Hydrostatics — Pressure & Fluid Forces
Hydrostatics — Pressure & Fluid Forces
Civil engineers design dams, retaining walls, and tanks by calculating hydrostatic pressure and the resultant force on surfaces.
$$P = \rho g h$$
where $\rho$ = fluid density (kg/m³), $g = 9.81$ m/s², $h$ = depth below the surface (m). Pressure increases linearly with depth.
$$F = \frac{1}{2} \rho g h^2 \cdot w$$
where $h$ = height of the fluid and $w$ = width of the surface. The force acts at $\frac{2}{3}h$ from the surface.
A dam face is 8 m high and 20 m wide. Water is at full height. Find the total hydrostatic force.
$$F = \frac{1}{2}(1{,}000)(9.81)(8^2)(20) = \frac{1}{2}(1{,}000)(9.81)(64)(20) = 6{,}278 \text{ kN}$$
It acts at $\frac{2}{3}(8) = 5.33$ m below the surface.
A rectangular tank is 3 m deep, filled with water. Find the pressure at the bottom.
$$P = 1{,}000 \times 9.81 \times 3 = 29{,}430 \text{ Pa} = 29.4 \text{ kPa}$$
A retaining wall holds back water 5 m deep. Width = 1 m (per unit length). Find the overturning moment about the base.
$F = \frac{1}{2}(1{,}000)(9.81)(25)(1) = 122.6$ kN. Acts at $\frac{5}{3} = 1.667$ m above the base.
$$M = 122.6 \times 1.667 = 204.4 \text{ kN·m per meter of wall}$$
Practice Problems
Show Answer Key
1. $P = 1{,}000 \times 9.81 \times 12 = 117.7$ kPa
2. $F = \frac{1}{2}(1{,}000)(9.81)(36)(4) = 706$ kN
3. $h = P/(\rho g) = 200{,}000/(1{,}025 \times 9.81) = 19.9$ m
4. Average width = 7.5 m; $F = \frac{1}{2}(1{,}000)(9.81)(100)(7.5) = 3{,}679$ kN (using integration for triangular shape)
5. $y_{cp} = \frac{2h}{3}$ from the surface
6. $F = \frac{1}{2}(1{,}000)(9.81)(4)(1) = 19.62$ kN/m
7. $P_{abs} = 101.3 + 49.05 = 150.4$ kPa
8. $F = \rho g \bar{h} \cdot A_{\text{projected}} = 1{,}000(9.81)(2)(3 \times 4) = 235.4$ kN
9. $F \propto h^2$: at 6 m force is $4\times$ the force at 3 m
10. $F_1 = \frac{1}{2}\rho g (49)(8)$, $F_2 = \frac{1}{2}\rho g (9)(8)$; net $= \frac{1}{2}(9{,}810)(40)(8) = 1{,}570$ kN
11. Vertical depth to centroid $= \frac{5\sin 60°}{2} = 2.165$ m; $F = \rho g (2.165)(5 \times 2) = 212.6$ kN
12. Overturning moment $= 500 \times 4 = 2{,}000$ kN·m; Stabilising $= 800 \times 3 = 2{,}400$ kN·m; Factor of safety $= 1.2$ — marginally stable