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

(a) Tempering of hardened steel

Purpose: tempering follows quenching to relieve internal stresses, reduce brittleness, restore toughness and ductility, stabilise dimensions, and set the hardness–strength balance needed in service.

Principle: quenched steel is mainly body-centred tetragonal martensite, a carbon-supersaturated and highly strained phase, often with some retained austenite. Reheating below the lower critical temperature A₁ (727°C for plain carbon steel) lets carbon diffuse out of the lattice and form carbides, so the structure moves towards ferrite plus fine cementite (tempered martensite). A higher temperature gives coarser carbides, lower hardness and higher toughness.

Heating requirement: steel is held at a chosen temperature, normally 150–650°C and always below A₁, for about an hour per 25 mm section, then cooled in air.

Microstructural changes with temperature:

• 100–250°C: carbon precipitates as ε-carbide, and tetragonality falls.
• 200–300°C: retained austenite decomposes to ferrite and cementite.
• 250–400°C: ε-carbide gives way to Fe₃C and the martensite lattice becomes cubic ferrite.
• 400–650°C: cementite coarsens and spheroidises, giving maximum toughness and minimum hardness.

Typical use: 150–250°C for cutting tools, 350–450°C for springs, 500–650°C for shafts and structural parts. Tempering between about 260–370°C can embrittle some steels and is avoided.

The schematic is a plot of hardness against tempering temperature: a falling curve from the as-quenched value, with the as-quenched structure at the top and spheroidised cementite at the bottom.

(b) Shaft diameter

Allowable shear stress τ = 500/6 = 83.33 MPa; allowable tensile stress σ = 700/6 = 116.67 MPa. For a ductile shaft use the maximum shear stress theory.

Equivalent torque: Tₑ = √(M² + T²) = √(2500² + 8000²) = 8381.5 N·m.

Then 16Tₑ/(πd³) = τ gives

d³ = 16 × 8381.5/(π × 83.33 × 10⁶) = 5.122 × 10⁻⁴ m³, so d = 0.0800 m.

Check by the maximum normal stress theory: Mₑ = ½(M + Tₑ) = 5440.8 N·m, and 32Mₑ/(πd³) = σ gives d = 78.0 mm. The shear-stress criterion gives the larger value and governs.

Shaft diameter ≈ 80 mm.

(c) Coefficient of fluctuation of speed

A four-stroke cycle is 4π rad, and ω = 2π × 250/60 = 26.18 rad/s.

Mean torque Tₘ = P/ω = 20 000/26.18 = 763.9 N·m. Work per cycle = Tₘ × 4π = 9600 J.

Let W_e = 3W_c, with W_e − W_c = 9600 J. So W_c = 4800 J and W_e = 14 400 J.

Each stroke spans π rad. A triangular expansion curve of area W_e has peak torque T_e = 2W_e/π = 9167 N·m. Suction and exhaust contribute nothing, and compression is a negative triangle with peak 3056 N·m.

Only the part of the expansion triangle above the mean line gives surplus energy. The similar triangle above Tₘ has height T_e − Tₘ and base π(T_e − Tₘ)/T_e, so

ΔE = ½π(T_e − Tₘ)²/T_e = ½ × (28 800 − 2400)²/28 800 = 12 100 J.

(A full cumulative-energy check over the cycle gives the same maximum fluctuation.)

I = mk² = 1500 × 0.6² = 540 kg·m².

δ = ΔE/(Iω²) = 12 100/(540 × 26.18²) = 0.0327.

Coefficient of fluctuation of speed ≈ 0.0327 (3.27%).