Merchant's circle and cutting force analysis

Why it matters

Cutting forces size the machine-tool motor, spindle bearings, tool holder and fixture, and they decide deflection, chatter and accuracy. Merchant's circle turns two forces you can measure with a dynamometer (cutting and thrust) into the forces you cannot measure directly — on the shear plane and on the rake face — and from them the shear stress, friction coefficient and energy of cutting.

Key ideas

Assumptions of Merchant's analysis (orthogonal cutting): sharp tool with no flank contact; deformation on a single thin shear plane; continuous chip without BUE; uniform shear stress on the shear plane; plane strain (width of cut much larger than t); the chip is a free body in equilibrium under two equal and opposite resultants.

The three force pairs. The single resultant R between tool and chip can be resolved three ways:

• Along and normal to the cutting velocity: cutting force Fc (main, power-consuming) and thrust force Ft (normal to the machined surface). These are what a dynamometer measures.
• Along and normal to the rake face: friction force F and normal force N. F/N = μ = tan β, where β is the friction angle.
• Along and normal to the shear plane: shear force Fs and normal force Ns. Fs/As gives the shear flow stress of the work material.

Because all three pairs resolve the same R, their tips lie on a circle of diameter R — Merchant's circle. The angle between R and Fc is (β − α).

Merchant's shear angle relation. Assuming the shear angle takes the value that minimises the cutting force (with shear stress independent of normal stress), Merchant obtained 2φ + β − α = 90°. It explains the trends (larger rake or lower friction → larger φ → lower force) but usually overestimates φ; Lee–Shaffer's slip-line solution gives φ + β − α = 45°. Use whichever relation the question names.

Effect of rake angle on thrust. If β < α, the thrust force becomes negative — the tool is pulled into the work. This happens with large positive rake and good lubrication.

Specific cutting energy u = Fc/(w·t) is the energy per unit volume removed; it rises as t falls (size effect), because ploughing and friction become a larger share.

Formulas

• F = Fc·sin α + Ft·cos α, N = Fc·cos α − Ft·sin α, μ = tan β = F / N = (Ft + Fc·tan α) / (Fc − Ft·tan α)
• Fs = Fc·cos φ − Ft·sin φ, Ns = Fc·sin φ + Ft·cos φ
• R = √(Fc² + Ft²); Fc = R·cos(β − α); Ft = R·sin(β − α); Fs = R·cos(φ + β − α)
• As = w·t / sin φ; τs = Fs / As; σs = Ns / As
  • Fc, Ft, F, N, Fs, Ns, R in N; w = width of cut (mm); t = uncut chip thickness (mm); As in mm²; stresses in N/mm² = MPa; α rake, β friction angle, φ shear angle (degrees).
• Fs = τs·w·t / sin φ; Fc = Fs·cos(β − α) / cos(φ + β − α); Ft = Fs·sin(β − α) / cos(φ + β − α)
  • Use when τs is known and you need the forces.
• 2φ + β − α = 90° (Merchant); φ + β − α = 45° (Lee–Shaffer)
• P = Fc·V (W, with V in m/s); u = Fc / (w·t) (in N/mm², which equals MJ/m³; 1 N/mm² = 0.001 J/mm³)
• Power split: shear power Fs·Vs, friction power F·Vchip; Fc·V = Fs·Vs + F·Vchip.

Worked examples

Example 1 (standard). Orthogonal cutting: α = 10°, t = 0.25 mm, w = 2.5 mm, chip thickness 0.60 mm, Fc = 900 N, Ft = 450 N, V = 2 m/s. Find μ, the shear stress, the power and the specific energy.

1. r = 0.25/0.60 = 0.4167; tan φ = 0.4167 × 0.9848/(1 − 0.4167 × 0.1736) = 0.4423 → φ = 23.86°.
2. F = Fc·sin α + Ft·cos α = 900 × 0.1736 + 450 × 0.9848 = 156.3 + 443.2 = 599.4 N.
3. N = Fc·cos α − Ft·sin α = 886.3 − 78.1 = 808.2 N → μ = 599.4/808.2 = 0.742 (β = 36.6°).
4. Fs = Fc·cos φ − Ft·sin φ = 900 × 0.9145 − 450 × 0.4045 = 823.0 − 182.0 = 641.0 N.
5. As = w·t / sin φ = 2.5 × 0.25 / 0.4045 = 1.545 mm² → τs = 641.0/1.545 = 415 MPa.
6. P = Fc·V = 900 × 2 = 1800 W; u = 900/(2.5 × 0.25) = 1440 N/mm² = 1.44 J/mm³.

Example 2 (GATE level). A steel with shear flow stress τs = 400 MPa is cut orthogonally with α = 10°, t = 0.2 mm, w = 3 mm, μ = 0.5. Using Merchant's relation, find φ, Fc and Ft, and the power at V = 1.5 m/s.

1. β = tan⁻¹ 0.5 = 26.57°.
2. 2φ + β − α = 90° → φ = (90 + 10 − 26.57)/2 = 36.72°.
3. Fs = τs·w·t / sin φ = 400 × 3 × 0.2 / 0.5979 = 240 / 0.5979 = 401.4 N.
4. β − α = 16.57°; φ + β − α = 53.28°; cos 53.28° = 0.5979.
5. Fc = Fs·cos(β − α)/cos(φ + β − α) = 401.4 × 0.9585 / 0.5979 = 643.6 N.
6. Ft = Fs·sin(β − α)/cos(φ + β − α) = 401.4 × 0.2851 / 0.5979 = 191.4 N.
7. P = Fc·V = 643.6 × 1.5 = 965 W. (Check: with Merchant's relation φ + β − α = 90° − φ, so cos(φ + β − α) = sin φ — a quick shortcut.)

Common mistakes

• Using β = φ − α or other invented relations; the friction angle comes from F/N, or from Merchant's relation 2φ + β − α = 90°.
• Mixing up which pair a formula needs: F and N use the rake angle α; Fs and Ns use the shear angle φ.
• Using t·w as the shear-plane area — it must be divided by sin φ.
• Taking Ft as always positive: for β < α it is negative.
• Unit slips in specific energy: N/mm² is numerically MJ/m³, and 1 N/mm² = 0.001 J/mm³.
• Applying Merchant's relation when the question gives the measured chip thickness — then compute φ from r instead.

For GATE PI

• The classic 2-mark NAT: given Fc, Ft, α, and chip data, find μ, Fs, τs, or the power split between shear and friction.
• Given τs and μ, use Merchant's relation to predict φ and then the forces and power.
• Specific energy and power from Fc and V; MRR-based power estimates.
• Conceptual MCQs on the assumptions of Merchant's theory, the sign of thrust force, and the Merchant vs Lee–Shaffer relations.

Quick check

1. Fc = 1000 N, Ft = 500 N, α = 0°. What is μ?
2. α = 10°, β = 30°. What shear angle does Merchant's relation predict?
3. Write As in terms of w, t and φ.
4. When does the thrust force become negative?
5. Fc = 800 N at V = 2.5 m/s: cutting power?

Answers: 1. With α = 0, μ = Ft/Fc = 0.5. 2. φ = (90 + 10 − 30)/2 = 35°. 3. As = w·t / sin φ. 4. When β < α. 5. 2000 W.