Material properties and processing is 7–11 questions and reads more like a vocabulary test than the areas around it: fewer long calculations, more questions that turn on one precise word such as yield, ultimate, hardness or toughness.
Updated Oct 29, 2026 · Not affiliated with NCEES.
ExamFE Mechanical
questions on the exam7–11
worked problems here15
of 14 knowledge areas9
Checked
What the area covers
The specification lists ten topics. Grouped:
Properties: chemical, electrical, mechanical, physical and thermal, and reading them off a stress-strain diagram.
Metals: ferrous and nonferrous alloys, and what alloying and processing change.
Engineered materials: polymers and composites.
Processing: manufacturing processes, phase diagrams, phase transformations and heat treating.
Selection: choosing a material for a load, temperature or environment.
Degradation and failure: corrosion, and thermal, fatigue, fracture and creep failure.
15 materials problems, every option explained
0 / 15 answered · 0 correct
Question 1 of 15 · Material Properties and Processing · 7–11 items
A large structural steel plate has a fracture toughness of KIC = 65 MPa·√m and carries a nominal tensile stress of 250 MPa. Assuming a geometry factor Y = 1.12, what critical surface crack length would cause the component to fail by rapid brittle fracture?
Given
Fracture toughness KIC = 65 MPa·√m
Nominal stress σ = 250 MPa
Geometry factor Y = 1.12
Find
Critical surface crack length a (mm)
Solution
KIC = Yσ√(πac) → ac = (1/π)(KIC/(Yσ))2
ac = (1/π)(65/(1.12 × 250))2 = 0.01715 m ≈ 17.2 mm
FE Reference Handbook, Materials Science: fracture toughness.
This applies Y twice, (KIC/(Y2σ))2/π; the geometry factor multiplies the stress only once.
ac = (1/π)(KIC/(Yσ))2 = (1/π)(65/(1.12 × 250))2 = 0.01715 m ≈ 17.2 mm.
This leaves out the geometry factor Y = 1.12, which makes a surface crack more severe than an embedded one.
This is 2ac; a surface crack length is a, not the 2a used for a through-crack in an infinite plate.
This forgets to divide by π; K = Yσ√(πa) puts π under the square root with a.
Question 2 of 15 · Material Properties and Processing · 7–11 items
A high-carbon steel component with complex geometry and sharp internal fillets is heated to the austenite region and subjected to a severe water quench to maximize hardness. Post-treatment inspection reveals extensive surface and internal cracking. Which of the following best explains the primary metallurgical cause of this quench cracking?
Solution
Quench cracking: thermal gradients plus the volume increase of austenite → martensite, transforming at different times in surface and core, create residual tensile stress concentrated at sharp corners.
Hydrogen embrittlement is delayed cracking from absorbed hydrogen (plating, welding); a plain water quench is not its source.
A severe water quench suppresses pearlite; soft pearlite would make the part less prone to cracking, not more.
Decarburization softens the surface and lowers its hardenability; it does not create the stresses that crack a quenched part.
Grain-boundary carbides form during slow cooling or holding in a sensitizing range, not in a rapid quench.
Austenite expands on transforming to martensite; in a severe quench the surface transforms first and the core later, so the expansion mismatch sets up tensile stresses that crack at sharp fillets.
Question 3 of 15 · Material Properties and Processing · 7–11 items
A high-performance machine shaft subjected to severe cyclic bending and occasional impact loading requires an optimum balance of high ultimate strength and high toughness. Which sequence of heat treatments will best achieve the necessary microstructural characteristics for this application?
Solution
Quench and temper: austenitize → quench (martensite, maximum hardness) → temper (tempered martensite, strength with toughness).
Untempered martensite is very strong but brittle, a poor choice for a shaft that sees impact.
Heating below the eutectoid temperature never forms austenite, so air cooling cannot harden the steel.
This is a full anneal: coarse pearlite gives ductility but too little strength for the shaft.
Spheroidizing is done below the austenite range; quenching from there cannot form martensite, so the strength gain is lost.
Quenching from austenite forms hard martensite, and tempering trades some of that hardness for the toughness needed under impact and cyclic loads.
Question 4 of 15 · Material Properties and Processing · 7–11 items
A structural carbon steel component is bolted to a stainless steel 304 fitting in continuously submerged seawater. To mitigate galvanic corrosion of the carbon steel, which strategy is most effective based on the galvanic potentials below?
Metal / alloy
Anodic index (V) vs SCE
Magnesium alloy
−1.75
Zinc
−1.20
Carbon steel
−0.85
Aluminum 2024-T4
−0.60
Stainless steel 304 (passive)
−0.08
Solution
In a galvanic couple the more active metal is the anode. A sacrificial anode must be more active than the metal it protects: Zn (−1.20 V) < steel (−0.85 V) < Al 2024 (−0.60 V) < SS 304 (−0.08 V).
A metallic bond only improves the galvanic circuit; the steel stays the anode and corrodes faster.
At −0.60 V aluminum 2024 is more noble than carbon steel, so the steel would remain the anode.
Coating only the anode leaves small holidays facing a large bare cathode, which concentrates attack at the defects.
Zinc (−1.20 V) is more active than carbon steel (−0.85 V), so it corrodes in place of the steel and protects the whole couple.
Question 5 of 15 · Material Properties and Processing · 7–11 items
A carbon steel bolt with an exposed surface area of 2 cm2 connects two copper plates with a total exposed surface area of 200 cm2 immersed in seawater. The corrosion rate is limited by oxygen reduction on the copper, which has a limiting current density of 10 μA/cm2. Assuming uniform corrosion of the bolt, a steel density of 7.87 g/cm3, and oxidation to Fe2+ (M = 55.85 g/mol), what is the approximate corrosion penetration rate of the bolt? (Use F = 96,500 C/mol.)
FE Reference Handbook, Materials Science: corrosion (Faraday's law).
This uses the 10 μA/cm2 cathodic density directly on the bolt; the large cathode to small anode area ratio multiplies it by 100.
This is off by a factor of 10, typically from a cm → mm conversion slip in the penetration rate.
This assumes oxidation to Fe3+ (n = 3); the stem specifies Fe2+, n = 2.
All cathodic current from 200 cm2 of copper (2 mA) is drawn from a 2 cm2 bolt, so ia = 1 mA/cm2 and the penetration rate is about 11.6 mm/yr.
This uses n = 1 electron per iron atom; Fe → Fe2+ releases two electrons.
Question 6 of 15 · Material Properties and Processing · 7–11 items
Sections of an austenitic stainless steel (Standard Type 304) pipe system used in a chloride-rich environment were recently welded. Within a few months, severe intergranular corrosion is observed strictly in the heat-affected zones (HAZ) of the welds. Which of the following material selection changes would most effectively prevent this specific corrosion mode in future installations without requiring post-weld heat treatment?
Solution
Weld decay: at 425–870 °C (800–1600 °F) chromium carbides form on grain boundaries and deplete chromium next to them. Low-carbon (L) or stabilized grades prevent it.
A coating is not a material selection change and does not remove the chromium-depleted zones that cause the attack.
Martensitic grades have lower chromium and poorer corrosion resistance, and they are prone to HAZ cracking.
More carbon forms more chromium carbide at grain boundaries and worsens sensitization.
This is a welding-procedure change, not a material change, and the HAZ still passes through the sensitizing range.
With 0.03 % carbon or less, too little carbon is available to form chromium carbides in the HAZ, so the grain boundaries keep their chromium.
Question 7 of 15 · Material Properties and Processing · 7–11 items
Which material property represents the amount of energy per unit volume that a material can absorb before rupturing?
Solution
Modulus of toughness = ∫σ dε from 0 to fracture strain; modulus of resilience = σy2/(2E).
Hardness is resistance to surface indentation and says nothing directly about energy to fracture.
Toughness is the total area under the stress-strain curve up to fracture, the energy per unit volume absorbed before rupture.
Resilience counts only the elastic part of the curve, energy that is recovered on unloading; it stops at yield, not at rupture.
Stiffness is resistance to elastic deformation (the modulus), not absorbed energy.
Question 8 of 15 · Material Properties and Processing · 7–11 items
How does a continuous cooling transformation (CCT) diagram differ from a time-temperature-transformation (TTT) diagram in predicting the microstructural evolution of alloy steels?
Solution
TTT (isothermal transformation): quench to a temperature and hold. CCT: cool continuously at a given rate; curves shift down and to the right relative to TTT.
CCT start and finish curves are measured transformation kinetics; nucleation is built into them.
A TTT diagram is built from isothermal holds, while a CCT diagram follows cooling curves, which is how real parts are quenched.
Transformations during continuous cooling start later and at lower temperatures than in isothermal holds, so the diagrams differ.
This reverses the definitions: TTT means isothermal transformation.
Question 9 of 15 · Material Properties and Processing · 7–11 items
In a fiber-reinforced composite material, what is the primary role of the matrix phase?
Solution
Fiber composite: fibers carry load; matrix binds, distributes and transfers load, and protects the fibers.
Weight savings come from the whole composite, and the matrix is not there to lower mass.
Density is a consequence of the constituents, not a design role; the matrix is usually the lighter phase.
The matrix holds the fibers in place, protects them, and passes load into them by shear at the interface.
In a fiber composite the stiff, strong fibers carry most of the load.
Question 10 of 15 · Material Properties and Processing · 7–11 items
What is the primary mechanism by which precipitation hardening strengthens an alloy?
Precipitation hardening adds a second phase; it relies on solute atoms, not on removing them.
Larger grains lower strength (Hall-Petch); precipitation hardening is a separate mechanism.
Fewer grain boundaries means fewer barriers to slip and lower strength.
Fine, coherent precipitates formed during aging act as obstacles that dislocations must cut through or bow around, raising strength.
Question 11 of 15 · Material Properties and Processing · 7–11 items
What is the primary factor that determines the glass transition temperature (Tg) of a polymer?
Solution
Tg rises with backbone stiffness, bulky or polar side groups and cross-linking; flexible chains (Si–O, C–O) have low Tg.
Tg is where chain segments gain enough mobility to rotate; stiff backbones and bulky side groups raise it, flexible chains lower it.
Density follows from packing and composition; it is not what sets segmental mobility.
Crystallinity affects how pronounced the glass transition is, but Tg is a property of the amorphous regions.
Molecular weight affects Tg only weakly at high values, through chain ends; it is not the main factor.
Question 12 of 15 · Material Properties and Processing · 7–11 items
Which microstructural mechanism in advanced ceramic-matrix composites is most effective at improving fracture toughness?
Solution
Ceramic-matrix composites gain toughness from crack deflection, fiber bridging and pull-out at the crack wake.
Finer grains raise strength somewhat, but grain boundaries alone do little for ceramic fracture toughness.
Crystallinity does not by itself blunt or shield cracks in a brittle matrix.
Fibers or whiskers deflect the crack along tortuous paths and bridge it behind the tip, so more energy is needed to extend it.
Uniform grains reduce flaw size scatter, but they do not add a toughening mechanism.
Question 13 of 15 · Material Properties and Processing · 7–11 items
In high-temperature applications, why is creep resistance often prioritized over static yield strength during material selection?
Solution
Above about 0.4 Tm, creep (time-dependent strain under constant stress) can cause failure below the yield strength.
At high temperature a part can keep stretching over time at stresses below yield, so long-term deformation or rupture governs instead of the short-term yield strength.
Thermal conductivity is unrelated to the choice between creep and yield criteria.
Yield strength still limits short-term load capacity; creep adds a time-dependent limit on top of it.
Yield strength and elastic modulus are separate properties; strong alloys are not inherently less stiff.
Question 14 of 15 · Material Properties and Processing · 7–11 items
How does phase transformation induced toughening improve the fracture resistance in ceramics?
Solution
Transformation toughening (e.g., ZrO2): tetragonal → monoclinic near the crack tip, volume expansion puts the tip in compression.
Transformation toughening raises fracture toughness without relying on lower hardness.
The ceramic remains a ceramic; only its crystal structure changes.
In partially stabilized zirconia, stress at the crack tip triggers the tetragonal-to-monoclinic change; its 3–5 % volume increase squeezes the crack shut.
Porosity lowers strength and is not the mechanism; transformation toughening works through a phase change.
Question 15 of 15 · Material Properties and Processing · 7–11 items
Which heat treatment process involves heating a material to a suitable temperature, holding it at that temperature, and then cooling it very slowly to reduce hardness and increase ductility?
Full annealing heats steel into the austenite range, holds it, and cools it slowly in the furnace, producing coarse pearlite with low hardness and high ductility.
Case hardening enriches and hardens the surface layer; it raises hardness rather than lowering it.
Work hardening is strengthening by plastic deformation; it involves no heating cycle and lowers ductility.
Quenching is the opposite: rapid cooling that forms hard, brittle martensite.
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Common traps
Most of this area sits in the Materials Science/Structure of Matter chapter of the FE Reference Handbook; Callister and Rethwisch's Materials Science and Engineering: An Introduction covers the same ground. Where candidates slip:
Three strengths on one curve. Yield, ultimate and fracture strength are different points. For metals without a sharp yield point, yield is read with the 0.2 % offset line, parallel to the elastic slope.
Ductility has two measures. Percent elongation and percent reduction of area are both ductility, and they are not interchangeable.
The lever rule reads the opposite arm. On a two-phase field, the fraction of one phase is the length of the tie line on the far side of the overall composition, divided by the whole tie line.
An endurance limit is a steel habit. Most steels show a stress below which fatigue life is very long; aluminum and copper alloys keep losing strength with cycles, so their fatigue strength is quoted at a stated number of cycles.
Thermoplastic or thermoset. Thermoplastics soften on reheating and can be remelted; thermosets cross-link and do not. The Handbook's polymer section uses this classification.
Heat capacity per what. Specific heat is per unit mass; heat capacity is for the whole body. Mixing the two breaks the units.
Read the phase diagram at the right temperature
Phase fractions change with temperature. A question about a slowly cooled steel just below the eutectoid line and one at room temperature can give different answers on the same diagram.
Concrete, asphalt, soils and wood are civil materials and live in the civil discipline's materials area. Corrosion and chemistry of process equipment belong to FE Chemical's process design.
Yes, by name in the specification. Expect questions on what a process does to the material, such as cold work raising strength and lowering ductility, rather than on machine settings.
How much materials science is on the FE Mechanical exam?
7–11 questions, the same range as heat transfer. That is fewer than any of the mechanics areas, fluid mechanics or thermodynamics, and more than measurements and controls.
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