Common Data:
A cylinder is turned on a lathe with orthogonal machining principle. Spindle rotates at $200 r p m$ . The axial feed rate is $0.25 m m$ per revolution. Depth of cut is $0.4 m m$ The rake angle is $10^{o}$ . In the analysis it is found that the shear angle is ${27.75}^{o}$ .

The thickness of the produced chip is

0.511 m m

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0.528 m m

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0.818 m m

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0.846 m m

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Solution

The correct option is A $0.511 m m$
Given,
$N = 200 r p m$
$Feed (t_{1}) = uncut chip thickness$
$= 0.25 m m / r e v$
Depth $= 0.4 m m$
Rake angle,
$α = 10^{o}$
Shear angle,
$ϕ = {27.75}^{o}$

We Know
$t a n ϕ = \frac{c o s α}{\frac{t_{2}}{t_{1}} - s i n α}$
The thickness of produced chip:
$t_{2} = t_{1} [\frac{c o s α}{t a n ϕ} + s i n α]$
$= 0.25 [\frac{c o s 10^{o}}{t a n {27.75}^{o}} + s i n 10^{o}]$
$= 0.511 m m$

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Similar questions

Common Data:
A cylinder is turned on a lathe with orthogonal machining principle. Spindle rotates at $200 r p m$ . The axial feed rate is $0.25 m m$ per revolution. Depth of cut is $0.4 m m$ The rake angle is $10^{o}$ . In the analysis it is found that the shear angle is ${27.75}^{o}$ .

In the above problem, the coefficient of friction at the chip tool interface obtained using Earnest and Merchant theory is