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 Merchant’s first solution, the shear strength of the material is closest to

The correct answer is 722 MPa.

Step-by-Step Explanation:

1. Identify the given parameters from the problem and the provided analysis diagram:
Axial force, $F_x=1259\text{N}$
Tangential cutting force, $F_z=1601\text{N}$
Measured chip thickness after machining, $t_c=0.3\text{mm}$
Rake angle, $\alpha =10°$
Axial feed rate, $fN=100\text{mm/min}$
Rotational speed of the spindle, $N=1000\text{rpm}$
Wall thickness of the cylindrical tube (depth of cut), $d=5\text{mm}$
For orthogonal turning, the principal tool approach angle is $\lambda =90°$.

2. Determine the cutting force ($F_c$) and feed force ($F_t$):
The cutting force is equal to the tangential force:

$F_c=F_z=1601\text{N}$

The feed force is related to the axial force by:

$F_t=\frac{F_x}{\sin \lambda }=\frac{1259}{\sin 90°}=1259\text{N}$

3. Calculate the feed per revolution ($f$) and uncut chip thickness ($t$):
The feed per revolution is calculated from the feed rate and spindle speed:

$f=\frac{100\text{mm/min}}{1000\text{rpm}}=0.1\text{mm/rev}$

The uncut chip thickness ($t$) is given by:

$t=f\sin \lambda =0.1\times \sin 90°=0.1\text{mm}$

The width of the cut ($b$) is:

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

4. Calculate the chip thickness ratio ($r$) and the shear angle ($\varphi$):
The chip thickness ratio is:

$r=\frac{t}{t_c}$

The shear angle is computed using the relation:

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

Substituting the values:

$\tan \varphi =\frac{0.33\times \cos 10°}{1-0.33\times \sin 10°}=\frac{0.33\times 0.9848}{1-0.33\times 0.1736}=\frac{0.3250}{0.9427}=0.3447$

Taking the arctangent:

$\varphi ={\tan }^{-1}\left(0.3447\right)\approx 19.02°$

5. Calculate the shear strength (${\tau }_s$) of the material:
The shear force ($F_s$) acting along the shear plane is given by Merchant's circle relations:

$F_s=F_c\cos \varphi -F_t\sin \varphi$

The shear plane area ($A_s$) is given by:

$A_s=\frac{bt}{\sin \varphi }$

Thus, the shear strength (${\tau }_s$) is:

${\tau }_s=\frac{F_s}{A_s}=\frac{(F_c\cos \varphi -F_t\sin \varphi )\sin \varphi }{bt}$

Substituting the numerical values:

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

${\tau }_s=\frac{(1601\times 0.9454-1259\times 0.3259)\times 0.3259}{0.5}$

${\tau }_s=\frac{(1513.58-410.31)\times 0.3259}{0.5}$

${\tau }_s=\frac{1103.27\times 0.3259}{0.5}\approx 719.12\text{MPa}$

This value is closest to 722 MPa.