Machine Design Miscellaneous


  1. Two shafts A and B are made of the same material. The diameter of shaft B is twice that of shaft A The ratio of power which can be transmitted by shaft A to that of shaft B is









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    TA =
    π
    × d3τ
    16

    TB =
    π
    × (2d)3τ
    16

    PA
    =
    TA
    =
    1
    PBTB8

    Correct Option: C

    TA =
    π
    × d3τ
    16

    TB =
    π
    × (2d)3τ
    16

    PA
    =
    TA
    =
    1
    PBTB8


  1. Torque to weight ratio for a circular shaft transmitting power is directly proportional to the









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    T =
    π
    d3τ
    16

    w = π
    d2
    ρl
    4

    T
    ∝ d
    w

    Correct Option: B

    T =
    π
    d3τ
    16

    w = π
    d2
    ρl
    4

    T
    ∝ d
    w



  1. A large uniform plate containing a rivet hole subjected to uniform uniaxial tension of 95 MPa. The maximum stress in the plate is










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    Max Stress =
    Avg. Force
    Min. Area

    =
    95 × 10 × t
    = 100 MPa
    (10 - 0.5)t

    Correct Option: A

    Max Stress =
    Avg. Force
    Min. Area

    =
    95 × 10 × t
    = 100 MPa
    (10 - 0.5)t


  1. The outside diameter of a hollow shaft that is twice its inside diameter the ratio of its torque carrying capacity to that of a solid shaft of the same material and the same outside diameter is









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    Thollow =
    π
    d4 -
    d4
    ......(1)
    3216d

    Tsolid =
    π
    d3τ .......(2)
    16

    By dividing (1) with (2)
    Thollow
    =
    15
    Tsolid16

    Correct Option: A

    Thollow =
    π
    d4 -
    d4
    ......(1)
    3216d

    Tsolid =
    π
    d3τ .......(2)
    16

    By dividing (1) with (2)
    Thollow
    =
    15
    Tsolid16



  1. The expected life of a ball bearing subjected to a load of 9800 N and working at 1000 rpm is 3000 hours. What is the expected life of the same bearing for a similar load of 4900 N and speed of 2000 rpm?









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    L (P)3 = constant

    L1
    =
    P2
    3
    L2P1

    1000 × 60 × 3000
    =
    4900
    3
    2000 × 60 × t29800

    t2 = 12000 hours

    Correct Option: B

    L (P)3 = constant

    L1
    =
    P2
    3
    L2P1

    1000 × 60 × 3000
    =
    4900
    3
    2000 × 60 × t29800

    t2 = 12000 hours