User:Eas4200c.f08.aero6.behnke/HW5

Continuing the Roadmap
The kinematic assumption, the 4 zero strain components, and the 4 zero stress components can be found on page 70 of C.T. Sun's "Mechanics of Aircraft Structures."

$$\varepsilon -\sigma $$ Relationship

You can rewrite $$\varepsilon -\sigma $$ Relationship as follows

$$\begin{Bmatrix} \varepsilon_{ij}\end{Bmatrix}_{6x1} = \begin{bmatrix} A & 0 \\ 0 & B \end{bmatrix}\begin{Bmatrix} \sigma _{ij} \end{Bmatrix}$$

You can also rewrite $$\sigma -\varepsilon $$ Relationship as follows

$$\begin{Bmatrix} \sigma_{ij}\end{Bmatrix}_{6x1} = \begin{bmatrix} A^{-1} & 0 \\ 0 & B^{-1} \end{bmatrix}\begin{Bmatrix} \varepsilon _{ij} \end{Bmatrix}$$

In order to prove that the above matrices are inverses of one another, multiply them together. If an identity matrix is produced, than the matrices are inverses of one another.

$$\sigma_{11} = 0 ;\sigma_{23} = 2G\epsilon_{23} = 0 ;\sigma_{31} = 2G\epsilon_{31} ;\sigma_{12} = 2G\epsilon_{12}\,$$
 * \sigma_{22} = 0
 * \sigma_{33} = 0

Stresses on a unit cube


Figure 1 only has stresses that have x-components acting on the block. Therefore, we can use this figure to sum all the forces in the x-direction. This procedure will lead to the following equation:


 * $$0 = \frac{d\sigma_{xx}(x)}{dx}+\frac{d\sigma_{yx}(y)}{dy}+\frac{d\sigma_{zx}(z)}{dz}\!$$

The same can be done for the stresses acting in the y- and z-directions. However, it is the same procedure as the stresses in the x-direction. Therefore, in the calculations below only the x-component stresses have been used to obtain the above formula.

The procedure to obtain the above formula follows. The formulas produced from the y- and z-component stresses are also recorded and are obtained in a similar fashion.


 * $$\Sigma F_{x} = 0 = \sigma_{xx}(x+dx)(dydz)-\sigma_{xx}(x)(dydz)+\sigma_{yx}(y+dy)(dxdz)-\sigma_{yx}(y)(dxdz)+\sigma_{zx}(z+dz)(dxdy)-\sigma_{zx}(z)(dxdy)\!$$


 * $$0 = dydz[\sigma_{xx}(x+dx)=\sigma_{xx}(x)]+dxdz[\sigma_{yx}(y+dy)-\sigma_{yx}(y)]+dxdy[\sigma_{zx}(z+dz)-\sigma_{zx}(z)]\!$$

Using the following theorem and neglecting the squared terms


 * $$g(x+dx) = g(x) + \frac{dg(x)}{dx}dx + \frac{d^{2}g(x)}{dx^{2}}dx^{2} + ...\!$$


 * $$0 = dydz[\sigma_{xx}(x)+\frac{d\sigma_{xx}(x)}{dx}dx-\sigma_{xx}(x)]+dxdz[\sigma_{yx}(y)+\frac{d\sigma_{yx}(y)}{dy}dy-\sigma_{yx}(y)]+dxdy[\sigma_{zx}(z)+\frac{d\sigma_{zx}(z)}{dz}dz-\sigma_{zx}(z)]\!$$


 * $$0 = dydz[\frac{d\sigma_{xx}(x)}{dx}dx]+dxdz[\frac{d\sigma_{yx}(y)}{dy}dy]+dxdy[\frac{d\sigma_{zx}(z)}{dz}dz]\!$$

We can then divide this equation by $$dxdydz\!$$ to obtain:


 * $$0 = \frac{d\sigma_{xx}(x)}{dx}+\frac{d\sigma_{yx}(y)}{dy}+\frac{d\sigma_{zx}(z)}{dz}\!$$

The same thing could be done for the stresses in the y- and z-direction to obtain the following equations:


 * $$0 = \frac{d\sigma_{yy}(y)}{dy}+\frac{d\sigma_{xy}(x)}{dx}+\frac{d\sigma_{zy}(z)}{dz}\!$$


 * $$0 = \frac{d\sigma_{zz}(z)}{dz}+\frac{d\sigma_{xz}(x)}{dx}+\frac{d\sigma_{yz}(y)}{dy}\!$$

Bidirectional Bending
Read section 4.1 and 4.2

$$ M_y = \int_A z\sigma_{xx} dA = \int_A x\sigma_{xx} dy dz \, $$ $$ M_z = \int_A y\sigma_{xx} dA = \int_A y\sigma_{xx} dy dz \, $$

Moment of Inertia tensor: $$ I_{y}, I_{z}, I_{yz} \,$$

$$ I_y = \int_A z^2 dA \, $$
 * $$ I_z = \int_A y^2 dA \, $$

$$ I_{yz} = \int_A y z dA \, $$

$$ \sigma_{xx} = E \epsilon_{xx} = \frac {I_y M_z - I_{yz}M_y}{I_yI_z - I_{yz}^2 }y + \frac {I_zM_y - I_{yz} M_z}{I_yI_z - I_{yz}^2 }z \, $$

Recall

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

Note the denominator the determinant of the matrix $$\, \begin{bmatrix} I _{y} & I _{yz}\\ I _{zy} & I _{z}\\ \end{bmatrix}$$

$$ D = I_y I_z - I_{yz}^2 \, $$

$$ D = I_{22} I_{33} - I_{23}^2 \, $$

NACA Airfoil Problem
Part 1 Results:



Part 2 Results:



As expected, the inclusion of the skin and walls in the calculation resulted in an increase in $$I_{22}$$, $$I_{33}$$, and $$I_{23}$$. The largest increase can be seen in $$I_{33}$$, while there is very little increase in $$I_{22}$$. This is because the additional elements are a greater distance from the z-axis than the y-axis.

The addition of the skin and partitions also results in slightly lower maximum stresses. This change is due to the increased moment of inertia.

Team Members
Felipe OrtegaEas4200c.f08.aero6.ortega 11:55, 6 November 2008 (UTC)

Kris Loper Eas4200c.f08.aero6.loper 00:32, 7 November 2008 (UTC)

Eduardo Villalba Eas4200c.f08.aero6.villalba 19:58, 7 November 2008 (UTC)

Matt Inman Eas4200c.f08.aero6.inman 20:05, 7 November 2008 (UTC)

Marlana Behnke Eas4200c.f08.aero6.behnke 20:06, 7 November 2008 (UTC)

Scott Chastain Eas4200c.f08.aero6.chastain 21:00, 7 November 2008 (UTC)