A dam has a parabolic shape z/z_0 = (x/x_0)^2 as shown in Fig. E2.8a, with x_0 = 10 ft and z_0 = 24 ft. The fluid is water, γ = 62.4 lbf/ft^3, and atmospheric pressure may be omitted. Compute the forces F_H and F_V on the dam and their line of action. The width of the dam is 50 ft.

Question

A dam has a parabolic shape [latex]z/z_0 = (x/x_0)^2[/latex] as shown in Fig. E2.8a, with [latex]x_0 = 10[/latex] ft and [latex]z_0 = 24[/latex] ft. The fluid is water, [latex]\gamma = 62.4 lbf/ft^3[/latex], and atmospheric pressure may be omitted. Compute the forces [latex]F_H and F_V[/latex] on the dam and their line of action. The width of the dam is 50 ft.

Accepted Answer

• System sketch: Figure E2.8b shows the various dimensions. The dam width is b = 50 ft.

• Approach: Calculate [latex]F_H[/latex] and its line of action from Eqs. (2.26) and (2.29). Calculate [latex]F_V[/latex] and its line of action by finding the weight of fluid above the parabola and the centroid of this weight.

[latex]F = p_aA + \gamma h_{CG}A = (p_a + \gamma h_{CG})A = p_{CG}A[/latex] (2.26)

[latex]F = \gamma h_{CG}A y_{CP} = - \frac{I_{xx} \sin heta}{h_{CG}A} x_{CP} = - \frac{I_{xy} \sin heta}{h_{CG}A}[/latex] (2.29)

• Solution steps for the horizontal component: The vertical projection of the parabola lies along the z axis in Fig. E2.8b and is a rectangle 24 ft high and 50 ft wide. Its centroid is halfway down, or [latex]h_{CG}[/latex] = 24/2 = 12 ft. Its area is [latex]A_{proj} = (24 ft)(50 ft) = 1200 ft^2[/latex]. Then, from Eq. (2.26),

[latex]F_H = \gamma h_{CG} A_{proj} = \left(62.4\frac{lbf}{ft^3} ight)(12 ft)(1200 ft^2) = 898,560 lbf \approx 899 imes 10^3 lbf[/latex]

The line of action of [latex]F_H[/latex] is below the centroid of [latex]A_{proj}[/latex], as given by Eq. (2.29):

[latex]y_{CP, proj} = - \frac{I_{xx} \sin heta}{h_{CG} A_{proj}} = - \frac{(1/12)(50 ft)(24 ft)^3 \sin 90^{\circ}}{(12 ft)(1200 ft^2)} = -4 ft[/latex]

Thus [latex]F_H[/latex] is 12 + 4 = 16 ft, or two-thirds of the way down from the surface (8 ft up from the bottom).

• Comments: Note that you calculate [latex]F_H[/latex] and its line of action from the vertical projection of the parabola, not from the parabola itself. Since this projection is vertical, its angle θ = 90°.

• Solution steps for the vertical component: The vertical force [latex]F_V[/latex] equals the weight of water above the parabola. Alas, a parabolic section is not in Fig. 2.13, so we had to look it up in another book. The area and centroid are shown in Fig. E2.8b. The weight of this parabolic amount of water is

[latex]F_V = \gamma A_{section}b = \left(62.4\frac{lbf}{ft^3} ight) \left[\frac{2}{3}(24 ft)(10 ft) ight] (50 ft) = 499,200 lbf \approx 499 imes 10^3 lbf[/latex]

This force acts downward, through the centroid of the parabolic section, or at a distance 3[latex]x_0[/latex]/8 = 3.75 ft over from the origin, as shown in Figs. E2.8b,c. The resultant hydrostatic force on the dam is

[latex]F = (F^2_H + F^2_V)^{1/2} = [(899E3 lbf)^2 + (499E3 lbf)^2]^{1/2} = 1028 imes 10^3 lbf at \angle 29^{\circ}[/latex]

This resultant is shown in Fig. E2.8c and passes through a point 8 ft up and 3.75 ft over from the origin. It strikes the dam at a point 5.43 ft over and 7.07 ft up, as shown.

• Comments: Note that entirely different formulas are used to calculate [latex]F_H and F_V[/latex]. The concept of center of pressure CP is, in the writer’s opinion, stretched too far when applied to curved surfaces.

Question 2.8: A dam has a parabolic shape z/z_0 = (x/x_0)^2 as shown in Fi...

Related Answered Questions