Manufacturing · Chapter 5 of 10 · Intermediate
Theory of Metal Cutting
A cutting tool does not slice metal, it shears it. Understanding the chip, the shear plane, and the forces explains why machining takes the power it does and leaves the finish it leaves.
Readiness check
This chapter uses trigonometry and force balance. Tick only what you can do closed-notes.
- Work with tangent, cotangent, and angles.
- Resolve a force into components.
- Recall shear stress and shear strain.
- Compute power as force times velocity.
- Recall material removal as volume per time.
The core idea
In cutting, the tool concentrates deformation onto a thin shear plane; the chip forms by intense shear there, and the geometry of that plane sets the forces and power.
r = t₀/tctan φ = r cosα/(1 − r sinα)P = Fc·V = u·MRRAs the tool advances, metal ahead of it shears along a narrow plane inclined at the shear angle φ, then slides up the rake face as a chip. The chip comes off thicker and shorter than the layer removed, and measuring that thickness ratio gives the shear angle and the (very large) shear strain. The cutting force times the speed is the power, which also equals the specific cutting energy times the material removal rate.
The skills, taught in order
Cutting theory is the mechanics of one shear plane. Five skills cover chip formation, the shear angle, shear strain, forces, and power.
5.1 Chip formation
The orthogonal model treats cutting as two-dimensional shear along a plane. The chip that results depends on the material and conditions.
| Chip type | Conditions | Consequence |
|---|---|---|
| Continuous | ductile metal, high speed, sharp tool | good finish |
| Built-up edge | low speed, high friction | poor finish, edge welds on |
| Discontinuous | brittle material, low speed | rough surface, lower force |
| Serrated | hard, low-conductivity (titanium) | cyclic forces, tool stress |
5.2 Chip-thickness ratio and shear angle
Because volume is conserved, the chip is thicker (tc) than the uncut layer (t₀), and the ratio r = t₀/tc < 1. The shear angle follows from r and the rake angle α: tan φ = r cosα/(1 − r sinα). A larger shear angle means a thinner chip and lower force.
5.3 Shear strain
The strain on the shear plane is γ = cot φ + tan(φ − α), typically 2 to 5, far larger than any tensile test reaches. This explains why cutting generates so much heat and why chips are so heavily worked.
5.4 Cutting forces
Merchant's circle resolves the resultant cutting force into the cutting force Fc (along the velocity, doing the work) and the thrust force Ft (perpendicular), and relates them to the shear-plane and friction forces. Merchant's analysis predicts the shear angle that minimises energy: φ = 45° + α/2 − β/2, where β is the friction angle.
5.5 Power and specific energy
The cutting power is P = Fc·V. It also equals the specific cutting energy u (energy per unit volume, a material property) times the material removal rate: P = u·MRR. This lets you size a machine's spindle motor from the cut.
Engineering connection: these relations set spindle power, tool forces, and surface finish, and they feed straight into the process calculations of Chapter 6.
Worked example 1: shear angle and strain
In orthogonal cutting with a 10° rake angle, the uncut chip thickness is 0.2 mm and the chip comes off 0.5 mm thick. Find the chip-thickness ratio, the shear angle, and the shear strain.
- ProblemFind r, φ, and γ for the cut in Figure 1.
- Given / findα = 10°, t₀ = 0.2 mm, tc = 0.5 mm. Find the chip ratio, shear angle, and shear strain.
- AssumptionsOrthogonal cutting, thin shear plane, volume conserved.
- ModelChip ratio from the thicknesses, shear angle from the geometry relation, shear strain from φ and α.
- Equationsr = t₀/tc tan φ = r cosα/(1 − r sinα) γ = cot φ + tan(φ − α)
- Solver = 0.2/0.5 = 0.40. tan φ = 0.40 cos10°/(1 − 0.40 sin10°) = 0.394/0.931 = 0.423, so φ = 22.9°. γ = cot 22.9° + tan(22.9° − 10°) = 2.36 + 0.23 = 2.6.
- CheckA shear strain of 2.6 dwarfs the ~0.25 a tensile test reaches at fracture, which is why chips are so hot and hard. A larger rake angle would raise φ, thin the chip, and lower the strain and force.
- ConclusionMeasuring the chip thickness is enough to recover the whole shear-plane geometry. The huge strain explains machining's heat, the central challenge of tool life in the next chapter.
Worked example 2: cutting power and force
A cut runs at 2 m/s with a 0.2 mm uncut chip thickness and a 4 mm width, on a steel of specific cutting energy u = 2.8 J/mm³. Find the material removal rate, the cutting power, and the cutting force.
- ProblemFind the MRR, cutting power, and cutting force for the cut in Figure 2.
- Given / findV = 2 m/s = 2000 mm/s, t₀ = 0.2 mm, w = 4 mm, u = 2.8 J/mm³. Find MRR, P, Fc.
- AssumptionsOrthogonal cut, all cutting energy goes through this zone, specific energy constant.
- ModelRemoval rate is the cut cross-section times speed; power is u times MRR; cutting force is power over speed.
- EquationsMRR = V·t₀·w P = u·MRR Fc = P/V
- SolveMRR = 2000 × 0.2 × 4 = 1600 mm³/s. P = 2.8 × 1600 = 4480 W = 4.48 kW. Fc = 4480/2 = 2240 N (using V in m/s).
- CheckEquivalently Fc = u × (t₀w) = 2800 N/mm² × 0.8 mm² = 2240 N, matching. The 4.5 kW sets the minimum spindle motor; a real machine adds friction and idle losses.
- ConclusionPower and force fall straight out of the specific cutting energy and the removal rate. This is how you check whether a machine can take a proposed cut, the bridge to selecting feeds and speeds in Chapter 6.
Misconceptions and diagnostics
| Mistake | Symptom | Diagnostic question | Correction |
|---|---|---|---|
| Cutting seen as scraping | Forces and heat mis-estimated | "Where does the metal shear?" | Metal shears on the shear plane; the chip forms by shear, not scraping. |
| Chip ratio inverted | Shear angle comes out wrong | "Is r = t₀/tc less than one?" | The chip is thicker than the cut, so r < 1. |
| Mixing speed units | Force off by a factor of 1000 | "Is V in m/s for P = FcV?" | Keep V in m/s when combining with power in watts. |
| Ignoring rake angle | Shear angle and force mispredicted | "Did I include α?" | A higher rake angle raises the shear angle and lowers the force. |
Practice ladder
With a 0° rake angle, the uncut chip is 0.15 mm and the chip is 0.30 mm. Find the chip ratio and shear angle.
Show answer
r = 0.5. tan φ = 0.5 cos0°/(1 − 0.5 sin0°) = 0.5, so φ = 26.6°. With zero rake the relation simplifies to tan φ = r.
For the Worked Example 2 cut, what spindle power is needed if the speed doubles to 4 m/s at the same depth and width?
Show answer
MRR doubles to 3200 mm³/s, so P = 2.8 × 3200 = 8960 W ≈ 9 kW. The cutting force stays 2240 N (same cross-section), but the power doubles with speed.
A cut removes 1200 mm³/s of aluminium with specific energy 0.7 J/mm³ at 3 m/s. Find the power and cutting force.
Show answer
P = 0.7 × 1200 = 840 W. Fc = P/V = 840/3 = 280 N. Aluminium's low specific energy is why it machines so easily compared with steel.
Observe a real machining operation (a lathe, a mill, a drill). Estimate the cut cross-section and speed, and use a specific cutting energy to estimate the power, then compare with the machine's rating.
What good work looks like
A removal rate estimated from the cut, a specific energy chosen for the material, a power figure, and a sanity check against the machine's spindle power.
Working with AI, and proving it yourself
Use AI as an examiner, not a solver
Portfolio task
Analyse one orthogonal-style cut: from chip measurements get the shear angle and strain, and from the cut size and speed get the power and force.
Retrieval and spaced review
Closed notes. Answer out loud, then reveal.
1. How is metal removed in cutting?
By intense shear along a thin shear plane; the chip forms there and slides up the rake face.
2. Write the chip ratio and shear-angle relation.
r = t₀/tc; tan φ = r cosα/(1 − r sinα).
3. Why is the shear strain so large?
γ = cot φ + tan(φ − α) is typically 2 to 5, far above tensile-test strains, hence the heat.
4. Give two expressions for cutting power.
P = Fc·V and P = u·MRR.
5. What does a built-up edge do?
It welds material to the tool tip at low speed, worsening the finish.
Textbook mapping
| Item | Mapping |
|---|---|
| Primary source | Kalpakjian and Schmid, Manufacturing Engineering and Technology, Chapter 21 (Fundamentals of Machining) |
| Cross-reference | Groover, Ch. 21 · DeGarmo, machining theory chapters |
| Core topics | 5.1 Chip formation · 5.2 Shear angle · 5.3 Shear strain · 5.4 Cutting forces · 5.5 Power and specific energy |
| Engineering connection | Sets spindle power, tool forces, and surface finish. |
| Read next | Chapter 6: Machining Processes and Tool Life. |