Drawing Uniaxial Compression and Completing Mohr's Circle

[vela]

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?

 

[Precursor]

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]

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.

 

[Precursor]

So I use trigonometry to find those values?

 

[Precursor]

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

 

[vela]

I think it would help you immensely to go back and read up on Mohr's circle to understand what it represents.

 

[pichet]

Your Mohr's circle is almost complete. Try not to forget to draw the sigma-tau reference before you draw a circle. Then you will see that the circle is on the left side of the reference (Uniaxial compression). Note that the right part of the circle touches the origin. I think now the problem solved. If you want stresses at X’Y’, just rotates the horizontal line through the angle you want (it doesn’t matter whether the angle is given since, generally, we often need the principle stresses and maximum shear stress which are on the horizontal and vertical lines)