Drawing Uniaxial Compression and Completing Mohr's Circle

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SUMMARY

The discussion focuses on drawing Mohr's Circle for a uniaxial compression scenario where \(\sigma_{x} = -p\) MPa, with \(\sigma_{y}\) and \(\tau_{xy}\) both equal to 0. Participants clarify that uniaxial compression implies no shear stress and that the center of Mohr's Circle is located at \((-p/2, 0)\) with a radius of \(p/2\). The maximum shear stress occurs at a 90-degree rotation on Mohr's Circle, corresponding to a 45-degree rotation of the element. The conversation emphasizes the importance of accurately sketching the stress state and understanding the relationships between axial and shear stresses.

PREREQUISITES
  • Understanding of Mohr's Circle and its application in stress analysis
  • Knowledge of uniaxial compression and its implications on stress states
  • Familiarity with axial and shear stress concepts
  • Basic trigonometry for angle calculations in stress transformations
NEXT STEPS
  • Study the derivation and application of Mohr's Circle for different stress states
  • Learn how to calculate principal stresses and maximum shear stresses using Mohr's Circle
  • Explore the effects of biaxial and triaxial stress states on Mohr's Circle
  • Review trigonometric relationships in the context of stress transformations
USEFUL FOR

Students and professionals in civil engineering, mechanical engineering, and materials science who are involved in stress analysis and design, particularly those focusing on uniaxial and biaxial stress states.

  • #31
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?
 
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  • #32
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?
 
  • #33
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.
 
  • #34
So I use trigonometry to find those values?
 
  • #35
So if I rotate on the Mohr's circle by 90 degrees I will reach the max and min shear stresses?
 
  • #36
I think it would help you immensely to go back and read up on Mohr's circle to understand what it represents.
 
  • #37
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)
 

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