User:Eas4200c.f08.blue.a/Homework5

October 20, 2008
Reading page 70 in textbook

$$\left\{\epsilon_{ij} \right\}_{6x1} = \begin{bmatrix} \bar{A}_{3x3} & o\\ 0 & \bar{B}_{3x3} \end{bmatrix}_{6x6}\left\{\sigma_{ij} \right\}_{6x1}$$

The $$\sigma$$ - $$\epsilon$$ relationship: Assuming diagonal coefficient is not zero

$$\left\{\sigma_{ij} \right\}_{6x1} = \begin{bmatrix} \bar{A}^{-1}_{3x3} & 0\\ 0 & \bar{B}^{-1}_{3x3} \end{bmatrix}_{6x6}\left\{\epsilon_{ij} \right\}_{6x1}$$

If $$\begin{bmatrix} \bar{A}^{-1}_{3x3} & 0\\ 0 & \bar{B}^{-1}_{3x3} \end{bmatrix}_{6x6}$$ we can varify that the relationship by $$\bar{c}^{-1}\bar{c} = \bar{I}$$ then $$\bar{c}^{-1}\bar{c} = \begin{bmatrix} \bar{A}^{-1}\bar{A} & 0 \\ 0 & \bar{B}^{-1}\bar{B} \end{bmatrix} = \bar{I}$$ therefore,

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

Road map

Equilibrium Equation for Stress in a nonuniform stress field (section 2.4 in the book)

Consider a one dimensional case as a model.For a nonuniform field, the distribution is not constant, therefore $$f\left(x \right)\neq constant$$

and showing the shear forces acting on this



October 27, 2008
Bidirectional Bending



$$M_{y}=\int_{aA}^{}{z\sigma _{xx}dA}$$

$$M_{z}=\int_{aA}^{}{y\sigma _{xx}dA}$$

Moment of Inertia Tensors:

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

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

$$I_{yz}=\int_{A}^{}{zydA}$$

Recall that, by Hooke's Law :

$$\sigma _{xx}=E\varepsilon _{xx}$$

With the above equations this becomes:

$$\sigma _{xx}=(\frac{I_{yy}M_{z}-I_{yz}M_{y}}{I_{yy}I_{zz}-(I_{yz})^{2}})y$$+$$(\frac{I_{zz}M_{y}-I_{yz}M_{z}}{I_{yy}I_{zz}-(I_{yz})^{2}})z$$

Where the denominator of both terms is the determinant of bottom right corner of the Moment of inertia matrix (shown below):

$$I=\begin{bmatrix} I_{11} &I_{22} &I_{13} \\ I_{21}&I_{22} &I_{23} \\ I_{31}& I_{32} & I_{33} \end{bmatrix}$$

In addition, the location of the neutral axis can be found by setting $$\sigma _{xx}=0$$.

$$\sigma _{xx}=(\frac{I_{yy}M_{z}-I_{yz}M_{y}}{I_{yy}I_{zz}-(I_{yz})^{2}})y$$+$$(\frac{I_{zz}M_{y}-I_{yz}M_{z}}{I_{yy}I_{zz}-(I_{yz})^{2}})z=M_{yy}+M_{zz}=0$$

Therefore, $$z=(-\frac{M_{y}}{M_{z}})y=\tan (\beta) y$$

October 29, 2009
Continuing from the equation of equilibrium in terms of $$\bar{\sigma}$$

The goal is to find $$\frac{\partial \sigma_{yx}}{\partial y}+\frac{\partial \sigma_{zx}}{\partial z}=0$$ (equation 3.14 in textbook) Lets use indicial notation $$\frac{\partial \sigma_{21}}{\partial x_{2}}+\frac{\partial \sigma_{31}}{\partial x_{3}}=0$$

Recall from section 2.4 that

$$\frac{\partial \sigma_{xx}}{\partial x}+\frac{\partial \tau_{yx}}{\partial y}+\frac{\partial \tau_{zx}}{\partial z}=0$$ (equation 2.21 in textbook)

It follows, using the indicial notation that the three equilibrium equations become

$$\frac{\partial \sigma_{11}}{\partial x_{1}}+\frac{\partial \tau_{21}}{\partial x_{2}}+\frac{\partial \tau_{31}}{\partial x_{3}}=0$$ (equation 2.21 in textbook)

$$\frac{\partial \tau_{12}}{\partial x_{1}}+\frac{\partial \sigma_{22}}{\partial x_{2}}+\frac{\partial \tau_{32}}{\partial x_{3}}=0$$ (equation 2.22 in textbook)

$$\frac{\partial \tau_{13}}{\partial x_{1}}+\frac{\partial \tau_{23}}{\partial x_{2}}+\frac{\partial \sigma_{33}}{\partial x_{3}}=0$$ (equation 2.23 in textbook)

Therefore the summation notation of the equilibrium equations is as follows

$$\sum_{i=1}^{3}{\frac{\partial \sigma_{ij}}{\partial x_{i}}}=0$$ for $$j=1,2,3$$

October 31, 2008
Goal to determine the four zero stress components.

Obtain the equations resulting from

$$\sum_{i=1,j=1}^3\frac{\delta \sigma_{ij}}{\delta x_i}=0$$

which would result in

$$\frac{\delta \sigma_{11}}{\delta x_{1}}+\frac{\delta \sigma_{12}}{\delta x_{2}}+\frac{\delta \sigma_{13}}{\delta x_{3}}=0$$

$$\frac{\delta \sigma_{21}}{\delta x_{1}}+\frac{\delta \sigma_{22}}{\delta x_{2}}+\frac{\delta \sigma_{23}}{\delta x_{3}}=0$$

$$\frac{\delta \sigma_{31}}{\delta x_{1}}+\frac{\delta \sigma_{32}}{\delta x_{2}}+\frac{\delta \sigma_{33}}{\delta x_{3}}=0$$

One Dimensional model problem:

$$\sum F_2=0=-\sigma_w A+\sigma_{x+dx}A+f(x)dx$$

$$0=A[\sigma_{x+dx}-\sigma_x]+f(x)dx$$

Taylor series expansion $$=\frac{d \sigma}{dx}+$$ higher order terms

Now neglecting the higher order terms

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

So for all the forces in the X direction the sum is

$$\sum Fx =0= dy dz [-\sigma_{xx}(x,y,z)+\sigma_{xx}(x+dx),y,z)]+dzdx[-\sigma_{yx}(x,y,z)+\sigma_{yx}(x,y+dy,z)]+dxdy[\sigma_{zx}(x,y,z)+\sigma_{zx}(x,y,z+dz)]=0$$

which will yield

$$0=dx dy dz (\frac{\delta \sigma_{xx}}{\delta_x}+\frac{\delta \sigma_{xy}}{\delta_y}+\frac{\delta \sigma_{xz}}{\delta_z})$$

So for all the forces in the Y direction the sum is

$$\sum Fy =0= dy dz [-\sigma_{yx}(x,y,z)+\sigma_{yx}(x+dx),y,z)]+dzdx[-\sigma_{yy}(x,y,z)+\sigma_{yy}(x,y+dy,z)]+dxdy[\sigma_{yz}(x,y,z)+\sigma_{yz}(x,y,z+dz)]=0$$

which will yield

$$0=dx dy dz (\frac{\delta \sigma_{yx}}{\delta_x}+\frac{\delta \sigma_{yy}}{\delta_y}+\frac{\delta \sigma_{yz}}{\delta_z})$$

So for all the forces in the Z direction the sum is

$$\sum Fy =0= dy dz [-\sigma_{zx}(x,y,z)+\sigma_{zx}(x+dx),y,z)]+dzdx[-\sigma_{zy}(x,y,z)+\sigma_{zy}(x,y+dy,z)]+dxdy[\sigma_{zz}(x,y,z)+\sigma_{zz}(x,y,z+dz)]=0$$

which will yield

$$0=dx dy dz (\frac{\delta \sigma_{zx}}{\delta_x}+\frac{\delta \sigma_{zy}}{\delta_y}+\frac{\delta \sigma_{zz}}{\delta_z})$$

NACA Airfoil Matlab Code
Note 1: The code here uses x and y as the reference axes( as used in the textbook as opposed to y and z).

Note 2: The bending moment code is still being worked on at this time. What has been developed to this point is included and will be fully functional in a few days.

Note 3. Initially, i had planned on only having 1 m-file, but after adding more functionality to this project, i have split everything up into individual m-files ( for easier understanding).



Team Members That Contributed To This Report
Mark Barry Eas4200c.f08.blue.a 14:16, 5 November 2008 (UTC)

Clay Robertson eas4200c.f08.blue.f 13:42, 6 November 2008 (UTC)

Marc Pelletier EAS4200c.f08.blue.b 05:18, 7 November 2008 (UTC)

Jeffrey Shea EAS4200c.f08.Blue.E 19:24, 6 November 2008 (UTC)

Christian Garabaya Eas4200c.f08.blue.d 08:25, 7 November 2008 (UTC)

Tom Sexton EAS4200c.f08.blue.c 22:27, 6 November 2008 (UTC)