Materials Science and Manufacturing Engineering Miscellaneous
Direction: The following data relate to an orthogonal turning process: back rake angle = 15 deg. width of cut = 2 mm chip thickness = 0.4 mm, feed rate = 0.2 mm/rev
- The shear angle is
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α = 15°
W = 2 mm
t2 = 0.4mm
fr = 0.2 mm/rev = t1r = t1 = 0.2 = 1 t2 0.4 2
r = 0.5φ = tan-1 
r cos ∝ 
1 - r sin∝ φ = tan-1 0.5 cos 15 1 - 0 sin 15
φ = 0Correct Option: C
α = 15°
W = 2 mm
t2 = 0.4mm
fr = 0.2 mm/rev = t1r = t1 = 0.2 = 1 t2 0.4 2
r = 0.5φ = tan-1 r cos ∝ 1 - r sin∝ φ = tan-1 0.5 cos 15 1 - 0 sin 15
φ = 0
- If the cutting force and the thrust force are 900 N and 810 N, the mean strength in MPa
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Fc = 900 N,
Ft = 810 Ntanβ = FT + tanα Fc 1 - FT tanα Fc tanβ = 810 + tan 15° 900 1 - 810 tan 15° 900
β = 56.98°Fs = Rcos(φ + β - ∝) Fc Rcos(β - ∝)
Fs = 394.56 NCs = Fs × sinφ = 394.56 × sin 29 Wt1 2 × 0.2
Cs = 478 mpaCorrect Option: B
Fc = 900 N,
Ft = 810 Ntanβ = FT + tanα Fc 1 - FT tanα Fc tanβ = 810 + tan 15° 900 1 - 810 tan 15° 900
β = 56.98°Fs = Rcos(φ + β - ∝) Fc Rcos(β - ∝)
Fs = 394.56 NCs = Fs × sinφ = 394.56 × sin 29 Wt1 2 × 0.2
Cs = 478 mpa
Direction: In an orthogonal cutting test, the following observations were made
Cutting force = 1200 N
Thrust force = 500 N
Tool rake angle = zero
Cutting speed = 1 m/s
Depth of cut = 0.8 mm
Chip thickness = 1.5 mm
- Chip speed along the tool rake face will be
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NA
Correct Option: B
NA
- In an orthogonal machining operation, the tool life obtained is 10 min at a cutting speed of 100 m/min, while at 75 m/min cutting speed, the tool life is 30 min. The value of index (n) in the Taylor's tool life equation is
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Ti = 10 min V1 = 100 m /min
T2 = 75 min V2 = 30 m /minn = In V2 = In 75 V1 100 In T1 In 10 T2 30 n = - 0.2876 - 1.09
n = 0.262Correct Option: A
Ti = 10 min V1 = 100 m /min
T2 = 75 min V2 = 30 m /minn = In V2 = In 75 V1 100 In T1 In 10 T2 30 n = - 0.2876 - 1.09
n = 0.262
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
- The shear angle and shear force, respectively, are
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∝ = 0; t1 -0.25 mm t2 = 0.75 mm
w = 2.5 mm Fc = 950 N, FT = 475 Nt1 = r = 0.25 = 1 = 0.333 t2 0.75 3
tanφ = r ⇒ φ = tan-1(0.333)
φ = 18.435°tanβ = FT = 475 fτ 950
β = tan-1(0.5)
b = 26.56°Fs = FcRcos(φ + β - ∝) RCos(β - ∝) Fs = 950Rcos(45°) Cos(26.56°)
Fs = 75InCorrect Option: C
∝ = 0; t1 -0.25 mm t2 = 0.75 mm
w = 2.5 mm Fc = 950 N, FT = 475 Nt1 = r = 0.25 = 1 = 0.333 t2 0.75 3
tanφ = r ⇒ φ = tan-1(0.333)
φ = 18.435°tanβ = FT = 475 fτ 950
β = tan-1(0.5)
b = 26.56°Fs = FcRcos(φ + β - ∝) RCos(β - ∝) Fs = 950Rcos(45°) Cos(26.56°)
Fs = 75In
