What is the difference between direct and indirect extrusion?

In direct (forward) extrusion the ram pushes the billet through a die at the far end of the container, so the billet slides along the container wall and friction adds a large force that falls as the billet shortens. In indirect (backward) extrusion the die is carried on a hollow ram (or the container moves with the billet), so there is no billet–container sliding; force is lower and independent of billet length. Indirect extrusion is limited by the strength and size of the hollow ram.

Why is there a limit to the reduction per pass in wire drawing but not in extrusion?

In drawing the force is applied by pulling the reduced wire, so the drawing stress must be less than the flow stress of the exiting wire, or it necks and breaks. Ideally this gives ln(A₀/A_f) ≤ 1, a maximum reduction of about 63%; friction and redundant work cut this to roughly 30–45% in practice. In extrusion the billet is pushed under compressive stress, so the product is not loaded in tension and very large ratios are possible, limited mainly by press capacity.

How does die angle affect the drawing or extrusion force?

A small die angle gives a long contact length, so friction work is high. A large angle shortens contact but increases redundant work — internal shearing as the metal turns sharply at entry and exit. Total force is therefore minimum at an optimum die angle, which grows with the reduction. Too large an angle with a small reduction also promotes centre-burst cracks.

Calculate the ideal ram force to extrude a 100 mm diameter aluminium billet to a 25 mm rod if the average flow stress is 60 MPa.

Extrusion ratio R = (100/25)² = 16, ln 16 = 2.773. Ideal pressure p = Ȳ ln R = 60 × 2.773 = 166.4 MPa. Ram force F = p × A₀ = 166.4 × (π/4 × 100²) = 166.4 × 7854 ≈ 1.31 × 10⁶ N, about 1.31 MN. Real presses must allow more, since Johnson's formula and container friction typically raise the pressure to about twice the ideal value.

What is hydrostatic extrusion and when is it used?

In hydrostatic extrusion the billet is surrounded by a pressurised fluid in the container, and the fluid pressure pushes it through the die. There is no billet–container friction, the die is well lubricated by the fluid, and the high hydrostatic pressure increases ductility. It is used for brittle or hard-to-work materials and for fine, accurate products, but sealing the high-pressure fluid makes it costly.

What is piping (tailpipe) in extrusion and how is it avoided?

Near the end of a direct extrusion stroke, the metal flow pattern draws the oxidised, contaminated outer skin and back face of the billet into the centre of the product, giving a central defect called piping. It is avoided by not extruding the last part of the billet (leaving a butt or discard of roughly 10–15%), by using a dummy block slightly smaller than the container to leave a skull, and by controlling friction and temperature so flow is more uniform.

Why are drawn wires stronger than the original rod?

Wire drawing is normally done cold, so the metal strain hardens: dislocation density rises and the flow stress increases with the accumulated true strain ln(A₀/A_f). The grains are also elongated along the wire axis, giving a fibrous texture. This is why intermediate annealing is needed after several passes to restore ductility before further drawing.

Extrusion and wire drawing

Direct, indirect, hydrostatic and impact extrusion, wire and tube drawing, extrusion pressure, drawing stress with friction, and the maximum reduction per pass.

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Why it matters

Aluminium window sections, heat-sink profiles, copper tubes, seamless steel tube blanks, electrical wire and welding wire are all made by pushing metal through a die (extrusion) or pulling it through one (drawing). The two processes look alike but have a key difference: an extruded product is pushed, so its exit stress can exceed the flow stress, while a drawn wire is pulled, so the drawing stress must stay below its own strength. That single idea sets the limits of each process.

Key ideas

Extrusion. A billet in a container is forced through a die opening of the desired cross-section by a ram. The metal is under high compressive hydrostatic stress, so even brittle-ish alloys can take very large reductions (extrusion ratios of 10–100 are common in hot aluminium extrusion) in one stroke.

Direct (forward) extrusion — ram and product move in the same direction. The whole billet slides along the container wall, so friction adds a large extra force that falls as the billet shortens. A short discard (butt) is left.
Indirect (backward or reverse) extrusion — the die is on a hollow ram and moves into the billet, or the container moves with the billet; there is no relative movement between billet and container, so friction and force are lower and independent of billet length, but the hollow ram limits the product size.
Hydrostatic extrusion — the billet is surrounded by a pressurised fluid; container friction is eliminated and brittle materials can be extruded.
Impact extrusion — a punch strikes a slug in a die, and metal flows backward around the punch (toothpaste tubes, battery cans); usually cold.
Hot vs cold — aluminium, copper and steel billets are usually hot extruded; cold extrusion gives finish and strength for small parts. Hollow sections are made with mandrels or with porthole/bridge dies, where metal splits and re-welds inside the die.

Extrusion defects. Centre burst (chevron cracking) — internal arrow-shaped cracks from tensile stresses at the centre line when die angle is high and reduction is low. Piping (tailpipe) — surface oxides and impurities drawn into the centre of the product at the end of the stroke; avoided by leaving a butt. Surface cracking (fir-tree) — at too high a temperature, speed or friction.

Wire, rod and tube drawing. The material is pulled through a converging die (half-angle typically 6°–15°) of tungsten carbide or diamond. Because the pulling stress is applied to the already-reduced product, the drawing stress must be less than the flow stress of the exiting wire, or the wire will neck and break. This limits the reduction per pass; fine wire needs many dies in series (multi-pass machines with intermediate annealing). Tube drawing uses a fixed plug, floating plug or moving mandrel to control the bore. Lubrication (soap powder, oil, copper coating) is essential. Drawn products have excellent finish, close tolerance and increased strength.

Die angle trade-off. A small die angle gives a long contact length and high friction; a large angle gives severe redundant (internal shear) work. There is an optimum angle where total drawing or extrusion stress is minimum, and it increases with reduction.

Work components. Real pressure = ideal (homogeneous) deformation + friction + redundant work. The ideal value Ȳ·ln(A₀/A_f) is a lower bound. Empirical formulas (Johnson's) and friction models (Sachs' slab analysis) correct for the rest.

Maximum reduction in drawing. With no friction and no strain hardening, drawing stress Ȳ·ln(A₀/A_f) equals Ȳ when ln(A₀/A_f) = 1, i.e. A₀/A_f = e, a maximum area reduction of 1 − 1/e = 63%. In practice friction and redundant work limit reductions to about 30–45% per pass; strain hardening raises the exit strength and helps slightly.

Formulas

R_x = A₀ / A_f — extrusion ratio; A₀ = billet area, A_f = product area.

r = (A₀ − A_f) / A₀ — fractional reduction in area (drawing).

ε = ln(A₀ / A_f) — true strain.

p_ideal = Ȳ × ln(A₀/A_f) — ideal extrusion pressure (MPa); Ȳ = average flow stress (MPa).

p = Ȳ × (a + b·ln R_x) — Johnson's empirical extrusion pressure; a ≈ 0.8 and b ≈ 1.2–1.5 (take values from the question).

p_direct ≈ Ȳ × (a + b·ln R_x + 2L/D₀) — direct extrusion with sticking friction along the remaining billet length L in a container of diameter D₀ (Groover's form).

F = p × A₀ (extrusion ram force, N) and P = F × v_ram (W).

σ_d,ideal = Ȳ × ln(A₀/A_f) — ideal drawing stress (MPa).

σ_d = Ȳ × ((1 + B)/B) × [1 − (A_f/A₀)^B], B = μ·cot α — Sachs drawing stress with Coulomb friction; μ = die friction coefficient, α = die half-angle.

F_d = σ_d × A_f, P = F_d × v_f — drawing force (N) and power (W); v_f = exit (drawing) speed.

r_max (ideal) = 1 − 1/e ≈ 0.632.

Worked examples

Example 1 (standard — direct extrusion). An aluminium billet 100 mm in diameter is hot extruded to a 40 mm round bar. Average flow stress at temperature = 50 MPa; use Johnson's formula with a = 0.8, b = 1.5. Ignore container friction. Find the extrusion pressure, ram force and power at a ram speed of 10 mm/s.

R_x = (100/40)² = 6.25, ln R_x = 1.833.
p = 50 × (0.8 + 1.5 × 1.833) = 50 × 3.549 = 177.4 MPa. (Ideal: 50 × 1.833 = 91.6 MPa — real pressure is almost twice the ideal.)
A₀ = (π/4) × 100² = 7854 mm², F = 177.4 × 7854 = 1.394 × 10⁶ N.
P = 1.394 × 10⁶ × 0.010 = 13.9 kW.

Answer: p ≈ 177 MPa, F ≈ 1.39 MN, P ≈ 13.9 kW. With sticking container friction on a 200 mm billet, Groover's form adds 2L/D₀ = 4 to the bracket and the pressure at the start of the stroke rises to about 377 MPa — which is why indirect extrusion is used for long billets.

Example 2 (GATE level — wire drawing with friction). Wire is drawn from 5 mm to 4 mm diameter through a die of half-angle 6°, μ = 0.1, average flow stress 300 MPa. Drawing speed 2 m/s. Find the drawing stress, force and power, and check the pass is feasible.

A_f/A₀ = (4/5)² = 0.64.
B = μ cot α = 0.1 / tan 6° = 0.951.
σ_d = 300 × (1.951/0.951) × [1 − 0.64^0.951] = 212.9 MPa (ideal would be 300 × ln(1/0.64) = 133.9 MPa).
A_f = (π/4) × 4² = 12.57 mm², F = 212.9 × 12.57 = 2675 N.
P = 2675 × 2 = 5.35 kW.
Check: σ_d = 213 MPa < 300 MPa, so the wire will not break.

Answer: σ_d ≈ 213 MPa, F ≈ 2.68 kN, P ≈ 5.35 kW; feasible.

Example 3 (limit). Ideal maximum reduction in area per pass: r = 1 − 1/e = 0.632 → 63.2% (diameter ratio d_f/d₀ = e^(−1/2) = 0.607).

Common mistakes

Using diameter ratio instead of area ratio in ln(A₀/A_f). For round sections, A₀/A_f = (d₀/d_f)².
Multiplying extrusion pressure by the product area; ram force uses the billet (container) area A₀.
Multiplying drawing stress by the entry area; drawing force uses the exit area A_f.
Thinking indirect extrusion has no friction at all — die friction remains; only container friction is removed.
Forgetting to check σ_d < Ȳ of the exit wire in drawing problems.
Using the die full angle in B = μ cot α; α is the half-angle.

For GATE PI

Expect numericals on ideal and empirical extrusion pressure and ram force, drawing stress and force with and without friction, drawing power, and the maximum possible reduction in one pass. One-mark questions test direct vs indirect extrusion, which defect is centre burst, why drawing has a reduction limit that extrusion does not, and the effect of die angle. Practise converting diameters to area ratios and keeping MPa × mm² = N.

Quick check

Billet 80 mm diameter extruded to 20 mm. Extrusion ratio?
Which needs less ram force for a long billet, direct or indirect extrusion?
Ideal maximum area reduction per drawing pass?
Drawing stress 250 MPa, exit diameter 2 mm. Drawing force?
Which defect appears as internal arrow-shaped cracks?

Answers: 1. (80/20)² = 16. 2. Indirect. 3. About 63%. 4. 250 × 3.142 ≈ 785 N. 5. Centre burst (chevron cracking).

Interview questions

All Casting, Forming and Joining interview questions

Try answering each one aloud before you open it.

1.What is the difference between direct and indirect extrusion?Concept
In direct (forward) extrusion the ram pushes the billet through a die at the far end of the container, so the billet slides along the container wall and friction adds a large force that falls as the billet shortens. In indirect (backward) extrusion the die is carried on a hollow ram (or the container moves with the billet), so there is no billet–container sliding; force is lower and independent of billet length. Indirect extrusion is limited by the strength and size of the hollow ram.
2.Why is there a limit to the reduction per pass in wire drawing but not in extrusion?Concept
In drawing the force is applied by pulling the reduced wire, so the drawing stress must be less than the flow stress of the exiting wire, or it necks and breaks. Ideally this gives ln(A₀/A_f) ≤ 1, a maximum reduction of about 63%; friction and redundant work cut this to roughly 30–45% in practice. In extrusion the billet is pushed under compressive stress, so the product is not loaded in tension and very large ratios are possible, limited mainly by press capacity.
3.How does die angle affect the drawing or extrusion force?Concept
A small die angle gives a long contact length, so friction work is high. A large angle shortens contact but increases redundant work — internal shearing as the metal turns sharply at entry and exit. Total force is therefore minimum at an optimum die angle, which grows with the reduction. Too large an angle with a small reduction also promotes centre-burst cracks.
4.Calculate the ideal ram force to extrude a 100 mm diameter aluminium billet to a 25 mm rod if the average flow stress is 60 MPa.Concept
Extrusion ratio R = (100/25)² = 16, ln 16 = 2.773. Ideal pressure p = Ȳ ln R = 60 × 2.773 = 166.4 MPa. Ram force F = p × A₀ = 166.4 × (π/4 × 100²) = 166.4 × 7854 ≈ 1.31 × 10⁶ N, about 1.31 MN. Real presses must allow more, since Johnson's formula and container friction typically raise the pressure to about twice the ideal value.
5.What is hydrostatic extrusion and when is it used?Concept
In hydrostatic extrusion the billet is surrounded by a pressurised fluid in the container, and the fluid pressure pushes it through the die. There is no billet–container friction, the die is well lubricated by the fluid, and the high hydrostatic pressure increases ductility. It is used for brittle or hard-to-work materials and for fine, accurate products, but sealing the high-pressure fluid makes it costly.
6.What is piping (tailpipe) in extrusion and how is it avoided?Concept
Near the end of a direct extrusion stroke, the metal flow pattern draws the oxidised, contaminated outer skin and back face of the billet into the centre of the product, giving a central defect called piping. It is avoided by not extruding the last part of the billet (leaving a butt or discard of roughly 10–15%), by using a dummy block slightly smaller than the container to leave a skull, and by controlling friction and temperature so flow is more uniform.
7.Why are drawn wires stronger than the original rod?Concept
Wire drawing is normally done cold, so the metal strain hardens: dislocation density rises and the flow stress increases with the accumulated true strain ln(A₀/A_f). The grains are also elongated along the wire axis, giving a fibrous texture. This is why intermediate annealing is needed after several passes to restore ductility before further drawing.

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