GATE Mechanical Engineering (ME) 2019 Shift 2 Solved Paper

Question : 62 of 65

Marks: +1, -0

In an orthogonal machining with a single point cutting tool of rake angle 10o , the uncut chip thickness and the chip thickness are 0.125 mm and 0.22 mm, respectively. Using Merchant’s first solution for the condition of minimum cutting force, the coefficient of friction at the chip-tool interface is ___________(round off to two decimal places).

Your Answer:

Validate

Solution:

uncut chip thickness

(t) = 0.125

mm
Cut chip thickness

(t_{c})=0.22 {mm}

So, ship thickness ratio

=\;{t}/{t_{c}}=\;{0.125}/{0.22}=0.568

Rake angle

(α)=10^{∘}

\tan ϕ=\;{r \cos α}/{1-r \sin α}=\;{0.568 × \cos 10}/{1-0.568 × \sin 10}=0.6206

Or shear angle

(φ)=31.82^{∘}

According to Merchant's theory, we have

2 φ+β-α=\;{π}/{2}

So,

β=90^{∘}-63.64+10=36.36^{∘}

Now,

\tan β=µ

(coefficient of friction)
Therefore, coefficient of friction

=\tan 36.36^{∘}=0.74

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