Centre of pressure below centroid proof

Total pressure is the resultant force exerted by a static fluid on a surface, acting normal to it. Centre of pressure is the point on the surface where this resultant force acts.

Derivation for a vertical plane surface

Let a plane surface of area $A$ be immersed vertically in a liquid of density $\rho$. Let $\bar h$ = depth of centroid G and $h^*$ = depth of centre of pressure P below the free surface.

 free surface ===========================
        |                |
        | h-bar          | h*
  +-----G------+         |
  | ---strip-- | dh at h |
  |        P   |  <------'
  +------------+

Total pressure: for a horizontal strip of area $dA$ at depth $h$, $dF = \rho g h\, dA$.

$$F = \rho g \int h\, dA = \rho g A \bar h$$

Centre of pressure: moment of all strip forces about the free surface equals the moment of $F$:

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

By the parallel axis theorem, $I_o = I_G + A\bar h^2$, where $I_G$ is the second moment of area about the horizontal axis through G. Therefore

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

Proof that P is below G

• $I_G$ is a second moment of area, always positive.
• $A$ and $\bar h$ are positive for a submerged surface.

So $\dfrac{I_G}{A\bar h} > 0$ and hence $h^* > \bar h$: the centre of pressure always lies below the centroid.

Physical reason: pressure increases linearly with depth, so the lower part of the surface carries more force than the upper part, pulling the resultant below the centroid.

Note: as the surface goes deeper ($\bar h$ large), $I_G/(A\bar h) \to 0$ and P approaches G. Example: rectangle $b \times d$ with top edge at the surface: $h^* = d/2 + \frac{bd^3/12}{bd \cdot d/2} = \frac{2d}{3}$, which is below $\bar h = d/2$.