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. A disc of 200 mm diameter is blanked from a strip of an aluminum alloy of thickness 3.2 mm. The material shear strength to fracture is 150 MPa. The blanking force (in kN) is








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    Blanking force = πDtτ
    = π × 200 × 3.2 × 150 = 301.59 KN

    Correct Option: B

    Blanking force = πDtτ
    = π × 200 × 3.2 × 150 = 301.59 KN

  1. Consider the following statements : The size of the heat affected zone (HAZ) will increase with
    1. Increased starting temperature
    2. Increased welding speed
    3. Increased thermal conductivity of the base metal
    4. Increase in base metal thickness
    Which of the statements given above are correct?








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    H A Z increases due to higher temperature & higher heat dissipation due to higher thermal conductivity

    Correct Option: B

    H A Z increases due to higher temperature & higher heat dissipation due to higher thermal conductivity

  1. The current in Amperes used in resistance spot welding of plain carbon steel sheets (1 to 3 mm thick) lies within the range








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    Very high current is used in air gap.

    Correct Option: D

    Very high current is used in air gap.

  1. Match List-I (welding defects)with List-II (Causes) and select the correct answer using the codes given below the lists:
    List-I
    A. Spatter
    B. Distortion
    C. Slag inclusion
    D. Porosity
    List-II
    1. Damp electrodes
    2. Arc blow
    3. Improper cleaning in multi pass welding
    4. Poor joint selection








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    Spotter - Arc blow
    Distortion - Poor joint selection
    Slag Inclusion - Improper cleaning in multipass welding
    Porosity - Damp electrodes

    Correct Option: D

    Spotter - Arc blow
    Distortion - Poor joint selection
    Slag Inclusion - Improper cleaning in multipass welding
    Porosity - Damp electrodes

  1. It is require to punch a hole of diameter 10 mm on a sheet of thickness 3 mm. The shear strength of the work material is 420 MPa. The required punch force is








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    Punching force = πDtτ
    = π × 10 × 3 × 420 = 39.58 KN

    Correct Option: B

    Punching force = πDtτ
    = π × 10 × 3 × 420 = 39.58 KN