User:Eas4200c.f08.carbon.orear/HW7/LectureNotes

=Homework 7 Lecture Notes=

Roadmap
We have developed a Mini-plan as follows:

For single-cell sections, we have two classifications. There are sections with and without stringers.

Without stringers, the analysis can be summarized (and will be reviewed in more detail as we continue) that:

Shear flow is a constant for an arbitrarily shaped single-cell design. We want to know if this can resist the shear force Vx. From our analysis, the z-resultant of shear flow will equal zero.

With stringers, q (shear flow) in every section is not the same, though it is constant within each individual skin panel. In this way, it is "piecewise constant." We can superimpose the results due to their linearity, and our z-resultants will be nonzero.

Open C-section


Mean Value Theorem (MVT)

$$\int_A y dA = \bar y \int_A dA = \bar y A$$

$$\int_A z dA = \bar z \int_A dA = \bar z A$$

Where $$A=\sum_{i=1}^4 A_i$$

Equation 5.5 (from Lecture 34-3) can be particularized to: $$q(s) = (k_{yz}Q_z-k_zQ_y)V_z$$ Since all components on the RHS of the equation are independent of s. Note that for $$Q_z, Q_y$$ all areas are concentrated on the stringer.

=>shear flow q(s) is constant between 2 stringers, but q(s) would jump when crossing a stringer.

Steps to solve problems:

1) Find $$(\bar y, \bar z)$$

2) Find $$I_y, I_z, I_{yz}$$

3) Find $$k_y, k_z, k_{yz}$$

4) Follow path "s" to find $$q_{12}, q_{23}, q_{34}$$

$$q_{12}=(k_{yz}Q_z^{12}-k_zQ_y^{12})V_z$$

$$Q_z^{12} = y_1A_1$$, where y is the y-coordinate of stringer 1 with respect to the origin at C.

$$Q_y^{12} = z_1A_1$$


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

Single Cells With and Without Stringers
Single Cell without Stringers where q is a constant shear flow

Q: Can the set up shown above resist shear in the z direction? A: No

$$ R^z=R_{AB}^z + R_{BA}^z$$ Where $$R^z$$ is the total resultant in the z direction, $$R_{AB}^z$$ is the resultant of q in AB, and $$R_{BA}^z$$ is the total resultant of q in BA.

$$R_{AB}^z = -qA'B' = -R_{BA}^z$$ $$R^z=0$$

Single Cells with Stringers (neglect contribution of web to bending)

$$ q_{12}, q_{23}, q_{31}$$ do not equal each other, while $$q_{ij}(s)$$ is constant within each skin panel. $$q(s)$$ is a piecewise function with respect to s. The function is assumed to be piecewise because we neglect the contribution of stringers to bending.

The presence of stringers in the cell makes the shear flow non constant.

Principle of Superpostion due to Linearity

For this you add q as a contant, like in the figure for a single cell without stringers, to all the different q's in the figure for a single cell with stringers.

$$q_{12}=q +\tilde q_{12}$$ $$q_{23}=q +\tilde q_{23}$$ $$q_{ij}=q +\tilde q_{ij}$$ - this is the general form

Analysis Algorythem Method: 1) Data: $$ V_y, V_z$$ 2) Solve problem for $$\tilde q_{12}, \tilde q_{23} (\tilde q_{31}=0)$$ 3) Moment equation: Take the moment about any point in plane (x,y) 3.1) Superpostion $$q_{ij}=q + \tilde q_{ij}$$ 3.2) Select point O in plane (x,y)

Solving Problem 2 Using Euler Cut Principle


$$\sum F_{x} = 0$$
 * $$=\int_{A_3}^{} [\sigma_{xx} (x+dx) - \sigma_{xx} (x)] dA_3$$
 * $$[-\tilde{q}_{23} - \tilde{q}_{43} + \tilde{q}_{31}] dx$$

Taking the Taylor Series of $$\displaystyle [\sigma_{xx} (x+dx) - \sigma_{xx} (x)]$$ is $$\frac{d \sigma_{xx}}{dx} + h.o.t.$$
 * $$\tilde{q}_{31}=\tilde{q}_{23} + \tilde{q}_{13} + q^{(3)}$$

$$q^{(3)} = - \int_{A_3} \frac{d \sigma_{xx}}{dx} dA_3 $$ is the contribution to shear flow by stringer 3.

Recall $$V_y=\frac{dM_z}{dx}, V_z=\frac{dM_y}{dx}$$

From Lecture 34-3: $$q^{(3)}=-(k_yV_y-k_{yz}V_z)Q_z^{(3)}-(k_zV_z-k_{yz}V_y)Q_y^{(3)}$$

where $$Q_z^{(3)}=\int_{A_3}ydA_3, Q_y^{(3)}=\int_{A_3}zdA_3$$

Stringer 2: $$\tilde q_{23}=\tilde q_{12}-\tilde q_{24}+q^{(3)} = q^{(3)}$$

$$q^{(2)}$$ is computed as in (1).

$$Q_z^{(2)}=y_2A_2$$

$$Q_y^{(2)}=z_2A_2$$

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

$$Q_{y}^{(3)} = \int_{A_{3}}^{}{zdA_{3}}$$

$$Q_{y}^{(2)} = \int_{A_{2}}^{}{zdA_{2}}$$

Resuperposition of Problem 2
The basic formula for shear flow along the skin:

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

Following this formula, several equations can be found:


 * $$q_{12} = \tilde{q}_{12} + q_{1}$$, where $$\tilde{q}_{12}$$ is known as zero and $$q_{1}$$ is unknown


 * $$q_{23} = \tilde{q}_{23} + q_{1} - q_{2}$$


 * $$q_{31} = \tilde{q}_{31} + q_{1} - q_{3}$$


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


 * $$q_{43} = \tilde{q}_{43} + q_{3} - q_{4}$$


 * $$q_{41} = \tilde{q}_{41} + q_{4}$$

Looking at all these equations, there are 3 unknowns, and thus we need 3 equations.

1) The first equation would be one equation; finding the moments of Vy and Vz, and $${q_{12}, q_{23}, q_{31}, q_{24}, q_{43}, q_{41}}$$ about any convenient point.  The point about which the moment is taken should be where the lines of action of $$V_{y}$$ and $$V_{z}$$ intersect.

Because each of the three sections are combined, if one section twists by a certain angle, then the other two must also twist by the same angle. Therefore,

2) $$\theta _{1} = \theta _{2}$$ 3) $$\theta _{2} = \theta _{3}$$

Solving Problem 2 for each cell
To solve problem 2, follow each step for each cell:


 * Follow the path "$$s_{i}$$"
 * Write the equilibrium of each stringer on the path "$$s_{i}$$"

There are two ways to write the equilibrium:

1) Method shown in previous section, using FBD 2) The consequence of the first method is shown below:



The equilibrium equation is: $$\tilde{q}_{j6} = \tilde{q}_{3j} - \tilde{q}_{j5} - \tilde{q}_{j7}  q^{j} $$



The equilibrium equation is: $$-\tilde{q}_{6j} = \tilde{q}_{3j} - \tilde{q}_{j5} - \tilde{q}_{j7}  q^{j} $$

Note:

Consider what would happen if the cell walls were cut such that one stringer was isolated. Theoretically, it cannot be solved.



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



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

This means that $$q^{(3)} = 0$$, but this is not true!

A second example is shown below:



$$\tilde{q}_{24} = \tilde{q}_{23} = \tilde{q}_{12} = 0$$

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

This shows that $$q^{(2)} = 0$$ which also is not true!

Using the peace sign, which was popular in the '60s:




 * $$\tilde{q}_{15} = \tilde{q}_{54} = \tilde{q}_{42} = \tilde{q}_{21} = 0$$

$$\tilde{q}_{15} = \tilde{q}_{31} + \tilde{q}_{21} + q^{1}$$


 * $$\Rightarrow \tilde{q}_{31} = -q^{(1)}$$


 * $$\tilde{q}_{53} = q^{(5)}$$


 * $$\tilde{q}_{43} = q^{(4)}$$


 * $$\tilde{q}_{23} = q^{(2)}$$


 * $$\tilde{q}_{15} = \tilde{q}_{31} + \tilde{q}_{21} + q^{1}$$

$$\tilde{q}_{31} = \tilde{q}_{53} + \tilde{q}_{43} + \tilde{q}_{23} + q^{(3)}$$

$$\Rightarrow \tilde{q}_{31} = q^{(5)} + q^{(4)} + q^{(2)} - q^{(1)}$$

There are four unknowns, so there will need to be four equations described in the previous section:

1) Moment equation

2) $$\theta _{1} = \theta _{2}$$

3) $$\theta _{2} = \theta _{3}$$

4) $$\theta _{3} = \theta _{4}$$

A second method of cuts for the peace sign is shown below:



$$\tilde{q}_{53} = \tilde{q}_{43} = \tilde{q}_{23} = \tilde{q}_{31} = 0$$

$$\tilde{q}_{31} = \tilde{q}_{53} + \tilde{q}_{43} + \tilde{q}_{23} + q^{(3)}$$

This leads to $$q^{(3)} = 0$$ which is not true.

The "Not Possible" zero result
Isolation of stringer 3 would tell us that

0 = q(1) + q(2) + q(4)

and this result is NOT POSSIBLE. The actual result is:

0 = q(1) + q(2) + q(3) + q(4)

0 = Σq(e) where the summation is from 1 to 4.

We then see that

q(e) = nzQz(e) + nyQy(e)

nz = -(kyVy-kyzVz)

ny = -(kzVz-kyzVy)



Shear Buckling
Plotting the buckling shape can be done by the following steps. This will also express the coefficients $$\begin{Bmatrix} C_{22} & C_{13} & C_{31} & C_{33} \end{Bmatrix}$$ in terms of $$\displaystyle C_{11}$$ with the aspect ratio $$\vartheta = 1.5$$. Equations from the wiki article by Dr. Vu-Quoc on plate buckling will be used in the following steps.

1. Find $$\displaystyle \lambda$$ for $$\vartheta = 1.5$$

$$  \displaystyle \lambda =  \left[ \frac{\vartheta^4}{81 (1 + \vartheta^2)^4} \left\{ 1	 +	 \frac{81}{625} +	 \frac{81}{25} \left(	   \frac{1 + \vartheta^2}{1 + 9 \vartheta^2}	 \right)^2 +	 \frac{81}{25} \left(	   \frac{1 + \vartheta^2}{9 + \vartheta^2}	 \right)^2 \right\} \right]^{1/2} $$ (Equation 30 for wiki article)

Substituting yields $$\displaystyle \lambda = .028753$$

2. Evaluate numerically $$\displaystyle \mathbf K _{5 \times 5}$$ using Equation 26.

$$  \displaystyle \left[ \begin{array}{lllll} \frac{\lambda (1 + \vartheta^2)^2}{\vartheta^2} &	 \frac{4}{9} &	 0	 &	 0	 &	 0	 \\	 \frac{4}{9} &	 \frac{16 \lambda (1 + \vartheta^2)^2}{\vartheta^2} &	 - \frac{4}{5} &	 - \frac{4}{5} &	 \frac{36}{25} \\	 0	 &	 - \frac{4}{5} &	 \frac{\lambda (1 + 9 \vartheta^2)^2}{\vartheta^2} &	 0	 &	 0	 \\	 0	 &	 - \frac{4}{5} &	 0	 &	 \frac{\lambda (9 + \vartheta^2)^2}{\vartheta^2} &	 0	 \\	 0	 &	 \frac{36}{25} &	 0	 &	 0	 &	 \frac{\lambda (9 + 9 \vartheta^2)^2}{\vartheta^2} \end{array} \right] \left\{ \begin{array}{l} C_{11} \\	 C_{22} \\	 C_{13} \\	 C_{31} \\	 C_{33} \end{array} \right\} =  \left\{ \begin{array}{l} 0	 \\	 0	 \\	 0	 \\	 0	 \\	 0     \end{array} \right\} $$

3. Using indicial notation, $$\displaystyle \mathbf K = [\mathbf K _{ij}]$$. Truncating $$\displaystyle \mathbf K _{5 \times 5}$$ into a $$\displaystyle \mathbf K _{4 \times 4}$$ using $$\displaystyle \mathbf K _{22}$$ through $$\displaystyle \mathbf K _{55}$$, or the lower right hand corner of $$\displaystyle \mathbf K _{5 \times 5}$$, yields

$$\displaystyle \mathbf \bar K _{4 \times 4} = \left[ \begin{array}{llll} \frac{16 \lambda (1 + \vartheta^2)^2}{\vartheta^2} &	 - \frac{4}{5} &	 - \frac{4}{5} &	 \frac{36}{25} \\	 - \frac{4}{5} &	 \frac{\lambda (1 + 9 \vartheta^2)^2}{\vartheta^2} &	 0	 &	 0	 \\	 - \frac{4}{5} &	 0	 &	 \frac{\lambda (9 + \vartheta^2)^2}{\vartheta^2} &	 0	 \\	 \frac{36}{25} &	 0	 &	 0	 &	 \frac{\lambda (9 + 9 \vartheta^2)^2}{\vartheta^2} \end{array} \right] \left\{ \begin{array}{l} C_{22} \\	 C_{13} \\	 C_{31} \\	 C_{33} \end{array} \right\} =  \left\{ \begin{array}{l} \frac{-4 C_{11}}{9} \\	 0	 \\	 0	 \\	 0     \end{array} \right\} $$

4. Solving $$\begin{Bmatrix} C_{22} & C_{13} & C_{31} & C_{33} \end{Bmatrix}$$ in terms of $$\displaystyle C_{11}$$, with $$\displaystyle \mathbf K^{-1}$$ being the inverse of $$\displaystyle \mathbf K$$

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

$$U_{z} = C_{11} sin(\frac{\pi x}{a}) * sin (\frac{\pi y}{b}) + C_{22} sin(\frac{2 \pi x}{a}) * sin (\frac{2 \pi y}{b}) $$
 * $$ + C_{31} sin(\frac{\pi x}{a}) * sin (\frac{3 \pi y}{b}) + C_{31} sin(\frac{3\pi x}{a}) * sin (\frac{\pi y}{b}):$$
 * $$+ C_{33} sin(\frac{3\pi x}{a}) * sin (\frac{3\pi y}{b})$$

Finally, set $$\displaystyle C_{11} = 1$$, and then plot $$\displaystyle U_{z} (x,y)$$

Continuing lecture 39-2, Part 2

Stringer 1.

The equilibrium equation, or the FBD method, yields the equation for stringer 1 to be

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

With $$\tilde{q}_{12}$$ and $$\tilde{q}_{41}$$ being unknown, and $$\tilde{q}_{31} = 0$$, due to a cut in the panel between stringer 1 and stringer 3.

Canceling out $$\tilde{q}_{31}$$,

$$\tilde{q}_{12} = -\tilde{q}_{41} - q^{(1)} $$ (Eq. 1)

Stringer 2

Again the equilibrium equation about stringer 2

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

This time with $$\tilde{q}_{23} = 0$$ yielding

$$\tilde{q}_{24} = \tilde{q}_{12} - q^{(2)} $$ (Eq. 2.)

With $$\tilde{q}_{24} $$ and $$ \tilde{q}_{12} $$ being unknown.

Stringer 4

The equilibrium equation about stringer 4 gives

$$\tilde{q}_{41} = \tilde{q}_{24} - \tilde{q}_{34} - q^{(4)} $$ (Eq. 3)

Substituting Eq. 2 into Eq. 3 and canceling out $$\tilde{q}_{34} = 0$$ due to a cut.

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

Subtracting $$\tilde{q}_{41}$$ from both sides

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

Bidirectional Bending
The revisiting of this topic has already been covered in Team Carbon's previous HW.

Wiki Comparison
In concluding this course, our group, Team Carbon, has unanimously concluded that the Wikiversity/Mediawiki work we have done far outweighs a system like e-Learning.

By best summary, e-learning is a tool to interface student and teacher. Mediawiki is a tool to interface student to student, team to team, instructor to class, and the world can view and benefit from our work in "the spirit of copyleft."

Mediawiki excels in the following sense: It is a teamwork based cooperative communication and learning tool. The capability of logging all changes in history prevents vandalism of others' work. Account privacy is maintained, yet work can be shared! Furthermore, any individual can challenge and question and correct the mistakes of another!

E-learning loses work year after year, semester after semester, and Mediawiki retains this work for the future and for future students. Mediawiki can be useful in our employment and in industry communication. Other media can be inserted to our work. And there are many more benefits out there for students to yet discover.

Team Carbon has chosen the following 5 best features of the Wiki-system for this class.

1) Saving work for future reference.

2) Help for employment/resume building.

3) Individuals can pace their work and contribute in multiple manners.

4) Anti-vandalism history log.

5) Example set for future students to build on.