User:Eas4200c.f08.aero6.inman/HW6

Compute $$\sigma_{xx}$$ in Segments BF and EH and Plot the Bending Stress Distribution
The problem statement provided that the Airfoil was experiencing the following moments:
 * $$M_y = -1250 N \cdot m$$
 * $$M_z = 500 N \cdot m$$

Using the MATLab Code presented in HW5 we can obtain the values for $$I_y$$, $$I_z$$, and $$I_{yz}$$, and now we can use EQN 4.29 from C.T. Sun to solve for every value of $$\sigma_{xx}$$ that we need.


 * $$\sigma_{xx} = \frac{I_yM_z-I_{yz}M_y}{I_yI_z-I_{yz}^2}y + \frac{I_zM_zy-I_{yz}M_z}{I_yI_z-I_{yz}^2}z$$ EQN 4.29

Figures 2 and 3 show the calculated bending stress distributions plotted as a function of $$y$$, and the dotted line on each represents the average bending stress experienced by the specified segment.





Figure 2 is both in tension and compression because the neutral axis passes through segment BF. Figure 3 is in tension because $$\sigma_{xx}$$ is always positive.

Case 1
Case 1 specified that the values $$m$$ and $$n$$ where to both be set equal to $$1$$ and then $$u_z$$ was to be calculated along the Rectangular, simply supported plate shown in Figure 4 above, using the following relationship:
 * $$u_z = c_{mn}sin\left(\frac{m \pi x}{a}\right)\sin\left(\frac{n \pi y}{b}\right)$$

where
 * $$c_{mn} = 1\!$$
 * $$a = 1.5\!$$
 * $$b = 1\!$$

the result is shown on the right in Figure 5.

Case 2


Case 2 specified that the values $$m$$ and $$n$$ where to be set equal to $$2$$ and $$1$$ respectively, and then $$u_z$$ was to be calculated along the Rectangular, simply supported plate shown in Figure 4.

It can be easily seen that the value $$m$$ directly correlates to the number of half wave lengths along $$x$$.

Find the Period of the Function $$u_z$$
Given the relationships
 * $$u_z = sin\left(\frac{m \pi x}{a}\right)$$
 * and
 * $$sin\left(\frac{m \pi (x+T)}{a}\right)=sin\left(\frac{m \pi x}{a}\right)$$

it is easily seen that
 * $$sin\left(\frac{m \pi x}{a}+\frac{m \pi T}{a}\right)=sin\left(\frac{m \pi x}{a}\right)$$
 * and therfore
 * $$\frac{m \pi T}{a} = 2 \pi n$$ where $$n = 0,1,2,...$$

Hence
 * $$T = \frac{2a}{m}n$$

Plot Relationship between $$k_c$$ and $$a/b$$


$$k_c = \left(\frac{mb}{a}+\frac{a}{mb}\right)^2 = \left(m\left(\frac{a}{b}\right)^{-1}+\frac{1}{m}\left(\frac{a}{b}\right)\right)^2$$


 * $$m = 1$$
 * $$m = 2$$
 * $$m = 3$$
 * $$m = 4$$

Plot Relationship between $$\sigma_{xx,cr}$$ and Aspect Ratio for Simply Supported and Clamped Boundary Conditions



 * Simply Supported
 * Clamped

Contribution of Airfoil Skin in Resistance to Bending
The range in which the critical buckling stress of the skin is bounded is much lower than the average stress in the airfoil skin. It is approximately 1% of this average. This shows that the skin has a very small and for most purposes negligible affect on the airfoil's critical buckling stress.

Notes on C.T. Sun's chapter 3.3 Bars with Circular Cross-sections
Beginning with the equation of a circle, a stress function must be derived for the its contour. Figure 9 displays the geometry of the stress function. $$x^{2} + y^{2} = a^{2}$$

$$\Phi = C(\frac{x^{2}}{a^{2}} + \frac{y^{2}}{a^{2}} -1)$$ Implementation of the Prandtl compatibility equation yields the constant term. $$C = -\frac{1}{2}a^{2}G\Theta$$ Thus the torque may now be expresses as: $$T = 2C\iint_{A}(\frac{x^{2}}{a^{2}} + \frac{y^{2}}{a^{2}} - 1)dxdy$$
 * $$ = 2C\iint_{A}(\frac{r^{2}}{a^{2}} - 1)dA$$
 * $$ = 2C(\frac{I_{p}}{a^{2}} - A)$$

Where $$I_{p} = \iint_{A}r^{2}dA = \frac{1}{2}\pi a^{4}$$

Since $$a^{2}A = 2I_{p}$$ then $$T = -\frac{2CI_{p}}{a^{2}} = \Theta GI_{p}$$

$$\tau_{xz} = \frac{\delta \Phi}{\delta y} = 2C\frac{y}{a^{2}} = -G\Theta y$$

$$\tau_{yz} = -\frac{\delta \Phi}{\delta x} = -2C\frac{x}{a^{2}} = G\Theta x$$

$$t_{z} = \tau_{xz}n_{x} + \tau_{yz}n_{y} = -G\Theta\frac{xy}{r} + G\Theta\frac{xy}{r} = 0\!$$

The tangential shear stress perpendicular to the radial direction is equal to $$\tau_{z}$$ in magnitude. Eliminating $$\Theta$$ with the previous equation the shear stress can be expressed in terms of torque.

$$\tau = \frac{Tr}{J}$$

As mentioned in previous sections of chapter 3, the warping of a beam in the axial direction is a function of y and z. Relating the following equations:

$$\gamma_{yx} = \frac{\theta_{yx}}{G} = \frac{\delta u_{x}}{\delta y} - \Theta z$$

$$\gamma_{zx} = \frac{\theta_{zx}}{G} = \frac{\delta u_{x}}{\delta z} + \Theta y$$

$$\theta_{yx} = -G\Theta z\!$$

$$\theta_{zx} = G\Theta y\!$$

Since $$\frac{\delta u_{x}}{\delta y} = 0$$

$$\frac{\delta u_{x}}{\delta z} = 0$$

Thus $$U_{x}(y,z) = 0 = w\!$$

Notes on C.T. Sun's chapter 5.1 Flexural Shear Flow in Open Thin-walled Sections
When considering the bending mode of an open thin-walled beam, the derivations from previous chapters are still accurate as long as the span of the beam is much greater than the depth. The shear on the other hand may be quite complex to derive and a change of coordinates is useful. As practiced in chapter 3 where thin-walled circular cross-sections were analyzed, a s-n coordinate plane will reduce the shear components of the thin-wall into just $$\Tau_{s}\!$$ since the shear in the normal direction of the cross-section $$\Tau_{n}\!$$ may be assumed equal to zero.

Chapter 5.1.1
This chapter deals with symmetric thin-walled cross sections. From figure 10 we stated that $$\Tau_{n}\!$$ may be assumed to be zero. We can also derive a formula from figure 10 when we balance the forces in the x-direction and assume that $$V_{z} \ne 0, V_{y} = 0.$$
 * $$\iint\limits_{A_{s}} \, \Delta \sigma_{xx}\, dA = -q_{s}\Delta x$$

This equation can be rewritten as


 * $$\iint\limits_{A_{s}} \, \frac{d \sigma_{xx}}{dx}\, dA = -q_{s}$$

If we note that $$\sigma_{xx} = \frac{M_{y}z}{I_{y}}$$ and that $$\frac{dM_{y}}{dx} = V_{z}$$ and substitute them into the above equation we will obtain


 * $$q_{s} = -\frac{V_{z}}{I_{y}} \iint\limits_{A_{s}} \, z\, dA$$

where $$\iint\limits_{A_{s}} \, z\, dA = A_{s}z_{c} = Q$$ and therefore $$q_{s} = -\frac{V_{z}Q}{I_{y}}$$

Example 5.2 from C.T. Sun "Mechanics of Aircraft Structures"
A four-stringer thin-walled channel beam with cross-section shown in Figure 11. Assume $$A_3 = A_2\,$$ and $$A_4 = A_1 \,$$ ; then there is symmetry about the y-axis and we can use the equation $$q_s = -\frac{V_z\,Q}{I_y}$$.


 * Expressing the equation above differently,


 * $$q_i = -\frac{V_z\,Q_i}{I_y}$$


 * where,
 * $$I_y = 2h^2\left(A_1 + A_2 \right)$$ and
 * $$Q_i = \sum_{k=1}^i {z_k A_k}$$, and $$A_k \,$$ is a stringer and $$z_k\,$$ is its corresponding vertical position.


 * so for $$q_1\,$$,
 * $$Q_1 = A_1 h\,$$
 * $$q_1 = -\frac{V_z A_1 h}{2 h^2 \left( A_1 + A_2\right)} = -\frac{V_z A_1}{2 h \left( A_1 + A_2\right)}$$


 * for $$q_2\,$$,
 * $$Q_2 = A_1 h+ A_2h\,$$
 * $$q_2 = -\frac{V_z \left(A_1 + A_2\right)h}{2 h^2 \left( A_1 + A_2\right)} = -\frac{V_z}{2 h}$$


 * for $$q_3\,$$,
 * $$Q_3 = A_1 h+ A_2 h- A_3 h\,$$
 * however $$A_3 = A_2\,$$, so
 * $$q_3 = -\frac{V_z \left(A_1 h+ A_2 h- A_2 h\right)}{2 h^2 \left( A_1 + A_2\right)} = -\frac{V_z A_1}{2 h \left( A_1 + A_2\right)} = q_1$$


 * and for $$q_4\,$$,
 * $$Q_4 = A_1 h+ A_2 h- A_3 h- A_4 h\,$$
 * however $$A_3 = A_2\,$$ and $$A_4 = A_1\,$$, so
 * $$q_4 = -\frac{V_z \left(0\right)}{2 h^2 \left( A_1 + A_2\right)} = 0$$

Therefore, the shear flow can be expressed as
 * $$q_{i+1}=q_i - \frac{V_z}{I_y}z_{i+1}A_{i+1}$$

We can conclude that $$\sum F_y = bq_3-bq_1 = bq_1-bq_1 = 0$$, so the resultant force in the y-direction goes away since there is no applied force in that direction. Also, the resultant shear flow in the z-direction must equal the transverse shear force, $$V_z=-2hq_2\,$$.

Chapter 5.1.2
This chapter deals with asymmetric thin-walled cross sections. The bending stress that is derived for this case is as follows.


 * $$\sigma_{xx} = (k_{y}M_{z} - k_{yz}M_{y})y + (k_{z}M_{y} - k_{yz}M_{z})z\!$$

where $$k_{y} = \frac{I_{y}}{D},  k_{z} = \frac{I_{z}}{D},   k_{yz} = \frac{I_{yz}}{D}$$    and    $$D = I_{y}I_{z} - I_{yz}^{2}$$

We can write the above formula in a more beautiful matrix form as follows:

$$\sigma_{xx} = \begin{bmatrix} z & y \end{bmatrix}\begin{bmatrix} k_{z} & -k_{yz}     \\ -k_{yz} & k_{y} \end{bmatrix}\begin{bmatrix} M_{y}     \\ M_{z} \end{bmatrix}$$

Substituting $$\sigma_{xx}, \frac{dM_{y}}{dx} = V_{z},  \frac{dM_{z}}{dx} = V_{y}$$  into    $$\iint\limits_{A_{s}} \, \frac{d \sigma_{xx}}{dx}\, dA = -q_{s}$$ we will obtain:


 * $$q_{s} = -(k_{y}V_{y} - k_{yz}V_{z})Q_{z} - (k_{z}V_{z} - k_{yz}V_{y})Q_{y}\!$$

Like $$\sigma_{xx}$$ we can rewrite $$q_{s}$$ as the following:

$$q_{s} = \begin{bmatrix} Q_{z} & Q_{y} \end{bmatrix}\begin{bmatrix} -k_{y} & k_{yz}     \\ k_{yz} & -k_{z} \end{bmatrix}\begin{bmatrix} V_{y}     \\ V_{z} \end{bmatrix}$$

When we have a symmetric force acting on a uniform bar we can take $$Q_{z} = 0\!$$ and $$I_{yz} = 0\!$$ which will leave us with the following:

$$q_{s} = -k_{z}V_{z}Q_{y} = -\frac{V_{z}Q_{y}}{I_{y}}$$

Example 5.3 from C.T. Sun "Mechanics of Aircraft Structures"
Given - The stringer web beam shown in figure ? has a shear flow produced by vertical load Vz (does not equal zero) and horizontal load Vy = 0 can be solved by considering the two loads seperately.

The moments and product of inertia are

$$I_{y} = 4Ah^{2}\,$$

$$I_{z} = 2Ah^{2}\,$$

$$I_{yz} = -2Ah^{2}\,$$

$$k_{y} = \frac{1}{Ah^{2}}\,$$

$$k_{z} = \frac{1}{2Ah^{2}}\,$$

$$k_{yz} = \frac{1}{-2Ah^{2}}\,$$

For shear flow across the top surface (q1), only one stringer is involved. Given that

$$Q_{y} = Ah\,$$

$$Q_{z} = -Ah\,$$

then the shear flow can be written as

$$ q_{1} = k_{yz}V_{z}Q_{z} - k_{z}V_{z}Q_{y} = 0\,$$

For q2 (the shear flow parallel to the z-axis)

$$Q_{y} = 2Ah\,$$

$$Q_{z} = -Ah\,$$

then the shear flow can be written as

$$ q_{2} = -\frac{V_z{}}{2h}$$

Note that the negative sign is due to the fact that the direction of the shear flow is in the positive z direction, rather than an assumed negative z direction.

also q 3 is the shear flow on the bottom surface given as

$$ q_{3} = 0\,$$

2-D Case discussed in class


$$dz = ds \cos \theta \,$$

$$ny = \cos \theta \,$$

$$dy = ds \sin \theta \,$$

$$nz = \sin \theta \,$$

$$\sum F_{y} = 0 = - \sigma_{yy}(ds n_{y}) - \sigma_{yz}(ds n_{z}) + t_{y}ds \,$$

similarily

$$t_{y} = 0 $$

and

$$\sum F_{y} = 0 $$

Contributing Team Members
Felipe OrtegaEas4200c.f08.aero6.ortega 20:03, 13 November 2008 (UTC)

Matt Inman Eas4200c.f08.aero6.inman 22:17, 13 November 2008 (UTC)

Kris Loper Eas4200c.f08.aero6.loper 17:22, 16 November 2008 (UTC)

Marlana Behnke Eas4200c.f08.aero6.behnke 19:32, 20 November 2008 (UTC)

Eduardo Villalba Eas4200c.f08.aero6.villalba 20:21, 20 November 2008 (UTC)

Scott Chastain Eas4200c.f08.aero6.chastain 07:02, 21 November 2008 (UTC)