(a) What is the purpose of tempering of hardened steel? Explain the principle of tempering using suitable schematics including…

(a) Tempering is a heat treatment given immediately after hardening (quenching). Its purpose is to relieve internal stresses, reduce brittleness, improve toughness and ductility, stabilise the size and structure, and adjust the hardness/strength to the required service condition.

Principle: hardened steel consists mainly of tetragonal martensite, a supersaturated and highly strained solution of carbon in α-iron, often with retained austenite. On reheating to a temperature below the lower critical temperature A₁, carbon atoms diffuse and precipitate as carbides. The structure gradually becomes tempered martensite: fine cementite particles in a ferrite matrix. Higher tempering temperature gives coarser carbides, lower hardness and higher toughness.

Heating requirement: tempering is done below A₁; for plain carbon steel A₁ ≈ 727°C, but practical tempering range is 150–650°C. It is held at the chosen temperature and then cooled, usually in air.

Schematic: As-quenched martensite (BCT, stressed) → reheat 150–650°C, below A₁, hold → air cool → tempered martensite: α-ferrite + fine Fe₃C.

Microstructural stages:

• 100–250°C: carbon rejects as ε-carbide; low-temperature tempered martensite.
• 200–300°C: retained austenite transforms to ferrite and cementite.
• 250–400°C: ε-carbide converts to Fe₃C; martensite loses tetragonality.
• 400–650°C: carbide coarsening/spheroidisation; toughness increases.

Low tempering (150–250°C) is used for cutting tools; medium (350–450°C) for springs; high (500–650°C) for structural parts needing toughness.

(b) Use the maximum shear stress theory for a ductile shaft. Allowable shear stress: τ_allow = 500/6 = 83.33 MPa. Allowable tensile stress: σ_allow = 700/6 = 116.67 MPa.

Given: M = 2500 N-m, T = 8000 N-m.

Equivalent torque: Te = √(M² + T²) = √(2500² + 8000²) = 500√281 N-m = 8381.53 N-m.

For a solid circular shaft under equivalent torque: τ = 16 Te/(π d³) ≤ τ_allow. So d³ = 16 Te/(π τ_allow) = 16 × 500√281 / [π × (500/6) × 10⁶] = 96√281/(π × 10⁶) m³. Thus d = ∛(96√281/π)/100 m = 0.08001 m ≈ 80.01 mm.

Check by maximum normal stress theory: Me = ½[M + √(M² + T²)] = ½(2500 + 500√281) = 1250 + 250√281 = 5440.76 N-m. σ = 32 Me/(π d³) ≤ σ_allow. This gives d ≈ 78.03 mm.

The shear criterion is more conservative and governs. Diameter of shaft ≈ 80 mm.

(c) For a four-stroke engine, one cycle = 2 revolutions = 4π rad. Mean angular speed: ω = 2πN/60 = 2π × 250/60 = 25π/3 rad/s.

Mean torque: T_m = P/ω = 20,000 ÷ (25π/3) = 2400/π = 763.94 N-m.

Work per cycle: W = T_m × 4π = (2400/π) × 4π = 9600 J.

Let W_c = work done on gases during compression, W_e = work done by gases during expansion. Given W_e = 3W_c and W_e − W_c = 9600 J. So 2W_c = 9600 ⇒ W_c = 4800 J, W_e = 14400 J.

Each stroke is π rad. For triangular turning moment, peak expansion torque: T_e = 2W_e/π = 28800/π = 9167.3 N-m. peak compression torque: T_c = −2W_c/π = −9600/π = −3055.8 N-m.

The mean torque line is T_m = 2400/π. The maximum energy fluctuation occurs during the expansion stroke between the two points where the triangular curve crosses the mean torque line.

Take θ from start of expansion. Rising side: T = (2T_e/π)θ = (57600/π²)θ. Set T = T_m: (57600/π²)θ = 2400/π ⇒ θ₁ = π/24. By symmetry, θ₂ = π − π/24 = 23π/24.

Net expansion work relative to mean: W_e − T_mπ = 14400 − 2400 = 12000 J.

At each end, the small area below the mean line is: D = T_m(π/24) − ½(57600/π²)(π/24)² = (2400/π)(π/24) − (28800/π²)(π²/576) = 100 − 50 = 50 J.

Total area below mean within expansion = 2D = 100 J. Therefore surplus area above mean: ΔE = 12000 + 100 = 12100 J.

Moment of inertia: I = m k² = 1500 × (0.6)² = 540 kg-m².

Coefficient of fluctuation of speed: δ = ΔE/(Iω²) = 12100/[540 × (25π/3)²] = 12100/(37500π²) = 121/(375π²) = 0.03269.

δ ≈ 0.0327 = 3.27%.