Engineering Materials and Manufacturing

GATE ME syllabus for materials and manufacturing, with the iron-carbon diagram and heat treatment, casting (Chvorinov's rule, gating), metal forming, welding, machining (Merchant's theory, Taylor's tool life), non-traditional machining, metrology and CIM, with solved examples.

Syllabus

  • Engineering materials: structure and properties, phase diagrams, heat treatment, stress-strain diagrams.
  • Casting, forming and joining: casting types, patterns, moulds and cores, solidification, riser and gating design; plastic deformation and yield criteria, hot and cold working, forging, rolling, extrusion, drawing, sheet metal operations, powder metallurgy; welding, brazing, soldering, adhesive bonding.
  • Machining and machine tools: mechanics of machining, cutting tools, tool geometry and materials, tool life and wear, machining economics, non-traditional machining, jigs and fixtures, abrasive machining, NC/CNC programming.
  • Metrology and inspection: limits, fits and tolerances, linear and angular measurement, comparators, interferometry, surface finish, CMMs.
  • Computer integrated manufacturing: CAD/CAM, additive manufacturing.

Key concepts and formulas

Iron-carbon diagram. Eutectoid at about 0.76–0.8% C and 727 °C (austenite → pearlite); eutectic at 4.3% C and about 1147 °C (liquid → ledeburite). Steels have up to about 2.1% C. Heat treatment: annealing (soft), normalising (air cooled, finer grains), hardening (quenching to martensite), tempering (reduces brittleness).

Casting.

  • Chvorinov's rule: solidification time t = B(V/A)²; the riser must solidify after the casting, so its modulus V/A must be larger.
  • Gating: continuity A₁V₁ = A₂V₂; velocity at the sprue base V = √(2gh). Pouring time of a top gate t = AmH/(Ag√(2gH)).

Forming. Rolling: maximum draft Δhmax = μ²R. True strain ε = ln(L/L₀). Wire drawing and extrusion forces grow with ln(A₀/Af). Blanking force = perimeter × thickness × shear strength.

Welding. Heat input = ηVI/v (per unit length). Arc power V × I. Resistance welding heat H = I²Rt.

Machining.

  • Chip thickness ratio r = t₁/t₂ (< 1). Shear angle: tan φ = r cos α/(1 − r sin α), with rake angle α.
  • Merchant's theory: 2φ + β − α = 90° (β = friction angle).
  • Taylor's tool life: VTⁿ = C. Typical n: about 0.1–0.15 for HSS, 0.2–0.4 for carbides, higher for ceramics.
  • Material removal rate in turning MRR = πDfdN (f = feed per revolution, d = depth of cut).
  • Machining time for turning: L/(fN).

Metrology. Hole basis system: the lower deviation of the hole is zero. Clearance, transition and interference fits. Tolerance = upper limit − lower limit. Statistical (root-sum-square) tolerance stacking is tighter than the worst case.

Non-traditional machining. EDM (spark erosion, conductive materials), ECM (electrolytic dissolution, no tool wear), USM (brittle materials), LBM, EBM, AJM, WJM.

Worked example

Orthogonal cutting with rake angle 10°, uncut chip thickness 0.3 mm and chip thickness 0.75 mm: find the shear angle. With Taylor's equation VT0.25 = 300, find the tool life at 150 m/min. Then compare the solidification times of a cube of side 100 mm and a sphere of the same volume.

import math

a, t1, t2 = math.radians(10), 0.3, 0.75
r = t1 / t2
phi = math.degrees(math.atan(r * math.cos(a) / (1 - r * math.sin(a))))
print(f"chip ratio = {r:.2f}, shear angle = {phi:.2f} deg")
n, C, V = 0.25, 300, 150
print(f"tool life at {V} m/min = {(C / V) ** (1 / n):.1f} min")
side = 0.1
V_ = side ** 3
m_cube = V_ / (6 * side ** 2)
rs = (3 * V_ / (4 * math.pi)) ** (1 / 3)
m_sph = rs / 3
print(f"modulus cube = {m_cube * 1000:.3f} mm, sphere = {m_sph * 1000:.3f} mm, t_sphere/t_cube = {(m_sph / m_cube) ** 2:.3f}")
Output
chip ratio = 0.40, shear angle = 22.94 deg
tool life at 150 m/min = 16.0 min
modulus cube = 16.667 mm, sphere = 20.678 mm, t_sphere/t_cube = 1.539

Practice questions

GATE-styleNAT · numerical answer

In Taylor's equation VT0.5 = C, the tool life is 25 min at 100 m/min. What is the tool life (min) at 125 m/min?

Show answer and explanationAnswer: 16

C = 100 × 250.5 = 500. At 125 m/min, T0.5 = 500/125 = 4, so T = 16 min.

GATE-styleNAT · numerical answer

Two castings have the same volume. Casting A has twice the surface area of casting B. What is the ratio of their solidification times tA/tB by Chvorinov's rule?

Show answer and explanationAnswer: 0.25

t ∝ (V/A)². With A doubled, (V/A) halves, so tA/tB = (1/2)² = 0.25.

GATE-styleNAT · numerical answer

The maximum possible draft in rolling with a roll radius of 300 mm and coefficient of friction 0.1 is (in mm):

Show answer and explanationAnswer: 3

Δhmax = μ²R = 0.01 × 300 = 3 mm.

GATE-styleNAT · numerical answer

In orthogonal cutting, the chip thickness ratio is 0.5 and the rake angle is 0°. What is the shear angle in degrees? Give the answer to two decimal places.

Show answer and explanationAnswer: 26.57

With α = 0, tan φ = r = 0.5, so φ = tan⁻¹(0.5) ≈ 26.57°.

GATE-styleMCQ · one correct answer

Which non-traditional process involves no tool wear because metal is removed by electrolytic dissolution?

Show answer and explanationAnswer: B

Electrochemical machining dissolves the workpiece anodically, and the tool (cathode) does not wear. EDM tools do wear by sparking.