User:Eas4200c.f08.vqcrew.E/Homework6

Traction Force
The traction force is defined as the product of the components of the stress tensor matrix, $$ \begin{bmatrix}\sigma\end{bmatrix} \;$$, and the components of the normal vector, $$  \begin{Bmatrix}n\end{Bmatrix} \;$$.

This relation can be derived using Figures #-#. The normal vector, $$ \begin{Bmatrix}n\end{Bmatrix} \;$$, is a unit vector with magnitude equal to one.


 * $$ \begin{Bmatrix}t\end{Bmatrix}_{3x1} = \begin{bmatrix}\sigma\end{bmatrix}_{3x3} \begin{Bmatrix}n\end{Bmatrix}_{3x1} \;$$, where $$\begin{Vmatrix}n\end{Vmatrix} = 1\;$$

The traction force can then broken into its components in the y and z directions.


 * $$\vec{t} = t_y\vec{j} + t_z\vec{k}\;$$


 * $$\vec{n} = n_y\vec{j} + n_z\vec{k}\;$$

The differential lengths of the section and the components of the normal vector can be found using the surface length, ds, and the angle between the normal vector and the y-axis, $$\theta\;$$.


 * $$dz = ds(cos\theta)\;$$


 * $$dy = ds(sin\theta)\;$$


 * $$n_y = cos\theta\;$$


 * $$n_z = sin\theta\;$$

The sum of the forces in the y and z directions must be zero, so an expression relating the traction force, normal force and stresses can be derived. The depth of the section is set to be equal to one. Below is the development for the component of the traction force in the y-direction. The development of the component in the z-direction is reserved for the homework section.


 * $$\sum F_y = 0 = -\sigma_{yy}(dz)(1) - \sigma_{yz}(dy)(1) + t_y(ds)(1)\;$$


 * $$ 0 = -\sigma_{yy}(ds)(n_y) - \sigma_{yz}(ds)(n_z) + t_y(ds)\;$$


 * $$t_y = \sigma_{yy}(n_y) + \sigma_{yz}(n_z)\;$$

The components of the traction force can be written in matrix form.


 * $$\begin{Bmatrix}t_y\\t_z\end{Bmatrix} = \begin{bmatrix}\sigma_{yy} & \sigma_{yz}\\ \sigma_{zy} & \sigma_{zz}\end{bmatrix}\begin{Bmatrix}n_y\\ n_z \end{Bmatrix}\;$$

The three dimensional case can be expanded from the two dimesional case.


 * $$\begin{Bmatrix}t_1\\t_1\\t_3\end{Bmatrix} = \begin{bmatrix}\sigma_{11} & \sigma_{12} & \sigma_{13}\\ \sigma_{11} & \sigma_{12} & \sigma_{13}\\ \sigma_{11} & \sigma_{12} & \sigma_{13}\end{bmatrix}\begin{Bmatrix}n_1\\ n_2\\n_3 \end{Bmatrix}\;$$

This result then can be compressed into a compacet notation. This brings the derivation back to the origin point with the original definition of the traction force in a general notation.


 * $$t_i = \sum_{j=1}^3\sigma_{ij}n_j$$ where i = 1, 2, 3


 * $$\begin{Bmatrix}t_i\end{Bmatrix}_{3x1} = \begin{bmatrix}\sigma_{ij}\end{bmatrix}_{3x3} \begin{Bmatrix}n_{j}\end{Bmatrix}_{3x1}\;$$