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Stress and strain. young modulus

  1. Nov 15, 2009 #1
    in the following question, how do i find the changle in AB, the change in AD and the change in AC?

    http://lh4.ggpht.com/_H4Iz7SmBrbk/Sv_hYT83rYI/AAAAAAAAB8s/7l5YRXMxVfQ/s912/Capture.JP [Broken]


    what i have done is used the following
    [tex]\sigma[/tex]xx=E*[tex]\epsilon[/tex]xx

    [tex]\delta[/tex]AB=[tex]\epsilon[/tex]AB*AB



    since AB is on the x axis,
    [tex]\epsilon[/tex]AB=[tex]\epsilon[/tex]xx=[tex]\sigma[/tex]xx/E=150000000/200000000000=3/400

    [tex]\delta[/tex]AB=[tex]\epsilon[/tex]AB*AB=(3/400)*100*10^-5=75[tex]\mu[/tex]m



    but, 1st of all this is wrong! the correct answer is meant to be 60[tex]\mu[/tex]m, and second of all it seems too simple, why would [tex]\nu[/tex]=0.3 be given if i dont use it?
     
    Last edited by a moderator: May 4, 2017
  2. jcsd
  3. Nov 15, 2009 #2

    Mapes

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    The equation [itex]\sigma=E\epsilon[/itex] only applies to uniaxial loading of a long, thin object (like a rod or a beam). Do you know of a more general equation for 2-D or 3-D loading configurations?

    (This http://john.maloney.org/Papers/Generalized%20Hooke%27s%20Law%20%283-12-07%29.pdf" [Broken] about stresses, strains, and constitutive equations may be helpful.)
     
    Last edited by a moderator: May 4, 2017
  4. Nov 15, 2009 #3
    dont think so, what equation can i use?
     
  5. Nov 15, 2009 #4

    Mapes

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    Did you look at the link? Let me put it this way: strain in a certain direction can arise from a load in that direction, but it can also arise from lateral contraction caused by a load in another direction. Know what I mean?
     
  6. Nov 15, 2009 #5

    Maples is right.
    You should be using the equations for PLANE stress/strain and not AXIAL stress/strain. For axial strain ONLY, you would solve [tex]\epsilon[/tex]xx=[tex]\sigma[/tex]xx/E.

    For plane strain, use
    [tex]\epsilon[/tex]xx=[tex]\sigma[/tex]xx/E - [tex]\upsilon[/tex]/E*[tex]\sigma[/tex]yy
    and comparably
    [tex]\epsilon[/tex]yy=[tex]\sigma[/tex]yy/E - [tex]\upsilon[/tex]/E*[tex]\sigma[/tex]xx.

    When you [positively] load a member in direction x, it elongates in that direction. In order to hold the volume of the material constant**, the object will "thin" in the other two directions y and z. The amount that they thin is described by Poisson's Ratio [tex]\upsilon[/tex].
    **Note: The volume of the material can change slightly, but for loading within the normal range of the material's limits, the change will be imperceptible.

    Hope this gets you on your way.
     
  7. Nov 15, 2009 #6
    okay got is, hookes law right??
     
  8. Nov 15, 2009 #7
    for questions a and b it worked perfectly since AB and AD lie on the X and Z axes and i have all the info i need for them, for c) i need to find the change in the length in AC which is on a differend axis system, lets call it n-y-t. now i know that the angle between them is 53.12 (atan(100/75) ) but i want the other angle,- 90+53.13 since i want the strain which goes along AC and i have the transformations equations for stress/strains
    so i looked for the strain[tex]\epsilon[/tex]t

    [tex]\epsilon[/tex]t=[tex]\epsilon[/tex]x*cos2(143.13)+[tex]\epsilon[/tex]z*sin2(143.13)=2.86*10-4

    [tex]\delta[/tex]AB=[tex]\epsilon[/tex]t*AB=2.86*10-4*125*10-3=35.75[tex]\mu[/tex]m
    which is way off, anyone see where im going wrong?
     
  9. Nov 15, 2009 #8

    Mapes

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    Instead of messing around with different coordinate systems, why not just look at the hypotenuse of the right triangle?
     
  10. Nov 15, 2009 #9
    i learned that that wouldnt work since these are not vectors, but i figured it out anyways, i was using the wrong angle.

    (looking at the answers pythagoras wont work here --> 60^2+20.6^2 is not 60.4^2)
     
  11. Nov 15, 2009 #10

    Mapes

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    I'm talking about the hypotenuse of the strained lengths: the hypotenuse of 100.060mm and 75.0206mm is 125.0604mm.
     
  12. Nov 15, 2009 #11
    nice
     
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