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Shear and principal stress

  1. Oct 4, 2011 #1
    id like to think i did this right, but i just want to make sure. i cant do symbols so let d equal sigma.

    the stress components are d_xx= 200 MPa, d_xy= 50 MPa, d_yy=-100 MPa. I need to find the max shear stress, and the principal stresses, and then a_s and a_p.
    heres what i got:
    max shear= 158
    principal stresses= 108 and 208
    a_s=-35.8 degrees
    a_p=9.2 degrees

    am i right?; im not sure if i need to do anything different since d_yy is negative? i think i used the right equations but not sure, and what would my units be for the stresses ? MPa?

    lastly, it asks "denote them on the stress element" (them meaning the things i just found). theres the generic 2-D square with sigma_yy going up from the top, sigma_xx from the right, and sigma_xy parallel to the square. what am i supposed to do about that?

    thanks a ton in advance
  2. jcsd
  3. Oct 4, 2011 #2


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    You have a sign error in one of your principle stresses as far as I can tell.

    Just out of curiosity, what exactly are you calling a_s and a_p?

    Just doing it really quickly, I get [itex]\sigma_1 = -108\; \mathrm{MPa}[/itex] and [itex]\sigma_2 = 208\; \mathrm{MPa}[/itex], which comes from rotating the element counter-clockwise by [itex]9.2^{\circ}[/itex].

    Principle stresses arise from the fact that for any compatible state of stress, you can rotate the element in some way such that there are only normal stresses on the element. The shear stresses are therefore zero by definition when you rotate it into its principle orientation and the units for everything remain the same as the units you originally had. I can't see your paper, but I assume that they gave you a stress element originally, and you have to draw its new configuration and label the values appropriately.
  4. Oct 4, 2011 #3
    thanks ill go check into my signs. and a_s is the associated angle of the shear (alpha sub s).
    there is a diagram of a square showing the perpendicular and parallel forces acting on it, but how can i redraw the new configuration?
  5. Oct 5, 2011 #4


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    Just rotate the figure by however many degrees you need to get rid of shear entirely.
  6. Oct 5, 2011 #5
    so for the principal angle, i just rotate the square 9 degrees counter clockwise? do i do anything for the shear angle? and on this new figure, would the forces now be perpendicular to the shape or to the original axis?
  7. Oct 5, 2011 #6


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    There is no shear when an element is in its principle orientation. That is the point. You have found the orientation where only normal stresses act on the element. I would suggest making sure you understand the fundamental concepts such as this first, as without it you have no intuition as to whether your math is correct or not. The math here is extraordinarily easy and fast if you know linear algebra. It is just an eigenvalue problem.
  8. Oct 5, 2011 #7
    im still not sure what exactly my drawing should look like? i know that the 2 angles should be 45 degrees apart which they are. so ill rotate the drawing counter clockwise 9 degrees? and then have my new stress values coming out of it? or the old stress values? im still not getting it, so can you walk me thru it so i understand it for my exam coming up.. i think you can do it without eigen values? but how would you do it? what exactly should my drawing look like?
  9. Oct 5, 2011 #8


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    You don't need eigenvalues, true, but it is easier to calculate all the stuff that way rather than with the formulas you find in your book.

    Anyway, I imagine they have a square with the original stresses labeled. The whole purpose behind the concept of principal stresses is that you find an orientation for that same element where there are no shear stresses and instead you just have a rotated element with the principal stresses on each face.

    I don't really know how else to describe it without just drawing a picture and giving you the answer, which I am not about to do. =/
  10. Nov 3, 2011 #9
    I think it is too late for me to reply to this thread, I could not hold myself.

    You are right, there is nothing wrong with your solution. The only things remaining are to make that stress diagrams.

    Rotate the stress element diagram by 9.2 degrees and put the values of principal stresses. You will not need to put the shear stress values as it is zero on principal planes.
    Last edited: Nov 3, 2011
  11. Nov 4, 2011 #10
    Guys I'm sorry, I made a calculation error during my previous post. I get σ1=208 MPA and σ2=-108MPA;θ1=9.2°. So, there was an error in calculation which was initially posted.
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