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. In a single pass rolling process using 410 mm diameter steel rollers, a strip of width 140 mm and thickness 8 mm undergoes 10% reduction of thickness. The angle of bite in radians is








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    ∆H = D(1 – cosα)

    ⇒ cosα = 1 -
    ∆H
    		 = 1 -
    0.1 × 8
    D
    		410

    ⇒ α = 3.57°
    ∴ α = 3.57 ×
    π
    		 = 0.062 radians
    180

    Correct Option: C

    ∆H = D(1 – cosα)

    ⇒ cosα = 1 -
    ∆H
    		 = 1 -
    0.1 × 8
    D
    		410

    ⇒ α = 3.57°
    ∴ α = 3.57 ×
    π
    		 = 0.062 radians
    180

  1. A solid cylinder of diameter 100 mm and height 50 mm is forged between two frictionless flat dies to a height of 25 mm. The percentage change in diameter is








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    From incompressibility,
    πd1² h1 = πd2² h2

    ⇒ d2 = d2 × √
    h1
    h2

    ⇒ d2 = 100 × √
    50
    		 = 141.42
    25

    Percentage change in diameter =
    d2 - d1
    		 × 100 = 41.42%
    d1

    Correct Option: D

    From incompressibility,
    πd1² h1 = πd2² h2

    ⇒ d2 = d2 × √
    h1
    h2

    ⇒ d2 = 100 × √
    50
    		 = 141.42
    25

    Percentage change in diameter =
    d2 - d1
    		 × 100 = 41.42%
    d1

  1. The maximum possible draft in cold rolling of sheet increases with the








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    ∆h = µ2R, µ ↑ ∆h ↑

    Correct Option: A

    ∆h = µ2R, µ ↑ ∆h ↑

  1. The thickness of a metallic sheet is reduced from an initial value of 16 mm to a final value of 10 mm in one single pass rolling with a pair of cylindrical rollers each of diameter of 400 mm. The bite angle in degree will be








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    tanα = √(t1 - t2) / R = √(16 - 10) / 200
    tanα = 0.173
    ∴ Bite angle, α = tan-1(0.173) = 9.936°

    Correct Option: D

    tanα = √(t1 - t2) / R = √(16 - 10) / 200
    tanα = 0.173
    ∴ Bite angle, α = tan-1(0.173) = 9.936°

  1. A 4 mm thick sheet is rolled with 300 mm diameter rolls to reduce thickness without any change in its width. The friction coefficient at the work-roll interface is 0.1. The minimum possible thickness of the sheet that can be produced in a- single pass is








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    Given, Initial thickness of sheet, ti = 4 mm = 4 × 10–3 m
    Diameter of roll, D = 300 mm = 0.3 m
    Frictional coefficient at work-roll interface, μ = 0.1
    Let tf = Final thickness of sheet after rolling
    For single pass without slipping minimum possible thickness,
    ti - tf = μ²R
    μ = √(ti - tf) / R
    0.1 = √(4 × 10–3 - tf) / 0.15
    or tf = 2.5 mm

    Correct Option: C

    Given, Initial thickness of sheet, ti = 4 mm = 4 × 10–3 m
    Diameter of roll, D = 300 mm = 0.3 m
    Frictional coefficient at work-roll interface, μ = 0.1
    Let tf = Final thickness of sheet after rolling
    For single pass without slipping minimum possible thickness,
    ti - tf = μ²R
    μ = √(ti - tf) / R
    0.1 = √(4 × 10–3 - tf) / 0.15
    or tf = 2.5 mm