Relative energies in cutting

An orthogonal cutting operation is being carried out in which $t_{\omicron }=0.010$ mm, V = 125 m/min, $α = 10^{°}$ , and the width of cut = 6 mm. It is observed that $t_{c} = 0.014$ mm, $F_{c}= 55$ kg, and $F_{t }= 25$ kg. Calculated the percentage of the total energy that is dissipated in friction at the tool-chip interface.

Question

Relative energies in cutting

An orthogonal cutting operation is being carried out in which [latex]t_{\omicron }=0.010[/latex] mm, V = 125 m/min, [latex]α = 10^{°} [/latex] , and the width of cut = 6 mm. It is observed that [latex]t_{c} = 0.014[/latex] mm,[latex] F_{c}= 55 [/latex] kg, and [latex]F_{t }= 25 [/latex]kg. Calculated the percentage of the total energy that is dissipated in friction at the tool-chip interface.

Accepted Answer

The percentage of energy can be expressed as

[latex]\frac{Friction \ energy}{Total \ energy} =\frac{FV_{c}}{F_{c}V} =\frac{Fr}{F_{c}} ,[/latex]

where

[latex]r=\frac{t_{\omicron }}{t_{c}} =\frac{0.010}{0.014} =0.7143[/latex]

[latex]F=R \ \sin \beta [/latex]
[latex]F_{c}=R \ \cos (\beta-\alpha ) [/latex]

and

[latex]R=\sqrt{F^{2}_{t}+F^{2}_{c}} =\sqrt{25^2+55^2} = 60.41 \ kg[/latex]

Thus

[latex]55 = 60.41 \ cos (\beta - 10°)[/latex]

from which we find that

[latex]\beta = 34°[/latex] and [latex] F =60.41 \ sin34°=31.5 \ kg[/latex]

Therefore, the percentage of friction energy is calculated as

Percentage[latex]=\frac{\left(31.5\right) \left(0.7143\right) }{55} =0. 444 \ = \ 44 .4 [/latex]%

and similarly, the percentage of shear energy is calculated as 55.6%.

Question 8.1: Relative energies in cutting An orthogonal cutting operation...

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