Overview of Elasticity of Materials/Isotropic Response

This section covers linear isotropic response.

We now understand tensor relations and have established two tensors that we need to elasticity; ${\displaystyle {\boldsymbol {\sigma }}}$ and ${\displaystyle {\boldsymbol {\varepsilon }}}$. In this chapter we will learn how to relate these to each other. In this particular lecture we will begin studying elasticity in the case of isotropic elasticity where we assume that all directions have the same material response.

Modulus of Elasticity

In earlier chapters we learned that Hook's Law can relate uniaxial loading to uniaxial strain:

${\displaystyle \sigma _{11}=E\ \varepsilon _{11}}$

Here, ${\displaystyle E}$ is the modulus of elasticity. Note that we picked the ${\displaystyle x_{1}}$ direction arbitrarily. It could have easily been ${\displaystyle x_{2}}$ or ${\displaystyle x_{3}}$, or any other direction in this crystal, and found the same material response, ${\displaystyle E}$. This is what comes from working with an isotropic solid.

Poisson's Ratio

When we load stress in the ${\displaystyle x_{1}}$ direction, there is also a natural contraction in the transverse directions, ${\displaystyle x_{2}}$ and ${\displaystyle x_{3}}$. This is expressed as Poisson's Ratio, ${\displaystyle \nu }$.

${\displaystyle \varepsilon _{22}=\varepsilon _{33}=-\nu \varepsilon _{11}=-{\nu \ \sigma _{11} \over E}}$
For perfectly incompressible material, which means it conserves volume, we will have a Poison's Ratio of 0.50. However, most real materials such as common metals have a Poisson's Ratio of around 0.33.

For linear elastic, isotropic materials, normal stresses will not induce shear strain and shear stress will not cause normal stress. In linear elasticity the various contributions to strain can be superimposed. We can use the two above equations for the Modulus of Elasticity and Poisson's Ratio to determine that if a triaxial normal stress is applied, the strain will be:

{\displaystyle {\begin{aligned}\varepsilon _{11}&={\sigma _{11} \over E}-{\nu \sigma _{22} \over E}-{\nu \sigma _{33} \over E}\quad {\begin{cases}\varepsilon _{11}&={1 \over E}\left[\sigma _{11}-\nu \left(\sigma _{22}\sigma _{33}\right)\right]\\\varepsilon _{22}&={1 \over E}\left[\sigma _{22}-\nu \left(\sigma _{33}\sigma _{11}\right)\right]\\\varepsilon _{33}&={1 \over E}\left[\sigma _{33}-\nu \left(\sigma _{11}\sigma _{22}\right)\right]\end{cases}}\end{aligned}}}
Moving to shear relations, the shear stress and strain are related by the shear modulus, ${\displaystyle G}$.

${\displaystyle \sigma _{12}=G\gamma _{12}=G2\varepsilon _{12}}$

Again, due to the isotropic nature this relation holds in all the other directions as well. These three constants of proportionality are sufficient to describe isotropic linear response. They take typical values (for common engineering metals)...

{\displaystyle {\begin{aligned}50<\ &E<200\quad GPa\\25<\ &G<75\quad GPa\\0.25<\ &\nu <0.35\end{aligned}}}

There are many other useful relations that we will summarize here. The Block Modulus, ${\displaystyle K}$, is the ratio of the hydrostatic pressure to the dilatation it produces. ${\textstyle K={-p \over \Delta }={\sigma _{m} \over \Delta }={1 \over \beta }}$, where ${\textstyle -p}$ is the hydrostatic pressure and ${\displaystyle \beta }$ is the compressibility.

Adding together our earlier equations for the triaxial normal strain, we get:

{\displaystyle {\begin{aligned}\underbrace {\varepsilon _{11}+\varepsilon _{22}+\varepsilon _{33}} _{\Delta }&={1-2\nu \over E}(\underbrace {\sigma _{11}+\sigma _{22}+\sigma _{33}} _{3\sigma _{m}})\\\Delta &={1-2\nu \over E}3\sigma _{m}\end{aligned}}}

Applying this to our equations for the Block Modulus results in:

${\displaystyle K={\sigma _{m} \over \Delta }={E \over 3(1-2\nu )}}$

These next solutions are given without proof, because several advanced topics are used to show the relationships.

<TABLE> Moduli Relationships

The stress-strain relationships can be expressed in compact tensor notation.

${\displaystyle \varepsilon _{ij}={1+\nu \over E}\sigma _{ij}-{\nu \over E}\sigma _{kk}\delta _{ij}}$

Looking specifically at a normal strain, ${\displaystyle \varepsilon _{22}}$ for example:

${\displaystyle \varepsilon _{22}={1+\nu \over E}\sigma _{22}-{\nu \over E}(\sigma _{11}+\sigma _{22}+\sigma _{33})\delta _{22}}$

Similarly, for a shear strain, ${\displaystyle \varepsilon _{12}}$, gets us:

${\displaystyle \varepsilon _{12}={1+\nu \over E}\sigma _{12}-{\nu \over E}(\sigma _{11}+\sigma _{22}+\sigma _{33})\delta _{12}}$

From ${\textstyle G={E \over 2(1+\nu )}}$, we find that:

{\displaystyle {\begin{aligned}{1+\nu \over E}={1 \over 2G}\\\varepsilon _{12}={1 \over 2G}\ \sigma _{12}\\\sigma _{12}=2G\ \varepsilon _{12}\end{aligned}}}

For a given stress state:

${\displaystyle {\boldsymbol {\sigma }}=\left({\begin{matrix}\sigma _{11}&\sigma _{12}&\sigma _{13}\\\sigma _{21}&\sigma _{22}&\sigma _{23}\\\sigma _{31}&\sigma _{32}&\sigma _{33}\end{matrix}}\right)}$

We can only write the strain tensor as:

${\displaystyle {\boldsymbol {\varepsilon }}=\left({\begin{matrix}{1 \over E}(\sigma _{11}-\nu (\sigma _{22}+\sigma _{33}))&{\sigma _{12} \over 2G}&{\sigma _{13} \over 2G}\\{\sigma _{12} \over 2G}&{1 \over E}(\sigma _{22}-\nu (\sigma _{33}+\sigma _{11}))&{\sigma _{23} \over 2G}\\{\sigma _{13} \over 2G}&{\sigma _{23} \over 2G}&{1 \over E}(\sigma _{33}-\nu (\sigma _{11}+\sigma _{22}))\end{matrix}}\right)}$

So we have an expression for strain in terms of the applied load, the stress. This can certainly be useful for predicting the deformation to anticipate when loading in application. There is another application we are interested in. Given we deform a part, what stress does it feel? This answers questions such as, "How much can I bend this part before I reach the yield strength?" Basically we want to take our existing solution and invert it.

To do this, we can take our first triaxial strain equation and rearrange it.

{\displaystyle {\begin{aligned}\varepsilon _{11}&={\sigma _{11} \over E}-{\nu \sigma _{22} \over E}-{\nu \sigma _{33} \over E}\\&\quad +0={\nu \sigma _{11} \over E}-{\nu \sigma _{11} \over E}\\&={\sigma _{11} \over E}+{\nu \sigma _{11} \over E}-{\nu \sigma _{11} \over E}-{\nu \sigma _{22} \over E}-{\nu \sigma _{33} \over E}\\\varepsilon _{11}&={1+\nu \over E}\sigma _{11}-{\nu \over E}(\sigma _{11}+\sigma _{22}+\sigma _{33})\end{aligned}}}

We can additionally take our earlier equation for ${\displaystyle \Delta }$, and rearrange it to get:

{\displaystyle {\begin{aligned}\Delta &={1-2\nu \over E}3\sigma _{m}\\\varepsilon _{11}+\varepsilon _{22}+\varepsilon _{33}&={1-2\nu \over E}(\sigma _{11}+\sigma _{22}+\sigma _{33})\\\sigma _{11}+\sigma _{22}+\sigma _{33}&={E \over 1-2\nu }(\varepsilon _{11}+\varepsilon _{22}+\varepsilon _{33})\end{aligned}}}

Then combining the two above equations we get:

{\displaystyle {\begin{aligned}\varepsilon _{11}={1+\nu \over E}\sigma _{11}-{\nu \over E}\left({E \over 1+2\nu }(\varepsilon _{11}+\varepsilon _{22}+\varepsilon _{33})\right)\\\sigma _{11}=\varepsilon _{11}\left({E \over 1+\nu }\right)+{E \over 1+\nu }{\nu \over E}{E \over 1-2\nu }(\varepsilon _{11}+\varepsilon _{22}+\varepsilon _{33})\\\sigma _{11}=\left({E \over 1+\nu }\right)\varepsilon _{11}+{\nu E \over (1+\nu )(1-2\nu )}(\varepsilon _{11}+\varepsilon _{22}+\varepsilon _{33})\end{aligned}}}

Thinking about shear stress is simpler:

{\displaystyle {\begin{aligned}\sigma _{12}=2G\ \varepsilon _{12}\\\sigma _{12}=2\ {E \over 2(1+\nu )}\ \varepsilon _{12}\\\sigma _{12}=\left({E \over 1+\nu }\right)\varepsilon _{12}\end{aligned}}}

Taking the above two equations gives us the tensor expression for stress in terms of strain, which is:

${\displaystyle \sigma _{ij}={E \over 1+\nu }\varepsilon _{ij}+\underbrace {\nu E \over (1+\nu )(1-2\nu )} _{\lambda }\varepsilon _{kk}\delta _{ij}}$

Here, ${\displaystyle \lambda }$ is the Lamé Constant. We can analyze these results and extract useful equations. Let's begin by extracting the deviatoric and hydrostatic contributions to stress.

${\displaystyle \sigma _{ij}'=2G\ \varepsilon _{ij}'}$

We can show this is true by considering shear and normal components.

Shear:

${\displaystyle \sigma _{12}'=\sigma _{12}={E \over (1+\nu )}\varepsilon _{12}=2G\ \varepsilon _{12}=2G\ \varepsilon _{12}'}$

Normal:

{\displaystyle {\begin{aligned}\sigma _{11}'&={2\sigma _{22}-\sigma _{22}-\sigma _{33} \over 3}\\&={1 \over 3}\left[2\ (2G\varepsilon _{11}+\lambda \varepsilon _{11}+\lambda \varepsilon _{22}+\lambda \varepsilon _{33})-(2G\varepsilon _{22}+\lambda \varepsilon _{11}+\lambda \varepsilon _{22}+\lambda \varepsilon _{33})-(2G\varepsilon _{33}+\lambda \varepsilon _{11}+\lambda \varepsilon _{22}+\lambda \varepsilon _{33})\right]\\&={1 \over 3}(4G\varepsilon _{11}-2G\varepsilon _{22}-2G\varepsilon _{33})\\&=2G{2\varepsilon _{11}-\varepsilon _{22}-\varepsilon _{33} \over 3}\\&=2G\varepsilon _{11}'\end{aligned}}}

The relationship between the hydrostatic stress and the mean strain is:

${\displaystyle \sigma _{ii}={E \over 1-2\nu }\varepsilon _{kk}=3K\varepsilon _{kk}}$

Note that we've seen this expression several times before!

Simplified Cases

Plane stress ${\displaystyle \sigma _{33}=0}$ .

Once again looking at our earlier triaxial strain equations, let's start by adding the first two together.

{\displaystyle {{\begin{aligned}\varepsilon _{11}&={1 \over E}[\sigma _{11}-\nu \sigma _{22}]={1 \over E}\sigma _{11}-{\nu \over E}\sigma _{22}\\+\nu (\varepsilon _{22}&={1 \over E}[\sigma _{22}-\nu \sigma _{11}]=-{\nu \over E}\sigma _{11}+{1 \over E}\sigma _{22})\end{aligned}} \over \varepsilon _{11}+\nu \varepsilon _{22}=\left({1 \over E}-{\nu ^{2} \over E}\right)\sigma _{11}}{\begin{aligned}\\\\\rightarrow \sigma _{11}&={E \over 1-\nu ^{2}}(\varepsilon _{11}+\nu \varepsilon _{22})\\\sigma _{22}&={E \over 1-\nu ^{2}}(\varepsilon _{22}+\nu \varepsilon _{11})\end{aligned}}}

Plane stress encountered either as loaded sheet, or more likely a pressure vessel.

<Double Check... Much Greater than or Less than???>

Another simplification is Plane Strain, where ${\displaystyle \varepsilon _{33}=0}$ , typically when one dimension is much greater than the other two so ${\displaystyle \varepsilon _{11}}$  and ${\displaystyle \varepsilon _{22}}$  are much greater than ${\displaystyle \varepsilon _{33}}$ , such as a long rod where strain along the length of the rod is constrained. Here, our third triaxial strain equation rearranges to:

{\displaystyle {\begin{aligned}\varepsilon _{33}&={1 \over E}\left[\sigma _{33}-\nu (\sigma _{11}+\sigma _{22})\right]=0\\\sigma _{33}&=\nu (\sigma _{11}+\sigma _{22})\end{aligned}}}

Note that ${\displaystyle \sigma _{33}}$  does not equal zero just because ${\displaystyle \varepsilon _{33}}$  equals zero. Substituting this ${\displaystyle \sigma _{33}}$  equation into the triaxial strain equations yields us:

{\displaystyle {\begin{aligned}\varepsilon _{11}&={1 \over E}\left[(1-\nu ^{2})\sigma _{22}-\nu (1+\nu )\sigma _{11}\right]\\\varepsilon _{22}&={1 \over E}\left[(1-\nu ^{2})\sigma _{22}-\nu (1+\nu )\sigma _{11}\right]\end{aligned}}}