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I have a question regarding Solid Mechanics and Mohr's circle.

In my test today, we had a plane stress state represented in a Mohr circle which was not centered at the origin. The question asked to find the orientation of the planes subjected to a pure shear stress state. However, I thought that when you have a pure shear stress state, all planes passing through the vicinity of the point fulfill the pure shear condition ($\sigma_x=-\sigma_y$, so that the center of Mohr's circle is $C=\frac{\sigma_x+\sigma_y}2=0$). In this case, since the center is not zero, there would not be such a plane. Is this reasoning correct? I really struggled with this course and I want to understand this concept which will be crucial for me in the future. All help is greatly appreciated.

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  • $\begingroup$ For the state of stress you describe, if you consider a surface oriented at an angle of 45 degrees to the principal directions, the normal stress will be zero and the shear stress will be maximum. This is what is commonly referred to as a state of "pure shear." $\endgroup$ – Chet Miller Jan 25 '18 at 3:06
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It looks like your professor made a mistake. In pure shear, $\sigma_x=\sigma_y=0$. The $x$-coordinate of the center of Mohr's circle for plane stress is given by $\sigma_{avg}=\frac{\sigma_x+\sigma_y}{2}$, which is an invariant (i.e., does not change with coordinate transformation). If a pure shear stress state exists, then the $x$-coordinate of the center of the circle is $0$. Equivalently, if the $x$-coordinate of the center of the circle is not $0$, then a pure shear stress state does not exist.

The question would make sense if it instead asked for the maximum shear stress, or the angle corresponding to that value. Or perhaps your professor threw in a trick question to check your understanding.

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  • $\begingroup$ Thank you for your explanation. This is what I had in mind, but now I am more sure about my ideas. Was my reasoning correct then? As you said, if a pure shear stress state exists, then the x-coordinate of Mohr's Circle must be zero; if we have this case, then we can rotate our tensor by changing the system of reference, in which case $\sigma_x$ and $\sigma_y$ will no longer be zero, but by symmetry with respect to the y-axis (shear component axis), $\sigma_x = - \sigma_y$. $\endgroup$ – Bee Dec 31 '17 at 8:41
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    $\begingroup$ That is correct. Something interesting: if you consider an element rotated at 45 degrees relative to the element in the pure shear stress state (which is also the state of maximum shear), the new element is in the principal stress state. Since $\sigma_x=-\sigma_y$, this principal stress state is equibiaxial. (P.S. One more thing: I would be careful to distinguish between the concepts of rotation and coordinate transformation. See boris-belousov.net/2016/05/31/change-of-basis for an explanation.) $\endgroup$ – jrhsy Dec 31 '17 at 14:55
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    $\begingroup$ Minor correction to above: "is equibiaxial" should be "has equal and opposite tensile and compressive stresses, and no shear stresses." This equivalence between the pure shear stress state and the tensile-compressive stress state may be useful in later courses on structural mechanics. $\endgroup$ – jrhsy Dec 31 '17 at 15:07

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