Force on end of half-filled horizontal tank

Data: circular end, $D=2$ m ($R=1$ m). Upper half oil ($\rho_o=880$, $\gamma_o=8.633$ kN/m³), lower half water ($\gamma_w=9.81$ kN/m³); the top of the tank is at atmospheric pressure. Depths are measured from the top.

   _______ top, p = 0
  /  oil  \
 |---------|  interface at depth 1 m
  \ water /
   -------

For a semicircle of radius $R$: $A=\dfrac{\pi R^{2}}{2}=1.5708\ \text{m}^2$, centroid $\dfrac{4R}{3\pi}=0.4244$ m from the diameter, $I_G=\left(\dfrac{\pi}{8}-\dfrac{8}{9\pi}\right)R^{4}=0.1098\ \text{m}^4$.

Oil part (upper semicircle)

Centroid depth $\bar h_o=1-0.4244=0.5756$ m.

$$F_o=8.633(0.5756)(1.5708)=7.805\ \text{kN}$$ $$h_{cp,o}=0.5756+\frac{0.1098}{1.5708(0.5756)}=0.5756+0.1214=0.697\ \text{m (from top)}$$

Water part (lower semicircle)

Pressure at the interface: $p_i=8.633(1)=8.633$ kPa, equivalent to $0.880$ m of water. Centroid is 0.4244 m below the interface.

$$F_w=(8.633+9.81\times0.4244)(1.5708)=12.797\times1.5708=20.10\ \text{kN}$$ $$h_{cp,w}=1+0.4244+\frac{0.1098}{1.5708(0.880+0.4244)}=1.4244+0.0536=1.478\ \text{m (from top)}$$

Net force and position

$$F=7.805+20.100=27.91\ \text{kN}$$ $$\bar h=\frac{7.805(0.697)+20.100(1.478)}{27.91}=1.260\ \text{m below the top}$$

Answer: net hydrostatic force $\approx27.9$ kN, acting 1.26 m below the top of the tank (0.74 m above the bottom) on the vertical centre line.