Merchant circle and shear angle relation

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.

 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.