Merchant’s machinability constant is

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BPSC Lecturer ME Held on July 2016 (Advt. 35/2014)
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  1. ϕ + β -  α 
  2. 2ϕ - β + α 

  3. 2ϕ + β - α 
  4. 2ϕ - β -  α 

Answer (Detailed Solution Below)

Option 3 : 2ϕ + β - α 

Detailed Solution

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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

  • 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 

Now F= f× A2

where, F= Shear stress, A= Area of shear plane = 

 

differentiate w.r.t β 

 

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

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

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