Total pressure and centre of pressure derivation

Total pressure is the resultant force exerted by a static fluid on a surface in contact with it; it acts normal to the surface. Centre of pressure is the point on the surface through which this resultant force acts.

Setup

Consider a plane surface of any shape, area $A$, immersed vertically in a liquid of specific weight $\rho g$. Let $\bar h$ be the depth of its centroid G and $h^*$ the depth of the centre of pressure P, both measured from the free surface.

  free surface
 ---------------------------------
      |          |
      | h-bar    | h*
      v          |
  +---G------+   |
  |  [strip] |<- dh at depth h
  |       P  |   v
  +----------+

Total pressure

Take a thin horizontal strip of width $b$ and thickness $dh$ at depth $h$. Pressure on it is $p = \rho g h$, so

$$dF = \rho g h\, (b\, dh) = \rho g h\, dA$$ $$F = \int \rho g h\, dA = \rho g \int h\, dA = \rho g A \bar h$$

since $\int h\, dA = A\bar h$ (first moment of area about the free surface).

Centre of pressure

Take moments about the free surface line. The moment of $dF$ is $dF\cdot h = \rho g h^2 dA$. Sum of moments equals the moment of the resultant:

$$F\, h^* = \rho g \int h^2 dA = \rho g\, I_o$$

where $I_o$ is the second moment of area about the free surface. By the parallel axis theorem $I_o = I_G + A\bar h^2$. So

$$\begin{aligned} h^* &= \frac{\rho g (I_G + A\bar h^2)}{\rho g A \bar h} \\ &= \bar h + \frac{I_G}{A\bar h} \end{aligned}$$

Results

• $F = \rho g A \bar h$ (independent of the shape; depends only on area and centroid depth).
• $h^* = \bar h + \dfrac{I_G}{A\bar h}$. Because $I_G/(A\bar h)$ is always positive, the centre of pressure lies below the centroid; the gap reduces as depth increases.
• Example: rectangle $b \times d$ with top at the surface: $\bar h = d/2$, $I_G = bd^3/12$, so $h^* = 2d/3$.