Materials Science and Manufacturing Engineering Miscellaneous


Materials Science and Manufacturing Engineering Miscellaneous

Materials Science and Manufacturing Engineering

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.

  1. 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


  1. The values of shear angle and shear strain, respectively, are









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    Here, r = cutting ratio,

    n =
    t1
    =
    0.5
    = 0.714
    t20.7

    ∴  tanφ =
    rcosα
    1 − rsinα

    where, φ = shear angle
    and α = rake angle
    ∴  φ = 40.2º
    Shear strain = cot φ + tan (φ – α) = 1.65

    Correct Option: D

    Here, r = cutting ratio,

    n =
    t1
    =
    0.5
    = 0.714
    t20.7

    ∴  tanφ =
    rcosα
    1 − rsinα

    where, φ = shear angle
    and α = rake angle
    ∴  φ = 40.2º
    Shear strain = cot φ + tan (φ – α) = 1.65



  1. 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 = ECEA
    hP
    =
    tan15° + Cot8°
    hQtan30° + cot8°

    Correct Option: B

    Hmax =
    f
    tan(ψ) + cot(ψ)

    ψ = SCEA
    ψ1 = ECEA
    hP
    =
    tan15° + Cot8°
    hQtan30° + cot8°


  1. 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.



  1. 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.