and F_{2} , and dimensions d_{1} and d_{2} are shown in Figure 4. Using (6.16), we find the magnitude of the total force on the gate due to the hydrostatic gage pressure:
F_{total}=\sum\limits_{i=1}^{N}{\underbrace{F_{i}}_{force of each standard distribution} } (6.16)
R = F_{1} + F_{2} = 2649 kN + 883 kN
R = 3532 kN
To find the magnitude of force F_{A} , we use the equilibrium equation (5.5C), selecting the moment center at B:
\sum{M_{z} } =0 (5.5C)
\sum{M_{z @ B} } =0\left(\curvearrowleft + \right)
2649 kN (10.0 m – 5.0 m) + 883 kN(10.0 m – 3.33 m) – F_{A} (10 m)=0
F_{A} = 1913 kN
Check One check of the total force R and its location would be to solve the problem using an alternative approach, such as integrating the pressure distribution. We can check equilibrium by summing moments about a point other than B.
