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Consider the simple case of a 2D stress element with its corresponding Mohr's circle:

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Now, for stress transformation using Mohr's circle, we rotate the diameter AB by twice the angle of rotation of element θ:

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How to geometrically derive the following stress transformation equations from the above Mohr's circle:

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My attempt: enter image description here

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Where am I making a mistake?

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  • $\begingroup$ Can you do dot product of vectors? Original and rotated stresses should be in equilibrium. $\endgroup$
    – Tomáš Létal
    Commented Jun 12 at 18:36
  • $\begingroup$ @Tomáš Létal How to derive it using plain trigonometry from Mohr's circle? $\endgroup$
    – Utility ZC
    Commented Jun 13 at 4:34
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    $\begingroup$ Try a simplest example of pure tension $\sigma_x\neq 0, \sigma_y=\tau_{xy}=0$. If you rotate this by 90°, you will also get pure tension, only this time in the $y$ direction. However, if you do just a 45° rotation, only the shear component will be nonzero, a situation, which is clearly achieved by 90° rotation in Mohr's circle. $\endgroup$
    – Tomáš Létal
    Commented Jun 13 at 10:47
  • $\begingroup$ @Tomáš Létal Please check the edited question $\endgroup$
    – Utility ZC
    Commented Jun 14 at 4:22
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    $\begingroup$ I don't see a mistake, so maybe your notation is a bit different. If you switch $\sigma_x$ with $\sigma_y$ and $\sigma_x'$ and $\sigma_y'$ in your figure, it could flip the sign of $\frac{\sigma_x-\sigma_y}{2}$ in the end result of the derivation. $\endgroup$
    – Tomáš Létal
    Commented Jun 14 at 7:20

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