Question

Common Data:
In an orthogonal machining operation:
Uncut thickness $=0.5mm$
Cutting speed $=20m/min$
Rake angle $={15}^o$
Width of cut $=5mm$
Chip thickness $=07mm$
Thrust force $=200N$
Cutting force $=1200N$
Assume Merchant's theory.

The percentage of total energy dissipated due to friction at the tool-chip interface is

• A. $30$%
• B. $42$%
• C. $58$%
• D. $70$%

Answer 1

The correct option is A $30$%

Frictional force:
$F_f=F_{C}sin\alpha +F_{T}cos\alpha$
$=1200sin\left({15}^o\right)+200cos\left({15}^o\right)$
$=503.8N$

Velocity of the chip:
$V_f=\frac{V_C\sin \varphi }{cos(\varphi -\alpha )}$
$=\frac{V_{C}sin\left({40.2}^o\right)}{cos(40.2-15)}=0.713V$
Total power consumption:
$P=F_C{V}_C=1200V_C$

Power consumption due to friction
$P_f=F_f{V}_f$
$=503.8\times 0.713V_C$
$=359.21V_C$
Percentage of energy dissipated due to friction
$\frac{P_f}{P}\times 100=\frac{359.21V_C}{1200V_C}\times 100=30\%$