Manufacturing Processes: Casting, Metal Working, Joining, Powder Metallurgy and Non-Destructive Testing
2. Hot, warm and cold working: rolling, forging, extrusion, drawing and sheet forming
Hot working is done above the recrystallisation temperature (roughly above 0.6 T_m): the metal recrystallises dynamically or between passes, so flow stress is low, large strains are possible, cast structure and porosity are broken down, but surface oxidation and dimensional control are worse. Cold working, below about 0.3 T_m, strain-hardens the metal, raises its strength, lowers its ductility, gives good surface finish and tolerance, and elongates grains into a fibre texture; annealing between stages restores ductility. Warm working (about 0.3-0.5 T_m) sits between. The ideal work of deformation per unit volume is w = σ̄ε, which is what the ideal forming loads below are built on; friction and redundant work raise the real loads.
| Process | Key relation | Typical defects |
|---|---|---|
| Flat rolling | Draft Δh = h₀ − h_f; maximum draft μ²R; contact length L = √(RΔh); roll force ≈ σ̄ w L (plus friction) | Edge cracks, wavy edges, centre buckles, alligatoring |
| Open-die forging (upsetting a cylinder) | Mean pressure p̄ ≈ σ̄(1 + 2μr/(3h)): the friction hill | Barrelling, laps and cold shuts, internal cracks |
| Extrusion (direct or indirect) | Ideal pressure p = σ̄ ln R, R = A₀/A_f the extrusion ratio | Centre burst (chevron cracks), surface (fir-tree) cracking, piping |
| Wire and rod drawing | Ideal draw stress σ_d = σ̄ ln(A₀/A_f); ideal limit ln(A₀/A_f) = 1, a 63% reduction per pass | Centre burst, seams, die marks |
| Sheet forming: blanking, bending, deep drawing, stretching | Blanking force = shear strength × perimeter × thickness; limiting draw ratio about 2; springback after bending | Earing (planar anisotropy), wrinkling, tearing, orange peel, stretcher strains (Lüders bands) |
The limits in each process are metallurgical as much as mechanical. In drawing, the pulled product must carry the draw stress without yielding, so the ideal reduction per pass is capped where σ_d = σ̄. In deep drawing the wall must carry the load of drawing in the flange; a high normal anisotropy r̄ (width strain over thickness strain in a tensile test) resists thinning and raises the limiting draw ratio, while planar variation of r causes earing. Stretcher strains appear when a low-carbon steel with a yield-point elongation is formed; temper rolling (a light skin pass) removes them for a time, until strain ageing brings the yield point back. Forming-limit diagrams plot the major and minor surface strains at which a sheet necks, and a press shop keeps its strains below that curve.
3. Joining: soldering, brazing, welding and welding metallurgy
Soldering and brazing join without melting the base metal: a filler with a lower melting point wets the surfaces and is drawn into a narrow joint gap by capillarity, and bonds by forming a thin interdiffusion or intermetallic layer. By convention soldering uses fillers melting below 450 °C (tin-based solders, historically tin-lead, now lead-free Sn-Ag-Cu) and brazing uses fillers melting above 450 °C but below the base metal’s solidus (copper, brass, silver and aluminium-silicon fillers). Fluxes dissolve and prevent oxide so the filler can wet. Welding joins by fusion (arc, gas, laser, electron beam, electroslag, resistance) or in the solid state (friction, friction stir, diffusion, explosive). The arc processes differ in how they shield the pool: shielded metal arc (flux-coated electrode), gas metal arc (MIG/MAG, consumable wire with inert or active gas), gas tungsten arc (TIG, non-consumable tungsten electrode), and submerged arc (granular flux, high deposition). The heat input per unit length, H = ηVI/v, together with preheat and plate thickness, sets the cooling rate of the weld and heat-affected zone.
A fusion weld is a small casting on a large chill. The fusion zone solidifies epitaxially from the partially melted grains of the base metal and grows as columnar grains towards the centreline, with segregation between them. Outside it, the heat-affected zone (HAZ) is base metal heated into different ranges: in a steel, the coarse-grained region next to the fusion line (grain growth at high peak temperature), a fine-grained normalised region, an intercritical region and a subcritical tempered region. In a hardenable steel, fast cooling of the coarse-grained HAZ produces martensite; with hydrogen from moist consumables and restraint stresses, this is the recipe for hydrogen-induced (cold) cracking, which may appear hours after welding. Susceptibility is gauged by the carbon equivalent, CE = C + Mn/6 + (Cr + Mo + V)/5 + (Ni + Cu)/15 (IIW); above roughly 0.4-0.45 preheat and low-hydrogen practice are needed. Preheat lowers the cooling rate, and post-weld heat treatment tempers the martensite and relieves stress.
| Defect | Steels | Aluminium alloys |
|---|---|---|
| Cold (hydrogen) cracking | Main risk in hardenable steels: HAZ martensite + hydrogen + stress | Not a problem: no martensite |
| Solidification (hot) cracking | Promoted by S and P forming low-melting films; Mn ties up S as MnS | Serious in 6xxx and 2xxx alloys welded without a suitable filler (Al-Si or Al-Mg) |
| Porosity | CO, N₂ or H₂ from poor shielding or damp consumables | Hydrogen porosity, the most common defect: moisture, oils, hydrated oxide |
| HAZ property loss | Hardening and embrittlement; grain coarsening | Softening of work-hardened and age-hardened alloys (overageing, dissolution of precipitates) |
| Other | Lamellar tearing in rolled plate under through-thickness stress; slag inclusions; lack of fusion; undercut | Oxide film inclusions and lack of fusion (tenacious Al₂O₃, high conductivity) |
4. Powder metallurgy: powders, compaction and sintering
Metal powders are made by atomisation — a molten stream broken up by water jets (irregular particles, cheap, for iron and steel) or gas jets (spherical particles, for superalloys and titanium) — by chemical reduction of oxides (sponge iron powder, tungsten and molybdenum by hydrogen reduction), by electrolysis (dendritic copper and iron of high purity), by decomposition of carbonyls (fine spherical nickel and iron), and by mechanical comminution and mechanical alloying (for brittle materials and oxide-dispersion-strengthened alloys). Powders are characterised by particle size and distribution, shape, apparent and tap density, flow rate (the Hall flowmeter) and compressibility. They are blended, usually with a lubricant such as zinc stearate, and compacted: uniaxial pressing in a rigid die (fast, but friction with the die wall leaves density gradients, so height-to-diameter ratio is limited), cold isostatic pressing in a flexible mould under fluid pressure (uniform density), or hot isostatic pressing, which compacts and sinters at once to full density. The green compact has only enough strength to be handled.
Sintering heats the compact below the melting point of the main constituent (roughly 0.7-0.9 T_m), usually in a protective or reducing atmosphere, and the driving force is the reduction of surface energy. In the initial stage necks form and grow between particles, by surface diffusion, grain-boundary diffusion, volume diffusion and evaporation-condensation; only mechanisms that remove matter from the particle contacts (grain-boundary and volume diffusion) bring the centres together and shrink the compact. In the intermediate stage the pores form an interconnected network along grain edges and most densification occurs; in the final stage pores are isolated and spherical, and densification slows. Liquid-phase sintering (cemented carbides with a cobalt binder, tungsten heavy alloys, bronze bearings) adds a small amount of a low-melting phase that wets the solid, speeds rearrangement and solution-reprecipitation, and gives near-full density. Shrinkage links the densities: for isotropic shrinkage, (1 − ΔL/L₀)³ = ρ_green/ρ_sintered. Controlled porosity is a feature, not a flaw, in self-lubricating bearings and filters.
5. Non-destructive testing
| Method | Principle | Finds | Limitations |
|---|---|---|---|
| Dye penetrant | A low-viscosity dye seeps into surface-breaking cracks by capillarity; after cleaning, a developer draws it out | Surface-breaking cracks, porosity, laps in any non-porous material | Surface-open defects only; not for porous parts |
| Magnetic particle | A crack in a magnetised part causes flux leakage that gathers iron particles | Surface and near-surface cracks | Ferromagnetic materials only; the crack must lie across the flux |
| Eddy current | An AC coil induces currents whose change in impedance reveals defects, conductivity or thickness changes | Surface and near-surface cracks, sorting alloys, coating thickness, tube inspection | Conductors only; depth limited by the skin depth δ = 1/√(πfμσ) |
| Ultrasonic | Pulses of 1-10 MHz sound reflect from interfaces; depth = vt/2 in pulse-echo | Internal planar defects (cracks, laminations, lack of fusion), thickness | Needs a couplant and skilled interpretation; coarse-grained cast structures scatter the beam |
| Radiography (X-ray, gamma) | Differential absorption of penetrating radiation recorded on film or a detector | Volumetric defects: porosity, shrinkage, inclusions | Radiation hazard; planar cracks seen only if aligned with the beam; sensitivity checked with an image-quality indicator |
| Acoustic emission | Sensors pick up elastic waves released by growing cracks, yielding or phase changes under load | Active, growing defects anywhere in a large structure (pressure vessels during proof testing) | Passive: detects only defects that are growing while the structure is loaded; does not size them |
Key takeaways
- Chvorinov t_s = B(V/A)²; a riser needs about 1.2 times the casting modulus (cylinder H = D has modulus D/6). Top-gated fill time V/(A_g√(2gh)); sprue tapered as √(h).
- Hot working above recrystallisation, cold working strain-hardens. Rolling Δh_max = μ²R, L = √(RΔh); drawing σ_d = σ̄ ln(A₀/A_f), ideal limit 63%; extrusion p = σ̄ ln R; blanking F = τ × perimeter × t.
- Soldering below 450 °C, brazing above 450 °C but below the base-metal solidus; heat input H = ηVI/v; CE = C + Mn/6 + (Cr + Mo + V)/5 + (Ni + Cu)/15 sets the need for preheat.
- Steel welds crack cold (HAZ martensite + H + stress) or hot (S, P); aluminium welds suffer hydrogen porosity, hot cracking without filler, and HAZ softening.
- Powders by atomisation, reduction, electrolysis, carbonyl; sintering is driven by surface energy, densifies by boundary and volume diffusion; (1 − ΔL/L₀)³ = ρ_green/ρ_sintered. NDT: penetrant surface-open, MPI ferromagnetic, eddy current conductors, UT planar, RT volumetric, AE growing cracks.
Practice questions (17)
Attempt each one before opening the answer. Every explanation names the tempting wrong option as well as the right one, because that is where marks are lost.
A 60 mm cube casting solidifies in 4 min. By Chvorinov’s rule (exponent 2), how long, in minutes, will a 120 mm cube of the same alloy take in the same mould material?
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Answer: 16
For a cube V/A = a³/(6a²) = a/6, so the moduli are 10 mm and 20 mm. t ∝ (V/A)², giving t = 4 × (20/10)² = 16 min. Equivalently B = 4/10² = 0.04 min/mm² and 0.04 × 20² = 16. Scaling with volume (eight times) gives 32 min; scaling with size alone gives 8 min.A cylindrical side riser with height equal to diameter is to feed a 60 mm cube casting. If the riser modulus must be 1.2 times the casting modulus, what riser diameter, in mm, is required? (Treat all surfaces of the riser as cooling surfaces.)
Numerical answer — type the value.
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Answer: 72
Casting modulus = a/6 = 10 mm, so the riser needs 12 mm. For H = D, V = πD³/4 and A = πD²/2 + πD² = 3πD²/2, so V/A = D/6. D/6 = 12 gives D = 72 mm. Check: V/A = (π × 72³/4)/(1.5π × 72²) = 72/6 = 12 mm, and it freezes (1.2)² = 1.44 times later than the cube. Using D/4 (forgetting the end faces) gives 48 mm, which would freeze too early.A mould cavity of 0.006 m³ is top-gated through an ingate of 4 cm² area, with a sprue height of 0.2 m. Neglecting losses, what is the filling time, in seconds (to one decimal place)? Take g = 9.81 m/s².
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Answer: 7.6
v = √(2gh) = √(2 × 9.81 × 0.2) = 1.981 m/s; Q = A_g v = 4 × 10⁻⁴ × 1.981 = 7.924 × 10⁻⁴ m³/s; t = V/Q = 0.006/7.924 × 10⁻⁴ = 7.57 s. In a top-gated mould the head at the gate stays constant, which is why the formula is this simple; bottom gating would take longer. Leaving the area in cm² gives a time 10⁴ times too short.Which of the following statements about casting defects are correct?
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Answer: A — Hot tears form near the solidus when contraction is hindered by the mould or cores; B — Hydrogen porosity in aluminium castings arises because hydrogen is much less soluble in the solid than in the liquid; D — Chills placed at a heavy section help promote directional solidification towards the riser
Hot tears are contraction cracks in the weak, nearly solid mush (A); hydrogen rejected on freezing forms the pores in aluminium (B); chills lower the local modulus so the heavy section freezes before the riser (D). C is wrong: a misrun is metal freezing before the cavity fills — too low a pouring temperature, too slow a pour or too thin a section — and riser size has nothing to do with it.What is the maximum possible draft, in mm, when rolling a strip with rolls of 300 mm radius if the coefficient of friction is 0.1?
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Answer: 3
The strip is gripped only if the bite angle satisfies tan α ≤ μ. The two rolls together reduce the thickness by Δh = 2R(1 − cos α) ≈ Rα², so with α ≈ μ, Δh_max = μ²R = 0.01 × 300 = 3 mm. Check the angle: α ≈ √(Δh/R) = √(0.01) = 0.1 rad, whose tangent is about 0.1 = μ. Using μR gives 30 mm.A strip 500 mm wide is rolled with a draft of 4 mm on rolls of 300 mm radius. Taking the mean flow stress in the roll gap as 200 MPa and neglecting friction, what is the roll separating force, in MN (to two decimal places)?
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Answer: 3.46
Projected contact length L = √(RΔh) = √(300 × 4) = √1200 = 34.64 mm. F = σ̄ w L = 200 × 500 × 34.64 = 3.464 × 10⁶ N = 3.46 MN. Check the units: MPa × mm² = N. Using the arc length or the draft itself as the contact length is the usual error; friction would raise the real force above this ideal value.A wire is drawn from 10 mm to 8 mm diameter. If the mean flow stress is 300 MPa, what is the ideal drawing stress, in MPa (to one decimal place)?
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Answer: 133.9
σ_d = σ̄ ln(A₀/A_f) = 300 × ln(10²/8²) = 300 × ln 1.5625 = 300 × 0.4463 = 133.9 MPa. Equivalently 2σ̄ ln(D₀/D_f) = 600 × ln 1.25 = 600 × 0.2231 = 133.9. The reduction in area is 36%, well inside the ideal limit of 63%. Using the diameter ratio without the factor 2 gives 66.9 MPa.A hole of 50 mm diameter is to be blanked in a sheet 2 mm thick whose shear strength is 300 MPa. What is the blanking force, in kN (to one decimal place)?
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Answer: 94.2
F = τ × (π d) × t = 300 × π × 50 × 2 = 94 248 N = 94.2 kN. The sheared area is the perimeter times the thickness, 314.2 mm², not the circular face area πd²/4 = 1963 mm², which would give 589 kN.A cylinder of 100 mm diameter is upset between flat dies to a height of 25 mm. The flow stress is 100 MPa and the coefficient of friction 0.2. Using p̄ = σ̄(1 + 2μr/(3h)), what is the forging load at that height, in MN (to two decimal places)?
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Answer: 0.99
r = 50 mm, h = 25 mm: p̄ = 100 × (1 + 2 × 0.2 × 50/(3 × 25)) = 100 × (1 + 20/75) = 126.7 MPa. Area = π × 50² = 7854 mm², so F = 126.7 × 7854 = 9.95 × 10⁵ N = 0.99 MN. Without friction the load would be 0.785 MN; the friction hill adds 27%. Using the diameter for r doubles the friction term and gives 1.20 MN.Centre burst (chevron cracking) in extrusion and drawing is most likely when:
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Answer: A — The die angle is high and the reduction per pass is small
With a large die angle and a small reduction, the deformation zone is shallow and does not penetrate to the centre, which is left in hydrostatic tension and ruptures in a series of internal cones. Lowering the die angle or increasing the reduction per pass makes the deformation reach the axis and cures it. Heating and good lubrication do not create the tensile centre that causes it.An arc weld is made at 25 V and 200 A with a travel speed of 5 mm/s and an arc efficiency of 0.8. What is the heat input, in kJ/mm?
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Answer: 0.8
H = ηVI/v = 0.8 × 25 × 200/5 = 4000/5 = 800 J/mm = 0.8 kJ/mm. Without the efficiency the arc energy is 1.0 kJ/mm; the 0.8 factor accounts for heat lost from the arc before it reaches the work. Reporting 800 answers in J/mm.A steel contains 0.20% C, 1.20% Mn, 0.50% Cr, 0.20% Mo, 0.30% Ni and 0.15% Cu. What is its IIW carbon equivalent (to two decimal places)?
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Answer: 0.57
CE = C + Mn/6 + (Cr + Mo + V)/5 + (Ni + Cu)/15 = 0.20 + 0.20 + 0.70/5 + 0.45/15 = 0.20 + 0.20 + 0.14 + 0.03 = 0.57. Above about 0.45 a steel needs preheat and low-hydrogen practice to avoid HAZ cold cracking. Putting Ni and Cu over 5 with Cr and Mo gives 0.63.Which of the following statements about welding metallurgy and weld defects are correct?
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Answer: A — Hydrogen-induced cold cracking needs a susceptible microstructure, hydrogen and tensile stress together; B — Preheating a hardenable steel lowers the cooling rate of the HAZ; D — Sulphur and phosphorus promote solidification (hot) cracking in steel welds
Cold cracking needs all three factors at once — remove any one and it does not occur (A). Preheat reduces the temperature gradient and so the cooling rate, avoiding hard martensite (B). S and P form low-melting films at the grain boundaries of the solidifying weld metal (D). C is reversed: the weld thermal cycle overages or dissolves the strengthening precipitates, so the HAZ of an age-hardened alloy is SOFTER than the base metal.The process in which a filler metal melting above 450 °C but below the solidus of the base metals is drawn into the joint by capillary action is:
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Answer: A — Brazing
Brazing is defined by a filler liquidus above 450 °C and below the base-metal solidus, with capillary flow into the gap. Soldering uses fillers below 450 °C; fusion welding melts the base metal; friction welding is a solid-state process with no filler at all.An iron powder compact has a green density of 6.0 g/cm³ and sinters to 7.2 g/cm³. Assuming isotropic shrinkage and no mass loss, what is the linear shrinkage, in percent (to two decimal places)?
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Answer: 5.90
Mass is conserved, so V_s/V_g = ρ_g/ρ_s = 6.0/7.2 = 0.8333. For isotropic shrinkage L_s/L_g = 0.8333^(1/3) = 0.9410, so the linear shrinkage is 1 − 0.9410 = 5.90%. Check: 0.9410³ = 0.8333. The volumetric shrinkage is 16.7%; dividing it by three (5.56%) is only the small-strain approximation.In pulse-echo ultrasonic testing of steel (longitudinal wave velocity 5900 m/s), a defect echo returns 10 µs after the initial pulse. How deep is the defect, in mm?
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Answer: 29.5
The pulse travels to the defect and back, so d = vt/2 = 5900 × 10 × 10⁻⁶/2 = 0.0295 m = 29.5 mm. Forgetting the round trip gives 59 mm, twice the true depth.Which NDT methods could be used to detect a crack lying about 5 mm below the surface of a thick austenitic stainless steel plate?
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Answer: A — Ultrasonic testing; B — Radiography with the beam aligned with the crack plane
A buried planar crack is found by ultrasonics (A) and by radiography when the beam runs along the crack plane, so the crack presents its full depth to the beam (B). Magnetic particle inspection needs a ferromagnetic material, and austenitic stainless steel is not (C is out); dye penetrant detects only surface-breaking defects (D is out).