Bulk forming: forging, rolling, extrusion and drawing

Why it matters

Crankshafts, connecting rods, rails, sheet and plate, aluminium window sections, wire and tubes are all made by bulk forming. These processes save material compared with machining and give a continuous grain flow that improves fatigue strength. Engineers must estimate forces and power to choose a press, mill or draw bench, and know the limits (maximum draft, maximum drawing reduction) to plan the number of passes.

Key ideas

Bulk forming changes the shape of a thick workpiece (billet, bar, slab) by large plastic deformation, mostly under compressive stresses. Volume is constant during plastic deformation: A₀·L₀ = A_f·L_f.

Hot, warm and cold working. Hot working is done above the recrystallisation temperature (roughly 0.6 of the absolute melting temperature): flow stress is low, large strains are possible and the grain structure is refined, but surface finish and tolerances are poorer because of scale. Cold working is done near room temperature: higher forces, strain hardening, better finish and accuracy, but limited ductility. The flow stress in cold work is modelled as σ_f = K·εⁿ; for hot work strain-rate sensitivity matters more than strain hardening.

Forging. Compressive shaping between dies.

• Open-die forging (upsetting, drawing out) uses flat or simple dies. Friction at the die faces makes the pressure rise towards the centre and causes barrelling, so the average pressure is above the flow stress.
• Impression (closed) die forging uses shaped cavities; excess metal escapes as flash, whose narrow land builds back-pressure to fill the cavity. The peak force occurs at the end when flash forms.
• Defects: laps and cold shuts (metal folding), incomplete filling, internal cracks.

Rolling. Material is drawn into the gap between two rotating rolls by friction.

• Entry speed is below roll surface speed and exit speed is above it; at the neutral (no-slip) point the strip and roll surface speeds are equal, and friction reverses direction there.
• Bite condition: the strip can be drawn in only if μ ≥ tan α (α = angle of bite), which gives the maximum draft Δh_max = μ²·R.
• The roll force acts over the projected contact length L = √(R·Δh). Smaller rolls, thinner draft and lower friction reduce force. Cluster and four-high mills use small work rolls backed by larger rolls to reduce force and roll deflection.
• Defects: wavy edges and zipper cracks (roll bending), edge cracks, alligatoring.

Extrusion. A billet in a container is pushed through a die. Direct (forward) extrusion has friction between billet and container; indirect (backward) extrusion moves the die and avoids this friction, so its force is lower. Hot extrusion of aluminium produces complex constant-section profiles; cold impact extrusion makes cans and tubes. The extrusion ratio R = A₀/A_f. Defects: centre burst (chevron cracking), piping, surface cracking.

Drawing. Rod, wire or tube is pulled through a converging die. Because the drawn product must carry the drawing stress, that stress must stay below the product's flow stress. For an ideal, rigid–perfectly plastic material this gives a maximum true strain of 1 per pass, i.e. a maximum area reduction of 1 − e⁻¹ = 63.2%; friction and redundant work lower this in practice (about 30–45% per pass). Strain hardening raises the limit slightly.

Ideal work. Frictionless, homogeneous deformation gives the minimum force. Real forces are higher because of friction and redundant (non-useful) shear; empirical factors from your data book are used for design.

Formulas

ε = ln(A₀/A_f) = ln(h₀/h_f) (plane strain rolling uses thickness)

• ε = true strain (dimensionless), A = area (m²), h = thickness (m).

σ̄_f = K·εⁿ / (1 + n)

• σ̄_f = average flow stress over the strain range (Pa), K = strength coefficient (Pa), n = strain-hardening exponent (from data book).

Δh_max = μ²·R and L = √(R·Δh)

• μ = roll–strip friction coefficient, R = roll radius (m), Δh = h₀ − h_f (m), L = projected contact length (m).

F = σ̄_f·w·L (rolling, friction neglected)

• F = roll separating force (N), w = strip width (m).

T = 0.5·F·L (per roll), P = 2·T·ω = 2π·N·F·L / 60 (both rolls)

• T = torque (N·m), ω = roll speed (rad/s), N = roll speed (rpm), P = power (W).

p_avg = σ_f·(1 + 2μr / (3h)), F = p_avg·π·r²

• open-die upsetting of a cylinder (sliding friction), r = current radius (m), h = current height (m), σ_f = flow stress at that strain (Pa).

F = K_f·σ_f·A

• simple impression-die forging estimate, A = projected area including flash (m²), K_f = shape factor from your data book (roughly 6–12 for impression dies with flash, higher for more complex shapes).

σ_d = σ̄_f·ln(A₀/A_f), F_d = σ_d·A_f (ideal drawing)

• σ_d = drawing stress at exit (Pa), F_d = drawing force (N).

p = σ̄_f·ln R (ideal extrusion), F = p·A₀

• p = ram pressure (Pa), R = A₀/A_f, A₀ = billet area (m²). Real pressure is higher; use an empirical relation such as p = σ̄_f·(a + b·ln R) with a, b from your data book.

Worked examples

Example 1 – flat rolling force and power (standard). A 300 mm wide strip is rolled from 25 mm to 20 mm thickness. Roll radius R = 250 mm, roll speed N = 50 rpm, μ = 0.15. Flow curve: K = 275 MPa, n = 0.15. Neglect friction in the force estimate.

1. Draft Δh = 5 mm. Check bite: Δh_max = μ²R = 0.15² × 250 = 5.625 mm ≥ 5 mm, so the pass is feasible.
2. ε = ln(25/20) = 0.2231.
3. σ̄_f = K·εⁿ/(1 + n) = 275 × 0.2231^0.15 / 1.15 = 191.0 MPa.
4. L = √(R·Δh) = √(250 × 5) = 35.36 mm.
5. F = σ̄_f·w·L = 191.0 × 300 × 35.36 = 2.025 × 10⁶ N.
6. Torque per roll T = 0.5·F·L = 0.5 × 2.025 × 10⁶ × 0.03536 = 35.8 kN·m.
7. Power for both rolls P = 2·T·ω = 2 × 35.8 × 10³ × (2π × 50/60) = 3.75 × 10⁵ W. F ≈ 2.03 MN, P ≈ 375 kW

Example 2 – wire drawing and its limit (GATE level). A rigid–perfectly plastic wire with flow stress 300 MPa is drawn from 10 mm to 8 mm diameter. Neglect friction.

1. A₀/A_f = (10/8)² = 1.5625; ε = ln 1.5625 = 0.4463.
2. σ_d = σ_f·ε = 300 × 0.4463 = 133.9 MPa (below 300 MPa, so the pass is possible).
3. A_f = (π/4) × 8² = 50.27 mm².
4. F_d = σ_d·A_f = 133.9 × 50.27 = 6 730 N.
5. Limit: σ_d = σ_f when ln(A₀/A_f) = 1, so maximum reduction = 1 − e⁻¹ = 63.2%. F_d ≈ 6.73 kN; ideal maximum reduction 63.2% per pass

Example 3 – open-die upsetting with friction. A cylinder 50 mm in diameter and 50 mm high is upset to 25 mm height. Flow stress (perfectly plastic) = 300 MPa, μ = 0.1. Find the force at the end of the stroke.

1. Constant volume: r_f² × 25 = 25² × 50 → r_f = 25 × √2 = 35.36 mm.
2. p_avg = σ_f·(1 + 2μr/(3h)) = 300 × (1 + 2 × 0.1 × 35.36 / (3 × 25)) = 300 × 1.0943 = 328.3 MPa.
3. F = p_avg·π·r_f² = 328.3 × π × 35.36² = 1.289 × 10⁶ N. F ≈ 1.29 MN

Common mistakes

• Using the strip's final thickness instead of the draft in L = √(R·Δh), or using roll diameter instead of radius.
• Multiplying flow stress by the draft (or by a volume) to get a "force"; force is always stress × area.
• In extrusion, multiplying the ram pressure by the product area; the ram acts on the billet area A₀.
• Using engineering strain instead of true strain ln(A₀/A_f).
• In upsetting, using the initial radius and height instead of the current values at the instant of interest.
• Forgetting the bite check before calculating rolling force.

For GATE ME

Common questions: maximum draft from μ and R, contact length, roll force, torque and power; neutral-point reasoning; ideal drawing stress and the 63% reduction limit; extrusion pressure from the extrusion ratio; average pressure and force in upsetting with friction; volume constancy to find final dimensions; and hot versus cold working comparisons. Practise carrying units (N, mm, MPa) consistently through multi-step force and power calculations.

Quick check

1. What is the maximum draft for μ = 0.1 and R = 300 mm?
2. Where along the roll gap are strip and roll speeds equal?
3. What is the ideal maximum area reduction in one drawing pass of a perfectly plastic material?
4. Why does indirect extrusion need less force than direct extrusion?
5. Why is the average upsetting pressure higher than the flow stress?

Answers: 1. 3 mm. 2. At the neutral (no-slip) point. 3. 63.2%. 4. There is no friction between the billet and the container. 5. Die–workpiece friction raises the pressure towards the centre.