Question

In an orthogonal cutting operation, the 0.250 in wide tool has a rake angle of 5º. The lathe is set so the chip thickness before the cut is 0.010 in. After the cut, the deformed chip thickness is measured to be 0.027 in. Calculate (a) the shear plane angle and (b) the shear strain for the operation.

Answer 1

(a) Shear plane angle will be 21.9°

(b) Shear train will be 2.7913 radian

Step-by-step explanation:

We have given rake angle
`\alpha =5^(\circ)`

Thickness before cut
`t_1=0.01inch`

Thickness after cutting
`t_2=0.027inch`

Now ratio of thickness before and after cutting
`r=(t_1)/(t_2)=(0.01)/(0.027)=0.3703`

(a) Shear plane angle is given by
`tan\Phi =(rcos\alpha )/(1-rsin\alpha )`, here
`\alpha` is rake angle.

So
`tan\Phi =(0.3703* cos5^(\circ))/(1-sin5^(\circ))=0.4040`

`\Phi =tan^(-1)0.4040`

`\Phi =21.9^(\circ)`

(b) Shear strain is given by
`\gamma =tan(\Phi -\alpha )+cot\Phi`

So
`\gamma =tan(21.9 -5 )+cot21.9=2.7913radian`