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Complex Loading Systems

  1. Sep 3, 2013 #1
    1. The problem statement, all variables and given/known data

    A tubular column supports a mass as shown in the diagram below. A
    strain gauge is placed transversely on the outside diameter of the bar to
    act as a load measuring device.

    The column has an outside diameter of 50 mm and an inside diameter of
    40 mm. The modulus of elasticity of the tube material is 250 GN m–2 and
    the poison’s ratio is 0.33.

    Untitled.png

    Q. Sketch a graph of the expected strain against the applied load, for a
    load range from 0 to 500 kg. Make load the horizontal axis on the
    graph and strain the vertical axis.


    2. Relevant equations

    Force,N = M.G

    Area,A = ∏/4 . (Do2 - Di2) = ∏/4 (0.05m2) - (0.04m2) = 7.0685834 x10-4 m2

    Normal Stress, σ = F / A

    Normal Strain, ε = deformation in length / Original Length = l - l0 / l0

    Young's Modulus, E = σ / ε = 250 GN m–2 = 250x10-9 N m-2

    ε = σ / E

    Poisson's Ration, V = 0.33

    Transversal Strain, εt = -νε

    3. The attempt at a solution


    Calculate Force, N, for range 0kg to 500kg using using F=MG

    Calculate Normal Stress for the range 0kg to 500kg using σ = F/A

    Calculate Normal Strain for the range 0kg to 500kg using ε = σ / E

    Calculate Transverse Strain for the range 0kg to 500kg using εt = -νε

    Plot a graph using Transverse Strain εt Vs Force, N

    Capture.PNG


    Does this look anywhere near correct?
     
  2. jcsd
  3. Sep 5, 2013 #2
    Anyone?
     
  4. Sep 5, 2013 #3

    SteamKing

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    Staff Emeritus
    Science Advisor
    Homework Helper

    E should be 250*10^9 N/m^2
     
  5. Sep 10, 2013 #4
    ok so i had another go, changing the modulus of elasticity. The graph looks much the same as before.

    Does this look correct?




    Capture.PNG
     
  6. Sep 10, 2013 #5
    Please show us your numbers for 500 kg. Also, aren't you asked to plot the transverse strain, rather than the axial strain? Also, the axial strains are negative, and the transverse strains are positive.
     
  7. Sep 16, 2013 #6
    Load(Kg) | Load (N) | Normal Stress | Normal Strain | transverse strain | longitudinal strain

    100kg | 981N | 1387831.118 | 5.55132E-06 | 1.83194E-06 | -5.55132E-06
    200kg | 1962N | 2775662.235 | 1.11026E-05 | 3.66387E-06 | -1.11026E-05
    300kg | 2943N | 4163493.353 | 1.6654E-05 | 5.49581E-06 | -1.6654E-05
    400kg | 3924N | 5551324.47 | 2.22053E-05 | 7.32775E-06 | -2.22053E-05
    500kg | 4905N | 6939155.588 | 2.77566E-05 | 9.15969E-06 | -2.77566E-05​



    here are my numbers for 100Kg to 500Kg. apologies for the formatting.
     
  8. Sep 16, 2013 #7
    Graph looks like this


    Capture.PNG
     
  9. Sep 16, 2013 #8
    I spot checked your numbers, and they look good. What sign convention do you use for stress; is compressive stress regarded as positive or negative?
     
  10. Sep 17, 2013 #9
    Stresses that result in extension, or tensile stresses, are defined as positive, and
    compressive stresses as negative in sign.
     
  11. Sep 17, 2013 #10
    If this is the case, then the normal stresses in your table should be negative, and the normal strains should be the same as the longitudinal strains.
     
  12. Sep 20, 2013 #11
    ok I've had a rethink and changed the table. I'm a little confused as I believe the transverse strains should be positive in value. however it is showing as negative.


    Load(Kg) | Load (N) | Normal Stress | Normal Strain | transverse strain | longitudinal strain

    -100 | -981 | -1387831.118 | -5.55132E-06 | -1.83194E-06 | -5.55132E-06
    -200 | -1962 | -2775662.235 | -1.11026E-05 | -3.66387E-06 | -1.11026E-05
    -300 | -2943 | -4163493.353 | -1.6654E-05 | -5.49581E-06 | -1.6654E-05
    -400 | -3924 | -5551324.47 | -2.22053E-05 | -7.32775E-06 | -2.22053E-05
    -500 | -4905 | -6939155.588 | -2.77566E-05 | -9.15969E-06 | -2.77566E-05
     
    Last edited: Sep 20, 2013
  13. Sep 20, 2013 #12
    The transverse strain should be opposite in sign to the longitudinal strain. If the cylinder compresses axially, it must expand in the radial and circumferential directions. That's what your Poisson Ratio equation says.

    Chet
     
  14. Sep 23, 2013 #13
    ok. So i changed the signs. I now have these figures and the graph is below. Does this now look ok?



    Load(Kg)| Load(N) | Normal Stress | Normal Strain | Transverse Strain | Longitudinal Strain
    100 | 0981 | -1387831.118 | -5.55132E-06 | 1.83194E-06 | -5.55132E-06
    200 | 1962 | -2775662.235 | -1.11026E-05 | 3.66387E-06 | -1.11026E-05
    300 | 2943 | -4163493.353 | -1.66540E-05 | 5.49581E-06 | -1.6654E-05
    400 | 3924 | -5551324.470 | -2.22053E-05 | 7.32775E-06 | -2.22053E-05
    500 | 4905 | -6939155.588 | -2.77566E-05 | 9.15969E-06 | -2.77566E-05


    e222.PNG
     
  15. Sep 23, 2013 #14
    Looks good.

    Chet
     
  16. Sep 23, 2013 #15
    Great Thank you. The next part to the question has me a bit confused.

    Show with the aid of a diagram how the orientation of the strain gauge could be changed to provide a greater strain for the same load. Sketch a graph of strain against applied load when the strain gauge is in
    the new position.


    I'm thinking that maybe the strain gauge could be mounted at a 45° angle. But would this give greater strain for the same load? Or am I over thinking this and should it just be mounted vertically?
     
  17. Sep 23, 2013 #16
    Have you learned how to express the strain in an arbitrary direction in terms of the components of strain in the principal directions? Is the question asking about the absolute magnitude of the strain in an arbitrary direction (i.e., irrespective of its sign) or about the actual strain in an arbitrary direction? I can see why you are confused.
     
  18. Sep 24, 2013 #17
    I have nothing in my learning material that details anything like that.


    here is the question:

    1.PNG
    2.PNG
     
  19. Sep 24, 2013 #18
    Well, OK, if you're talking about the absolute magnitude of the strain, and not whether it is extensional or compressive, then you can mount the strain gage in the vertical direction. That will give you the maximum absolute magnitude. However, if you're including the sign of the strain, the maximum value of strain is in the circumferential direction.
     
  20. Sep 24, 2013 #19
    Many Thanks. You are a great help :)
     
  21. Sep 26, 2013 #20
    the next part of the question asks to Calculate the change in:
    (i) the outside diameter (Do)
    (ii) the inside diameter (Di)
    (iii) the column length (l)


    I have got the following answers:

    (i) = 0.000457984 mm
    (ii) = 0.000366387 mm
    (iii) = -0.005551324 mm

    do these answers look reasonable?
     
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