In an orthogonal cutting operation, if the friction coefficient between the tool and the chip is decreased while the material and the rake angle are kept constant, then the shear angle will increase.

True

False

The Correct Answer is False.

Explanation:

It is logical that the cutting force increases as the depth of cut increases and rake angle decreases. Deeper cuts remove more material, thus requiring a higher cutting force. As the rake angle, α, decreases, the shear angle, φ, decreases and hence shear energy dissipation and cutting forces increase.

True

Explanation:

- The shear plane angle is given by the following formula (Merchant's Equation):

∅ = 45 + α/2 - β/2

Where, α: Rake Angle

β: Friction angle

- It is seen from the relation above that shear plane angle depends on friction angle (β) and rake angle (α).

- The friction angle (β) is related by the friction coefficient (u) with the following relation:

u = tan (β)

β = arctan (u)

- The expression becomes:

∅ = 45 + α/2 - arctan (u) / 2

- Its given that rake angle (α) remains constant, while friction coefficient (u) between the tool and the chip is decreased.

- For the equation to hold true then shear plane angle (∅) must increase.