Mechanical and structural analysis miscellaneous


Mechanical and structural analysis miscellaneous

Mechanics and Structural Analysis

  1. The buckling load P = Pcr for the column AB in the figure, as KT approaches infinity, become
    α
    π2EI
    L2

    where α is equal to









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    Since stiffness KT ≈ ∞,
    it act as fixed end. At the top end there will be only axial deformation. so it also cuts as fixed end.

    P = 4.
    π2EI
    l2

    ∴ α = 4

    Correct Option: D

    Since stiffness KT ≈ ∞,
    it act as fixed end. At the top end there will be only axial deformation. so it also cuts as fixed end.

    P = 4.
    π2EI
    l2

    ∴ α = 4


  1. The bending moment diagram for a beam is given below: The shear force at sections aa' and bb' respectively are of the magnitude.









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    Consider left of α – α'
    Shear force = 0
    (Since there is uniform bending moment) Consider right bb'
    Shear force = change in bending moment
    = (200 – 150) kNm = 50 kNm

    Correct Option: C

    Consider left of α – α'
    Shear force = 0
    (Since there is uniform bending moment) Consider right bb'
    Shear force = change in bending moment
    = (200 – 150) kNm = 50 kNm



  1. kN 75. A cantilever beam of length l width b and depth d is loaded with a concentrated vertical load at the tip. If yielding starts at a load P, the collapse load shall be









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    Shape factor =
    MP
    ie,
    Plastic moment
    MeElastic moment

    = 1.5 for rectangular.
    Ultimate load = yield point × shape factor yield point = P
    ∴ ultimate load = P ×1.5 = 1.5 P

    Correct Option: B

    Shape factor =
    MP
    ie,
    Plastic moment
    MeElastic moment

    = 1.5 for rectangular.
    Ultimate load = yield point × shape factor yield point = P
    ∴ ultimate load = P ×1.5 = 1.5 P


  1. A three hinged parabolic arch ABC has a span of 20 m and a central rise of 4 m. The arch has hinges at the ends at the centre. A train of two point loads of 20 kN and 10 kN, 5 m apart, crosses this arch from left to right, with 20 kN load leading. The maximum thrust induced at the supports is









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    Max thrust occurs when 20 kN coal reaches point B.
    This is found out by drawing ILD.
    Max Thrust = 20 × 1.25 + 10 × 1.25/2
    = 31.25 kN

    Correct Option: C


    Max thrust occurs when 20 kN coal reaches point B.
    This is found out by drawing ILD.
    Max Thrust = 20 × 1.25 + 10 × 1.25/2
    = 31.25 kN



  1. In a two dimensional stress analysis, the state of stress at a point is shown below.
    If σ = 120 Mpa and τ = 70 MPa,
    then σx and σy are respectively
    AB = 4
    BC = 3
    AC = 5









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    = 67.5 MPa

    = 213.3 MPa

    Correct Option: C


    = 67.5 MPa

    = 213.3 MPa