FIGURE 1. Hydrostatic thrust on plane FIGURE 2. Pressure prism,
surface.
P = wyc*A (3)ZPC= ^T- +ZCA (4)
AZCA- Compute the required values, and solve the equations in step 1
Thus A =^1 / 2 (5)(6) = 15 ft^2 (1.39 m^2 ); yCA = 2 + 4 sin 60° = 5.464 ft (166.543 cm); ZCA = 2
esc 60° + 4 = 6.309 ft (192.3 cm); ICAIA = (bf/36)/(bd/2) = </^2 /18 = 2 ft^2 (0.186 m^2 ); P =
62.4(5.464)(15) = 5114 Ib (22.747 N); zpc = 2/6.309 + 6.309 = 6.626 ft (201.960 cm);
ypc = 6.626 sin 60° = 5.738 ft (174.894 cm). By symmetry, the pressure center lies on the
centroidal axis through C.
An alternative equation for P is
P=PnA (5)Equation 3 shows that the mean pressure occurs at the centroid of the area. The above two
steps constitute method 1 for solving this problem. The next three steps constitute method
2.- Now construct the pressure "prism" associated with the area
In Fig. 2, construct the pressure prism associated with area CDE. The pressures are as fol-
lows: at apex,/? = 2w; at base,/? = (2 + 6 sin 60°)w = 7.196w.
The force P equals the volume of this prism, and its action line lies on the centroidal
plane parallel to the base. For convenience, resolve this prism into a triangular prism and
rectangular pyramid, as shown. - Determine P by computing the volume of the pressure prism
Thus, P = Aw[2 +^2 / 3 (5.196)] =Aw(2 + 3.464) = 15(62.4)(5.464) = 5114 Ib (22,747 N). - Find the location of the resultant thrust
Compute the distance h from the top line to the centroidal plane. Then find ypc. Or, h =
[2(^2 / 3 )(6) + 3.464(^3 / 4 )(6)]/5.464 = 4.317 ft (131.582 cm); ypc = 2 + 4.317 sin 60° = 5.738 ft
(174.894cm).
Surface