- #1
Zachary Liu
- 17
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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.
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 [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.
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.
Ok i'll give this one more try. Previous post is not a calculation example, you dont 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 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.
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.
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?