Question: Determine the theoretically minimum possible shear strain in orthogonal turning with a cutting tool of zero orthogonal rake angle. [ESE 2009]

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Solution: Metal cutting processes are employed to remove excess material from workpiece to impart desired shape and finish. This material removal takes place by shearing when a wedge shaped cutting tool compresses a layer of workpiece material. The cutting strain developed due to shearing is termed as shear strain in machining. This shear strain (ε) in orthogonal machining can be expressed easily in terms of orthogonal shear angle (β_O) and orthogonal rake angle (γ_O) of the cutting tool, as given below.

\[\varepsilon  = \cot {\beta _o} + \tan ({\beta _o} – {\gamma _o})\]

The objective now is to calculate theoretically minimum possible shear strain. In this particular case, rake angle (γ_O) is given as zero. Substituting γ_O = 0°, the above expression for shear strain can be reduced to its simplest form as given below.

\[\varepsilon  = \cot {\beta _o} + \tan {\beta _o}\]

In this expression, the only variable is the orthogonal shear angle (β_O). In order to find out minimum shear strain (ε), let us first find out first order differentiation and second order differentiation of ε with respect to β_O.

First order differentiation:

\[\frac{{d\varepsilon }}{{d{\beta _o}}} = \frac{d}{{dx}}\left( {\frac{2}{{\sin 2{\beta _o}}}} \right)\]

\[\frac{{d\varepsilon }}{{d{\beta _o}}} = \frac{{0 – 2.2.\cos 2{\beta _o}}}{{{{\left( {\sin 2{\beta _o}} \right)}^2}}}\]

\[\frac{{d\varepsilon }}{{d{\beta _o}}} =  – \frac{{4\cos 2{\beta _o}}}{{{{\left( {\sin 2{\beta _o}} \right)}^2}}}\]

Second order differentiation:

\[\frac{{{d^2}\varepsilon }}{{d{\beta _o}^2}} =  – \frac{{{{\left( {\sin 2{\beta _o}} \right)}^2}\left( { – 8\sin 2{\beta _o}} \right) – \left( {4\cos 2{\beta _o}} \right)\left( {4\cos 2{\beta _o}\sin 2{\beta _o}} \right)}}{{{{\left( {\sin 2{\beta _o}} \right)}^4}}}\]

\[\frac{{{d^2}\varepsilon }}{{d{\beta _o}^2}} = \frac{{8{{\left( {\sin 2{\beta _o}} \right)}^2} – 16{{\left( {\cos 2{\beta _o}} \right)}^2}}}{{{{\left( {\sin 2{\beta _o}} \right)}^3}}}\]

Shear angle for minimum cutting strain

Now for the shear strain to be either minimum or maximum, the first order derivative has to be zero. Substituting the (dε/dβ_o) = 0, the following can be found.

\( – \frac{{4\cos 2{\beta _o}}}{{{{\left( {\sin 2{\beta _o}} \right)}^2}}} = 0\)

or, \(\cos 2{\beta _o} = 0\)

or, \(\cos 2{\beta _o} = \cos 90\) [∵ 0° < β_o < 90°]

or, \({\beta _o} = 45\)

Checking whether minimum or maximum

Now the shear strain (ε) will be minimum if the value of second order derivative (d²ε/dβ_o²) at β_o = 45° is greater than 0 (i.e. positive).  If it is negative then ε will have maximum value at β_o = 45°. So let us calculate (d²ε/dβ_o²) value at β_o = 45° using the expression derived earlier for second order differentiation.

\[\frac{{{d^2}\varepsilon }}{{d{\beta _o}^2}} = \frac{{8{{\left( {\sin 90} \right)}^2} – 16{{\left( {\cos 90} \right)}^2}}}{{{{\left( {\sin 90} \right)}^3}}}\]

\[\frac{{{d^2}\varepsilon }}{{d{\beta _o}^2}} = \frac{{8 – 0}}{1} = 8 > 0\]

Value of minimum shear strain

Hence, shear strain (ε) is minimum at the shear angle of β_o = 45°, when orthogonal rake angle of the cutting tool is zero. The value of this minimum shear strain (ε_min) can be calculated easily from the basic expression of shear strain as mentioned at the beginning. Minimum shear strain in this case is 2.0 as calculated below:

\({\varepsilon _{\min }} = \cot 45 + \tan 45\)

\({\varepsilon _{\min }} = 1 + 1 = 2\)

It is worth mentioning that minimum shear strain (ε_min) will deviate from this value if orthogonal rake angle (γ_o) of the cutting tool is not zero (γ_O ¹ 0°). Since rake angle can be either negative, zero or positive, so separate but similar calculation is desired to get minimum shear strain for non-zero rake angle.