Why stress is defined on 3 surface

[kidsasd987]

https://www.google.ca/search?q=stre...m=isch&q=stress+tensor&imgrc=xK8-tGSPxXha7M%3
This is a simple question.

-why stress is defined on 3 surfaces only instead of the entire 6 surfaces? I believe 3 surfaces are enough to explain the stress of a small cube. At static equilibrium, the opposite surfaces will experience the same stress but with the opposite direction. Is this right?

 

[Chestermiller]

https://www.google.ca/search?q=stre...m=isch&q=stress+tensor&imgrc=xK8-tGSPxXha7M%3
This is a simple question.

-why stress is defined on 3 surfaces only instead of the entire 6 surfaces? I believe 3 surfaces are enough to explain the stress of a small cube. At static equilibrium, the opposite surfaces will experience the same stress but with the opposite direction. Is this right?

Yes, if the state of stress is homogeneous. If not, the stresses on one side of a cube can be slightly different from the corresponding stresses on the other side of the small cube.

Using a cube is just a way to get you started understanding the stress tensor. But really, the stress tensor is defined at each point in a material. Within this context, it is better to learn the advanced treatment in which the Cauchy Stress Relationship is used to determine the "traction vector" on a differential area of arbitrary orientation at a given location in a medium.

Chet

 

[kidsasd987]

Yes, if the state of stress is homogeneous. If not, the stresses on one side of a cube can be slightly different from the corresponding stresses on the other side of the small cube.

Using a cube is just a way to get you started understanding the stress tensor. But really, the stress tensor is defined at each point in a material. Within this context, it is better to learn the advanced treatment in which the Cauchy Stress Relationship is used to determine the "traction vector" on a differential area of arbitrary orientation at a given location in a medium.

Chet

Thank you for your answer.

I understood it this way. if we find the moment then it will end up that the opposite surface will have stress components with the same magnitude but with the opposite directions because moment should be 0.so, I assume this is true
*iff the cube is under static equilibrium.
Also, I have another question.

I looked up tensor briefly, and I have a question.
Angle Beta is preserved, and I don't know how it should be. Is this a tensor property that preserved for any type of coordinate system?

if so, could you tell me that how I can mathmatically and physically interpret this.

 

[DrDu]

Yes, if the state of stress is homogeneous. If not, the stresses on one side of a cube can be slightly different from the corresponding stresses on the other side of the small cube.

I think it is implicit in these pictures of cubes that the cube is infinitesimally small, so that this difference becomes negligible also if stress in the material is non-homogeneous.

 

[Chestermiller]

I think it is implicit in these pictures of cubes that the cube is infinitesimally small, so that this difference becomes negligible also if stress in the material is non-homogeneous.

Well, this is how they teach it to get us started. But, when we do differential force balances to obtain the Stress-Equilibrium equations in solid mechanics or the Navier Stokes equations in fluid mechanics, these stresses are taken to change from one side of the differential cube to the other, and these changes become key terms in the final differential force balance equations.