Question

In orthogonal metal cutting, cutting speed is 2 m/s and depth of cut is 0.5 mm. If chip thickness is 0.75 mm, the chip velocity in m/s will be:
1. 1.33
2. 2.75
3. 2.15
4. 1.75

Answer 1

Correct Answer - Option 1 : 1.33

Concept:

Chip thickness ratio / Cutting ratio (r):

It is the ratio of chip thickness before cut (t1) to the chip thickness after cut (t2).

\(r=\frac{Chip\;thickness\;before\;cut\;(t_1)}{Chip\;thickness\;after\;cut\;(t_2)}\Rightarrow \frac{uncut\;chip\;thickness}{chip\;thickness}\)

chip thickness after the cut (t2) is always greater than the chip thickness before the cut (t1), ∴ r is always < 1, i.e. the uncut chip thickness value is less than the chip thickness value.

Assuming discharge to be constant:

t1b1V = t2b2Vc

\(\frac{t_1}{t_2}=\frac{V_c}{V}\;\;\;\;(\because b_1=b_2)\)

as t2 > t1, ∴ V > Vc i.e. cutting velocity is greater than the chip velocity.

[NOTE: In an orthogonal metal cutting depth of cut = uncut chip thickness]

Calculation:

Given:

V_c = 2 m/s, depth of cut = t₁ = 0.5 mm, t₂ = 0.75 mm.

\(\frac{t_1}{t_2}=\frac{V_c}{V}\;\;\;\;(\because b_1=b_2)\)

\(\frac{0.5}{0.75}=\frac{V_c}{2}\)

∴ V_c = 1.33 m/s