Q1 (a) Define coordination number for any crystal structure.

From Unit 1: Crystallography

• Coordination number is the number of nearest neighboring atoms directly touching a central atom in the crystal lattice.
• Example: FCC = 12, BCC = 8, SC = 6.

Q1 (b) How ceramics differ from alloys?

From Unit 6: Advanced Materials

• Ceramics: Inorganic, non-metallic, ionic/covalent bonding, brittle, high hardness, high melting point.
• Alloys: Metallic mixtures, metallic bonding, ductile, good conductivity.

Q1 (c) How is Young’s modulus important for engineering materials?

From Unit 2: Deformation

• Young’s modulus measures stiffness; higher value means material resists elastic deformation better.
• Critical for structural applications to avoid excessive deflection.

Q1 (d) How does Vickers indenter differ from Brinell indenter?

From Unit 3: Fracture & Testing

• Vickers: Diamond pyramid indenter, square impression, suitable for hard/thin materials.
• Brinell: Steel/tungsten ball, circular impression, better for softer materials.

Q1 (e) Illustrate the fatigue limit with the help of S-N curve.

From Unit 3: Fracture

• S-N curve shows stress vs cycles to failure.
• Fatigue limit: Horizontal asymptote (ferrous materials) below which no failure occurs even at infinite cycles.

Q1 (f) For what types of materials does normal stress theory of failure apply and why?

From Unit 3: Fracture

• Applies to brittle materials (cast iron, ceramics).
• Failure occurs when maximum normal stress exceeds ultimate tensile strength.

Q1 (g) Draw a characteristic unary phase diagram by showing phases on it.

From Unit 4: Phase Diagrams

• Unary diagram: Single component, pressure vs temperature.
• Shows solid, liquid, gas phases and triple/sublimation points.

Q1 (h) Illustrate Peritectic reaction with respect to Fe-C system.

From Unit 4: Phase Diagrams

• Peritectic: Liquid + Solid1 → Solid2 at constant temperature.
• In Fe-C: δ-ferrite + Liquid → Austenite at 1493°C.

Q1 (i) Show Austempering process on an isothermal TTT curve.

From Unit 5: Phase Transformations

• Rapid cooling to bainite region, hold isothermally.
• Results in bainitic structure (no pearlite/martensite).

Q1 (j) Differentiate between brass and bronze on the basis of their composition.

From Unit 6: Advanced Materials

• Brass: Cu + Zn alloy.
• Bronze: Cu + Sn (or other elements like Al, Si) alloy.

Q2: Unit cell parameters and seven crystal systems

From Unit 1: Crystallography

• Explain a, b, c, α, β, γ.
• Draw and list parameters for cubic, tetragonal, orthorhombic, hexagonal, rhombohedral, monoclinic, triclinic.

Q3 (a): True vs Engineering stress-strain

From Unit 2: Deformation

• Engineering: Nominal area/load.
• True: Instantaneous area/load; accounts for necking.

Q3 (b): Rockwell vs Brinell hardness

From Unit 3: Testing

• Rockwell: Direct reading, minor/major load, suitable for finished parts.
• Brinell: Ball indenter, measures diameter, better for coarse materials.

Q4: NDT advantages and methods

From Unit 3: Testing

• Advantages: Non-destructive, in-service inspection.
• Magnetic particle: Detects surface/sub-surface defects in ferromagnetic materials.
• Eddy current: Conductivity changes detect cracks in conductors.

Q5: Cooling of 0.5% C steel using Fe-C diagram

From Unit 4 & 5

• Hypoeutectoid steel.
• Liquid → Austenite → Proeutectoid ferrite + Pearlite at room temperature.

Q6 (a) Why is heat treatment needed in case of alloys? Differentiate between isothermal TTT curve and continuous cooling curve. (10)

From Unit 5: Phase Transformations & Heat Treatment

• Heat treatment needed: To control microstructure, improve mechanical properties (strength, toughness, ductility), relieve stresses, refine grains in alloys.
• TTT (Isothermal): Transformation at constant temperature, two distinct noses (pearlite & bainite).
• CCT (Continuous Cooling): Realistic cooling rates, curves shifted right, overlapping regions, no separate bainite nose in many steels.

Q6 (b) How is flame hardening different from carburizing? (5)

From Unit 5: Heat Treatment

• Flame hardening: Surface hardening by rapid heating (oxy-acetylene flame) + quenching; no composition change, for medium carbon steels.
• Carburizing: Diffusion of carbon into surface at high temperature; increases surface carbon content for hardening on quenching.

Q7 Write notes on:

From Unit 6: Advanced Engineering Materials

(i) Stainless steels and their types. (5)

• Fe-Cr alloys (>10.5% Cr) with corrosion resistance.
• Types: Ferritic (magnetic, good ductility), Austenitic (non-magnetic, excellent corrosion), Martensitic (hardenable, magnetic), Duplex, Precipitation hardening.

(ii) Types of cast iron. (5)

• Gray (flake graphite, good machinability), White (cementite, hard/brittle), Ductile (nodular graphite, tough), Malleable (tempered white, ductile), Compacted graphite.

(iii) Nickel based super-alloys. (5)

• High temperature strength, creep/oxidation resistance.
• Applications: Turbine blades, jet engines.
• Strengthening: Solid solution (Co, Cr, Mo), precipitation (γ’ phase Ni3(Al,Ti)).

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