User:Eas4200c.f08.radsam/HW7

=Homework #7=

=Class Notes=

November 17th - Asymmetric Cross-Section Shear Flow Analysis
Mean Value Theorem (MVT) analysis of Figure 1:

$$\int_A ydA=\bar{y}\int_A dA=\bar{y}A\,$$

Similarly:

$$\int_A zdA=\bar{z}\int_A dA=\bar{z}A\,$$

To find the total area (sans skin and spar webs):

$$A=\sum_{i=1}^{4}A_i\mbox{cf. HW5}\,$$

$$q(s)=(k_{yz}Q_z(s)-k_zQ_y(s))V_z(5.5)\,$$

Since:


 * 1) $$V_z\,$$
 * 2) $$K_{yz}\,$$, $$K_z\,$$
 * 3) $$Q_z\,$$, $$Q_y\,$$

Are all independent of $$s\,$$.

And #3 is independent because all areas are concentrated on stringers. Shear flow $$q(s)\,$$ is constant between 2 stringers, but $$q(s)\,$$ would increment (jump) when crossing a stringer.

The solution to this is as follows:

Step 1: Find $$(\bar{y}_c\,, \bar{z}_c)\,$$ and $$(y_i,z_i)i=1~\mbox{to}~4\,$$ Step 2: Find $$I_y, I_z, I_{yz}\,$$ Step 3: Find $$k_y,k_z,k_yz\,$$ Step 4: Follow path "$$s\,$$" to find $$q_{12},q_{23},q_{34}\,$$ For example: $$q_{12}=(k_{yz}Q_z^{12}-k_zQ_y^{12})V_z\,$$ $$Q_z^{12}=y_1A_1\,$$

Where $$y_1\,$$ is the coordinate of stringer 1.

$$Q_y^{12}=z_1A_1\,$$

Note: $$Q_z^{23}=y_1A_1+y_2A_2\,$$

In Figure 2, $$A=A_{\hat{s}}+A_{\hat{s},L}\,$$

Contribution by radsam.a

Mini Plan
S) Single-cell sections   S.1) w/o stringers S.2) w/ stringers

M) Multi-cell sections   M.1) w/o stringers M.2) w/ stringers

S) Single-cell. S.1) Without stringers $$q=\mbox{constant shear flow}\,$$

$$R^Z=R_{AB}^Z+R_{BA}^Z\,$$

$$R_{AB}^2={-q}\bar{A'B'}=R_{BA}^2\,$$

$$\Rightarrow R^Z=0\,$$

S.2) With stringers (Neglected contribution of web to bending)



$$q_{12}=q_{23}=q_{31}\,$$

but $$q_{ij}(s)=\mbox{constant with in each panel}\,$$

$$\Rightarrow q(s)\,$$ is piecewise constant for s.

$$R^Z=V_Z=0\,$$ Presence of stringers ==> non-constant shear flow



Principle of Superposition due to linearity

$$q_{ij}=\tilde {q}_{ij} + q_{k}\,$$

$$q_{12}=\tilde {q}_{12} + q_{1}\,$$

$$q_{23}=\tilde {q}_{23} + q_{1}\,$$

$$q_{31}=\tilde {q}_{31} + q_{1}\,$$

Analysis Algorithm
Observation : One unknown q => need one equation for one unknown. Method : Data: Vy,Vz 1) Solve P2 for $$\tilde {q}_{12}~,~\tilde {q}_{23}(\tilde {q}_{31} = 0)$$    2) Moment equation: Take moment about any point in plane(y,z) 2.1) Superposition: $$q_{ij}=\tilde {q}_{ij} + q_{k}\,$$       2.2) Select point in plane(y,z).

Contribution by radsam.r, images by radsam.a

November 21st - Multi-cell
3) Back to superposition:

$$q_{ij}=\tilde {q}_{ij} + q_{k}\,$$

M) Multi-cell  M.1) Without stringers

$$R^Z=\sum_{i=1}^{N_{cells}}R_i^Z\,$$

Where $$N_{cells}=\mbox{number of cells}\,$$

$$R_i^Z=0\,{\Rightarrow}R^Z=0\,$$ M.2) With stringers

$$R^Z=R^{Z_1}+R^{Z_2}\,$$

Recall $$R^{Z_1}=\sum_{i=1}^{N_{cells}}R_i^{Z_1}=0\,$$ ==> $$R^Z=R^{Z_2}=V_Z~{\ne}~0\,$$.

Contributed by radsam.r, images by radsam.a

November 24th - Stringer Equilibrium
Solving the problem: Equilibrium of each stringer

 Stringer #3 



$$\sum F_{x}=0=\int_{A_{3}} [\sigma_{xx}(x+dx)-\sigma_{xx}(x)]dA_{3}\,$$

The terms:

$$[\sigma_{xx}(x+dx)-\sigma_{xx}(x)]dA_{3}\,$$

Can be expanded in a Taylor series as:

$$\frac{\partial \sigma_{xx}}{dx}dx+h.o.t\,$$

$$[-\tilde{q}_{23} -\tilde{q}_{43} +\tilde{q}_{31}]dx\,$$

Now, observing the contribution to shear flow by stringer 3:

$$q^{3}=-\int_{A_{3}} \frac{\partial \sigma_{xx}}{dx} A_{3}\,$$

And recalling:

$$V_{y}=\frac{\partial d M_{z}}{dx}\,$$

And

$$V_{z}=\frac{\partial d M_{y}}{dx}\,$$

$$q^{3}=-(KyVy-KyzVz)Qz^{3}-(KzVz-KyzVy)Qy^{3}\,$$

$$Qz^{3}=\int_{A_{3}} ydA_{3}\,$$

$$Qy^{3}=\int_{A_{3}} zdA_{3}\,$$

Now for stringer 2:

$$\tilde {q}_{23}=\tilde {q}_{12} - \tilde {q}_{24} + \tilde {q}^{4}\,$$

Where the first term and second term on the right equal the shear flow in and shear flow out respectively:

$$Qy^{2}=\int_{A_{2}} zdA_{2}\,$$

Now for stringer 4:

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

$$Qy^{4}=\int_{A_{4}} zdA_{4}\,$$

Superposition

$$q_{ij}=\tilde {q}_{ij} + q_{k}\,$$

$$q_{12}=\tilde {q}_{12} + q_{1}\,$$

Where the first term and second term to the right are the known and unknown parameters of the equation, therefore:

$$q_{23}=\tilde {q}_{23} + q_{1}\,$$

$$q_{31}=\tilde {q}_{31} + q_{1}\,$$

$$q_{24}=\tilde {q}_{24} + q_{2}\,$$

$$q_{43}=\tilde {q}_{43} + q_{2} - q_{3}\,$$

$$q_{41}=\tilde {q}_{41} + q_{3}\,$$

Where we have 3 unknowns:

$$q_{1}, q_{2}, q_{3}\,$$

Thus, we need 3 equations:

1) Moment Equations: Take moments of $$V_{y} V_{z}\,$$ and $$[q_{12}, ... q_{41}]\,$$ about any convenient point.

(This point usually lies where lines of action of $$V_{y} V_{z}\,$$ intersect)

2) $$\theta_{1}=\theta_{2}\,$$

3) $$\theta_{2}=\theta_{3}\,$$

Contribution by radsam.z, images by radsam.a

December 1st - Equilibrium on Stringers
When calculating the shear flow on the various elements, equilibrium must be obtained in all the stringers.

2 ways to compute
 * Complete method from previous lecture.
 * Consequence of 1st method.

Question: What if we cut cell walls and ISOLATE one stringer? Isolating any one of the stringers would result in a wrong answer because shear must flow through all the stringers. If one is isolated, all components of the shear flow for that given stringer will be equal to zero. If we look at Figure 3, we may notice the cuts on all the members that connect stringer 3 to the rest of the structure, therefore the shear flow in/out of stringer 3 is zero. $$\tilde{q_{23}} = \tilde{q_{31}} =\tilde{q_{34}} = 0\,$$

Figure 4 shows a detail of stringer 3, along with the direction of shear flows. If we were to compute the shear flow on stringer 3 and solved for $$\bar q_{31}\,$$, the following expression would result: $$\tilde{q_{31}} = \tilde{q_{23}} - \tilde{q_{34}} + q^{(3)}\,$$

However, we know that $$\tilde{q_{31}} = \tilde{q_{23}} = \tilde{q_{34}} = 0\,$$, therefore $$q^{(3)}\,$$ must be equal to zero. This answer is NOT true and therefore we conclude that when calculating the shear flows, none of the stringers must be isolated.        Contribution by radsam.d, images by radsam.a

December 3rd - Continuation of Multicell Example
Now we need to look at the total shear for the entire structure:


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


 * $$\displaystyle q_{12} = \tilde{q}_{12} + q_1 = 0 + q_1$$


 * $$\displaystyle q_{24} = \tilde{q}_{24} + q_2 = 0 + q_2$$


 * $$\displaystyle q_{43} = \tilde{q}_{43} + q_2 - q_3 = q^{(4)} + q_4$$


 * $$\displaystyle q_{31} = \tilde{q}_{31} + q_1 = q^{(2)} - q^{(4)} + q^{(3)} + q_3$$


 * $$\displaystyle q_{41} = \tilde{q}_{41} + q_4 = 0 + q_4$$

The method used above is the superposition method. It was used above in the mini-plan for multi-cell stringers, and this problem is a good example of its application.

Notice how there are now three unknowns that must be solved for: $$q_{1}, q_{2}, q_{3}\,$$

Thus, we need three equations to solve for these unknowns. The ones we will use are the moment equation, and the compatibility equations for rigid bodies under torsion.

1. Momentum Equation: Equate the moment generated by $$V_z\,$$ and $$V_y\,$$ to the moment produced by the shear flows. Since $$V_z\,$$ and $$V_y\,$$ are given, and the moment arm about which they are applied is zero, the left hand side usually goes to zero.

$$V_{z}*0 = \sum{q_{ij}'h_{z}}$$

2. $$\theta _{1}=\theta _{2}\,$$

3. $$\theta _{2}=\theta _{3}\,$$

In class, the situation was discussed where a specific string would be isolated, thus interrupting the shear flow. Does this method still yield a useful solution in this configuration?

If we were to make cuts such that stringer three was isolated, our new equation would be:

$$\tilde{q}_{23}=\tilde{q}_{31}=\tilde{q}_{34}=0$$

Upon summing the forces coming in and out of stringer 3:

$$\tilde{q}_{31}=\tilde{q}_{23}-\tilde{q}_{34}+{q}^{(3)}$$

Which yields $$q^{(3)}=0\,$$, which is not possible.

In conclusion, the superposition method works well for multi-cell structures as long as the cuts are chosen wisely. A stringer cannot be isolated using this method because a component of the shear flow is essentially stranded. Additionally, good choices for summing the moment about a point will make this method even more efficient.

Contribution by radsam.s

December 5th - Shear Buckling

 * Refer to "Plate Buckling" article.

Steps

1) Find $$\lambda \,$$ for $$\upsilon = 1.5\,$$ (Eqn. 30 in "Plate Buckling"):
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$$  \displaystyle \lambda =  \left[ \frac{\vartheta^4}{81 (1 + \vartheta^2)^4} \left\{ 1	 +	 \frac{81}{625} +	 \frac{81}{25} \left(	   \frac{1 + \vartheta^2}{1 + 9 \vartheta^2}	 \right)^2 +	 \frac{81}{25} \left(	   \frac{1 + \vartheta^2}{9 + \vartheta^2}	 \right)^2 \right\} \right]^{1/2} $$
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2) Evaluate num. $$K_{5x5}\,$$ (Eqn. 26):
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$$  \displaystyle \left[ \begin{array}{lllll} \frac{\lambda (1 + \vartheta^2)^2}{\vartheta^2} &	 \frac{4}{9} &	 0	 &	 0	 &	 0	 \\	 \frac{4}{9} &	 \frac{16 \lambda (1 + \vartheta^2)^2}{\vartheta^2} &	 - \frac{4}{5} &	 - \frac{4}{5} &	 \frac{36}{25} \\	 0	 &	 - \frac{4}{5} &	 \frac{\lambda (1 + 9 \vartheta^2)^2}{\vartheta^2} &	 0	 &	 0	 \\	 0	 &	 - \frac{4}{5} &	 0	 &	 \frac{\lambda (9 + \vartheta^2)^2}{\vartheta^2} &	 0	 \\	 0	 &	 \frac{36}{25} &	 0	 &	 0	 &	 \frac{\lambda (9 + 9 \vartheta^2)^2}{\vartheta^2} \end{array} \right] \left\{ \begin{array}{l} C_{11} \\	 C_{22} \\	 C_{13} \\	 C_{31} \\	 C_{33} \end{array} \right\} =  \left\{ \begin{array}{l} 0	 \\	 0	 \\	 0	 \\	 0	 \\	 0     \end{array} \right\} $$
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3) $$K = [K_{ij}]\,$$:
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$$\begin{bmatrix} K_{22} & K_{23} & K_{24} & K_{25}\\ K_{32} & K_{33} & K_{34} & K_{35}\\ K_{42} & K_{43} & K_{44} & K_{45}\\ K_{52} & K_{53} & K_{54} & K_{55} \end{bmatrix} \begin{Bmatrix} C_{22}\\ C_{13}\\ C_{31}\\ C_{33} \end{Bmatrix} = \begin{Bmatrix} -\frac{4}{g}C_{11}\\ 0\\ 0\\ 0 \end{Bmatrix}$$
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Solve $$\lbrace C_{22}, ~ C_{13}, ~ C_{31}, ~ C_{33} \rbrace\,$$ in terms of $$C_{11}\,$$


 * $$\bar K^{-1}\,$$ is inverse of $$\bar K\,$$

Therefore, to solve for all C's, multiply the following:

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

$$u_z = C_{11} \sin\left( \frac{\pi x}{a}\right) \sin\left(\frac{\pi y}{b} \right) + C_{22} \sin\left(\frac{2\pi x}{a} \right) \sin\left(\frac{2\pi y}{b} \right)$$

$$.+ C_{13} \sin\left(\frac{\pi x}{a} \right) \sin \left(\frac{3\pi y}{b} \right) + C_{31} \sin\left(\frac{3\pi x}{a} \right)\sin\left(\frac{\pi y}{b} \right)$$

$$.+ C_{33} \sin\left(\frac{3\pi x}{a} \right)\sin\left(\frac{3\pi y}{b} \right)$$

 Stringer 1 

$$\tilde{q_{12}} = \tilde{q_{31}} + \tilde{q_{41}} + q^{(1)}// \tilde{q_{31}} = 0 \,$$

 Stringer 2 

$$\tilde{q_{24}} = \tilde{q_{12}} + \tilde{q_{23}} + q^{(2)}// \tilde{q_{23}} = 0$$

 Stringer 3 

$$\tilde{q_{41}} = \tilde{q_{24}} + \tilde{q_{34}} + q^{(4)}// \tilde{q_{34}} = 0$$

Contribution by radsam.d

December 8th - Analysis of Bidirectional Bending in Random Cross Sections
Pg 40.3 continued.

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

...not true or possible! Since:

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

$$0=\sum_{e=1}^{4} q^{e}\,$$

pg 38-2:

$$q^{e}=n_{z}Q_{z}^{e}+n_{y}Q_{y}^{e}\,$$

$$n_{z}:= -(k_{y}V_{y}-k_{yz}V_{z})\,$$

$$n_{y}:=-(k_{z}V_{z}-k_{zy}V_{y})\,$$

$$\sum_{e=1}^{4} q^{e}=n_{z} \sum_{e=1}^{4} Q_{z}^{e}+n_{y} \sum_{e=1}^{4} Q_{y}^{e}\,$$

Where the sums above are equal to zero!

$$Q_y=(\hat {z})=\int_{A(\hat {z})}zdA\,$$

$$Q_y=\int_{A}zdA=0\,$$

Recalling the Mean Value Theorem, the above would simply be:

$$Q_y=Z_c\int_{A}dA\,$$

Where:

$$Z_c=0\,$$

HW #7: NACA Airfoil - "Back of envelope" verification


 * Looking at Figure 2, we can see that:

$$\bar B \bar E=\bar F \bar H=\frac{1}{2}(BE+FH)\,$$

Verify magnitude of $$I_y, I_z, I_{yz} \Rightarrow\,$$ (Section pg. 34-4) Angle B of N.A. $$\sigma_{xx}\,$$ @ each stringer

Contribution by radsam.z

= Read and Report =

Section 4.2 - Bidirectional Bending


When considering beams with arbitrarily shaped cross-sections, as shown in Figure 1, we place the external loads passing through the shear center/center of twist if we are avoiding torsion.

Studying the bidirectional bending in this cross-section, we obtain the longitudinal displacement as a function of $$x,y,z\,$$:

$$u=u_{0}(x)+z\psi_{y}(x)+y\psi_{z}(x)\,$$  (4.22a)

$$v=v_{0}(x)\,$$    (4.22b)

$$w=w_{0}(x)\,$$    (4.22c)

Where:

$$\psi_{y}, \psi_{z}\,$$

Are rotations of the cross-section about the $$y\,$$ and $$z\,$$ axes.

The strains can be defined as:

$$\epsilon_{xx}=\frac{\partial u}{\partial x} = \frac{du_{0}}{dx}+z\frac{d\psi_{y}}{dx}+y\frac{d\psi_{z}}{dx}\,$$   (4.23a)

$$\gamma_{xy}=\frac{\partial v}{\partial x}+\frac{\partial u}{\partial y}=\frac{dv_{0}}{dx}+\psi_{z}\,$$  (4.23b)

$$\gamma_{xz}=\frac{\partial w}{\partial x}+\frac{\partial u}{\partial z}=\frac{dw_{0}}{dx}+\psi_{y}\,$$  (4.23c)

However, recalling the assumptions:

$$\gamma_{xy}=\gamma_{xz}=0\,$$

We get:

$$\psi_{z}=-\frac{dv_{0}}{dx}\,$$

$$\psi_{y}=-\frac{dw_{0}}{dx}\,$$

Which we would substitute into equation 4.23c, and this yields:

$$\epsilon_{xx}=\frac{du_{0}}{dx}-y\frac{d^{2}v_{0}}{dx^{2}}-z\frac{d^{2}w_{0}}{dx^{2}}\,$$   (4.24)

Recalling:

$$\frac{du_{0}}{dx}=0\,$$ ...if... $$N_{x}=0\,$$

$$\epsilon_{xx}=-y\frac{d^{2}v_{0}}{dx^{2}}-z\frac{d^{2}w_{0}}{dx^{2}}\,$$          (4.25)

The Bending moments about the y and z axes can be written as:

$$M_{y}=\iint_{A}z\sigma_{xx}dA=-E\iint_{A}(yz\frac{d^{2}v_{0}}{dx^{2}}+z^{2}\frac{d^{2}w_{0}}{dx^{2}})dA=-EI_{yz}\frac{d^{2}v_{0}}{dx^{2}}-EI_{y}\frac{d^{2}w_{0}}{dx^{2}}\,$$  (4.26)

$$M_{z}=\iint_{A}y\sigma_{xx}dA=-EI_{z}\frac{d^{2}v_{0}}{dx^{2}}-EI_{yz}\frac{d^{2}w_{0}}{dx^{2}}\,$$ (4.27)

Where:

$$I_{y}=\iint_{A}z^{2}dA\,$$(4.28a)

is the moment of inertia about y-axis.

$$I_{z}=\iint_{A}y^{2}dA\,$$ (4.28b)

is the moment of inertia about z-axis.

$$I_{yz}=\iint_{A}yzdA\,$$ (4.28c)

is the product of inertia.

Matrix Form:

$$\begin{Bmatrix} M_{y} \\ M_{z} \end{Bmatrix}= \begin{vmatrix} I_{y} & I_{yz} \\ I_{yz} & I_{z} \end{vmatrix} \begin{Bmatrix} -E\chi_{z} \\ -E\chi_{y} \end{Bmatrix}\,$$

$$\epsilon_{xx}=-y\chi_{y}-z\chi_{z}\,$$

$$\epsilon_{xx}=\begin{bmatrix} z & y \end{bmatrix} \begin{Bmatrix} -\chi_{z} \\ -\chi_{y} \end{Bmatrix}\,$$

$$\sigma_{xx}=\Epsilon \epsilon_{xx}=\epsilon \begin{bmatrix} z & y \end{bmatrix} \begin{Bmatrix} -\chi_{z} \\ -\chi_{y} \end{Bmatrix} \,$$

$$I^{-1}=\frac{1}{D}\begin{bmatrix} I_{z} & -I_{yz} \\ -I_{yz} & I_{y} \end{bmatrix}\,$$

Where:

$$D=I_{y}I_{z}-(I_{yz})^{2}\,$$

Shear flow:

$$q=-\int_{A} \frac{d\sigma_{xx}}{dx} dA\,$$

$$\frac{d\sigma_{xx}}{dx}=\begin{bmatrix} z & y \end{bmatrix} I^{-1}\begin{Bmatrix} \frac{dM_{y}}{dx} \\ \frac{dM_{z}}{dx} \end{Bmatrix}\,$$

$$V_{y}=\frac{dM_{y}}{dx}\,$$

$$V_{z}=\frac{dM_{z}}{dx}\,$$

$$Q_{y}=\int_{A}zdA\,$$

$$Q_{z}=\int_{A}ydA\,$$

Contribution by radsam.z

=Homework Problems and Solutions=

=Homework #7 Contributors= Eas4200c.f08.radsam.a. 01:16, 7 December 2008 (UTC) Eas4200c.f08.radsam.d 21:05, 8 December 2008 (UTC) Eas4200c.f08.radsam.m 22:33, 8 December 2008 (UTC) Eas4200c.f08.radsam.r 18:34, 9 December 2008 (UTC) Eas4200c.f08.radsam.z 19:35, 9 December 2008 (UTC) Eas4200c.f08.radsam.s 20:27, 9 December 2008 (UTC)