Materials Science and Manufacturing Engineering Miscellaneous
- Assertion (A) : A diamond tool is used for USM of glass workpiece.
Reason (R) : Diamond is harder than glass
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Soft tool material must be used in order to avoid tool wear.
Correct Option: D
Soft tool material must be used in order to avoid tool wear.
- BUE formation.... A... the cutting force and... B.. the surface finish
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BUE increases the cutting force due to decrease in apparent rake angle and spoils the surface finish.
Correct Option: A
BUE increases the cutting force due to decrease in apparent rake angle and spoils the surface finish.
- In HSS, the tungsten can be substituted by
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Tungsten is available in T-series & M- series.
Correct Option: C
Tungsten is available in T-series & M- series.
- In an orthogonal machining with a single point cutting tool of rake angle 10°, the uncut chip thickness and the chip thickness are 0.125 mm and 0.22 mm, respectively. Using Merchant’s first solution for the condition of minimum cutting force, the coefficient of friction at the chip-tool interface is _____ (round off to two decimal places).
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t = 0.125 mm tc = 0.22 mm
rake angle, α = 10°
Chip thickness ratio , r = t tc r = 0.125 = 0.568 0.22 tanφ = rcosα 1 - rsinα tanφ = 0.568cos10° 1 - 0.568sin10°
Shear angle φ = 31.83°2φ + β - α = π = 90° 2
⇒ 2 × 31.83 + β – 10 = 90°
Friction angle β = 36.33°
∵ tan β = μ
⇒ Coefficient of friction, μ = tan36.33 = 0.7355°
μ = 0.7355 ≈ 0.74
Correct Option: B
t = 0.125 mm tc = 0.22 mm
rake angle, α = 10°
Chip thickness ratio , r = t tc r = 0.125 = 0.568 0.22 tanφ = rcosα 1 - rsinα tanφ = 0.568cos10° 1 - 0.568sin10°
Shear angle φ = 31.83°2φ + β - α = π = 90° 2
⇒ 2 × 31.83 + β – 10 = 90°
Friction angle β = 36.33°
∵ tan β = μ
⇒ Coefficient of friction, μ = tan36.33 = 0.7355°
μ = 0.7355 ≈ 0.74
- In orthogonal turning of a cylindrical tube of wall thickness 5 mm, the axial and the tangential cutting forces were measured at 1259 N and 1601 N, respectively. The measured chip thickness after machining was found to be 0.3 mm. The rake angle was 10° and the axial feed was 100 mm/min. The rotational speed of the spindle was 1000 rpm. Assuming the material to be perfectly plastic and merchant’s first solution, the shear strength of the martial is closest to
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For, orthogonal turning,
λ = 90° , Fc = 1601 N
Chip thickness ‘tc ’ = 0.3, Ft = 1259 N
N = 1000 rpm, α = 10 mf = 100 mm / min = 100 mm / rev 1000 f = 0.1 mm = t rev r = t = 0.1 = 1 tc 0.3 3 Fa = τs × bt sinφ ⇒ τs = Fs sinφ 6 × t tanφ = rcosα 1 - rsinα
⇒ φ = 19.19°
Fs = Fccosφ - Ftsinφ
Fs = 1601 × cos19.19° - 1259 × sin19.19° = 1098.20 Nτs = 1098.20 × sin19.19° = 721.96 ≈ 722 MPa 5 × 0.1
Correct Option: C
For, orthogonal turning,
λ = 90° , Fc = 1601 N
Chip thickness ‘tc ’ = 0.3, Ft = 1259 N
N = 1000 rpm, α = 10 mf = 100 mm / min = 100 mm / rev 1000 f = 0.1 mm = t rev r = t = 0.1 = 1 tc 0.3 3 Fa = τs × bt sinφ ⇒ τs = Fs sinφ 6 × t tanφ = rcosα 1 - rsinα
⇒ φ = 19.19°
Fs = Fccosφ - Ftsinφ
Fs = 1601 × cos19.19° - 1259 × sin19.19° = 1098.20 Nτs = 1098.20 × sin19.19° = 721.96 ≈ 722 MPa 5 × 0.1
