- Get link
- Other Apps

### Featured Post

- Get link
- Other Apps

Hello everyone! Since the beginning of our blog/fanpage, we have mentioned the concepts of stress and strain multiple times. However, we never really discuss the origin of these concepts. Today we are going to talk about the basic meaning of stress and strain, the relations between stress and strain, and the energy changes involved in deformation. Most of the materials in this episode could be found in Lev Landau's classic textbooks --

To find the relative changes of adjacent volume elements, we first evaluate the changes in length of a tiny segment within the deformed material. Consider a tiny segment with length $\text{d}l$. The square of its length could be easily expressed in tensor form $\text{d}l^2=\text{d}x_{i}^2$. After the deformation, the square of its length changes to $\text{d}l'^2=\text{d}x_{i}'^2$. We can easily relate $\text{d}x_{i}'$ to our displacement vector $u_{i}$:

Since strain tensor is a symmetric tensor, we can easily diagonalize it into another form with only 3 principal values $u^{(1)}, u^{(2)}, u^{(3)}$. By substituting $\text{d}l^2 = \delta_{ik}\text{d}x_i\text{d}x_k$, we can get the following:

Since we define our stress tensor from $F_i = \frac{\partial \sigma_{ik}}{\partial x_k}$, it is evident that this definition of stress tensor is not unique. To give our stress tensor a unique definition, we will have to introduce the thermodynamics involved in elastic deformation.

Once we know how to express work, the expression of the changes in internal energy and Helmholtz free energy become quite evident:

But how can we determine the functionality between Helmholtz free energy and the strain tensor? (i.e., $F(u_{ik})=?$) We can reason as followed. First of all, since internal forces always arise from the deformation of an elastic body, the stress tensor should be zero if the strain tensor vanishes. That implies the functionality of Helmholtz free energy does not contain the first-order term of the strain tensor. Secondly, since Helmholtz free energy is a scalar, the Helmholtz free energy can only be expressed in terms of the invariants of a strain tensor. As we showed in our previous episode, there are two second-order invariant of a strain tensor: $u_{ii}^2$ and $u_{ij}u_{ij}=u_{ij}^2$. Therefore, it is reasonable to guess the functionality between Helmholtz free energy and strain tensor to be something like this:

Since in the function of Helmholtz free energy, the strain tensor occur twice: $u_{ll}$ and $u_{ik}$. The former term is only related to the relative volume changes of a volume element within the material -- or the simple compression/expansion. It would be nice if we can separate the deformation associated with compression/expansion and shear:

To find our stress tensor, we need to know the differential of Helmholtz free energy:

Note that $u_{ii}=\frac{1}{3K}\sigma_{ii}$. By rearranging the aforementioned formula, we can also calculate strain tensor if we know the stress tensor:

We have mentioned that $\sigma_{ik}n_k=p_i$ on the surface. Since the stress tensor is a constant tensor throughout the body, we can easily deduce that $\sigma_{zz}=p$, and other $\sigma_{ik}=0$. We can then deduce $u_{ik}$ from $u_{ik}=\frac{1}{2\mu}(\sigma_{ik}-\frac{1}{3}\sigma_{ll}\delta_{ik})+\frac{1}{9K}\sigma_{ll}\delta_{ik}$:

From this result we can derive the formula for Young's modulus and the Poisson ratio. For Young's modulus, it is defined as the ratio between a uniaxial stress and its corresponding parallel strain:

*Course of Theoretical Physics*.### The Strain Tensor

When it comes to elastic theory, there are always some forces and deformations. If the deformed material can return to its original initial state after the removal of external forces, the deformation is called elastic. Otherwise, it is called plastic deformation. To quantify deformation, an instinctive way is to define a displacement vector $\textbf{u}$.

$\textbf{u}=\textbf{r'}-\textbf{r}$

Or written in tensor:

$u_{i}=x_{i}'-x_{i}$

That seems like a very fair representation of the deformation. However, it can be very deceiving sometimes. For example, the tip of a deflected beam may experience a huge displacement, but the real deformation between the adjacent segments are much less dramatic. In a more extreme case, if we move the rod horizontally by 1m, the displacement vector is non-zero, but the rod is not deformed at all. Therefore, what we really care about is the relative changes between adjacent volume elements.

To find the relative changes of adjacent volume elements, we first evaluate the changes in length of a tiny segment within the deformed material. Consider a tiny segment with length $\text{d}l$. The square of its length could be easily expressed in tensor form $\text{d}l^2=\text{d}x_{i}^2$. After the deformation, the square of its length changes to $\text{d}l'^2=\text{d}x_{i}'^2$. We can easily relate $\text{d}x_{i}'$ to our displacement vector $u_{i}$:

$\text{d}l'^2=\text{d}x_{i}'^2=(\text{d}x_i+\text{d}u_i)^2=\text{d}x_i^2+\text{d}u_i^2+2\text{d}x_i\text{d}u_i$

We can easily transform the differential of our displacement vectors into the differential of our coordinate system by
$\text{d}u_i=\text{d}x_k\frac{\partial u_i}{\partial x_k}$

Substitute it back to our equations of $\text{d}l'^2$ and we will get:

$\text{d}l'^2=\text{d}x_i^2+2\text{d}x_i\text{d}x_k\frac{\partial u_i}{\partial x_k}+\text{d}x_i\text{d}x_k\frac{\partial u_l}{\partial x_i}\frac{\partial u_l}{\partial x_k}\\=\text{d}x_i^2+\text{d}x_i\text{d}x_k(\frac{\partial u_i}{\partial x_k}+\frac{\partial u_k}{\partial x_i})+\text{d}x_i\text{d}x_k\frac{\partial u_l}{\partial x_i}\frac{\partial u_l}{\partial x_k}\\=\text{d}l^2+2\text{d}x_i\text{d}x_k(\frac{1}{2}(\frac{\partial u_i}{\partial x_k}+\frac{\partial u_k}{\partial x_i}+\frac{\partial u_l}{\partial x_i}\frac{\partial u_l}{\partial x_k}))\\=\text{d}l^2+2\text{d}x_i\text{d}x_k u_{ik}$

And we get our definition of strain tensor. Usually the magnitude of a strain tensor is small, and thus we can ignore the second order term:

$u_{ik}=\frac{1}{2}(\frac{\partial u_i}{\partial x_k}+\frac{\partial u_k}{\partial x_i})$

It is evident that the strain tensor is symmetric from its definition.Since strain tensor is a symmetric tensor, we can easily diagonalize it into another form with only 3 principal values $u^{(1)}, u^{(2)}, u^{(3)}$. By substituting $\text{d}l^2 = \delta_{ik}\text{d}x_i\text{d}x_k$, we can get the following:

$\text{d}l'^2=(\delta_{ik}+2u_{ik})\text{d}x_i\text{d}x_k=(1+2u^{(1)})\text{d}x_{1}^2+(1+2u^{(2)})\text{d}x_{2}^2+(1+2u^{(3)})\text{d}x_{3}^2$

Thus we can see that if a strain tensor is diagonalized, its principle values correspond to the relative changes in length on a certain axis.
$\text{d}x_1'=\sqrt{1+2u^{(1)}}\text{d}x_1,\quad\frac{\text{d}x_1'-\text{d}x_1}{\text{d}x_1}=\sqrt{1+2u^{(1)}}-1\approx u^{(1)}$

And the trace of the strain tensor corresponds to the relative changes in size of a unit volume:
$\text{d}V'=\text{d}x_1'\text{d}x_2'\text{d}x_3'=\text{d}V(1+u^{(1)})(1+u^{(2)})(1+u^{(3)})\approx (1+u^{(1)}+u^{(2)}+u^{(3)})\text{d}V$

$\text{d}V'=\text{d}V(1+u_{ii})$

### The Stress Tensor

After introducing the strain tensor as the relative changes in length following a deformation, we will move on to the stress tensor. The concept of stress tensor originates from the Newton's law of motion. When we deform an elastic material and it reaches equilibrium, there will be some internal forces within the elastic material. Not only do these forces must balance our external forces on the surface of the material, all these internal forces must cancel out each other as well. That means the volume integral of the internal forces over the whole material must equal to the surface integral of the internal forces over the material surface:
$\int F_i \text{d}V = \oint_A F_i \text{d}S$

From vector calculus, this means that we can always express the internal forces as the divergence of something, and that something is our stress tensor:
$F_i = \frac{\partial \sigma_{ik}}{\partial x_k},\quad\int F_i \text{d}V =\int\frac{\partial \sigma_{ik}}{\partial x_k}\text{d}V= \oint_A \sigma_{ik}\text{d}f_k$

If there is an external force $\textbf{P}$, the force balance is also required on the surface:
$\sigma_{ik}n_{k}=P_{i}\quad\text{(on the surface)}$

$n_{k}$ is just the unit vector normal to the surface.Since we define our stress tensor from $F_i = \frac{\partial \sigma_{ik}}{\partial x_k}$, it is evident that this definition of stress tensor is not unique. To give our stress tensor a unique definition, we will have to introduce the thermodynamics involved in elastic deformation.

### The Thermodynamics of Elastic Deformation

When we deform an elastic body, the external forces only acts on the surface. That means, it is in fact the internal forces that do all the work to bring every part of the body to their new positions. When there are forces and displacements, there are always works. We can easily derive work in terms of stress and strain:(the surface integral can be ignored if we extend the elastic material to infinity)
$\int\delta W\text{d}V=\int F_i\delta u_i\text{d}V=\int\frac{\partial\sigma_{ik}}{\partial x_k}\delta u_i\text{d}V$

$=\oint\sigma_{ik}\delta u_{i}\text{d}f_k-\int\sigma_{ik}\frac{\partial\delta u_{i}}{\partial x_k}\text{d}V$

$=-\frac{1}{2}\int\sigma_{ik}(\frac{\partial\delta u_{i}}{\partial x_k}+\frac{\partial\delta u_{k}}{\partial x_i})\text{d}V=-\frac{1}{2}\int\sigma_{ik}\delta(\frac{\partial u_{i}}{\partial x_k}+\frac{\partial u_{k}}{\partial x_i})\text{d}V$

$=-\int\sigma_{ik}\delta u_{ik}\text{d}V$

Thus we get $\delta W = -\sigma_{ik}\delta u_{ik}$.Once we know how to express work, the expression of the changes in internal energy and Helmholtz free energy become quite evident:

$\text{d}U=\delta q-\delta W = T\text{d}S+\sigma_{ik}\text{d}u_{ik}$

$\text{d}F=\delta q-\delta W = -S\text{d}T+\sigma_{ik}\text{d}u_{ik}$

This implies the stress tensor can be uniquely defined as the differential of internal energy or Helmholtz free energy.
$\sigma_{ik}=\frac{\partial U}{\partial u_{ik}}|_{S}=\frac{\partial F}{\partial u_{ik}}|_{T}$

But how can we determine the functionality between Helmholtz free energy and the strain tensor? (i.e., $F(u_{ik})=?$) We can reason as followed. First of all, since internal forces always arise from the deformation of an elastic body, the stress tensor should be zero if the strain tensor vanishes. That implies the functionality of Helmholtz free energy does not contain the first-order term of the strain tensor. Secondly, since Helmholtz free energy is a scalar, the Helmholtz free energy can only be expressed in terms of the invariants of a strain tensor. As we showed in our previous episode, there are two second-order invariant of a strain tensor: $u_{ii}^2$ and $u_{ij}u_{ij}=u_{ij}^2$. Therefore, it is reasonable to guess the functionality between Helmholtz free energy and strain tensor to be something like this:

$F = F_0+\frac{1}{2}\lambda u_{ll}^2+\mu u_{ik}^2$

in which $\lambda$ and $\mu$ are called the "Lame coefficient."Since in the function of Helmholtz free energy, the strain tensor occur twice: $u_{ll}$ and $u_{ik}$. The former term is only related to the relative volume changes of a volume element within the material -- or the simple compression/expansion. It would be nice if we can separate the deformation associated with compression/expansion and shear:

$u_{ik}=\frac{1}{3}\delta_{ij}u_{ll}+(u_{ik}-\frac{1}{3}\delta_{ij}u_{ll})$

Substitute this back to the Helmholtz free energy, and we will get:
$F = F_0+\frac{1}{2}\lambda u_{ll}^2+\mu (\frac{1}{3}\delta_{ij}u_{ll}+(u_{ik}-\frac{1}{3}\delta_{ij}u_{ll}))(\frac{1}{3}\delta_{ij}u_{ll}+(u_{ik}-\frac{1}{3}\delta_{ij}u_{ll}))$

$= F_0+\frac{1}{2}(\lambda+\frac{2}{3}\mu)u_{ll}^2+\mu(u_{ik}-\frac{1}{3}\delta_{ij}u_{ll})^2$

$= F_0+\frac{1}{2}Ku_{ll}^2+\mu(u_{ik}-\frac{1}{3}\delta_{ij}u_{ll})^2$

in which $K=\lambda+\frac{2}{3}\mu$ is called the bulk modulus, and $\mu$ is called the shear modulus.To find our stress tensor, we need to know the differential of Helmholtz free energy:

$\text{d}F=Ku_{ll}\text{d}u_{ll}+2\mu(u_{ik}-\frac{1}{3}\delta_{ik}u_{ll})(\text{d}u_{ik}-\frac{1}{3}\delta_{ik}\text{d}u_{ll})$

$=Ku_{ll}\delta_{ik}\text{d}u_{ik}+2\mu(u_{ik}\text{d}u_{ik}-\frac{1}{3}u_{ll}\text{d}u_{ll})$

$=[Ku_{ll}\delta_{ik}+2\mu(u_{ik}-\frac{1}{3}u_{ll}\delta_{ik})]\text{d}u_{ik}$

Therefore we can easily derive stress tensor to be:
$\sigma_{ik}=Ku_{ll}\delta_{ik}+2\mu(u_{ik}-\frac{1}{3}u_{ll}\delta_{ik})$

From this, it is evident that the stress tensor is also always symmetric.Note that $u_{ii}=\frac{1}{3K}\sigma_{ii}$. By rearranging the aforementioned formula, we can also calculate strain tensor if we know the stress tensor:

$u_{ik}=\frac{1}{2\mu}(\sigma_{ik}-\frac{1}{3}\sigma_{ll}\delta_{ik})+\frac{1}{9K}\sigma_{ll}\delta_{ik}$

### Homogeneous Extension of a Rod

After we successfully derived the strain tensor and the stress tensor, we should move on to a basic case: homogeneous extension of a rod along its long axis. By saying homogeneous extension, we means the stress tensor and strain tensor are both a constant tensor throughout the whole elastic material. The pressure exerted on the side is equal to $p$.We have mentioned that $\sigma_{ik}n_k=p_i$ on the surface. Since the stress tensor is a constant tensor throughout the body, we can easily deduce that $\sigma_{zz}=p$, and other $\sigma_{ik}=0$. We can then deduce $u_{ik}$ from $u_{ik}=\frac{1}{2\mu}(\sigma_{ik}-\frac{1}{3}\sigma_{ll}\delta_{ik})+\frac{1}{9K}\sigma_{ll}\delta_{ik}$:

$u_{ik}=\frac{1}{2\mu}(\sigma_{ik}-\frac{p}{3}\delta_{ik})+\frac{p}{9K}\delta_{ik}$

Or:
$u_{xx}=u_{yy}=-\frac{1}{3}(\frac{1}{2\mu}-\frac{1}{3K})p,\quad u_{zz}=\frac{1}{3}(\frac{1}{\mu}+\frac{1}{3K})p,\quad u_{xy}=u_{yz}=u_{xz}=0$

It's worth to note that there are strain components in the x & y direction even though the pressure is only in the z direction.From this result we can derive the formula for Young's modulus and the Poisson ratio. For Young's modulus, it is defined as the ratio between a uniaxial stress and its corresponding parallel strain:

$E = \frac{p}{u_{zz}}=\frac{9K\mu}{3K+\mu}$

As for the Poisson ratio, it is defined as the ratio of transverse strain to axial strain when a material is subject to a uniaxial stress:
$\nu=\frac{-u_{xx}}{u_{zz}}=\frac{1}{2}\frac{3K-2\mu}{3K+\mu}$

Since Young's modulus and Poisson ratio is a lot more easier to measure than bulk modulus or shear modulus, the following two formula is a lot more useful:
$\mu = \frac{E}{2(1+\nu)}$

$K=\frac{E}{3(1-2\nu)}$

And we can substitute these back to the stress-strain relationship we just derived:
$u_{ik}=\frac{1}{E}[(1+\nu)\sigma_{ik}-\nu\sigma_{ll}\delta_{ik}]$

$\sigma_{ik}=\frac{E}{1+\nu}[u_{ik}+\frac{\nu}{1-2\nu}u_{ll}\delta_{ik}]$

It is also quite frequent to see them in matrix form:
$\begin{bmatrix}u_{xx}\\u_{yy}\\u_{zz}\\u_{yz}\\u_{zx}\\u_{xy}\end{bmatrix}=\frac{1}{E}\begin{bmatrix}1& -\nu & -\nu & 0 & 0 &0 \\ -\nu& 1 & -\nu & 0 & 0 &0 \\-\nu & -\nu & 1 & 0 & 0 &0 \\ 0&0 &0 &1+\nu &0 &0 \\ 0& 0 & 0 &0 & 1+\nu & 0\\ 0& 0 & 0 &0 &0 &1+\nu \end{bmatrix}\begin{bmatrix}\sigma_{xx}\\\sigma_{yy}\\\sigma_{zz}\\\sigma_{yz}\\\sigma_{zx}\\\sigma_{xy}\end{bmatrix}$

$\begin{bmatrix}\sigma_{xx}\\\sigma_{yy}\\\sigma_{zz}\\\sigma_{yz}\\\sigma_{zx}\\\sigma_{xy}\end{bmatrix}=\frac{E}{(1+\nu)(1-2\nu)}\begin{bmatrix}1-\nu& \nu & \nu & 0 & 0 &0 \\ \nu& 1-\nu & \nu & 0 & 0 &0 \\\nu & \nu & 1-\nu & 0 & 0 &0 \\ 0&0 &0 &1-2\nu &0 &0 \\ 0& 0 & 0 &0 & 1-2\nu & 0\\ 0& 0 & 0 &0 &0 &1-2\nu\end{bmatrix}\begin{bmatrix}u_{xx}\\u_{yy}\\u_{zz}\\u_{yz}\\u_{zx}\\u_{xy}\end{bmatrix}$

And these two are the Hooke's law for isotropic materials.### Some Wrap Up

So in this episode, we first derived strain tensor from the displacement vector. We then know that stress tensor are linked to the internal forces by a divergence, but that does not give our stress tensor a unique definition. Next we explore the work in elastic deformation, and how Helmholtz free energy help define a unique and symmetric stress tensor. Finally, we talked about the homogeneous extension of a rod, and derived the relations between Young's modulus, Poisson ratio, bulk modulus, shear modulus, and the Hooke's law for isotropic materials. I hope you guys enjoy so far, and we will try to solve some problems from the equations we derived here. Stay tuned!

## Comments

## Post a Comment