Question 1 of 4: Casting, Forming and Deformation Fundamentals (Ten Short Sub-Parts)

Question 1: Casting, Forming and Deformation Fundamentals (Ten Short Sub-Parts) (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. Ten independent 2-mark sub-parts spanning casting, phase diagrams and bulk/sheet deformation. Numeric data: (b) Pb-Sn alloy, $C_0=20$ wt%Sn at 250°C, Figure 1; (d) cylinder $h/a=3$ solidifies in 5 min in sand; (f) 1008 steel cold-working exponent $n=0.25$ (Table 6, Schey); (i) AA1100 billet, $D_0=100$ mm → $D_f=25$ mm, cold extrusion.

Find. A short, complete answer to each of (a)–(j).

[Figure not reproduced: Figure 1 (redrawn) — Pb–Sn phase diagram with the 250°C tie line read from the exam's own printed figure: $C_\alpha=14.6$, $C_0=20$, $C_L=33.6$ wt%Sn. See the official exam paper.]

1. (a) Turbulence in gravity sand casting. A highly turbulent pouring stream entrains air and mold-cavity gas into the melt, tears and folds the surface oxide film into the bulk liquid (forming oxide/dross inclusions), and erodes the sand mold wall (sand inclusions). Both effects nucleate gas and non-metallic-inclusion porosity in the solidified casting and roughen the surface, so gating systems are designed to keep the Reynolds number low (laminar, non-turbulent filling) — e.g. bottom-gating, tapered sprues, and filters.
2. (b) Pb–Sn lever rule at 250°C, $C_0=20$ wt%Sn. Reading Figure 1 at 250°C places $C_0$ inside the two-phase $\alpha+L$ field, bounded by the solidus ($C_\alpha=14.6$ wt%Sn) and liquidus ($C_L=33.6$ wt%Sn). $$W_\alpha=\frac{C_L-C_0}{C_L-C_\alpha}=\frac{33.6-20}{33.6-14.6}=\frac{13.6}{19.0}=0.716\qquad W_L=\frac{C_0-C_\alpha}{C_L-C_\alpha}=\frac{20-14.6}{19.0}=0.284$$ $$\boxed{W_\alpha \approx 71.6\%\ \text{solid }\alpha,\qquad W_L\approx 28.4\%\ \text{liquid}}$$
3. (c) Role of a riser. A riser is a reservoir of molten metal, open to the casting, that feeds liquid metal into the casting cavity as the casting shrinks during cooling and solidification, so the resulting voids concentrate in the (sacrificial) riser rather than as internal shrinkage porosity in the part. Two design considerations: (i) the riser must remain liquid and feed after the section it serves has solidified — by Chvorinov's rule this means the riser's own $V/A$ modulus must exceed that of the casting section it feeds; (ii) the riser must hold enough liquid volume to supply the full volumetric solidification shrinkage of that section (a few percent of its volume for most alloys), while being placed on the last-to-freeze, thickest region and kept as small as practical to limit yield loss (metal that is ultimately cut off and remelted).
4. (d) Chvorinov's rule, $h/a=3\to h/a=1$, $n=2$. Including top and bottom, total surface area of a cylinder of radius $a$ and height $h$ is $A=2\pi a^2+2\pi ah$, and $V=\pi a^2h$. $$\frac{V}{A}=\frac{a^2h}{2a^2+2ah}=\frac{ah}{2(a+h)}$$ For $h/a=3$ (let $a=1$): $V/A=\dfrac{1(3)}{2(1+3)}=\dfrac{3}{8}=0.375$. For $h/a=1$: $V/A=\dfrac{1(1)}{2(1+1)}=\dfrac{1}{4}=0.25$. Since $t=C(V/A)^2$, the constant $C$ cancels between the two cylinders: $$t_{h/a=1}=t_{h/a=3}\left(\frac{(V/A)_{h/a=1}}{(V/A)_{h/a=3}}\right)^2=5\ \text{min}\times\left(\frac{0.25}{0.375}\right)^2=5(0.667)^2$$ $$\boxed{t_{h/a=1}=2.22\ \text{min}\approx 133\ \text{s}}$$ The shorter, squatter cylinder has a smaller $V/A$ ratio (more surface per unit volume once the extra end area is counted), so it loses heat faster and solidifies sooner.
5. (e) Expendable vs. permanent patterns. An expendable pattern is consumed (melted out or burned out) in producing a single casting and can never be reused — e.g. the wax pattern in investment (lost-wax) casting, or the polystyrene foam pattern in lost-foam casting, each destroyed as the metal is poured. A permanent pattern is a durable master shape (wood, metal or plastic) that is reused hundreds or thousands of times to form a new, disposable sand mold for every casting — e.g. an aluminum or hardwood pattern used repeatedly in conventional gravity sand casting. (Note this is distinct from a permanent mold, such as a die-casting die, which is itself the reusable cavity rather than the master used to form one.)
6. (f) Necking strain, 1008 steel. For a power-law hardening material $\sigma=K\varepsilon^n$, the Considère criterion places the onset of necking at true strain $\varepsilon_{true}=n$. Table 6 (cold-worked 1008) gives $n=0.25$. $$\varepsilon_{true}=n=0.25\qquad e_{eng}=e^{\varepsilon_{true}}-1=e^{0.25}-1$$ $$\boxed{e_{eng}\approx 0.284\ (28.4\%)}$$
7. (g) Hardness and springback. Yes. Springback is governed by the ratio of elastic recovery to the imposed elastic-plastic bend, $R_i/R_f=4(R_iY/Et)^3-3(R_iY/Et)+1$ (formula sheet), which increases monotonically with the yield strength $Y$ for a fixed modulus $E$, thickness $t$ and radius $R_i$. A harder sheet has a higher yield strength (hardness and $\sigma_y$ scale together for a given alloy/temper), so it stores more elastic strain energy at the same bend geometry and springs back more on unloading than a softer, lower-hardness sheet of the same material class.
8. (h) Hydrogen porosity in aluminum. Liquid aluminum dissolves atomic hydrogen readily (roughly 20× more soluble in the liquid than in the solid just below the melting point), sourced from moisture in the furnace atmosphere, damp charge/tooling, hydrated surface oxide, or (in welding) moisture in the shielding gas, flux or on the joint/electrode surface. On solidification the solubility drops sharply; hydrogen that cannot diffuse out in time is rejected ahead of the solidification front and nucleates as trapped gas bubbles (porosity). Prevention: keep melting/welding materials and atmosphere dry, degas the melt (inert-gas purging, rotary or vacuum degassing) before pouring, minimize melt turbulence/superheat, and in welding use clean, dry filler/base-metal surfaces with well-shielded, low-moisture gas coverage and controlled cooling rate.
9. (i) AA1100 cold extrusion strain, $D_0=100$ mm $\to D_f=25$ mm. Cold extrusion is treated as an area-reduction, constant-volume process; the (true) strain is $$\varepsilon=\ln\!\left(\frac{A_0}{A_f}\right)=\ln\!\left(\frac{D_0}{D_f}\right)^2=2\ln\!\left(\frac{100}{25}\right)=2\ln(4)$$ $$\boxed{\varepsilon \approx 2.77}$$
10. (j) Cold rolling advantages over warm/hot rolling. (1) Much tighter dimensional tolerance and a smooth, oxide-scale-free surface finish, since cold rolling runs below the recrystallization temperature and never forms the hot-rolling scale that must later be pickled off. (2) The strain hardening accumulated during cold rolling raises the yield and tensile strength (and hardness) of the finished sheet well above the hot-rolled/annealed condition, so a high-strength temper can be produced directly off the mill without a separate strengthening heat treatment.