Materials Science and Manufacturing Engineering Miscellaneous


Materials Science and Manufacturing Engineering Miscellaneous

Materials Science and Manufacturing Engineering

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 90 m/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 at the chip-tool interface in N is









  1. View Hint View Answer Discuss in Forum

    NA

    Correct Option: B

    NA


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









  1. View Hint View Answer Discuss in Forum

    NA

    Correct Option: A

    NA



  1. Which of the following process can be used for welding of Aluminum alloys?
    P. Submerged arc welding
    Q. Gas metal arc welding
    R. Electroslag welding
    S. Gas tungsten arc welding









  1. View Hint View Answer Discuss in Forum

    To remove the ceramic layer of aluminium oxide, TIG is used.

    Correct Option: B

    To remove the ceramic layer of aluminium oxide, TIG is used.


Direction: A weld is made using MIG welding process with the following welding parameters:
Current: 200 A: Voltage: 25 V; Welding speed: 18 cm/min; Wire diameter: 1.2 mm; Wire feed rate: 4 m/min. Thermal efficiency of the process: 65%

  1. The area of cross-section of weld bead in mm² is









  1. View Hint View Answer Discuss in Forum

    Volume of weld

    =
    π
    d2 × V
    4

    =
    π
    (1.2)2 × 4000
    4

    = 4521.6 mm3/min
    Area of weld =
    4521.6
    =
    4521.6
    = 25.12mm2
    speed18 × 10

    Correct Option: B

    Volume of weld

    =
    π
    d2 × V
    4

    =
    π
    (1.2)2 × 4000
    4

    = 4521.6 mm3/min
    Area of weld =
    4521.6
    =
    4521.6
    = 25.12mm2
    speed18 × 10



  1. The heat input per unit length of the weld in kJ/cm is









  1. View Hint View Answer Discuss in Forum

    Current = 200A
    Voltage = 25 V
    dw = 1.2 mm
    Fr = 4 m/min
    = 4000 mm/min
    V = 18 cm/min
    = 180 mm/min.
    ηTh= 65%
    Heat Input per unit length

    =
    V × I × η
    =
    200 × 25 × 0.65
    = 10.833 kJ/cm
    speed18/60

    Correct Option: C

    Current = 200A
    Voltage = 25 V
    dw = 1.2 mm
    Fr = 4 m/min
    = 4000 mm/min
    V = 18 cm/min
    = 180 mm/min.
    ηTh= 65%
    Heat Input per unit length

    =
    V × I × η
    =
    200 × 25 × 0.65
    = 10.833 kJ/cm
    speed18/60