Drawing Uniaxial Compression and Completing Mohr's Circle

In summary: No, the angle of rotation is 0 degrees. What does that mean for the orientation of the element?It means that the element must be rotated so that the normal stress is at (\sigma_{y},+\tau_{xy}) instead of (\sigma_{y},-\tau_{xy}). Does that make sense?Yes, that makes sense.How would I go about finding the orientation of the element if I wanted it to be at (\sigma_{y},+\tau_{xy})?If you wanted the element to be at (\sigma_{y},+\tau_{xy}), you'd find the x-coordinate and y-coordinate of (\
  • #1
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Homework Statement
Sketch the element for the stress state indicated and then draw Mohr's circle.

Given: Uniaxial compression, i.e. [tex]\sigma_{x} = -p[/tex] 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 [Broken]

What am I missing on the Mohr's circle? Did I even go about this correctly?
 
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  • #2
Any ideas?
 
  • #3
First, tell us what [itex]\sigma_y[/itex] and [itex]\tau_{xy}[/itex] are equal to.
 
  • #4
The thing is, they don't provide [tex]\sigma_{y}[/tex] or [tex]\tau_{xy}[/tex], which is why I was confused.
 
  • #5
Well, the problem says the compression is uniaxial. What does uniaxial mean?
 
  • #6
vela said:
Well, the problem says the compression is uniaxial. What does uniaxial mean?

That would mean having a single axis, so [tex]\sigma_{y}[/tex] is not involved here. But how about [tex]\tau_{xy}[/tex]? I simply assumed it existed, as you can see in my drawing of the Mohr's circle.
 
  • #7
I'd take it to be 0 as well.
 
  • #8
vela said:
I'd take it to be 0 as well.

If [tex]\tau_{xy} = 0[/tex] then there wouldn't even be a circle. Would it be a straight line?
 
  • #9
No, you always get a circle. The two points you know are on the circle will be [itex](\sigma_y,\tau_{xy}) = (0,0)[/itex] and [itex](\sigma_x,-\tau_{xy})=(-p,0)[/itex]. Now you go about the same procedure as before and find the location of the center of the circle, its radius, etc.
 
  • #10
So in my sketch I should remove the shear stress arrows?
 
  • #11
Sure, or label them as being equal to 0.
 
  • #12
I found the centre to be [tex](\frac{-p}{2},0)[/tex] and the radius to be [tex]\frac{p}{2}[/tex]. Is this correct.
 
  • #13
Yes, that's correct.
 
  • #14
How do I find the line X'Y' since I don't know the angle [tex]\theta[/tex]?
 
  • #15
What are X, Y, X', and Y' supposed to denote?
 
  • #16
vela said:
What are X, Y, X', and Y' supposed to denote?

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

Y:([tex]\sigma_{y},+\tau_{xy}[/tex])

X':([tex]\sigma_{x}',-\tau_{xy}'[/tex])

Y':([tex]\sigma_{y}',+\tau_{xy}'[/tex])

There are equations used to solve for X' and Y', but one of the variables is [tex]\theta[/tex], which isn't given.
 
  • #17
OK. Did the problem ask you to find the axial and shear stresses for some plane?
 
  • #18
vela said:
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, [tex]\theta=0[/tex] because the angle between the line XY and the x-axis is 0.
 
  • #19
What points on the circle correspond go the orientation when the shear stress is maximized?
 
  • #20
vela said:
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?
 
  • #21
OK, I'm still not clear on exactly what you're trying to do with (X', Y') and θ.
 
  • #22
vela said:
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.
 
  • #23
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.
 
  • #24
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., [tex]\sigma_{x}=\sigma_{y}=-p[/tex] MPa.

I found the radius to be 0. Does that mean it's simply a point, instead of a cricle?
 
  • #25
Yes.
 
  • #26
For the case of uniaxial compression([itex]\sigma_{x}=-p[/itex]) 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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  • #27
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?
 
  • #28
Here is my Mohr's circle for this case.

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

The centre point corresponds to [itex](\frac{-p}{2},0)[/itex]

Therefore, the maximum shear stress occurs when the normal stress is [itex]\frac{-p}{2}[/itex]. So does that mean the angle of rotation is 90 degrees clockwise?
 
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  • #29
Yes, you rotate by 90 degrees on Mohr's circle (clockwise or counterclockwise doesn't really matter).
 
  • #30
So will it look something like this:

[PLAIN]http://img190.imageshack.us/img190/8683/cive4.jpg [Broken]
 
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  • #31
No, that's not right. First, what are [itex]\sigma'_x[/itex], [itex]\sigma'_y[/itex], and [itex]\tau'_{xy}[/itex] equal to? Second, since you have to rotate by 90 degrees on Mohr's circle, that means [itex]2\theta=90^\circ[/itex], so [itex]\theta=45^\circ[/itex]. What do you suppose this 45 degrees corresponds to?
 
  • #32
I believe there are formulas to solve for [itex]\sigma_x[/itex]', [itex]\sigma_y[/itex]', and [itex]\tau_{xy}[/itex]'.

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?
 
<h2>1. What is uniaxial compression?</h2><p>Uniaxial compression is a type of stress that occurs when a material is being compressed in only one direction. This means that the force being applied to the material is perpendicular to the cross-sectional area of the material.</p><h2>2. How is uniaxial compression represented in a drawing?</h2><p>Uniaxial compression is typically represented in a drawing by a single arrow pointing towards the material. The arrow represents the direction and magnitude of the compressive force being applied to the material.</p><h2>3. What is Mohr's circle and how is it used in uniaxial compression?</h2><p>Mohr's circle is a graphical method used to determine the principal stresses and maximum shear stress for a given stress state. In uniaxial compression, Mohr's circle is used to plot the stress state of the material and determine the maximum compressive stress and the direction in which it occurs.</p><h2>4. How do you complete Mohr's circle for uniaxial compression?</h2><p>To complete Mohr's circle for uniaxial compression, you first need to plot the initial stress state of the material, which can be represented by a single point on the circle. Then, using the radius of the circle as a reference, you can plot the maximum and minimum principal stresses. Finally, the angle between the radius and the horizontal axis represents the direction of the maximum compressive stress.</p><h2>5. Why is it important to understand uniaxial compression and Mohr's circle?</h2><p>Understanding uniaxial compression and Mohr's circle is important for engineers and scientists because it allows them to analyze and predict the behavior of materials under different stress states. This information is crucial in designing structures and materials that can withstand various types of loading conditions.</p>

1. What is uniaxial compression?

Uniaxial compression is a type of stress that occurs when a material is being compressed in only one direction. This means that the force being applied to the material is perpendicular to the cross-sectional area of the material.

2. How is uniaxial compression represented in a drawing?

Uniaxial compression is typically represented in a drawing by a single arrow pointing towards the material. The arrow represents the direction and magnitude of the compressive force being applied to the material.

3. What is Mohr's circle and how is it used in uniaxial compression?

Mohr's circle is a graphical method used to determine the principal stresses and maximum shear stress for a given stress state. In uniaxial compression, Mohr's circle is used to plot the stress state of the material and determine the maximum compressive stress and the direction in which it occurs.

4. How do you complete Mohr's circle for uniaxial compression?

To complete Mohr's circle for uniaxial compression, you first need to plot the initial stress state of the material, which can be represented by a single point on the circle. Then, using the radius of the circle as a reference, you can plot the maximum and minimum principal stresses. Finally, the angle between the radius and the horizontal axis represents the direction of the maximum compressive stress.

5. Why is it important to understand uniaxial compression and Mohr's circle?

Understanding uniaxial compression and Mohr's circle is important for engineers and scientists because it allows them to analyze and predict the behavior of materials under different stress states. This information is crucial in designing structures and materials that can withstand various types of loading conditions.

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