How to Determine Tensile and Compression using 3D Element: Tips and Techniques

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Discussion Overview

The discussion revolves around methods for determining tensile and compressive states using 3D elements in the context of finite element analysis (FEA). Participants explore the application of Mohr's circle for stress analysis, particularly in relation to a new material model for ice, which exhibits different strengths under tension and compression. The conversation touches on theoretical concepts, practical applications, and the complexities of stress states.

Discussion Character

  • Exploratory
  • Technical explanation
  • Debate/contested
  • Mathematical reasoning

Main Points Raised

  • Some participants question the applicability of hydrostatic pressure in all cases for determining tensile and compressive states.
  • Others emphasize the need to define an orientation to analyze stress states, as stress is a tensor that varies with angles.
  • A participant discusses implementing a new material model for ice, highlighting its differing strengths under tension and compression.
  • There are multiple references to the use of Mohr's circle for calculating principal stresses and determining tensile versus compressive states.
  • Some participants express skepticism about the simplicity of defining positive and negative stresses, suggesting that real-world scenarios may complicate these definitions.
  • One participant argues that materials do not fail under compression, raising questions about the relationship between principal stresses and failure modes.
  • Another participant points out that shear stresses may exist in directions other than those defined by principal stresses, which could influence failure points.
  • Discussion includes the variability of material properties, particularly for ice, under different stress conditions and environmental factors.

Areas of Agreement / Disagreement

Participants do not reach a consensus on the definitions of tensile and compressive states or the sufficiency of using Mohr's circle for failure analysis. There are competing views on the interpretation of stress states and their implications for material failure.

Contextual Notes

Limitations include the dependence on specific definitions of stress states, the complexity of real-world material behavior, and unresolved mathematical steps related to the application of Mohr's circle.

Who May Find This Useful

Readers interested in finite element analysis, material science, and the mechanics of materials, particularly in relation to ice and other complex materials.

Zachary Liu
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Hi, if I'm using the 3d element, I'm wondering how to detect the tensil and compression for a known stress state? the hydrostatic pressure p has been used before, but i don't think it is correct to all the cases.
 
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Can you elaborate more on what you are trying to do?So you're running finite element analysis(3d element?) or you're referring to something else?
Basically, if you know the stress state, you still need to define an orientation to find stresses; stress is a tensor, it varies with angles. After that, use a 3d mohr circle to calculate them at any orientation. Positive means tensile, negative means compressive.
 
Yes! I'm implementing a new material model on ice, which has different strength on compression and tensil. The stress state is known according to my routine, the question is how to detect the tensil or compression state and invoke damage or fracture mechanics.

could tell more about the 3D mohr circle?

Emreth said:
Can you elaborate more on what you are trying to do?So you're running finite element analysis(3d element?) or you're referring to something else?
Basically, if you know the stress state, you still need to define an orientation to find stresses; stress is a tensor, it varies with angles. After that, use a 3d mohr circle to calculate them at any orientation. Positive means tensile, negative means compressive.
 
Wikipedia?
http://en.wikipedia.org/wiki/Stress_(mechanics )

From your stress state, find the principal stresses and then calculate 3d mohr circle, its there in wiki. Like i said, positive values mean tensile, negative means compressive.
 
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No offense, but you should already know Mohr's circle like the back of your hand if you're doing any kind of structural analysis. This is really basic stuff.
 
You might treat this problem too simple.

I don't understand what do you mean the 'positive' and 'negative'. Does it mean the relative value of the three principle stress? please specify!

As far as I know, the 3D Mohr circle should be obtained after the derivation of 3 principle stress. And organize the three princle stress in a sequence according to the value.



Emreth said:
Wikipedia?
http://en.wikipedia.org/wiki/Stress_(mechanics )

From your stress state, find the principal stresses and then calculate 3d mohr circle, its there in wiki. Like i said, positive values mean tensile, negative means compressive.
 
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No offence at all, but you should know that 'no reply doesn't mean acknowledgment'! I don't think you know much more than me on the Mohr's circle stuff.

Brian_C said:
No offense, but you should already know Mohr's circle like the back of your hand if you're doing any kind of structural analysis. This is really basic stuff.
 
You do have to know it very well to do advanced analysis like FEA, etc.
Anyway, if you know the six stresses sigmaxx,sigmayy,sigmazz,sigmaxy,sigmaxz,sigmayz (if isotropic, they define the stress state) at any orientation, you can calculate the principal stresses using the formula in wiki. Then build the 3d mohr circle.
Look at the mohr diagram (sigman) axis. They plotted it so that everything is in the positive, but its not always the case. It can sit closer to origin so sometimes, the stresses can go in negative(compressive) for certain orientations.
 
Okay, let's have more discussion.

Say we have a typical cylinder triaxial test. The three principal stresses, \sigma_{1},\sigma_2,\sigma_3, are coincidentally the same as \sigma_x,\sigma_y,\sigma_z( axis z follows the symmetry axis of cylinder). Assuming compression is negative and \sigma_1=\sigma_2\le\sigma_3. One case might be \sigma_1=\sigma_2\le\sigma_3<0, a kind of 'unilaterial' case, then it should in compression according to your 'guess' if I'm right.

then how about if \sigma_3 is far smaller than other two stresses (absolutely value) ? Obviously, the cylinder will show elongation in this case! And the tensil fracture is probably happening!

So I still don't think your 'positive' and 'negative' criteria is sufficient.
 
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  • #10
My guess?What are you talking about? The negative positive is BY DEFINITION compressive and tensile.
I told you these are tensors, stress will be like this in one direction, like that in other direction. Why don't you do a calculation instead of speculating? Solve a problem with a cylinder under pressure or something, do all the calculations.
 
  • #11
I think we are not talking about the DEFINITION, as it could be found in any textbook.

The real physical mechanics is of great concerning.

I have shown my calculation example in previous post all ready.

All the principlal stress \sigma_1,\sigma_2,\sigma_3 are 'negative', so this element should be in 'compression' according to definition.

but the thing is the cylinder is in elongation on z-axis. Fracture happens. Please explain!
 
  • #12
Ok i'll give this one more try. Previous post is not a calculation example, you don't really calculate, just assume stresses.
Say you have those as your principal stresses, they are all negative. As you change the orientation, shear stresses will appear within the planes along that orientation.(look at the stress transformation in wiki) Wherever you have the largest shear will be your failure point since materials don't fail under compression, as long as stresses are not equal, you'll always get some shear along some directions. Now, if the three princ. stresses are not just all negative but also equal, you'll get pure hydrostatic stress state where you won't have any shear stresses in any direction and the mohr circle becomes a point. You cannot have failure in that case.
 
  • #13
One of the big problems that ground engineers and scientists face is that the properties of ground materials, including the ice you mentioned, vary greatly with the applied stress.

Since bulk material in the real world can be very big bulks the interior of such a mass can have significantly different properties from the outer layers.

Ice does not have the pore fluid of normal soils, but often has entrained air. This air causes slip weakness to shear.
Further real world ice often overlies weaker material such as water, whereas normally encountered soils usually overlie stronger harder materials.
The foundation is as always all important.
 
  • #14
Now the discussion is becoming more interesting.

Yes, the point or say my intention of the original post is to seek one method to detect the 'failure', the physical mechanics behind. However, I just interpreted as 'tensil' and 'compression' because i also believe that ' material do not fail in compression'.

As you said, you might be able to obtain the principle stress whenever you know the stress stae and this is quite easy in any FEM code. But it is still not sufficient to detect the 'failue' as the shear stress in the principle stress space is ZERO! But it does exist in other direction. I think you have already pointed it out.

Thanks for your discussion.

Emreth said:
Ok i'll give this one more try. Previous post is not a calculation example, you don't really calculate, just assume stresses.
Say you have those as your principal stresses, they are all negative. As you change the orientation, shear stresses will appear within the planes along that orientation.(look at the stress transformation in wiki) Wherever you have the largest shear will be your failure point since materials don't fail under compression, as long as stresses are not equal, you'll always get some shear along some directions. Now, if the three princ. stresses are not just all negative but also equal, you'll get pure hydrostatic stress state where you won't have any shear stresses in any direction and the mohr circle becomes a point. You cannot have failure in that case.
 
  • #15
Yes, Studiot. The strength of ice material is highly dependent on the confiment and also strain rate, salinity, granuar size etc.. It is not isotropic in most cases.

Anyway, I came to this forum with the hope to find a more efficient way to detect material's failure or damage based on the stress state.

Studiot said:
One of the big problems that ground engineers and scientists face is that the properties of ground materials, including the ice you mentioned, vary greatly with the applied stress.

Since bulk material in the real world can be very big bulks the interior of such a mass can have significantly different properties from the outer layers.

Ice does not have the pore fluid of normal soils, but often has entrained air. This air causes slip weakness to shear.
Further real world ice often overlies weaker material such as water, whereas normally encountered soils usually overlie stronger harder materials.
The foundation is as always all important.
 
  • #16
What I don't know is whether you are trying to develop your model for laboratory test specimens or bulk material mass.

You did quote a result from the Mohr-Coulomb yield function.

Do you understand the derivation of this and the way yield functions (which is what you seek) are built up?
 
  • #17
I don't know why you are keeping asking me questions.

I'm using more complicate yield function, say Hill yield function, the cap yielding and also the Tsai-wu yield if you know.

If you know better, then why not say your solution. thanks!


Studiot said:
What I don't know is whether you are trying to develop your model for laboratory test specimens or bulk material mass.

You did quote a result from the Mohr-Coulomb yield function.

Do you understand the derivation of this and the way yield functions (which is what you seek) are built up?
 

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