Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)
Notes on this paper
National Exam 04-BS-11, Properties of Materials — May 2019 sitting (the cover page and every page footer read “04-BS-11, May2019”). 3 hours, closed-book examination (Casio/Sharp calculator only). Notes on the paper state that candidates are to
attempt five, and only five, questions, with only the first five appearing in the answer book marked and
all questions of equal value. All seven questions are solved below for completeness.
Reference texts: Callister & Rethwisch, Materials Science and Engineering:
An Introduction, 9th ed. (crystal structure and Miller indices; tensile testing and true strain;
hardness testing; solid solutions and grain size; phase diagrams and the lever rule; dislocations and
cold work; polymer molecular weight and viscoelastic behaviour; TTT diagrams and heat treatment;
fracture/fatigue).
Page-1 data used below: atomic masses (g/mol) H 1.01, C 12.01,
Mo 95.94; $N_A=0.602\times10^{24}$ mol$^{-1}$; cold work $CW=(A_0-A_f)/A_0$; grain size
$N=2^{n-1}$. Fig 1 (Al–Si diagram), Fig 2 (cold work vs. properties, iron and copper) and
Fig 3 (isothermal diagram, 0.8% C steel) are printed in the paper; the values used below were read off them.
Given. Fig 1 (Al–Si): pure Al melts at $660^{\circ}$C, pure Si at
$1430^{\circ}$C; eutectic at $C_E=12.6$ wt% Si, $T_E=577^{\circ}$C; terminal $\alpha$ solubility
$1.65$ wt% Si and $\beta$ composition $\approx99$ wt% Si at $T_E$. Alloy
$C_0=20$ wt% Si. Slow (equilibrium) cooling.
Find. Liquidus (start) and end-of-solidification temperatures; room-temperature
microstructure; fraction of eutectic microconstituent; why the alloy casts better than pure Al.
Approach
Since $C_0=20\%>C_E=12.6\%$ the alloy is hypereutectic: primary $\beta$ (almost pure
Si) forms first, on the silicon-side liquidus. The start temperature is read off that printed liquidus at
$20\%$ Si. The printed curve is strongly convex, so it must be followed, not replaced by a straight
line from the eutectic point to pure Si. Solidification ends at the eutectic isotherm. The eutectic
fraction follows from the lever rule at $T_E^{+}$ between primary $\beta$ and eutectic-composition
liquid.
Confirm the alloy is hypereutectic. $C_0=20\%>C_E=12.6\%$, so the alloy lies on the
Si-rich side of the eutectic: primary $\beta$ (Si) forms first, not primary $\alpha$.
Solidification-start temperature (liquidus). Reading the printed silicon-side
liquidus of Fig 1 at $20\%$ Si (digitised; the curve rises about $14^{\circ}$C per %Si just
above the eutectic):
$$\boxed{T_{\text{start}} \approx 685^{\circ}\text{C}}$$
Primary $\beta$ (Si) crystals begin to form here. (A straight chord from $(12.6\%,\,577^{\circ}\text{C})$
to $(100\%,\,1430^{\circ}\text{C})$ would give only $\approx649^{\circ}$C, which is $\approx36^{\circ}$C too
low.)
Solidification-complete temperature. As primary $\beta$ grows, the remaining liquid
is enriched in Al and slides down the liquidus to the eutectic point. The last liquid then freezes
isothermally by the eutectic reaction $L\rightarrow\alpha+\beta$:
$$\boxed{T_{\text{complete}} = T_E = 577^{\circ}\text{C}}$$
Eutectic fraction, by the lever rule at $T_E^{+}$. Just above $T_E$ the constituents
are primary $\beta$ ($C_\beta\approx99\%$ Si) and liquid of eutectic composition
($C_E=12.6\%$). All of that liquid becomes the eutectic microconstituent:
$$W_{\text{eutectic}} = \frac{C_\beta - C_0}{C_\beta - C_E} = \frac{99-20}{99-12.6} = \frac{79}{86.4}$$
$$\boxed{W_{\text{eutectic}} \approx 0.914\ (91\%)}$$
$$W_{\text{primary }\beta} = \frac{C_0-C_E}{C_\beta-C_E} = \frac{7.4}{86.4} \approx 0.086\ (9\%)$$
Room-temperature microstructure. The solvus lines on both sides are close to the pure
components ($\alpha$ holds only $1.65\%$ Si even at $577^{\circ}$C, far less at room temperature;
$\beta$ is essentially pure Si). On slow cooling the structure formed at $577^{\circ}$C therefore persists:
coarse, blocky/polyhedral primary $\beta$ (Si) particles ($\approx9\%$ by mass) in a
matrix of $\alpha+\beta$ eutectic ($\approx91\%$), i.e. fine Si flakes/needles in
Al-rich $\alpha$. Only a little extra Si is rejected from $\alpha$ during cooling, which does not change
these fractions noticeably.
Why 20% Si is a better casting alloy than pure aluminum.
(1) Fluidity: silicon has a very large latent heat of fusion (about four times that of Al per
kg), so the melt stays liquid longer and fills the thin walls and water passages of an engine block.
Most of the alloy (the 91% eutectic) also freezes at one temperature, $577^{\circ}$C, which feeds well.
(2) Shrinkage and hot tearing: Si expands on freezing, which offsets aluminum’s
$\approx6$–$7\%$ solidification shrinkage. Pure Al shrinks heavily, forms coarse columnar grains and
is prone to hot tearing and shrinkage porosity. (3) Service properties: the hard primary Si
particles give the wear resistance a cylinder bore needs (this is the basis of the hypereutectic A390
family, $\approx17\%$ Si). Si also lowers thermal expansion, raises hardness, stiffness and strength,
and slightly lowers density. Pure Al is soft, wears badly and expands too much. Note that the 20% Si
liquidus ($\approx685^{\circ}$C) is slightly above pure Al’s melting point, so the benefit
is castability and service properties, not a lower pouring temperature.
[Figure not reproduced: Fig. Q4 — Al–Si phase diagram redrawn from the printed Fig 1 (liquidus digitised), with the 20% Si alloy, its liquidus intercept ($\approx685^{\circ}$C) and the $577^{\circ}$C eutectic isotherm marked. See the official exam paper.]
