How Hard Water Pushes on a Dam

Water pressure grows with depth, so the bottom of a dam is pushed far harder than the top. To get the total force, slice the dam's face into thin horizontal strips, each at its own depth and pressure, and add up pressure times area.

How it works

Hydrostatic pressure and force

In one sentence

The force of a fluid of density $\rho$ on a vertical plate is $F = \int_a^b \rho g\,d(y)\,w(y)\,dy$ , where $d(y)$ is the depth and $w(y)$ the plate's width at height $y$ .

General formula

{\htmlData{sym=0}{F}} = \int_0^{\htmlData{sym=3}{{H}}} {\htmlData{sym=1}{\rho}} {\htmlData{sym=2}{g}}\,({\htmlData{sym=3}{H}} - {\htmlData{sym=4}{y}})\,{\htmlData{sym=5}{w(y)}}\,d{\htmlData{sym=4}{y}}, \qquad {\htmlData{sym=5}{w(y)}} = {\htmlData{sym=6}{b}} + 2{\htmlData{sym=7}{k}}{\htmlData{sym=4}{y}}

Pressure is the same everywhere along a horizontal strip, so each strip is easy: pressure times area. The integral adds strips at every depth.

What the symbols mean

$F$
Total force
Newtons of push on the whole face.
$\rho$
Water density
1000 kg per cubic meter.
$g$
Gravity
9.81 m/s².
$H$
Water depth
How deep the reservoir is at the dam.
$y$
Strip height
How far above the bottom a strip of the face sits.
$w(y)$
Face width
How wide the dam's face is at height y.
$b$
Bottom width
Width of the face at the riverbed.
$k$
Side slope
How much each side widens per meter of height.

At depth $d$ below the surface, water pressure is $\rho g d$ , the same in every direction.
A thin strip of the dam's face at height $y$ has area $w(y)\,dy$ and sits at depth $H - y$ .
Its force is pressure times area, $\rho g(H - y)\,w(y)\,dy$ . Adding the strips gives the integral.
Force grows with the square of the water depth (faster for a widening face): a flood that raises a rectangular reservoir 10% raises the force 21%.

Learn the math

Hydrostatic Pressure and ForceCalc II · Lesson 14, Applications of Integrals