Overview of Elasticity of Materials/Introducing Strain

This section introduces strain and show tensor symmetry of strain tensor. We also will discuss special subsets of stress and strain including dilatation and deviatoric stresses and strains.

Average Strain

Figure 1: a) Linear elongation b) Shear deformation.

Infinitesimal Strain

While average strain generally looks at the strain of a volume, we will not consider how a point on an elastic body moves and how points near it do also.

Figure 2: 2D strain at a point where $P$ and $Q$ are infinitesimally close to each other and $P'$ and $Q'$ are the displacements, respectively.

We will begin in 2D. Say there is a point ($P$) on an elastic body that is located at coordinate $\{x_{1},\ x_{2}\}$ such as in Figure 2 (this notation will be more convenient when we want to work with tensors). If we deform the body, then $P$ is displaced to $P'$ which has the coordinates $\{x_{1}+u_{1},\ x_{2}+u_{2}\}$. We call $\mathbf {u}$ the displacement vector.

Looking at Figure 2, there is a point infinitesimally close to $P$, called $Q$, with coordinates $\{x_{1}+dx_{1},\ x_{2}+dx_{2}\}$. When $P$ is displaced to $P'$ by the deformation, $Q$ is similarly displaced to $Q'$ with coordinates $\{x_{1}+u_{1}+dx_{1},\ x_{2}+u_{2}+dx_{2}\}$. Thinking critically, the displacement experienced on a body depends on the position on the body. Therefore $\mathbf {u} =u(x_{1},\ x_{2})$. This allows us to use the chain rule to express infinitesimal displacements. Now define the following terms:

${\begin{aligned}\operatorname {d} \!u_{1}&={\partial u_{1} \over \partial x_{1}}\operatorname {d} \!x_{1}+{\partial u_{1} \over \partial x_{2}}\operatorname {d} \!x_{2}\end{aligned}}$           [3]

${\begin{aligned}\\\operatorname {d} \!u_{2}&={\partial u_{2} \over \partial x_{1}}\operatorname {d} \!x_{1}+{\partial u_{2} \over \partial x_{2}}\operatorname {d} \!x_{2}\end{aligned}}$           [4]

${\begin{matrix}e_{11}&={\partial u_{1} \over \partial x_{1}}\qquad e_{12}&={\partial u_{1} \over \partial x_{2}}\\e_{21}&={\partial u_{2} \over \partial x_{1}}\qquad e_{22}&={\partial u_{2} \over \partial x_{2}}\end{matrix}}$           [5-8]

This allows us to write our infinitesimal displacements using Einstein Notation:

${\begin{aligned}du_{i}=e_{ij}\ dx_{j}\end{aligned}}$           [9]

Displacement Tensors

Figure 3: Displacement tensors $P$ and $Q$ after deformations.

What is the physical significance of this? This is easier to see looking in special directions. Considering the points $P=\{x_{1},\ x_{2}\}$, $Q_{1}=\{x_{1}+dx_{1},\ x_{2}\}$ where $dx_{2}=0$ and $Q_{2}=\{x_{1},\ x_{2}+dx\}$ where $dx_{1}=0$ as seen in Figure 3. Then after the deformation:

${\begin{aligned}P'&=\{x_{1}+u_{1},\ x_{2}+u_{2}\}\\{Q_{1}}'&=\{x_{1}+u_{1}+dx_{1}+du_{1},\ x_{2}+u_{2}+du_{2}\}\\{Q_{2}}'&=\{x_{1}+u_{1}+du_{1},\ x_{2}+u_{2}+dx_{2}+du_{2}\}\end{aligned}}$

How do we interpret this? In the case of $Q_{1}$, we have $\operatorname {d} \!x_{2}=0$. Thus, based on Equations 3 & 5 and Equations 4 & 7 expressing the infinitesimal displacements, we can infer that $\operatorname {d} \!u_{1}=e_{11}\operatorname {d} \!x_{1}$, and $\operatorname {d} \!u_{2}=e_{21}\operatorname {d} \!x_{1}$.

This tells us that $e_{11}$ is an expression of uniaxial extension in the $x_{1}$ direction and $e_{21}$ is a rotation of $Q_{1}$ around the point $P$. Similarly, in the case of $Q_{2}$, we can again combine Equation 3 with Equation 6 and Equation 4 with Equation 8 which yields $\operatorname {d} \!u_{1}=e_{12}\ \operatorname {d} \!x_{2}$, and $\operatorname {d} \!u_{2}=e_{22}\ \operatorname {d} \!x_{2}$. Thus, $e_{22}$ is a uniaxial extension in the $x_{2}$ direction, and $e_{12}$ is a rotation of $Q_{2}$ around point $P$. These $\mathbf {e}$ are our displacement tensors.

The Strain Tensor

Let's return to $P$, displacing to $P'$. What is the relationship between $x_{i}$, and ${x_{i}}'$?

${x_{1}}'=x_{1}+u_{1}$           [10]

From Equations 3, 5-8:

${\begin{aligned}\operatorname {d} \!u_{1}&=e_{11}\operatorname {d} \!x_{1}+e_{12}\operatorname {d} \!x_{2}\end{aligned}}$           [11]

Integrating, we have

${\begin{aligned}\\\int _{0}^{u_{1}}\operatorname {d} \!u_{1}&=e_{11}\int _{0}^{x_{1}}\operatorname {d} \!x_{1}+e_{12}\int _{0}^{x_{2}}\operatorname {d} \!x_{2}\\u_{1}&=e_{11}x_{1}+e_{12}x_{2}\\{x_{1}}'&=x_{1}+e_{11}x_{1}+e_{12}x_{2}\end{aligned}}$

Which can be rewritten as

${\begin{aligned}{x_{1}}'&=\left(1+e_{11}\right)x_{1}+e_{12}x_{2}\end{aligned}}$           [12]

Figure 4: Visual representation of the displacement tensor $(\mathbf {e} )$. Here, the displacement tensor $(\mathbf {e} )$ has been broken up into the strain tensor $({\boldsymbol {\varepsilon }})$, and the rotation tensor $({\boldsymbol {\omega }})$.

In a similar fashion, we can also prove that

${\begin{aligned}{x_{2}}'=e_{21}x_{1}+(1+e_{22})x_{2}\end{aligned}}$           [13]

Looking at our tensor, we can see that our deformations also have translations and rotations. We are not interested in these because they do not tell us about material response such as dilatation (change in volume) or distortion (change in shape). Translations and rotations are a part of the field of mechanics called dynamics. Here we are interested in small scale elastic deformations. Our $e_{12}\neq e_{21}$, but we know that our stress tensor is symmetric as $\sigma _{12}=\sigma _{21}$. We can therefore rewrite our displacement tensor as a combination of a symmetric and antisymmetric tensor.

$\underbrace {\left({\begin{matrix}e_{11}&e_{12}\\e_{21}&e_{22}\end{matrix}}\right)} _{\mathbf {e} }=\underbrace {\left({\begin{matrix}e_{11}&{1 \over 2}(e_{12}+e_{21})\\{1 \over 2}(e_{21}+e_{12})&e_{22}\end{matrix}}\right)} _{\boldsymbol {\varepsilon }}+\underbrace {\left({\begin{matrix}0&{1 \over 2}(e_{12}+e_{21})\\{1 \over 2}(e_{21}+e_{12})&0\end{matrix}}\right)} _{\boldsymbol {\omega }}$           [14]

Here, ${\boldsymbol {\varepsilon }}$ is the strain tensor and ${\boldsymbol {\omega }}$ is the rotation tensor. This can be seen schematically in Figure 4. In the scope of this text, we are only interested in ${\boldsymbol {\varepsilon }}$, but it is generally still worth remembering that displacement includes both shear and rotation components:

${\begin{aligned}u_{i}=\varepsilon _{ij}x_{j}+\omega _{ij}x_{j}\end{aligned}}$           [15]

The strain tensor maps the irrotational displacement at a point to an imaginary plane, with normal in any direction that cuts through the point. Because the strain tensor $({\boldsymbol {\varepsilon }})$ is a tensor, it must transform in the same manner as the stress tensor did in earlier sections of this text. As a reminder:

${\begin{aligned}{\sigma _{kl}}'&=a_{ki}\ a_{\ell j}\ \sigma _{ij}\end{aligned}}$           [17]

${\begin{aligned}{\varepsilon _{k\ell }}'&=a_{ki}\ a_{\ell j}\ \varepsilon _{ij}\end{aligned}}$           [18]

Average Engineering Strain

Note that when we first started to look at this subject we defined shear strain as $\gamma =\tan \theta$, which is asymmetric. In terms of our strain tensor, this would be $\gamma _{ij}=2\varepsilon _{ij}$. (It must be rotated back so each side had an angle of ${\textstyle {1 \over 2}\theta }$.) You may frequently see a matrix written as:

$\left({\begin{matrix}\varepsilon _{x}&\gamma _{xy}\\\gamma _{yx}&\varepsilon _{y}\end{matrix}}\right)$

It is sometimes useful to write it this way. However, it is not a tensor, because it does not transform the same as Equations 17 & 18 do, due to its asymmetry. Textbooks generally like using this "average engineering strain" ${\textstyle (\gamma _{ij})}$, but we will not be using this here unless absolutely necessary.

Generalizing 2D to 3D

The results we found for our 2D strain tensor can easily be generalized to 3D by writing them in Einstein Notation and using "3" in the place of "2" in the implicit sums. Equations 3 & 4 are

$\operatorname {d} \!u_{j}={\partial u_{j} \over \partial x_{i}}\operatorname {d} \!x_{i}$           [19]

Which in 3D expresses 3 equations $j=1,\ 2,\ 3$, and expands to:

$\operatorname {d} \!u_{j}={\partial u_{j} \over \partial x_{1}}\operatorname {d} \!{x_{1}}+{\partial u_{j} \over \partial x_{2}}\operatorname {d} \!{x_{2}}+{\partial u_{j} \over \partial x_{3}}\operatorname {d} \!{x_{3}}$

The displacement tensor is:

$e_{ij}={\partial u_{i} \over \partial x_{j}}$           [20]

The strain tensor is:

$\varepsilon _{ij}={1 \over 2}\left(e_{ij}+e_{ji}\right)$           [21]

The rotation tensor is:

$\omega _{ij}={1 \over 2}\left(e_{ij}-e_{ji}\right)$           [22]

Which gives us the displacement:

$u_{i}=\varepsilon _{ij}x_{j}+\omega _{ij}x_{j}=e_{ij}x_{j}$           [23]

And the new displaced coordinates are:

$x_{j}'=x_{j}+u_{j}$           [24]

Now that we have a symmetric strain tensor with properties analogous to stress, we can examine the other properties using similar methods as when we analyzed stress. For small strains where $\Delta \approx \varepsilon _{11}+\varepsilon _{22}+\varepsilon _{33}=I_{1}$, we define the mean strain as:

${\begin{aligned}{\varepsilon _{11}+\varepsilon _{22}+\varepsilon _{33} \over 3}={\varepsilon _{kk} \over 3}=\varepsilon _{m}\end{aligned}}$

Thus,

$\varepsilon _{m}\approx {\Delta \over 3}$           [25]

The total strain tensor can be broken into dilatation and deviatoric components:

$\varepsilon _{ij}={\varepsilon _{ij}}'+\varepsilon _{m}=\left(\varepsilon _{ij}-{\Delta \over 3}\delta _{ij}\right)+{\Delta \over 3}\delta _{ij}$           [26]

In a similar fashion, we also have deviatoric stresses and hydrostatic stresses which are analogous to the deviatoric and dilatation strains. The hydrostatic, or mean, stress is:

$\sigma _{m}={\sigma _{kk} \over 3}={\sigma _{11}+\sigma _{22}+\sigma _{33} \over 3}$           [27]

Therefore, the deviatoric stress can be deduced because:

${\begin{aligned}\sigma _{ij}&={\sigma _{ij}}'+{1 \over 3}\delta _{ij}\sigma _{kk}\\{\sigma _{ij}}'&={\sigma _{ij}}+\sigma _{m}\delta _{ij}\\\\{\sigma _{ij}}'&=\left({\begin{matrix}\sigma _{11}-\sigma _{m}&\sigma _{12}&\sigma _{13}\\\sigma _{12}&\sigma _{22}-\sigma _{m}&\sigma _{23}\\\sigma _{13}&\sigma _{23}&\sigma _{33}-\sigma _{m}\end{matrix}}\right)\end{aligned}}$           [28-31]

The principal components of ${\sigma _{ij}}'$ break down to:

${\sigma _{1}}'={2 \over 3}\left({\sigma _{1}-\sigma _{2} \over 2}+{\sigma _{1}-\sigma _{3} \over 2}\right)$           [32]

And we know that these are just the maximum shear stresses:

${\sigma _{1}}'={2 \over 3}\left({\sigma _{12}}^{MAX}+{\sigma _{13}}^{MAX}\right)$           [33]

Keep in mind that we took this from:

$0=\left(\sigma '\right)^{3}-J_{1}\left(\sigma '\right)^{2}-J_{2}\sigma '-J_{3}$           [34]

Where:

${\begin{aligned}J_{1}&=(\sigma _{11}-\sigma _{m})+(\sigma _{22}-\sigma _{m})+(\sigma _{33}-\sigma _{m})\\J_{2}&={1 \over 6}\left[\left(\sigma _{11}-\sigma _{22}\right)^{2}+\left(\sigma _{22}-\sigma _{33}\right)^{2}+\left(\sigma _{33}-\sigma _{11}\right)^{2}+6\left(\sigma _{12}^{2}+\sigma _{23}^{2}+\sigma _{31}^{2}\right)\right]\\J_{3}&=\det |{\boldsymbol {\sigma }}'|\end{aligned}}$           [35-38]