Tool wear, Taylor's tool life equation and machinability

Why it matters

A worn tool raises cutting forces and temperature, spoils surface finish and size, and eventually breaks. Tool changes cost machine time and money, so a production engineer has to predict how long a tool will last at a given speed and feed — Taylor's equation is the tool for that, and it is the basis of the machining-economics calculations in the next topic.

Key ideas

Modes of tool failure

• Gradual wear — the normal, predictable end of a tool's life (flank and crater wear).
• Plastic deformation of the edge — when the edge softens at excessive temperature.
• Brittle fracture / chipping — mechanical or thermal shock, interrupted cuts, brittle tool materials.

Wear mechanisms

• Abrasion — hard particles (carbides, oxides, scale) plough the tool surfaces; dominant at low and moderate speeds and the main cause of flank wear.
• Adhesion (attrition) — micro-welds between chip and tool are torn away, taking tool material; linked with BUE at low speed.
• Diffusion — at high interface temperatures atoms (e.g. carbon, cobalt, tungsten) diffuse from tool to chip, weakening the tool surface; main cause of crater wear on carbide at high speed.
• Oxidation / chemical wear — oxide formation at the edges of the contact, often causing notch wear at the depth-of-cut line.

Types (locations) of wear

• Flank wear — a wear land on the flank, measured as the average land width VB. It controls dimensional accuracy and finish, so it is the usual life criterion.
• Crater wear — a depression on the rake face a short distance behind the edge (where the temperature peaks), measured by its depth KT. Deep craters weaken the edge.
• Notch wear at the depth-of-cut line, nose wear, and chipping.

Flank wear curve. VB against cutting time shows three stages: rapid initial (break-in) wear, a long steady region of nearly uniform wear rate, and an accelerating final region before failure. Tools are changed before the final stage.

Tool life criteria. Tool life T is the cutting time until a chosen criterion is reached — for example a uniform flank wear of 0.3 mm or a maximum of 0.6 mm (ISO 3685 style criteria for carbide), a crater depth limit, loss of finish or size, or catastrophic failure. Always state the criterion.

Taylor's tool life equation. Experiments at several speeds give straight lines on a log–log plot of V against T: V·Tⁿ = C. Speed dominates tool life; a small speed increase cuts life sharply. Typical n values: HSS about 0.1–0.15, cemented carbide about 0.2–0.3, ceramics about 0.4–0.6 (take exact values from data). C is the cutting speed that gives a tool life of 1 minute. The extended form adds feed and depth of cut, with speed the strongest effect, then feed, then depth.

Machinability is the ease with which a material can be machined, judged by tool life (most common), surface finish, cutting force/power and chip form. The machinability index compares the speed for a fixed tool life (e.g. 60 min) with that of a reference material (free-cutting steel, AISI 1112, = 100 %). Machinability improves with additions of lead, sulphur (MnS inclusions) or bismuth, with annealed microstructures and with high thermal conductivity; it falls with hardness, work-hardening rate and abrasive inclusions.

Formulas

• V·Tⁿ = C
  • V = cutting speed (m/min); T = tool life (min); n = Taylor exponent (dimensionless); C = Taylor constant (m/min, the speed for T = 1 min).
• n = ln(V₂/V₁) / ln(T₁/T₂) (from two tests)
• T₂ / T₁ = (V₁ / V₂)^(1/n)
• V·Tⁿ·f^a·d^b = K (extended Taylor equation; f feed in mm/rev, d depth of cut in mm; exponents typically a > b and both fitted from tests)
• Machinability index (%) = V₆₀(material) / V₆₀(reference) × 100

Worked examples

Example 1 (standard). A carbide tool gives a life of 60 min at V = 100 m/min and 12 min at V = 150 m/min. Find n and C, and the tool life at 120 m/min.

1. n = ln(V₂/V₁) / ln(T₁/T₂) = ln(1.5) / ln(5) = 0.4055 / 1.6094 = 0.252.
2. C = V₁·T₁ⁿ = 100 × 60^0.252 = 280.5 m/min. (Check: 150 × 12^0.252 = 280.5.)
3. At 120 m/min: T = (C / V)^(1/n) = (280.5/120)^(1/0.252) = 2.3375^3.968 = ≈ 29.1 min.

Example 2 (GATE level). The extended equation for a tool–work pair is V·T^0.25·f^0.5·d^0.3 = K. A job is run at V = 120 m/min, f = 0.25 mm/rev, d = 2 mm with a tool life of 40 min. To raise output, the feed is increased to 0.30 mm/rev and the depth to 2.5 mm. What speed keeps the tool life at 40 min, and by what percentage does the material removal rate change?

1. With T fixed, V·f^0.5·d^0.3 is constant: V₂ = V₁·(f₁/f₂)^0.5·(d₁/d₂)^0.3.
2. V₂ = 120 × (0.25/0.30)^0.5 × (2/2.5)^0.3 = 120 × 0.9129 × 0.9353 = 102.5 m/min.
3. MRR ∝ V·f·d: before 120 × 0.25 × 2 = 60; after 102.5 × 0.30 × 2.5 = 76.8 (units of mm·m/min, i.e. cm³/min).
4. MRR rises by ≈ 28 % at the same tool life — raising feed and depth is a better way to boost output than raising speed, because tool life is least sensitive to them.

Example 3 (sensitivity). With n = 0.25, a 20 % increase in speed changes tool life by (1/1.2)^(1/0.25) = 1.2⁻⁴ = 0.482 — tool life falls by about 52 %.

Common mistakes

• Swapping the ratio in n: the speed ratio and the life ratio go opposite ways (higher speed, shorter life), so n = ln(V₂/V₁)/ln(T₁/T₂).
• Raising to the power n instead of 1/n when solving for T.
• Giving C without units — it is a speed (m/min) with T in minutes; if T is in seconds C changes.
• Assuming tool life is in machining hours or number of parts without converting.
• Treating BUE as a wear type; it is a deposit that causes adhesive wear and edge chipping when it breaks off.
• In the extended equation, keeping T fixed but forgetting that the other variables then trade off against V only — write the ratio form before substituting.

For GATE PI

• Two-test problems: find n and C, then predict tool life or speed.
• Percentage change in tool life for a given change in speed; extended Taylor with feed and depth.
• Combining tool life with machining time (number of parts per tool, tool changes per batch).
• MCQs on wear mechanisms and their locations (diffusion → crater, abrasion → flank), wear-curve stages and machinability.

Quick check

1. V = 100 m/min, n = 0.25, C = 400 m/min. Tool life?
2. Which wear mechanism mainly causes crater wear on carbide at high speed?
3. If n = 0.5 and speed is doubled, what happens to tool life?
4. What does C physically represent in V·Tⁿ = C?
5. Which is the usual tool-life criterion in finish turning?

Answers: 1. T = (400/100)^4 = 256 min. 2. Diffusion. 3. It falls to (1/2)² = 1/4. 4. The cutting speed that gives a 1-minute tool life. 5. A limiting flank wear land width VB.