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Stress and Strain Coursework Question

  1. Oct 13, 2014 #1
    I am struggling with a question in my coursework, and would appreciate some guidance.

    The question is:

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

    Material Properties :

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


    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

    (Figure 1 is attached)

    I have worked out question 1 i) and ii) to be 7.07Pa and 3.125Pa respectively, but I do not know where to begin with the rest of this question.

    Thanks,
    Pete
     

    Attached Files:

  2. jcsd
  3. Oct 13, 2014 #2

    billy_joule

    User Avatar
    Science Advisor

    Show your working and relevant equations. What strain equations are you familiar with?
     
  4. Oct 13, 2014 #3
    So for the first question, I used Stress = Force / Area.

    Working:
    706.86mm2 / 5000N = 7.07Pa
    1600mm2 / 5000N = 3.125Pa

    I think I have figured out the 2nd question using Strain = Stress / Youngs Modulus

    Working:
    7.07Pa / 200GN = 3.535 x 10^-11
    3.125Pa / 200GN = 1.5625 x 10*-11

    Question c) I have attempted as: (F x length) / E

    For the cylindrical part of the component:
    = (7.07 x 60) / 200
    = 2.121 x 10^-9

    For the cuboid component:
    = (3.125 x 60) / 200
    = 9.375 x 10^-10

    Total change in length = 3.0575 x 10^-9

    But I am really struggling to find anything relating to a change in diameter when under compression. I know it is something to do with Poissons ratio, but not sure how to apply it.

    I forgot, there is another part to the question:

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

    Appreciate your help.

    Pete
     
  5. Oct 13, 2014 #4
    In terms of Poisson's ratio, how is the extensional strain in the radial direction related to the compressional strain in the axial direction?

    Chet
     
  6. Nov 14, 2014 #5
    Pete. Did you get a reply to the rest of this question.
     
  7. Nov 14, 2014 #6

    SteamKing

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    Staff Emeritus
    Science Advisor
    Homework Helper

    These values for the stresses are wrong. They're not even written correctly according to the formulas you show.

    1 Pascal ≠ 1 Newton / 1 mm2

    You need to check the definition of the Pascal unit and adjust your stress calculations accordingly.
     
  8. Nov 20, 2014 #7
    I have had another stab at it, and I would really appreciate it someone could sense check my answers? I haven't shown all the working on here, and the questions are in the attached files.

    Q1 a)
    Stress = Force / Area
    i) 5000N / 706.68mm2 = -7.07MPa
    ii) 5000N / 1600mm2 -3.125MPa

    b)
    i) Strain = Stress / E = -7.07MPa / 200GNm-2 = -3.535 x 10^-5
    ii) -3.125MPa / 200GNm-2 = -1.5625 x 10^-5

    c)
    Compression = Strain x Original Length
    Cylinder = -3.535 x 10^-5 x 60 = -2.121 x 10^-3
    Cuboid = -9.375 x 10^-5 x 60 = -9.375 x 10^-4
    Total change in length = -3.0585 x 10^-3 mm

    d) Using Poissons Ratio @ 0.32 = 3.3936 x 10^-4 mm increase in diameter when compressed

    e) 2 x 10^-4 mm increase in dimension when compressed

    f) Stress = Force / Area
    i) 7000 / 706.86 = 9.9029MPa
    Shear Strain = Shear Stress / Modulus of Rigidity = 9.0929MPa / 90 x 10^9 = 1.1003 x 10-4
    ii) 7000 / 1600 = 4.375MPa
    4.375MPa / 90 x 10^9 = 4.86 x 10^-5

    Thanks,
    Pete
     

    Attached Files:

  9. Feb 24, 2016 #8
    I would really like to see know how you got 3.3936 x 10^-4 for question d?
    I got 4.8936x10^-4 when using Change in diameter = - original diameter x Poisons ratio x (change in length / original length)?
    Ive been stuck on this question for some time and would appreciate some help.
     
  10. Jun 8, 2017 #9
     
  11. Jun 9, 2017 #10
    This thread is almost 3 years old, and I'm closing it.
     
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