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In an orthogonal cutting test on mild Steel, the following data were obtained
Cutting speed : 40 m/min
Depth of cut : 0.3 mm
Tool rake angle : +5º
Chip thickness : 1.5 mm
Cutting force : 900 N
Thrust force : 450 N
Using Merchants analysis, the friction angle during the machining will be
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- 26.6°
- 31.5°
- 45°
- 63.4°
Correct Option: B
| μ = | = | ||
| N | FCcosα − FTsinα |
| = | = | = 0.614 | ||
| 900cos5º −450sin5º | 857.35 |
∴ λ = tan−1 = 31.5º
Alternative method
From Merchants' Circle
Given: FC = 900 N
FT = 450 N
Now
| tan(λ − α) = | = | ||
| FC | 900 |
| ∴ (λ − α) = tan−1 |  |  | |
| 2 |
[where, α = Rake angle]
Friction angle, λ = 26.5 + 5 = 31.50
