Extension of an elastic block of Neo-Hookean material

[bobinthebox]

I'm studying elasticity from classical Gurtin's book, and my professor gave us the following example, during lecture. Unfortunately, this is not present in our references, so I'm posting it here the beginning of the solution, and I will highlight at the end my questions. First I need to state the problem clearly:


> Let us consider a body $B$ which has as reference configuration a cube $[0,b] \times [0,b] \times [0,h]$ of incompressible, Neo-Hookean material, without body forces acting on the body.

In the following, $S$ will be the first Piola-Kirchoff stress tensor, and $f$ will indicate the homogeneous deformation.

Boundary conditions on the lateral surfaces:

- $S n = 0$ (i.e. contact forces at the boundary are 0, i.e. I have no traction)
- - - - - - - - - - - -
Boundary conditions on the bottom surface:

- $u(0)=f(0)-0 = 0$ (i.e. origin is fixed during the deformation)
- $u_3(p)= (f(p)-p)\cdot e_3 = 0$ (No displacement along e_3)
- $S(-e_3) \cdot e_1 = S(-e_3) \cdot e_2 = 0$ (Body is constrained to a frictionless plane)

- - - - - - - - - - - - -

Boundary conditions on the top surface:

- $u_3(p)=\delta$ (we allow displacement of an amount $\delta$ on direction $e_3$
- $S(e_3) \cdot e_1 = S(e_3) \cdot e_2 = 0$


- - - - - - - - - - - - -

**Attacking the problem**

To solve it, we make an ansatz, i.e. that the deformation is going to be homogeneous. The field equation is $\operatorname{div}(S)=0$, since there are no body forces. Also, since the material is Neo-Hookean, we have $T=\mu B - \pi I$, and hence the first Piola $S$ is:
$$S = \mu F - \pi F^{-T}$$


Here's where I start having big troubles: I will write in bold where I can't understand

Considering the polar decomposition of the deformation gradient $F$, $F=RU$, we notice that $R=I$ (**why is that**?) and hence $F=U$ and so $$F=\lambda_1 e_1 \otimes e_1 + \lambda_2 e_2 \otimes e_2 + \lambda_3 e_3 \otimes e_3$$ where $\lambda_i$ are the principal stretches. Therefore, $\lambda_3 = \frac{h+\delta}{h}$ (**why this ratio?**)

Also, by invariance w.r.t rotation around $e_3$ we have $\lambda_1 = \lambda_2 = \lambda$. How can I see that I have invariance under rotations about $e_3$? Also, why does this imply that the two principal stretches are the same?

 

[Chestermiller]

You are saying that you have a uni-axial homogeneous deformation in the 3 direction, and no constraint in the 1 and 2 directions, right?

 

[bobinthebox]

You are saying that you have a uni-axial homogeneous deformation in the 3 direction, and no constraint in the 1 and 2 directions, right?

@Chestermiller, thanks for your reply. Yes, given those boundary conditions I think I am in the situation you wrote.

 

[bobinthebox]

@Chestermiller I have just changed my first post, so maybe it's more clear what are the points where I need a clarification.

 

[Chestermiller]

In the situation you have, if you drew differential position vectors between pairs of adjacent material points in the 1, 2, and 3 directions of the original configuration of your body, do you think they would change direction in the final configuration of your body?

 

[bobinthebox]

@Chestermiller

No, I don't think they would change direction. So after the deformation I would not see any rotation, hence $R=I$ and hence $F=U$, which means that I have only a stretch, right?

However, I still can't see why $\lambda_3 = \frac{h+\delta}{h}$, could you explain me why?

 

[Chestermiller]

@Chestermiller

No, I don't think they would change direction. So after the deformation I would not see any rotation, hence $R=I$ and hence $F=U$, which means that I have only a stretch, right?

correct

However, I still can't see why $\lambda_3 = \frac{h+\delta}{h}$, could you explain me why?

What is the ratio of the final length to the initial length in the 3-direction of this homogeneous deformation.

 

[bobinthebox]

What is the ratio of the final length to the initial length in the 3-direction of this homogeneous deformation.

Well, precisely $\frac{h+ \delta }{h}$.

So, $\lambda_3$ has that form because it is the amount of extension.

One last question: why does the professor say that there's invariance w.r.t rotation around $e_3$? Is it possible to prove it formally? Why does this imply $\lambda_1 = \lambda_2$? @Chestermiller

 

[Chestermiller]

Suppose I rotated the sample 90 degrees about the 3 axis. Could you tell the difference?

 

[bobinthebox]

I see. No, I wouldn't be able to tell the difference. This implies that the amount of extension in direction $e_1$ and $e_2$ will be the same, hence $\lambda_1 = \lambda_2$.


@Chestermiller I think I got it, thanks :-)



I'm sorry but I have a very last question: my professor said to try to write which should be the boundary conditions in order to let the lateral surfaces of the body slight on a plane parallel to $e_1, e_3$. I honestly don't know how to move. I assume I should change also the way/direction in which the body is deformed, right?

 

[Chestermiller]

I see. No, I wouldn't be able to tell the difference. This implies that the amount of extension in direction $e_1$ and $e_2$ will be the same, hence $\lambda_1 = \lambda_2$.


@Chestermiller I think I got it, thanks :-)



I'm sorry but I have a very last question: my professor said to try to write which should be the boundary conditions in order to let the lateral surfaces of the body slight on a plane parallel to $e_1, e_3$. I honestly don't know how to move. I assume I should change also the way/direction in which the body is deformed, right?

I don't understand what he is asking for.