Von Mises Failure Criterion

The general strain energy density function in terms of stresses and strains, applies for all materials (timber, concrete, steel, rubber, etc.)

$\begin{equation*} U = \frac{1}{2}\bigg(\sigma_x\epsilon_x + \sigma_y\epsilon_y + \sigma_z\epsilon_z + \tau_{xy}\gamma_{xy} + \tau_{yz}\gamma_{yz} + \tau_{xz}\gamma_{xz}\bigg) \end{equation*}$

However, for any cube in space, you can find the orientation (pitch, yaw) that has only principal stresses, i.e. no shear stresses.
In this unique orientation, the strain energy density definition reduces to:

(1) $\begin{equation*} U = \frac{1}{2}\bigg(\sigma_{1}\epsilon_{1} + \sigma_{2}\epsilon_{2} + \sigma_{3}\epsilon_{3}\bigg) \end{equation*}$

where $\sigma_1$ , $\sigma_2$ , $\sigma_3$ refer to the stresses in the principal directions.
To express equation 1 in terms of only stresses, we can use the generalized version of Hooke’s law (Eqn. 1 in Section 8: Elastic Moduli and Poisson’s Ration, in terms of Lamé constants)

$\begin{equation*} \epsilon_{ij} = \frac{1}{2\mu}\left(\sigma_{ij} - \frac{\lambda}{3\lambda + 2\mu}\sigma_{kk}\delta_{ij}\right) \end{equation*}$

Using the same notation for principal stresses ( $\sigma_1$ , $\sigma_2$ , $\sigma_3$ ), we can find the strain in those directions as follows:

(*) $\begin{align*} \\ \tag{a}\epsilon_{1} &= \frac{1}{2\mu}\left(\sigma_{1} -\frac{\lambda}{3\lambda+2\mu}(\sigma_{1} + \sigma_{2} + \sigma_{3})\right)\\ \tag{b} \epsilon_{2} &= \frac{1}{2\mu}\left(\sigma_{2} - \frac{\lambda}{3\lambda+2\mu}(\sigma_{1} + \sigma_{2} + \sigma_{3})\right)\\ \tag{c} \epsilon_{3} &= \frac{1}{2\mu}\left(\sigma_{3} - \frac{\lambda}{3\lambda+2\mu}(\sigma_{1} + \sigma_{2} + \sigma_{3})\right)\\ \end{align*}$

Finding a common denominator for all terms and subsequently subtracting leaves us with:

(*) $\begin{align*} \\ \tag{a}\epsilon_{1} &= \frac{1}{2\mu(3\lambda + 2\mu)}\bigg((2\lambda + 2\mu)\sigma_{1} - \lambda\sigma_{2} - \lambda\sigma_{3}\bigg)\\ \tag{b}\epsilon_{2} &= \frac{1}{2\mu(3\lambda + 2\mu)}\bigg((2\lambda + 2\mu)\sigma_{2} - \lambda\sigma_{1} - \lambda\sigma_{3}\bigg)\\ \tag{c}\epsilon_{3} &= \frac{1}{2\mu(3\lambda + 2\mu)}\bigg((2\lambda + 2\mu)\sigma_{3} - \lambda\sigma_{1} - \lambda\sigma_{2}\bigg)\\ \end{align*}$

If we substitute (a), (b), and (c) into 1, we can get the strain energy density in terms of the Lamé constants and stresses.

$\begin{equation*} U = \frac{1}{2}\frac{1}{2\mu(3\lambda + 2\mu)}* \begin{align*} \bigg[&\bigg((2\lambda + 2\mu)\sigma_{1} - \lambda\sigma_{2} - \lambda\sigma_{3}\bigg)\sigma_{1} + \\ &\bigg((2\lambda + 2\mu)\sigma_{2} - \lambda\sigma_{1} - \lambda\sigma_{3}\bigg)\sigma_{2} + \\ &\bigg((2\lambda + 2\mu)\sigma_{3} - \lambda\sigma_{1} - \lambda\sigma_{3}\bigg)\sigma_{3}\bigg] \end{align*} \end{equation*}$

This reduces to:

(2) $\begin{equation*} U = \frac{1}{2\mu(3\lambda + 2\mu)}\bigg[(\lambda + \mu)(\sigma_{1}^2 + \sigma_{2}^2 + \sigma_{3}^2) - \lambda\sigma_{1}\sigma_{2} - \lambda\sigma_{1}\sigma_{3} - \lambda\sigma_{2}\sigma_{3}\bigg] \end{equation*}$

The key discovery of Von Mises was based on the observation that when metals, such as low-carbon steel, begin to yield, the atoms slide past each other, leading to plastic deformation, but the bonds between the metal atoms does not break (till significant deformation). This is due to the nature of typical metallic bonds, in which electrons are shared between metal atoms as a “sea” of electrons. However this does not apply for all metal alloys, as alloys exist that utilize a higher carbon content to produce “hard” crystal structures, such as cast iron, which experiences brittle failure rather than ductile failure.

Von Mises hypothesized that any stress can be divided into a hydrostatic stress, $\boldsymbol{\sigma_{\text{hydrostatic}}}$ or $\boldsymbol{\sigma_{\text{volumetric}}}$ , and a deviatoric stress, $\boldsymbol{\sigma_{\text{deviatoric}}}$ or $\boldsymbol{\sigma'}$ . The hydrostatic pressure acts equally on each side of the cube (think of it like pressure that a cube experiences underwater). He further deduced that yielding depends only on the deviatoric stress!

These concepts can be summarized in the following equations.

(3) $\begin{equation*} \boldsymbol{\sigma} = \boldsymbol{\sigma_{\text{hydrostatic}}} + \boldsymbol{\sigma'} \end{equation*}$

By definition, hydrostatic pressure only acts on the diagonal terms.

(d) $\begin{equation*} \boldsymbol{\sigma_{\text{hydrostatic}}} = \begin{bmatrix} \sigma_{\text{hydrostatic}} & 0 & 0 \\ 0 & \sigma_{\text{hydrostatic}} & 0 \\ 0 & 0 & \sigma_{\text{hydrostatic}} \\ \end{bmatrix} \end{equation*}$

For equation 3 to be true, the deviatoric stress $\boldsymbol{\sigma'}$ must be as follows:

$\begin{equation*} \boldsymbol{\sigma'} = \begin{bmatrix} \sigma_{11} - \sigma_{\text{hydrostatic}} & \sigma_{12} & \sigma_{13} \\ \sigma_{12} & \sigma_{22} - \sigma_{\text{hydrostatic}} & \sigma_{23} \\ \sigma_{13} & \sigma_{23} & \sigma_{33} - \sigma_{\text{hydrostatic}} \end{bmatrix} \end{equation*}$

If we substitute $\text{(d)}$ into Eqn. 2, we can find the hydrostatic strain energy density (where $\sigma_1 = \sigma_2 =\sigma_3 = \sigma_{\text{hydrostatic}}$ ):

$\begin{align*} U_{\text{hydrostatic}} &= \frac{1}{2\mu(3\lambda + 2\mu)}\bigg[(\lambda + \mu)3\sigma_{\text{hydrostatic}}^2 - 3\lambda\sigma_{\text{hydrostatic}}^2\bigg] \\ U_{\text{hydrostatic}} &= \frac{1}{2(3\lambda + 2\mu)}(3\sigma_{\text{hydrostatic}}^2) \end{align*}$

Also, $\sigma_{\text{hydrostatic}}$ in terms of principal stresses is:

$\begin{equation*} \sigma_{\text{hydrostatic}} = \frac{\sigma_1 + \sigma_2 + \sigma_3}{3} \end{equation*}$

Then the hydrostatic strain energy density in terms of principal stresses is:

(4) $\begin{equation*} U_{\text{hydrostatic}} = \frac{(\sigma_{1}^2 + \sigma_{2}^2 + \sigma_{3}^2)}{2(3\lambda + 2\mu)} \end{equation*}$

The deviatoric strain energy density, $U'$ , can be found by subtracting Eqn. 4 from Eqn. 2:

$\begin{align*} U' = &U - U_{\text{hydrostatic}} \\ U' = &\frac{1}{2\mu(3\lambda + 2\mu)}\bigg[(\lambda + \mu)(\sigma_1^2 + \sigma_2^2 + \sigma_3^2) - \lambda(\sigma_1\sigma_2 + \sigma_2\sigma_3 + \sigma_1\sigma_3)\bigg] \,- \\ &\frac{1}{2\mu(3\lambda + 2\mu)}\bigg[\mu\sigma_1^2 + \mu\sigma_2^2 + \mu\sigma_3^2\bigg] \\ U' = &\frac{1}{2\mu(3\lambda + 2\mu)}\bigg[\lambda(\sigma_1^2 + \sigma_2^2 + \sigma_3^2) - \lambda(\sigma_1\sigma_2 + \sigma_2\sigma_3 + \sigma_1\sigma_3)\bigg] \end{align*}$

For a uniaxial tension test, if $\sigma_{11} = \sigma_1 = \sigma_y$ and $\sigma_2 = \sigma_3 = 0$ :

$\begin{equation*} U' = \frac{\lambda}{2\mu(3\lambda + 2\mu)}[\sigma_1^2] = \frac{\lambda}{2\mu(3\lambda + 2\mu)}[\sigma_y^2] \end{equation*}$

If we set the deviatoric strain energy density from a uniaxial tension state at yield equal to the general deviatoric strain energy, we can get the envelope of principal stresses that will produce yielding, which is also know as the Von Mises yield criterion.

$\begin{align*} \frac{\lambda}{2\mu(3\lambda + 2\mu)}\sigma_y^2 &= \frac{\lambda}{2\mu(3\lambda + 2\mu)}\bigg[(\sigma_1^2 + \sigma_2^2 + \sigma_3^2 - (\sigma_1\sigma_2 +\sigma_2\sigma_3 + \sigma_1\sigma_3)\bigg] \\ \sigma_y^2 &= \frac{1}{2}\bigg[(\sigma_1 - \sigma_2)^2 + (\sigma_1 - \sigma_3)^2 + (\sigma_2 - \sigma_3)^2\bigg] \\ \sigma_y &= \sqrt{\frac{(\sigma_1 - \sigma_2)^2 + (\sigma_1 - \sigma_3)^2 + (\sigma_2 - \sigma_3)^2}{2}}\\ \end{align*}$