Solution :
Given :
External diameter of the hemispherical shell, D = 500 mm
Thickness, t = 20 mm
Internal diameter, d = D - 2t
= 500 - 2(20)
= 460 mm
So, internal radius, r = 230 mm
= 0.23 m
Density of molten metal, ρ = [tex]$7.2 \ g/cm^3$[/tex]
= [tex]$7200 \ kg/m^3$[/tex]
The height of pouring cavity above parting surface is h = 300 mm
= 0.3 m
So, the metallostatic thrust on the upper mold at the end of casting is :
[tex]$F=\rho g A h$[/tex]
Area, A [tex]$=2 \pi r^2$[/tex]
[tex]$=2 \pi (0.23)^2$[/tex]
[tex]$=0.3324 \ m^2$[/tex]
[tex]$F=\rho g A h$[/tex]
[tex]$=7200 \times 9.81 \times 0.3324 \times 0.3$[/tex]
= 7043.42 N