# User:Eas4200c.f08.aeris.krammer/HW5

Eml4500.f08 15:48, 11 November 2008 (UTC)

- Finished the matlab problem

Eas4200c.f08.aeris.holladay 03:03, 21 November 2008 (UTC)

## Strain Displacement Relationship and Equilibrium Equations (Stresses)

With the theory of elasticity, we first considered the kinematic assumptions (21.3) which resulted in realizing that the four strain components in the xx, yy, zz, and yz directions were equal to zero (21.3)and consequently the stresses in these directions were zero (22.4). Comparing this to the equations in page 70 of the book, we find that since the book has used the convention of an (x,y) coordinate plane in order to analyze the strain and stress, however we used a (y,z) coordinate system. These expressions are easily equated by using a cyclic permutations of the variables. Then, the stress strain relationship (tensorial or engineering) were established (23.2). Now, we will rewrite the stress strain relationship as shown below.

$[\varepsilon_{ij}]_{6x1}= \begin{bmatrix} A_{3x3} & 0_{3x3}\\ 0_{3x3} & B_{3x3}\\ \end{bmatrix}_{6x6} [\sigma_{ij}]_{6x1}= C[\sigma_{ij}]_{6x1}$

Where A and B are matrices with the values shown below and B is a diagonal matrix. The 6x6 block diagonal matrix will be referred to as the C matrix.

$A= \begin{bmatrix} \frac{1}{E} &\frac{-\nu }{E} &\frac{-\nu}{E} \\ \frac{-\nu }{E} &\frac{1}{E} &\frac{-\nu }{E} \\ \frac{-\nu }{E} &\frac{-\nu }{E} &\frac{1}{E} \\ \end{bmatrix}, B= \begin{bmatrix} \frac{1}{2G} &0 &0 \\ 0 &\frac{1}{2G} &0 \\ 0 &0 &\frac{1}{2G} \end{bmatrix}$

However, we will keep the C in the block diagonal form. To solve for the stress, we multiply both sides by the inverse of C. To invert a diagonal matrix, one must take the inverses of each value of the diagonal. It is the same for a block diagonal matrix as seen below. Invert C by inverting the diagonal matrices.

$[\sigma_{ij}]_{6x1}= \begin{bmatrix} A_{3x3}^{-1} & 0_{3x3}\\ 0_{3x3} & B_{3x3}^{-1}\\ \end{bmatrix}_{6x6} [\varepsilon_{ij}]_{6x1}= C^{-1}[\varepsilon_{ij}]_{6x1}$

$C^{-1}C= \begin{bmatrix} AA^{-1} & 0\\ 0 & BB^{-1}\\ \end{bmatrix} =I$

Where I is the identity matrix, a diagonal matrix with the diagonal values all equaling one. Then, we will continue on proving $\displaystyle \sigma_{xx}=\sigma_{yy}=\sigma_{zz}=\tau_{yz}=0$. First, we will apply the zero strain relations in which the strains in the xx, yy, zz, and yz directions are zero. Knowing that there are essentially six strain components since the strain tensor is a 3 by 3 symmetrical matrix, a 6 by 1 column matrix can be constructed where the first four rows are zero and the last two are non zero. Subscripts x, y, and z will be replaced with indices 1, 2, and 3 respectively. This gives the following expression.

$[\sigma_{ij}]= \begin{bmatrix} A^{-1} & 0\\ 0 & B^{-1}\\ \end{bmatrix} \begin{bmatrix} 0\\ 0\\ 0\\ 0\\ \varepsilon _{31}\\ \varepsilon _{12}\\ \end{bmatrix} = \begin{bmatrix} 0\\ 0\\ 0\\ 0\\ \sigma _{31}\\ \sigma _{12}\\ \end{bmatrix}$

Multiplying the top three rows of the 6 x 6 matrix to the top three rows of the strain tensor results in $\displaystyle \sigma_{11}=\sigma_{22}=\sigma_{33}$. The resulting last three rows are derived in the collapsible box below.

(16.2) Recall the road map step: equilibrium equation for stresses (section 4.2) in a nonuniform stress field. To understand the nonuniform stress state, let's consider a 1 dimensional case (a cantilever beam) as a model:

It is necessary to point out that the axial stress in the uniform case is the same for every value of x (i.e. $\displaystyle \sigma(x)=\sigma(x+dx)$) and F(x)=0. For the nonuniform case, axial stress DOES vary with x (i.e. $\sigma(x)\ne\sigma(x+dx)$) and F(x) is either a nonuniform function or constant. We will use this concept to derive the equilibrium equations for the stresses.

## Bidirectional Bending

Let us consider a bar with a noncircular cross section.
$\displaystyle Px(x)$ and $\displaystyle Py(x)$ are distributed loadings on the bar, $\displaystyle dA=dyzdz$ is an infinitely small area and $\displaystyle \sigma _{xx}$ is the normal bending stress.

The bending moments about the y and z axes are

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

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

The moment of inertia tensors are shown also in indicial notations

$\displaystyle I_{y}=I_{yy}=I_{22}$

$\displaystyle I_{z}=I_{zz}=I_{33}$

$\displaystyle I_{yz}=I_{23}$

And they are defined as

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

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

$I_{yz}=\int_{A}^{}{}\int y zdA$

Examining Hooke's Law we can see that

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

Where the denominator of the equation that is the determinant of the matrix

$\displaystyle D:= {I_{y}I_{z}-(I_{yz})^{2}} = {I_{22}{I_33}-(I_{23})^{2}}$

Is the determinant of

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

At the neutral axis the bending moment $\displaystyle \sigma _{xx}=0$

$\displaystyle\sigma _{xx}= m_{y}y+m_{z}z=0$

Solving the above equation, we get

$\displaystyle z=(-\frac{m_{y}}{m_{z}})y$

The above equation is the equation of a straight line.
Where $\displaystyle (-\frac{m_{y}}{m_{z}})= tan\beta$ is the slope of the equation.

Because the distribution of bending stresses is linear over this cross-section, it can be said that it is also linear with respect to the neutral axis. This means that the further away from the neutral axis we go, greater bending stress we experience.

## Zero Stress Components

Using indicial notation for $i = 1,2,3$ $\sum_{j=1}^{3}{\frac{\delta \sigma _{ij}}{\delta x_{j}}}=0$, the equations expanded out look like the following:

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

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

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

Analyzing the forces for an infinitesimally small square, like the following:

the sum of the forces, $\sum{F}=0=-\sigma _{x}A+\sigma _{x+\delta x}A+f(x)\delta x$.

This equation can also be written as follows:

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

after neglecting the higher order terms of its Taylor series expansion.

Now, summing all the forces in the x-direction:

$\sum Fx=[-\sigma_{xx}(x,y,z)+\sigma_{xx}(x+dx),y,z)]dydz+[-\sigma_{xy}(x,y,z)+\sigma_{xy}(x,y+dy,z)]dxdz+[\sigma_{xz}(x,y,z)+\sigma_{xz}(x,y,z+dz)]dxdy=0$

Summimg all the forces in the y-direction:

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

Summimg all the forces in the z-direction:

$\sum Fz=[-\sigma_{xz}(x,y,z)+\sigma_{xz}(x+dx),y,z)]dydz+[-\sigma_{yz}(x,y,z)+\sigma_{yz}(x,y+dy,z)]dxdz+[\sigma_{zz}(x,y,z)+\sigma_{zz}(x,y,z+dz)]dxdy=0$

The sum of the forces in each direction are also written as:

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

$(\frac{\delta \sigma_{xy}}{\delta_x}+\frac{\delta \sigma_{yy}}{\delta_y}+\frac{\delta \sigma_{yz}}{\delta_z})dxdydz=0$

$(\frac{\delta \sigma_{xz}}{\delta_x}+\frac{\delta \sigma_{yz}}{\delta_y}+\frac{\delta \sigma_{zz}}{\delta_z})dxdydz=0$

respectively.

## Equilibrium Equations for a non-symmetrical beamshape

The equilibrium equations are broken down into the x, y, and z directions of forces (i.e. ΣFx, ΣFy, and ΣFz respectively). In the x-direction, the ultimate goal equilibrium equation is:

$\frac{\delta \sigma_{yx}}{\delta_y}+\frac{\delta \sigma_{zx}}{\delta_z}=0$

Changing these terms into their indicial notation brings about $\frac{\delta \sigma_{21}}{\delta x_{2}}+\frac{\delta \sigma_{31}}{\delta x_{3}}=0$

Recalling that $\frac{\delta \sigma_{xx}}{\delta_x}+\frac{\delta \sigma_{yx}}{\delta_y}+\frac{\delta \sigma_{zx}}{\delta_z}=0$ we can also write the similarity equations for ΣFy, and ΣFz as

$\frac{\delta \sigma_{xy}}{\delta_x}+\frac{\delta \sigma_{yy}}{\delta_y}+\frac{\delta \sigma_{zy}}{\delta_z}=0$ and $\frac{\delta \sigma_{xz}}{\delta_x}+\frac{\delta \sigma_{yz}}{\delta_y}+\frac{\delta \sigma_{zz}}{\delta_z}=0$ respectively.

The conversion to indicial notation for each would be as follows:

$\frac{\delta \sigma_{12}}{\delta y_{2}}+\frac{\delta \sigma_{22}}{\delta y_{2}}+\frac{\delta \sigma_{32}}{\delta y_{2}} =0$ and $\frac{\delta \sigma_{13}}{\delta z_{2}}+\frac{\delta \sigma_{23}}{\delta z_{2}}+\frac{\delta \sigma_{33}}{\delta z_{2}} =0$ respectively.

## HW5 MatLAB Problem Statement

In this assignment we will modify the MatLAB code from the previous homework to include stringers attaching the spars to the flange, as can be seen in the figure below.

Given an input NACA airfoil (NACA 2415 shown in the above figure), we are to compute the centroid of the stringers (as well as the airfoil), the components of the moment of inertia tensor, the highest normal bending stress and location of this stress, and the ultimate bending moments of the airfoil.

We first compute this data neglecting the bending effects of the skin and the partition walls (spar webs). We are then to recompute the data, only this time NOT neglecting the bending effects of the skin or partition walls (or the stringers themselves). A comparison of the difference will then be made so we can access the contribution of these bending effects.

In order to make these calculations, we need to know some initial data on the stringers to be used. We first make an assumption on the bending moments, shown below.

$\ M_{y}=-125,000 N*cm$

$\ M_{x}=50,000 N*cm$

Airfoil with select points labeled (for reference).

We also need to know the cross sectional area of these stringers, which is given below.

$\ A_{B}=2 cm^{2}$

$\ A_{E}=2 cm^{2}$

$\ A_{H}=1 cm^{2}$

$\ A_{F}=1 cm^{2}$

For the second part of our calculations, we are to take into account the bending affects of the skin, partition walls, and stringers themselves. We can note that the stringers are essentially stamped out sheet metal which is then bent into the L-shape as seen in the first figure. Because of this, thickness of the stringers are uniform, and we let this thickness equal 5 mm, or 0.005 m. Knowing this, we can compute the width of the metal sheet used to manufacture the stringers, as seen below.

$\ \frac{\frac{0.0002}{2}}{0.5}=0.02m=2cm$

Because the thin skin would buckle under high normal bending stress acting on the plane of the skin, we will next find the highest normal bending stress in any of the stringers, and take a closer look at that stringer. Assuming the ratio of My and Mz are constant, we will then set this maximum normal bending stress to the ultimate stress for our material, steel 300M. Upon doing this, we are now able to compute the ultimate bending moments My and Mz for our full blown airfoil (which we will be imputing as the NACA 2415 airfoil).

The output from the program can be seen below. The dashed line is the neutral axis.
$\bar{M_{y,ult}}=2.7368*10^4 Nm$     $\bar{M_{z,ult}}=-1.0947*10^4 Nm$
For part 2 we take into account the spars and the skin in the stress analysis. Inherently we would assume a larger moments of inertia, slightly different centroid, and a larger maximum bending moment for steel.
$\bar{M_{y,ult}}=9.7364*10^4 Nm$     $\bar{M_{z,ult}}=-3.8946*10^4 Nm$
As expected, the moments of inertia increased. The point of maximum stress remained the same and the ultimate bending moments more than tripled.

## Contributions

Jin Yu Guan Eas4200c.f08.aeris.guan 05:30, 7 November 2008 (UTC)
Nelson Caceres Eas4200c.f08.aeris.caceres 05:40, 7 November 2008 (UTC)
Anett Krammer Eas4200c.f08.aeris.krammer 19:27, 7 November 2008 (UTC)