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: 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








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    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








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    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

  1. The voltage arc length characteristics of a DC arc is given by V = 20 + 40L, where L = arc length in cm. The power source characteristics can be approximated by a straight line. Open circuit is 80 V and short circuit current 1000 Amps. The optimum arc length in mm








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    Constant voltage Characteristics
    V = 20 + 40 L     ...(i)
    Vocv = 80 V,     Iscc = 1000 Amp.

    V = vocv −
    vocv
    		.I
    1scc

    V = 80 −
    80
    		I      ....(ii)
    1000

    Vtransfermer = Varc
    20 + 40L = 80 −
    80
    		L
    1000

    I =
    1000
    		(60 − 40L)
    80

    Power = V.I.
    For optimum Arc length
    dP
    		 = 0
    dL

    (20 + 40L)(60 − 40L)
    1000
    		.
    80

    Hence L = 0.5 cm = 5mm

    Correct Option: C

    Constant voltage Characteristics
    V = 20 + 40 L     ...(i)
    Vocv = 80 V,     Iscc = 1000 Amp.

    V = vocv −
    vocv
    		.I
    1scc

    V = 80 −
    80
    		I      ....(ii)
    1000

    Vtransfermer = Varc
    20 + 40L = 80 −
    80
    		L
    1000

    I =
    1000
    		(60 − 40L)
    80

    Power = V.I.
    For optimum Arc length
    dP
    		 = 0
    dL

    (20 + 40L)(60 − 40L)
    1000
    		.
    80

    Hence L = 0.5 cm = 5mm

  1. High alloy steel components are preheated before welding for reducing








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    Due to preheating, thermal conductivity will decrease so less thermal stressese and hence less distortion will take place.

    Correct Option: D

    Due to preheating, thermal conductivity will decrease so less thermal stressese and hence less distortion will take place.

  1. In TIG welding a.... A.... and.... B.... electrode is used








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    Non consumable base electrode is used.

    Correct Option: A

    Non consumable base electrode is used.