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Manufacturing · Chapter 3 of 10 · Intermediate

Bulk Deformation: Rolling, Forging, Extrusion

Push metal hard enough and it flows like thick clay, keeping its volume but changing shape. Rolling, forging, and extrusion all exploit that flow, and all need to know one thing: the flow stress.

Readiness check

This chapter builds on plastic deformation and uses logarithms. Tick only what you can do closed-notes.

Recall that metals deform plastically by dislocation glide.
Use true strain ε = ln(L/L₀).
Recall strain hardening (the flow curve).
Compute areas and force as stress times area.
Evaluate powers with fractional exponents.

0 or 1 weak itemsContinue with this chapter.

2 weak itemsReview strengthening and the flow curve in Materials Chapter 6.

3 or more weak itemsRevisit mechanical properties in Materials Chapter 5 first.

The core idea

Forming shapes metal by forcing it to flow plastically at constant volume; the force needed is the flow stress times the contact area, and the flow stress rises with strain.

Y = Kεⁿε = ln(L/L₀)Y_avg = Kεⁿ/(1+n)

Above the yield point a metal flows, and as it flows it work-hardens, so its flow stress Y follows Y = Kεⁿ. Each process imposes a true strain set by its geometry (the reduction in thickness, the extrusion ratio), and the work per unit volume is the area under the flow curve, captured by the average flow stress Y_avg. Multiply by the right area and you get the force; that one idea sizes rolling mills, forging presses, and extrusion rams alike.

The skill works when: you find the true strain from the geometry, the average flow stress from the flow curve, and multiply by the contact area.

The skill breaks down when: engineering strain is used instead of true strain, or the flow stress is taken as a single yield value rather than averaged.

The concept. The flow stress rises with strain along Y = Kεⁿ. The work per unit volume is the area under the curve up to the process strain, which the average flow stress Y_avg represents.

The skills, taught in order

Bulk forming is the flow stress applied to four geometries. Five skills cover the flow curve and the rolling, forging, extrusion, and drawing processes.

3.1 Flow stress and true strain

In the plastic range the flow stress is Y = Kεⁿ, where K is the strength coefficient and n the strain-hardening exponent. True strain ε = ln(L/L₀) is used because deformations are large and additive. The average flow stress over a process, Y_avg = Kεⁿ/(1+n), gives the work per unit volume when multiplied by ε.

3.2 Rolling

Flat rolling squeezes a slab between rolls, reducing its thickness by the draft Δh over a contact length L = √(RΔh). The true strain is ln(h₀/h_f), and the roll force is F = L·w·Y_avg (contact area times average flow stress). The maximum draft a roll can bite is limited by friction, μ²R.

3.3 Forging

Forging compresses a workpiece between dies, open-die (simple shapes, free spread) or impression-die (the dies carry the shape). Friction at the die faces raises the pressure toward the centre, so the average forging pressure exceeds the flow stress by a friction factor, and forces grow with reduction.

3.4 Extrusion

Extrusion pushes a billet through a die to a smaller, constant cross-section. The extrusion ratio R = A₀/A_f sets the true strain ε = ln R, and the ideal pressure is p = Y_avg·ln R, with friction and die geometry adding more. It makes rails, tubes, and complex profiles in one pass.

3.5 Wire and rod drawing

Drawing pulls stock through a die to reduce its diameter. Because the work is done by pulling, the drawing stress must stay below the exit material's flow stress, or the wire necks and snaps. This caps the reduction per pass, so wire is drawn in many small steps.

Aspect	Hot working	Cold working
Temperature	above recrystallization	room temperature
Force needed	lower	higher
Finish and tolerance	rough, oxide scale	good, precise
Resulting strength	no net hardening	strain-hardened, stronger

Engineering connection: rolling makes plate and structural shapes, forging makes high-strength crankshafts and gears, extrusion makes profiles and tubes, and drawing makes wire, each chosen by shape and strength needs.

Worked example 1: roll force in flat rolling

A 300 mm wide slab is rolled from 25 mm to 20 mm thick on rolls of 250 mm radius. The metal follows Y = Kεⁿ with K = 600 MPa, n = 0.20. Find the contact length, the true strain, the average flow stress, and the roll force.

Figure 1. A slab squeezed between two rolls. The contact length and width set the area, and the average flow stress sets the pressure; their product is the roll force.

ProblemFind L, ε, Y_avg, and the roll force for the slab in Figure 1.
Given / findh₀ = 25 mm, h_f = 20 mm, w = 300 mm, R = 250 mm, K = 600 MPa, n = 0.20. Find L, ε, Y_avg, F.
AssumptionsPlane-strain flat rolling, friction within the bite limit, average-pressure model.
ModelContact length from the geometry, true strain from the thickness ratio, average flow stress from the flow curve, force as area times pressure.
EquationsL = √(RΔh) ε = ln(h₀/h_f), Y_avg = Kεⁿ/(1+n) F = L·w·Y_avg
SolveΔh = 5 mm, L = √(250 × 5) = 35.4 mm. ε = ln(25/20) = 0.223. Y_avg = 600 × 0.223^0.2/1.2 = 370 MPa. F = 35.4 × 300 × 370 = 3.93×10⁶ N = 3.93 MN.
CheckNearly four meganewtons for a modest 5 mm reduction shows why rolling mills are massive. Using true strain (0.223) rather than engineering strain (0.20) matters because the flow stress is averaged over the real, logarithmic deformation.
ConclusionRoll force is contact area times average flow stress. To cut it you reduce the draft per pass (smaller L) or roll hot (lower Y), which is exactly what real schedules do.

Result. L = 35.4 mm, ε = 0.223, Y_avg = 370 MPa, roll force ≈ 3.93 MN.

Worked example 2: direct extrusion

A 100 mm diameter billet is extruded to 40 mm diameter. The metal follows Y = Kεⁿ with K = 400 MPa, n = 0.15. Find the extrusion ratio, the true strain, the ideal extrusion pressure, and the ram force.

Figure 2. A ram forces the billet through a die. The area reduction sets the extrusion ratio, whose logarithm is the true strain, and the pressure follows from the average flow stress.

ProblemFind the extrusion ratio, true strain, ideal pressure, and ram force for the billet in Figure 2.
Given / findD₀ = 100 mm, D = 40 mm, K = 400 MPa, n = 0.15. Find R, ε, p, F.
AssumptionsIdeal (frictionless) deformation work; round billet and product.
ModelExtrusion ratio from the areas, true strain as its log, ideal pressure as the work per volume, force as pressure times billet area.
EquationsR = A₀/A_f = (D₀/D)² ε = ln R, p = Y_avg·ε F = p·A₀
SolveR = (100/40)² = 6.25. ε = ln 6.25 = 1.833. Y_avg = 400 × 1.833^0.15/1.15 = 381 MPa, so p = 381 × 1.833 = 698 MPa. A₀ = π(50)² = 7854 mm², giving F = 698 × 7854 = 5.48×10⁶ N = 5.48 MN.
CheckThe true strain (1.83) is large, as a 6:1 reduction should be, and the ideal pressure of 698 MPa is the floor; real friction and die geometry push it higher. The huge force is why extrusion is usually done hot.
ConclusionThe extrusion ratio drives everything: its log is the strain, which sets the pressure and force. Higher ratios make finer profiles but demand more pressure, the central trade-off of extrusion.

Result. Ratio 6.25, ε = 1.833, ideal pressure 698 MPa, ram force ≈ 5.48 MN.

Misconceptions and diagnostics

Mistake	Symptom	Diagnostic question	Correction
Engineering instead of true strain	Force off, worse at large reductions	"Is the deformation large?"	Use true strain ε = ln(L/L₀) for forming.
Single flow-stress value	Work underestimated	"Did I average over the strain?"	Use Y_avg = Kεⁿ/(1+n), not the yield point.
Ignoring friction	Real force exceeds the estimate	"Is this the ideal or the real force?"	Ideal models are a lower bound; friction and geometry add to it.
Drawing reduction too large	Wire necks and breaks	"Is the draw stress below the exit flow stress?"	Limit reduction per pass; draw wire in many small steps.

Practice ladder

Level 1 · Direct skill

A bar is extruded from 80 mm to 32 mm diameter. Find the extrusion ratio and the true strain.

Show answer

R = (80/32)² = 6.25, ε = ln 6.25 = 1.83. Same ratio as the worked example, since only the diameter ratio matters.

Level 2 · Mixed concept

For the Worked Example 1 slab, by how much does the roll force change if the same metal is rolled hot, halving its average flow stress?

Show answer

Force scales directly with Y_avg, so halving it halves the force to about 1.97 MN. This is why heavy reductions are taken hot, then finished cold for tolerance and surface.

Level 3 · Independent problem

A slab is rolled from 30 mm to 24 mm on 300 mm radius rolls, width 400 mm, with Y_avg = 250 MPa. Find the roll force.

Show answer

Δh = 6 mm, L = √(300 × 6) = 42.4 mm. F = L·w·Y_avg = 42.4 × 400 × 250 = 4.24×10⁶ N = 4.24 MN. Larger rolls and width raise the force.

Level 4 · Transfer to real engineering

Pick a real formed product (an aluminium window frame, a railway rail, a forged spanner). Identify the process, estimate its true strain from the shape change, and comment on the force or temperature it needs.

What good work looks like

The process named, a true strain estimated from the geometry, and the force or hot-versus-cold choice justified from the flow stress.

Working with AI, and proving it yourself

Use AI as an examiner, not a solver

"Check that I used true strain and the average flow stress, not the yield point."

"Give me five formed parts; I will name the process and estimate the strain."

"Compute the roll force." Finding ε and Y_avg yourself is the skill.

"Is this hot or cold worked?" Reasoning from finish and strength is the point.

Portfolio task

Analyse one bulk-forming operation: find the true strain from the geometry, the average flow stress, and the force, and state whether it should run hot or cold.

Must include: true strain, Y_avg from a flow curve, a force estimate, and a hot/cold judgement.

Retrieval and spaced review

Closed notes. Answer out loud, then reveal.

1. Write the flow-stress law and the average flow stress.

Y = Kεⁿ; Y_avg = Kεⁿ/(1+n).

2. Give the rolling contact length and roll force.

L = √(RΔh); F = L·w·Y_avg.

3. What sets the strain in extrusion?

The extrusion ratio R = A₀/A_f; ε = ln R.

4. Why is wire drawn in many passes?

The draw stress must stay below the exit flow stress, capping reduction per pass.

5. Name two differences between hot and cold working.

Hot needs less force and gives a rough, scaled surface with no net hardening; cold needs more force and gives a precise, strain-hardened part.

TodayFinish this quiz and Levels 1 and 2 of the ladder.

+1 dayRe-derive the roll force from a blank page.

+3 daysOne rolling and one extrusion calculation.

+7 daysMove to sheet-metal forming, Chapter 4.

+30 daysConnect flow stress to machining forces, Chapter 5.

Textbook mapping

Item	Mapping
Primary source	Kalpakjian and Schmid, Manufacturing Engineering and Technology, Chapters 13 to 15 (rolling, forging, extrusion and drawing)
Cross-reference	Groover, Ch. 18 and 19 · DeGarmo, forming chapters
Core topics	3.1 Flow stress · 3.2 Rolling · 3.3 Forging · 3.4 Extrusion · 3.5 Drawing
Engineering connection	Plate and structural shapes, forged crankshafts, extruded profiles, and drawn wire.
Read next	Chapter 4: Sheet-Metal Forming.