Vertical force on cylinder with hemispherical ends

The figure values are only partly legible, so the reading used is stated. Pressures are gauge, $g=9.81$.

Assumed reading: above the plate AB the fluids are, from the top, air at $p_1=500$ kPa, oil ($S=0.8$) 3 m thick, then water 2 m thick down to AB. Below AB is water with $p_2=-360$ kPa at the plate level. The cylinder ($D=0.3$ m) extends 0.2 m above and 0.2 m below the plate and ends in hemispheres of radius $r=0.15$ m.

      air  p1 = 500 kPa
      oil  S=0.8   3 m
      water        2 m
   ---+===[ cap ]===+--- plate AB
         | cyl 0.2m|
         [ lower  ]
      water, p2 = -360 kPa

Pressures at the bases of the two hemispheres (flat plane at the end of each cylinder portion):

$$p_{plate,top}=500+0.8(9.81)(3)+9.81(2)=543.16\ \text{kPa}$$ $$p_u=543.16-9.81(0.2)=541.20\ \text{kPa},\qquad p_l=-360+9.81(0.2)=-358.04\ \text{kPa}$$

Vertical forces (the cylinder side has no vertical component). Projected area $A=\pi(0.15)^{2}=0.07069\ \text{m}^2$, hemisphere volume $V_h=\tfrac23\pi(0.15)^{3}=0.007069\ \text{m}^3$.

• Downward force on the upper hemisphere: $F_u=p_uA-\gamma_wV_h$
• Upward force on the lower hemisphere: the pressure there is lower, giving a net downward effect

$$F_{net}=(p_u-p_l)A-(\gamma_w+\gamma_w)V_h$$ $$F_{net}=(541.20+358.04)(0.07069)-2(9.81)(0.007069)$$ $$F_{net}=63.57-0.14=63.4\ \text{kN}$$

Answer: the fluid exerts a net vertical force of about 63.4 kN downward on the cylinder, and this is the load carried by the weld to the plate AB. The result is dominated by $(p_u-p_l)A$; use the same method with the exact figure heights.