For an orthogonal cutting operation, tool material is HSS, rake angle is 22°, chip thickness is 0.8 mm, speed is 48 m/min and red is 0.4 mm/rev. The shear plane angle (in degrees) is

19.24

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29.70

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56.00

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68.75

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Solution

The correct option is B 29.70
$α = 22^{\circ}, t_{2} = 0.8;$
$v = 48$ m\min ; $t_{1} = 0.4$

$r = t_{1} / t_{2} = 0.5$

$ϕ = {tan}^{- 1} (\frac{r cos α}{1 - r sin α})$

$= {tan}^{- 1} (\frac{0.5 cos 22^{\circ}}{1 - 0.5 sin 22^{\circ}})$

$= 29.7^{\circ}$

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For an orthogonal cutting operation, tool material is HSS, rake angle is 22°, chip thickness is 0.8 mm, speed is 48 m/min and red is 0.4 mm/rev. The shear plane angle (in degrees) is

The correct option is B 29.70α=22∘,t2=0.8; v=48 m\min ; t1=0.4 r=t1/t2=0.5 ϕ=tan−1(rcosα1−rsinα) =tan−1(0.5cos22∘1−0.5sin22∘) =29.7∘

The correct option is B 29.70
$α = 22^{\circ}, t_{2} = 0.8;$
$v = 48$ m\min ; $t_{1} = 0.4$

$r = t_{1} / t_{2} = 0.5$

$ϕ = {tan}^{- 1} (\frac{r cos α}{1 - r sin α})$

$= {tan}^{- 1} (\frac{0.5 cos 22^{\circ}}{1 - 0.5 sin 22^{\circ}})$

$= 29.7^{\circ}$