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What's the physical meaning of true shear strain and simple shear strain

  1. Sep 16, 2008 #1
    :smile:Hey guys. I just get some information about true shear strain and simple shear strain from the lecture of engineering mechanic material .And i just want to know the physical meaning of true shear strain and simple shear strain.As a result, i have found some reference books ,but i still can't get a clear answer.:frown: Can someone give me a hint or recommend somethings that i can read in order to get the answer,please?
     
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  3. Sep 16, 2008 #2

    Andy Resnick

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    Here's a link that may help:

    http://en.wikipedia.org/wiki/Simple_shear

    I am not familiar with the concept of "true shear", although I vaguely recall something about von Mises strain ellipsoids...

    Anyhow, let me know if the wiki article helped.
     
  4. Sep 26, 2008 #3
    mydogsmall,

    Engineering strain (or linear strain) assumes that the relative-displacements between the points in the unstrained configuration can be predicted based solely on the unstrained configuration. This is not the case and I'll explain why. Consider a 1D line segment, in which you want to apply a strain [tex]\epsilon[/tex] to. If this strain was a engineering strain a simple multiplication of the strain with the unstrained line-segment length will give the newly strained line-segment length. In the case of true strain, the displacements of the points along the line segment will depend continuously on the line lengths between the unstrained and strained states. So if one were to deform the line in separate, but equal infinitesimally small strain steps ( d[tex]\epsilon[/tex] ). Multiplying d[tex]\epsilon[/tex] by the instantaneous length will give that strain steps contribution to the overall displacement. After summing up the displacements from each step, the final line length using true strain will be larger than that of engineering strain. I hope this helps.

    modey3
     
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