Question 8 of 8: Hydrostatic Casting Forces on a Spherical Die-Cast Mold

Question 8: Hydrostatic Casting Forces on a Spherical Die-Cast Mold (20 marks)

Question text not reproduced: the examination questions are © Engineers and Geoscientists BC. Open the official past paper (linked at the top of this page) to read the question, then follow the worked solution below.

Given. Hollow spherical mold cavity, radius $R$, diameter $D=2R$, filled with molten zinc alloy of density $\rho$; $z=0$ at the top (fill point), $z=D$ at the bottom; gauge pressure referenced to zero at the top vent.

Find. $F_{top}$ and $F_{bottom}$ (vertical forces on the top and bottom hemispherical wall halves), each expressed as a multiple of the total metal weight $W$.

Approach. Parametrize the sphere by polar angle $\theta$ from the top ($z=R(1-\cos\theta)$), write the local hydrostatic gauge pressure $P(\theta)=\rho g z(\theta)$, project it onto the vertical direction, and integrate over each hemisphere.

1. Geometry and pressure. At polar angle $\theta$ (0 at top, $\pi$ at bottom): $z(\theta)=R(1-\cos\theta)$, surface element $dA=2\pi R^2\sin\theta\,d\theta$, outward-normal vertical component $n_z=-\cos\theta$ (positive $z$ downward), and gauge pressure $P(\theta)=\rho g R(1-\cos\theta)$.
2. Vertical force element. The liquid pushes outward on the wall with pressure $P(\theta)$; its vertical (downward-positive) component per unit area is $P(\theta)\,n_z$, so $$dF_z=P(\theta)(-\cos\theta)(2\pi R^2\sin\theta)\,d\theta=-2\pi\rho gR^3(1-\cos\theta)\cos\theta\sin\theta\,d\theta$$
3. Top half, $\theta:0\to\pi/2$ ($z:0\to R$). With $u=\cos\theta$: $\displaystyle\int_0^{\pi/2}(1-\cos\theta)\cos\theta\sin\theta\,d\theta=\int_0^1(1-u)u\,du=\tfrac16$, so $$F_{top}=-2\pi\rho gR^3\left(\tfrac16\right)=-\dfrac{\pi\rho gR^3}{3}$$ (negative $\Rightarrow$ net force is upward: the dome-shaped top half is pushed up and outward by the melt.)
4. Bottom half, $\theta:\pi/2\to\pi$ ($z:R\to D$). Same substitution gives $\displaystyle\int_{\pi/2}^{\pi}(1-\cos\theta)\cos\theta\sin\theta\,d\theta=-\tfrac56$, so $$F_{bottom}=-2\pi\rho gR^3\left(-\tfrac56\right)=\dfrac{5\pi\rho gR^3}{3}$$ (positive $\Rightarrow$ net force is downward, onto the bottom cup.)
5. Express as a multiple of the total weight. $W=\rho g V_{sphere}=\rho g\left(\tfrac43\pi R^3\right)$, so $$\dfrac{F_{top}}{W}=\dfrac{-\pi\rho gR^3/3}{(4/3)\pi\rho gR^3}=-\dfrac14,\qquad \dfrac{F_{bottom}}{W}=\dfrac{5\pi\rho gR^3/3}{(4/3)\pi\rho gR^3}=\dfrac54$$ $$\boxed{F_{top}=\dfrac{W}{4}\ (\text{upward}),\qquad F_{bottom}=\dfrac{5W}{4}\ (\text{downward})}$$ Check: $F_{bottom}-F_{top}=5W/4-W/4=W$ — the net vertical force balances the total weight, as it must for a static fluid. ✓

The top hemisphere is pushed outward and upward with only $W/4$ of net force, and the local gauge pressure falls to zero right at the fill hole — so near the top of the casting, hydrostatic pressure alone gives very little contact pressure holding the metal against the mold wall, risking poor surface conformity, porosity, and a weak as-cast surface finish exactly where gravity is least helpful. Superimposing an external pressure (as in high-pressure die casting) adds a uniform pressure term on top of the hydrostatic profile everywhere in the melt, guaranteeing a healthy minimum contact pressure even at the top of the cavity. This is appropriate and is precisely what external-pressure die casting is used for: it improves mold–metal conformity, reduces shrinkage porosity and gas entrapment, and improves as-cast dimensional fidelity and surface finish — at the cost of higher mold-clamping-force requirements, since the bottom half's force (already $5W/4$ under gravity alone) grows further with the applied pressure.