Question 6.2.7: Figure 1 shows a cross section through a rectangular gate th...
PRESSURE ON A RECTANGULAR WATER GATE
Figure 1 shows a cross section through a rectangular gate that is h = 30 ft high and w = 8 ft wide (where the width dimension is perpendicular to the yz plane). The gate is subjected to a load generated by fresh water stored to a depth of d = 25 ft in a reservoir behind the gate. The pressure that the water applies to the gate varies linearly, as indicated in Figure 2. Determine the magnitude of the total force F exerted on the gate by the water and its location with respect to the hinge.
Goal Find the magnitude of the total force of the water on the gate and the location of the centroid of the pressure distribution with respect to the hinge.
Given Information about the geometry of the rectangular water gate and the depth of the fresh water.
Assume The weight of the gate is negligible (we aren’t given a weight), the hinge is frictionless, the fluid is static, and the system is in equilibrium.
Draw Based on the information given in the problem and our assumptions, we draw a free-body diagram (Figure 2) of the gate. We have selected positive z acting downward, meaning that the x axis acts into the page, and locate the origin of the coordinate system at the top of the water. B_{y} and B_{z} are the forces of the hinge acting on the gate, and C_{y} is the force due to the sill pushing on the bottom of the gate.
Formulate Equations and Solve The pressure distribution varies according to
p(z)=\gamma _{wat}z
The specific weight of water, \gamma _{wat} = ρg = 62.4 lb/ft³. Therefore, p(z) = 62.4z in units of lb/ft².
From (6.18), the total force of the water acting on the gate is
Total force in z direction = F_{z}=\int_{surface area}^{}{\int{pdxdy} } (6.18)
F=\int_{0}^{25 ft}{\int_{0}^{8 ft}{p(z) dxdz} \int_{0}^{25 ft }{\int_{0}^{8 ft}{\left\lgroup62.4\frac{lb}{ft^{3} } \right\rgroup zdxdz }} } = \int_{0}^{25 ft}{\left\lgroup62.4\frac{lb}{ft^{3} } \right\rgroup \left[xz\right] ^{8 ft}_{0}dz }
F=\left\lgroup62.4\frac{lb}{ft^{3} } \right\rgroup\left(8 ft\right)\int_{0}^{25 ft}{zdz}=\left\lgroup499.2 \frac{lb}{ft^{3}}\right\rgroup \left[\frac{z^{2} }{2} \right] ^{25 ft}_{0} = 156,000 lb
F = 156 kip
We now calculate Z_{C}, the distance to the pressure center, by rewriting (6.20B) in terms of z, instead of y.
X_{C}=\frac{\int_{surface area}^{}{\int{xpdxdy} }}{F_{z}} \underbrace{=}_{substituting in from (6.18)} \frac{\int_{surface area}^{}{\int{xpdxdy} }}{\int_{surface area}^{}{\int{pdxdy} }} (6.20B)
Z_{C}= \frac{\int_{0}^{25 ft}{\int_{0}^{8 ft}{zp\left(z\right) dxdz} } }{F} = \frac{\int_{0}^{25 ft}{\int_{0}^{8 ft}{ \left\lgroup62.4\frac{lb}{ft^{3} }\right\rgroup z^{2} dxdz} } }{F}
Z_{C}=\frac{ \left\lgroup62.4\frac{lb}{ft^{3} }\right\rgroup \left(8 ft\right) \int_{0}^{25 ft}{z^{2}dz} } {F}= \frac{\left\lgroup499.2\frac{lb}{ft^{2} }\right\rgroup \left[\frac{z^{3} }{3} \right] ^{25 ft}_{0} }{F}
\frac{2,600,000 lb.ft}{156,000 lb} =156,000 lb
Z_{C} is measured with respect to the top of the water. We find Z_{hinge} (the distance between the hinge and the centroid) using
Z_{hinge}= Z_{C} – distance from top of water to hinge
=16.67 ft – 10 ft =6.67 ft
free-body diagram of the gate is shown in Figure 3.
Check We can use Appendix C to check our results because the pressure distribution can be modeled as a standard line load distribution. We multiply the pressure distribution by the width of the gate to calculate the force per unit length along the height of the gate.
\omega _{max}=(62.4 lb/ft^{3} )(25 ft)(8 ft)= 12,480 lb/ft
Based on the triangular distribution in Appendix C, we find
F=\frac{(25 ft)(12,480 lb/ft)}{2} =156,000 lb
Z_{C}=\frac{2(25 ft)}{3} =16.67 ft
Z_{hinge}=16.67 ft – 10 ft = 6.67 ft
Yes, our answer checks!
