Theory Introduction

A comprehensive description of the possible stress on a finite element will include six unique factors. These are the three normal stresses, and the three shear stresses. Because the finite element is in rotational equilibrium, the shears will exist in pairs, giving rise to symmetry in the stress tensor.

$$ \sigma= \begin{bmatrix} \sigma_{xx} & \tau_{xy} & \tau_{xz}\\ \tau_{xy} & \sigma_{yy} & \tau_{yz}\\ \tau_{xz} & \tau_{yz} & \sigma_{zz}\\ \end{bmatrix} $$

The principle stresses are given by the tensor’s eigenvalues, $\sigma_1, \sigma_2, \sigma_3$

For our application to beam problems, the stress tensor will take on a more specific form based on the internal forces. (The dashes indicate repetitions of the cross diagonal terms, according to the symmetry of the matrix)

$$ \sigma= \begin{bmatrix} \frac{N_x}{A}+\frac{M_yz}{I_y}-\frac{M_zy}{I_z} & -\frac{T_xz}{J}-\frac 32\frac {V_y}A(1-\frac{y^2}{c_y^2}) & \frac{T_xy}{J}+\frac 32\frac {V_z}A(1-\frac{z^2}{c_z^2}) \\

& 0 & 0 \\

& 0 & 0 \end{bmatrix} $$

A comprehensive description of the deformation of a finite element will include six unique factors. These includes the three normal strains, and three strains which describe shearing. These shear strains can be thought of as generated by shear on one face, in the direction of a second face, and occurring symmetrically with a shear on the second face in the direction of the second face.

$$ \varepsilon

\begin{bmatrix} \varepsilon_{xx} & \varepsilon_{xy} & \varepsilon_{xz}\\

• & \varepsilon_{yy} & \varepsilon_{yz}\\
• & - & \varepsilon_{zz}\\ \end{bmatrix} = \begin{bmatrix} \frac{\partial u}{\partial x} & \frac12(\frac{\partial u}{\partial y}+\frac{\partial v}{\partial x}) & \frac12(\frac{\partial u}{\partial z}+\frac{\partial w}{\partial x}) \\

& \frac{\partial v}{\partial y} & \frac12(\frac{\partial v}{\partial z}+\frac{\partial w}{\partial y}) \\

&

& \frac{\partial w}{\partial z} \\ \end{bmatrix} $$

The stress and strain tensors are related to one another by Hooke’s Law

$$ \bm\varepsilon=\frac{1+\nu}{E}\bm\sigma-\frac \nu E tr(\sigma)\bm I+\alpha\Delta T\bm I $$

$$ \bm\sigma=\frac{E}{1+\nu}\bm\varepsilon+\frac{E\nu tr(\bm\varepsilon)}{(1+\nu )(1-2\nu )}\bm I-\frac{E}{1-2\nu }\alpha\Delta T\bm I $$

Theses stress and strain tensors give rise to Von Mises stress, which we utilized earlier to predict yielding.

$$ \sigma_{vm}= \sqrt{ \frac{(\sigma_1-\sigma_2)^2}2 + \frac{(\sigma_2-\sigma_3)^2}2 + \frac{(\sigma_1-\sigma_3)^2}2 } $$

$$ \sigma_{vm} = \sqrt{ \frac{(\sigma_x-\sigma_y)^2}2 + \frac{(\sigma_x-\sigma_z)^2}2 + \frac{(\sigma_y-\sigma_z)^2}2 + 3\tau_{xy}^2 + 3\tau_{xz}^2 + 3\tau_{yz}^2} $$

Stress and Strain Tensor Application

Traction Vector from Stress Tensor

The traction vector describes the particular force vector (with units $N/m^2$) applied at the point $(x,y,z)$, at which $\sigma$ is evaluated, on the face described by the normal vector $\vec{n}$

$$ \vec{T}=\sigma\cdot \vec{n} $$

Stress from Internal Forces

$$ \sigma

\begin{bmatrix} \frac{N_x}{A}+\frac{M_yz}{I_y}-\frac{M_zy}{I_z} & -\frac{T_xz}{J}-\frac 32\frac {V_y}A(1-\frac{y^2}{c_y^2}) & \frac{T_xy}{J}+\frac 32\frac {V_z}A(1-\frac{z^2}{c_z^2}) \\

& 0 & 0 \\

& 0 & 0 \end{bmatrix} $$

Strain from Deformation

$$ \varepsilon

\begin{bmatrix} \frac{\partial u}{\partial x} & \frac12(\frac{\partial u}{\partial y}+\frac{\partial v}{\partial x}) & \frac12(\frac{\partial u}{\partial z}+\frac{\partial w}{\partial x}) \\

& \frac{\partial v}{\partial y} & \frac12(\frac{\partial v}{\partial z}+\frac{\partial w}{\partial y}) \\

&

& \frac{\partial w}{\partial z} \\ \end{bmatrix} $$