Materials Science and Manufacturing Engineering Miscellaneous


Materials Science and Manufacturing Engineering Miscellaneous

Materials Science and Manufacturing Engineering

  1. A strip with cross-sectional area 150 mm × 4.5 mm is being rolled with 20% reduction of area using 450 mm diameter rollers. The angle subtended by the deformation zone at the roll centre is (in radians)









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    Bite angle ; θ = tan-1
    R∆h
    { R - (∆h / 2) }

    From question, we have 20% reduction in cross sectional area
    150 × 4.5 × 0.8 = 150 × h
    ∆h = 3.6 mm
    h = 4.5 – 3.6 = 0.9 mm
    Given , R =
    450
    = 225 mm
    2

    θ = tan-1
    225 × 0.9
    { 225 - (0.9/ 2) }

    θ = 3.62° = 0.06 rad

    Correct Option: D

    Bite angle ; θ = tan-1
    R∆h
    { R - (∆h / 2) }

    From question, we have 20% reduction in cross sectional area
    150 × 4.5 × 0.8 = 150 × h
    ∆h = 3.6 mm
    h = 4.5 – 3.6 = 0.9 mm
    Given , R =
    450
    = 225 mm
    2

    θ = tan-1
    225 × 0.9
    { 225 - (0.9/ 2) }

    θ = 3.62° = 0.06 rad


  1. A wire of 0.1 mm dia is drawn from a rod of 15 mm diameter dies giving reductions of 20%, 40% and 80% are available. For minimum error in the final size, the number of stages and reduction at each stage respectively would be









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    % reduction =
    Diameter reduced in draw
    Diameter before draw

    for option (a) =
    d0 - d1
    = 0.8
    d0

    d1 = 2d0
    Given d0 = 15 mm , we get d1 = 3 mm
    Similarly, after 2nd pass, we get d2 = 0.6 mm and after 3rd pass, we get d3 = 0.12 mm For option (b), 1st 3 stages are same as in option (a), in 4th pass we have
    d3 - d4
    = 0.2
    d3

    ⇒ d4 = 0.8d3
    ⇒ d4 = 0.8 × 0.12 = 0.096 mm
    For option (c), first two stages are same as in option (a), in 3rd pass we have
    d3 - d4
    = 0.4
    d3

    ⇒ d3 = 6 , d2 = 0.6 × 0.6 = 0.36 mm
    For 4th pass we have
    d3 - d4
    = 0.4
    d3

    ⇒ d4 = 0.6 , d3 = 0.6 × 0.36 = 0.216 mm
    For 5th stage, we have
    d5 - d4
    = 0.2
    d4

    ⇒ d5 = 0.8 , d4 = 0.8 × 0.216 = 0.1728 mm
    Output in option (b) is closest to the desired diameter of 0.1 mm

    Correct Option: B

    % reduction =
    Diameter reduced in draw
    Diameter before draw

    for option (a) =
    d0 - d1
    = 0.8
    d0

    d1 = 2d0
    Given d0 = 15 mm , we get d1 = 3 mm
    Similarly, after 2nd pass, we get d2 = 0.6 mm and after 3rd pass, we get d3 = 0.12 mm For option (b), 1st 3 stages are same as in option (a), in 4th pass we have
    d3 - d4
    = 0.2
    d3

    ⇒ d4 = 0.8d3
    ⇒ d4 = 0.8 × 0.12 = 0.096 mm
    For option (c), first two stages are same as in option (a), in 3rd pass we have
    d3 - d4
    = 0.4
    d3

    ⇒ d3 = 6 , d2 = 0.6 × 0.6 = 0.36 mm
    For 4th pass we have
    d3 - d4
    = 0.4
    d3

    ⇒ d4 = 0.6 , d3 = 0.6 × 0.36 = 0.216 mm
    For 5th stage, we have
    d5 - d4
    = 0.2
    d4

    ⇒ d5 = 0.8 , d4 = 0.8 × 0.216 = 0.1728 mm
    Output in option (b) is closest to the desired diameter of 0.1 mm



  1. Calculate the bite angle when rolling plates 12 mm thick using work rolls 600 mm diameter and reducing the thickness by 3 mm









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    D = 600 mm , R =
    D
    = 300 mm
    2

    hi = 12 mm
    ∆h = 3 mm
    θ = tan-1
    R∆h
    { R - (∆h / 2) }

    θ = tan-1
    300 × 3
    { 300 - (3/ 2) }

    θ = 5.73°

    Correct Option: D

    D = 600 mm , R =
    D
    = 300 mm
    2

    hi = 12 mm
    ∆h = 3 mm
    θ = tan-1
    R∆h
    { R - (∆h / 2) }

    θ = tan-1
    300 × 3
    { 300 - (3/ 2) }

    θ = 5.73°


  1. A cylindrical riser of 6 cm diameter and 6 cm height has to be designed for a sand casting mould for producing a steel rectangular plate casting of 7 cm × 10 cm × 2 cm dimensions having the total solidification time of 1.36 minute. The total solidification time (in minute) of the riser is _____.









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    V
    =
    7 × 10 × 2
    = 0.673 cm
    Acasting2(7 × 10 + 7 × 2 + 2 × 10)

    V
    =
    (π / 4)d3
    ATop riser{ (π / 4)d2 + πd2 }

    V
    =
    d
    =
    6
    = 1.25
    ATop riserS5

    V
    =
    (π / 4)d3
    =
    d
    = 1 cm
    ABottom riser{ (π / 2)d2 + πd2 }6

    tTop riser
    =
    (V / A)2Top riser
    =
    1.2
    2
    tcasting(V / A)2casting0.673

    tTop riser = 4.133 min
    tBottom riser
    =
    (V / A)2Bottom riser
    =
    1
    2
    tcasting(V / A)2casting0.673

    tBottom riser = 3 min

    Correct Option: B

    V
    =
    7 × 10 × 2
    = 0.673 cm
    Acasting2(7 × 10 + 7 × 2 + 2 × 10)

    V
    =
    (π / 4)d3
    ATop riser{ (π / 4)d2 + πd2 }

    V
    =
    d
    =
    6
    = 1.25
    ATop riserS5

    V
    =
    (π / 4)d3
    =
    d
    = 1 cm
    ABottom riser{ (π / 2)d2 + πd2 }6

    tTop riser
    =
    (V / A)2Top riser
    =
    1.2
    2
    tcasting(V / A)2casting0.673

    tTop riser = 4.133 min
    tBottom riser
    =
    (V / A)2Bottom riser
    =
    1
    2
    tcasting(V / A)2casting0.673

    tBottom riser = 3 min



  1. A cube shaped casting solidifies in 5 min. The solidification time in min for a cube of the same material, which is 8 times heavier than the original casting, will be









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    t ∝ (V/A)2 and m2 = 8m1,
    ρV2 = 8ρV1
    ⇒ V2 = 8V1
    a2 = 2a1 ⇒ t2 = 4t1

    Correct Option: B

    t ∝ (V/A)2 and m2 = 8m1,
    ρV2 = 8ρV1
    ⇒ V2 = 8V1
    a2 = 2a1 ⇒ t2 = 4t1