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PE Metallurgical practice questions

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Question 1Materials properties and behavior

In the face-centered cubic crystal structure, what is the coordination number (number of nearest-neighbor atoms) of each atom?

  1. 4
  2. 8
  3. 6
  4. 12 Correct answer.

Explanation

Each atom in an FCC crystal touches 12 nearest neighbors: four in its own close-packed plane, four in the plane above, and four in the plane below. This gives FCC its close-packed atomic arrangement.

Question 2Materials properties and behavior

The atomic packing factor of the body-centered cubic structure is approximately:

  1. 0.68 Correct answer.
  2. 0.34
  3. 0.74
  4. 0.52

Explanation

The BCC unit cell contains two atoms and has an atomic packing factor of about 0.68. This is lower than the close-packed FCC and HCP structures because BCC atoms are not in the densest possible arrangement.

Question 3Materials properties and behavior

The Rockwell C hardness scale uses which type of indenter?

  1. Hardened steel ball
  2. Square-based pyramid
  3. Diamond cone Correct answer.
  4. Tungsten carbide ball

Explanation

The Rockwell C scale uses a diamond spheroconical indenter with a 150 kgf major load. It is suited to hard materials such as hardened steels.

Question 4Materials properties and behavior

On a tensile stress–strain curve, the slope of the initial linear portion represents:

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

Explanation

The initial linear region is elastic, and its slope is the elastic modulus (Young's modulus), a measure of stiffness. Yield strength marks the onset of plastic deformation, while toughness is related to the area under the curve.

Question 5Materials properties and behavior

A typical Poisson's ratio value for metallic alloys is closest to:

  1. 0.15
  2. 0.50
  3. 0.05
  4. 0.33 Correct answer.

Explanation

Most metals have Poisson's ratios in the range of about 0.25 to 0.35, with 0.33 being a common representative value. A value near 0.50 indicates nearly incompressible behavior.

Question 6Materials properties and behavior

The 0.2% offset method is used to determine which property from a tensile test?

  1. Elastic modulus
  2. Fracture strain
  3. Modulus of resilience
  4. Yield strength Correct answer.

Explanation

A line parallel to the elastic slope is drawn from 0.2% strain, and its intersection with the stress–strain curve defines the 0.2% offset yield strength. This is used when the material does not show a distinct yield point.

Question 7Materials properties and behavior

Ductility of a metal in a tensile test is most commonly expressed as:

  1. Poisson's ratio
  2. Modulus of resilience
  3. Percent elongation Correct answer.
  4. Hardness

Explanation

Percent elongation and percent reduction in area are the standard measures of ductility from a tensile test. They indicate how much plastic strain the metal can sustain before fracture.

Question 8Materials properties and behavior

The ideal c/a ratio for a hexagonal close-packed unit cell is approximately:

  1. 1.00
  2. 1.63 Correct answer.
  3. 1.41
  4. 1.22

Explanation

For ideal HCP packing of equal spheres, c/a equals the square root of 8/3, which is approximately 1.633. Many real HCP metals deviate somewhat from this ideal value.

Question 9Materials properties and behavior

Fick's first law describes steady-state diffusion. Which expression correctly represents the diffusion flux J?

  1. J = cD
  2. J = -D (dc/dx) Correct answer.
  3. J = -D (dT/dx)
  4. J = D (d²c/dx²)

Explanation

Fick's first law states that flux is proportional to the concentration gradient, with a negative sign because diffusion occurs down the gradient. The proportionality constant is the diffusion coefficient D.

Question 10Materials properties and behavior

For a diffusion-controlled process in a solid, increasing the temperature from 500 °C to 600 °C typically causes the diffusion coefficient to:

  1. Increase then decrease
  2. Remain unchanged
  3. Increase exponentially Correct answer.
  4. Decrease linearly

Explanation

The diffusion coefficient follows an Arrhenius relationship, D = D0 exp(-Q/RT), so it increases exponentially with absolute temperature. The activation energy Q controls the sensitivity to temperature.

Question 11Materials properties and behavior

According to the Hall–Petch relationship, the yield strength of a polycrystalline metal increases as:

  1. Solute content increases
  2. Dislocation density decreases
  3. Grain size decreases Correct answer.
  4. Grain size increases

Explanation

Hall–Petch gives yield strength proportional to d^(-1/2), where d is grain size. Smaller grains provide more grain-boundary barriers to dislocation motion and raise strength.

Question 12Materials properties and behavior

In substitutional solid-solution strengthening, which characteristic of the solute atom contributes most to strengthening?

  1. Same crystal structure as the solvent
  2. Size mismatch with the solvent Correct answer.
  3. A significant atomic size difference from the solvent
  4. Same valence as the solvent

Explanation

A size mismatch between solute and solvent atoms creates local elastic strain fields that interact with dislocations and impede their motion. This is the primary source of solid-solution strengthening.

Question 13Materials properties and behavior

During artificial aging of an aluminum alloy, overaging causes a decrease in strength because:

  1. Solute atoms evaporate
  2. Precipitates fully dissolve
  3. The matrix recrystallizes
  4. Precipitates coarsen Correct answer.

Explanation

Overaging coarsens the precipitates and increases their spacing, and the particles may lose coherency with the matrix. Dislocations then bypass them more easily, so strength decreases.

Question 14Materials properties and behavior

For an edge dislocation, the Burgers vector is:

  1. At 45° to the line
  2. Undefined in direction
  3. Parallel to the dislocation line
  4. Perpendicular to the line Correct answer.

Explanation

An edge dislocation has an extra half-plane of atoms, and its Burgers vector lies perpendicular to the dislocation line. In a screw dislocation, by contrast, the Burgers vector is parallel to the line.

Question 15Materials properties and behavior

In a binary phase diagram, the lever rule is used to determine:

  1. Phase fractions Correct answer.
  2. Number of components
  3. Alloy melting point
  4. Cooling rate

Explanation

The lever rule uses the compositions of the two coexisting phases and the alloy composition to calculate the relative amounts of each phase. It is applied within two-phase regions at equilibrium.

Question 16Materials properties and behavior

The eutectoid reaction in the iron–carbon system occurs at approximately:

  1. 0.76 wt% C and 1148 °C
  2. 0.02 wt% C and 727 °C
  3. 0.76 wt% C and 727 °C Correct answer.
  4. 2.14 wt% C and 1148 °C

Explanation

The eutectoid point is at about 0.76 wt% C and 727 °C, where austenite transforms into ferrite and cementite to form pearlite. The eutectic point occurs at a much higher carbon content and temperature.

Question 17Materials properties and behavior

The transformation of austenite to martensite is best described as:

  1. An order–disorder reaction
  2. Diffusionless and displacive Correct answer.
  3. A spinodal decomposition
  4. Diffusion-controlled and reconstructive

Explanation

Martensite forms by a cooperative shear-like displacement of atoms without long-range diffusion. Carbon remains trapped in supersaturated solid solution, producing a hard, strained structure.

Question 18Materials properties and behavior

On an S–N fatigue curve for steel, the horizontal plateau at high cycles represents:

  1. The ultimate strength
  2. The yield strength
  3. The endurance limit Correct answer.
  4. The creep limit

Explanation

The endurance limit is the stress amplitude below which steel can survive a very large number of cycles without fatigue failure. It appears as a horizontal asymptote on the S–N curve.

Question 19Materials properties and behavior

In a creep test at constant load and temperature, the steady-state creep rate is the:

  1. Minimum steady creep rate Correct answer.
  2. Rate during tertiary creep
  3. Initial instantaneous strain
  4. Strain at fracture

Explanation

After primary creep, the creep rate reaches a minimum and remains nearly constant during secondary (steady-state) creep. This minimum rate is commonly used to compare creep resistance.

Question 20Materials properties and behavior

When two dissimilar metals are coupled in a corrosive electrolyte, the metal that corrodes is the one that is:

  1. Cathodic in the couple
  2. More noble
  3. Anodic in the couple Correct answer.
  4. More passive

Explanation

In a galvanic couple, the anodic member undergoes oxidation and corrodes while the cathodic member is protected. The more active metal is generally anodic when coupled.

Question 21Materials properties and behavior

A single crystal is loaded in tension. The Schmid factor for a slip system is maximized when the angle between the tensile axis and the slip plane normal, and the angle between the tensile axis and the slip direction, are:

  1. 30° and 30°
  2. 0° and 0°
  3. 45° and 45° Correct answer.
  4. 90° and 90°

Explanation

The Schmid factor is m = cos φ cos λ, where φ is the angle to the slip plane normal and λ is the angle to the slip direction. Its maximum value of 0.5 occurs when both angles are 45°.

Question 22Materials properties and behavior

In a diffusion couple between two metals of different intrinsic diffusivities, inert markers at the interface are observed to:

  1. Move perpendicular to the diffusion direction
  2. Move toward the faster-diffusing species Correct answer.
  3. Move toward the slower-diffusing species
  4. Remain fixed at the original interface

Explanation

This is the Kirkendall effect: the faster-diffusing species leaves a net vacancy flux in the opposite direction, causing the lattice on the faster-diffusing side to contract. The markers therefore shift toward the side of the faster diffuser.

Question 23Materials properties and behavior

For a valid plane-strain fracture toughness (KIC) measurement, the specimen thickness must be:

  1. Less than the grain size
  2. Equal to the specimen width
  3. Thick enough for plane strain Correct answer.
  4. Very thin relative to the plastic zone

Explanation

A valid KIC requires a thickness large enough to maintain plane-strain conditions and small-scale yielding at the crack tip. Thin specimens tend to give plane-stress toughness values rather than the conservative KIC.

Question 24Materials properties and behavior

In diffusional creep of a polycrystalline metal, Coble creep differs from Nabarro–Hering creep in that its creep rate depends on:

  1. Inverse square of grain size
  2. Directly with grain size
  3. Inverse cube of grain size Correct answer.
  4. Inverse square root of grain size

Explanation

Nabarro–Hering creep is controlled by lattice diffusion and varies as 1/d², while Coble creep is controlled by grain-boundary diffusion and varies as 1/d³. Coble creep therefore becomes more important at finer grain sizes and lower temperatures.

Question 25Materials properties and behavior

In FCC metals, a decrease in stacking-fault energy generally promotes which deformation mechanism over dislocation cross-slip?

  1. Grain boundary sliding
  2. Deformation twinning Correct answer.
  3. Cleavage fracture
  4. Diffusional creep

Explanation

Low stacking-fault energy increases the separation of partial dislocations, which inhibits cross-slip and favors planar slip and deformation twinning. High stacking-fault energy metals cross-slip more easily.

Question 26Materials properties and behavior

Which atomic packing factor is characteristic of a body-centered cubic (BCC) unit cell?

  1. 0.68 Correct answer.
  2. 0.52
  3. 0.34
  4. 0.74

Explanation

The BCC cell contains two atoms whose spheres touch along the body diagonal, giving an atomic packing factor of about 0.68. The more densely packed structures, FCC and HCP, reach 0.74, while simple cubic is 0.52.

Question 27Materials properties and behavior

In face-centered cubic metals, plastic slip occurs most readily on which crystallographic plane?

  1. {0001}
  2. {100}
  3. {110}
  4. {111} Correct answer.

Explanation

Slip occurs on the most densely packed plane, which in the FCC lattice is the octahedral {111} plane containing three close-packed directions. This gives four independent {111} planes and twelve slip systems in total.

Question 28Materials properties and behavior

Toughness of a material is best described as:

  1. The resistance of the surface to indentation
  2. The energy absorbed before fracture Correct answer.
  3. The stress at which permanent deformation begins
  4. The maximum stress the material sustains before necking

Explanation

Toughness measures the energy a material absorbs up to the point of fracture, combining strength and ductility. It is represented by the area under the stress–strain curve rather than by a single stress value.

Question 29Materials properties and behavior

Cold working a copper wire raises its strength mainly because:

  1. Solute atoms are added to the lattice
  2. Second-phase particles precipitate at grain boundaries
  3. Grain size decreases during the deformation
  4. Dislocation density rises and dislocations obstruct each other Correct answer.

Explanation

Strain hardening occurs because deformation multiplies dislocations, which then intersect, tangle and form barriers to further motion. A higher stress is therefore needed to continue slip.

Question 30Materials properties and behavior

A pipeline steel shows a Charpy absorbed energy that falls from 100 J at 20 °C to 10 J at −40 °C. For safe low-temperature service, the steel should be selected so that its ductile-to-brittle transition temperature is:

  1. Well above the lowest service temperature
  2. Well below the lowest service temperature Correct answer.
  3. Unrelated to the service temperature range
  4. Exactly equal to normal room temperature

Explanation

Below its transition temperature a ferritic steel absorbs little impact energy and can fail by cleavage. Keeping the transition temperature well below the coldest service temperature ensures the material remains on the upper shelf, where it is tough.

Question 31Materials properties and behavior

Which statement best explains why pure metals are generally better thermal conductors than ceramics?

  1. Ceramics contain no mobile charge carriers at any temperature
  2. Metals have stronger ionic bonds that transmit vibration more efficiently
  3. Free electrons in metals transport heat in addition to lattice vibrations Correct answer.
  4. Metals have lower melting points, so heat passes through them faster

Explanation

Heat in a solid is carried by lattice vibrations (phonons) and, where present, by free electrons. Metals contribute a large electronic term on top of the phonon term, giving them high thermal conductivity.

Question 32Materials properties and behavior

A metallurgist must measure the hardness of a carburized case only about 0.05 mm deep without influence from the softer core. Which test is most appropriate?

  1. Vickers test at 10 kgf
  2. Knoop microhardness test at 100 gf Correct answer.
  3. Rockwell C test at 150 kgf
  4. Brinell test with 10 mm ball, 3000 kgf

Explanation

To isolate a shallow surface layer, the indentation must be much smaller than the layer thickness, which requires a low-force microindentation test. A Knoop indenter at about 100 gf produces a very shallow impression confined to the case.

Question 33Materials properties and behavior

Copper is FCC with an atomic radius of 0.1278 nm and an atomic weight of 63.55 g/mol. Using Avogadro's number as 6.022 × 10^23, the theoretical density of copper is closest to:

  1. 8.9 g/cm³ Correct answer.
  2. 10.1 g/cm³
  3. 7.2 g/cm³
  4. 6.3 g/cm³

Explanation

For FCC, the lattice parameter is a = 4R/√2 = 0.3615 nm, giving a cell volume of 4.72 × 10^-23 cm³ with four atoms per cell. Dividing the mass of four atoms by that volume yields about 8.9 g/cm³.

Question 34Materials properties and behavior

A steel has a plane-strain fracture toughness of 50 MPa√m and will operate at a stress of 500 MPa. Taking Y = 1.0 and K = Yσ√(πa), the largest tolerable crack length is closest to:

  1. 3.2 mm Correct answer.
  2. 10 mm
  3. 1.0 mm
  4. 100 mm

Explanation

Rearranging the fracture mechanics relation gives a = (1/π)(K/σ)². Substituting K = 50 MPa√m and σ = 500 MPa gives a = (0.1)²/π = 3.2 × 10^-3 m.

Question 35Metals and alloys

Which of the following metals has a face-centered cubic crystal structure at room temperature?

  1. Chromium
  2. Vanadium
  3. Magnesium
  4. Aluminum Correct answer.

Explanation

Aluminum retains the face-centered cubic structure at room temperature. Chromium and vanadium are body-centered cubic, and magnesium is hexagonal close-packed. Crystal structure controls slip behavior and ductility.

Question 36Metals and alloys

Which reaction defines a eutectic transformation on cooling?

  1. A liquid solidifying into two solids at once Correct answer.
  2. A solid phase splitting into two solids
  3. A liquid separating into two immiscible liquids
  4. A liquid and a solid forming one solid

Explanation

A eutectic reaction converts a single liquid into two solid phases simultaneously at a fixed temperature and composition. A solid splitting into two solids is eutectoid, liquid plus solid forming one solid is peritectic, and one liquid separating into two liquids is monotectic.

Question 37Metals and alloys

Which cast iron contains graphite in the form of spheroids rather than flakes?

  1. Ductile iron Correct answer.
  2. Malleable iron
  3. White iron
  4. Gray iron

Explanation

Ductile iron is treated with magnesium or cerium so that graphite grows as spheroids, which blunt stress concentration and raise toughness. Gray iron contains flake graphite, white iron contains cementite rather than free graphite, and malleable iron contains temper carbon clusters.

Question 38Metals and alloys

What is the main purpose of tempering a quenched steel?

  1. To complete the austenite-to-martensite change
  2. To dissolve coarse alloy carbides
  3. To raise hardness above the as-quenched value
  4. To restore toughness and relieve stress Correct answer.

Explanation

Tempering reheats martensite below the lower critical temperature so that carbon can precipitate and residual stresses relax. This trades some hardness for improved toughness and ductility. Tempering does not increase hardness above the as-quenched level.

Question 39Metals and alloys

Which aluminum alloy designation is age-hardenable?

  1. 1100
  2. 7075 Correct answer.
  3. 3003
  4. 5052

Explanation

The 7xxx series, such as 7075, is a heat-treatable Al-Zn-Mg-Cu alloy that responds to solution treatment and aging. The 1xxx, 3xxx, and 5xxx series are not strengthened by precipitation hardening.

Question 40Metals and alloys

Brass is primarily an alloy of copper with which element?

  1. Nickel
  2. Beryllium
  3. Tin
  4. Zinc Correct answer.

Explanation

Brass is the copper-zinc alloy family. Bronze is primarily copper-tin, cupronickel is copper-nickel, and beryllium copper is a separate precipitation-hardenable alloy.

Question 41Metals and alloys

The recrystallization temperature of a pure metal is approximately what fraction of its absolute melting temperature?

  1. 0.4 Tm Correct answer.
  2. 0.1 Tm
  3. 0.2 Tm
  4. 0.8 Tm

Explanation

Recrystallization typically occurs in the range of about 0.3 to 0.5 of the absolute melting temperature. It is not a fixed temperature but depends on purity, strain, and time. Solutes and second phases generally raise the recrystallization temperature.

Question 42Metals and alloys

Which stainless steel family is ferromagnetic and hardened mainly by cold work rather than by heat treatment?

  1. Ferritic Correct answer.
  2. Martensitic
  3. Austenitic
  4. Duplex

Explanation

Ferritic stainless steels are body-centered cubic, ferromagnetic, and not hardenable by quenching because they do not transform to austenite on heating. They gain strength mainly through cold work. Martensitic grades are also ferromagnetic but are hardened by heat treatment.

Question 43Metals and alloys

On an isothermal transformation diagram for eutectoid steel, the nose of the C-curve indicates the...

  1. The equilibrium eutectoid temperature
  2. The temperature at which martensite starts
  3. The temperature of the shortest incubation time Correct answer.
  4. The temperature where bainite becomes unstable

Explanation

The nose of the C-curve is the temperature at which the incubation time before transformation is shortest and the transformation rate is fastest. It sets the critical cooling rate needed to miss the pearlite and bainite regions. The martensite start temperature and the eutectoid temperature are separate features.

Question 44Metals and alloys

The Jominy end-quench test is used to evaluate which property of steel?

  1. Weldability
  2. Hardenability Correct answer.
  3. Resistance to tempering
  4. Creep strength

Explanation

The Jominy test quenches one end of a standard bar and measures hardness along its length. The hardness profile indicates how deeply the steel can harden, which is its hardenability. It does not directly measure temper resistance, weldability, or creep strength.

Question 45Metals and alloys

Which sequence correctly describes precipitation hardening of an aluminum alloy?

  1. Age, solution treat, quench
  2. Solution treat, age, quench
  3. Quench, age, solution treat
  4. Solution treat, quench, age Correct answer.

Explanation

Precipitation hardening first dissolves the alloying elements at a high solution temperature, then quenches to retain a supersaturated solid solution, and finally ages to form fine precipitates. Aging after quenching provides the strengthening precipitates. Quenching after aging would not produce the required supersaturated condition.

Question 46Metals and alloys

At the eutectic point of a binary system under constant pressure, how many degrees of freedom remain?

  1. Zero Correct answer.
  2. Two
  3. One
  4. Three

Explanation

With pressure fixed, the Gibbs phase rule gives F = C - P + 1. At a binary eutectic, C equals 2 and P equals 3, so F equals 0 and the eutectic temperature and composition are fixed.

Question 47Metals and alloys

In steel welding, increasing the carbon equivalent generally...

  1. Raises toughness in the weld metal
  2. Raises hardenability and cold-cracking risk Correct answer.
  3. Lowers the martensite start to room level
  4. Lowers hardenability and cold-cracking risk

Explanation

Carbon equivalent combines carbon with other alloying elements that increase hardenability. Higher values promote martensite formation in the heat-affected zone and raise susceptibility to hydrogen-assisted cold cracking, so preheat and limit controls are used.

Question 48Metals and alloys

Sensitization of austenitic stainless steel is caused by...

  1. Loss of nickel by selective evaporation
  2. Formation of sigma phase throughout the core
  3. Growth of ferrite at prior austenite boundaries
  4. Chromium carbide at grain boundaries Correct answer.

Explanation

In the sensitization range, chromium carbides precipitate at grain boundaries and deplete the adjacent matrix of chromium. The chromium-depleted zones then lose corrosion resistance. Low-carbon grades or stabilized grades are used to limit this effect.

Question 49Metals and alloys

The main alloying element in aluminum alloy 2024 is...

  1. Silicon
  2. Magnesium
  3. Zinc
  4. Copper Correct answer.

Explanation

Alloy 2024 is an Al-Cu-Mg alloy in the 2xxx series and is strengthened by precipitation hardening. Zinc dominates the 7xxx series, magnesium is important in 5xxx and 6xxx alloys, and silicon is central to many casting alloys.

Question 50Metals and alloys

Spheroidizing annealing is performed mainly to...

  1. Coalesce carbides to improve machinability Correct answer.
  2. Maximize hardness for improved wear resistance
  3. Refine the austenite grain size
  4. Produce a martensitic surface layer

Explanation

Spheroidizing holds steel near the lower critical temperature so that lamellar or network carbides coalesce into spheroids. The soft ferrite-carbide microstructure improves machinability and formability. It is not a hardening treatment.