Question

The following data relate to an orthogonal turning process: back rake angle $=15deg.$ width of cut $=2mm$ chip thickness $=0.4mm$ feed rate $=0.2mm/rev$

If the cutting force and the thrust force are $900N$ and $810N$ the shear strength in $MPa$

• A. $137.94$
• B. $477.91$
• C. $500.58$
• D. $635.84$

Answer 1

The correct option is B $477.91$

$F_c=900,F_T=810$

$\beta =\alpha +ta{n}^{-1}\frac{F_T}{F_C}$

$=15+ta{n}^{-1}\frac{810}{900}={56.98}^o$

$F_s=\frac{F_c}{cos(\beta -\alpha )}cos(\varphi +\beta -\alpha )$

$=\frac{900}{cos{41.98}^o}cos\left({70.98}^o\right)$

$=39456N$

${\tau }_s=\text{mean shear strength}$

$=\frac{F_s}{A_0}\times sin\varphi =\frac{394.56}{2\times 0.2}\times sin{29}^o$

$=478MPa$