Strength Of Materials Miscellaneous


Strength Of Materials Miscellaneous

Strength Of Materials

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











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    σ =
    P
    A

    P = σ A = 95 × 106 × 10 × 10–2 t
    σ =
    P
    =
    95 × 106 × 10 × 10–2 t
    = 190 MPa
    A(10 × 10-2 - 5 × 10–2)t

    Correct Option: B


    σ =
    P
    A

    P = σ A = 95 × 106 × 10 × 10–2 t
    σ =
    P
    =
    95 × 106 × 10 × 10–2 t
    = 190 MPa
    A(10 × 10-2 - 5 × 10–2)t


  1. A rod of length 20 mm is stretched to make a rod of length 40 mm. Subsequently, it is compressed to make a rod of final length 10 mm. Consider the longitudinal tensile strain as positive and compressive strain as negative. The total true longitudinal strain in the rod is









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    Volume remain same L1 = 20 mm, L2 = 40 mm, L3 = 10mm
    A1 L1 = A2 L2 = A3 L3

    A1
    =
    L3
    A3L1

    True strain = ln
    A1
    = ln
    L3
    = -0.693
    A3L1

    Correct Option: B

    Volume remain same L1 = 20 mm, L2 = 40 mm, L3 = 10mm
    A1 L1 = A2 L2 = A3 L3

    A1
    =
    L3
    A3L1

    True strain = ln
    A1
    = ln
    L3
    = -0.693
    A3L1



  1. An elastic body is subjected to a tensile stress X in a particular direction and a compressive stress Y in its perpendicular direction. X and Y are unequal in magnitude. On the plane of maximum shear stress in the body there will be









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    NA

    Correct Option: C

    NA


  1. The three-dimensional state of stress at a point is given by

    The shear stress on the x-face in y-direction at the same point is then equal to









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    Correct Option: C



  1. A point mass of 100 kg is dropped onto a massless elastic bar (cross-sectional area = 100 mm2, length = 1 m, Young's modulus = 100 GPa) from a height H of 10 mm as shown (Figure is not to scale). If g = 10 m/s2, the maximum compression of the elastic bar is ______ mm.









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    m = 100 kg , h = 10 mm, L = 1m
    E = 100 GPa, g = 10m/s2 A = 100 mm2

    δ =
    WL
    1 + √
    2EAh
    AEwL

    δ =
    100 × 10 × 1
    [ 1 + √1 + { (2 × 100 × 109 × 100 × 10-6 × 0.01) / 1000 } ]
    100 × 10-6 × 100 × 109

    δ = 1.517 mm

    Correct Option: C

    m = 100 kg , h = 10 mm, L = 1m
    E = 100 GPa, g = 10m/s2 A = 100 mm2

    δ =
    WL
    1 + √
    2EAh
    AEwL

    δ =
    100 × 10 × 1
    [ 1 + √1 + { (2 × 100 × 109 × 100 × 10-6 × 0.01) / 1000 } ]
    100 × 10-6 × 100 × 109

    δ = 1.517 mm