User:Eas4200c.f08.blue.a/Homework6

Lecture Notes
Below are the lecture notes for HW #6. They are divided by date with a summary of each section at the top.

Intro To Torsional Analysis
This lecture covered deriving the shear stress matrices for a given load F.

$$ F = \frac{P}{A} $$

The shear stress can then be computed in terms of x,y,and z coordinates as well as the force F.

These results will be used in our discussion of torsional analysis.

Torsional Analysis 1
This lecture continued the discussion of torsional analysis that was done in previous lectures.

Continuing Torsional Analysis:

Dimensional analysis, from lecture 29, is applied to the differential stress formula from lecture 27 to show that the differential of stress over distance x has the dimensions of force over volume.

$$\left[\frac{\delta \sigma _{ij}}{\delta x_{i}} \right]=\frac{F}{L^{3}}=\frac{force}{volume}$$

In equation 3.14 in Mechanics of aircraft structures by C.T. Sun, it is shown that the of partial derivatives of shear force in the x and y directions is equal to zero.

$$\frac{\partial \tau _{xz}}{\partial x}+\frac{\partial \tau _{yz}}{\partial y}=\frac{\partial \sigma _{xz}}{\partial x}+\frac{\partial \sigma _{yz}}{\partial y}=0$$

From the roadmap described in lecture 16, the Prandtl stress function $$\phi $$ can be found using the following formulas (Sun 3.15):

$$\sigma _{yx}=\frac{\delta \phi }{\delta z}$$ $$\sigma _{zx}=\frac{-\delta \phi }{\delta y}$$

As can be seen, the Prandtl stress function plays the role of a potential function. $$\left(\sigma _{yx},\sigma _{zx} \right)$$ is a component of the "gradient" of the Prandtl function with respect to y and z.

Recall that the gradient of a scalar function f(x,y,z) is simply the partial differential of f in each direction (x, y, and z) or (i, j, and k).

$$\vec{\bigtriangledown} f(x,y,z)=\frac{\partial f}{\partial x}\hat{i}+\frac{\partial f}{\partial y}\hat{j}+\frac{\partial f}{\partial z}\hat{k} $$

In lecture 27, the class found that the sum of the partial derivative with respect to y of stress in the yx direction and the partial derivative with respect to z of the stress in the zx direction results in zero. Now we can replace the stresses with the Prandtl stress function equivalents.

$$\frac{\partial \sigma _{yx}}{\partial y}+\frac{\partial \sigma _{xz}}{\partial z}=0$$

$$\frac{\partial }{\partial y}\left(\frac{\partial \phi }{\partial z} \right)+\frac{\partial }{\partial z}\left(\frac{-\partial \phi }{\partial y} \right)=0$$

$$\frac{\partial ^{2}\phi }{\partial y\partial z}-\frac{\partial ^{2}\phi }{\partial z\partial y}=0$$

Since $$\phi$$ is continuous and smooth then the second mixed derivative is interchangable.

Now the Prandtl stress function $$\phi (x,y)$$ can be introduced in terms of the shear stress as in equation 3.15 (Sun, 70).

$$\tau _{xz}=\frac{\partial \phi }{\partial y}$$ $$\tau _{yz}=-\frac{\partial \phi }{\partial x}$$

It follows that:

$$\gamma _{xz}=\frac{\partial w}{\partial x}+\frac{\partial u}{\partial z}=\frac{\partial w}{\partial x}-\theta y$$

$$\gamma _{yz}=\frac{\partial w}{\partial y}+\frac{\partial v}{\partial z}=\frac{\partial w}{\partial y}+\theta x$$

$$\frac{\partial \gamma _{yz}}{\partial x}-\frac{\partial \gamma _{xz}}{\partial y}=2\theta $$

$$\gamma _{yz}=\frac{1}{G}\tau _{yz}$$

$$\gamma _{xz}=\frac{1}{G}\tau _{xz}$$

$$\frac{\partial \tau _{yz}}{\partial x}-\frac{\partial \tau _{xz}}{\partial y}=2G\theta $$

$$\frac{\partial ^{2}\phi }{\partial x^{2}}+\frac{\partial ^{2}\phi }{\partial y^{2}}=-2G\theta $$

$$\frac{\partial ^{2}\phi }{\partial y^{2}}-\frac{\partial ^{2}\phi }{\partial z^{2}}=-2G\theta=\bigtriangledown ^{2}\phi $$

This equation for the Prandtl Equation is equivalent to step F of the Torsional analysis Roadmap.

To find a function for the Prandtl Function, a two dimensional case is first looked at:



In the above image, a cross section of a non-uniform bar is shown with a force (t) pu.



This image represents the stress vector (t) on an infinitesimal surface of a non-uniform beam.

Torsional Analysis 2
This lecture covered how to obtain the stress matrices

$$\vec{t}=t_{y}\vec{j}+t_{z}\vec{k}$$

It can be infered that:

$$dz=dscos\theta $$  $$n_{y}=cos\theta $$

$$dy=dssin\theta $$  $$n_{z}=sin\theta $$

Summing the forces in the y directon:

$$\sum{F_{y}}=0=-\sigma _{yy}(dz)-\sigma _{yz}(dy)+t_{y}(ds)$$

and using the above relations:

$$t_{y}=\sigma _{yy}n_{y}+\sigma _{yz}n_{z}$$

Similarly,

$$\sum{F_{z}}=0=-\sigma _{zz}(dy)-\sigma _{yz}(dz)+t_{z}(ds)$$

and again, substituting, we find:

$$t_{z}=\sigma _{zz}n_{z}+\sigma _{yz}n_{y}$$

In matrix form this becomes:

$$\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}$$

Note: $$\sigma _{zy}= \sigma _{yz}$$ (because $$\sum{M_{x}}=\sum{M_{y}}=\sum{M_{z}}=0$$), but in order to keep with the unified notation, and to be compatible with the indicial notation, the difference is retained.

The generalized form is as follows:

$$\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}$$

ShorthandSymbollically it would written as

$$\begin{Bmatrix} t_{i} \end{Bmatrix}_{3x1}

=\begin{bmatrix} \sigma _{ij} \end{bmatrix}_{3x3}

\begin{Bmatrix} n_{j} \end{Bmatrix}_{3x1}$$

Plate Buckling 1
This lecture covered two cases of buckling for different cross sections.

Torsional Analogy: Continued from last lecture, there are two types of buckling

Plate Buckling 2
This lecture covered finding the shear stress in a thin walled cross section.

$$r=(y^2+z^2)^{1/2}$$

$$T=2\int \phi dA=2c(\frac{J}{a^2}-\pi a^2)$$

$$J=\int_A r^2 dA=\frac{1}{2}\pi a^4$$

$$T= GJ\theta$$

$$\sigma_{yz}=\frac{\delta \phi}{\delta z}=-G \theta_z$$

$$\sigma_{zy}=\frac{\delta \phi}{\delta y}= -G\theta_y$$

$$\tau=\frac{T r}{J}$$

$$u_x(x,y)=0$$

i,e no warping

since

$$\gamma_{yx}=\frac{\delta_{yx}}{G}=\frac{\delta_u}{\delta_y}-\theta_z$$

$$\gamma_{zx}=\frac{\delta_{zx}}{G}=\frac{\delta_u}{\delta_z}-\theta_y$$

$$\sigma_{yx}=-G\theta_z$$

$$\sigma_{zx}=G\theta_y$$

therefore

$$\delta u=(G+1)\theta_z\ \delta y=(G+1)\theta_y \delta z$$

since there is no warping

$$\delta y= /delta z = 0$$

therefore

$$u=o$$

Torsional Analysis and Plate Buckling Combined
This lecture finalizes the discussion of torsional analysis on a non uniform object.



The figure above shows an unsymmetric thin-walled section. $$\displaystyle y_c$$ and $$\displaystyle z_c$$ are measured from the centroid of the shaded area. As labeled in the figure, the thickness $$\displaystyle t_s$$ can be a function of s, the curvilinear coordinate.

A general equation for thin-walled sections is the following (5.1):
 * $$\iint_{A_s} \frac{d\sigma_{xx}}{dx} dA = -q_s$$

A comparison of equations for symmetric and unsymmetric sections: (Note: in the following symmetric cases, the section is symmetric about the y-axis)

Symmetric: $$\displaystyle \sigma_{xx} = \frac{M_yz}{I_y}$$

Unsymmetric: $$\displaystyle \sigma_{xx} = (k_yM_z - k_{yz}M_y)y + (k_zM_y - k_{yz}M_z)z$$

where
 * $$\displaystyle k_y = \frac {I_y}{D}$$
 * $$\displaystyle k_{yz} = \frac {I_{yz}}{D}$$
 * $$\displaystyle k_z = \frac {I_z}{D}$$
 * $$\displaystyle D = I_yI_z - I_{yz}^2$$

The above equation for an unsymmetric section can be rewritten in matrix form to become
 * $$\displaystyle \sigma_{xx} = \begin{bmatrix}

y & z \end{bmatrix} \begin{bmatrix} k_y & -k_{yz} \\ -k_{yz} & k_z \end{bmatrix} \begin{Bmatrix} M_z \\ M_y \end{Bmatrix} = \begin{bmatrix} z & y \end{bmatrix} \begin{bmatrix} k_z & -k_{yz} \\ -k_{yz} & k_y \end{bmatrix} \begin{Bmatrix} M_y \\ M_z \end{Bmatrix} $$

The equation for an unsymmetric section can also be particularized to become the equation for a symmetric section (considering $$\displaystyle M_z$$ to be equal to zero). With a symmetric section, $$\displaystyle I_{yz}$$ is equal to zero.


 * $$\displaystyle M_z = 0 $$
 * $$\displaystyle I_{yz} = 0 $$
 * $$\displaystyle\therefore D = I_yI_z$$
 * $$\displaystyle k_y = \frac {I_y}{I_yI_z} = \frac {1}{I_z}$$
 * $$\displaystyle k_{yz} = \frac {I_{yz}}{I_yI_z} = 0$$
 * $$\displaystyle k_z = \frac {I_z}{I_yI_z} = \frac {1}{I_y}$$
 * $$\displaystyle\therefore \sigma_{xx} = (k_yM_z - k_{yz}M_y)y + (k_zM_y - k_{yz}M_z)z = (0-0)y + (k_zM_y - 0)z = \frac{M_yz}{I_y}$$

Continuing with our comparisons of equations between symmetric and unsymmetric sections:

Symmetric: $$\displaystyle q(s) = -\frac {V_zQ_y}{I_y}$$
 * $$Q_y = \int_{A_s} z\,dA = z_cA_s$$

Unsymmetric: $$\displaystyle q(s) = -(k_yV_y - k_{yz}V_z)Q_z - (k_zV_z - k_{yz}V_y)Q_y$$
 * $$Q_y = \int_{A_s} z\,dA$$
 * $$Q_z = \int_{A_s} y\,dA$$

The above equation for an unsymmetric section can be rewritten in matrix form to become
 * $$\displaystyle q(s) = -\begin{bmatrix}

Q_z & Q_y \end{bmatrix} \begin{bmatrix} k_y & -k_{yz} \\ -k_{yz} & k_z \end{bmatrix} \begin{Bmatrix} V_y \\ V_z \end{Bmatrix} = -\begin{bmatrix} Q_y & Q_z \end{bmatrix} \begin{bmatrix} k_z & -k_{yz} \\ -k_{yz} & k_y \end{bmatrix} \begin{Bmatrix} V_z \\ V_y \end{Bmatrix} $$

As with the equation for $$\displaystyle \sigma_{xx}$$, the equation for $$\displaystyle q(s)$$ with an unsymmetric section can be particularized to become the equation with a symmetric section. Similar to the case for $$\displaystyle I_{yz}$$ proven above, $$\displaystyle Q_z$$ is equal to zero due to symmetry about the y-axis. Again, $$\displaystyle I_{yz}$$ is also equal to zero.
 * $$\displaystyle Q_z = \int_{A_s} y\,dA = 0$$
 * $$\displaystyle k_y = \frac {I_y}{I_yI_z} = \frac {1}{I_z}$$
 * $$\displaystyle k_{yz} = \frac {I_{yz}}{I_yI_z} = 0$$
 * $$\displaystyle k_z = \frac {I_z}{I_yI_z} = \frac {1}{I_y}$$
 * $$\displaystyle \therefore q(s) = -(k_yV_y - k_{yz}V_z)Q_z - (k_zV_z - k_{yz}V_y)Q_y = -(k_yV_y-0)0 - (k_zV_z - 0)Q_y = -\frac {V_zQ_y}{I_y}$$



When dealing with stringer-web sections, it is assumed that the thickness $$\displaystyle t$$ of the skin and spar web is very small, therefore their areas are neglected in the computation of $$\displaystyle I_y, I_z, I_{yz}, Q_y, Q_z$$. The only areas used are the areas of the stringers (as shown in Fig. 2 to the left).

With a section that is symmetric about the y-axis, the areas will be set such that $$\displaystyle A_3 = A_2$$ and $$\displaystyle A_4 = A_1$$. However, for the unsymmetric case, the areas will be set such that $$\displaystyle A_1 = A,\,\, A_2 = 2A,\,\, A_3 = 3A,\,\, A_4 = 4A$$.

Matlab Code and Results
The Matlab code for this homework covers a plate buckling analysis for the NACA Airfoil. First, the compressive stresses are computed in the X direction so they can be compared to the buckling stress. An analysis is done comparing the stress in each direction for a variety of aspect ratios. This leads to the final comparison of the compressive stresses to buckling stress.

Matlab Code
This code is always being modified to optimize its performance and meet the requirements of the class. As revisions are made, they will be posted.

The m-file used in this HW Report is called NACA_Airfoil_Buckling_Stress.

Note: The NACA_Airfoil_Generator m-file is needed when running the Bucking Stress Program.

Now, here is the m-file used to calculate the Plate Buckling Characteristics:

Team Members that Contributed to This Report
Mark Barry Eas4200c.f08.blue.a 20:20, 19 November 2008 (UTC)

Thomas Sexton EAS4200c.f08.blue.c 00:15, 21 November 2008 (UTC)

Christian Garabaya Eas4200c.f08.blue.d 01:32, 20 November 2008 (UTC)

Clay Robertson eas4200c.f08.blue.f 00:21, 21 November 2008 (UTC)

Jeffrey Shea Eas4200c.f08.blue.e 23:10, 19 November 2008 (UTC)

Marc Pelletier EAS4200c.f08.blue.b 15:27, 21 November 2008 (UTC)

Wei-Teck Lee Eas4200c.f08.blue.g 05:40, 21 November 2008 (UTC)