Machine Design Miscellaneous


  1. A 60 mm long and 6 mm thick fillet weld carries a steady load of 15 kN along the weld. The shear strength of the weld material is equal to 200 MPa. The factor of safety is









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    P = .707 t × Le × τs.

    15 × 103
    = τs
    .707 × 6 × 60

    ∴ τs = 58.94 MPa
    Given shear strength of material = 200 MPa
    FOS =
    200
    = 3.4
    58.94

    Correct Option: B

    P = .707 t × Le × τs.

    15 × 103
    = τs
    .707 × 6 × 60

    ∴ τs = 58.94 MPa
    Given shear strength of material = 200 MPa
    FOS =
    200
    = 3.4
    58.94


  1. Endurance limit of a beam subjected to pure bending decreases with









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    NA

    Correct Option: A

    NA


  1. Fatigue life of a material for a fully reversed loading condition is estimated from σa = 1100 N–0.15 where σa is the stress amplitude in MPa and N is the failure life in cycles. The maximum allowable stress amplitude (in MPa) for a life of 1 × 105 cycles under the same loading condition is_______(correction to two decimal places).









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    σmax - σmin
    = 1100N-0.15
    2

    σmax - (- σmin)
    = 1100N-0.15
    2

    max
    = 1100N-0.15(N = 105)
    2

    ∴ We get, σmax = 195.61 N

    Correct Option: A

    σmax - σmin
    = 1100N-0.15
    2

    σmax - (- σmin)
    = 1100N-0.15
    2

    max
    = 1100N-0.15(N = 105)
    2

    ∴ We get, σmax = 195.61 N


  1. A machine element has an ultimate strength (σu) of 600 N/mm², and endurance limit (σen) of 250 N/mm². The fatigue curve for the element on a log-log plot is shown below. If the element is to be designed for a finite life of 10000 cycles, the maximum amplitude of a completely reversed operating stress is ______N/mm².









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    log Sf - log (0.8 × 600)
    =
    4 - 3
    log 250 - log (0.6 × 600)6 - 3

    log Sf - log (480)
    =
    1
    log 250 - log (360)3

    Sf = 386.19 MPa

    Correct Option: A

    log Sf - log (0.8 × 600)
    =
    4 - 3
    log 250 - log (0.6 × 600)6 - 3

    log Sf - log (480)
    =
    1
    log 250 - log (360)3

    Sf = 386.19 MPa


  1. At a critical point in a component, the state of stress is given as σxx = 100 MPa, σyy = 220 MPa, σxy = σyx = 80 M Pa and al l ot her st r ess components are zero. The yield strength of the material is 468 MPa. The factor of safety on the basis of maximum shear stress theory is_____ (round off to one decimal place).









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    Given, σx = 100 MPa
    σy = 220 MPa
    σxy = 80 MPa

    = 160 ± √55² + 80²
    σ1, 2 = 160 ± 97.0824
    ⇒ σ1 = 257.08 Mpa, σ2 = 62.9 Mpa
    As per maximum shear stress theory

    Syt
    =
    σ1
    =
    257.08

    2N22

    N =
    468
    = 1.8
    257.08

    Correct Option: A

    Given, σx = 100 MPa
    σy = 220 MPa
    σxy = 80 MPa

    = 160 ± √55² + 80²
    σ1, 2 = 160 ± 97.0824
    ⇒ σ1 = 257.08 Mpa, σ2 = 62.9 Mpa
    As per maximum shear stress theory

    Syt
    =
    σ1
    =
    257.08

    2N22

    N =
    468
    = 1.8
    257.08