Zinc (HCP) is brittle at room temperature while aluminium (FCC) is highly ductile, even though both have an atomic packing factor of 0.74. The reason is that

HCP has fewer independent slip systems than FCC

zinc has a higher density than aluminium

HCP has a lower coordination number than FCC

zinc contains more interstitial impurities

On the iron–carbon diagram, the carbon content that separates steel from cast iron is

2.11 per cent — the maximum solubility of carbon in austenite

0.77 per cent — the eutectoid composition

2.11 per cent — the maximum solubility of carbon in austenite

4.3 per cent — the eutectic composition

0.008 per cent — the maximum solubility of carbon in ferrite

Which of the following are correct comparisons of cold working with hot working?

Cold working raises yield strength and reduces ductility

Cold working gives better dimensional accuracy and surface finish

Hot working permits much larger deformation per pass at lower force

Cold working raises yield strength and reduces ductility

Cold working gives better dimensional accuracy and surface finish

Hot working leaves large residual stresses in the product

Hot working permits much larger deformation per pass at lower force

A welder needs to reduce distortion and narrow the heat-affected zone on a thin steel sheet, keeping the same welding current and voltage. The correct action is to

preheat the plate to a higher temperature

switch to submerged-arc welding

Which heat treatment produces martensite in a plain carbon steel?

Quenching — cooling fast enough to bypass the pearlite transformation

Full annealing — slow cooling in the furnace

Quenching — cooling fast enough to bypass the pearlite transformation

Normalising — cooling in still air

Tempering — reheating below the eutectoid temperature

Engineering Materials, and Casting, Forming and Joining

Section 4 of the Mechanical Engineering paper is the widest of its three discipline sections — eight named sub-items against Section 3's four — and it rewards a different habit from the two before it. There is less to derive and more to connect, and the questions that carry two marks are almost always about a mechanism rather than a formula. Engineering materials starts from the crystal: only three lattices matter, and the two close-packed ones — FCC and HCP — have the same packing factor of 0.7405 while behaving completely differently, because FCC has far more slip systems and so is ductile where HCP is not. That is the shape of most questions in this sub-item: the property follows from the structure, not from the composition alone. The iron–carbon diagram is the single densest source of marks, and three points on it carry most of them — the eutectoid at 0.77 per cent carbon and 727 °C, the eutectic at 4.3 per cent and 1147 °C, and the 2.11 per cent solubility limit that separates steel from cast iron. Casting has one quantitative law, Chvorinov’s t = B(V/A)², and its consequences run through riser design, chills and directional solidification. Forming divides on temperature: cold working raises strength and consumes ductility, hot working does neither because recrystallisation keeps up. And joining is where a process is chosen by its heat input, because everything a weld does wrong — distortion, residual stress, a coarse heat-affected zone — is heat that went somewhere other than the joint.

Structure First: Two Lattices That Pack Identically and Behave Nothing Alike

The three lattices, with packing factors derived rather than quoted
Lattice	Atoms per cell	Coordination	Packing factor	Ductility
BCC — α-iron, chromium, tungsten	2	8	0.6802	Moderate; brittle when cold
FCC — γ-iron, aluminium, copper	4	12	0.7405	High — 12 slip systems
HCP — zinc, magnesium, titanium (α)	6	12	0.7405 — the same	Low — 3 slip systems

🎯 Identical packing, opposite ductility — and why that is the whole lesson

Compute the packing factors rather than looking them up, because doing it once makes the comparison memorable. For BCC, the body diagonal is 4R and equals a√3, so a = 4R/√3; two atoms per cell gives APF = 2(4πR³/3)/a³ = 0.6802. For FCC, the face diagonal is 4R = a√2, so a = 4R/√2; four atoms give 0.7405. HCP works out to exactly the same 0.7405, because both are close-packed structures — they differ only in the stacking sequence, ABCABC against ABAB. And here is the fact worth carrying: the two structures with identical density behave completely differently under load. FCC has twelve close-packed slip systems, so a dislocation always has a favourably oriented plane available and the metal deforms plastically before it fractures — aluminium and copper draw into wire, and austenitic stainless steel stays tough at cryogenic temperatures. HCP has essentially three, all on the basal plane, so a grain oriented unfavourably cannot slip and cracks instead — which is why magnesium and zinc are brittle at room temperature and why titanium has to be worked hot. The general lesson is the one this section keeps repeating: a property follows from the structure and its defects, not from the composition or the density, and any question that asks you to compare two metals is usually asking which structure they have.

The iron–carbon diagram carries more marks per square centimetre than anything else in the section, and three coordinates hold most of them. The eutectoid sits at 0.77 per cent carbon and 727 °C, where austenite decomposes into the ferrite-plus-cementite lamellar mixture called pearlite — a steel at exactly that composition is fully pearlitic, below it is hypoeutectoid with proeutectoid ferrite, and above it hypereutectoid with proeutectoid cementite. The eutectic sits at 4.3 per cent and 1147 °C, giving ledeburite. And 2.11 per cent is the maximum carbon austenite will dissolve, which is the line that defines the boundary between steel and cast iron — not a convention but a solubility limit, and that is why the two are worked by completely different processes. Heat treatment is then a set of moves on that diagram, and the useful way to hold them is by what each does to the cooling rate: annealing cools slowly in the furnace for maximum softness, normalising cools in air for a finer pearlite and better uniformity, quenching cools fast enough to bypass pearlite entirely and trap the carbon in the hard, brittle body-centred-tetragonal phase called martensite, and tempering then reheats that martensite to trade some hardness back for toughness. Martensite is the one product on the list that is not an equilibrium phase and so does not appear on the diagram at all — it appears on a TTT curve, which is the diagram redrawn against time.

Casting: One Law, and What Shape Does to It

Chvorinov’s rule says the solidification time of a casting is t = B(V/A)², where V/A — volume over cooling surface area — is called the modulus and B is a mould constant. The rule is worth taking seriously because everything else in casting practice follows from it. Compare three shapes of the same volume, 1000 cm³. A cube of side 10 cm has A = 600 cm² and modulus 1.667 cm. A cylinder with height equal to diameter has d = 10.84 cm, A = 553.6 cm² and modulus 1.806 cm. A sphere has r = 6.204 cm, A = 483.6 cm² and modulus 2.068 cm. Squaring the ratios, the cylinder takes 1.175 times and the sphere 1.539 times as long as the cube to freeze — the same metal, the same volume, 54 per cent more solidification time from shape alone, because a sphere has the least surface area for its volume of any shape there is.

🧠 Everything a foundry does is Chvorinov’s rule applied backwards

Read the modulus as a design tool and three separate pieces of foundry practice collapse into one idea. Riser design: a riser must stay liquid until the casting has frozen, so it must have a larger modulus than the section it feeds — and since a sphere has the largest modulus per unit volume, that is why risers are made stubby and near-spherical rather than long and thin. A tall thin riser holds the same metal and freezes first, which makes it worse than no riser at all: it becomes a shrinkage cavity of its own instead of feeding one. Chills: a metal insert placed against a heavy section adds cooling area locally, cutting the modulus there and freezing that part early, which is how a thick hub is made to solidify before the thin rim that feeds it. Directional solidification: the whole aim is to arrange moduli so the casting freezes progressively towards the riser, so that whatever shrinks last is drawing on metal that is still liquid. And the square in the rule is what makes all of this so leveraged — halving a section’s modulus cuts its freezing time to a quarter, so small geometric changes produce large process changes. That is also the honest limit of the rule: it assumes a single dominant cooling direction into a semi-infinite mould, so it works well for plates and blocks and poorly for thin fins and long cores, where it under-predicts the time because heat has nowhere convenient to go.

Forming Divides on Temperature; Joining Divides on Heat Input

Every bulk-forming process is the same trade in different geometry, and the first question is always whether it is cold or hot. Cold working, below the recrystallisation temperature, work-hardens the metal: strength and hardness rise, ductility falls, residual stresses are left behind, and the surface finish and dimensional accuracy are good. Hot working, above it, recrystallises as fast as it deforms, so the metal never hardens, forces stay low, huge deformations are possible in one pass — and the finish and accuracy are poor because of oxidation and thermal contraction. Neither is better; they answer different questions. One piece of algebra matters throughout: true strain ε = ln(1 + e) is what you must use for large deformation, because engineering strain stops being additive. An engineering strain of 0.5 is a true strain of 0.405, and an engineering strain of 1.0 — doubling the length — is a true strain of 0.693, not 1. Using engineering strain in a forming-load calculation overstates the deformation by a growing margin.

🎯 In rolling, the friction you can get decides how much you can reduce

Rolling looks like a matter of squeezing hard enough, and it is not. The rolls can only pull the strip in if friction at the entry is enough to overcome the horizontal component of the roll pressure, and working that condition through gives a maximum draft Δh_max = μ²R — the largest thickness reduction the pass can bite. Note what it depends on: the coefficient of friction squared and the roll radius, and not on the mill’s power, the strip’s strength or the reduction you would like. With rolls of R = 250 mm: at μ = 0.1 the maximum draft is 2.5 mm; at μ = 0.2 it is 10 mm; at μ = 0.3 it is 22.5 mm. A three-fold change in friction gives a nine-fold change in what the mill can do in one pass, which is why roll surface condition and lubricant choice are not housekeeping details but the thing that sets the pass schedule — and why a mill that starts slipping is not fixed by turning up the pressure. Two design consequences follow directly. Larger rolls bite deeper, since Δh_max is linear in R, but larger rolls also need more force and more powerful drives, so the choice is a genuine trade rather than a free improvement. And for thin strip you want small rolls to keep the roll force down, but small rolls cannot bite, which is exactly the contradiction that four-high and cluster mills exist to resolve: small work rolls to do the rolling, big backup rolls to stop them bending.

Joining is chosen by heat input, because most of what a weld does wrong is heat that ended up somewhere other than the fusion zone. The arc energy per unit length is H = ηVI/v, so travel speed is the lever a welder actually has — a faster pass at the same current puts less heat into the plate. High heat input means a wide heat-affected zone with coarse grains and lower toughness, more distortion because more metal expanded and then contracted, and more residual stress locked in. Low heat input means a narrow HAZ but a risk of lack of fusion and, in hardenable steels, an untempered martensitic zone that cracks. Read the common processes in that light. Resistance spot welding generates H = I²Rt at the interface itself and dumps almost nothing into the surrounding sheet, which is why car bodies are assembled with thousands of spots. Submerged-arc welding is the opposite extreme, running enormous heat input to lay thick beads fast, which is fine on heavy plate and unusable on thin sheet. Electron-beam and laser welding achieve very deep, very narrow welds precisely by keeping the total energy small and the power density enormous. And brazing and soldering do not melt the parent metal at all — a distinction that matters because it means they cannot produce a HAZ, and the joint strength comes from the filler and the fit rather than from fusion.

Key takeaways

FCC and HCP have the identical packing factor 0.7405 and opposite ductility — twelve slip systems against three. Property follows from structure, not from density or composition.
Three points carry the iron–carbon diagram: eutectoid 0.77 % C at 727 °C, eutectic 4.3 % at 1147 °C, and 2.11 % as austenite’s solubility limit — the line that defines steel against cast iron.
Martensite is not on the equilibrium diagram — it needs a TTT curve, because it is a product of cooling rate rather than of composition and temperature.
Chvorinov: t = B(V/A)². At 1000 cm³, moduli are 1.667 (cube), 1.806 (cylinder) and 2.068 cm (sphere), so the sphere takes 1.539× as long. Risers are stubby because a large modulus is the whole requirement.
Maximum draft in rolling is μ²R, so a three-fold friction change is a nine-fold capability change: 2.5, 10 and 22.5 mm at μ = 0.1, 0.2 and 0.3 on 250 mm rolls. Mill power is not in the formula.
Use true strain for large deformation: ε = ln(1 + e), so e = 0.5 is ε = 0.405 and e = 1.0 is ε = 0.693, not 1.
Welding is chosen by heat input H = ηVI/v. High input widens the HAZ and adds distortion; travel speed is the lever. Spot welding puts I²Rt at the interface only; brazing does not melt the parent metal and so has no HAZ.

Practice questions (8)

Attempt each one before opening the answer. Every explanation names the tempting wrong option as well as the right one, because that is where marks are lost.

Zinc (HCP) is brittle at room temperature while aluminium (FCC) is highly ductile, even though both have an atomic packing factor of 0.74. The reason is that
1. HCP has fewer independent slip systems than FCC
2. zinc has a higher density than aluminium
3. HCP has a lower coordination number than FCC
4. zinc contains more interstitial impurities
Show answer
Answer: A — HCP has fewer independent slip systems than FCC
Fewer slip systems. Plastic deformation happens by dislocations gliding on close-packed planes, and what decides ductility is how many such planes are available in orientations a general stress can activate. FCC has twelve independent close-packed slip systems, so almost any loading finds a favourably oriented plane and the metal yields before it fractures. HCP has essentially three, all on the basal plane, so a grain oriented unfavourably cannot slip at all and cracks instead. Option C is a factual error worth correcting: both FCC and HCP have coordination number 12 — they are both close-packed, differing only in stacking sequence, ABCABC against ABAB, and that difference in stacking is precisely what gives one twelve slip systems and the other three. Option B is true and irrelevant; density says nothing about deformation mechanism. Option D introduces a compositional explanation for a structural fact — pure zinc is brittle and pure aluminium ductile, so impurities are not the cause. The reason this comparison is worth carrying is what it teaches generally: two structures can be geometrically identical in packing and mechanically opposite, so a property question is usually a structure question. It also explains why titanium and magnesium must be formed hot: heating activates additional non-basal slip systems that room temperature does not reach.
On the iron–carbon diagram, the carbon content that separates steel from cast iron is
1. 0.77 per cent — the eutectoid composition
2. 2.11 per cent — the maximum solubility of carbon in austenite
3. 4.3 per cent — the eutectic composition
4. 0.008 per cent — the maximum solubility of carbon in ferrite
Show answer
Answer: B — 2.11 per cent — the maximum solubility of carbon in austenite
2.11 per cent. The boundary is not a convention but a solubility limit: 2.11 per cent is the most carbon austenite will dissolve, at 1147 °C. Below it an alloy can be heated into a fully austenitic single phase and then hot-worked — forged, rolled, drawn — which is what makes it a steel. Above it, some carbon must remain as a separate carbide phase at every temperature, the alloy cannot be made single-phase, and it is too brittle to work plastically, so it must be cast into shape — hence cast iron. Option A, 0.77 per cent, is the eutectoid, where austenite decomposes wholly into pearlite; it separates hypoeutectoid from hypereutectoid steel, which is a real and different boundary. Option C, 4.3 per cent, is the eutectic at the same 1147 °C, giving ledeburite; it separates hypoeutectic from hypereutectic cast iron. Option D is ferrite’s solubility limit at room temperature, and it is the reason nearly all the carbon in a slowly cooled steel ends up as cementite rather than in solution. Getting these four apart is worth the effort because each is the answer to a different question, and a question naming a temperature usually tells you which: 727 °C points at the eutectoid, 1147 °C at either the eutectic or the 2.11 per cent limit depending on whether it asks about liquid.
A spherical casting and a cubical casting have the same volume of 1000 cm³ and are poured in the same mould material. The ratio of the sphere’s solidification time to the cube’s is ______ (round to three decimals)
Numerical answer — type the value.
Show answer
Answer: 1.539
1.539. Chvorinov’s rule is t = B(V/A)², so with the same mould constant and the same volume the ratio is just the square of the modulus ratio. The cube has side a = 1000^(1/3) = 10 cm, area 6 × 100 = 600 cm² and modulus 1000/600 = 1.6667 cm. The sphere has r = (3 × 1000/4π)^(1/3) = 6.2035 cm, area 4πr² = 483.60 cm² and modulus 1000/483.60 = 2.0678 cm. The ratio of times is (2.0678/1.6667)² = 1.5393. The physical reason is that a sphere has the least surface area of any shape for a given volume, so it has the largest modulus and the least area through which to lose heat — same metal, same volume, 54 per cent longer to freeze from shape alone. That single fact is why risers are made stubby and near-spherical rather than long and thin: a riser must stay liquid until the section it feeds has frozen, so it needs a larger modulus than that section, and a tall thin riser holding the same metal would freeze first and become a shrinkage cavity of its own rather than feeding one. The square in the rule is what makes this so leveraged: halving a section’s modulus cuts its freezing time to a quarter, which is exactly how a chill works — it adds local cooling area, drops the modulus there, and forces a heavy section to solidify before the feeder that serves it.
A rolling mill has rolls of 250 mm radius and the coefficient of friction between roll and strip is 0.2. The maximum possible draft in one pass, in millimetres, is ______
Numerical answer — type the value.
Show answer
Answer: 10
10 mm. The maximum draft is Δh_max = μ²R = 0.04 × 250 = 10 mm. Read what governs it and what does not. The condition comes from bite: the rolls can only draw the strip in if entry friction overcomes the horizontal component of roll pressure, so the limit involves the coefficient of friction squared and the roll radius — and involves the mill’s power, the strip’s yield strength and the reduction the schedule wants not at all. That is the examinable point, because it means a mill that starts slipping cannot be fixed by pressing harder. The quadratic in μ is dramatic: on the same rolls, μ = 0.1 allows only 2.5 mm and μ = 0.3 allows 22.5 mm, so a three-fold change in friction is a nine-fold change in capability. This is why roll surface condition, roughening and lubricant selection are not housekeeping details but the things that set the pass schedule — and why over-lubricating a mill reduces what it can do. Two design consequences follow. Bigger rolls bite deeper, since the limit is linear in R, but bigger rolls need more force and stiffer housings, so it is a trade rather than a free win. And thin strip wants small rolls to keep roll force and flattening down, yet small rolls cannot bite — which is exactly the contradiction four-high and cluster mills exist to resolve, with small work rolls doing the rolling and large backup rolls preventing them from bending.
Which of the following are correct comparisons of cold working with hot working?
1. Cold working raises yield strength and reduces ductility
2. Cold working gives better dimensional accuracy and surface finish
3. Hot working leaves large residual stresses in the product
4. Hot working permits much larger deformation per pass at lower force
Show answer
Answer: A — Cold working raises yield strength and reduces ductility; B — Cold working gives better dimensional accuracy and surface finish; D — Hot working permits much larger deformation per pass at lower force
A, B and D. A is the definition of work hardening: deforming below the recrystallisation temperature multiplies dislocations, which then obstruct each other, so strength and hardness rise while the remaining ductility is consumed. B follows from doing the work cold — no oxidation, no scale, no thermal contraction on cooling — which is why final sizing passes and finished bright bar are cold operations even when the bulk reduction was hot. D is the other half of the same physics: above the recrystallisation temperature the metal re-forms its grain structure as fast as deformation damages it, so it never hardens, flow stress stays low, and enormous reductions are possible in a single pass. C is the false one, and it is false in a useful way. Residual stresses come from non-uniform plastic deformation being locked in by surrounding elastic material — which is exactly what cold working leaves behind, and it is why cold-drawn bar warps when machined asymmetrically and why cold-formed parts are stress-relieved. Hot working leaves comparatively little residual stress, because recrystallisation relieves it as it forms. Getting C the right way round matters practically: it is the reason a cold-rolled sheet springs back and a hot-rolled one does not, and the reason stress-relief annealing exists as a separate step after cold forming rather than after hot forming.
A bar is stretched so that its length doubles. The true (logarithmic) strain is ______ (round to three decimals)
Numerical answer — type the value.
Show answer
Answer: 0.693
0.693. Doubling the length is an engineering strain of e = ΔL/L₀ = 1.0, and the true strain is ε = ln(1 + e) = ln 2 = 0.6931. The two diverge as soon as deformation is large, and the reason to use the true measure is that only true strains are additive. Check it: stretch a bar by 50 per cent and then by 50 per cent again, and the total engineering strain is not 1.0 but 1.5 × 1.5 − 1 = 1.25, whereas the true strains add exactly — ln 1.5 + ln 1.5 = 0.4055 + 0.4055 = 0.8109 = ln 2.25 ✓. That additivity is what makes true strain the right variable for a multi-pass forming schedule, where each pass has to build on the last. Two more properties are worth having. True strain is symmetric in reversal: compressing to half the length gives ln 0.5 = −0.693, the exact negative of stretching to double, while engineering strains give +1.0 and −0.5. And for a constant-volume plastic deformation, ε = ln(A₀/A) in terms of area as readily as length, which is the form actually used for wire drawing and extrusion, where the reduction is quoted as an area ratio. For small strains the distinction does not matter — e = 0.01 gives ε = 0.00995 — which is why elastic analysis never bothers with it.
A welder needs to reduce distortion and narrow the heat-affected zone on a thin steel sheet, keeping the same welding current and voltage. The correct action is to
1. increase the travel speed
2. decrease the travel speed
3. preheat the plate to a higher temperature
4. switch to submerged-arc welding
Show answer
Answer: A — increase the travel speed
Increase the travel speed. Heat input per unit length is H = ηVI/v, so with V and I fixed the only lever left is v — and H is inversely proportional to it. Less heat per millimetre of joint means a narrower heat-affected zone with finer grains and better toughness, less metal expanding and then contracting so less distortion, and lower residual stress locked in. Option B does the opposite and is the answer if you have the formula upside down. Option C raises the total thermal input and widens the HAZ further; preheating is a legitimate technique but it addresses a different problem — it slows the cooling rate to avoid a brittle untempered martensitic zone in hardenable steels, at the deliberate cost of more distortion. Confusing "reduce heat input" with "control cooling rate" is the conceptual trap here, and the two often pull in opposite directions. Option D goes the wrong way hardest: submerged-arc welding is the high-heat-input process, designed to lay thick beads quickly on heavy plate, and is unusable on thin sheet for exactly this reason. The honest caveat on A is that speed cannot be raised indefinitely — too little heat gives lack of fusion, undercut, and in hardenable steels a quench-cracked HAZ. The processes chosen when a genuinely narrow weld is required are electron-beam and laser welding, which achieve it not by moving faster but by making the power density enormous while the total energy stays small.
Which heat treatment produces martensite in a plain carbon steel?
1. Full annealing — slow cooling in the furnace
2. Quenching — cooling fast enough to bypass the pearlite transformation
3. Normalising — cooling in still air
4. Tempering — reheating below the eutectoid temperature
Show answer
Answer: B — Quenching — cooling fast enough to bypass the pearlite transformation
Quenching. Martensite forms only when cooling is fast enough to miss the nose of the TTT curve entirely, so the carbon has no time to diffuse out of solution and the face-centred austenite shears into a supersaturated body-centred-tetragonal structure. It is very hard and very brittle, and — the point worth carrying — it is not an equilibrium phase and therefore does not appear anywhere on the iron–carbon diagram. Only a TTT or CCT curve, which redraws the transformation against time, can show it. Options A and C are both slow enough to form pearlite, differing in fineness: full annealing in the furnace gives coarse pearlite and maximum softness, normalising in air gives finer pearlite, a little more strength and better uniformity. Option D is the process applied after quenching, not instead of it: tempering reheats existing martensite so that some carbon precipitates as fine carbides, trading hardness back for toughness. So a hardened part is always the sequence quench then temper, and untempered martensite is generally too brittle to use — a fact with a direct welding consequence, since the heat-affected zone of a weld on hardenable steel is quenched by the surrounding cold plate and can end up untempered, which is why preheat is specified for exactly those steels.