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zloy xaker [14]
3 years ago
14

In an orthogonal cutting operation, if the friction coefficient between the tool and the chip is decreased while the material an

d the rake angle are kept constant, then the shear angle will increase.
True
False
Engineering
2 answers:
levacccp [35]3 years ago
5 0

Answer:

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.

Alexus [3.1K]3 years ago
3 0

Answer:

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.                          

                             

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