# Finite bending of an elastic block - Equilibrium equations

• bobinthebox
In summary, the conversation revolved around the computation of principal stresses for a Neo-Hookean material, as well as solving for equilibrium equations in the reference configuration using the first Piola-Kirchoff stress. The speaker also mentioned a potential error in the book's equations and asked for clarification.
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.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

## 1. What is finite bending of an elastic block?

Finite bending of an elastic block refers to the deformation of a solid block due to the application of external forces. This deformation is characterized by changes in the shape and size of the block, and it is limited by the elastic properties of the material.

## 2. What are the equilibrium equations for finite bending of an elastic block?

The equilibrium equations for finite bending of an elastic block are based on the principles of statics and mechanics of materials. These equations include the sum of forces in the x and y directions, as well as the sum of moments about a point. They are used to determine the internal stresses and deformations in the block.

## 3. How do the material properties affect the finite bending of an elastic block?

The material properties, such as Young's modulus and Poisson's ratio, play a crucial role in the finite bending of an elastic block. These properties determine the stiffness and strength of the material, which in turn affects the amount of deformation and stress the block can withstand before reaching its elastic limit.

## 4. What are the applications of studying finite bending of an elastic block?

Studying finite bending of an elastic block has several applications in engineering and science. It can be used to analyze the behavior of structures, such as beams and bridges, under different loading conditions. It is also important in the design and optimization of mechanical components, such as springs and plates.

## 5. What are the limitations of the equilibrium equations for finite bending of an elastic block?

The equilibrium equations for finite bending of an elastic block assume that the material is homogeneous, isotropic, and behaves in a linear-elastic manner. In reality, materials may exhibit non-linear behavior or have varying properties throughout the block, which can lead to discrepancies between theoretical predictions and experimental results.

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