Materials Science and Manufacturing Engineering Miscellaneous


Materials Science and Manufacturing Engineering Miscellaneous

Materials Science and Manufacturing Engineering

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









  1. View Hint View Answer Discuss in Forum

    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?









  1. View Hint View Answer Discuss in Forum

    = √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



  1. At 1000°C the crystallographic structure of iron is _______.









  1. View Hint View Answer Discuss in Forum

    FCC → due to Austenitic structure.

    Correct Option: D

    FCC → due to Austenitic structure.


  1. At the last hammer stroke the excess material from the finishing cavity of a forging die is pushed into _______.









  1. View Hint View Answer Discuss in Forum

    Gulter is provided in finishing stroke.

    Correct Option: A

    Gulter is provided in finishing stroke.



  1. The relationship between true strain (εT) and engineering strain (εE) in a uniaxial tension test is given as









  1. View Hint View Answer Discuss in Forum

    εtrue = LfLo
    dL
    = In1 +
    ∆L
    LLo

    But
    = εL ⇒ εtrue = In(1 + εE)
    Lo

    Correct Option: C

    εtrue = LfLo
    dL
    = In1 +
    ∆L
    LLo

    But
    = εL ⇒ εtrue = In(1 + εE)
    Lo