Question

Orthogonal turning of a mild steel tube with a tool of rake angle ${10}^{\circ }$ carried out at a feed of 0.
14 mm/rev. If the thickness of the chip produced is 0.28 mm, the values of shear angle and shear strain will be respectively

• A. ${28}^{\circ }{20}^{{}^{\prime }}\text{and}2.19$
• B. ${22}^{\circ }{20}^{{}^{\prime }}\text{and}3.53$
  
• C. ${24}^{\circ }{30}^{{}^{\prime }}\text{and}3.53$
  
• D. ${37}^{\circ }{20}^{{}^{\prime }}\text{and}5.19$

Answer 1

The correct option is A ${28}^{\circ }{20}^{{}^{\prime }}\text{and}2.19$

Feed = unit thickness of orthogonal turning

Chip thickness ratio:
$r=\frac{t_1}{t_2}=\frac{0.14}{0.28}=0.5$

Shear plane angle:
$tan\varphi =\frac{rcos\alpha }{1-rsin\alpha }$

$tan\varphi =\frac{0.5cos{10}^{\circ }}{1-\left(0.5sin{10}^{\circ }\right)}$

$\varphi ={28.33}^{\circ }={28}^{\circ }{20}^{\prime }$
Shear strain:
$\gamma =cot\varphi +tan(\varphi -\alpha )$
$\gamma =2.1858=2.19$