User:Eas4200c.f08.vqcrew.f/hw6

Flexural Shear Flow in Thin Walled Sections





In dealing with Aerospace Structures it is sometimes easier to examine idealized cases to have an easier look at the forces acting on the structure. For this case we would like to examine an aircraft wing and idealize the wing as a cantilever beam. In Fig ##. a nonuniform stress field can be seen acting on the beam.

The cantilever beam can then be broken up even more into two parts to examine the internal shear force and bending moments acting on a specific spot on the beam a distance x from the end. This image can be seen in Fig##. The beam has been cut at a distance x, and separated for a view of the internal forces.

Unsymmetrical Thin-walled Cross-Section

Now that we have examined a simple cantilever beam, we can now broaden our scopes to an unsymmetrical case. Fig ##. shows a sample unsymmetrical case over a small area, A_s, of the cross-section. The cross-section shown displays that a varialbe thickness and area along the path s. Examining the small section, s, of the cross section gives us the following relationship for the general case for unsymmetrical cross-sections:

$$\int_{A_s} \frac {d\sigma_{xx}}{dx} \,dA = -q(s)$$



Now that we have a brief introduction to unsymmetrical cross sections, it will help to examine the similarities and differences between symmetrical cross-sections about the z-axis and unsymmetrical cross-sections.

For Symmetrical cross-sections about the z-axis the relations come straight forward.


 * $$\displaystyle \sigma_{xx} = \frac{M_y z}{I_y}$$


 * $$q(s) = - \frac{V_z Q_y}{I_y}$$


 * $$Q_y = \int_{A_s} z \,dA = Z_c A_s$$

Now that we have the equations for a symmetrical case, lets examine the unsymmetrical case and see the differences:


 * $$\displaystyle \sigma_{xx} = (k_y M_z - k_{yz} M_y)y + (k_z M_y - k_{yz} M_z)z$$

Where the coefficients for k determined by the following equations:


 * $$k_y = \frac{I_y}{I_y I_z - (I_{yz})^2}\;$$


 * $$k_z = \frac{I_z}{I_y I_z - (I_{yz})^2}\;$$


 * $$k_{yz} = \frac{I_{yz}}{I_y I_z - (I_{yz})^2}\;$$

The equation for the stress can also be looked at in matrix form.


 * $$\left[\displaystyle \sigma_{xx} \right]_{1x1} = \left[ z y \right] \begin{bmatrix} k_y & -k_{yz} \\ -k_{yz} & k_z \end{bmatrix} \begin{Bmatrix} M_y \\ M_z \end{Bmatrix}$$

This can then be simplified to a 1x2 matrix and a 2x1 matrix forming the correct 1x1 matrix from $$\sigma_{xx}$$.


 * $$\left[\displaystyle \sigma_{xx} \right]_{1x1} = \left[ z y \right]_{1x2} \begin{bmatrix} k_z M_y - k_{yz} M_z \\ -kyz M_y + k_y M_z \end{bmatrix}_{2x1}$$

Now look at the shear flow:


 * $$q(s) = -(k_y V_y - k_{yz} V_z)Q_z - (k_z V_z - k_{yz} V_y) Q_y$$

Where:


 * $$ Q_z = \int_{A_s} y \,dA$$


 * $$ Q_y = \int_{A_s} z \,dA$$