The length of chip over one revolution of work piece is 72mm. The cutting speed is 12.5m/min the tangential force is 410N and feed force is 170N.

i) Coefficient of friction on a rake face

ii) Thickness of chip

iii) Velocity of shear

iv) Velocity of chip along the tool face

Mumbai University > Mechanical Engineering > Sem 4 > Production Process II

Marks: 10M

D=38mm

α=33°

f = 0.15mm rev

l2 = 72mm

vc = 12.5m/min

Fc = 410N

Ft = 170N

To find

i) µ

ii) t2

iii) Vs

iv) Vf

i) µ ;μ=$\frac{F}{N}$

F= FtCos γ+ FcSin γ

=170 Cos 33 + 410 Sin33

=356.88N

N= Fc Cos γ- Ft Sin γ

=410 Cos 33 – 170 Sin33

 = 251.27N

∴ µ=1.46

ii) t2

$r_{c} =\frac{t_{1}}{t_{2}} = \frac{l_{2}}{l_{1}} =\frac{72}{π ×D}=.603 \ \ \ \ here l_{1}=π.D$

∴$t_2$=0.249mm

iii) $V_{s}$

we know that by sine rule; $ \frac{V_c}{Sin[90-(∅-∝)]} =\frac{V_S}{Sin(90-∝)} ⇛\frac{12.5}{Sin[90-(36.98-33)]}=\frac{V_{S}}{Sin(90-33)}$

Therefore Vs=10.51m/min

iv) $V_{f}$

we know by sine rule; $\frac{V_{f}}{Sin ∅}=\frac{V_{S}}{Sin(90-∝)} ⇛\frac{V_{f}}{Sin 36.98}=\frac{10.51}{Sin(90-33)}$

Therefore $V_f$=7.54m/min