37 solved GATE ME questions on Engineering Materials, drawn from 13 exam years and grouped by year. Every question shows the official answer and a step-by-step solution.
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Connect crystal structure and phase diagrams to mechanical properties and heat-treatment outcomes. These selected foundations cover crystallography, the iron-carbon system, and basic material testing. They are not a complete syllabus; composites, polymers, and advanced alloys need separate study.
Our study notes and original examples support the PYQs below; they are not official exam questions or a replacement for the current syllabus.
Common metallic structures: BCC (e.g., α-iron), FCC (e.g., γ-iron, aluminum), and HCP (e.g., zinc). BCC has 2 atoms per unit cell, FCC has 4. Atomic packing factor is 0.68 for BCC and 0.74 for FCC. Point defects (vacancies), line defects (dislocations), and planar defects (grain boundaries) control plastic behavior.
Check yourself: Which crystal structure does the metal have at the temperature in question?
A binary phase diagram shows stable phases as functions of composition and temperature. In a two-phase region, the lever rule gives weight fractions: fraction of phase α = (C₀ - C_β)/(C_α - C_β), where C₀ is overall composition and C_α, C_β are the compositions at the phase boundaries (tie-line endpoints).
Check yourself: Are you reading compositions from the correct tie-line endpoints at the given temperature?
Key phases: ferrite (α, BCC, low C solubility ≈ 0.022 wt%), austenite (γ, FCC, up to 2.14 wt% C), and cementite (Fe₃C, 6.67 wt% C). Eutectoid at 727 °C and 0.76 wt% C: austenite transforms to pearlite (ferrite + cementite). Hypoeutectoid steels (< 0.76% C) have proeutectoid ferrite; hypereutectoid steels have proeutectoid cementite.
Check yourself: Is the carbon content above or below the eutectoid composition?
Quenching austenite produces martensite (hard, brittle, BCT). Tempering reduces hardness and increases toughness. Annealing (slow cooling) produces equilibrium phases. Normalizing (air cooling) produces a finer pearlite than annealing. The TTT diagram shows transformation products as functions of time and temperature.
Check yourself: Was the cooling rate fast enough to suppress equilibrium transformations?
Tensile test yields: elastic modulus E (slope), yield strength σ_y (0.2% offset), ultimate tensile strength σ_ut, and percent elongation (ductility). Hardness tests (Brinell, Rockwell, Vickers) measure resistance to indentation. Hardness correlates with tensile strength for steels. Impact tests (Charpy, Izod) measure energy absorbed during fracture.
Check yourself: Does the test measure strength, ductility, hardness, or toughness?
Original teaching example · not a PYQ
Original example: a plain-carbon steel has 0.40 wt% C. At just below the eutectoid temperature (727 °C), determine the weight fractions of proeutectoid ferrite and pearlite. Use the Fe-C diagram values: ferrite boundary at 0.022 wt% C, eutectoid at 0.76 wt% C.
Proeutectoid ferrite ≈ 48.8%; pearlite ≈ 51.2%.
An FCC unit cell has a lattice parameter a = 0.361 nm (aluminum). Calculate the atomic radius assuming atoms touch along the face diagonal.
r ≈ 0.1276 nm.
In FCC, the face diagonal = 4r = a√2. So r = a√2/4 = 0.361 x 1.4142/4 = 0.5105/4 ≈ 0.1276 nm.
Supplemental reading, not an official GATE reading list or an endorsement of these notes.
Free supplemental lectures covering crystal structures, bonding, phase diagrams, and material properties. Provides the atomic-level foundation for engineering materials.
This page shows 37 recent questions from the released archive, newest first. For older questions and complete papers, browse all GATE ME papers. Questions can carry more than one subject tag; counts are not marks weightage.
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