# What is Merchant’s circle diagram? (cutting force analysis)

Let’s begin with their assumptions.

## Assumptions for Merchant’s circle diagram

Contents
The assumptions for the merchant circle diagram are as follows.
• The cutting tool must be sharp.
• There is no contact between the cutting tool and the clearance edge.
• The chip formation is in continuous form and without a built-up edge.
• The thickness of the uncut chip is constant.
• The workpiece moves with a uniform velocity during the cutting process.
• The width of the workpiece is always lesser than the cutting tool width.
• Shear occurs on a shear plane where the stress is uniformly distributed.

## Merchant’s circle diagram

The above figure shows a merchant circle diagram which is easy to determine the cutting force relations in metals.

## Merchant’s circle equation and derivation

From the above merchant circle diagram, the following force components are mentioned below.
• The diameter of the circle and the resultant force is denoted by $F$.
• The cutting force is denoted by $F_{z}$.
• The thrust force is denoted by $F_{x}$.
The above cutting force and thrust force are analyzed using a dynamometer. After analyzing these forces, they are plotted under a suitable scale.
Force ($F$) is the resultant of cutting force($F_{z}$) and thrust force($F_{x}$) .
From the above diagram, the following angles are denoted as follows.
where,
$$\displaystyle \begin{array}{l}\alpha =\text{ Rake angle}\\\beta \text{ = Shear angle}\\\gamma \text{ = Friction angle}\end{array}$$
The frictional force $(P)$ and normal force $(N)$ are acting against the cutting tool which is mutually perpendicular to each other. The frictional force is the resisting force created along the tool rack face to resist the chip flow. Hence by taking the ratio of these two forces, the coefficient of friction between the tool and chip is obtained.

The coefficient of friction is denoted as $\displaystyle \mu$.
$$\displaystyle \mu =\frac{P}{N}\text{ }\to (1)$$
The ratio of above two forces is also expressed in the form of trignometry relation.
$$\displaystyle \mu =\text{ tan}\gamma \text{ }\to (2)$$
From the above diagram, the chip thickness ratio is denoted as $r$.
$$\displaystyle \text{r = }\frac{{{{t}_{1}}}}{{{{t}_{2}}}}\text{ }\to (3)$$
where,$$\displaystyle \begin{array}{l}{{t}_{1}}=\text{ h}\sin \beta \\{{t}_{2}}=\text{ h}\cos (\beta -\alpha )\end{array}$$
($t_{1}$ and $t_{2}$ are thickness of chip before and after cutting)
Therefore,    $$\displaystyle (3)\Rightarrow \text{ r = }\frac{{h\sin \beta }}{{h\cos (\beta -\alpha )}}$$
$$\displaystyle \text{r = }\frac{{\sin \beta }}{{\cos (\beta -\alpha )}}\text{ }\to \text{(4)}$$
We know that,$\displaystyle [\because \text{ cos (A-B) = cosA cosB + sinA sinB }\!\!]\!\!\text{ }$
Therefore,$$\displaystyle \text{(4)}\Rightarrow \text{ r = }\frac{{\sin \beta }}{{\cos \beta \cos \alpha +\sin \beta \sin \alpha }}$$
$$\displaystyle r\cos \beta \cos \alpha +r\sin \beta \sin \alpha =\sin \beta \text{ }\to (5)$$
by dividing equation $(5)$ by $\displaystyle \sin \beta \text{ }$
$$\displaystyle \frac{{r\cos \beta \cos \alpha }}{{\sin \beta }}+\frac{{r\sin \beta \sin \alpha }}{{\sin \beta }}=1$$
$$\displaystyle \frac{{r\cos \alpha }}{{\tan \beta }}=1-r\sin \alpha$$
Therefore, the value of $\displaystyle {\tan \beta }$ is,$$\displaystyle \tan \beta =\frac{{r\cos \alpha }}{{1-r\sin \alpha }}\text{ }\to \text{(6)}$$
From the merchant diagram we know that,
$$\displaystyle \begin{array}{l}P=\text{ }Friction\text{ }force\\P={{F}_{x}}cos\alpha +{{F}_{z}}sin\alpha \\N=Normal\text{ }force\\N={{F}_{z}}cos\alpha +{{F}_{x}}sin\alpha ~\\F=Resultant\text{ }force\\F=\sqrt{{{{F}_{z}}^{2}+{{F}_{x}}^{2}}}\\{{F}_{z}}=Cutting\text{ }force\\{{F}_{z}}\text{ }=Fcos(\gamma -\alpha )\\{{F}_{s}}\text{ }=Shear\text{ }force\\{{F}_{s}}\text{ }=Fcos\theta \end{array}$$

where,      $\displaystyle \theta =\beta +\gamma -\alpha$
$$\displaystyle \therefore F=\frac{{{{F}_{s}}}}{{\cos \theta }}\text{ }\to (7)$$
substitute $(7)$ in $F_{z}$ equation,$$\displaystyle \therefore {{F}_{z}}=\frac{{{{F}_{s}}\cos (\gamma -\alpha )}}{{\cos \theta }}$$
$$\displaystyle {{F}_{z}}=\frac{{{{F}_{s}}\cos (\gamma -\alpha )}}{{\cos (\beta +\gamma -\alpha )}}$$
$\displaystyle [\because \theta =\beta +\gamma -\alpha ]$
From $(1)$ and $(2)$, $$\displaystyle \mu =\tan =\frac{P}{N}\text{ }$$
$$\displaystyle \mu =\frac{{{{F}_{x}}\cos \alpha +{{F}_{z}}\sin \alpha }}{{{{F}_{z}}\cos \alpha -{{F}_{x}}\sin \alpha }}$$
And the coefficient of friction,$$\displaystyle \mu =\frac{{{{F}_{x}}+{{F}_{z}}\tan \alpha }}{{{{F}_{z}}-{{F}_{x}}\tan \alpha }}$$
By trignometric relations, the value of $F_{s}$ and $F_{n}$ is obtained from the merchant circle diagram.$$\displaystyle {{F}_{s}}={{F}_{z}}cos~\beta -{{F}_{x}}sin\beta$$ $$\displaystyle {{F}_{n}}={{F}_{z}}\sin ~\beta +{{F}_{x}}\cos \beta$$
and,$$\displaystyle F={{F}_{z}}\sin ~\alpha +{{F}_{x}}\cos \alpha$$ $$\displaystyle N={{F}_{z}}cos~\alpha -{{F}_{x}}sin\alpha$$
In some special cases , when rake angle is zero $( \displaystyle \alpha = 0)$, then $F$ and $N$ becomes,$$\displaystyle \begin{array}{l}F={{F}_{x}}\\N={{F}_{z}}\end{array}$$
A dynamometer is used directly to measure cutting force and its thrust force in this special case.

## How to draw Merchant’s circle diagram?

Consider the condition of an orthogonal cutting operation for the given feed, tool geometry, and depth of cut. Now we have only the values of two forces $\displaystyle ({{F}_{x}}\text{ and }{{F}_{z}})$, which is measured by using a dynamometer. By using the merchant’s circle diagram and its equations we are going to determine the other force relations simply and efficiently.
• Determine the value of thrust force $F_{x}$.
• As the rake angle of the tool is known, draw the chip and tool in the orthogonal plane.
• Draw the force $F_{z}$ as a horizontal component and the other force $F_{x}$ as a vertical component.
• For the forces, $F_{x}$ and $F_{z}$ draw their resultant force $F$. The value of resultant force $F$ is calculated by a formula.$$\displaystyle F=\sqrt{{{{F}_{z}}^{2}+{{F}_{x}}^{2}}}$$
• Mark the midpoint of the resultant force $F$ and draw the circle.
• Extend the tool’s rake surface by extending a line along with it and mark it as friction force $P$.
• Draw the normal force $N$, which is normal to the friction force $P$. The friction coefficient $\displaystyle \mu$ is obtained by a formula.$$\displaystyle \mu =\frac{P}{N}$$
• Draw the shear plane by using a shear angle $\displaystyle \beta$.
• The value of shear angle $\displaystyle \beta$ is obtained by using a formula, which is already derived.$$\displaystyle \tan \beta =\frac{{r\cos \alpha }}{{1-r\sin \alpha }}$$
• Draw $F_{s}$ and $F_{n}$ as intercepts.

## Advantages of Merchant’s circle diagram

The advantages of the merchant’s circle diagram are as follows.
• MCD is easy to use and analyze.
• Other force relations can be easily solved.
• By using MCD, friction at the tool surface and yield shear strength can be easily determined.

## Limitations of Merchant’s circle diagram

Merchant’s circle diagram also has some limitations.
• Other than orthogonal cutting, the merchant’s circle diagram is not used since it is based on a single shear plane principle.
• The friction coefficient obtained by the formula which is derived from MCD gives only the approximate value.