Finite bending of an elastic block - Equilibrium equations

AI Thread Summary
The discussion focuses on the finite bending of a Neo-Hookean elastic block, with the author deriving the Cauchy stress tensor and attempting to solve the equilibrium equations in cylindrical coordinates. The author computes the principal stresses, but finds discrepancies between their results and those in a referenced book. A key point of confusion arises regarding the equilibrium equations, particularly the presence of the term involving pressure (π) in the equations derived. The author suspects errors in both their calculations and the book's presentation, particularly in the application of the first Piola-Kirchhoff stress in the reference configuration. Ultimately, the discussion highlights the complexities of deriving equilibrium equations for elastic materials and the importance of careful differentiation.
bobinthebox
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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.Bob
 

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bobinthebox said:
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.Bob
I get their answer when I make the substitutions.
 
@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}##
 
bobinthebox said:
@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.
 
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
 
This has been discussed many times on PF, and will likely come up again, so the video might come handy. Previous threads: https://www.physicsforums.com/threads/is-a-treadmill-incline-just-a-marketing-gimmick.937725/ https://www.physicsforums.com/threads/work-done-running-on-an-inclined-treadmill.927825/ https://www.physicsforums.com/threads/how-do-we-calculate-the-energy-we-used-to-do-something.1052162/
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