Hydrostatic forces on submerged surfaces

The clearwell at Northside has a steel sluice gate in its wall, 1.5 m wide and 2.0 m tall, that isolates the pump suction channel for maintenance. Its top edge sits 3.0 m below the normal water level. Before a contractor replaces the gate's hinge pins, Hana Lindqvist wants to know the force the water exerts on the gate, where that force acts, and how hard the hoist must pull to hold the gate shut. These are the classic questions of hydrostatic forces on submerged surfaces, and they apply equally to dam faces, tank walls, inspection hatches and the curved walls of storage tanks.

Pressure distribution on a plane surface

On a submerged plane surface, the pressure at each point acts perpendicular to the surface and increases linearly with depth. The total force is the sum of the pressure times area over the whole surface. Because the pressure varies, the force does not act at the centroid of the surface but lower down, where the pressure is greater. Working in gauge pressure is natural here, because atmospheric pressure acts on the dry side of the gate as well and cancels.

Let the plate be inclined at angle \(\theta\) to the free surface, and measure distance \(y\) along the plate from the line where its plane meets the surface. A point at distance \(y\) is at depth \(h = y\sin\theta\), so the pressure there is \(\gamma y \sin\theta\). Integrating over the area gives a force proportional to the first moment of area, which equals the area times the distance to its centroid. The result is simple and worth memorizing: the resultant force equals the pressure at the centroid times the area.

The center of pressure

To find where the resultant acts, take moments about the surface line. The moment of the distributed pressure equals the moment of the resultant, which brings in the second moment of area about the surface line. Using the parallel-axis theorem, \(I_x = I_{xc} + A y_c^2\), gives the location of the center of pressure along the plate.

Here \(I_{xc}\) is the second moment of area about a horizontal axis through the centroid. The center of pressure always lies below the centroid, by a distance \(I_{xc}/(y_c A)\) along the plate. That distance shrinks as the surface is submerged more deeply, because the pressure becomes more uniform across the plate, which is why deep gates have their resultant almost at the centroid. For symmetric shapes, such as rectangles and circles, the center of pressure lies on the vertical centerline. For a rectangle of width \(b\) and height \(h\), \(I_{xc} = bh^3/12\); for a circle of diameter \(D\), \(I_{xc} = \pi D^4/64\).

The program below repeats this calculation, checks it by summing the pressure over two thousand thin horizontal strips, and applies the same formulas to a circular access hatch on an inclined wall. The strip summation is a useful habit: it shows that the formulas are nothing more than integrals done in advance, and it works for shapes that have no tabulated properties.

# Hydrostatic force on the clearwell sluice gate (vertical rectangle)
# and on an inclined circular access hatch
import math
rho, g = 998.0, 9.81
gam = rho * g

def plane_force(hc, A, Ixc, sin_theta):
    """Resultant force, and distance along the plate from the centroid down to the point where it acts."""
    F = gam * hc * A
    yc = hc / sin_theta
    ycp = yc + Ixc / (yc * A)
    return F, ycp - yc

# Vertical rectangular gate 1.5 m wide x 2.0 m tall, top edge 3.0 m below the surface
b, h, top = 1.5, 2.0, 3.0
A = b * h; Ixc = b * h**3 / 12; hc = top + h / 2
F, e = plane_force(hc, A, Ixc, 1.0)
print(f"Sluice gate: F = {F/1000:.1f} kN acting {e*100:.1f} cm below the centroid")
print(f"  depth of the point of action = {hc + e:.3f} m")

# Hinged at the top edge: force needed at the bottom edge to hold it shut
M = F * (h / 2 + e)
print(f"  holding force at the bottom edge = {M / h / 1000:.1f} kN")

# Circular hatch, diameter 0.9 m, on a wall inclined 60 degrees, centroid 4.2 m deep
D, theta, hcc = 0.9, math.radians(60), 4.2
A = math.pi * D**2 / 4; Ixc = math.pi * D**4 / 64
F, e = plane_force(hcc, A, Ixc, math.sin(theta))
print(f"Inclined hatch: F = {F/1000:.1f} kN, acting {e*1000:.1f} mm below the centroid along the wall")

# Pressure prism check for the vertical gate: integrate p dA numerically
n = 2000; Fn = 0.0; Mn = 0.0
for i in range(n):
    y = top + (i + 0.5) * h / n
    dF = gam * y * b * h / n
    Fn += dF; Mn += dF * y
print(f"Numerical check: F = {Fn/1000:.1f} kN, depth of action = {Mn/Fn:.3f} m")

Sluice gate: F = 117.5 kN acting 8.3 cm below the centroid
  depth of the point of action = 4.083 m
  holding force at the bottom edge = 63.6 kN
Inclined hatch: F = 26.2 kN, acting 10.4 mm below the centroid along the wall
Numerical check: F = 117.5 kN, depth of action = 4.083 m

The pressure prism

For rectangular surfaces with one edge horizontal, a geometric picture is often quicker. Draw the pressure distribution across the plate as a solid: its base is the plate and its height at each point is the local pressure. The volume of this pressure prism equals the resultant force, and the force passes through the prism's centroid. For a gate reaching up to the free surface, the prism is a triangular wedge and the force acts two thirds of the way down. For a submerged gate like the sluice gate, the prism is a trapezoid, which can be split into a rectangle (uniform pressure from the top-edge depth) and a triangle (the extra pressure below), each with an obvious centroid.

The prism method also handles a surface with liquid on both sides. If water stands 5 m deep on one side of a gate and 2 m on the other, the net pressure is uniform below the lower surface, and the net force is the difference between the two prisms. This is how canal lock gates and the divided cells of a filter gallery are checked.

Forces on curved surfaces

Tank walls, pipe walls and dome roofs are curved, and the pressure on them changes direction as well as magnitude, so direct integration becomes awkward. The trick is to isolate a block of fluid bounded by the curved surface and by vertical and horizontal planes, and to apply equilibrium to that block. Two results follow.

The horizontal component of the force on a curved surface equals the force on the vertical projection of that surface, and acts through the center of pressure of the projection. The vertical component equals the weight of the fluid directly above the curved surface, up to the free surface, real or imaginary, and acts through the centroid of that volume. If the fluid is below the surface, as on the underside of a submerged dome, the vertical force acts upward and equals the weight of the fluid that would fill the space above it to the free surface level.

On a circular surface, every pressure force is normal to the surface and therefore passes through the center of the circle, so the resultant does too. Radial gates on spillways exploit this: the water force passes through the trunnion pin, so it exerts no moment that the hoist must overcome, and the hoist lifts only the gate's weight and friction.

Design notes from the station

Hydrostatic loads are the most predictable loads in structural design, because water density is known precisely and depth is controlled. Codes still apply load factors to them, mainly to cover overfilling. Hana checks three conditions for every tank: full to the overflow, empty with groundwater pressing on the outside, and the worst case during construction or testing. An empty clearwell in spring, with the water table high, can float out of the ground if its weight and anchors do not exceed the uplift, which is a buoyancy problem of the kind we study next.

Gate seals and hatch gaskets also depend on these forces. A seal that is pushed onto its seat by water pressure works better as the depth increases, so engineers try to arrange gates to close with the water pressure, not against it. When a gate must open against full head, as the sluice gate does, its hoist must overcome both the water load's moment and the friction in its guides, which can add 20 to 40 percent to the force calculated from hydrostatics alone.