Question

In an orthogonal cutting operation, the tool has a rake angle = 12°. The chip thickness before the cut = 0.32 mm and the cut yields a deformed chip thickness = 0.65 mm. Calculate: a-the shear plane angle. b-the shear strain for the operation.

Answer 1

The shear plane angle and shear strain are 28.21° and 2.155 respectively.

Step-by-step explanation:

(a)

Orthogonal cutting is the cutting process in which cutting direction or cutting velocity is perpendicular to the cutting edge of the part surface.

Given:

Rake angle is 12°.

Chip thickness before cut is 0.32 mm.

Chip thickness is 0.65 mm.

Calculation:

Step1

Chip reduction ratio is calculated as follows:

`r=(t)/(t_(c))`

`r=(0.32)/(0.65)`

r = 0.4923

Step2

Shear angle is calculated as follows:

`tan\phi=(rcos\alpha)/(1-rsin\alpha)`

Here,
`\phi` is shear plane angle, r is chip reduction ratio and
`\alpha` is rake angle.

Substitute all the values in the above equation as follows:

`tan\phi=(rcos\alpha)/(1-rsin\alpha)`

`tan\phi=(0.4923cos12^(\circ))/(1-0.4923sin12^(\circ))`

`tan\phi=(0.48155)/(0.8976)`

`\phi=28.21^(\circ)`

Thus, the shear plane angle is 28.21°.

(b)

Step3

Shears train is calculated as follows:

`\gamma=cot\phi+tan(\phi-\alpha)`

`\gamma=cot28.21^(\circ)+tan(28.21^(\circ)-12^(\circ))`
`\gamma = 2.155`.

Thus, the shear strain rate is 2.155.