Question

Assume that in orthogonal cutting the rake angle is 15° and the coefficient of friction is 0.15. a. Determine the percentage change in chip thickness when the coefficient of friction is doubled. Justify your answer. b. Determine the percentage change in chip thickness when the rake angle is increased to 25o . Justify your answer.

Answer 1

Δr=20.45 %

Step-by-step 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

`r=(tan\phi)/(cos\alpha +sin\alpha\ tan\phi)`

`r=(tan48.08^(\circ))/(cos15^(\circ) +sin15^(\circ)\ tan48.08^(\circ))`

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

`r'=(tan\phi')/(cos\alpha +sin\alpha\ tan\phi')`

`r'=(tan44.15^(\circ))/(cos49.15^(\circ) +sin49.15^(\circ)\ tan44.15^(\circ))`

r'=0.70

Percentage change

`\Delta r=(r-r')/(r)* 100`

`\Delta r=(0.88-0.70)/(0.88)* 100`

Δr=20.45 %