Pressure on circular plate (2.5 m)

The plate is not vertical because its depth range (3 − 1 = 2 m) is less than its diameter, so it is inclined. Use the formulas for an inclined plane surface.

Given: $d = 2.5$ m, $S = 0.9$ so $\rho = 900$ kg/m³, depths of top and bottom points $= 1$ m and $3$ m.

  free surface =====================
         :  1 m
         :     * top point
         :    / \
         :   /   \   plate, d = 2.5 m
         :  /  G  \   inclined at theta
         :   \   /
         :    \ /
         :     * bottom point (3 m)

Inclination and centroid depth

$$\begin{aligned} \sin\theta &= \frac{3 - 1}{2.5} = 0.8000 \\ \bar h &= \frac{1 + 3}{2} = 2\ \text{m} \end{aligned}$$

Area and second moment of area

$$\begin{aligned} A &= \frac{\pi}{4}(2.5)^2 = 4.9087\ \text{m}^2 \\ I_G &= \frac{\pi d^4}{64} = \frac{\pi (2.5)^4}{64} = 1.9175\ \text{m}^4 \end{aligned}$$

Total pressure

$$\begin{aligned} F &= \rho g A \bar h = 900 \times 9.81 \times 4.9087 \times 2 \\ &= 86\,679\ \text{N} = 86.68\ \text{kN} \end{aligned}$$

Centre of pressure

$$\begin{aligned} h^* &= \bar h + \frac{I_G \sin^2\theta}{A \bar h} \\ &= 2 + \frac{1.9175 \times 0.6400}{4.9087 \times 2} \\ &= 2 + 0.1250 = 2.125\ \text{m} \end{aligned}$$

Answer: total pressure $F \approx 86.68$ kN; centre of pressure at a depth of $2.125$ m below the free surface (on the plate's vertical diameter, below the centroid).