- #1

- 17

- 0

You are using an out of date browser. It may not display this or other websites correctly.

You should upgrade or use an alternative browser.

You should upgrade or use an alternative browser.

- Thread starter Zachary Liu
- Start date

- #1

- 17

- 0

- #2

- 96

- 2

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

- 17

- 0

could tell more about the 3D mohr circle?

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

- 96

- 2

Wikipedia?

http://en.wikipedia.org/wiki/Stress_(mechanics [Broken])

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.

http://en.wikipedia.org/wiki/Stress_(mechanics [Broken])

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

- 245

- 0

- #6

- 17

- 0

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.

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.

Wikipedia?

http://en.wikipedia.org/wiki/Stress_(mechanics [Broken])

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

- 17

- 0

- #8

- 96

- 2

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

- 17

- 0

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.

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

- 96

- 2

I told you these are tensors, stress will be like this in one direction, like that in other direction. Why dont you do a calculation instead of speculating? Solve a problem with a cylinder under pressure or something, do all the calculations.

- #11

- 17

- 0

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

- 96

- 2

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 dont 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 wont have any shear stresses in any direction and the mohr circle becomes a point. You cannot have failure in that case.

- #13

- 5,439

- 9

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

- 17

- 0

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.

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 dont 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 wont have any shear stresses in any direction and the mohr circle becomes a point. You cannot have failure in that case.

- #15

- 17

- 0

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.

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

- 5,439

- 9

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

- 17

- 0

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!

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?

Share: