Hydrostatic pressure is the simplest pressure problem in civil and mechanical design — and one of the easiest to get wrong because the units jump around. Engineers routinely flip pcf and psi, forget to include the centroid offset when computing overturning moment, or apply the wrong load factor. This article walks the math, calls out the traps, and runs two full examples.
The governing equation
For an incompressible fluid at rest, pressure varies linearly with depth:
p(z) = γ · z
where γ is the unit weight (force per unit volume) and z is the depth below the free surface. Water at room temperature has γ = 62.4 lb/ft³ (9.81 kN/m³). Saturated soil might be 110–130 pcf. Crude oil ~55 pcf. Density and gravity matter — on Mars your retaining wall would need to be very different.
Force on a vertical wall
For a wall of width b retaining fluid to depth H, the pressure distribution is triangular. The resultant force per unit length is the area of the pressure triangle, and the full force is that times b:
F = (1/2) · γ · H² · b
The resultant acts at the centroid of the triangle, which sits H/3 above the base:
̅z = H/3 (measured up from base)
Overturning moment about the toe is therefore:
MOT = F · (H/3) = (1/6) · γ · H³ · b
Example 1 — Concrete retaining wall, 12 ft tall
- H = 12 ft of water (worst-case saturated case behind wall)
- γ = 62.4 pcf
- b = 1 ft (per foot of wall length)
Force per linear foot of wall:
F = 0.5(62.4)(12)²(1) = 0.5(62.4)(144) = 4,493 lb/ft
Overturning moment per linear foot:
MOT = 4,493 · (12/3) = 4,493 · 4 = 17,972 ft-lb/ft
That overturning moment is what your stem reinforcement and footing-toe pressure check must resist. For a 12 ft saturated condition, the heel needs to carry significant uplift and the soil bearing under the toe needs to stay below allowable.
Example 2 — Cylindrical steel tank, 6 m diameter, 5 m tall
- D = 6 m, H = 5 m
- Fluid: water, γ = 9.81 kN/m³
Maximum hydrostatic pressure (at the bottom):
pmax = 9.81(5) = 49.1 kPa
This pressure generates hoop tension in the tank wall. For a thin-walled cylinder:
σhoop = p · r / t
For a 6 mm wall: σhoop = 49.1(3,000)/6 = 24,550 kPa = 24.6 MPa at the base. Compare to allowable stress of the steel grade. A36 yields at 248 MPa, allowable around 150 MPa under tension — plenty of margin here.
Common mistakes
- Forgetting the ½. The force is the area of a triangle, not a rectangle. pmax·H gives you 2x the actual force.
- Resultant at the base or mid-height. The triangle’s centroid is at H/3 from the base, not H/2.
- Mixing fluid heights with soil heights. If you have a partial saturation condition, split the diagram into a soil-only Rankine triangle plus a water-only hydrostatic triangle. Add the forces, do the moments about a common point.
- Surcharge on top. A uniform surcharge q on the surface adds a rectangular pressure q·Ka (for soil) or q (for fluid) to the wall, which is a separate load case.
- Using nominal pressures with strength design. ASCE 7 load factor for fluid load F is 1.4 (LRFD), not 1.6. It’s a separate load category — check Section 2.3.
Buoyancy and uplift
Don’t forget the vertical: a submerged structure with a hollow chamber will see uplift equal to the weight of displaced fluid. Empty tanks at high groundwater is a classic failure mode. Buoyant force = γfluid · Vdisplaced. Tie-downs sized for this case have to use a safety factor of 1.5–2.0 against uplift.
Designing the wall itself
Once you have the lateral force and moment, you’re into reinforced concrete or steel plate design. For RC stems, our RC column calculator handles flexure + axial interaction. For steel tank ring stiffeners or framing, the beam calculator handles ring bending and the column buckling calculator sizes vertical compression members under fluid-induced loads.
Closing
Hydrostatic loads collapse to one equation: p = γz. Force is ½γH² per unit width, acting at H/3 from the base. The errors are always in the unit conversions and in forgetting that a triangle isn’t a rectangle. Slow down on those two things and you’ll never miss.