A bar of 75 mm diameter is reduced to 73 mm by a cutting tool while cutting orthogonally- If the rake angle is 15- and shear angle is 0-3328 radian- The mean length of cut chip is mm-

76.137 Length of uncut chip,l=π/2(D_i+D_f) Length of uncut chip, l=π (75+73)/2=232.48 mm Φ=0.3328 radian = 19.068^∘ tan 19.068^∘ =r cos15^∘/1 r sin 15^∘ r=l_c/ ...

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Abbreviation Used in Analysis of Metal Cutting & Cutting Ratio

A bar of 75 m...

Question

A bar of 75 mm diameter is reduced to 73 mm by a cutting tool while cutting orthogonally. If the rake angle is $15^{\circ}$ and shear angle is 0.3328 radian. The mean length of cut chip is _____ mm.

76.137

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Solution

The correct option is A 76.137
76.137

Length of uncut chip, $l = \frac{π}{2} (D_{i} + D_{f})$
Length of uncut chip, $l = \frac{π (75 + 73)}{2} = 232.48 m m$
$Φ = 0.3328 r a d i a n = {19.068}^{\circ}$
$t a n {19.068}^{\circ} = \frac{r c o s 15^{\circ}}{1 - r s i n 15^{\circ}}$
$r = \frac{l_{c}}{l}$
$0.3275 = \frac{l_{c}}{232.48}$
$l_{c} = 76.137 m m$

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A bar of 75 mm diameter is reduced to 73 mm by a cutting tool while cutting orthogonally. If the rake angle is 15∘ and shear angle is 0.3328 radian. The mean length of cut chip is _____ mm. 76.137

The correct option is A 76.13776.137 Length of uncut chip,l=π2(Di+Df) Length of uncut chip, l=π(75+73)2=232.48mm Φ=0.3328radian=19.068∘ tan19.068∘=rcos15∘1−rsin15∘ r=lcl 0.3275=lc232.48 lc=76.137mm

A bar of 75 mm diameter is reduced to 73 mm by a cutting tool while cutting orthogonally. If the rake angle is $15^{\circ}$ and shear angle is 0.3328 radian. The mean length of cut chip is _____ mm.

76.137