Rolling: mechanics, roll force and power

Why it matters

Around 90% of all steel passes through rolling mills, turning cast slabs and billets into plate, strip, rails, beams and bars. Sizing a mill stand — its roll force, motor power and the largest reduction it can take in one pass — uses a handful of formulas that GATE asks almost every year in some form.

Key ideas

Flat rolling geometry. A strip of thickness h₀ and width w enters between two rolls of radius R and leaves at h_f. The draft is Δh = h₀ − h_f. The arc of contact subtends the bite (contact) angle α, and its horizontal projection is the contact length L ≈ √(R·Δh). Width barely changes in flat rolling of wide strip (plane strain), so by volume constancy h₀·v₀ = h_f·v_f: the strip speeds up as it thins.

Neutral point and slip. The roll surface speed v_r lies between the entry and exit speeds of the strip. At entry the roll moves faster than the strip, and friction drags the strip in; after the neutral (no-slip) point the strip moves faster than the roll, and friction acts backward. The pressure distribution therefore peaks at the neutral point (friction hill). Forward slip = (v_f − v_r)/v_r.

Bite condition. The rolls can pull the strip in only if friction overcomes the push-back of the normal force: tan α ≤ μ. This gives the maximum draft Δh_max = μ²·R. Larger rolls and rougher (higher-friction) rolls allow larger reductions; hot rolling has high μ (around 0.2–0.7, sticking), cold rolling with lubricant has low μ (around 0.02–0.1). If the required total reduction is larger, it is split into several passes.

Roll force. The simplest estimate treats the contact zone as a plane-strain compression: F = Ȳ·w·L, where Ȳ is the average flow stress over the pass. Refinements multiply by the plane-strain factor 1.155 (von Mises) and by a friction factor such as (1 + μL/(2h_avg)). For cold rolling, Ȳ comes from the Hollomon law using the true strain ln(h₀/h_f); for hot rolling, from the hot flow stress at the working strain rate. Roll force is reduced by smaller rolls (shorter L), lower friction, front and back tension, and higher temperature.

Torque and power. The roll force acts at a moment arm a = λ·L from the roll centre line, with λ ≈ 0.5 for hot rolling and ≈ 0.45 for cold rolling. Torque per roll T = F·λL. Both rolls are driven, so total power P = 2·T·ω. With λ = 0.5 this becomes P = F·L·ω = 2πN·F·L/60. Power is not F × strip speed: F is a vertical force, and the strip does not move vertically.

Roll flattening and deflection. High forces elastically flatten the rolls (increasing effective R and L) and bend them (thicker strip at the centre). Mills use small work rolls backed by large back-up rolls (four-high, cluster/Sendzimir mills), and crowned rolls to compensate bending.

Mills and products. Two-high (reversing or non-reversing), three-high, four-high, cluster, tandem (several stands in series, same mass flow through all). Shape rolling uses grooved rolls for rails, beams and bars; ring rolling and thread rolling are variants. Hot rolling breaks down cast structure; cold rolling gives finish, tolerance and strength.

Defects. Wavy edges (rolls bend, edges thinner and longer), zipper cracks in the centre (insufficient ductility or excessive crown), edge cracks, alligatoring (splitting of the slab end, from non-uniform deformation), and surface scale or laps.

Formulas

Δh = h₀ − h_f — draft (mm).

L ≈ √(R·Δh) — contact length (mm); R = roll radius (mm).

Δh_max = μ²·R — maximum draft in one pass; equivalently α_max = tan⁻¹ μ.

Δh = 2R(1 − cos α) — exact relation between draft and bite angle.

h₀·w₀·v₀ = h_f·w_f·v_f — volume (mass) flow constancy; v = strip velocity (m/s).

ε = ln(h₀/h_f), Ȳ = K·εⁿ / (1 + n) — true strain and average flow stress for cold rolling (MPa).

F = Ȳ·w·L — roll force (N, with Ȳ in MPa and w, L in mm).

T = F·λ·L — torque per roll (N·m with L in m); λ ≈ 0.5 hot, 0.45 cold.

P = 2·T·ω = 2πN·F·L / 60 (λ = 0.5) — total power for both rolls (W); N = roll speed (rpm), F in N, L in m.

Forward slip = (v_f − v_r) / v_r.

Worked examples

Example 1 (standard — force and power). A 300 mm wide strip is cold rolled from 25 mm to 22 mm with rolls of 250 mm radius at 50 rpm. The metal has K = 275 MPa and n = 0.15. Find the roll force and the power (λ = 0.5).

1. Δh = 3 mm, ε = ln(25/22) = 0.1278.
2. Ȳ = 275 × 0.1278^0.15 / 1.15 = 175.6 MPa.
3. L = √(250 × 3) = 27.39 mm.
4. F = 175.6 × 300 × 27.39 = 1.443 × 10⁶ N.
5. P = 2π × 50 × 1.443 × 10⁶ × 0.02739 / 60 = 2.07 × 10⁵ W.

Answer: F ≈ 1.44 MN, P ≈ 207 kW.

Example 2 (GATE level — maximum draft and passes). A 30 mm thick slab is to be hot rolled to 18 mm with rolls of 300 mm radius; μ = 0.1. Find the maximum draft per pass, the bite angle and the minimum number of passes.

1. Δh_max = μ²R = 0.1² × 300 = 3 mm.
2. α_max = tan⁻¹(0.1) = 5.71° (check: cos α = 1 − 3/600 = 0.995 → α = 5.73°, consistent).
3. Total draft = 30 − 18 = 12 mm; passes = 12/3 = 4.

Answer: 3 mm per pass, α ≈ 5.7°, at least 4 passes.

Example 3 (exit speed). Strip enters at 1.0 m/s and 20 mm thick, leaves at 16 mm, width unchanged. v_f = 1.0 × 20/16 = 1.25 m/s. If the roll surface speed is 1.2 m/s, forward slip = (1.25 − 1.2)/1.2 = 4.2%.

Common mistakes

• Computing power as F × v. Use torque × angular speed for both rolls.
• Using roll diameter instead of radius in L = √(RΔh) or Δh_max = μ²R.
• Using the draft on one side (Δh/2) in the contact-length formula; L uses the full draft with radius R.
• Forgetting that power counts two rolls.
• Using the final flow stress in F = Ȳ·w·L; the formula uses the average over the pass.
• Mixing mm and m in the power formula.

For GATE PI

Expect numericals on maximum draft or minimum roll radius from the bite condition, number of passes, contact length, roll force, torque and power, exit velocity and forward slip, and the neutral point. One-mark questions test where the neutral point is, why back-up rolls are used, and how friction, roll size and tension affect force. Practise a full force–torque–power calculation with units checked at each line.

Quick check

1. μ = 0.12, R = 400 mm. Maximum draft?
2. Where is the roll pressure highest?
3. Draft 4 mm, R = 225 mm. Contact length?
4. Why do cold mills use small work rolls?
5. Strip leaves at 2.5 m/s, rolls at 2.4 m/s. Forward slip?

Answers: 1. 0.0144 × 400 = 5.76 mm. 2. At the neutral point. 3. √(225 × 4) = 30 mm. 4. Shorter contact length → lower roll force and power. 5. 0.1/2.4 ≈ 4.2%.