Drawing Uniaxial Compression and Completing Mohr's Circle

[Precursor]

Homework Statement
Sketch the element for the stress state indicated and then draw Mohr's circle.

Given: Uniaxial compression, i.e. \sigma_{x} = -p MPa

The attempt at a solution

Below I have the sketch and a partially complete Mohr's circle:

[PLAIN]http://img710.imageshack.us/img710/6001/civek.jpg

What am I missing on the Mohr's circle? Did I even go about this correctly?

 

[Precursor]

Any ideas?

 

[vela]

First, tell us what \sigma_y and \tau_{xy} are equal to.

 

[Precursor]

The thing is, they don't provide \sigma_{y} or \tau_{xy}, which is why I was confused.

 

[vela]

Well, the problem says the compression is uniaxial. What does uniaxial mean?

 

[Precursor]

Well, the problem says the compression is uniaxial. What does uniaxial mean?

That would mean having a single axis, so \sigma_{y} is not involved here. But how about \tau_{xy}? I simply assumed it existed, as you can see in my drawing of the Mohr's circle.

 

[vela]

I'd take it to be 0 as well.

 

[Precursor]

I'd take it to be 0 as well.

If \tau_{xy} = 0 then there wouldn't even be a circle. Would it be a straight line?

 

[vela]

No, you always get a circle. The two points you know are on the circle will be (\sigma_y,\tau_{xy}) = (0,0) and (\sigma_x,-\tau_{xy})=(-p,0). Now you go about the same procedure as before and find the location of the center of the circle, its radius, etc.

 

[Precursor]

So in my sketch I should remove the shear stress arrows?

 

[vela]

Sure, or label them as being equal to 0.

 

[Precursor]

I found the centre to be (\frac{-p}{2},0) and the radius to be \frac{p}{2}. Is this correct.

 

[vela]

Yes, that's correct.

 

[Precursor]

How do I find the line X'Y' since I don't know the angle \theta?

 

[vela]

What are X, Y, X', and Y' supposed to denote?

 

[Precursor]

What are X, Y, X', and Y' supposed to denote?

These 4 variables are points on the Mohr's circle denoted by:
X:(\sigma_{x},-\tau_{xy})

Y:(\sigma_{y},+\tau_{xy})

X':(\sigma_{x}',-\tau_{xy}')

Y':(\sigma_{y}',+\tau_{xy}')

There are equations used to solve for X' and Y', but one of the variables is \theta, which isn't given.

 

[vela]

OK. Did the problem ask you to find the axial and shear stresses for some plane?

 

[Precursor]

OK. Did the problem ask you to find the axial and shear stresses for some plane?

That's part b of the question, which askes me to determine the maximum shear stresses that exist and to identify the planes on which they act by drawing the orientation of the element for these normal stresses.

But actually, \theta=0 because the angle between the line XY and the x-axis is 0.

 

[vela]

What points on the circle correspond go the orientation when the shear stress is maximized?

 

[Precursor]

What points on the circle correspond go the orientation when the shear stress is maximized?

Would that be the points where the circle is at the highest and lowest in the y direction?

 

[vela]

OK, I'm still not clear on exactly what you're trying to do with (X', Y') and θ.

 

[Precursor]

OK, I'm still not clear on exactly what you're trying to do with (X', Y') and θ.

2θ is what separates the lines XY an X'Y'. I think since θ = 0, there isn't an X'Y' line.

 

[vela]

When you draw Mohr's circle, typically you start with the axial and shear stresses for a given orientation of the element, so you know where the points X and Y lie on the diagram. Where X' and Y' lie depend on what you're trying to find. For instance, if you're interested in the principal axes, you'd choose to have X'Y' lie on the horizontal axis. If you wanted to find where the shear stress is maximized, you'd choose X'Y' so that it was vertical.

 

[Precursor]

For part c of the question, it asks me to sketch the element for the stress state and draw the Mohr's circle for the case of a biaxial compressive stress, i.e., \sigma_{x}=\sigma_{y}=-p MPa.

I found the radius to be 0. Does that mean it's simply a point, instead of a cricle?

 

[vela]

Yes.

 

[Precursor]

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

 

[vela]

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?

 

[Precursor]

Here is my Mohr's circle for this case.

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

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?

 

[vela]

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

 

[Precursor]

So will it look something like this:

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

 

[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)