Question

Minimum shear strain in orthogonal turning with a cutting tool of zero rake angle is

• A. $0.0$
• B. $0.5$
• C. $1.0$
• D. $2.0$

Answer 1

The correct option is D $2.0$

During orthogonal machining.
$\text{Shear strain,}\gamma =cot\varphi +tan(\varphi +\alpha )$
$\text{Cutting tool having rake angle,}\alpha =0^o$
$\text{then,}\gamma =cot\varphi +tan\varphi$
For minimum value of shear strain
$\frac{d\gamma }{\varphi }=\frac{d}{d\varphi }(cot\varphi +tan\varphi )$
$\frac{d\gamma }{\varphi }=-cose{c}^2\varphi +se{c}^2\varphi =0$\)
$\Rightarrow -\frac{1}{si{n}^2\varphi }+\frac{1}{co{s}^2\varphi }=0$
$\frac{co{s}^2\varphi -si{n}^2\varphi }{co{s}^2\varphi si{n}^2\varphi }=0$
$\Rightarrow cos2\varphi =0=cos\frac{\pi }{2}=cos\frac{3\pi }{2}$
$\varphi -\frac{\pi }{4}$
$\frac{d^2\gamma }{d{\varphi }^2}=2cosec\varphi (-cosec\varphi cot\varphi )+2sec\varphi \left(sec\varphi tan\varphi \right)$
$\frac{d^2\gamma }{d{\varphi }^2}=2cose{c}^2\varphi cot\varphi +2se{c}^2\varphi tan,\varphi$
${\begin{array}{c}\frac{d^2\gamma }{d{\varphi }^2}\end{array}|}_{at\varphi =\frac{\pi }{4}}>0$
$\text{Therefore}\gamma \text{is minimum at}\varphi =\frac{\pi }{4}$
Therefore minimum shear strain $\gamma$ is given by
$\gamma =cot\frac{\pi }{4}+tan\frac{\pi }{4}=2^o$