# Mohr's Circle

For the case of uniaxial compression($\sigma_{x}=-p$) I am asked to determine the maximum shear stresses, and to draw the orientation of the element for these normal stresses. Below I have what I think it should be. Is it correct?

[PLAIN]http://img404.imageshack.us/img404/116/cive2.jpg [Broken]

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vela
Staff Emeritus
Homework Helper
No, that's not correct. First, what are the axial and shear stresses equal to when the shear stress is maximized? What angle do you have to rotate by on Mohr's circle to reach those points? How does that translate to the orientation of the element?

Here is my Mohr's circle for this case.

[PLAIN]http://img710.imageshack.us/img710/4182/cive3.jpg [Broken]

The centre point corresponds to $(\frac{-p}{2},0)$

Therefore, the maximum shear stress occurs when the normal stress is $\frac{-p}{2}$. So does that mean the angle of rotation is 90 degrees clockwise?

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vela
Staff Emeritus
Homework Helper
Yes, you rotate by 90 degrees on Mohr's circle (clockwise or counterclockwise doesn't really matter).

So will it look something like this:

[PLAIN]http://img190.imageshack.us/img190/8683/cive4.jpg [Broken]

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vela
Staff Emeritus
Homework Helper
No, that's not right. First, what are $\sigma'_x$, $\sigma'_y$, and $\tau'_{xy}$ equal to? Second, since you have to rotate by 90 degrees on Mohr's circle, that means $2\theta=90^\circ$, so $\theta=45^\circ$. What do you suppose this 45 degrees corresponds to?

I believe there are formulas to solve for $\sigma_x$', $\sigma_y$', and $\tau_{xy}$'.

Would the 45 degress correspond to the rotation of the element?

vela
Staff Emeritus
Homework Helper
Yes, it's the angle through which the element is rotated. The stresses you should be able to read off of Mohr's circle. There's no need to resort to formulas for this problem.

So I use trigonometry to find those values?

So if I rotate on the Mohr's circle by 90 degrees I will reach the max and min shear stresses?

vela
Staff Emeritus