--- title: "Merchant circle and shear angle relation" subject: "Manufacturing and Production Processes" chapter: 7 key: "semester_4/manufacturing_and_production_processes:merchant-circle-force-relations-derivation" source: https://www.bctengineer.com/q/semester_4/manufacturing_and_production_processes/merchant-circle-force-relations-derivation site: BCT Engineering --- # Merchant circle and shear angle relation Manufacturing and Production Processes · Chapter 7: Material Removal Processes: “ Chip-forming” - Asked in: year not given, 8 marks - Video: [Merchant Circle Diagram](https://www.youtube.com/watch?v=vTmQu9zxtuA) — Nikhil Pandya - Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14 - Page: https://www.bctengineer.com/q/semester_4/manufacturing_and_production_processes/merchant-circle-force-relations-derivation - Whole chapter: https://www.bctengineer.com/md/semester_4/manufacturing_and_production_processes/chapter-7 ## Question Draw the Merchant circle diagram for orthogonal cutting and derive the expressions for the shear force, the normal force on the shear plane, the friction force and the normal force on the tool face in terms of the cutting force $F_c$ and thrust force $F_t$. Derive Merchant's relation for the shear angle. ## Short answer Force components are resolved from measured Fc and Ft using shear and rake angles. Shear force Fs = Fc cos(φ) - Ft sin(φ), friction force F = Fc sin(α) + Ft cos(α). Merchant's relation: 2φ + β - α = 90°, giving φ = 45° + α/2 - β/2, obtained by minimizing cutting force assuming constant shear stress. ## Answer **Orthogonal cutting:** the cutting edge is perpendicular to the cutting velocity, so the process can be treated in two dimensions. Symbols: $\alpha$ rake angle, $\phi$ shear angle, $\beta$ friction angle ($\tan\beta=\mu$), $t_0$ uncut chip thickness, $t_c$ chip thickness, $b$ width of cut. #### Merchant circle The resultant force $R$ on the chip from the tool is balanced by an equal and opposite force from the workpiece. A circle drawn with $R$ as its diameter contains all the force components. ```text Chip Tool \ shear plane (angle phi) \ /| \ / | Rake angle alpha ----O----+-----> V (work moves right) Pairs of components of the same resultant R: (Fc , Ft) - measured by dynamometer (Fs , Fn) - along / normal to shear plane (F , N ) - along / normal to tool face All lie on a circle of diameter R through O. ``` In the diagram, $F_c$ and $F_t$ are measured by a dynamometer (horizontal and vertical); $F_s$ and $F_n$ act along and normal to the shear plane; $F$ and $N$ act along and normal to the tool face. Each pair forms a right triangle with the common hypotenuse $R$, so all six components end on the circle. The angles are: $R$ makes $(\beta-\alpha)$ with $F_c$, $(\phi+\beta-\alpha)$ with $F_s$, and $\beta$ with $N$. #### Force relations Resolving $F_c$ and $F_t$ along the shear plane (inclined at $\phi$ to the cutting direction): $$ F_s = F_c\cos\phi - F_t\sin\phi $$ $$ F_n = F_c\sin\phi + F_t\cos\phi $$ Resolving along and normal to the tool face (rake angle $\alpha$): $$ F = F_c\sin\alpha + F_t\cos\alpha $$ $$ N = F_c\cos\alpha - F_t\sin\alpha $$ Coefficient of friction: $\mu = F/N = \tan\beta$. Chip thickness ratio and shear angle: $$ r = \frac{t_0}{t_c} = \frac{\sin\phi}{\cos(\phi-\alpha)}\quad\Rightarrow\quad \tan\phi = \frac{r\cos\alpha}{1 - r\sin\alpha} $$ #### Merchant's relation Shear stress on the shear plane, whose area is $A_s = b\,t_0/\sin\phi$: $$ \tau = \frac{F_s}{A_s} = \frac{R\cos(\phi+\beta-\alpha)\,\sin\phi}{b\,t_0} $$ Since $F_c = R\cos(\beta-\alpha)$, eliminating $R$ gives $$ F_c = \frac{\tau\,b\,t_0\cos(\beta-\alpha)}{\sin\phi\,\cos(\phi+\beta-\alpha)} $$ Merchant assumed that the shear angle adjusts itself so that the cutting force (and so the power) is a minimum, with $\tau$ constant. Setting $dF_c/d\phi = 0$ means the denominator $\sin\phi\cos(\phi+\beta-\alpha)$ is a maximum: $$ \cos\phi\cos(\phi+\beta-\alpha) - \sin\phi\sin(\phi+\beta-\alpha) = \cos(2\phi+\beta-\alpha) = 0 $$ $$ 2\phi + \beta - \alpha = 90^\circ \quad\Rightarrow\quad \phi = 45^\circ + \frac{\alpha}{2} - \frac{\beta}{2} $$ #### Meaning A larger rake angle or lower friction increases the shear angle, which reduces the chip thickness and the cutting force. Experimental results agree only roughly with the formula, so Lee and Shaffer's relation $\phi = 45^\circ + \alpha - \beta$ is sometimes used instead. ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. What is the idea of the Merchant circle? A. All pairs of force components (Fc,Ft), (Fs,Fn), (F,N) are sides of right triangles on the common resultant R as diameter B. It shows that Fc always equals Ft C. It shows that the shear force equals the friction force D. It shows that R is the sum of Fs and F Answer: A. The resultant R is the common hypotenuse, so all component pairs lie on a circle of diameter R. (Key point: Merchant circle) 2. What is the shear force on the shear plane in terms of Fc and Ft? A. Fs = Fc cos(phi) + Ft sin(phi) B. Fs = Fc cos(phi) - Ft sin(phi) C. Fs = Fc sin(phi) + Ft cos(phi) D. Fs = Fc sin(alpha) + Ft cos(alpha) Answer: B. Resolving Fc and Ft along the shear plane gives Fs = Fc cos(phi) - Ft sin(phi). (Key point: Shear force) 3. Which pair gives the friction force F and normal force N on the tool face? A. F = Fc cos(alpha) - Ft sin(alpha); N = Fc sin(alpha) + Ft cos(alpha) B. F = Fc cos(phi) - Ft sin(phi); N = Fc sin(phi) + Ft cos(phi) C. F = Fc sin(alpha) + Ft cos(alpha); N = Fc cos(alpha) - Ft sin(alpha) D. F = Fc sin(alpha) - Ft cos(alpha); N = Fc cos(alpha) + Ft sin(alpha) Answer: C. Resolving along and normal to the tool face with rake alpha gives these two relations; mu = F/N = tan(beta). (Key point: Tool face forces) 4. What does Merchant assume to get the shear angle? A. The shear angle adjusts so that friction is zero B. The shear angle adjusts to maximise the chip thickness C. The shear angle equals the rake angle D. The shear angle adjusts so cutting force (power) is a minimum, with constant shear stress Answer: D. Setting dFc/dphi = 0 maximises sin(phi)cos(phi+beta-alpha). (Key point: Merchant's assumption) 5. What is Merchant's relation for the shear angle? A. phi = 45 deg + alpha/2 - beta/2 B. phi = 45 deg + alpha - beta C. phi = 45 deg - alpha/2 + beta/2 D. phi = 90 deg - alpha - beta Answer: A. cos(2phi + beta - alpha) = 0 gives 2phi + beta - alpha = 90 deg. 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