Question

The following data was observed during orthogonal cutting a steel work piece of 60 mm diameter.
Cutting speed = 90 m/min,
Feed= 0.15 mm/rev,
Depth of cut 5 mm,
Length of chip /rev = 80.8 mm,
Rake angle = 10${}^{\circ }$,
Clearance angle = 8${}^{\circ }$
Tangential force = 2200 N.
Thrust force 1200 N

The energy lost in friction is

• A. 1414 W
• B. 1005 W
• C. 707 W
• D. 503 W

Answer 1

The correct option is B 1005 W

$\text{Length of chip/rev}=80.8mm,$

$\text{Length of uncut chip}=\pi d$
$=\pi \times 60=188.49=188.5\text{mm}$

$r=\frac{L_2}{L_1}=\frac{80.8}{188.5}=0.428$

$\tan \varphi =\frac{r\cos \alpha }{1-r\sin \alpha }=0.455$

$\varphi ={24.48}^{\circ }$

$V_f=\frac{V_c\sin \varphi }{\cos (\varphi -\alpha )}$

$=\frac{90\sin 24.48}{\cos 14.48}=38.51\text{m/min}=0.643\text{m/sec}$

Energy lost in friction:
$R=\sqrt{(2200{)}^2+(1200{)}^2}=2506\text{N}$

$\beta =\alpha +{\tan }^{-1}\left(\frac{F_T}{F_c}\right)$

$=10+{\tan }^{-1}\left(\frac{1200}{2200}\right)={38.61}^{\circ }$

$\text{Friction power}=F\times V_f$

$=R\sin \beta \times 0.643\text{m/sec}$
$=2506\sin \left(38.61\right)\times 0.643$
$=1005.3\text{W}$