Answer:
Δr=20.45 %
Explanation:
Given that
Rake angle α = 15°
coefficient of friction ,μ = 0.15
The friction angle β
tanβ = μ
tanβ = 0.15
β=8.83°
2φ + β - α = 90°
φ=Shear angle
2φ + 8.833° - 15° = 90°
φ = 48.08°
Chip thickness r given as
[tex]r=\dfrac{tan\phi}{cos\alpha +sin\alpha\ tan\phi}[/tex]
[tex]r=\dfrac{tan48.08^{\circ}}{cos15^{\circ} +sin15^{\circ}\ tan48.08^{\circ}}[/tex]
r=0.88
New coefficient of friction ,μ' = 0.3
tanβ' = μ'
tanβ' = 0.3
β'=16.69°
2φ' + β' - α = 90°
φ'=Shear angle
2φ' + 16.69° - 25° = 90°
φ' = 49.15°
Chip thickness r' given as
[tex]r'=\dfrac{tan\phi'}{cos\alpha +sin\alpha\ tan\phi'}[/tex]
[tex]r'=\dfrac{tan44.15^{\circ}}{cos49.15^{\circ} +sin49.15^{\circ}\ tan44.15^{\circ}}[/tex]
r'=0.70
Percentage change
[tex]\Delta r=\dfrac{r-r'}{r}\times 100[/tex]
[tex]\Delta r=\dfrac{0.88-0.70}{0.88}\times 100[/tex]
Δr=20.45 %
