Question 9.5: A parabolic dam’s shape is given by z/zo = (x/xo)² , where x...
A parabolic dam’s shape is given by z/z_o = (x/x_o)² , where x_o = 10 ft and z_o = 24 ft. The dam is 50 ft wide (into the page). Find the resultant force on the dam due to the water pressure and its line of action (Figure 9.23).
Given: Geometry of dam; water depth.
Find: Resultant force due to pressure; line of action.
Assume: No relative motion of fluid elements (hydrostatics); fluids have constant, uniform density, and it is appropriate to evaluate densities at 68 ̊F.
We look up the density of water at this temperature and find that (ρg) = 62.4 lbf/ft³ . We find separately the horizontal and vertical components of the resultant force on the dam surface.
To find the horizontal component F_H , we consider the vertical projection of the curved dam surface, a rectangle that is 24 ft high and 50 ft wide (into the page). This projected surface has an area A = (24)(50) = 1200 ft² , and its centroid is halfway down, at a depth of h_c = 12 ft. So,
F_H = ρgh_c A= (62.4 lbf/ft³ )(12 ft)(1200 ft²)
= 899,000 lbf
= 899 kips.
This force acts at the centroid of the pressure prism on the projected vertical surface, at z_H = h/3 = 8 ft from the bottom.
The vertical component F_V can be interpreted as the weight of fluid above
the curved surface. We consider the properties of a parabolic section, as shown in the sketch in Figure 9.24, to find this value.
= ρg\left\lgroup\frac{2}{3}x_oz_o \right\rgroup ( 50 ft)
= (62.4 lbf/ft³)[ \frac{2}{3} (24 ft)(10 ft)](50 ft)
= 499,000 lbf
= 499 kips.
This force acts at the x coordinate of the centroid of the volume of fluid, V.
From our sketch, we see that this is 3x_o /8 = 3.75 ft from the origin indicated in Figure 9.23.
The resultant normal force on the surface of the parabolic dam is
F_R= \sqrt{F^{2}_{H}+F^{2}_{V}}= \sqrt{(899 \textrm{ k})^2+(499 \textrm{ k})^2}
= 1028 kips.
The line of action of this force (Figure 9.25) passes through the point (x_{_V} , z_{_H} ) = (3.75 ft, 8 ft) and has slope equal to
tan^{-1} (F_V /F_H ) = tan^{-1} (499/899) = 29 ̊.
