User:Eas4200c.f08.aeris.guan/hw6/117

Deriving Traction Force Equation cont...
Recall that the equation we are trying to derive is the following. $$\displaystyle [t]_{3x1}=[\sigma]_{3x3}[n]_{3x1}$$ Traction force is the stress applied to a lateral surface. For example, stresses due to aerodynamic forces on a wing such as drag or even friction can be a traction force. The goal is to essentially derive the three components of the traction force matrix from the stress tensor and the surface's normal vector. This expression will be obtained by geometrical manipulation. Let's consider the following figure to illustrate an infinitesimally small surface element.  In the diagram, ds is the length of the lateral surface, n is the unit normal vector of the lateral surface, and the base and height of this triangular element is dy and dz respectively. Now, in order to relate the traction force to the stress tensor components, the angle of the lateral surface with respect to the coordinate system must be utilized. First, the base and height of the triangle will be written in terms of the length of the lateral surface. Thus, it is simple to see the following by simple geometry.  $$\displaystyle dz=dscos(\theta)$$ $$\displaystyle dy=dssin(\theta)$$ Similarly, the normal vector can be written in terms of the angle $$\theta$$. Realize that the magnitude of the normal vector is one, the following can be written.  $$||\vec{n}||=1$$ $$\displaystyle n_y=cos(\theta)$$ $$\displaystyle n_z=sin(\theta)$$ To continue with the derivation, the next logical step is to write the equilibrium equations of the forces in the z and y directions. To simplify things, the depth in and out of the page will be assumed to be one (the area of the base and the height will be the base and the height respectively). Although this is the derivation for the 2D case, the process here can be extrapolated with another axis and equations for the 3D case can also be derived. Let's equate forces in the z and y directions.  $$\sum {F}_y=0=-\sigma_{yy}dz-\sigma_{yz}dy +t_yd_s$$ Substituting the base and height will the lateral surface length and its respect direction cosine, we get the following equation. $$\displaystyle =-\sigma_{yy}ds n_y-\sigma_{yz}ds n_z +t_yd_s$$ It is possible to divide out the ds terms resulting in... $$\displaystyle t_y= \sigma_{yy}n_y+\sigma_{yz}n_z ... (1)$$ The sum of the forces in the z direction can be found in the below collapsible box. Once the equations 1 and 2 are obtained, this can be arranged into a matrix notation. In book the subscripts are different in that it uses the (x,y) coordinate system. Using cyclic permutation, we have the following equation.  $$\begin{bmatrix} t_y\\ t_z\\ \end{bmatrix} = \begin{bmatrix} \sigma_{yy} & \sigma_{yz}\\ \sigma_{zy} & \sigma_{zz}\\ \end{bmatrix} \begin{bmatrix} n_y\\ n_z\\ \end{bmatrix}...(3) $$

The above equation is true because $$\sigma_{yz}=\sigma_{zy}$$. Equation 3 is written this way so that the first subscript indicates the row of the equation. In indicial notation, the equation is rewritten as the following.  $$\begin{bmatrix} t_1\\ t_2\\ \end{bmatrix} = \begin{bmatrix} \sigma_{11} & \sigma_{12}\\ \sigma_{21} & \sigma_{22}\\ \end{bmatrix} \begin{bmatrix} n_1\\ n_2\\ \end{bmatrix}...(3') $$

This can be projected into the third dimension. Generalizing this equation into the 3D case in indicial notation yields the following. $$ \begin{bmatrix} t_1\\ t_2\\ t_3\\ \end{bmatrix} = \begin{bmatrix} \sigma_{11} & \sigma_{12}& \sigma_{13}\\ \sigma_{21} & \sigma_{22}& \sigma_{23}\\ \sigma_{31} & \sigma_{32}& \sigma_{33}\\ \end{bmatrix} \begin{bmatrix} n_1\\ n_2\\ n_3\\ \end{bmatrix}...(4) $$ The advantage of using the indicial notation is that it can be written in a compact way that still provides all the information of the matrix notation.  $$t_i=\sum_{j=1}^3{\sigma_{ij}n_j},i=1,2,3$$ Now, the original equation is derived. $$\displaystyle [t_i]_{3x1}=[\sigma_{ij}]_{3x3}[n_i]_{3x1} $$     From the road map, we have the following three equations. $$\displaystyle T=2\int\int_{A}^{}{}\phi dA$$ $$\displaystyle T = GJ\theta $$ $$\displaystyle J=\frac{-4}{\nabla^{2}\phi }\int \int _{A}\phi dA$$ Combining these three equations and solving for the Lagrangian of phi, we have the following equation. $$ \frac{\partial^2{\phi}}{\partial{y}^2}+\frac{\partial^2{\phi}}{\partial{z}^2}=-2G\theta $$ where $$\phi$$ is constant on the lateral surface of the bar.