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Shear strain

  1. Nov 7, 2008 #1

    This page shows how to find shear strain in three dimensions.

    I understand how they found the shear strains as x and y components from dividing the change in length by the original length.

    But from the line "From the figure, it is also evident that..." I cannot understand how they combined the strains from the x and y axes to find the functions of tan.

    Also, how after I have found the functions of tan, how do I use approximation of small angles to find that the strains actually simply equals the angle itself? I have seen some other websites using sin instead of tan.

  2. jcsd
  3. Nov 7, 2008 #2


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    The tangent is the rise over the run: from the figure, [itex]\frac{\Delta x}{y}[/itex].

    For small angles, [itex]\alpha\simeq\tan \alpha\simeq\sin\alpha[/itex]. Is this where you got stuck, or is it in the derivation of this expression?
  4. Nov 8, 2008 #3
    I thought strain should be the change in length over original length. So shouldn't the strain along the x-axis be Delta-x over X instead of Delta-x over Y?
  5. Nov 8, 2008 #4


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    As discussed right under that image, the normal strain is the change in length divided by the original length. The shear strain can be defined as either the angular change of an originally right angle (i.e., [itex]\alpha[/itex]) or as the change in length divided by the orthogonal length (i.e., [itex]\frac{\Delta x}{y}[/itex]). For small angles, it's the same thing.
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