An orthogonal turning operation is carried out under the following conditions; rake angle = 5°, spindle rotational speed = 400 rpm; axial feed = 0.4 m/min and radial depth of cut = 5 mm. the chip thickness $t_{c},$ is found to be 3 mm. The shear angle (in degree) in this turning process is
18.88

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Solution

The correct option is A 18.88
Rake angle $(α) = 5^{\circ}$
N = 400 rpm
Axial feed = 0.4 m/min
$= \frac{0.4 \times 1000}{400} = 1 m m / r e v .$

Radial depth of cut = 5 mm
chip thickness: $t_{c} = 3 m m$

r = chip thickness ratio
$= \frac{t}{t_{c}} = (\frac{1}{3})$

$\Rightarrow t a n ϕ = \frac{r c o s α}{1 - r s i n α} = \frac{\frac{1}{3} c o s (5^{\circ})}{1 - \frac{1}{3} s i n (5^{\circ})}$

$t a n ϕ = 0.342$

or $ϕ = t a n^{- 1} (0.342) = {18.88}^{\circ}$

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An orthogonal turning operation is carried out under the following conditions; rake angle = 5°, spindle rotational speed = 400 rpm; axial feed = 0.4 m/min and radial depth of cut = 5 mm. the chip thickness tc, is found to be 3 mm. The shear angle (in degree) in this turning process is 18.88

The correct option is A 18.88Rake angle (α)=5∘ N = 400 rpm Axial feed = 0.4 m/min =0.4×1000400=1 mm/rev. Radial depth of cut = 5 mm chip thickness: tc=3 mm r = chip thickness ratio =ttc=(13) ⇒ tanϕ=r cos α1−r sin α=13cos(5∘)1−13sin(5∘) tan ϕ=0.342 or ϕ=tan−1(0.342)=18.88∘

An orthogonal turning operation is carried out under the following conditions; rake angle = 5°, spindle rotational speed = 400 rpm; axial feed = 0.4 m/min and radial depth of cut = 5 mm. the chip thickness $t_{c},$ is found to be 3 mm. The shear angle (in degree) in this turning process is
18.88