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Complex Loading Systems And Loaded Beams And Cylinders

  1. Apr 4, 2016 #1
    Hello this is my first post here so basically I've had my best crack at all the questions all I am really after is a bit or re assurance as to my answers and if any are wrong were I have gone wrong essentially. I've tried to include all my working out were possible.

    1. The problem statement, all variables and given/known data

    The component shown in Fig 1 is made from a material with the following properties and is subjected to a compressive force of 5kN.

    fig1-jpg.74374.jpg

    Material Properties :
    Young’s Modulus of Elasticity – 200 GNm - 2
    Modulus of Rigidity – 90 GNm - 2
    Poisons ratio – 0.32

    2. Relevant equations

    Calculate :
    (a) The stress in :
    (i) the circular section
    (ii) the square section

    (b) The strain in :
    (i) The circular section
    (ii) The square section

    (c) The change in length of the component

    (d) The change in diameter of the circular section

    (e) The change in the 40mm dimension on the square section

    (f) If the same component were subjected to a shear force of 7 kN as shown in FIG 2, calculate the shear strain in :

    (i) The circular section
    (ii) The square section

    3. The attempt at a solution

    Calculate :
    (a) The stress in :
    (i) the circular section


    Stress = Force / Area
    Area of circular section = πr2
    π(15)2 = 706.8583mm2
    Answer : Stress = 5000 / 706.86 = - 7.07MPa

    (ii) the square section

    Stress = Force / Area
    Area of the square section = length x height
    40 x 40 = 1600mm2
    Answer : Stress = 5000 / 1600 = - 3.125MPa

    (b) The strain in :
    (i) The circular section


    ε = dl / l0 = σ / E
    σ = F / A = 7.07MPa
    E = 200GN / m - 2 = 200000MPa
    7.07 / 200000 = 0.00003535
    Answer: Strain = - 3.535 x 10 - 5


    (ii) The square section

    ε = dl / l0 = σ / E

    σ = F / A = 3.125MPa
    E = 200GN / m - 2 = 200000MPa
    3.125 / 200000 = 0.000015625

    Answer: Strain = - 1.5625 x 10 - 5


    (c) The change in length of the component

    ε = dl / l0

    For the Cylinder Section
    ε = 3.535 x 10 - 5
    l0 = 60
    dl = ε x l0
    dl = 3.535 x 10 - 5 x 60 = 0.002121
    = - 2.121 x 10 - 3

    For the Square Section

    ε = 1.5625 x 10 - 5
    l0 = 60
    dl = ε x l0
    dl = 1.5625 x 10 - 5 x 60 = 0.005625
    = - 5.625 x 10 - 3

    Adding the two changes together
    - 2.121 x 10 - 3 + - 5.625 x 10 - 3
    = - 0.007746

    Answer Change in Length = - 7.75mm

    (d) The change in diameter of the circular section

    Change in Diameter = - original diameter x poisions ratio x (Change in length / Original length)
    - 30 x 0.32 x ( - 7.75 / 60 ) = 1.24

    Answer Change in diameter = 1.24mm

    (e) The change in the 40mm dimension on the square section

    Poisions ratio (v) = εx / εz = - εy / εz
    εx = εy = - vεz
    = - 0.32 x -1.5625 x 10 - 5
    = 0.000005
    Δx = εxx0 = 0.000005 x 40 = 0.0002

    Δx = εyy0 = 0.000005 x 40 = 0.0002
    Answer Change in dimension on the square section
    = 2 x 10 - 4

    (f) If the same component were subjected to a shear force of 7 kN as shown in FIG 2, calculate the shear strain in :

    eng2-jpg.75688.jpg

    (i) The circular section

    Stress = 7000 / 706.86 = 9.9029MPa
    Shear Strain = Shear Stress / Modulus of Rigidity
    Modulus of Rigidity = 90GN/m - 2
    = 90000MPa
    Shear Strain = 9.9029 / 90000 = 1.1003 x 10 - 4

    Answer Shear Strain = 1.1003 x 10 - 4

    (ii) The square section

    Stress = 7000 / 1600 = 4.375MPa
    Shear Strain = Shear Stress / Modulus of Rigidity
    Modulus of Rigidity = 90GN/m - 2
    = 90000MPa
    Shear Strain = 4.375 / 90000 = 4.86 x 10 - 5

    Answer Shear Strain = 4.86 x 10 - 5




















     
  2. jcsd
  3. Apr 5, 2016 #2
    * Edit
    2. Relevant equations
    Stress = Force / Area
    ε = dl / l0 = σ / E
    Shear Strain = Shear Stress / Modulus of Rigidity
    Poisions ratio (v) = εx / εz = - εy / εz
     
  4. Apr 7, 2016 #3
    Anyone ?
     
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