Direction of max strain from Mohr's circle

  • Thread starter GBA13
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  • #1
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Homework Statement


Hi,

I am doing some things with Mohr's circle. I am trying to find the direction of max strain but I am a bit confused. How can I do this?

Homework Equations




The Attempt at a Solution


I always assumed that the max strain occurs at 90O from the normal axis or 45O in real life. But I used a online calculator and got an angle of about 40O so that means that it isn't always at 45 like I thought.

Can anyone lend a hand?

Thanks!
 

Answers and Replies

  • #2
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You do know that the angle plotted on the Mohr's circle diagram is 2θ, not θ? So the maximum shear stress is at 2θ=90 degrees, indicating that the maximum shear stress occurs on a plane at 45 degrees to the maximum and minimum principal directions of strain. We're not familiar with your online calculator, so we don't know what it is doing. Does it give zero shear stress at 2θ=0 and 2θ=180 degrees?

Chet
 
  • #3
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Hi Chet. Yes I am aware of that thanks. I have found that the calculator uses the equation tan2θ = - (εxx - εyy) / 2εxy. Is that a valid equation? I though, as you said, that the direction of max shear strain is at 45 degrees - I thought that didn't change.
 
  • #4
20,854
4,544
Hi Chet. Yes I am aware of that thanks. I have found that the calculator uses the equation tan2θ = - (εxx - εyy) / 2εxy. Is that a valid equation? I though, as you said, that the direction of max shear strain is at 45 degrees - I thought that didn't change.
In the analysis that they are using, θ is not the angle that the plane of interest makes with the maximum principal stress. It is the angle that the plane of interest makes with the x coordinate direction (which is not a principal direction). See this link and the figure in the section entitled "Drawing Mohr's Circle." http://en.wikipedia.org/wiki/Mohr's_circle

Chet
 

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