Materials Science and Manufacturing Engineering Miscellaneous
Direction: In an orthogonal machining operation:
Uncut thickness = 0.5 mm
Cutting speed = 20 m/min
Rake angle = 15°
Width of cut = 5 mm
Chip thickness = 0.7 mm
Thrust force = 200 N
Cutting force = 1200 N
Assume Merchants theory.
- The percentage of total energy dissipated due to friction at the tool-chip interface is
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Total heat generated, Q1 = Fc × V = 24000 J/min
Heat dissipated due to friction, Q2 = F × r × v
Frictional force, F = Fc sin α + Ff cos α = 503.76 N
Q2 =7193.69 J/min.∴ Q2 = 0.299; 30% Q2 Correct Option: A
Total heat generated, Q1 = Fc × V = 24000 J/min
Heat dissipated due to friction, Q2 = F × r × v
Frictional force, F = Fc sin α + Ff cos α = 503.76 N
Q2 =7193.69 J/min.∴ Q2 = 0.299; 30% Q2
- The values of shear angle and shear strain, respectively, are
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Here, r = cutting ratio,
n = t1 = 0.5 = 0.714 t2 0.7 ∴ tanφ = rcosα 1 − rsinα
where, φ = shear angle
and α = rake angle
∴ φ = 40.2º
Shear strain = cot φ + tan (φ – α) = 1.65Correct Option: D
Here, r = cutting ratio,
n = t1 = 0.5 = 0.714 t2 0.7 ∴ tanφ = rcosα 1 − rsinα
where, φ = shear angle
and α = rake angle
∴ φ = 40.2º
Shear strain = cot φ + tan (φ – α) = 1.65
- Two tools P and Q have signatures 5°-50-60-6°- 8°-30°-0 and 5°-5°-7°-7°-8°-15°-0 (both ASA) respectively. They are used to turn components under the same machining conditions. If hp and hQ denote the peak-to-valley heights of surfaces produced by the tools P and Q, the ratio hp /hQ will be
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Hmax = f tan(ψ) + cot(ψ)
ψ = SCEA
ψ1 = ECEAhP = tan15° + Cot8° hQ tan30° + cot8° Correct Option: B
Hmax = f tan(ψ) + cot(ψ)
ψ = SCEA
ψ1 = ECEAhP = tan15° + Cot8° hQ tan30° + cot8°
- A 600 mm x 30 mm flat surface of a plate is to be finish machined on a shaper. The plate has been fixed with the 600 mm side along the tool travel direction. If the tool over-travel at each end of the plate is 20 mm, average cutting speed is 8 m/min, feed rate is 0.3 mm/stroke and the ratio of return time to cutting time of the tool is 1: 2, the time required for machining will be
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Length travelled in forwarded stroke = 640 mm
Number of strokes = 100
Time for cutting = 8 min
Return time = 4 min
Total time =12 min.Correct Option: B
Length travelled in forwarded stroke = 640 mm
Number of strokes = 100
Time for cutting = 8 min
Return time = 4 min
Total time =12 min.
- The figure below shows a graph which qualitatively relates cutting speed and cost per piece produced.
The three curves 1,2 and 3 respectively represent
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Machining cost = [Machining time × Direct Labour Cost]
So as cutting speed increases, machining time decreases and therefore machining cost decreases.Correct Option: A
Machining cost = [Machining time × Direct Labour Cost]
So as cutting speed increases, machining time decreases and therefore machining cost decreases.