Mechanics of orthogonal cutting and Merchant's circle

Why it matters

Every turning, milling and drilling calculation – spindle power, tool deflection, fixture clamping force, tool temperature – starts from the cutting force. The orthogonal (2-D) model and Merchant's circle turn two dynamometer readings and a chip-thickness measurement into shear angle, friction coefficient, shear stress and energy, and they explain why positive rake, sharp tools and good lubrication lower power. This is one of the highest-yield numerical topics in manufacturing.

Key ideas

Orthogonal vs oblique cutting. In orthogonal cutting the cutting edge is perpendicular to the cutting velocity and wider than the chip, so the deformation is two-dimensional (plane strain). In oblique cutting the edge is inclined (inclination angle i ≠ 0) and the chip flows sideways. Real turning is oblique, but orthogonal analysis is the standard model; turning a thin-walled tube end-on with a straight edge, or plain-turning with a 90° approach angle, approximates it, with depth of cut playing the role of width b and feed the role of uncut thickness t.

Tool angles that matter here. Rake angle α – between the rake face and the reference plane normal to the cutting velocity (positive rake makes a sharper, weaker wedge). Clearance angle – between the flank and the machined surface; it prevents rubbing.

Chip formation. The work material shears along a narrow shear plane inclined at the shear angle φ to the cutting direction; the chip then slides up the rake face with friction.

• Uncut thickness t becomes chip thickness t_c > t, so the chip thickness ratio r = t/t_c < 1. Its inverse, the chip reduction coefficient, is > 1.
• A large φ means a short shear plane, thin chip, less strain and lower force. Larger rake and lower friction raise φ.
• Chip types: continuous (ductile material, high speed, positive rake – good finish), continuous with built-up edge (ductile material at low speed, adhesion – poor finish), discontinuous (brittle material such as grey cast iron, or low speed with negative rake), serrated/segmented (titanium, hardened steel – localised shear bands).

Forces – the Merchant circle. Measured forces are the cutting force F_c (along the velocity, it sets power) and the thrust force F_t (normal to it, it deflects the tool and work). The resultant R can be resolved two other 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. All six components are drawn on a circle with R as the diameter – Merchant's circle diagram.

Shear-angle relations. Merchant minimised cutting energy assuming the shear stress is constant: 2φ + β − α = 90°. Lee and Shaffer (slip-line theory) obtained φ + β − α = 45°. Both show φ rises with rake angle and falls with friction; real values differ, so use whichever relation the question names.

Velocities. Cutting velocity V, chip velocity V_c along the rake face and shear velocity V_s along the shear plane form a closed triangle.

Energy. Power P = F_c·V. Most of it (often well over half) is spent in shearing, the rest in rake-face friction; almost all becomes heat in the chip, tool and work. Specific cutting energy u = F_c/(b·t) is roughly constant for a material and is used to estimate machine power.

Formulas

r = t / t_c = sin φ / cos(φ − α) tan φ = r·cos α / (1 − r·sin α) r = chip thickness ratio (dimensionless), t = uncut chip thickness (m), t_c = chip thickness (m), φ = shear angle, α = rake angle. By volume constancy r also equals chip length / uncut length = V_c / V.

F = F_c·sin α + F_t·cos α N = F_c·cos α − F_t·sin α μ = tan β = F / N Rake-face friction and normal forces (N). Signs as written are for positive rake.

F_s = F_c·cos φ − F_t·sin φ F_n = F_c·sin φ + F_t·cos φ Shear-plane forces (N).

A_s = b·t / sin φ τ_s = F_s / A_s Shear-plane area (m²) and shear stress (Pa); b = width of cut (m).

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

F_c = τ_s·b·t·cos(β − α) / [sin φ·cos(φ + β − α)] Cutting force from shear stress and angles.

2φ + β − α = 90° (Merchant) φ + β − α = 45° (Lee–Shaffer)

γ = cot φ + tan(φ − α) Shear strain (dimensionless).

V_c = V·sin φ / cos(φ − α) = r·V V_s = V·cos α / cos(φ − α) Chip and shear velocities (m/s).

P = F_c·V u = F_c / (b·t) P = F_s·V_s + F·V_c Power (W), specific energy (N/mm² = 10⁻³ J/mm³), and the split into shear and friction power.

Worked examples

Example 1 (standard) – full orthogonal analysis. Given: α = +10°, t = 0.2 mm, b = 2.5 mm, t_c = 0.5 mm, V = 2 m/s, F_c = 800 N, F_t = 400 N.

1. r = 0.2/0.5 = 0.4; tan φ = r cos α/(1 − r sin α) = 0.3939/0.9305 ⇒ φ = 22.9°.
2. F = 800 sin 10° + 400 cos 10° = 138.9 + 393.9 = 532.8 N; N = 800 cos 10° − 400 sin 10° = 787.8 − 69.5 = 718.4 N.
3. μ = 532.8/718.4 = 0.742 (β = 36.6°).
4. F_s = 800 cos 22.94° − 400 sin 22.94° = 736.6 − 155.9 = 580.8 N; A_s = 2.5 × 0.2/sin 22.94° = 1.283 mm²; τ_s = 453 MPa.
5. γ = cot 22.94° + tan 12.94° = 2.364 + 0.230 = 2.59.
6. V_c = 0.4 × 2 = 0.8 m/s; V_s = 2 cos 10°/cos 12.94° = 2.02 m/s.
7. P = 800 × 2 = 1600 W; check: F_s·V_s + F·V_c = 1173.7 + 426.3 = 1600 W ✓ (73 % in shear). u = 800/0.5 = 1600 N/mm² = 1.6 J/mm³.

Example 2 (GATE level) – predict forces with Merchant's theory. Given: α = 10°, μ = 0.5, τ_s = 400 MPa, b = 2 mm, t = 0.2 mm.

1. β = tan⁻¹ 0.5 = 26.57°.
2. Merchant: φ = 45° + α/2 − β/2 = 45 + 5 − 13.28 = 36.72°.
3. F_c = τ_s·b·t·cos(β − α)/[sin φ·cos(φ + β − α)] = 400 × 0.4 × cos 16.57°/(sin 36.72° × cos 53.28°) = 153.35/(0.5979 × 0.5979) = 429 N.
4. F_t = F_c·tan(β − α) = 429 × 0.2975 = 128 N.
5. Predicted chip thickness: r = sin 36.72°/cos 26.72° = 0.669 ⇒ t_c = 0.2/0.669 = 0.30 mm.

Common mistakes

• Swapping the sin/cos terms in F and N, or forgetting the sign change for negative rake.
• Using r = t_c/t (that is the chip reduction coefficient, > 1).
• Finding F_s as F_c/cos φ – the thrust force must be included: F_s = F_c cos φ − F_t sin φ.
• Calculator in radians when angles are in degrees.
• Forgetting that in turning the feed is the uncut thickness and the depth of cut is the width (for a 90° approach angle).
• Applying Merchant's relation when the question gives Lee–Shaffer, or vice versa.
• Specific energy units: N/mm² equals MPa, which is 10⁻³ J/mm³, not J/mm³.

For GATE ME

This topic produces several numericals: shear angle from chip thickness, friction coefficient from F_c and F_t, shear stress and shear-plane area, shear strain, chip and shear velocities, power and specific energy, and force prediction from Merchant or Lee–Shaffer. Conceptual items cover chip types and BUE, the effect of rake and friction on φ, and orthogonal vs oblique cutting. Practise the full chain in Example 1 until it takes under five minutes.

Quick check

1. If t = 0.25 mm and t_c = 0.5 mm, what is r?
2. Write Merchant's shear-angle relation.
3. With F_c = 1000 N, F_t = 500 N and α = 0°, what is μ?
4. Which force sets the cutting power?
5. Which chip type forms when machining grey cast iron?

Answers: 1. 0.5; 2. 2φ + β − α = 90°; 3. μ = F_t/F_c = 0.5; 4. Cutting force F_c; 5. Discontinuous.