Normal Stress Vectors: Sketch 1 vs Sketch 2

[person123]

TL;DR Summary

I'm wondering the direction the normal stress should be shown on a deformed material element.

Let's say you have a material element with normal and shear stress. These stresses were computed using stress transformation. When the material deforms, should the normal stress vectors remain normal to the surface (sketch 1) or parallel to the other surface (sketch 2)? Which would be more accurate? Thanks!

 

[Chestermiller]

A normal stress, by definition, is the component of the stress vector that is normal to the surface. A shear stress is, by definition, a component of the stress vector that is tangent to the surface.

I think you may be confusing the components of the 2nd order stress tensor with the components of the stress vector (first order tensor) on a surface of arbitrary orientation.

 

[person123]

Maybe this would be the question asked more precisely:
Let's say you take the traction vector on that plane. In which case would the sum of the vectors be equal to the traction vector?

 

[Chestermiller]

Maybe this would be the question asked more precisely:
Let's say you take the traction vector on that plane. In which case would the sum of the vectors be equal to the traction vector?

Yes.

 

[person123]

Does anyone have ideas on this more precise version of the question?

 

[Chestermiller]

Does anyone have ideas on this more precise version of the question?

Are you familiar with the Cauchy stress relationship?

 

[person123]

Yes. For 3 dimensions, take a matrix whose elements are the stresses on a differential cube in xyz orientation ( rows and columns indicating the plane and direction of stress). That's the cauchy stress tensor. When multiplied by a vector normal to a surface, the output is the traction vector for that surface. That can be decomposed into normal and shear stresses. I'm still unsure how this would work for an already deformed material because there would be another transform for deforming the material.

 

[Chestermiller]

Yes. For 3 dimensions, take a matrix whose elements are the stresses on a differential cube in xyz orientation ( rows and columns indicating the plane and direction of stress). That's the cauchy stress tensor. When multiplied by a vector normal to a surface, the output is the traction vector for that surface. That can be decomposed into normal and shear stresses. I'm still unsure how this would work for an already deformed material because there would be another transform for deforming the material.

That would only be for the components in a change of coordinate system. The stress tensor, the normal to a surface, and the corresponding traction vector are invariant entities, independent of any specific coordinate system and of the components resolved on that coordinate system. Thinking of them in terms of a matrix with rows and columns automatically implies a specific coordinate system.

Do yourself a favor and learn about dyadic tensor notation, which treat tensors as such invariant entities. For a good development of this, see Appendix B of Transport Phenomena by Bird et al. Or see the brief tutorial I presented in post #12 of the following thread: https://www.physicsforums.com/threads/so-tension-is-not-a-force.960204/#post-6089855

You seem to be focusing on using an embedded material coordinate system in describing the deformed material. Is this the case?

 

[Chestermiller]

It just dawned on me...Is it possible that you are talking about situations involving small displacements and small strains, like deformation of solids?

 

[person123]

My apology for the late response.

It just dawned on me...Is it possible that you are talking about situations involving small displacements and small strains, like deformation of solids?

Yes. I'm thinking of solids specifically. I'm also thinking of small displacements in order to ensure stress and strain relations are linear.

That would only be for the components in a change of coordinate system. The stress tensor, the normal to a surface, and the corresponding traction vector are invariant entities, independent of any specific coordinate system and of the components resolved on that coordinate system. Thinking of them in terms of a matrix with rows and columns automatically implies a specific coordinate system.

Yes, although I did have a sense that the stress tensor is independent of coordinate system, I realized that using a matrix does imply a specific coordinate system.

Using dyadics, the stress tensor is a sum of dyadics. Instead of doing matrix multiplication, the tensor is dotted with the normal vector, producing the traction vector. Using the equation you had in your post: $$(\bf v_1 v_2)\cdot v3=v_1(v_2\cdot v_3)$$ In order to ensure that this product is nonzero, $v_2$ must not be orthogonal to $v_3$, or for unit basis vectors they must be equivalent.

For the question I have though, I'm thinking specifically about change of coordinate systems though.

I thought about my question a bit more. I want to use a bit more of a specific, concrete example. Let's say there's a rod under pure tension in the x direction and you apply a 45 degree cut. There would be a normal and shear stress both half of the normal stress in the x direction. If the material deforms under tension however, the angle changes (let's say to a 30 degree angle). In order to maintain force balance, this would be equivalent to a 30 degree cut:

This seems to mean that if you take a section of a material under some stress, the stresses on that section would change if the elasticity modulus changes.

 

[Chestermiller]

If the strains and displacements are small, the changes in orientation of the plane and of the edges of the material elements are considered negligible. If the strains are large, we are dealing with non-linear material behavior, and you need to learn about the mathematics of large deformation behavior (google Cauchy-Green deformation tensor). If the displacements are large but the strains are small, this can be dealt with mathematically with the kinematics of the deformation, but with linear material behavior.