Set up boundary conditions for a simple elasticity problem

[bobinthebox]

TL;DR Summary

Given an elastic block, describe a possible set of boundary conditions so that the block can slight on a plane parallel to $e_1 e_3$ direction

[Mentor Note -- Thread moved to the ME forum to get better views]

Let's consider an incompressible block of Neo-Hookean material. Let the initial reference geometry be described by $B=[0,b] \times [0,b] \times [0,h]$. The professor gave me the following task:

Set up the boundary conditions of your problem so that the lateral surfaces can move on a plane parallel to the $e_1 e_3$ plane. Since you have to work in the reference, use first Piola-Kirchoff stress tensor $S$.

Of course there can be many possibilities. I was thinking to the case when the block is extended in the $e_1$ direction by a load factor of #\delta#.

I came up with the following B.Cs. Here $u(p) = x(p)-p$ is the displacement corresponding to a deformation map #x#

• $S_{22}=0$ so that I have no normal stress in the lateral faces with normal $e_2$.
• $e_3 \cdot S(-e_1) = e_2 \cdot S(e_1)=0$ so that I am constraining the block to do not move
• $e_2 \cdot S(e_1) = e_3 \cdot S(e_1)=0$ same as above, but on the the face with normal $e_1$
• $u_1(p)=\delta$ for $p \in \text{bottom face}$ i.e. displacement equal to $\delta$ in the $e_1$ direction##.

Below an image of how I image the first Piola stresses for this problem.

Do you think that the problem is set up in the proper way? I am afraid that I loose the incompressibility in this way

 

[Chestermiller]

Summary:: Given an elastic block, describe a possible set of boundary conditions so that the block can slight on a plane parallel to $e_1 e_3$ direction

[Mentor Note -- Thread moved to the ME forum to get better views]

Let's consider an incompressible block of Neo-Hookean material. Let the initial reference geometry be described by $B=[0,b] \times [0,b] \times [0,h]$. The professor gave me the following task:Of course there can be many possibilities. I was thinking to the case when the block is extended in the $e_1$ direction by a load factor of #\delta#.

I came up with the following B.Cs. Here $u(p) = x(p)-p$ is the displacement corresponding to a deformation map #x#

• $S_{22}=0$ so that I have no normal stress in the lateral faces with normal $e_2$.
• $e_3 \cdot S(-e_1) = e_2 \cdot S(e_1)=0$ so that I am constraining the block to do not move

What makes you think that this constrains the block to do not move?

• $e_2 \cdot S(e_1) = e_3 \cdot S(e_1)=0$ same as above, but on the the face with normal $e_1$

What is this supposed to mean?

• $u_1(p)=\delta$ for $p \in \text{bottom face}$ i.e. displacement equal to $\delta$ in the $e_1$ direction##.

What is the displacement condition on the top face?

 

[FEAnalyst]

Set up the boundary conditions of your problem so that the lateral surfaces can move on a plane parallel to the e1e3 plane.

Is that the only requirement ? Think how you would constrain this model in Finite Element Analysis. I would start from $u_{2}=0$ on those blue faces. Then you could use $u_{3}=0$ for bottom face so that the cube can't move upwards/downwards.

 

[bobinthebox]

First of all, thanks as usual @Chestermiller for your answer.

What makes you think that this constrains the block to do not move?

That's what I thought: $S_{22}=0$ implies that I do not have traction on the lateral faces. And then, also $e_3 \cdot S(-e_1) = e_2 \cdot S(-e_1)=0$ implies no shearing on the face with normal $-e_1$.

And $e_2 \cdot S(e_1) = e_3 \cdot S(e_1)=0$ means no shearing on the face with normal $e_1$.

What is the displacement condition on the top face?

I'm sorry, I meant:

$u_1(p)=\delta$ for $p$ in the orange face, i.e. the one with outer normal $e_1$, and $u_1(p)=0$ for $p$ in the face with outer normal $-e_1$.

I think that the block will be pulled in direction $e_1$, and since there's no shear on the lateral surfaces, those two will be parallel to $e_1 e_3$. But I have the feeling that you don't agree with me.

 

[Chestermiller]

First of all, thanks as usual @Chestermiller for your answer.That's what I thought: $S_{22}=0$ implies that I do not have traction on the lateral faces.

You do not have a normal component of traction on the faces normal to e2.

And then, also $e_3 \cdot S(-e_1) = e_2 \cdot S(-e_1)=0$ implies no shearing on the face with normal $-e_1$.

And $e_2 \cdot S(e_1) = e_3 \cdot S(e_1)=0$ means no shearing on the face with normal $e_1$.

It means that there are no shear components of traction on these faces.

I'm sorry, I meant:

$u_1(p)=\delta$ for $p$ in the orange face, i.e. the one with outer normal $e_1$, and $u_1(p)=0$ for $p$ in the face with outer normal $-e_1$.

I think that the block will be pulled in direction $e_1$,

It will be stretched in the direction e1

and since there's no shear on the lateral surfaces, those two will be parallel to $e_1 e_3$. But I have the feeling that you don't agree with me.

There are no shear (and normal) stresses on the surfaces normal to e2 and e3.

What you seem to be describing is "uniaxial extension in the e1 direction with no lateral constraint."

 

[bobinthebox]

Is that the only requirement ? Think how you would constrain this model in Finite Element Analysis. I would start from $u_{2}=0$ on those blue faces. Then you could use $u_{3}=0$ for bottom face so that the cube can't move upwards/downwards.

I've never done a FEA, I'm still math student.
So, $u_2=0$ on the lateral surfaces makes perfectly sense: no displacement along the $e_2$ direction on the lateral faces. And with $u_3=0$ I have correctly no vertical displacement.

Given those conditions only, I still can't determine the deformation, right? I have to give one more condition, I guess. For instance, I can say $u_1(p) = \delta$ on the orange face. Would this be okay? @FEAnalyst

 

[Chestermiller]

I've never done a FEA, I'm still math student.
So, $u_2=0$ on the lateral surfaces makes perfectly sense: no displacement along the $e_2$ direction on the lateral faces. And with $u_3=0$ I have correctly no vertical displacement.

I don't think you meant that the lateral displacements are zero.

 

[bobinthebox]

I don't think you meant that the lateral displacements are zero.

I got confused, you're right. So, with the conditions I imposed after your first message, the block will be stretched along direction $e_1$, as you said. But this implies a displacement along $e_2$, because the two lateral surfaces will be shrunk a little bit, right?

I think this has to happen also because of the incompressibility constraint.

 

[Chestermiller]

I got confused, you're right. So, with the conditions I imposed after your first message, the block will be stretched along direction $e_1$, as you said. But this implies a displacement along $e_2$, because the two lateral surfaces will be shrunk a little bit, right?

I think this has to happen also because of the incompressibility constraint.

Yes, that is all correct. But that doesn’t mean that there are stresses on the lateral faces.

 

[FEAnalyst]

Given those conditions only, I still can't determine the deformation, right? I have to give one more condition, I guess. For instance, I can say $u_1(p) = \delta$ on the orange face. Would this be okay? @FEAnalyst

Yes, to solve it as a static problem and avoid mechanism-like behavior of the model you will need an additional constraint. In displacement terms: $u_{1}=0$ for back face. This is how single element tests of uniaxial tension are done in FEA - roller support for each of the 3 orthogonal faces. This way you avoid rigid body motions and can stretch the cube (for example made of hyperelastic material). However, this approach won’t agree with your initial assumptions since you wanted the side faces to slide freely in one plane. Probably this assumption will have to be invalidated anyway.

 

[bobinthebox]

Yes, that is all correct. But that doesn’t mean that there are stresses on the lateral faces.

That's a tricky point for me: you're saying that $S_{12}=0$, i.e. shearing on lateral faces is 0. But during the deformation those faces are "shrunk", how is it possible that there's no stress on them? I think this is the crucial point I am missing. @Chestermiller

 

[Chestermiller]

That's a tricky point for me: you're saying that $S_{12}=0$, i.e. shearing on lateral faces is 0. But during the deformation those faces are "shrunk", how is it possible that there's no stress on them? I think this is the crucial point I am missing. @Chestermiller

When you stretch a rod, it shrinks in diameter without applying forces to the sides of the rod, right? Google Poisson's Ratio.

 

[bobinthebox]

Oh yes, now I see. Indeed Poisson's ratio quantifies the "amount of shrinking". This image also helps
Therefore on my block there's only $S_{11}$ which is acting, all the other components of PK stress are $0$.

To conclude this thread, I just need another check: if I want to constrain a face (like the one with normal $-e_1$) of the block to do not move, I'd say $e_3 \cdot S(-e_1) = e_2 \cdot S(-e_1)= e_1 \cdot S(-e_1)=0$. This is totally equivalent to $u_1=0$, right?

 

[Chestermiller]

Oh yes, now I see. Indeed Poisson's ratio quantifies the "amount of shrinking". This image also helps
Therefore on my block there's only $S_{11}$ which is acting, all the other components of PK stress are $0$.

To conclude this thread, I just need another check: if I want to constrain a face (like the one with normal $-e_1$) of the block to do not move, I'd say $e_3 \cdot S(-e_1) = e_2 \cdot S(-e_1)= e_1 \cdot S(-e_1)=0$. This is totally equivalent to $u_1=0$, right?

No way. You are confusing stresses with the kinematics of the motion that cause the stresses.

 

[bobinthebox]

I reflected a bit on your last useful message @Chestermiller.

It seems to me that $u_1=0$ on the surface with normal $-e_1$, which constrains that face to do not move, implies $e_3 \cdot S(-e_1) = e_2 \cdot S(-e_1)= e_1 \cdot S(-e_1)=0$. So first I notice that the conditon is $u_1=0$ on a face, and then I realize that this implies no shear and normal stresses on that face.

Is that correct ?

 

[Chestermiller]

I reflected a bit on your last useful message @Chestermiller.

It seems to me that $u_1=0$ on the surface with normal $-e_1$, which constrains that face to do not move, implies $e_3 \cdot S(-e_1) = e_2 \cdot S(-e_1)= e_1 \cdot S(-e_1)=0$. So first I notice that the conditon is $u_1=0$ on a face, and then I realize that this implies no shear and normal stresses on that face.

Is that correct ?

No. There is a normal force on the face. What do you get for the state of stress using Hooke's law?

 

[bobinthebox]

@Chestermiller For a Neo-Hookean material, I have that Cauchy is $T = - \pi I + \mu B$ where $B=F F^t$ and $F$ is the deformation gradient. Hence first Piola-Kirchoff stress is $S=\mu F - \pi F^{-t}$.

How does those relations imply the presence of a normal force on the face?

Thinking "physically" for a moment, now actually I think that a normal force must be present: if my conditions there are only $e_3 \cdot S(-e_1) = e_2 \cdot S(-e_1)=0$ (no shear stresses), then it could move! So there must be a normal stress in the opposite direction of the motion, namely the one with value $e_1 \cdot S(-e_1)$, right? @Chestermiller

 

[Chestermiller]

@Chestermiller For a Neo-Hookean material, I have that Cauchy is $T = - \pi I + \mu B$ where $B=F F^t$ and $F$ is the deformation gradient. Hence first Piola-Kirchoff stress is $S=\mu F - \pi F^{-t}$.

How does those relations imply the presence of a normal force on the face?

Thinking "physically" for a moment, now actually I think that a normal force must be present: if my conditions there are only $e_3 \cdot S(-e_1) = e_2 \cdot S(-e_1)=0$ (no shear stresses), then it could move! So there must be a normal stress in the opposite direction of the motion, namely the one with value $e_1 \cdot S(-e_1)$, right? @Chestermiller

I don't see how you can possibly study and understand this stuff without first having a course in Strength of Material (aka Mechanics of Materials). You, apparently have never had such a course.

What do you get for the stress tensor in component form? I get $$S_{11}=\mu\left(\lambda-\frac{1}{\lambda^2}\right)$$ where $$\lambda=1+\frac{\Delta u_1}{L}=F_{11}$$ where L is the undeformed length of the body in the 1 direction. I get zeros for all other components of the S tensor.

 

[bobinthebox]

@Chestermiller

Sorry, I really misunderstood your last question. Yes, that is precisely what I got, by exploiting symmetry. I got your $\lambda$ to be $\frac{b+\delta}{b}$, which is equivalent to yours ($b=1$).What is striking me is that to constraint the face with normal $-e_1$ to be fixed I need to impose conditions on first PK $e_2 \cdot S(-e_1)=e_3 \cdot S(-e_1)=0$ and on the displacement u, namely $u_1=0$.

Why can't I just impose the two conditions about no shear stress on that face? What happens if I drop the $u_1=0$ condition? I can't see physically what would be the problem.

I don't see how you can possibly study and understand this stuff without first having a course in Strength of Material (aka Mechanics of Materials). You, apparently have never had such a course.

You're right. I'm a mathematics student, and our course is quite focused on the math. Also, this part about elasticity has been the final topic and we just saw two quick examples about it.

 

[Chestermiller]

If you don't impose u at 0, how do you know that that end doesn't move $\delta$ also, and that the body has just translated as a rigid body without any stress?

 

[bobinthebox]

@Chestermiller Thanks a lot, $u_1 =0$ at that end is finally clear now.

For what concerns the $e_2 \cdot S(-e_1)=e_3 \cdot S(-e_1)=0$ conditions, what happens if I drop them and I leave $u_1=0$ there? I know they means no shear stress on that end face, but I don't see physically what could go wrong if I don't impose them.

Maybe because it could mean that that face is fixed and at the same time there are forces acting on that face in both $e_2$ and $e_3$ direction, which is certainly not the case if we want the face to be fixed.

 

[Chestermiller]

@Chestermiller Thanks a lot, $u_1 =0$ at that end is finally clear now.

For what concerns the $e_2 \cdot S(-e_1)=e_3 \cdot S(-e_1)=0$ conditions, what happens if I drop them and I leave $u_1=0$ there? I know they means no shear stress on that end face, but I don't see physically what could go wrong if I don't impose them.

Maybe because it could mean that that face is fixed and at the same time there are forces acting on that face in both $e_2$ and $e_3$ direction, which is certainly not the case if we want the face to be fixed.

The other two conditions would have to relate to what you do on the opposite face of the object. In this case, you are imposing zero shear stresses on that face, so this would be consistent with that.