# 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

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## Answers and Replies

Mentor
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}##

Mentor
@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