Question

In orthogonal turning of a cylindrical tube of wall thickness 5 mm, the axial and tangential cutting forces were measured as 1259 N and 1601 N, respectively. The measured chip thickness after machining was found to be 0.3 mm. The rake angle was 10° and axial feed was 100 mm/min. The rotational speed of the spindle was 1000 rpm. Assuming the material to be perfectly plastic and Merchants' first solution, the shear strength of the material is closest to

• A. 875 MPa
• B. 920 MPa
• C. 722 MPa
• D. 200 MPa

Answer 1

The correct option is C 722 MPa

Orthogonal turning,
$\lambda$ = 90°,
wall thickness = depth of cut, (d) = 5 mm (assumed)
Axial force,
$F_x=1259N;F_t=\frac{F_x}{\sin 90}$

$F_t=1259N$

Tangential force,

$F_z=1601N,F_c=F_t=1601N$

$t_c=0.3mm$

$\alpha =10{}^{\circ }$

$fN=100mm/min$

$N=1000rpmorf\times 1000=100$

or $f=0.1mm/rev.$

$t=f\sin \lambda =0.1\sin 90{}^{\circ }=0.1mm$

$b=\frac{d}{\sin \lambda }=\frac{5}{\sin 90}=5mm$

$r=\frac{t}{t_c}=\frac{0.1}{0.3}=0.33$

$\tan \varphi =\frac{r\cos \alpha }{1-r\sin \alpha }=\frac{0.33\cos 10{}^{\circ }}{1-0.33\sin 10{}^{\circ }}$

or $\varphi =19.02{}^{\circ }$

${\tau }_s=\frac{F_s}{A_s}=\frac{F_{s}sin\varphi }{bt}$

$=\frac{(F_c\cos \varphi -F_t\sin \varphi )\times \sin \varphi }{bt}$

$=\frac{(1601\cos 19.02-1259\sin 19.02)\times \sin 19.02}{5\times 0.1}$

$=719.12MPa\simeq 722MPa$

Option (c) is correct.