Expected Theory Que:

Explain the mechanism of chip formation in metal cutting. (7 marks)

Mechanics of the orthogonal cutting process:

For the purpose of the analysis presented below following Idealized conditions are assumed to exist during cutting:

1] The tool is perfectly sharp and there is no contact along the flank face.

2] The shear surface is a thin plane extending upward front the cutting edge.

3] The cutting edge is a straight line perpendicular to the motion and generates a plane surface as the work moves past it.

4] Width of tool is greater than width of work piece

5] Continuous type of chips are produced without build up edge.

6] There is no side movement of the chip in either direction.

7] The work piece moves with a uniform velocity related to the tool.

8] The stresses on the shear plane are uniformly distributed.

Chip Thickness Ratio:

• The ratio of the thickness of the chip before removal to the thickness after removal is termed as chip thickness ratio, chip thickness coefficient or cutting ratio.

• The reciprocal of this ratio is known as “Chip compression factor”

From Figure,

$\begin{aligned} \text { chip thickness Ratio }=r_{c} &=\frac{t}{t_{c}} \\ &=\frac{A B \cdot \sin \emptyset}{A B \cdot \cos (\emptyset-\gamma)} \\ &=\frac{\sin \emptyset}{\cos (\emptyset-\gamma)} \end{aligned}$

$\begin{aligned} \mathrm{NOTE} : \cos (A-B)=\cos A \cdot & \cos B+\sin A \cdot \sin B=\cos \emptyset \cdot \cos \gamma+\sin \emptyset \cdot \sin \gamma \\ &=\frac{1}{\cos \emptyset \cdot \cos \gamma+\sin \gamma} \end{aligned}$

$\begin{aligned} \cot \emptyset \cdot \cos \gamma &=\frac{1-r_{c} \sin \gamma}{r_{c}} \\ \therefore \tan \emptyset &=\frac{r_{c} \cos \gamma}{1-r_{c} \cdot \sin \gamma} \end{aligned}$

Shear Strain:

• The shearing strain that each under formed layers of the material undergoes as it passes through the shear plane during the process of chip formation can be calculated with reference to figure.

• The shearing strain defined as ratio of the displacement of the layer EE’ to the thickness of the layer KF’ is related to shearing and rake angle as follows:

$\epsilon=\frac{E E^{\prime}}{K F^{\prime}}=\frac{K E^{\prime}+K E}{K F^{\prime}}$

$\therefore \epsilon=\cot \emptyset+\tan (\emptyset-\gamma)$

Rate of strain in cutting:

$\epsilon^{*}=\frac{\varepsilon}{\Delta t}=\frac{E E^{\prime}}{K F^{\prime} \cdot \Delta t}=\frac{V_{s}}{K F^{\prime}}$

Where Vs = velocity of shear

$\triangle t$ = Time elapsed for the metal to travel a distance EE’ along the shear plane.

Velocity Ratio:

The three velocities which are of interest are:

1] The velocity of the tool relative to work or velocity of cutting (V).

2] The velocity of chip, relative to tool or chip flow velocity (Vf).

3] The velocity of chip relative to work or shear velocity (Vs).

From volume continuity,

$b t . V=b \cdot t_{c} \cdot V_{f}$

$\therefore \frac{V_{f}}{v}=\frac{t}{t_{c}}=r_{c}=\frac{\sin \emptyset}{\cos (\emptyset-\gamma)}$

From Figure,

$\frac{V_{s}}{\sin (90-\gamma)}=\frac{V}{\sin [90-(\emptyset-\gamma)]}$

$\frac{V_{s}}{\cos \gamma}=\frac{V}{\cos (\emptyset-\gamma)}$

$\frac{V_{s}}{V}=\frac{\cos \gamma}{\cos (\emptyset-\gamma)}$

Force Relationships:

1) If the chip is isolated as a free body it can be seen that the chip is in equilibrium under the action of only two forces R and R’ as shown in figure.

2) Force R is the force between the chip and the tool face and is a resultant of normal & tangential component of forces N & F on the tool face.

3) Similarly, R’ is the force between the chip and workpiece and is a resultant of normal & tangential components Ns & Fs on the shear plane.

4) For equilibrium R & R’ must be equal & opposite. Actually they may not be collinear but the small couple due to this can be neglected.

5) If the force R is assumed to act at the top tip instead of at its actual point of application on the tool face a compact force diagram shown in figure can be obtained. This type of plane was first suggested by Merchant (1942) and is known as Merchants diagram.

6) It is widely used for analysis of the orthogonal cutting process.

7) In this plot, circle AD is drawn from the top of the tool with AD=R

8) The resultant force R’ can then be resolved in any desired direction.

9) The three most convenient directions as shown in the figure are:

I. Along and at right angles to tool face F & N

II. Along and at right angles to the shear plane Fs & Ns

III. Horizontal & vertical directions Fp & Fq

10) Force Fp is the main cutting force on the tool and act in the direction of tool travel. Fq is the force at right angles to the cutting force.

11) Fp & Fq can be easily measured with the help of a two component force dynamometer.

12) Once Fp & Fq are known all other force components can easily be calculated from the force circle.

NOTE:

$\begin{array}{ll}{\text { 1. }} & {F=F_{p} \cdot \sin \gamma+F_{q} \cdot \cos \gamma} \\ {\text { 2. }} & {N=F_{p} \cdot \cos \gamma-F_{q} \cdot \sin \gamma} \\ {\text { 3. }} & {F_{s}=F_{p} \cdot \cos \emptyset-F_{q} \cdot \sin \emptyset} \\ {\text { 4. }} & {N_{s}=F_{p} \cdot \sin \emptyset+F_{q} \cdot \cos \emptyset=F_{s} \cdot \tan (\emptyset+\beta-\gamma)}\end{array}$

Stresses on the shear plane:

If ‘t’ is the depth of cut and ‘h’ its width assumed to remain unchanged during the shearing process, the area of the shear plane,

$A_{s}=\frac{b \cdot t}{\sin \emptyset}$

$\therefore$ Average shear stress on the shear plane

$\sigma_{A v g}=\frac{F_{s}}{A_{s}}=\frac{\left(F_{p} \cos \emptyset-F_{q} \sin \emptyset\right)}{b \cdot t / \sin \emptyset}$

Also, Average normal stress on the shear Plane

$\sigma_{A v g}=\frac{N_{s}}{A_{s}}=\frac{\left(F_{p} \sin \emptyset-F_{q} \cos \emptyset\right)}{b \cdot t / \sin \emptyset}$

Energy Considerations:

• When metal is cut in a two dimensional cutting operation the total energy will be consumed in the following ways:

I. As shear energy per unit volume Es on the shear plane.

II. As friction energy per unit volume Ef on the tool face.

III. As surface energy per unit volume $E_{A}$ required for the formation of new surface area.

IV. As kinetic energy per unit volume $E_{M}$ required to accelerate the chip

V. As chip curl energy Ec per unit volume required to curl the otherwise straight chip.

• It is observed that, in a typical cutting operation En & Em almost negligible while Ec is less than 5 percent. Therefore, the total energy Per unit volume is assumed to be made up only of Es & Ef

$\therefore$ $E=$Total Energy$=E_{s}+E_{f}$

• Between the two, Es is about 75% of the total energy. The values of Es & Ef can be calculated from the following relationships.

$E_{s}=\frac{F_{s} \cdot V_{s}}{b . t . V}=\frac{F_{s} \cdot \cos \gamma}{A_{s} \cdot \sin \emptyset \cdot \cos (\emptyset-\gamma)}=\tau . \epsilon$

$E_{f}=\frac{F_{s} \cdot V_{F}}{b \cdot t \cdot V}=\frac{F_{s} \cdot r_{c}}{b \cdot t}$

Merchant Theory:

1] This theory is based on the principle of minimum energy. According to this principle, angle $\phi$ will take a value such that total work done in cutting is a minimum.

2] As the work done depends only on the cutting force component Fp, then angle $\phi$ should assume such a value as to make Fp should be minimum for given $\beta$ and $\gamma$

3] From merchant’s circle diagram,

$\begin{aligned} F_{p} &=R \cdot \cos (\beta-\gamma) \\ F_{p} &=\frac{F_{s} \cdot \cos (\beta-\gamma)}{\cos \cdot \cot (\beta-\gamma)} \\ F_{p} &=\frac{z \cdot b \cdot t \cdot \cos (\beta-\gamma)}{\sin \emptyset \cdot \cos (\beta-\gamma)} \end{aligned}$

Where Z = mean shear stress

B = width of the chip

t = uncut chip thickness

4] If $\tau$ is the shear strength of the material being cut, shear will occur when the shear strength $\tau$ on the plane defined by angle $\theta$ be completely equal to the shear strength $\tau$ of the materials.

5] For a simple analytic if $\tau$ is assumed to be constant,

$F_{p}=\frac{\tau^{\prime} \cdot b t \cdot \cos (\beta-\gamma)}{\sin \emptyset \cdot \cos (\emptyset+\beta-\gamma)}$

For minimum Fp,

$\frac{d F_{p}}{d \emptyset}=0$

$\therefore \tau^{\prime} b t \cdot \cos (\beta-\gamma)[\cos \emptyset \cdot \cos (\emptyset+\beta-\gamma)-\sin \emptyset \cdot \sin (\emptyset+\beta-\gamma)]=0$

$\therefore \cos (2 \emptyset+\beta-\gamma)=0$

Which gives a minimum value,

$2 \emptyset+\beta-\gamma=\frac{7}{3}$

$\mathrm{Or}, \emptyset=\frac{\pi}{4}+\frac{\gamma}{2}-\frac{\beta}{2}$

Lee and Shaffer Theory:

1] Lee and shaffer’s relationship between $\phi$, $\gamma$ and is derived on the assumptions that:

A] The material being cut behaves like an ideal plastic which does not strain harden.

B] The shear plane represents a direction of maximum shear stresses in the material cut.

2] Any point in the material to be cut is initial unstressed when it is some distance away from the tool tip.

3] As the tip of the tool approaches the stress builds up till it reaches a value high energy to cause plastic deformation.

4] The material is then converted into chip and proceeds up the tool face. The stress in the chip gradually fall off to zero.

5] Assuming a possible slip line field to represent these conditions of plastic deformation Lee & Shaffer gave the following equation for shear angle.

$\emptyset=\frac{\pi}{4}+\gamma-\beta---A$

6] This value of $\phi$ (in equation A), however was found to be not correct for all values of $\gamma$ and $\beta$ and was modified by Lee & Shaffer to include the effect of a built-up edge to be made of two straight lines and a circular are with a subtended angle $\theta$ as shown in figure. It was shown that the plastic flow relationship should take the form.

$\emptyset=\frac{\pi}{4}+\gamma+\theta-\beta$