User:Eas4200c.f08.gator.edwards/ Wk5 Class Notes

Kinematic Assumptions

4 zero strain components

4 zero stress components

Rewrite $\ \epsilon -\sigma\$ relationship

$\left\{\epsilon_{ij} \right\}=\begin{bmatrix} \textbf A &0\\ 0& \textbf{B} \end{bmatrix}\left\{\sigma_{ij} \right\}$

$\left\{\sigma_{ij} \right\}=\begin{bmatrix} \textbf A^{-1} &0 \\ 0&\textbf B^{-1} \end{bmatrix}\left\{\epsilon_{ij} \right\}$

To verify they are inverses, multiply together to get identity matrix

$\textbf C^{-1} \textbf C= \textbf I$

$\textbf C^{-1} \textbf C= \begin{bmatrix} \textbf A^{-1}A &0 \\ 0&\textbf B^{-1}B \end{bmatrix}= \textbf I$

$\left\{\sigma_{ij} \right\}=\begin{bmatrix} \textbf A^{-1} &0 \\ 0&\textbf B^{-1} \end{bmatrix}\begin{bmatrix} 0\\ 0\\ 0\\ 0\\ \epsilon_{31}\\ \epsilon_{12} \end{bmatrix}$

$\sigma_{11}=0$

$\sigma_{22}=0$

$\sigma_{33}=0$

$\sigma_{23}=2G\epsilon_{23}=0$

$\sigma_{31}=2G\epsilon_{31}$

$\sigma_{12}=2G\epsilon_{12}$

Now we move on to part C of the roadmap: Equilibrium Equation for stresses

Bidirectional Bending

$M_{y}=\int_{A}z\sigma_{xx}dA$

Similarly for x.

Moment of Inertia: $I_{y},\ I_{z},\ I_{yz}$

$I_{y}=\int_{A}z^2dA$

$I_{z}=\int_{A}y^2dA$

$I_{yz}=\int_{A}yzdA$

$\sigma_{xx}=E\epsilon_{xx}$

   $\sigma_{xx}=\frac{I_{y}M_z-I_{yz}M_y}{I_yI_z-(I_{yz})^2}\ y\ +\frac{I_{z}M_y-I_{yz}M_z}{I_yI_z-(I_{yz})^2}\ z$

Notes on Determinants -team gator
A determinant is the scale factor for volume when matrix A is regarded as a linear transformation. To find the determinant of the 2 X 2 matrix A: $A=\begin{bmatrix} a & b\\ c & d \end{bmatrix}$ $det(A)=\begin{vmatrix} a &b \\ c& d \end{vmatrix}=ad-bc$

Recall:

$I=\begin{bmatrix} I_{11} &I_{12} &I_{13} \\ & \begin{vmatrix} I_{22} & I_{23}\\ & I_{33} \end{vmatrix} \\ & & \end{bmatrix}$

Note: the denominator of the above equation is the determinant of the I matrix.

$D:=I_yI_z-(I_{yz})^2 =I_{22}I_{33}-(I_{23})^2$

is the determinant of $\begin{bmatrix} I_{22} &I_{23} \\ I_{32}&I_{33} \end{bmatrix}$

Neutral Axis$\rightarrow$Bending Moment Stress = 0 $\ (\sigma_{xx}=0)$

Goal

Our goal is to prove:

     $\frac{\partial\sigma_{yx}}{\partial y}+\frac{\partial\sigma_{zx}}{\partial z}=0\ \ \ \ (1)$

Now let's use indicial notation for the above equation, where x=1, y=2, and z=3:

     $\frac{\partial\sigma_{21}}{\partial x_2}+\frac{\partial\sigma_{31}}{\partial x_3}=0\ \ \ \ (2)$

Recall:

    ${\partial \sigma_{xx} \over \partial x} + {\partial \sigma_{yx} \over \partial y} + {\partial \sigma_{zx} \over \partial z}=0\ \ \ \ (3)$

In indicial notation equation 3 becomes:

    $\sum_{i=1}^{3}\frac{\partial\sigma_{ij}}{\partial x_i}=0\ \ \ \ (4)$

Derivation of $\qquad \qquad \frac{\partial \sigma }{\partial x}+\frac{f(x)}{A}=0$

First take the sum of all of the forces in the x direction of our 1-D model.

$\sum{F_{x}}=0=-\sigma (x)A+\sigma (x+dx)A+f(x)dx$

$\displaystyle 0=A(-\sigma (x)+\sigma (x+dx))+f(x)dx$

The term multiplied by A is expanded using a Taylor series expansion. The higher order terms are ignored so it is replaced with the equivalent below.

$-\sigma (x)+\sigma (x+dx)\approx \frac{\partial \sigma }{\partial x}dx$

This is then substituted back into the equation.

$0=\frac{\partial \sigma }{\partial x}dx+f(x)dx$

The dx can be factored out and the final equation is shown below.

$\frac{\partial \sigma }{\partial x}+\frac{f(x)}{A}=0$

Derivation of Equation 4

Now, let's investigate the non-uniform stress field in 3-D, but without applied load and focusing on the x-direction only (i.e. w/o the other stress components to avoid cluttering the figure)

$\sum{F_{x}}=0=dydz\left[-\sigma _{xx}(x,y,z) + \sigma _{xx}(x+dx,y,z) \right]+dzdx\left[-\sigma _{yx}(x,y,z) + \sigma _{yx}(x,y+dy,z) \right]+dxdz\left[-\sigma _{zx}(x,y,z) + \sigma _{zx}(x,y,z+dz) \right]$

$0=dxdydz\left[\frac{\partial \sigma _{xx}}{\partial x} + \frac{\partial \sigma _{yx}}{\partial y} + \frac{\partial \sigma _{zx}}{\partial z}\right]$

$0=\frac{\partial \sigma _{xx}}{\partial x} + \frac{\partial \sigma _{yx}}{\partial y} + \frac{\partial \sigma _{zx}}{\partial z}$

 Note: $\sigma_{ij}$ is the normal stress normal to the i axis facet in the j axis direction.

$\Rightarrow \ Equation\ 3\ (and\ therefore\ 4)\ has\ been\ proved$

Prandtl Stress function Φ

The Prandtl stress function is a special case of the Morera stress functions, in which it is assumed that A=B=0 and C is a function of x and y only.

     $[\sigma_{yx}]=(\frac{\partial\Phi}{\partial z})$

      $[\sigma_{zx}]=(\frac{\partial\Phi}{\partial y})$

Φ acts as a potential function. ($\sigma_{yx}$, $\sigma_{zx}$) are components of Φ with respect to (y,z)

A brief mention should be made of the Maxwell stress functions. Maxwell stress functions are defined by assuming that the Beltrami stress tensor $\Phi_{mn}$ tensor is restricted to be of the form :

$\Phi_{ij}= \begin{bmatrix} A&0&0\\ 0&B&0\\ 0&0&C \end{bmatrix}$

Substituting the expressions for the stress into the Beltrami-Michell equations yields the expression of the elastostatic problem in terms of the stress functions:

$\nabla^2 A=\nabla^2 B=\nabla^2 C=\left( \frac{\partial^2 A}{\partial x^2}+ \frac{\partial^2 B}{\partial y^2}+ \frac{\partial^2 C}{\partial z^2}\right)/(2-\nu),$

These must also yield a stress tensor which obeys the specified boundary conditions.

The stress tensor which automatically obeys the equilibrium equation may now be written as:

$\sigma_x= \frac{\partial^2\Phi_{yy}}{\partial z \partial z} + \frac{\partial^2\Phi_{zz}}{\partial y \partial y} -2\frac{\partial^2\Phi_{yz}}{\partial y \partial z}$                

$\sigma_y= \frac{\partial^2\Phi_{xx}}{\partial z \partial z} +\frac{\partial^2\Phi_{zz}}{\partial x \partial x} -2\frac{\partial^2\Phi_{zx}}{\partial z \partial x}$                

$\sigma_z= \frac{\partial^2\Phi_{yy}}{\partial x \partial x} +\frac{\partial^2\Phi_{xx}}{\partial y \partial y} -2\frac{\partial^2\Phi_{xy}}{\partial x \partial y}$                

$\sigma_{xy}=-\frac{\partial^2\Phi_{xy}}{\partial z \partial z} -\frac{\partial^2\Phi_{zz}}{\partial x \partial y} +\frac{\partial^2\Phi_{yz}}{\partial x \partial y} +\frac{\partial^2\Phi_{zx}}{\partial y \partial z}$

$\sigma_{yz} =-\frac{\partial^2\Phi_{yz}}{\partial x \partial x} -\frac{\partial^2\Phi_{xx}}{\partial y \partial z} +\frac{\partial^2\Phi_{zx}}{\partial y \partial x} +\frac{\partial^2\Phi_{xy}}{\partial z \partial x}$

$\sigma_{zx}=-\frac{\partial^2\Phi_{zx}}{\partial y \partial y} -\frac{\partial^2\Phi_{yy}}{\partial z \partial x} +\frac{\partial^2\Phi_{xy}}{\partial z \partial y} +\frac{\partial^2\Phi_{yz}}{\partial x \partial y}$

References

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