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HKIE Professional Assessment - Materials Discipline study materials
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HKIE Professional Assessment - Materials Discipline practice questions

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Question 1Materials Science Fundamentals and Material Selection

In a tensile test on a ductile metal, which parameter is obtained directly from the slope of the linear elastic region of the engineering stress–strain curve?

  1. Young's modulus Correct answer.
  2. Fracture toughness
  3. Ultimate tensile strength
  4. Yield strength

Explanation

The elastic region is linear, and its slope (stress divided by strain) defines the modulus of elasticity or Young's modulus. Yield strength is read from the deviation from linearity, ultimate strength is the peak stress, and fracture toughness requires a separate fracture-mechanics test.

Question 2Materials Science Fundamentals and Material Selection

A steel component must resist repeated cyclic loading without failing. Which material property is the primary selection criterion?

  1. Electrical conductivity
  2. Hardness
  3. Thermal expansion coefficient
  4. Fatigue strength Correct answer.

Explanation

Fatigue strength describes the stress level a material can endure under repeated cyclic loading before fatigue failure occurs. Hardness relates to wear and indentation resistance, while conductivity and thermal expansion address unrelated functional needs.

Question 3Materials Science Fundamentals and Material Selection

Which crystal structure is characterised by a coordination number of 12 and an atomic packing factor of 0.74?

  1. Simple cubic
  2. Face-centred cubic Correct answer.
  3. Diamond cubic
  4. Body-centred cubic

Explanation

Face-centred cubic metals have 12 nearest neighbours and an atomic packing factor of 0.74, the densest arrangement for equal spheres. Body-centred cubic has a coordination of 8 and a packing factor of 0.68, simple cubic is 6 and 0.52, and diamond cubic is much more open.

Question 4Materials Science Fundamentals and Material Selection

In a binary phase diagram, what is the significance of the eutectic point?

  1. It marks the temperature at which a single solid phase melts completely.
  2. It is the composition and temperature at which two solid phases form simultaneously from the liquid. Correct answer.
  3. It is the point where the two liquid phases become completely miscible.
  4. It indicates the maximum solubility of one component in the other.

Explanation

At the eutectic point, liquid of a specific composition transforms on cooling into two distinct solid phases at a single fixed temperature. This invariant reaction is central to understanding microstructures such as lamellar eutectics.

Question 5Materials Science Fundamentals and Material Selection

Which hardening mechanism relies on solute atoms distorting the crystal lattice and impeding dislocation motion?

  1. Grain-boundary strengthening
  2. Solid-solution strengthening Correct answer.
  3. Precipitation hardening
  4. Strain hardening

Explanation

Solid-solution strengthening occurs when solute atoms substitute or occupy interstitial sites, creating local lattice strain that hinders dislocation movement. Grain-boundary strengthening, precipitation hardening and strain hardening operate through different microstructural features.

Question 6Materials Science Fundamentals and Material Selection

Which of the following best describes the purpose of a corrosion-resistant coating on a steel structure?

  1. It improves the electrical conductivity of the steel.
  2. It increases the yield strength of the steel substrate.
  3. It raises the melting temperature of the steel.
  4. It isolates the steel from the corrosive environment and disrupts the electrochemical circuit. Correct answer.

Explanation

A barrier coating prevents the electrolyte and oxygen from reaching the steel surface, interrupting the electrochemical corrosion cell. Coatings do not normally change bulk mechanical properties such as yield strength or melting temperature.

Question 7Materials Science Fundamentals and Material Selection

What does the term 'ductile-to-brittle transition temperature' refer to in body-centred cubic steels?

  1. The temperature at which the steel becomes superconducting
  2. The temperature range over which fracture mode changes from ductile to brittle as temperature falls Correct answer.
  3. The temperature above which the steel becomes fully liquid
  4. The temperature at which the steel stops conducting heat

Explanation

Body-centred cubic steels lose toughness at low temperature, changing from ductile tearing to brittle cleavage. The transition temperature characterises this change and is critical for cold-climate structural design.

Question 8Materials Science Fundamentals and Material Selection

Which material property is most relevant when selecting a material for a thermal barrier application in a gas turbine?

  1. High electrical conductivity
  2. Low thermal conductivity Correct answer.
  3. High density
  4. High magnetic permeability

Explanation

Thermal barrier coatings must impede heat flow, so low thermal conductivity is the key property. Electrical conductivity, magnetic permeability and high density do not serve the thermal insulation function.

Question 9Materials Science Fundamentals and Material Selection

A material shows a yield strength of 350 MPa and an ultimate tensile strength of 500 MPa. Which statement correctly interprets these values?

  1. The material fractures at 350 MPa.
  2. The material can sustain 500 MPa indefinitely without plastic deformation.
  3. The material begins to deform plastically at about 350 MPa and reaches its maximum engineering stress at 500 MPa. Correct answer.
  4. The material has a Young's modulus of 350 MPa.

Explanation

Yield strength marks the onset of plastic deformation, while ultimate tensile strength is the peak engineering stress before necking. Values beyond the yield point cause permanent deformation, and the modulus is unrelated to these stress magnitudes.

Question 10Materials Science Fundamentals and Material Selection

A steel shaft requires high surface hardness with a tough core. Which treatment is most appropriate?

  1. Electroplating with a soft metal
  2. Annealing to produce coarse pearlite
  3. Through-hardening by full austenitising and quenching
  4. Case carburising followed by quenching and tempering Correct answer.

Explanation

Case carburising enriches the surface in carbon, so subsequent quenching produces a hard martensitic case while the lower-carbon core remains tough. Through-hardening gives a more uniform hardness, annealing softens the whole part, and soft plating does not provide wear resistance.

Question 11Materials Science Fundamentals and Material Selection

Using the Hall–Petch relationship, what happens to the yield strength of a polycrystalline metal as its grain size is reduced?

  1. Yield strength increases because grain boundaries impede dislocation motion. Correct answer.
  2. Yield strength increases only if the crystal structure changes.
  3. Yield strength is unaffected because grain size does not influence slip.
  4. Yield strength decreases because more grain boundaries act as dislocation sources.

Explanation

The Hall–Petch equation shows yield strength rising with the inverse square root of grain size, because boundaries block dislocation pile-ups. Refining grains therefore strengthens the metal without changing its crystal structure.

Question 12Materials Science Fundamentals and Material Selection

In the iron–carbon system, which phase is a supersaturated interstitial solid solution of carbon in body-centred tetragonal iron formed by rapid quenching?

  1. Cementite
  2. Ferrite
  3. Martensite Correct answer.
  4. Austenite

Explanation

Martensite forms when austenite is quenched too quickly for carbon to diffuse out, trapping carbon in a body-centred tetragonal lattice. Ferrite is low-carbon body-centred cubic iron, cementite is the carbide Fe3C, and austenite is face-centred cubic.

Question 13Materials Science Fundamentals and Material Selection

Which statement correctly distinguishes hardness from toughness in material selection?

  1. Toughness is only relevant to polymers, while hardness applies only to metals.
  2. Hardness and toughness are identical properties measured in the same units.
  3. Hardness measures resistance to indentation, whereas toughness measures resistance to fracture under impact or flaw growth. Correct answer.
  4. Hardness measures resistance to fracture, whereas toughness measures resistance to indentation.

Explanation

Hardness quantifies resistance to localised plastic deformation such as indentation, while toughness reflects the energy absorbed before fracture and resistance to crack propagation. They are distinct properties and a hard material may still be brittle.

Question 14Materials Science Fundamentals and Material Selection

A galvanic couple is formed between copper and zinc in seawater. Which metal will corrode preferentially, and why?

  1. Both corrode at equal rates
  2. Zinc, because it is more active and becomes the anode Correct answer.
  3. Copper, because it is the more noble metal
  4. Neither corrodes because seawater is an electrolyte

Explanation

In a galvanic couple the more active metal, zinc, becomes the anode and corrodes while protecting the more noble copper cathode. This principle underpins sacrificial galvanic protection of steel by zinc.

Question 15Materials Science Fundamentals and Material Selection

Which of the following best explains why ceramics are typically brittle at room temperature?

  1. They contain a high density of mobile dislocations.
  2. They have very low melting temperatures.
  3. Their ionic or covalent bonding restricts dislocation motion, so cracks propagate with little plastic deformation. Correct answer.
  4. They possess a face-centred cubic crystal structure.

Explanation

Strong directional ionic and covalent bonds make dislocation glide difficult in ceramics, so stress concentrates at flaws and cracks propagate without blunting. This lack of plastic deformation produces brittle behaviour.

Question 16Materials Science Fundamentals and Material Selection

A polymer component must remain flexible at low temperature. Which property should be minimised in the selection?

  1. Glass transition temperature Correct answer.
  2. Elongation at break
  3. Molecular weight
  4. Impact strength

Explanation

Below the glass transition temperature a polymer becomes glassy and stiff, so a low glass transition temperature keeps it flexible at service temperature. High molecular weight and good elongation support toughness rather than limiting flexibility.

Question 17Materials Science Fundamentals and Material Selection

Which test method is appropriate for determining the plane-strain fracture toughness of a metallic material?

  1. Uniaxial tensile test
  2. Compact tension specimen test according to linear elastic fracture mechanics Correct answer.
  3. Rockwell hardness test
  4. Charpy impact test

Explanation

Plane-strain fracture toughness is measured using a pre-cracked specimen such as a compact tension geometry analysed with linear elastic fracture mechanics. Charpy testing gives impact energy, hardness gives indentation resistance, and tensile testing gives strength and ductility.

Question 18Materials Science Fundamentals and Material Selection

An engineer must select a material for a pressure vessel where weight is critical but strength must be high. Which figure of merit best guides the choice?

  1. Electrical resistivity
  2. Specific strength, the ratio of strength to density Correct answer.
  3. Melting temperature alone
  4. Coefficient of thermal expansion alone

Explanation

When weight is critical, performance depends on strength normalised by density, known as specific strength. Melting temperature, thermal expansion and resistivity do not capture the strength-to-weight trade-off.

Question 19Materials Science Fundamentals and Material Selection

Which phenomenon describes the slow, time-dependent plastic deformation of a metal under sustained load at elevated temperature?

  1. Corrosion fatigue
  2. Creep Correct answer.
  3. Quenching
  4. Fatigue

Explanation

Creep is time-dependent plastic strain that accumulates under constant stress, becoming significant at high homologous temperatures. Fatigue involves cyclic loading, corrosion fatigue combines cycling with environment, and quenching is a heat-treatment step.

Question 20Materials Science Fundamentals and Material Selection

Which statement correctly describes the role of a stabiliser in a polymer formulation?

  1. It slows degradation caused by heat, light or oxidation during service. Correct answer.
  2. It lowers the molecular weight to improve flow.
  3. It increases the glass transition temperature to a fixed value.
  4. It accelerates cross-linking to make the polymer rigid.

Explanation

Stabilisers such as antioxidants and UV absorbers retard degradation of the polymer chain caused by heat, light or oxygen. They are not primarily cross-linking agents, molecular-weight reducers or fixed glass-transition modifiers.

Question 21Materials Science Fundamentals and Material Selection

A component is made from a precipitation-hardenable aluminium alloy. Which sequence correctly describes the strengthening treatment?

  1. Electroplating followed by polishing
  2. Solution treatment, rapid quench, then ageing at an intermediate temperature Correct answer.
  3. Full austenitising, then slow furnace cooling
  4. Annealing, slow cooling, then cold rolling

Explanation

Precipitation hardening requires solution treatment to dissolve the alloying elements, a rapid quench to retain a supersaturated solid solution, and ageing to precipitate fine particles that impede dislocations. Annealing and slow cooling do not produce this fine dispersion.

Question 22Materials Science Fundamentals and Material Selection

Two steels have the same composition but one has a much finer prior-austenite grain size. Which consequence is expected after identical quenching?

  1. The finer-grained steel will have lower strength and lower toughness.
  2. Both steels will behave identically because composition is the same.
  3. The finer-grained steel will have a higher melting temperature.
  4. The finer-grained steel will have higher strength and improved toughness. Correct answer.

Explanation

Grain refinement increases strength through the Hall–Petch effect and also provides more boundaries to deflect cracks, improving toughness. Composition alone does not fix properties because microstructure, including grain size, strongly influences them.

Question 23Materials Science Fundamentals and Material Selection

In a galvanic corrosion cell, increasing the ratio of cathodic area to anodic area on a steel structure generally has which effect?

  1. It eliminates corrosion by making the steel cathodic everywhere.
  2. It has no effect on corrosion because the total current is fixed.
  3. It intensifies localised attack at the small anode, increasing corrosion rate per unit area. Correct answer.
  4. It reduces the corrosion rate by spreading the anodic current.

Explanation

When the cathode area is large relative to the anode, the anodic current is concentrated on a small area, raising the local anodic current density and accelerating attack. This is why small anodes coupled to large cathodes, such as a steel bolt in a copper plate, are particularly vulnerable.

Question 24Materials Science Fundamentals and Material Selection

Which selection strategy is most appropriate when choosing a material for a component subject to both high-cycle fatigue and a corrosive environment?

  1. Use a material with good fatigue strength and corrosion resistance, or apply a protective treatment compatible with cyclic loading. Correct answer.
  2. Maximise hardness alone, because hard materials never corrode.
  3. Choose a material with the highest thermal expansion coefficient.
  4. Select the cheapest available material regardless of environment.

Explanation

Corrosion fatigue is more severe than either mechanism alone, so the material must resist cyclic crack initiation and environmental attack. Hardness alone does not confer corrosion resistance, and cost or thermal expansion do not address the combined degradation.

Question 25Materials Science Fundamentals and Material Selection

Which statement correctly explains why fine pearlite is harder than coarse pearlite of the same steel composition?

  1. Fine pearlite has a lower carbon content than coarse pearlite.
  2. Fine pearlite has a smaller interlamellar spacing, so cementite lamellae impede dislocation motion more effectively. Correct answer.
  3. Fine pearlite has a larger interlamellar spacing, which eases dislocation motion.
  4. Fine pearlite contains no cementite phase.

Explanation

In fine pearlite the cementite lamellae are more closely spaced, so dislocations encounter more phase boundaries and are blocked more frequently, raising hardness. Coarse pearlite has wider spacing and permits easier dislocation motion.

Question 26Materials Science Fundamentals and Material Selection

In a uniaxial tensile test on a ductile metal, which point on the engineering stress–strain curve marks the onset of necking, after which deformation becomes localised?

  1. The 0.2% proof stress point
  2. The fracture point
  3. The proportional limit point
  4. The ultimate tensile strength point Correct answer.

Explanation

Necking begins at the maximum engineering stress, which is the ultimate tensile strength. Beyond this point the load-bearing capacity of the specimen falls as deformation concentrates in a local region. The proof stress and proportional limit lie on the early elastic portion, well before instability.

Question 27Materials Science Fundamentals and Material Selection

An alloy is required for a component that must resist elastic deflection under load while remaining as light as possible. Which material index should be maximised for a stiffness-limited, mass-minimising design of a bar in tension?

  1. ρ/E
  2. σy/ρ
  3. E·ρ
  4. E/ρ Correct answer.

Explanation

For a tie rod of fixed stiffness, mass scales with ρ/E, so minimising mass means maximising the specific stiffness E/ρ. This index captures the trade-off between Young's modulus and density directly. Strength-based indices such as σy/ρ apply only when yielding, not deflection, governs the design.

Question 28Materials Science Fundamentals and Material Selection

A carbon steel weldment is to be used in a cold climate where the service temperature approaches the ductile-to-brittle transition region of the steel. Which microstructural feature most improves resistance to brittle fracture in that service?

  1. A coarse Widmanstätten ferrite structure
  2. A high volume fraction of grain-boundary carbide films
  3. A fully martensitic untempered structure
  4. A fine ferrite grain size Correct answer.

Explanation

Grain refinement both lowers the transition temperature and raises toughness, because finer grains reduce the size of the microcracks that trigger cleavage and provide more barriers to crack propagation. Coarse structures and continuous grain-boundary carbides instead raise the transition temperature and promote brittle failure.

Question 29Materials Science Fundamentals and Material Selection

A polymer component must retain dimensional stability and load-bearing capability over years at ambient temperature under a sustained stress. Which viscoelastic effect should the designer specifically allow for?

  1. Thermal expansion mismatch with the mould
  2. Creep, described by an isochronous stress–strain curve Correct answer.
  3. Fatigue crack growth under cyclic loading
  4. Electrostatic charge accumulation

Explanation

Under sustained load, viscoelastic polymers deform progressively with time, and this creep is characterised by isochronous stress–strain curves or creep modulus data. Design must use the modulus appropriate to the service life rather than the short-term modulus. Fatigue, expansion and charging are separate phenomena not driven by sustained-load viscoelasticity.

Question 30Materials Science Fundamentals and Material Selection

A precipitation-hardenable alloy is solution treated, quenched and then aged. Which combination of microstructural conditions produces the maximum strengthening during ageing?

  1. A fine dispersion of coherent precipitates with an optimum particle spacing Correct answer.
  2. A continuous network of grain-boundary precipitates
  3. Complete dissolution of all alloying elements into a single-phase solid solution
  4. A coarse dispersion of widely spaced incoherent precipitates

Explanation

Peak hardness occurs when ageing produces a fine, dense dispersion of coherent precipitates whose spacing optimally impedes dislocation motion. Over-ageing coarsens and separates the particles, reducing the barrier to slip and softening the alloy. Coherent, closely spaced particles give the strongest resistance to dislocation bypass.

Question 31Metallic Materials: Structure, Properties, and Heat Treatment

In the iron–carbon equilibrium diagram, what is the maximum solubility of carbon in ferrite (α-Fe) at 723 °C?

  1. 0.022 wt% C Correct answer.
  2. 0.77 wt% C
  3. 2.11 wt% C
  4. 6.67 wt% C

Explanation

Ferrite has a body-centred cubic structure with very limited interstitial space, so its maximum carbon solubility is only about 0.022 wt% at the eutectoid temperature and falls further at room temperature. The larger values correspond to the eutectoid composition and to cementite, not to ferrite.

Question 32Metallic Materials: Structure, Properties, and Heat Treatment

Which microstructural constituent of slowly cooled plain carbon steel is a lamellar mixture of ferrite and cementite?

  1. Pearlite Correct answer.
  2. Martensite
  3. Austenite
  4. Bainite

Explanation

Pearlite forms when austenite of eutectoid composition decomposes at temperatures just below the eutectoid, producing alternating lamellae of ferrite and cementite. Martensite is a supersaturated single-phase structure, bainite is a non-lamellar acicular product, and austenite is the high-temperature face-centred cubic phase.

Question 33Metallic Materials: Structure, Properties, and Heat Treatment

Which quench medium generally produces the highest cooling rate at the surface of a hardened steel component?

  1. Still air
  2. Circulating oil
  3. Brine (salt water) Correct answer.
  4. Molten salt bath

Explanation

Brine combines high thermal conductivity with a violent vapour-blanket collapse, giving the fastest heat extraction of the common quenchants. Air, oil and molten salt all remove heat more slowly and are chosen when a milder cooling rate is required to limit distortion.

Question 34Metallic Materials: Structure, Properties, and Heat Treatment

A steel component is to be carburised. Which property of the base steel most directly governs the depth and uniformity of the case formed?

  1. Its coercive force
  2. Its carbon diffusivity at the carburising temperature Correct answer.
  3. Its density
  4. Its electrical resistivity

Explanation

Case depth in carburising is controlled by carbon diffusion, so the diffusion coefficient of carbon in austenite at the treatment temperature sets how far and how evenly carbon penetrates. Resistivity, coercive force and density are not the governing transport parameters.

Question 35Metallic Materials: Structure, Properties, and Heat Treatment

Which of the following best describes the crystal structure of cementite (Fe3C)?

  1. Orthorhombic Correct answer.
  2. Face-centred cubic
  3. Hexagonal close-packed
  4. Body-centred cubic

Explanation

Cementite is an orthorhombic intermetallic compound with a complex unit cell containing twelve iron and four carbon atoms. It is hard and brittle and is not a simple cubic or close-packed metallic structure.

Question 36Metallic Materials: Structure, Properties, and Heat Treatment

What is the primary purpose of tempering a quenched steel?

  1. To refine the prior austenite grain size before austenitising
  2. To convert retained austenite entirely into ferrite
  3. To relieve quenching stresses and restore toughness at some cost in hardness Correct answer.
  4. To increase the carbon content of the surface

Explanation

As-quenched martensite is very hard but brittle and internally stressed, so tempering precipitates fine carbides and relaxes residual stress, trading some hardness for toughness. It does not add carbon, refine grains before austenitising, or fully convert austenite to ferrite.

Question 37Metallic Materials: Structure, Properties, and Heat Treatment

Which alloying element is deliberately added to steel to stabilise the austenite phase and enable the austenitic stainless steels?

  1. Titanium
  2. Nickel Correct answer.
  3. Vanadium
  4. Niobium

Explanation

Nickel is a strong austenite stabiliser that lowers the A3 temperature and expands the austenite field, which is why 8–10 wt% Ni supports the austenitic structure of 18-8 stainless steels. Vanadium, niobium and titanium are strong carbide formers that promote ferrite.

Question 38Metallic Materials: Structure, Properties, and Heat Treatment

A plain carbon steel with 0.8 wt% C is heated to 780 °C and held until equilibrium is reached. Which phases are present?

  1. Ferrite and pearlite only
  2. Austenite only
  3. Ferrite plus austenite
  4. Austenite plus cementite Correct answer.

Explanation

The eutectoid temperature is about 723 °C and the eutectoid composition about 0.77 wt% C. At 780 °C a 0.8 wt% C steel lies just above the eutectoid and slightly hypereutectoid, so it consists of austenite with a small amount of proeutectoid cementite.

Question 39Metallic Materials: Structure, Properties, and Heat Treatment

Using the lever rule, estimate the mass fraction of pearlite in a slowly cooled 0.40 wt% C hypoeutectoid steel.

  1. 0.40
  2. 0.77
  3. 0.95
  4. 0.52 Correct answer.

Explanation

Applying the lever rule between the eutectoid composition 0.77 wt% C and the ferrite solubility 0.022 wt% C gives (0.40 − 0.022)/(0.77 − 0.022) ≈ 0.51. The result is close to 0.52, so roughly half the microstructure is pearlite with the balance proeutectoid ferrite.

Question 40Metallic Materials: Structure, Properties, and Heat Treatment

Which transformation product forms when austenite is isothermally held just above the martensite start temperature for a prolonged period?

  1. Upper bainite
  2. Lower bainite Correct answer.
  3. Coarse pearlite
  4. Untempered martensite

Explanation

At temperatures just above the martensite start range, diffusion is slow and the transformation product is lower bainite, consisting of fine carbide needles within acicular ferrite. Upper bainite forms at higher temperatures and pearlite requires higher temperatures still.

Question 41Metallic Materials: Structure, Properties, and Heat Treatment

Why does increasing the austenitising temperature generally increase the hardenability of a steel?

  1. It increases the carbon content of the steel
  2. It reduces the cooling rate required for full hardening
  3. It dissolves more carbide and coarsens the austenite grain size, retarding pearlite nucleation Correct answer.
  4. It raises the martensite start temperature

Explanation

A higher austenitising temperature dissolves more alloy carbide and coarsens the austenite grains, which reduces nucleation sites for pearlite and shifts the transformation curve to longer times. This delays diffusional decomposition of austenite, so thicker sections can harden.

Question 42Metallic Materials: Structure, Properties, and Heat Treatment

In the Jominy end-quench test, what does the distance from the quenched end at which a specified hardness is reached indicate?

  1. The carbon content of the steel
  2. The hardenability of the steel Correct answer.
  3. The tempering resistance of the steel
  4. The grain size of the steel

Explanation

The Jominy test applies a standard end quench and measures hardness along the specimen, so a greater distance to a given hardness means the steel resists diffusional transformation more strongly, that is, it has higher hardenability. Carbon content sets the maximum attainable hardness but not the hardness–distance profile.

Question 43Metallic Materials: Structure, Properties, and Heat Treatment

Two steels have identical carbon content, but one contains 1 wt% Cr and the other contains 1 wt% Ni. Which statement about their hardenability is most defensible?

  1. Both elements have negligible effect on hardenability
  2. Chromium raises hardenability more per unit addition than nickel does Correct answer.
  3. Both elements reduce hardenability relative to plain carbon steel
  4. Nickel raises hardenability more per unit addition than chromium does

Explanation

Hardenability multiplying factors show that chromium is a more potent hardenability agent than nickel at the same weight addition, because it both retards pearlite formation and forms alloy carbides. Nickel is a useful but weaker hardenability contributor.

Question 44Metallic Materials: Structure, Properties, and Heat Treatment

Which heat treatment is most appropriate for producing a tough, machinable structure in a hypoeutectoid forging before final machining?

  1. Normalise
  2. Austemper
  3. Martemper
  4. Full anneal Correct answer.

Explanation

A full anneal produces coarse, soft pearlite and ferrite with low hardness, which improves machinability and removes residual stress. Normalising gives a finer and somewhat harder structure, while austempering and martempering are hardening treatments.

Question 45Metallic Materials: Structure, Properties, and Heat Treatment

What is the main metallurgical reason for using a martempering (step-quench) treatment?

  1. To reduce thermal and transformation stresses during hardening Correct answer.
  2. To increase the carbon content of the surface
  3. To produce a fully pearlitic structure
  4. To eliminate the need for tempering

Explanation

Martempering quenches into a salt bath just above the martensite start temperature, equalises the section temperature, then cools slowly so that martensite forms with less thermal gradient and stress. It reduces distortion and cracking but does not change surface carbon or remove the need for tempering.

Question 46Metallic Materials: Structure, Properties, and Heat Treatment

During tempering of a quenched plain carbon steel, which sequence of carbide precipitation is observed as temperature rises from about 100 °C to 400 °C?

  1. Cementite first, then epsilon carbide
  2. Cementite only, with no intermediate carbide
  3. Graphite first, then epsilon carbide
  4. Epsilon carbide first, then cementite Correct answer.

Explanation

Low-temperature tempering first precipitates a fine transition carbide, often described as epsilon carbide, which is then replaced by the stable cementite as the temperature rises. This sequential precipitation controls the gradual softening observed during tempering.

Question 47Metallic Materials: Structure, Properties, and Heat Treatment

A steel shows temper embrittlement after slow cooling through the range 350–550 °C. Which combination of elements is most commonly associated with this phenomenon?

  1. Nickel and chromium with phosphorus or antimony Correct answer.
  2. Silicon and aluminium with nitrogen
  3. Copper and beryllium with oxygen
  4. Molybdenum and vanadium with sulphur

Explanation

Temper embrittlement is caused by segregation of impurity elements such as phosphorus, antimony, tin and arsenic to prior austenite grain boundaries, a process promoted by nickel and chromium. Molybdenum additions are used to suppress the effect.

Question 48Metallic Materials: Structure, Properties, and Heat Treatment

Which statement about the effect of cooling rate on the transformation temperature in continuous cooling is defensible?

  1. Faster cooling shifts the transformation to lower temperatures and finer products Correct answer.
  2. Faster cooling always produces a fully ferritic structure
  3. Faster cooling shifts the transformation to higher temperatures and coarser products
  4. Cooling rate has no effect on transformation temperature

Explanation

In continuous cooling, a faster rate depresses the temperature at which transformation occurs because there is less time for diffusion, so the products are finer. Very fast cooling can suppress diffusion entirely and produce martensite.

Question 49Metallic Materials: Structure, Properties, and Heat Treatment

What is the effect of a higher austenite grain size on the martensite plate size and the resulting toughness?

  1. Austenite grain size has no effect on martensite morphology
  2. Coarser austenite grains produce finer martensite and higher toughness
  3. Coarser austenite grains prevent martensite formation entirely
  4. Coarser austenite grains produce coarser martensite packets and lower toughness Correct answer.

Explanation

Prior austenite grain boundaries divide the martensite into packets and blocks, so a coarse austenite grain produces larger martensite units and longer effective slip lengths. This raises the ductile–brittle transition temperature and reduces toughness, which is why grain refinement is desirable.

Question 50Metallic Materials: Structure, Properties, and Heat Treatment

Which mechanism best explains the strengthening produced by cold working a low carbon steel?

  1. Solid solution hardening by interstitial carbon
  2. Precipitation of fine alloy carbides
  3. Grain coarsening during deformation
  4. Dislocation multiplication and interaction, raising the flow stress Correct answer.

Explanation

Cold working multiplies dislocations, and their mutual interaction and entanglement raise the stress required for further slip, which is strain hardening. The carbon content does not change, no new precipitate forms, and grains tend to elongate rather than coarsen.