Orthogonal and oblique cutting; chip formation

Why it matters

All metal cutting is controlled shearing: the tool pushes a layer of metal until it shears along a narrow zone and flows up the rake face as a chip. The chip's thickness, shape and speed tell you the shear angle, the strain and the energy being spent, so chip measurements are the starting point for every force, power and temperature calculation that follows.

Key ideas

Orthogonal cutting. The cutting edge is straight, perpendicular to the cutting velocity and to the feed direction (inclination angle λ = 0), and wider than the chip. The deformation is then two-dimensional (plane strain), the chip flows straight up the rake face, and only two force components act (cutting force Fc and thrust force Ft). Examples: turning the end of a thin tube with the edge square to the axis, planing a narrow plate, parting off.

Oblique cutting. The cutting edge is inclined to the cutting velocity (λ ≠ 0). The chip flows sideways at a chip-flow angle ηc ≈ λ (Stabler's rule), there are three force components (cutting, feed and radial), and the analysis is three-dimensional. Most practical operations — turning with an approach angle, milling with helical cutters, drilling — are oblique. Oblique cutting gives a longer effective edge, smoother entry and better chip control.

Model of chip formation. Merchant's model treats the deformation as simple shear on a single plane, the shear plane, inclined at the shear angle φ to the cutting velocity. The uncut layer of thickness t is converted into a chip of thickness tc > t. Because volume is conserved (and width barely changes in orthogonal cutting), the chip is shorter and slower than the uncut layer.

Chip thickness ratio r = t / tc (always < 1). Its reciprocal 1/r is the chip reduction coefficient ζ. A higher r means a larger shear angle, less strain, lower force and better machining.

Types of chips

• Continuous — ductile work, high speed, positive rake, small feed, good lubrication. Good finish but needs chip breakers for safety.
• Continuous with built-up edge (BUE) — ductile work at low-to-moderate speed with high friction: work metal welds to the rake face, grows and breaks off, spoiling the finish and changing the effective rake. BUE disappears at higher speed (higher temperature), with cutting fluid, or with larger rake.
• Discontinuous (segmented) — brittle materials (grey cast iron, brass), very low speed, large feed or negative rake. Fine chips, acceptable finish in cast iron, low power.
• Serrated (shear-localised) — hard-to-machine metals with low thermal conductivity (titanium, hardened steel) at high speed: saw-tooth chips from adiabatic shear bands.

Shear zones. Primary shear zone (along the shear plane, plastic deformation of the chip), secondary shear zone (chip–tool contact on the rake face, friction and heat), and a tertiary zone on the flank (rubbing on the machined surface).

Formulas

• r = t / tc = lc / l = Vc_chip / V
  • t = uncut chip thickness (mm), tc = chip thickness (mm), lc = chip length, l = length of uncut layer (same units), Vc_chip = chip velocity, V = cutting velocity (m/s or m/min). Based on volume constancy with constant width.
• tan φ = r·cos α / (1 − r·sin α)
  • φ = shear angle, α = orthogonal rake angle (degrees).
• γ = cot φ + tan(φ − α) = cos α / [sin φ·cos(φ − α)]
  • γ = shear strain (dimensionless).
• Vs = V·cos α / cos(φ − α) and Vc_chip = V·sin φ / cos(φ − α) = r·V
  • Vs = shear velocity along the shear plane. The three velocities form a closed triangle.
• ε̇ ≈ Vs / Δs
  • shear strain rate (1/s), Δs = thickness of the shear zone (m). Typical values are 10⁴–10⁶ s⁻¹.

Worked examples

Example 1 (standard). In orthogonal cutting with a tool of rake α = 10°, uncut chip thickness t = 0.25 mm, chip thickness tc = 0.60 mm and cutting speed V = 2 m/s. Find r, φ, γ, chip velocity and shear velocity.

1. r = t / tc = 0.25 / 0.60 = 0.417.
2. tan φ = r·cos α / (1 − r·sin α) = 0.4167 × 0.9848 / (1 − 0.4167 × 0.1736) = 0.4103 / 0.9277 = 0.4423 → φ = 23.9°.
3. γ = cot φ + tan(φ − α) = cot 23.86° + tan 13.86° = 2.261 + 0.247 = 2.51.
4. Chip velocity = r·V = 0.417 × 2 = 0.833 m/s.
5. Vs = V·cos α / cos(φ − α) = 2 × 0.9848 / cos 13.86° = 1.9696 / 0.9709 = 2.03 m/s.

Example 2 (GATE level). A tube is turned orthogonally (edge square to the tube axis) with rake α = 8°. A 200 mm length of uncut layer produces a chip 80 mm long. Cutting speed V = 150 m/min. Find the shear angle, shear strain, chip velocity and shear velocity.

1. r = lc / l = 80 / 200 = 0.40 (chip length measurement avoids measuring a rough chip thickness).
2. tan φ = 0.40 × cos 8° / (1 − 0.40 × sin 8°) = 0.3961 / 0.9443 = 0.4195 → φ = 22.8°.
3. γ = cot 22.76° + tan 14.76° = 2.384 + 0.263 = 2.65.
4. Chip velocity = r·V = 0.40 × 150 = 60 m/min. Check: V·sin φ / cos(φ − α) = 150 × 0.3868 / 0.9670 = 60.0 m/min.
5. Vs = 150 × cos 8° / cos 14.76° = 148.54 / 0.9670 = 153.6 m/min. Note that Vs > V: the material slides along the shear plane faster than the tool advances.

Common mistakes

• Inverting the chip thickness ratio: r = t/tc is less than 1. If you get r > 1 you have divided the wrong way.
• Using the length ratio the wrong way round: r = lc/l (chip length over uncut length), not l/lc.
• Using degrees in a calculator set to radians, or forgetting to take tan⁻¹ at the end.
• Calling every chip from steel "continuous" — at low speed with no fluid it carries a BUE.
• Assuming oblique cutting is rare; nearly all real operations are oblique, orthogonal cutting is the analysis model.
• Treating the shear plane as a material property; φ changes with rake, friction and speed.

For GATE PI

• NAT problems on chip thickness ratio from thickness, length or velocity data, then shear angle, shear strain and the velocity triangle.
• MCQs on chip types and the conditions that give each, and on BUE and how to suppress it.
• Distinguishing orthogonal and oblique cutting (number of force components, chip-flow direction, λ).
• The answers here feed straight into Merchant's circle problems in the next topic, so practise carrying 4 significant figures.

Quick check

1. t = 0.2 mm, tc = 0.5 mm. What is the chip reduction coefficient?
2. For r = 0.5 and α = 10°, what is the shear angle?
3. Which chip type is expected when machining grey cast iron?
4. In oblique cutting, roughly what is the chip-flow angle if the inclination angle is 10°?
5. Is the shear velocity smaller or larger than the cutting velocity for a typical positive-rake cut?

Answers: 1. 1/r = 0.5/0.2 = 2.5. 2. tan φ = 0.4924/0.9132 = 0.539, φ ≈ 28.3°. 3. Discontinuous. 4. About 10° (Stabler's rule ηc ≈ λ). 5. Larger.