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.

Here is the step-by-step derivation to find the shear strength of the material:

1. Identify the given parameters:
- Wall thickness of the cylindrical tube (which corresponds to the width of cut, $b$):
$b=5\text{mm}$
- Spindle speed, $N=1000\text{rpm}$
- Axial feed rate, $v_f=100\text{mm/min}$
- Tangential cutting force, $F_c=1601\text{N}$
- Axial cutting force (thrust force), $F_t=1259\text{N}$
- Rake angle, $\alpha ={10}^{\circ }$
- Chip thickness after machining, $t_2=0.3\text{mm}$

2. Calculate the uncut chip thickness ($t_1$):
In orthogonal turning of a cylindrical tube, the feed per revolution is equal to the uncut chip thickness:
$t_1=f=\frac{v_f}{N}$
$t_1=\frac{100}{1000}=0.1\text{mm}$

3. Calculate the chip thickness ratio ($r$):
$r=\frac{t_1}{t_2}=\frac{0.1}{0.3}=0.3333$

4. Determine the shear angle ($\phi$):
The relation between the shear angle, chip thickness ratio, and rake angle is:
$tan\phi =\frac{rcos\alpha }{1-rsin\alpha }$
Substituting the values:
$tan\phi =\frac{0.3333cos\left({10}^{\circ }\right)}{1-0.3333sin\left({10}^{\circ }\right)}$
$tan\phi =\frac{0.3333\times 0.9848}{1-0.3333\times 0.1736}\approx 0.3484$
Taking the arctangent:
$\phi ={19.21}^{\circ }$

5. Calculate the shear force ($F_s$):
The shear force along the shear plane is given by:
$F_s=F_{c}cos\phi -F_{t}sin\phi$
Substituting the calculated shear angle and given force values:
$F_s=1601cos\left({19.21}^{\circ }\right)-1259sin\left({19.21}^{\circ }\right)$
$F_s\approx 1601\times 0.9443-1259\times 0.3290$
$F_s\approx 1511.8-414.2=1097.6\text{N}$

6. Calculate the shear plane area ($A_s$):
The area of the shear plane is:
$A_s=\frac{b·t_1}{sin\phi }$
$A_s=\frac{5\times 0.1}{sin\left({19.21}^{\circ }\right)}\approx \frac{0.5}{0.3290}\approx 1.52{\text{mm}}^2$

7. Calculate the shear strength (${\tau }_s$):
The shear strength of the material is the shear stress on the shear plane:
${\tau }_s=\frac{F_s}{A_s}$
${\tau }_s=\frac{1097.6}{1.52}\approx 722.1\text{MPa}$

Thus, the shear strength of the material is closest to 722 MPa.