Merchant Cutting Simulator Back
Manufacturing Mechanics Simulator

Merchant Cutting Simulator — Orthogonal Cutting Force Analysis

Solve Merchant's equation φ = 45° + α/2 − β/2 for 2D orthogonal cutting. Vary tool rake angle α and tool-chip friction coefficient μ and watch the shear angle, friction angle, cutting force F_c and thrust F_t update in real time.

Parameters
Shear flow stress τ_s
MPa
Depth of cut t
mm
Rake angle α
deg
Friction coefficient μ

Friction angle β = atan(μ) and shear angle φ = 45° + α/2 − β/2. The sweep button slowly varies α from −15° to +30° so you can watch how the cutting and thrust forces respond.

Results
Shear angle φ
Friction angle β
Cutting force F_c / w
Thrust force F_t / w
2D orthogonal cutting model

Wedge tool with rake angle α removes a layer of thickness t. The chip forms by shearing along the shear plane (angle φ). Cutting force F_c (feed direction) and thrust F_t (perpendicular) are drawn as arrows.

Merchant force circle

Resultant R as the circle diameter. Three orthogonal decompositions are shown simultaneously: F_c/F_t (machine frame), F/N (rake face), F_s/N_s (shear plane). Angles α, β and φ are marked as arcs.

Theory & Key Formulas

Friction angle and shear angle from rake angle $\alpha$ and friction coefficient $\mu$:

$$\beta = \arctan(\mu),\qquad \phi = \frac{\pi}{4} + \frac{\alpha}{2} - \frac{\beta}{2}$$

Resultant force per unit width with depth $t$ and shear flow stress $\tau_s$:

$$\frac{R}{w} = \frac{\tau_s\,t}{\sin\phi\,\cos(\phi+\beta-\alpha)}$$

Cutting force (feed direction) and thrust force (perpendicular):

$$\frac{F_c}{w} = \frac{R}{w}\cos(\beta-\alpha),\qquad \frac{F_t}{w} = \frac{R}{w}\sin(\beta-\alpha)$$

Defaults $\tau_s=400$ MPa, $t=0.20$ mm, $\alpha=10°$, $\mu=0.50$ give $\beta=26.57°$, $\phi=36.72°$, $F_c/w \approx 214.5$ N/mm, $F_t/w \approx 63.8$ N/mm.

What is the Merchant cutting simulator?

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My shop manual says "use a larger rake angle for softer materials to reduce cutting force." Why does the rake angle of the tool change how much force is needed? I thought it just changed the direction of cutting.
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Great question. Metal cutting is essentially shearing — the chip peels off along a thin plane called the shear plane, inclined at the shear angle φ from the cutting direction. Merchant showed in 1945 that φ = 45° + α/2 − β/2, so a bigger rake angle α gives a larger φ, a shorter shear plane and less shear work. Move the rake slider from 10° to 30° in the sim and watch F_c drop. Then push α to −15° and see how the force shoots up. That's the same physics that drives all real-world rake-angle selection.
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So bigger rake is always better? Should I just crank everything to +30°?
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That's the trade-off. A large positive rake makes the cutting edge thinner (the included wedge angle is roughly 90° − α − clearance), which lowers force but also lowers strength. Hard or abrasive materials such as cast iron and hardened steel will chip the edge if α is too positive, so you actually use negative rakes (−5° to −10°) and accept the higher force as the price of edge integrity. Soft materials such as aluminium and copper happily run at +15–25°. The simulator lets you quickly check how much extra force a negative rake will cost you before you commit a setup.
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Does the friction coefficient really matter that much? Does cutting fluid actually change the force?
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Hugely. Drop μ from 0.5 to 0.3 in the simulator and β = atan(μ) falls from 26.6° to 16.7°, which pushes φ up from 36.7° to 41.7° and shaves roughly 10–20% off F_c. That is exactly why coatings such as TiN, TiAlN and DLC are worth their cost — they shave a few tenths off μ and gain you lower forces, lower temperatures and longer tool life all at once. The simulator gives you a quick way to estimate the payoff before you spend money on the insert.
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Why do we care about the thrust force F_t? Power only depends on F_c, right?
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Power is indeed F_c·v. F_t doesn't show up in that product, but it pushes the tool away from the workpiece. If the machine or workpiece is not stiff enough, F_t deflects them and you lose dimensional accuracy. Long slender shafts deflect under F_t and end up with a barrel-shaped profile after turning. So when you machine a thin-walled part, you want a positive rake and a low μ specifically to reduce F_t. The simulator lets you check how much F_t will grow if you switch to a negative rake.
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The force circle is pretty but why does R sit on a circle? What's the geometric trick here?
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Thales's theorem: any point on a circle whose diameter is R sees R at a right angle. So all three orthogonal decompositions of the same resultant — machine frame (F_c, F_t), rake face (F, N), shear plane (F_s, N_s) — can be packed onto one circle with R as the diameter. That single picture encodes every force-component relation you need, like F_s = F_c·cos φ − F_t·sin φ. Machining handbooks lean on it to back-calculate forces you didn't measure directly.

Merchant orthogonal cutting model

Merchant's orthogonal cutting model, developed by Eugene Merchant in 1944–1945, is a 2D analytical theory of metal cutting where the cutting edge is perpendicular to the cutting velocity — a configuration close to plunge turning or slotting on a milling machine. The model relates the cutting forces to the workpiece shear strength and the tool–chip friction. Its core assumptions are (1) the chip forms on a single, idealized shear plane of zero thickness, (2) tool–chip friction obeys Coulomb's law with coefficient μ, and (3) the shear flow stress τ_s is uniform along the shear plane.

This simulator takes four inputs — rake angle α, friction coefficient μ, depth of cut t and shear flow stress τ_s — solves Merchant's equation φ = 45° + α/2 − β/2 with β = atan(μ), and computes the cutting force F_c/w and thrust force F_t/w per unit width. It then draws an orthogonal cutting schematic and the Merchant force circle so you can see how rake-angle and friction changes propagate through the entire force balance at the cutting edge.

Physical model and key equations

In 2D orthogonal cutting the tool, with rake angle $\alpha$ measured from the cutting-velocity normal, takes a layer of thickness $t$ from the workpiece. The removed material flows into a chip after shearing along a single shear plane. The angle that this plane makes with the cutting velocity is the shear angle $\phi$. Coulomb friction at the tool–chip interface gives $F = \mu N$, equivalent to a friction angle $\beta = \arctan(\mu)$.

Merchant minimized the shear work per unit volume of removed material, $\partial W / \partial \phi = 0$, to obtain

$$\phi = \frac{\pi}{4} + \frac{\alpha}{2} - \frac{\beta}{2}$$

Assuming a uniform shear flow stress $\tau_s$ along the shear plane, the resultant force per unit width is

$$\frac{R}{w} = \frac{\tau_s\,t}{\sin\phi\,\cos(\phi+\beta-\alpha)}$$

and the cutting and thrust components in the machine frame are

$$\frac{F_c}{w} = \frac{R}{w}\cos(\beta-\alpha),\qquad \frac{F_t}{w} = \frac{R}{w}\sin(\beta-\alpha)$$

With defaults $\tau_s=400$ MPa, $t=0.20$ mm, $\alpha=10°$, $\mu=0.50$ this gives $\beta=26.57°$, $\phi=36.72°$, $R/w \approx 223.8$ N/mm, $F_c/w \approx 214.5$ N/mm and $F_t/w \approx 63.8$ N/mm. Measured forces are usually 80–90% of these predictions; Lee–Shaffer's slip-line solution $\phi = 45° + \alpha - \beta$ and Oxley's modified model give closer quantitative agreement at the cost of more parameters.

Real-world applications

Rake angle selection: Soft materials such as pure aluminium, copper and low-carbon steel call for positive rakes of +15° to +25° to keep cutting forces low. Hard, brittle or abrasive workpieces — cast iron, hardened steel, ceramic inserts — need negative rakes of −5° to −10° for edge strength, accepting the higher cutting force as a trade-off. The simulator quantifies that trade-off: sweep α and read off F_c at each setting.

Spindle power estimate: Sizing a spindle motor requires P = F_c·v. Once the cutting width w, depth t and speed v are fixed, this tool gives F_c/w directly and the power becomes P = (F_c/w)·w·v. For example with α=10°, μ=0.5, τ_s=400 MPa, t=0.2 mm, w=3 mm and v=100 m/min (=1.67 m/s), P = 214.5·3·1.67 ≈ 1.07 kW, suggesting a 1.5 kW motor at 0.7 efficiency.

Evaluating coolant and coatings: The main purpose of cutting fluids and tool coatings is to drop the tool–chip friction coefficient μ. Replacing an uncoated tool at μ=0.7 with a TiAlN-coated insert at μ=0.5 changes φ from 30.4° to 36.7° and F_c by roughly 15%. Plug the numbers in to make a quantitative case for the coating before committing to it.

Chip thickness ratio: The chip thickness $t_c = t\,\cos(\phi-\alpha)/\sin\phi$ gives the chip compression ratio $r_c = t/t_c = \sin\phi/\cos(\phi-\alpha)$. High $r_c$ (close to 1) means a short curled chip that is easy to clear; low $r_c$ produces long stringy chips that tangle in automated cells. Chip breakers on inserts are designed against this number.

Common misconceptions and pitfalls

The biggest pitfall is trusting Merchant's predicted shear angle as a quantitative truth. Real shear zones have finite thickness and a temperature-dependent flow stress, so measured φ is usually 5–10° smaller than the Merchant value. Lee–Shaffer's slip-line solution $\phi = 45° + \alpha - \beta$ and Oxley's modified theory match experiments better. Treat Merchant as a teaching model that captures the qualitative trends (α↑ ⇒ force↓, μ↑ ⇒ force↑) cleanly, not as a finished design tool.

The second pitfall is applying the 2D model to general 3D machining without correction. This simulator covers orthogonal cutting, in which the cutting edge lies perpendicular to the cutting velocity. Real turning and milling have an inclination angle that drives chip flow sideways and introduces a third force component F_r. Oblique cutting theory (Stabler's chip-flow rule η_c ≈ i) extends Merchant to that case but is beyond the scope of this tool.

The third trap is looking up μ from a sliding-friction handbook. In a real cut the tool–chip interface runs at 500–1000°C under GPa-level contact pressure, with adhesion and diffusion dominating over plain sliding. Effective μ values lie between 0.3 and 1.0, far above the 0.2 you'd read for clean steel-on-steel. Real practice measures F_c and F_t on a dynamometer and back-calculates μ from β = atan(F_t/F_c) + α.

Frequently asked questions

The metal-cutting tool takes a high-level view of machining (cutting speed, feed rate, depth of cut, Taylor tool life V·T^n = C) and is the right starting point for "how many kW does my spindle need". This Merchant tool zooms in on the force balance at the cutting edge: shear angle prediction, the three orthogonal force decompositions, and the rake-angle/friction sensitivity. Use Merchant to understand why the forces are what they are; use metal-cutting to size the machine.
Merchant's equation φ = 45° + α/2 − β/2 has no sign restriction on α. Negative α simply yields a smaller φ, a longer shear plane and a much larger force. Physically that is the cost of using negative rakes for edge strength on hard inserts (carbide and ceramics). Push the rake slider to α=−15° in the simulator and you'll see F_c jump by roughly 50% compared to α=+10°.
No, this simulator is strictly 2D orthogonal cutting. Oblique cutting introduces an inclination angle i along the cutting edge and a third force component F_r in the radial direction; Stabler's law η_c ≈ i predicts the chip flow direction. Most production turning and milling are oblique, but mastering the orthogonal case first is the conventional pedagogical route before tackling the 3D extension.
A reasonable rule of thumb is τ_s ≈ 0.577·σ_y (von Mises) or 0.5–0.7·UTS. Typical numbers: 400–500 MPa for medium-carbon steel (1045), 500–700 MPa for stainless (304), 200–300 MPa for aluminium alloys (2017), 600–800 MPa for titanium (Ti-6Al-4V). At cutting strain rates and temperatures, the effective τ_s is 1.0–1.5× the static value because of strain-rate hardening and thermal softening. Oxley's model evaluates τ_s dynamically from a Johnson–Cook constitutive law.