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

Answer:

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

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

$\tan\phi=\dfrac{r\cos\alpha}{1-r\sin\alpha}$

Put the value in to the formula

$\tan\phi=\dfrac{0.5\cos10}{1-0.5\sin10}$

$\tan\phi=0.539$

$\phi=\tan^{-1}(0.539)$

$\phi=28.32^{\circ}$

We need to calculate the shear velocity

Using formula of shear velocity

$v_{2}=\dfrac{v\cos\alpha}{\cos(\phi-\alpha)}$

Put the value into the formula

$v_{2}=\dfrac{3\times\cos10}{\cos(28.32-10)}$

$v_{2}=3.11\ m/s$

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

Using formula of energy dissipated

$\%d=\dfrac{P_{s}}{P}\times100$

$\%d=\dfrac{F_{s}\times v_{c}}{F_{c}\times v}\times100$

Put the value into the formula

$\%d=\dfrac{345.21\times3.11}{400\times3}\times100$

$d=89.46\%$

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