Finite bending of an elastic block - Equilibrium equations

[bobinthebox]

I am studying the finite bending of a rubber-like block, assuming Neo-Hookean response. In the following, $l_0$,$h$, $\bar{\theta}$ are parameters, while the variables are $r$ and $\theta$.

The Cauchy stress tensor is
$T= - \pi I + \mu(\frac{l_0^2}{4 \bar{\theta}^2 r^2} e_r \otimes e_r + \frac{4 \bar{\theta}^2}{l_0^2}r^2 e_{\theta} \otimes e_{\theta} - I)$

Now I need to solve $div(T)=0$, where the divergence has to be computed in cylindrical coordinates. The author says:

Since there are only two non-null principal stresses, $T_r$ and $T_θ$ , equilibrium becomes $\frac{\partial T_r}{\partial r} + \frac{T_r - T_{\theta}}{r}=0 \quad \frac{\partial T_{\theta}}{\partial \theta} = 0$

Question:
I assume $T_r$ means $e_r \cdot T e_r$, right? If so, I obtained

$T_r = - \pi + \mu \frac{l_0^2}{4 \bar{\theta}^2 r^2} -1$

$T_{\theta} = - \pi + \mu \frac{4 \bar{\theta}^2 r^2}{l_0^2} -1$

Unfortunately, the first equilibrium equation I obtain is different from the one of the book, which is attached to this message.

I obtain $\frac{\partial \pi}{ \partial r} + \mu \frac{4 \bar{\theta}^2 r}{l_0^2}=0$

I'd like to have a check about this, because I think I computed correctly the two principal stresses, so maybe there's a mistake in the book.

Best regards,
Bob

 

[Chestermiller]

Summary:: I computed the principal stresses for a Neo-Hookean material, but I think the equations I got are not correct, since the book writes a different one

I am studying the finite bending of a rubber-like block, assuming Neo-Hookean response. In the following, $l_0$,$h$, $\bar{\theta}$ are parameters, while the variables are $r$ and $\theta$.

The Cauchy stress tensor is
$T= - \pi I + \mu(\frac{l_0^2}{4 \bar{\theta}^2 r^2} e_r \otimes e_r + \frac{4 \bar{\theta}^2}{l_0^2}r^2 e_{\theta} \otimes e_{\theta} - I)$

Now I need to solve $div(T)=0$, where the divergence has to be computed in cylindrical coordinates. The author says:



Question:
I assume $T_r$ means $e_r \cdot T e_r$, right? If so, I obtained

$T_r = - \pi + \mu \frac{l_0^2}{4 \bar{\theta}^2 r^2} -1$

$T_{\theta} = - \pi + \mu \frac{4 \bar{\theta}^2 r^2}{l_0^2} -1$

Unfortunately, the first equilibrium equation I obtain is different from the one of the book, which is attached to this message.

I obtain $\frac{\partial \pi}{ \partial r} + \mu \frac{4 \bar{\theta}^2 r}{l_0^2}=0$

I'd like to have a check about this, because I think I computed correctly the two principal stresses, so maybe there's a mistake in the book.

Best regards,
Bob

I get their answer when I make the substitutions.

 

[bobinthebox]

@Chestermiller Thanks, I've just realized I forgot a factor of $\frac{1}{2}$ during differentiation.




However, I think there's another error when the author tries to solve the problem in the reference configuration, i.e. using first Piola - Kirchoff stress. See page 186 ( https://books.google.it/books?id=f3RrYLujw8oC&pg=PA179&lpg=PA179&dq=Finite+bending+of+an+incompressible+elastic+block&source=bl&ots=hDKzQSjxSv&sig=ACfU3U3fLyBQ9a3XuAl14bfvVHTrmfKs6Q&hl=it&sa=X&ved=2ahUKEwi61bzqxLfwAhVGi6QKHcWEBt8Q6AEwCXoECAUQAw#v=onepage&q=Finite bending of an incompressible elastic block&f=false)

The equilibrium equation is:

$Div(S) = \frac{\partial }{\partial x_1^0} S_{r1} e_r + \frac{\partial}{\partial x_2^0} S_{\theta 2} e_{\theta} = 0$

The problem is that I can't find the same result he wrote at equation (5.95): how is it possible that he has no $\pi$ in the r.h.s. of the equation? Applying the rule for the derivative of a quotient I should find also a term with $\pi$, not only $\frac{\partial \pi}{\partial x_1^0}$

 

[Chestermiller]

@Chestermiller Thanks, I've just realized I forgot a factor of $\frac{1}{2}$ during differentiation.




However, I think there's another error when the author tries to solve the problem in the reference configuration, i.e. using first Piola - Kirchoff stress. See page 186 ( https://books.google.it/books?id=f3RrYLujw8oC&pg=PA179&lpg=PA179&dq=Finite+bending+of+an+incompressible+elastic+block&source=bl&ots=hDKzQSjxSv&sig=ACfU3U3fLyBQ9a3XuAl14bfvVHTrmfKs6Q&hl=it&sa=X&ved=2ahUKEwi61bzqxLfwAhVGi6QKHcWEBt8Q6AEwCXoECAUQAw#v=onepage&q=Finite bending of an incompressible elastic block&f=false)

The equilibrium equation is:

$Div(S) = \frac{\partial }{\partial x_1^0} S_{r1} e_r + \frac{\partial}{\partial x_2^0} S_{\theta 2} e_{\theta} = 0$

The problem is that I can't find the same result he wrote at equation (5.95): how is it possible that he has no $\pi$ in the r.h.s. of the equation? Applying the rule for the derivative of a quotient I should find also a term with $\pi$, not only $\frac{\partial \pi}{\partial x_1^0}$

Sorry. I'm not familiar with this material.

 

[bobinthebox]

Nevermind. Maybe I got what I am missing:

when he writes $Div(S) = \frac{\partial }{\partial x_1^0} S_{r1} e_r + \frac{\partial}{\partial x_2^0} S_{\theta 2} e_{\theta} = 0$ maybe the first term is

$\frac{\partial}{\partial x_1^0} (S_{r1} e_r)$ i.e. the derivative of a product, rather than $\frac{\partial}{\partial x_1^0} (S_{r1}) e_r$

What do you think? @Chestermiller