User:Eas4200c.f08.aero.lee/HW5

= Stress - Strain Relation =

To obtain engineering stress results based on an assessment of the engineering strains, we analyze the previously given strain-stress relation equations represented in matrix form as:



\begin{Bmatrix} \epsilon _{xx}\\ \epsilon _{yy}\\ \epsilon _{zz}\\ \gamma _{yz}\\ \gamma _{zx}\\ \gamma _{xy}\\ \end{Bmatrix}=

\begin{bmatrix} \frac{1}{E}& \frac{-\nu }{E} & \frac{-\nu }{E} & 0 & 0 &0 \\ \frac{-\nu }{E}& \frac{1}{E} & \frac{-\nu }{E} & 0 &  0&0 \\ \frac{-\nu }{E}& \frac{-\nu }{E} &   \frac{1}{E}& 0 & 0 &0 \\ 0 & 0 & 0 & \frac{1}{G} & 0 & 0\\ 0 & 0 & 0 & 0 & \frac{1}{G}& 0\\ 0& 0 &0 &0 & 0 & \frac{1}{G} \end{bmatrix} \begin{Bmatrix} \sigma _{xx}\\ \sigma _{yy}\\ \sigma _{zz}\\ \sigma _{yz}\\ \sigma _{zx}\\ \sigma _{xy}\\ \end{Bmatrix} $$

The above matrices can be rearranged to solve directly for stress in terms of strain by inverting the 6x6 coefficient matrix. To do so, consider the 6x6 matrix as a composition of 3x3 matrices, with the upper-right and lower-left quadrants identical to the zero matrix. Such a composition could be represented as:



\begin{Bmatrix} \epsilon _{xx}\\ \epsilon _{yy}\\ \epsilon _{zz}\\ \gamma _{yz}\\ \gamma _{zx}\\ \gamma _{xy}\\ \end{Bmatrix}=

\begin{bmatrix} \bar{A} & \bar{0} \\ \bar{0} & \bar{B} \\ \end{bmatrix} \begin{Bmatrix} \sigma _{xx}\\ \sigma _{yy}\\ \sigma _{zz}\\ \sigma _{yz}\\ \sigma _{zx}\\ \sigma _{xy}\\ \end{Bmatrix} $$

where the matrices $$\bar{A}$$ and $$\bar{B}$$ are given as



\begin{Bmatrix} \bar{A}\\ \end{Bmatrix}=

\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}

\begin{Bmatrix} \bar{B}\\ \end{Bmatrix}=

\begin{bmatrix} \frac{1}{G} & 0 & 0\\ 0 & \frac{1}{G}& 0\\ 0 & 0 & \frac{1}{G} \end{bmatrix} $$

Then, the stress-strain relation can be determined by inverting the coefficient matrix.



\begin{Bmatrix} \sigma _{xx}\\ \sigma _{yy}\\ \sigma _{zz}\\ \sigma _{yz}\\ \sigma _{zx}\\ \sigma _{xy}\\ \end{Bmatrix}=

\begin{bmatrix} \bar{A}^{-1} & \bar{0} \\ \bar{0} & \bar{B}^{-1} \\ \end{bmatrix} \begin{Bmatrix} \epsilon _{xx}\\ \epsilon _{yy}\\ \epsilon _{zz}\\ \gamma _{yz}\\ \gamma _{zx}\\ \gamma _{xy}\\ \end{Bmatrix} $$

This solution is possible because of the structure of the coefficient matrix. The inversion of the coefficient matrix is similar to that of a diagonal matrix, whose inverse is solved by inverting each component of the diagonal (assuming nonzero components) while leaving the zero terms off-diagonal ignored. A confirmation of this solution can be found by premultiplying the $$ \bar{A} $$ and $$ \bar{B} $$ matrices by their inverses within the coefficient matrix and obtaining the Identity Matrix, $$ \bar{I} $$.



\begin{bmatrix} \bar{A}^{-1}\bar{A} & \bar{0} \\ \bar{0} & \bar{B}^{-1}\bar{B} \\ \end{bmatrix}=

\begin{Bmatrix} \bar{I}\\ \end{Bmatrix} $$

It is seen by this solution, since $$ \bar{B} $$ is a diagonal matrix, that for strains $$ \epsilon_{xx}, \epsilon_{yy}, \epsilon_{zz}, \gamma_{yz} $$, the corresponding stress components (i.e. $$ \sigma_{xx}, \sigma_{yy}, \sigma_{zz}, \sigma_{yz} $$) are identically zero. Thus the stress-strain relation reduces to



\begin{Bmatrix} \sigma _{zx}\\ \sigma _{xy}\\ \end{Bmatrix}=

\begin{bmatrix} G & 0 \\ 0 & G \\ \end{bmatrix} \begin{Bmatrix} \gamma _{zx}\\ \gamma _{xy}\\ \end{Bmatrix} $$

= Equilibrium Equation for Stresses in a Non-uniform Stress Field =

Consider a 1-D case as a model.



It is apparent from the uniform case that the axial loading is constant along the length of the member. Conversely, in the non-uniform case, the axial loading is non-constant along the length of the member. This results in a non-uniform stress and strain response of the member.

= Bidirectional Bending Recipe = The image below depicts a non-uniform bar under bidirectional bending



Looking at Figure 2, for this case, the bending moments about the y and z axis, respectively, can be expressed as follows:


 * $$M_y = \int_A z\sigma_{xx}\, dA \, \, =\, \int\int z \sigma_{xx} dydz$$


 * $$M_z = \int_A y\sigma_{xx}\, dA \, \, =\, \int\int y \sigma_{xx} dydz$$

where $$\sigma_{xx}$$ is the axial stress due to the bending moments about the y and z axes at the point (x,y).

Likewise, the moment of inertia tensor - $$I_y, I_z, I_{yz}$$, (which can be equivalently denoted as $$ I_{yy}, I_{zz}, I_{xy}$$ or $$I_{22}, I_{33}, I_{23}$$, respectively) - are defined by the expressions


 * $$\qquad I_y=\int_A z^2dA$$


 * $$\qquad I_z=\int_A y^2dA$$


 * $$\qquad I_{yz}= \int_A yzdA$$

Solving these two groups of expressions yields an expression for the axial due to the bending moments about y and z as


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

If one recalls that


 * $$\underline{I} = \begin{bmatrix} I_{11} & I_{12} & I_{13} \\ & I_{22} & I_{23}\\sym.& & I_{33} \end{bmatrix}$$

One can see that denominator of the $$\sigma_{xx}$$ expression is numerically the determinate of the lower right quadrant of the $$\underline{I}$$ matrix. Also, the symmetry of the $$\underline{I}$$ matrix can be seen by comparing the $$ I_{23}$$ and $$I_{32}$$ terms of the matrix.


 * $$I_{23} = I_{yz} = \int_A yz\ dA$$


 * $$I_{32} = I_{zy} = \int_A zy\ dA$$

since


 * $$\int_A yz\ dA = \int_A zy\ dA$$


 * $$\Rightarrow I_{23} = I_{32}$$

The neutral axis is an axis along which the axial stress due to bending moments is zero, i.e.


 * $$\qquad \sigma_{xx}=0$$

For the case when the y or z axis is an axis of symmetry, and hence $$ I_{yz} = 0$$ the above expression for the axial stress can be reduced to


 * $$\sigma_{xx} = M_yy+M_zz = 0$$

an expression for the neutral axis can be taken from this by solving for z in terms of y:


 * $$z=\left (\frac{-M_y}{M_z} \right ) y = \tan\beta\ y$$

where $$\beta$$ is the angle between the neutral and y axes as shown below in Figure 3.



= Equations of Equilibrium in Terms of $$\sigma$$=

The goal is to obtain an equilibrium equation such that $$ \frac{\partial \sigma_{yx}}{\partial y} + \frac{\partial \sigma_{zx}}{\partial z} = 0 $$, or using indicial notataion: $$ \frac{\partial \sigma_{21}}{\partial x_{2}} + \frac{\partial \sigma_{31}}{\partial x_{3}} = 0 $$.

Recall from previous material [Section 2.4, Sun],


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


 * $$ \frac{\partial \sigma_{yy}}{\partial y} + \frac{\partial \sigma_{zy}}{\partial z} + \frac{\partial \sigma_{xy}}{\partial x}= 0 $$


 * $$ \frac{\partial \sigma_{zz}}{\partial z} + \frac{\partial \sigma_{xz}}{\partial x} + \frac{\partial \sigma_{yz}}{\partial y}= 0 $$

It is seen that a cyclic permutation is possible such that


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

This notation is particularly powerful. So powerful in fact, that Einstein used it to formulate the Theory of Relativity.

= Using Taylor Series With Infinitesimal Elements = For the purpose of simplicity, we look at the derivation of equation (4) on pg 27-2 for the one-dimensional model first. Looking at the equilibrium of the infinitesimal figure:

$$\sum\ F_x = 0 = -\sigma_{(x)}\ A + \sigma_{(x + dx)}\ A + f_{(x)} dx\,$$ 

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

Next, we apply Taylor series to the term, $$\sigma_{(x + dx)}\ - \sigma_{(x)}$$:

$$\frac{\partial\sigma_{x}}{\partial\ x} dx + higher~order~terms\,$$

Knowing that the general Taylor Series Approximation for the first three series of terms is given as:

$$F_{(x + dx)} = F_{(x)} + \frac{\partial\ F_{(x)}}{\partial\ dx} dx + \frac{1}{2} \frac{\partial ^2 F_{(x)}}{\partial x^2} (dx)^2 + .~.~.~higher~order~terms\,$$

The higher order terms are neglected and yields the final equation of interest shown below:

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


 * where $$f_(x)$$ is a $$\frac{force}{length}\,$$ and has dimensional units of $$\frac{N}{m}\,$$

In the equation above, the applied load $$\frac{f_{(x)}}{A}\,$$ is a Body Force and has dimensional units of $$\frac{N}{m^3}\,$$

Now looking at the nonuniform stress field in 3D and without any applied load, we focus on the x-direction in an attempt to view a better representation of the stress componenents without any cluttering.



In Figure 4 above shows an infinitesimal elemental cube where the stress is not completely uniform. Since the infinitesimal cube is in equilibrium, summing the forces of the x-direction is observed to be the sum of stress components of each facet area multiplied by the respective infinitesimal area. Therefore, the sum of the forces in the x-direction is:

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

Here the terms can be separated and rewritten in terms of their respective Facets as:

Facets with normal X- $$ dy dz [ -\sigma_{xx}\ (x,y,z) + \sigma_{xx}\ (x+dx,y,z)]\,$$

Facets with normal Y- $$ dz dx [ -\sigma_{yx}\ (x,y,z) + \sigma_{yx}\ (x,y+dy,z)]\,$$

Facets with normal Z- $$ dx dy [ -\sigma_{zx}\ (x,y,z) + \sigma_{zx}\ (x,y,z+dz)]\,$$

The next step after is to take the Taylor Series expansion while neglecting the higher order terms.

Combining the terms from all facets, we derive the following equation:

$$0 = (dx dy dz) \left [ \frac{\partial\sigma_{xx}}{\partial\ x} + \frac{\partial\sigma_{yx}}{\partial\ y} + \frac{\partial\sigma_{zx}}{\partial\ z} \right ]\,$$

= Stringers =

In flight, an aircraft wing is subject to bending. This bending moment must be absorbed and transferred by some structure. It is known that the wing skin, because of its relative thinness, is prone to buckling when subjected to compressive loading. The spars, due to a small moment of inertia, are also relatively ineffective at carrying bending loads. Thus, it is desired to add an additional element to the wing structure.

Stringers are placed at the junction of the skin and spars in "nearly" L-shaped brackets. The stringers run the spanwise length of the wing along with the spars. By placing the stringers at the top and bottom of the wing, their moment of inertia about the wing centroid is maximized, allowing for a greater bending moment to be applied to the wing. The figure below provides a generic schematic of stringer placement for one spar - skin joint location.



= NACA Airfoil Matlab Project Continued =

The purpose of this project was to take the previously created Matlab code and add the option to include stringers to the modeled NACA airfoil. Results will show that it is sufficient to consider only the stringers when analyzing the bending moment effects on a wing.

= Contributing team members =


 * Jared Lee--Eas4200c.f08.aero.lee 18:05, 7 November 2008 (UTC)
 * Oliver Oyama--Eas4200c.fo8.aero.oyama 18:55, 7 November 2008 (UTC)
 * Gonzalo Barcia--Eas4200c.f08.aero.barcia 19:25, 7 November 2008 (UTC)
 * William Kurth--Eas4200c.f08.aero.kurth 20:37, 7 November 2008 (UTC)
 * Ray Strods-- Eas4200c.f08.aero.strods 21:01, 7 November 2008 (UTC)