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

  • Thread starter Zachary Liu
  • Start date
  • Tags
    Compression
In summary, the person is trying to determine how to detect tensile or compression stress in a known stress state. They need to use a 3d Mohr circle to calculate stresses at any orientation. Positive means tensile, negative means compressive. They need to know the principal stresses and the orientation in order to calculate the 3d mohr circle. If the principal stresses are different from the other two stresses, there is a chance for elongation and fracture.
  • #1
Zachary Liu
17
0
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.
 
Engineering news on Phys.org
  • #2
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.
 
  • #3
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.
 
  • #4
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.
 
Last edited by a moderator:
  • #5
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.
 
  • #6
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.
 
Last edited by a moderator:
  • #7
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.
 
  • #8
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.
 
  • #9
Okay, let's have more discussion.

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

then how about if [tex]\sigma_3[/tex] 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.
 
Last edited:
  • #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 [tex]\sigma_1,\sigma_2,\sigma_3[/tex] 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?
 

1. How does a 3D element measure tensile and compression?

A 3D element uses mathematical equations and algorithms to calculate the tensile and compression forces within a material. These equations are based on the material's mechanical properties and the applied loads.

2. What type of materials can be tested using 3D elements?

3D elements can be used to test a wide range of materials including metals, plastics, composites, and even biological tissues. As long as the material can be modeled accurately, it can be tested using 3D elements.

3. Are there any limitations to using 3D elements for tensile and compression testing?

While 3D elements are a powerful tool for testing materials, they do have some limitations. They may not be suitable for highly nonlinear materials or materials with complex microstructures. Additionally, the results may not be accurate if the model is not properly calibrated or if the element size is too large.

4. How do I interpret the results from a 3D element analysis?

The results from a 3D element analysis will typically include stress and strain values, as well as displacement and strain energy. These values can be compared to the material's yield strength and other mechanical properties to determine if it can withstand the applied loads without failing.

5. What are some tips for using 3D elements for tensile and compression testing?

Some tips for using 3D elements include properly calibrating the model, using a fine enough mesh to capture the material's behavior, and verifying the results with experimental data. It is also important to consider the boundary conditions and loading conditions to accurately simulate real-world scenarios.

Similar threads

  • Mechanical Engineering
Replies
1
Views
917
  • Mechanical Engineering
Replies
3
Views
936
Replies
5
Views
986
  • Engineering and Comp Sci Homework Help
Replies
15
Views
5K
  • Mechanical Engineering
Replies
8
Views
2K
  • Thermodynamics
Replies
20
Views
1K
Replies
5
Views
1K
  • Mechanical Engineering
Replies
6
Views
1K
  • Mechanical Engineering
Replies
11
Views
2K
Replies
2
Views
1K
Back
Top