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
  1. Autogeneous gas tungsten arc welding of a steel plate is carried out with welding current of 500 A. Voltage of 20 V and weld speed of 20 mm/ min. Consider the heat transfer efficiency from the arc to the weld pool as 90%. The heat input per unit length (in kJ/mm) is








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    Current = 500 A
    Voltage = 20 V
    V = 20 mm/min
    ηHT = 0.9

    Heat Input/length = ηHT
    V × I
    		 =
    0.9 × 20 × 500
    N20/60

    = 27 kJ/mm

    Correct Option: A

    Current = 500 A
    Voltage = 20 V
    V = 20 mm/min
    ηHT = 0.9

    Heat Input/length = ηHT
    V × I
    		 =
    0.9 × 20 × 500
    N20/60

    = 27 kJ/mm

  1. Group-I
    P. Arc welding
    Q. Friction welding
    R. Solid state Welding
    S. Laser welding
    Group-II
    1. Diffusion
    2. Polarity
    3. Focusing
    4. Kinetic energy








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    Arc welding - Polarity
    Friction welding - Kinetic energy
    Solid state welding - Diffusion
    Laser welding - Focusing

    Correct Option: D

    Arc welding - Polarity
    Friction welding - Kinetic energy
    Solid state welding - Diffusion
    Laser welding - Focusing

  1. Circular blanks of 35 mm dia are punched from a steel sheet of 2 mm thickness. If the clearance per side between the punch and die is to be kept as 40 microns. The sizes of punch and die should respectively be








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    Dia of Die = 35 mm
    Dia of Punch = 35 – 2 × 0.04 = 34.92mm

    Correct Option: B

    Dia of Die = 35 mm
    Dia of Punch = 35 – 2 × 0.04 = 34.92mm

Direction: A cup is to be drawn to a diameter of 70 mm with 35 mm depth from a 0.5 mm thick sheet metal. The cup is drawn in one operation. Assume σu = 430 MPa.

  1. The required blank diameter is








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    √d² + 4dh
    = √70² + (4 × 70 × 35)
    D = 121.24 mm

    Correct Option: C

    √d² + 4dh
    = √70² + (4 × 70 × 35)
    D = 121.24 mm

  1. A cylindrical cup of 48.5 mm diameter and 52 mm height has a corner radius of 1.5 mm. Cold rolled steel sheet of thickness 1.5 mm is used to produce the cup. Assume trim allowance as 2 mm per 25 mm of cup diameter. What is the blank size in mm?








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    = √d² + 4dh
    D = √(48.5)² + 4 × 48.5 × 52
    D = 111.53 mm
    Trim Allowance = 2/25 × 48.5 = 3.88 mm.
    Total blank size = 111.53 + 3.88 = 115.41 mm

    Correct Option: B

    = √d² + 4dh
    D = √(48.5)² + 4 × 48.5 × 52
    D = 111.53 mm
    Trim Allowance = 2/25 × 48.5 = 3.88 mm.
    Total blank size = 111.53 + 3.88 = 115.41 mm