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: Orthogonal turning is performed on a cylindrical workpiece with shear strength of 250 MPa. The following conditions are used: cutting velocity is 180 m/min, feed is 0.20 mm/rev, depth of cut is 3 mm, chip thickness ratio = 0.5. The orthogonal rake angle is 7°. Apply Merchants theory for analysis.

  1. The shear plane angle (in degrees) and the shear force respectively are








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    tanφ=
    rcosα
    		 =
    0.5 × cos7°
    		 = 0.503
    1 − rsinα1 − 0.5sin7°

    φ = 28°
    Now, shear force = shear stress × area
    F = 250 × AB width of cut

    ∴  AB =
    AC
    		 =
    0.2
    		 = 0.43
    sin28°sin28°

    where AC is incut chip thickness
    Now width ≌ depth of cut = 3 mm
    F = 250 × 0.43 ×3
    ≌ 320 N

    Correct Option: D

    tanφ=
    rcosα
    		 =
    0.5 × cos7°
    		 = 0.503
    1 − rsinα1 − 0.5sin7°

    φ = 28°
    Now, shear force = shear stress × area
    F = 250 × AB width of cut

    ∴  AB =
    AC
    		 =
    0.2
    		 = 0.43
    sin28°sin28°

    where AC is incut chip thickness
    Now width ≌ depth of cut = 3 mm
    F = 250 × 0.43 ×3
    ≌ 320 N

  1. In a single point turning tool, the side rake angle and orthogonal rake angle are equal. φ is the principal cutting edge angle and its range is 0° ≤ φ ≤ 90°. The chip flows in the orthogonal plane. The value of φ is closest to








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    Side rake angle is equal to orthogonal rake angle when principle cutting edge angle become 90° and corresponding approach angle SCEA = 0°

    Correct Option: D

    Side rake angle is equal to orthogonal rake angle when principle cutting edge angle become 90° and corresponding approach angle SCEA = 0°

Direction: A low carbon steel bar of 147 mm diameter with a length of 630 mm is being turned with uncoated carbide insert. The observed tool lives are 24 min and 12 min for cutting velocities of 90 m/min and 120 m/min respectively. The feed and depth of cut are 0.2 mm/ rev and 2 mm respectively. Use the unmachined diameter to calculate the cutting velocity.

  1. Neglect over-travel or approach of the tool. When tool life is 20 min, the machining time in min for a single pass is








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    V = πDN
    N =210.2 rpm
    ∴  Machining time,

    tm=
    L
    		 = .15min.
    N

    Correct Option: C

    V = πDN
    N =210.2 rpm
    ∴  Machining time,

    tm=
    L
    		 = .15min.
    N

  1. When tool life is 20 min, the cutting velocity in m/min is








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    Here, D = 147 mm, L = 630 mm,
    f = .2 mm/rev,
    d = 2 mm
    Tool life equation,
    VTn = constant
    Now,   V1T1n = V2T2n
    Here,   n = .415
    ∴  VTn = V1T1n
    ⇒  V =97.07 m/min

    Correct Option: B

    Here, D = 147 mm, L = 630 mm,
    f = .2 mm/rev,
    d = 2 mm
    Tool life equation,
    VTn = constant
    Now,   V1T1n = V2T2n
    Here,   n = .415
    ∴  VTn = V1T1n
    ⇒  V =97.07 m/min

  1. In orthogonal turning of low carbon steel pipe with principal cutting edge angle of 90°, the main cutting force is 1000 N and the feed force is 800 N. The shear angle is 25° and orthogonal rake angle is zero. Employing Merchants theory, the ratio of friction force to normal force acting on the cutting tool is








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    Here,   φ = shear angle = 25º
    λ = Friction angle,
    α =rake angle = 0º
    From Merchant’s theory, 2φ + λ − α = 90º
    ∴  λ = 90º – 50º = 40º

    ∴  μ =
    F
    		 =
    Friction force
    		= tan λ = .83
    NNormal force

    Correct Option: C

    Here,   φ = shear angle = 25º
    λ = Friction angle,
    α =rake angle = 0º
    From Merchant’s theory, 2φ + λ − α = 90º
    ∴  λ = 90º – 50º = 40º

    ∴  μ =
    F
    		 =
    Friction force
    		= tan λ = .83
    NNormal force