User:Eas4200c.f08.aero.e/HW7

=Week 13 - 11/24 through 11/28=

Solving Problem P2:
Equation of each stringer (Euler cut principle)

Stringer 3





$$ \sum F_x = 0 = \int_A [ \sigma_{xx}(x + dx) - \sigma_xx(x)]dA_3 + [-\tilde{q}_{23} - \tilde{q}_{43} + \tilde{q}_{31}] dx $$

$$ q^{3} = [ \sigma_{xx}(x + dx) - \sigma_xx(x)] $$

So

$$ \tilde{q}_{31} = \tilde{q}_{23} + \tilde{q}_{43} - q^{(3)} $$

The contribution to shear flow by stringer 3 is

$$ q^{(3)} = -\int_{A3} \frac{d\sigma_{xx}}{dx} dA_3 $$

Recall:

$$ V_y = \frac{dM_z}{dx} \;\; V_z = \frac{dM_x}{dx} $$

$$ q^{(3)} = -(K_yV_y - K_{yz}V_z)Q_z^{(3)} - (K_zV_z - K_{yz}V_y)Q_y^{(3)} $$

The equation above will be called Equation 1.

Stringer 2



$$ \tilde{q}_{23} = \tilde{q}_{12} - \tilde{q}_{24} + q^{(2)} $$

Where $$ \tilde{q}_{12} = \tilde{q}_{24} = 0 $$ because of the cuts.

Use Equation 1 to solve for $$\tilde{q}_{23} = q^{(2)} $$.

Stringer 4



$$ \tilde{q}_{43} = \tilde{q}_{24} - \tilde{q}_{41} + q^{(4)} $$

Where $$ \tilde{q}_{24} = \tilde{q}_{41} = 0 $$ because of the cuts.

Use Equation 1 to solve for $$\tilde{q}_{43} = q^{(4)} $$.

Superposition (again)

$$ q_{ij} = \tilde{q}_{ij} + q_k $$

This leads to

$$ q_{12} = \tilde{q}_{12} + q_1 $$

$$ q_{23} = \tilde{q}_{23} + q_1 - q_2 $$

$$ q_{31} = \tilde{q}_{31} + q_1 - q_3 $$

$$ q_{24} = \tilde{q}_{24} + q_2 $$

$$ q_{43} = \tilde{q}_{43} + q_3 - q_2 $$

$$ q_{41} = \tilde{q}_{41} + q_4 $$

The $$\tilde{q}$$s are known, and the qs are not. Witch leaves 3 unknowns; q1, q2, andq3. So 3 equations are needed. These are

1) Moment equation: Take the moment of Vy, Vz, and {q12,...., q41} about any convenient point (usually where lines of action Vy and Vz intersect).

2) Compatibility equation: $$ \theta_1 = \theta_2 $$

3) Compatibility equation: $$ \theta_2 = \theta_3 $$

--EAS4200C.F08.AERO.CLK04D 17:28, 9 December 2008 (UTC)

Wednesday
Thanksgiving Break

Friday
Thanksgiving Break

=Week 14 - 12/1 through 12/5=

Monday
As a continuation to problem (P) from last lecture

*Follow path "$$S_{i}$$"

*Equilibrium of each stringer on path $$S_{i}$$

There are two methods to solve the problem.

1) Complete method. (Free body diagram) see lecture 38-1



$$q_{j6}=q_{2j}-q_{j5}-q_{j8}+q^{(3)}$$

2) Method is a consequence of 1st method.



$$-q_{6j}=q_{2j}-q_{j5}-q_{j8}+q^{(3)}$$

--Eas4200c.f08.aero.arena 20:48, 9 December 2008 (UTC)

Wednesday
Exam 3

Friday
The following concepts will be applied to the NACA Airfoil and can be found in the Matlab section of this homework report.

Plotting buckling shape under shear expresion $$\left\{C_{22}, C_{13}, C_{31}, C_{33} \right\}$$ in terms of:

$$C_{11}$$ and for $$V=1.5$$

The following are the steps to follow:


 * 1) 1 Find λ for $$V=1.5$$


 * 1) 2 Evaluate num $$K\ddot{}_{5x5}$$


 * 1) 3 Remember that $$K\ddot{} = \left[K_{ij} \right]$$

Then we have the following matrix:

$$\begin{vmatrix} K_{22} & K_{23}&  K_{24}&  K_{25}\\ K_{32} & K_{33}&  K_{34}&  K_{35}\\ K_{42} & K_{43}&  K_{44}&  K_{45}\\ K_{52} & K_{53}&  K_{54}&  K_{55}\\ \end{vmatrix}\begin{Bmatrix} C_{22}\\ C_{13}\\ C_{31}\\ C_{33} \end{Bmatrix} = \begin{Bmatrix} -\frac{4}{9}C_{11}\\ 0\\ 0\\ 0\\

\end{Bmatrix} $$

So, we solve for: $$\left\{C_{22}, C_{13}, C_{31}, C_{33} \right\}$$ in terms of $$C_{11}$$

Now, we know that: $$K\ddot{}^{-1}$$ is the inverse of $$K\ddot{}$$

Then;

$$\begin{Bmatrix} C_{22}\\ C_{13}\\ C_{31}\\ C_{33} \end{Bmatrix} = K\ddot{}^{-1}\begin{Bmatrix} \frac{-4}{9}C_{11}\\ 0\\ 0\\ 0 \end{Bmatrix}$$

Now, we see that:

$$u_{z}=C_{11}\sin \left(\frac{\pi x}{a} \right)\sin \left(\frac{\pi y}{b} \right)+C_{22}\sin \left(\frac{2\pi x}{a} \right)\sin \left(\frac{2\pi y}{b} \right)+C_{13}\sin \left(\frac{\pi x}{a} \right)\sin \left(\frac{3\pi y}{b} \right)+C_{31}\sin \left(\frac{3\pi x}{a} \right)\sin \left(\frac{\pi y}{b} \right)+C_{33}\sin \left(\frac{3\pi x}{a} \right)\sin \left(\frac{3\pi y}{b} \right)$$

Finally, we set $$C_{11}=1$$ and plot $$u_{z}\left(x, y \right)$$

-- Now, we proceed to continue the lecture of 12/01/08.

So we answer the problem on meeting 39-2 (end of 12/01/08) in two parts:

Part1: Equilibrium of isolated stringer (Already done)

Part2: Closed cell, equilibrium of stringer.

Stringer 1:

$$q\tilde{}_{12} = q\tilde{}_{31}+q\tilde{}_{41}+q^{(1)}$$

Where $$q\tilde{}_{12} $$ and $$ q\tilde{}_{41} $$ are unknowns and $$ q\tilde{}_{31} = 0$$

Stringer 2:

$$q\tilde{}_{24} = q\tilde{}_{12}-q\tilde{}_{23}+q^{(2)}$$

Where $$ q\tilde{}_{23} = 0$$

Stringer 4:

$$q\tilde{}_{41} = q\tilde{}_{24}+q\tilde{}_{34}+q^{(4)}$$

Where $$q\tilde{}_{34}$$ and $$q\tilde{}_{24}= 0$$

Then, we have that:

$$q\tilde{}_{41} = \left(q\tilde{}_{41}+q^{(1)} +q^{(2)}+q^{(4)}\right)$$

The $$q\tilde{}_{41}$$'s cancel out and we finally obtain:

$$0=q^{(1)}+q^{(2)}+q^{4}=-q^{(3)}$$

(This problem will be continued on 12/08/08)

--Eas4200c.f08.aero.gabo28 07:31, 9 December 2008 (UTC)

=Week 15 - 12/8=

Monday
12/5 continued (page 40-3)

$$ 0 = q^{(1)} + q^{(2)} + q^{(4)} $$

"Not Possible"

What we really have is,

<p style="text-align:center;">$$ 0 = \sum_{e=1}^4 q^{(e)} \;\;\; \Rightarrow \;\;\; q^{(1)} + q^{(2)} + q^{(4)} = -q^{(3)} $$

Where,

<p style="text-align:center;">$$ q^{(e)} = n_zQ^{(e)}_z + n_yQ^{(e)}_y $$

<p style="text-align:center;">$$ n_z = -(K_yV_y - K_{yz}V_z) $$

<p style="text-align:center;">$$ n_y = -(K_zV_z - K_{yz}V_y) $$

<p style="text-align:center;">$$ \sum_{e=1}^4q^{(e)} = n_z \sum_{e=1}^4Q^{(e)}_z + n_y \sum_{e=1}^4Q^{(e)}_y $$

<p style="text-align:center;">$$ \sum_{e=1}^4Q^{(e)}_z = 0 $$

<p style="text-align:center;">$$ \sum_{e=1}^4Q^{(e)}_y = 0 $$

If we have,



<p style="text-align:center;">$$ Q_z(\hat{z}) = \int_{A(\hat{z})} z dA $$

If $$A(\hat{z}) = A$$

<p style="text-align:center;">$$ Q_y = \int_A z dA = 0 = z_c \int_A dA $$

This is true because, for A $$ \quad z_c = 0$$.

From Sect. 4.2, we have the approximate displacement expansions given by

<p style="text-align:center;">$$ u = u_0(x) + z\psi_y(x) + y\psi_z(x) $$

<p style="text-align:center;">$$ v = v_0(x) $$

<p style="text-align:center;">$$ w = w_0(x) $$

Where $$\psi_y$$ and $$\psi_z$$ are rotations of the cross-section about the y and z axes.

The corresponding strains are

<p style="text-align:center;">$$ \varepsilon_{xx} = \frac{\delta u}{\delta x} = \frac{du_0}{dx} + z \frac{d\psi_y}{dx} + y \frac{d\psi_z}{dx} $$

<p style="text-align:center;">$$ \gamma_{xy} = \frac{\delta v}{\delta x} + \frac{\delta u}{\delta y} = \frac{dv_0}{dx} + \psi_z $$

<p style="text-align:center;">$$ \gamma_{xz} = \frac{\delta w}{\delta x} + \frac{\delta u}{\delta z} = \frac{dw_0}{dx} + \psi_y $$

Simplifying with the assumption $$ \psi_{xy} = \psi_{xz} = 0 $$ yields the relations

<p style="text-align:center;">$$ \psi_z = -\frac{dv_0}{dx} $$

<p style="text-align:center;">$$ \psi_y = -\frac{dw_0}{dx} $$

Subbing this in to the bending strain and using the argument that $$ \frac{du_0}{dx} = 0 $$ if $$N_x = 0$$ yields

<p style="text-align:center;">$$ \varepsilon_{xx} = -y \frac{d^2v_0}{dx^2} - z \frac{d^2w_0}{dx^2} $$

The bending moments about the y and z axes, respectively, are defined as

<p style="text-align:center;">$$ M_y = \iint_{A} z \sigma_{xx} dA = -E \iint_A \left( yz \frac{d^2v_0}{dx^2} + z^2 \frac{d^2w_0}{dx^2} \right) $$

<p style="text-align:center;">$$ = -EI_{yz} \frac{d^2v_0}{dx^2} - EI_y \frac{d^2w_0}{dx^2} $$

<p style="text-align:center;">$$ M_z = \iint_{A} y \sigma_{xx} dA= -EI_{z} \frac{d^2v_0}{dx^2} - EI_{yz} \frac{d^2w_0}{dx^2} $$

Where

<p style="text-align:center;">$$ I_y = \iint_A z^2 dA $$ moment of inertia about y-axis

<p style="text-align:center;">$$ I_z = \iint_A y^2 dA $$ moment of inertia about z-axis

<p style="text-align:center;">$$ I_{yz} = \iint_A yz dA $$ product of inertia

And

<p style="text-align:center;">$$ \frac{d^2v_0}{dx^2} = X_y $$

<p style="text-align:center;">$$ \frac{d^2w_0}{dx^2} = X_z $$

Where Xy and Xz are the curvatures.

In matrix form this can be written

<p style="text-align:center;">$$ \begin{bmatrix} M_y \\ M_z \end{bmatrix} = -E \begin{bmatrix} I_y & I_{yz} \\ I_{yz} & I_z \end{bmatrix} \begin{bmatrix} X_z \\ X_y \end{bmatrix} $$

And

<p style="text-align:center;">$$ \varepsilon_{xx} = -yX_y - zX_z = \begin{bmatrix} z & y \end{bmatrix} \begin{bmatrix} -X_z \\ -X_y \end{bmatrix} $$

So plugging into the stress equation and solving for the matrix $$ \begin{bmatrix} -X_z \\ -X_y \end{bmatrix} $$ we get

<p style="text-align:center;">$$ \sigma_{xx} = E \varepsilon_{xx} = E \begin{bmatrix} z & y \end{bmatrix} \begin{bmatrix} -X_z \\ -X_y \end{bmatrix} = E \begin{bmatrix} z & y \end{bmatrix} \left( \frac{1}{E} \underline{I}^{-1} \begin{bmatrix} M_y \\ M_z \end{bmatrix} \right ) $$

Where

<p style="text-align:center;">$$ \underline{I}^{-1} = \frac{1}{D} \begin{bmatrix} I_z & -I_{yz} \\ -I_{yz} & I_y \end{bmatrix} $$

<p style="text-align:center;">$$ D = I_yI_z + (I_{yz})^2 $$

Shear flow

<p style="text-align:center;">$$ q = -\int_A \frac{d \sigma_{xx}}{dx}dA $$

<p style="text-align:center;">$$ \frac{d \sigma_{xx}}{dx} = \begin{bmatrix} z & y \end{bmatrix} \underline{I}^{-1} \begin{bmatrix} \frac{dM_y}{dx} \\ \frac{dM_z}{dx} \end{bmatrix} $$

Where

<p style="text-align:center;">$$ V_y = \frac{dM_y}{dx} $$

<p style="text-align:center;">$$ V_z = \frac{dM_z}{dx} $$

And

<p style="text-align:center;">$$ Q_y = \int_A zdA $$

<p style="text-align:center;">$$ Q_z = \int_A ydA $$

This leads directly to the shear flow in matrix form.

--Eas4200c.f08.aero.E 01:38, 9 December 2008 (UTC)

=Matlab= Results from code: (using 2000 segments) q_along_trailing_edge = -9.9964e+005 q_from_F_to_B = -2.8788e+006 q_from_B_to_E = -2.3337e+006 q_from_E_to_H = -1.8031e+006

--Eas4200c.f08.aero.r 19:45, 9 December 2008 (UTC)

=Mediawiki vs. E-leaning=

EAS4200C.Fall08.AERO.Watlington.VG 20:29, 9 December 2008 (UTC)