Question 6.3.3: Figure1a shows a cross section through a rectangular gate th...
HYDROSTATIC PRESSURE ON VERTICAL RESERVOIR GATE
Figure 1a shows a cross section through a rectangular gate that is 8 m high and 3 m wide (where the width dimension w is perpendicular to the plane of the page). The gate blocks the end of a freshwater channel and opens automatically when the water reaches a certain depth. It opens by rotating, as shown in Figure 1b. Determine the depth d at which the gate just begins to open.
Goal Find the depth of water that causes the gate to open.
Given Information about the dimensions of the rectangular water gate and the location of the hinge. The gate holds back fresh water (ρ =1000 kg/m³).
Assume The weight of the gate is negligible, the hinge is frictionless, the fluid is static, and the system is in equilibrium. We treat the system as planar because no fluid pressure acts on the sides of the gate.
Draw Included on the free-body diagram of the gate are the hydrostatic gage pressure (based on (6.22)), the forces at the frictionless hinge, and the force of the sill at B pushing on the bottom of the gate (Figure 2a).
P_{gage} = \rho_{liq} gh (6.22A)
P = \gamma _{liq} h (6.22B)
We then represent the hydrostatic pressure by an equivalent total force at the center of pressure (Figure 2b).
Comment Because we have chosen to work with gage pressure (as opposed to absolute pressure from (6.21)), we do not need to include the atmospheric pressure on the back side of the gate. If we had chosen to work with absolute pressure, the free-body diagram of the gate would look as in Figure 3.
p = p_{o} + \rho_{liq} gh (6.21A)
p = p_{o} + \gamma _{liq} h (6.21B)
Formulate Equations and Solve The pressure distribution is triangular, and we use the properties of standard shapes in Appendix C to find the magnitude and location of the total hydrostatic gage force
F_{tot}= \frac{p_{max} dw }{2} = \frac{\left(\rho gd\right)dw }{2} = \frac{\rho g d^{2} w}{2}
at a distance d/3 from the bottom of the gate. Just as the gate is about to open, the force B_{x} must be zero. To find B_{x} , we sum the moment about the z axis, using the hinge at A as the moment center. To simplify the calculation, we define an x*y* coordinate system with its origin at the hinge (Figure 2b).
\sum{M_{z @ A} } =0\left(\curvearrowleft + \right)
F_{tot}y^{\ast }- B_{x}\left(2 m\right)=0
y^{\ast }= \frac{B_{x}\left(2 m\right)}{F_{tot}}
Setting B_{x}= 0 gives y* = 0. This tells us that just as the gate is about to open, the total force F_{tot} must act through the hinge, which is located 2m from the base of the gate. Therefore when the gate is about to open, the location of the center of pressure is \frac{d}{3}=2 m
The gate opens when d = 6 m
Check We can check the result using an alternative approach to finding the depth at which the gate opens. Recognize that the gate opens when the moment created by the portion of the hydrostatic gage pressure above the hinge (which wants to open the gate) just equals the moment created by the hydrostatic gage pressure below the hinge (which wants to close the gate) (Figure 4). At larger depths, the opening moment will be greater than the closing moment, so the gate will open. This approach is a lot more work, because it means finding the total force and centroid for two separate distributed loads (as summarized in Figure 4).
