Common Data:
An orthogonal operation is carried out at $20 m / m i n$ cutting speed, using a cutting tool of rake angle $15$ degree. The chip thickness is $0.4 m m$ and uncut chip thickness is $0.2 m m$

The shear angle (in degrees) is

26.8

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27.8

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28.8

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29.8

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Solution

The correct option is C $28.8$
$V_{C} = 20 m / m i n, α = 15^{o}$
$t_{2} = 0.4 m m, t_{1} = 0.2 m m$

Chip thickness ratio:
$r = \frac{t_{1}}{t_{2}} = \frac{0.2}{0.4} = \frac{1}{2} = 0.5$

Shear plane angle:
$ϕ = t a n^{- 1} (\frac{r c o s α}{1 - 0.5 s i n α})$
$= t a n^{- 1} (\frac{0.5 c o s 15^{\circ}}{1 - 0.5 s i n 15^{\circ}}) = 29^{\circ}$

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Common Data: An orthogonal operation is carried out at 20m/min cutting speed, using a cutting tool of rake angle 15 degree. The chip thickness is 0.4mm and uncut chip thickness is 0.2mm The shear angle (in degrees) is

The correct option is C 28.8VC=20m/min,α=15o t2=0.4mm,t1=0.2mm Chip thickness ratio: r=t1t2=0.20.4=12=0.5 Shear plane angle: ϕ=tan−1(rcosα1−0.5sinα) =tan−1(0.5cos15∘1−0.5sin15∘)=29∘

Common Data:
An orthogonal operation is carried out at $20 m / m i n$ cutting speed, using a cutting tool of rake angle $15$ degree. The chip thickness is $0.4 m m$ and uncut chip thickness is $0.2 m m$

The shear angle (in degrees) is