Question

In a machining operation that approximates orthogonal cutting, the cutting tool has a rake angle =100 . The chip thickness ratio is 0.5. Calculate the shear plane angle and the shear velocity in the operation. Main Cutting force = 400N, thrust force = 200N. Calculate the percentage of the total energy dissipated into shear plane, if V = 3 m/s.

Answer 1

The percentage of the total energy dissipated into shear plane is 89.46%.

Step-by-step explanation:

Given that,

Rake angle = 10°

Thickness ratio= 0.5

Cutting Force = 400 N

Thrust force = 200 N

Speed =3 m/s

Suppose the shear force is 345.21 N.

We need to calculate the shear plane angle

Using formula shear angle

`\tan\phi=(r\cos\alpha)/(1-r\sin\alpha)`

Put the value in to the formula

`\tan\phi=(0.5\cos10)/(1-0.5\sin10)`

`\tan\phi=0.539`

`\phi=\tan^(-1)(0.539)`

`\phi=28.32^(\circ)`

We need to calculate the shear velocity

Using formula of shear velocity

`v_(2)=(v\cos\alpha)/(\cos(\phi-\alpha))`

Put the value into the formula

`v_(2)=(3*\cos10)/(\cos(28.32-10))`

`v_(2)=3.11\ m/s`

We need to calculate the percentage of the total energy dissipated into shear plane

Using formula of energy dissipated

`\%d=(P_(s))/(P)*100`

`\%d=(F_(s)* v_(c))/(F_(c)* v)*100`

Put the value into the formula

`\%d=(345.21*3.11)/(400*3)*100`

`d=89.46\%`

Hence, The percentage of the total energy dissipated into shear plane is 89.46%.