Mechanical and structural analysis miscellaneous


Mechanical and structural analysis miscellaneous

Mechanics and Structural Analysis

  1. A simply supported beam AB of span, L = 24 m is subjected to two wheel loads acting at a distance, d = 5 m apart as shown in the figure below. Each wheel transmits is a load, P = 3 kN and may occupy any position along the beam. If the beam is an I-section having section modulus, S = 16.2 cm3, the maximum bending stress (in GPa) due to the wheel loads is_______________









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    σ
    =
    m
    yI

    ⇒ σ =
    m
    . y =
    m
    Iz

    =
    28.5 × 106
    16.2 × 103

    (m = 3 × 9.5 = 28.5)
    = 1759.2 GPa

    Correct Option: D


    σ
    =
    m
    yI

    ⇒ σ =
    m
    . y =
    m
    Iz

    =
    28.5 × 106
    16.2 × 103

    (m = 3 × 9.5 = 28.5)
    = 1759.2 GPa


  1. A horizontal beam ABC is loaded as shown in the figure below. The distance of the point of contraflexure from end A (in m) is ____









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    Take RC at redundant
    By Marshall Reciprocal Theorem
    Deflection at C due to load B, ∆BC

    BC =
    10 ×(0.75)3
    +
    10 ×(0.75)3
    × 0.25
    3EI2EI

    =
    2.11
    EI

    Deflection at C due to redundant Rc, ∆cc
    cc =
    Rc ×(0.75)3
    =
    0.141Rc
    3EIEI

    ∴ ∆c = 0
    2.11
    -
    0.141Rc
    = 0
    EIEI

    Rc =
    2.11
    = 15 KN
    0.141


    Mx = 10x – 15 × (x – 0.25) = 0
    ⇒ = 10x – 15x + 3.75 = 0
    ⇒ x = 0.75m
    ∴ Distanced contraflexure from C = 0.75m
    ⇒ Distanced contraflexure from A = 1 – 0.75 = 0.25m

    Correct Option: A


    Take RC at redundant
    By Marshall Reciprocal Theorem
    Deflection at C due to load B, ∆BC

    BC =
    10 ×(0.75)3
    +
    10 ×(0.75)3
    × 0.25
    3EI2EI

    =
    2.11
    EI

    Deflection at C due to redundant Rc, ∆cc
    cc =
    Rc ×(0.75)3
    =
    0.141Rc
    3EIEI

    ∴ ∆c = 0
    2.11
    -
    0.141Rc
    = 0
    EIEI

    Rc =
    2.11
    = 15 KN
    0.141


    Mx = 10x – 15 × (x – 0.25) = 0
    ⇒ = 10x – 15x + 3.75 = 0
    ⇒ x = 0.75m
    ∴ Distanced contraflexure from C = 0.75m
    ⇒ Distanced contraflexure from A = 1 – 0.75 = 0.25m



  1. For formation of collapse mechanism in the following figure, the minimum value of Pu is cmp /L. mp and 3 mp denote the plastic moment capacities of beam sections as shown in this figure. The value of c is________.









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    Mechanism I
    3mp .θ + mp .(2 θ ) + mp θ

    P u ×
    L
    × θ
    4


    ⇒ 6mp .θ = P u ×
    L
    × θ
    4

    ⇒ P u = 24
    mp
    L

    Mechanism II

    1.Φ = 3θ
    ⇒ Φ = 3θ

    = 13.33
    mp
    L

    ∴ c = 13.33

    Correct Option: C

    Mechanism I
    3mp .θ + mp .(2 θ ) + mp θ

    P u ×
    L
    × θ
    4


    ⇒ 6mp .θ = P u ×
    L
    × θ
    4

    ⇒ P u = 24
    mp
    L

    Mechanism II

    1.Φ = 3θ
    ⇒ Φ = 3θ

    = 13.33
    mp
    L

    ∴ c = 13.33


  1. Two beams are connected by a linear spring as shown in the following figure. For a load P as shown in the figure, the percentage of the applied load P carried by the spring is ________.









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    ∆ =
    R
    =
    R
    × 2L3
    K3EI

    ∆ =K =
    3EI
    2L3

    ∆ =
    2RL3
    3EI


    P
    = R
    3

    ∴ % of applied load carried by spring
    =
    1
    × 100 = = 33.3%
    3

    Correct Option: A

    ∆ =
    R
    =
    R
    × 2L3
    K3EI

    ∆ =K =
    3EI
    2L3

    ∆ =
    2RL3
    3EI


    P
    = R
    3

    ∴ % of applied load carried by spring
    =
    1
    × 100 = = 33.3%
    3



  1. For the 2D truss with the applied loads shown below, the strain energy in the member XY is _______ kN-m. For member XY, assume AE=30kN, where A is cross section area and E is the modulus of elasticity.









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    RA × 3 + 10 × 9 = 0
    ⇒ RA = –30 kN
    RB = 35 kN
    Taking joint A

    Taking joint B

    Taking joint x
    Fx = 10 kN

    U =
    F2 × L
    2AE

    =
    102 × 3
    = 5 kNm
    2 × 3

    Correct Option: A


    RA × 3 + 10 × 9 = 0
    ⇒ RA = –30 kN
    RB = 35 kN
    Taking joint A

    Taking joint B

    Taking joint x
    Fx = 10 kN

    U =
    F2 × L
    2AE

    =
    102 × 3
    = 5 kNm
    2 × 3