User:EAS4200C.F08.WIKI.A/Homework5

=Completing the Stress Strain Relation =

kinematic assumption drop down

4 zero strain components

5 zero stress components

$$\varepsilon -\sigma $$ relation in two forms (Tensorial 23-2 or Engineering 23-1b)

Rewrite $$\varepsilon -\sigma $$ relation: Full (not diagonal)

$$\begin{Bmatrix} \varepsilon _{i,j} \end{Bmatrix}_{6x1}= \begin{vmatrix} \mathbf{A}_{3x3} & \mathbf{0}_{3x3}\\ \mathbf{0}_{3x3} & \mathbf{B}_{3x3} \end{vmatrix} \begin{Bmatrix} \sigma _{i,j} \end{Bmatrix}_{6x1}$$

$$\varepsilon -\sigma $$ relation

$$\begin{Bmatrix} \sigma _{i,j} \end{Bmatrix}= \begin{vmatrix} \mathbf{A}_{3x3}^{-1} & \mathbf{0}_{3x3}\\ \mathbf{0}_{3x3} & \mathbf{B}_{3x3}^{-1} \end{vmatrix} \begin{Bmatrix} \varepsilon _{i,j} \end{Bmatrix}$$

$$\mathbf{C}_{6x6}= \begin{vmatrix} \mathbf{A}_{3x3} & \mathbf{0}_{3x3}\\ \mathbf{0}_{3x3} & \mathbf{B}_{3x3} \end{vmatrix}$$

$$ \mathbf{C}_{6x6}\cdot \mathbf{C}^{-1}_{6x6}=\mathbf{I}_{6x6}$$

$$\mathbf{C}_{6x6}\cdot \mathbf{C}^{-1}_{6x6}=\begin{vmatrix} \mathbf{A_{3x3}A}_{3x3}^{-1} & \mathbf{0}_{3x3}\\ \mathbf{0}_{3x3} & \mathbf{B_{3x3}B}_{3x3}^{-1} \end{vmatrix}=\mathbf{I_{6x6}} $$

We are trying to prove that $$ \sigma _{xx}=\sigma _{yy}=\sigma _{zz}=\gamma _{yz}=0$$

$$\begin{Bmatrix} \sigma _{i,j} \end{Bmatrix}=

\begin{vmatrix} \mathbf{A}_{3x3}^{-1} & \mathbf{0}_{3x3}\\ \mathbf{0}_{3x3} & \mathbf{B}_{3x3}^{-1} \end{vmatrix}\begin{Bmatrix} 0\\ 0\\ 0\\ 0\\ \varepsilon _{31}\\ \varepsilon _{12} \end{Bmatrix}$$ $$\begin{matrix} \sigma _{11}=0\\ \sigma _{22}=0\\ \sigma _{33}=0\\ \sigma _{23}=2G\varepsilon _{23}\\ \sigma _{31}=2G\varepsilon _{31}\\ \sigma _{21}=2G\varepsilon _{12} \end{matrix}$$

Consider the 1D stress model



=Bidirectional Bending Recipe=



The following is a general recipe of equations used for solving a bidirectional bending problem.

The moment in the y-direction is

 $$\displaystyle M_y = \int_A z \sigma_{xx} dA $$

Similarly, the moment in the z-direction is

 $$\displaystyle M_z = \int_A y \sigma_{xx} dA $$

Where $$\displaystyle dA = dydz $$

The moment of intertia in the y-direction is

 $$\displaystyle I_y = \int_A z^2 dA $$

Similarly, the moment of intertia in the z-direction is

 $$\displaystyle I_z = \int_A y^2 dA $$

 $$\displaystyle I_{yz} = \int_A yz dA $$

The stress in the x-direction is

 $$\displaystyle \sigma_{xx} = E \epsilon_{xx} = {I_y M_z - I_{yz} M_y \over I_y I_z - (I_{yz})^2} y + {I_z M_y - I_{yz} M_z \over I_y I_z - (I_{yz})^2} z $$

Recall that

 $$\displaystyle I = \begin{bmatrix} I_{11} & I_{12} & I_{13} \\ I_{21} & I_{22} & I_{23} \\ I_{31} & I_{32} & I_{33} \end{bmatrix}$$

Note that the determinat is defined as

 $$\displaystyle D := I_y I_z - (I_{yz})^2 = I_{22} I_{33} - (I_{23})^2 $$

for $$\displaystyle \begin{bmatrix} I_{22} & I_{23} \\ I_{32} & I_{33} \end{bmatrix}$$

 $$\displaystyle I_{32} = I_{zy} = \int_A zy dA = I_{yz} = I_{23} $$

At the neutral axis, or where $$\displaystyle \sigma_{xx} = 0 $$

 $$\displaystyle \sigma_{xx} = m_y y + m_z z $$

Thus,

 $$\displaystyle z = (- {m_y \over m_z})y = (\tan \beta) y $$



=Elastic Bar Analysis Continued=

We will now continue the equation of equilibrium in terms of $$\sigma$$. The goal is to obtain the following formula.

 $$\displaystyle \frac{\partial \sigma_{yx}}{\partial y} + \frac{\partial \sigma _{zx}}{\partial z} = 0$$

In indicial notation the equation can be represented as follows;

 $$\displaystyle \frac{\partial \sigma_{21}}{\partial x_2} + \frac{\partial \sigma _{31}}{\partial x_3} = 0$$

Now we must recall equation 2.21 from the text which is the above equation but with the term $$\displaystyle \frac{\partial \sigma_{xx}}{\partial x}$$ added in.

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

This equation represented in indicial notation is represented as the following

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

All three cases ($$j= 1,2,3$$) are shown below.

$$\displaystyle j=1, \frac{\partial \sigma_{11}}{\partial x_1} +  \frac{\partial \sigma_{21}}{\partial x_2} + \frac{\partial \sigma_{31}}{\partial x_3}$$ <br \> <p style="text-align:center;">$$\displaystyle j=2, \frac{\partial \sigma_{12}}{\partial x_1}  + \frac{\partial \sigma_{22}}{\partial x_2}  +  \frac{\partial \sigma_{32}}{\partial x_3}$$ <br \> <p style="text-align:center;">$$\displaystyle j=3, \frac{\partial \sigma_{13}}{\partial x_1}  +  \frac{\partial \sigma_{23}}{\partial x_2}  +  \frac{\partial \sigma_{33}}{\partial x_3}$$

Noting that there are four zer stress components that come from the stress-strain relation, these equations reduce to the following.

<p style="text-align:center;">$$\displaystyle 0=\sigma_{11}=\sigma_{22}=\sigma_{33}=\sigma_{23}$$ <br \> <p style="text-align:center;">$$\displaystyle j=1, \frac{\partial \sigma_{21}}{\partial x_2} + \frac{\partial \sigma_{31}}{\partial x_3}$$ <br \> <p style="text-align:center;">$$\displaystyle j=2, \frac{\partial \sigma_{12}}{\partial x_1} +  \frac{\partial \sigma_{32}}{\partial x_3}$$ <br \> <p style="text-align:center;">$$\displaystyle j=3, \frac{\partial \sigma_{13}}{\partial x_1} + \frac{\partial \sigma_{23}}{\partial x_2}$$

We will now derive the equation of equilibrium in terms of $$\sigma$$. To do so we will call upon the one dimentional model problem that we previously looked at. For convenience it is depicted below.

[\Image:]

First we must sum the forces in the x direction. doing so results in the following equation.

<p style="text-align:center;">$$\sum{F_x} = 0 = -\sigma_{x}A + \sigma(x+dx)A + f(x)dx = A\left[\sigma(x+dx) - \sigma(x) \right] + f(x)dx$$

Taking the Taylor Series expansion of the part of the equation in the brackets (mutliplied by A) and ignoring the higher order terms yields the following.

<p style="text-align:center;"> $$\displaystyle f(x+dx) = f(x) + \frac{df(x)}{dx}dx + \frac{1}{2}\frac{d^2f}{dx^2}xdx^2....$$    Taylor Series Expansion <br \> <p style="text-align:center;">$$\displaystyle \frac{d\sigma}{dx} + \frac{f(x)}{A} = 0....$$   Result Ignoring Higher Order Terms


 * A is the applied load - Body force = force/volume
 * f(x) = force/length

Now, non-uniform stress field is in 3-D, but without an applied load, focusing on the x-direction only (i.e. without the other stress component to avoid cluttering the figure)

Below is an image of shear on a 3-D cube :



Note: the first part of the superscript refers to the normal direction of the cube face (facet), and the second part refers to the direction of shear.

$$\sum{F_x} = {\color{red}dydz[-\sigma_{xx}(x,y,z) + \sigma_{xx}(x+dx,y,z)]} + {\color{blue}dzdx[-\sigma_{yx}(x,y,z) + \sigma_{yx}(x,y+dy,z)]} + {\color{green}dxdy[-\sigma_{zx}(x,y,z) + \sigma{zx}(x,y,z+dz)]}$$
 * The red term has facets normal to the x-axis, the blue has facets normal to the y-axis, and the green has facets normal to the z-axis.

$$0 = $$( dxdydz )$$[\frac{d\sigma_{xx}}{dx} + \frac{d\sigma_{yx}}{dy} + \frac{d\sigma_{zx}}{dz}]$$

This ends Step C of the Road Map.

=Matlab Analysis= <p style="text-align:center;">

This picture represents the NACA airfoil with stringers attached. The analysis is done below.

Matlab Code
The Matlab Code used in the following analysis is provided below.

'Part 1' is commented out when doing the analysis for 'Part 2' and vice versa.

Stringer Analysis
The above shows the diagram of the NACA2415 airoil with only the stringers considered in the analysis and ignoring the contributions of the skin. The important results are given below.

I_22 =

3.5173e-005

I_23 =

9.5347e-007

I_33 =

6.6954e-007

beta =

1.5010

maxsigmaxx_ylocation =

0.4999

maxsigmaxx_zlocation =

-7.8576e-004

maxsigma_xx =

4.1620e+008

M_zult =

2.1625e+005

M_yult =

-8.6498e+004

Full NACA2415 Airfoil Analysis
The diagram above shows the centroid location and neutral axis of the airfoil when the skin is considered along with the stringers. The important results are as follows:

I_22combined =

3.5628e-005

I_23combined =

7.9850e-006

I_33combined =

1.2106e-004

beta =

0.1771

maxsigmaxx_ylocation =

0.4999

maxsigmaxx_zlocation =

-7.8576e-004

maxsigma_xx =

4.1620e+008

A close up of the difference in the centroids of the stringers and the total airfoil is provided below for easier analysis.

As seen, the location and magnitude of the max normal stress barely changes, which is expected due to the centroid barely changing with the addition of the skin to the analysis.

The highest bending stress is in a stringer occurs in stringer H due to its distance from the neutral axis. It's value is

maxsigma_xx1 =

3.3823e+008

This causes the ultimate bending moments experienced by the stringers to be:

M_zult =

3.1349e+003

M_yult =

-1.2539e+003

=Individual Contributions=

Christopher Pergola EAS4200C.F08.WIKI.A 20:06, 7 November 2008 (UTC)

Braden Barnes Eas4200c.f08.wiki.b 17:53, 4 November 2008 (UTC)

Eric Viale Eas4200c.f08.wiki.c 17:19, 7 November 2008 (UTC)

Jeff Huenink Eas4200c.f08.wiki.f 19:30, 7 November 2008 (UTC)

Chris Fontana Eas4200c.f08.WIKI.E 20:33, 7 November 2008 (UTC)

Michael Rolle Eas4200c.f08.wiki.d 21:36, 7 November 2008 (UTC)