Merchant’s machinability constant is

Explanation:

The relation between various forces have been worked out by Merchant with a large number of assumptions as follows

1. The chip behaves as a free body in stable equilibrium under the action of two equal opposite and collinear resultant forces.
2. A continuous chip without a built-up edge is produced.
3. The cutting velocity remains constant.
4. The cutting tool has a sharp cutting edge and it does not make any flank contact with the workpiece Merchant suggested a compact and easiest way of representing the various forces inside a circle having the vector F as the diameter.

where, γ = Rake angle of the tool, ϕ = Shear angle, β = Friction angle of tool face

According to modified merchant theory relation between rake angle γ, shear angle ϕ, and friction angle β as shown below

\(ϕ = \frac{\pi }{4} - \frac{β }{2} + \frac{\alpha }{2}\)

• shear will take place in a direction in which energy required for shearing is minimum. 
• shear stress is maximum at the shear plane and it remains constant.

we have proved that 

\({F_H} = \frac{{{F_s}{\rm{cos}}\left( {\beta ~- ~\alpha } \right)}}{{{\rm{cos}}\left( {ϕ~ +~ \beta~ -~ \alpha } \right)}}\)

Now Fs = fs × A2

where, Fs = Shear stress, A2 = Area of shear plane = \(\frac{{{b_1}{t_1}}}{{sinϕ }}\)

 \({F_H} = \frac{{{b_1}{t_1}{\rm{cos}}\left( {\beta ~- ~\alpha } \right)}}{{sinϕ {\rm{cos}}\left( {ϕ ~+ ~\beta ~-~ \alpha } \right)}}\)

differentiate w.r.t β 

 \(\frac{{d{F_H}}}{{d\beta }} = \; - {f_s} \times {b_1}{t_1}\cos \left( {\beta ~-~ \alpha } \right) \times \frac{{cosϕ \cos \left( {ϕ ~+ ~\beta ~-~ \alpha } \right) - sinϕ {\rm{sin}}\left( {ϕ ~+ ~\beta ~-~ \alpha } \right)}}{{{{\sin }^2}ϕ {{\cos }^2}\left( {\alpha ~+~ \alpha ~-~ \gamma } \right)}}~=~0\)

In order that ϕ will assume that value which required a minimum force to cut the material.

\(cos\phi \cos \left( {\phi + \beta - \alpha } \right) - sin\phi \sin \left( {\phi + \beta - \alpha } \right) = 0\)

\(\cos \left( {\phi + \phi + \beta - \alpha } \right) = 0\)

\(2\phi + \beta - \alpha = \frac{\pi }{2}\)

\(2\phi + \beta - \alpha = c\)