Mechanics of orthogonal cutting and Merchant's circle

Why it matters

Cutting forces size the machine-tool motor, spindle bearings, tool holder and fixture, and they set the power bill and the deflection that limits accuracy. Merchant's analysis of orthogonal cutting turns two dynamometer readings and a chip measurement into shear angle, friction coefficient, shear stress and power, and it is the basis of almost every machining numerical in GATE.

Key ideas

Orthogonal vs oblique cutting. In orthogonal cutting the cutting edge is perpendicular to the cutting velocity and wider than the uncut chip thickness, so the chip flows straight up the rake face and the problem is two-dimensional (plane strain). In oblique cutting the edge is inclined (inclination angle i ≠ 0) and the chip flows sideways; most real turning is oblique, but orthogonal theory captures the physics.

Geometry.

• Uncut chip thickness t (= feed per revolution in orthogonal turning of a tube end), width of cut b.
• Rake angle α: angle between the rake face and the normal to the cutting velocity (normal to the machined surface). Positive rake makes a sharper tool.
• Shear angle φ: angle between the shear plane (where the chip forms) and the cutting velocity direction.
• Chip thickness t_c > t, so the chip thickness ratio r = t/t_c < 1. Its inverse is the chip reduction coefficient.

Forces. A dynamometer measures the cutting force F_c (along the cutting velocity) and thrust force F_t (normal to it, along the feed direction in orthogonal cutting). Their resultant R can be resolved three ways:

• Along and normal to the rake face: friction force F and normal force N, with μ = F/N = tan β (β = friction angle).
• Along and normal to the shear plane: shear force F_s and normal force F_n.
• Along and normal to the cutting velocity: F_c and F_t.

Merchant's circle draws all three pairs as chords of one circle whose diameter is R. From it, the angle between R and F_c is (β − α), and between R and F_s is (φ + β − α).

Assumptions of Merchant's model: sharp tool (no flank contact), continuous chip with no built-up edge, plane strain, shear on a single thin plane, uniform stress on that plane, and constant friction on the rake face.

Merchant's shear-angle relation. Minimising cutting energy with respect to φ (taking shear strength as constant) gives 2φ + β − α = 90°. It shows that a larger rake angle and lower friction raise the shear angle, which shortens the shear plane, thins the chip and lowers forces and power. Measured shear angles are often smaller than Merchant predicts; the Lee–Shaffer relation φ + β − α = 45° is another model.

Velocities. Cutting velocity V, chip velocity V_c (up the rake face), shear velocity V_s (along the shear plane); they form a closed triangle.

Energy. Total power F_c · V splits into shear-plane power F_s · V_s and rake-face friction power F · V_c. Specific cutting energy u = F_c/(b·t) is a material property used to estimate power quickly.

Links. Higher cutting temperature from these energies drives tool wear and Taylor tool life (next topic); F_c sets spindle power in turning and milling.

Formulas

r = t / t_c = V_c / V = l_c / l

• t, t_c = uncut and chip thickness (m), l, l_c = uncut and chip length (m). Volume constancy.

tan φ = r · cos α / (1 − r · sin α)

F = F_c · sin α + F_t · cos α, N = F_c · cos α − F_t · sin α

• Friction and normal force on the rake face (N). μ = F/N = tan β.

F_s = F_c · cos φ − F_t · sin φ, F_n = F_c · sin φ + F_t · cos φ

• Shear and normal force on the shear plane (N).

R = √(F_c² + F_t²), F_c = R · cos(β − α), F_t = R · sin(β − α), F_s = R · cos(φ + β − α)

A_s = b · t / sin φ, τ_s = F_s / A_s

• A_s = shear-plane area (m²), τ_s = shear stress on shear plane (Pa).

2φ + β − α = 90° (Merchant)

V_c = V · sin φ / cos(φ − α), V_s = V · cos α / cos(φ − α)

• Velocities in m/s.

γ = cot φ + tan(φ − α) = cos α / [sin φ · cos(φ − α)]

• Shear strain in the chip (–).

P = F_c · V, u = F_c / (b · t)

• P = cutting power (W); u = specific cutting energy (J/m³; 1 N/mm² = 1 MJ/m³ = 10⁻³ J/mm³).

These relations hold for orthogonal cutting with a sharp tool; with significant flank wear part of F_t acts on the flank and the formulas overstate rake friction.

Worked examples

Example 1 (standard). In orthogonal cutting, F_c = 1000 N, F_t = 500 N, rake angle α = 10°, chip thickness ratio r = 0.5. Find φ, μ, F_s and F_n.

1. tan φ = 0.5 cos 10° / (1 − 0.5 sin 10°) = 0.4924 / 0.9132 = 0.5392 → φ = 28.33°.
2. F = 1000 sin 10° + 500 cos 10° = 173.6 + 492.4 = 666.1 N.
3. N = 1000 cos 10° − 500 sin 10° = 984.8 − 86.8 = 898.0 N.
4. μ = 666.1 / 898.0 = 0.742 (β = 36.57°).
5. F_s = 1000 cos 28.33° − 500 sin 28.33° = 880.2 − 237.3 = 642.9 N.
6. F_n = 1000 sin 28.33° + 500 cos 28.33° = 474.6 + 440.1 = 914.7 N.
7. Final: φ ≈ 28.3°, μ ≈ 0.74, F_s ≈ 643 N, F_n ≈ 915 N. Check: √(642.9² + 914.7²) = 1118 N = R ✓.

Example 2 (GATE level). Orthogonal cutting: uncut chip thickness 0.2 mm, width 3 mm, chip thickness 0.5 mm, rake angle 10°, cutting speed 2 m/s, F_c = 900 N, F_t = 450 N. Find shear angle, shear stress on the shear plane, chip and shear velocities, power split and specific energy. Compare φ with Merchant's prediction.

1. r = 0.2/0.5 = 0.4; tan φ = 0.4 × 0.9848 / (1 − 0.4 × 0.1736) = 0.3939 / 0.9305 = 0.4233 → φ = 22.94°.
2. F_s = 900 cos 22.94° − 450 sin 22.94° = 828.8 − 175.4 = 653.4 N.
3. A_s = b t / sin φ = 3 × 0.2 / 0.3898 = 1.539 mm²; τ_s = 653.4 / 1.539 = 424.5 N/mm² = 424.5 MPa.
4. V_c = r V = 0.4 × 2 = 0.80 m/s; V_s = V cos α / cos(φ − α) = 2 × 0.9848 / cos 12.94° = 2.021 m/s.
5. P = F_c V = 900 × 2 = 1800 W. Friction: F = 900 sin 10° + 450 cos 10° = 599.4 N; P_f = F V_c = 479.6 W. Shear: P_s = F_s V_s = 653.4 × 2.021 = 1320.4 W. Sum ✓ 1800 W.
6. u = F_c/(b t) = 900/(3 × 0.2) = 1500 N/mm² = 1.5 J/mm³.
7. Merchant: μ = 599.4/808.2 = 0.742 → β = 36.57°; φ_M = 45 + α/2 − β/2 = 45 + 5 − 18.28 = 31.7°, larger than the measured 22.9°, as is typical.

Common mistakes

• Using r = t_c/t (greater than 1) in the shear-angle formula.
• Writing F_c = R cos(φ − β); the correct angle between R and F_c is (β − α).
• Mixing up which pair of forces gives friction (rake face) and which gives shear (shear plane).
• Dividing shear force by b·t instead of the inclined shear-plane area b·t/sin φ.
• Forgetting that with negative rake sin α is negative, changing signs in every formula.
• Giving specific energy in J/mm³ without converting N/mm² (1 N/mm² = 0.001 J/mm³).
• Assuming Merchant's relation predicts measured shear angles exactly.

For GATE ME

Orthogonal-cutting numericals are a staple of the machining section: shear angle from chip ratio, friction coefficient from dynamometer forces, shear stress on the shear plane, chip/shear velocity, power and specific energy, or applying Merchant's relation to find φ from α and μ. Practise the full chain from F_c, F_t, r, α in one pass and memorise the angle relations on the circle.

Quick check

1. If t = 0.25 mm and t_c = 0.75 mm, what is the chip thickness ratio?
2. With zero rake angle, what is μ in terms of F_c and F_t?
3. Using Merchant's relation, find φ for α = 10° and β = 40°.
4. If V = 2 m/s and r = 0.4, what is the chip velocity?
5. Why does a larger shear angle reduce cutting force?

Answers: 1. 0.333; 2. μ = F_t / F_c; 3. φ = 45 + 5 − 20 = 30°; 4. V_c = r·V = 0.8 m/s; 5. The shear-plane area b·t/sin φ becomes smaller, so less force is needed to shear it.