Direction: During orthogonal machining of a mild steel specimen with a cutting tool of zero rake angle, the following data is obtained: Uncut chip thickness = 0.25 mm Chip thickness = 0.75 mm Normal force = 950 N Thrust force = 475 N
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The shear angle and shear force, respectively, are
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- 71.565°, 150.12 N
- 9.218°, 861.64 N
- 18.435°, 751.04 N
- 23.157°, 686.66 N
Correct Option: C
∝ = 0; t1 -0.25 mm t2 = 0.75 mm
w = 2.5 mm Fc = 950 N, FT = 475 N
| = r = | = | = 0.333 | |||
| t2 | 0.75 | 3 |
tanφ = r ⇒ φ = tan-1(0.333)
φ = 18.435°
| tanβ = | = | ||
| fτ | 950 |
β = tan-1(0.5)
b = 26.56°
| Fs = | ||
| RCos(β - ∝) |
| Fs = | ||
| Cos(26.56°) |
Fs = 75In
