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Analysis of Orthogonal Machining

In orthogonal...

Question

In orthogonal machining operation, the chip thickness and the uncut chip thickness are equal to $0.45 m m$ . If the tool rake and is $0$ deg. The shear plane angle is

45^{o}

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30^{o}

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18^{o}

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60^{o}

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Solution

The correct option is A $45^{o}$
$t_{1} = t_{2} = 0.45$
$\Rightarrow r = \frac{t_{1}}{t_{2}} = 1$
$α = 0$
$ϕ = t a n^{- 1} (\frac{1 c o s 0^{o}}{1 - 1 \times s i n 0^{o}})$
$= t a n^{- 1} (1) = 45^{o}$

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In orthogonal machining operation, the chip thickness and the uncut chip thickness are equal to 0.45mm. If the tool rake and is 0 deg. The shear plane angle is

The correct option is A 45ot1=t2=0.45 ⇒r=t1t2=1 α=0 ϕ=tan−1(1cos0o1−1×sin0o) =tan−1(1)=45o

In orthogonal machining operation, the chip thickness and the uncut chip thickness are equal to $0.45 m m$ . If the tool rake and is $0$ deg. The shear plane angle is

The correct option is A $45^{o}$
$t_{1} = t_{2} = 0.45$
$\Rightarrow r = \frac{t_{1}}{t_{2}} = 1$
$α = 0$
$ϕ = t a n^{- 1} (\frac{1 c o s 0^{o}}{1 - 1 \times s i n 0^{o}})$
$= t a n^{- 1} (1) = 45^{o}$