Materials Science and Manufacturing Engineering Miscellaneous

Materials Science and Manufacturing Engineering Miscellaneous

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Materials Science and Manufacturing Engineering

Materials Science and Manufacturing Engineering Miscellaneous

Direction: In Orthogonal turning of a bar of 100 mm diameter with a feed of 0.25 mm/rev, depth of cut of 4 mm and cutting velocity of 90m/min. it is observed that the main (tangential) cutting force is perpendicular to the friction force acting at the chip-tool interface. The main(tangential) cutting force is 1500 N.

  1. The normal force acting the chip-tool interface in N is








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    N = Fc .cosα – FT .sinα
    ∴ α = 0
    N = Fc = 1500 N

    Correct Option: B

    N = Fc .cosα – FT .sinα
    ∴ α = 0
    N = Fc = 1500 N

  1. The orthogonal rake angle of the cutting tool in degree is








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    D = 100 mm feed, f = 0.25 mm/rev d = 4 mm, V = 90m/min, Fc = 1500 N
    α = 0

    Correct Option: A

    D = 100 mm feed, f = 0.25 mm/rev d = 4 mm, V = 90m/min, Fc = 1500 N
    α = 0

  1. Minimum shear-strain in orthogonal turning with a cutting tool of zero rake angle is








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    Shear strain , τ =
    cosα
    sinφ.cos( φ - α )

    where, α = 0 = rake angle , φ = shear angle
    ∴ τ =
    1
    sinθcosθ

    ∴ T = sinφcosφ
    dt
    		 = cos ²φ - sin ²φ
    dφ

    or φ = 45°
    ∴ τmin =
    1
    { sin (π /4).cos(π /4) }

    ∴ τmin =
    1
    		 = 2
    { sin (1 /√2) × (1 /√2) }

    Correct Option: D

    Shear strain , τ =
    cosα
    sinφ.cos( φ - α )

    where, α = 0 = rake angle , φ = shear angle
    ∴ τ =
    1
    sinθcosθ

    ∴ T = sinφcosφ
    dt
    		 = cos ²φ - sin ²φ
    dφ

    or φ = 45°
    ∴ τmin =
    1
    { sin (π /4).cos(π /4) }

    ∴ τmin =
    1
    		 = 2
    { sin (1 /√2) × (1 /√2) }

  1. Friction at the tool-chip interface can be reduced by








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    By increasing the cutting speed, heat dissipation is increased so maximum temperature is decreased. Hence coefficient of friction is decreased.

    Correct Option: D

    By increasing the cutting speed, heat dissipation is increased so maximum temperature is decreased. Hence coefficient of friction is decreased.

  1. During orthogonal cutting of MS with a 10 deg rake angle tool, the chip thickness ratio was obtained as 0.4. The shear angle (in degrees) evaluated from this data is








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    tanφ =
    rcosα
    1 - rsinα

    where, r = 0.4 and α = 10°
    ∴ tanφ =
    0.4cos10°
    1 - 0.4sin10°

    or φ = 22.94°

    Correct Option: C

    tanφ =
    rcosα
    1 - rsinα

    where, r = 0.4 and α = 10°
    ∴ tanφ =
    0.4cos10°
    1 - 0.4sin10°

    or φ = 22.94°