text:
User:Eas4200c.f08.blue.a/Homework7

Mean Value Theorem and Stringers
Previously, we were discussing the problem with a set of 4 stringers with equal sizes versus a set of 4 stringers with varying sizes.

This leads to the discussion on the Mean Value Theorem. The Mean Value Theorem states that:

$$\int zdA = \bar z A$$

(neglect skin and sparwebs)

Since:
 * V2
 * KY2, K2
 * Q2, Qy

are all independent of 's' (the path), then shear flow will then be independent of q(s)

When crossing a stringer:

 * 1) Find (Yc, Xc)
 * 2) Find Kx, Kz, Kyz
 * 3) Find Iz, Iy, Izy
 * 4) Follow Path (s) to find:

q12, q23, q34

Where q12 is the shear flow in panel 12

where: (y1 is the y coordinate of stringer 1)

where:

Notice:

Single and Multi Cell Shear Flow
A plan for attacking the problem of solving for the shear flow in a multi-cell section with stringers was introduced.

This miniPLAN is shown below. S) Singe-cell sections                   S.1)  without stringers S.2) with stringers               M)  Multi-cell sections M.1) without stringers                   M.2)  with stringers

First, a single-cell section with no stringers(as in Figure 1) is analyzed to find the shear flow q through the entire section (S.1 from the miniPLAN). The section can be viewed as a thin walled cross section. As is shown below, a thin walled cross section cannot resist transverse shear (Vz).

Figure 1: Shear flow (q) in a single-cell section

Next, a single-cell section with stringers is analyzed as is shown in. The solution to this problem (P) can be analyzed by decomposing the problem into two parts: A section with stringers must be considered without stringers (as in problem P1 in Figure 3) and an open cross section with stringers (as in problem P2 in Figure 4).

Figure 2: Single-cell section with stringers

Figure 3: Single-cell section without stringers

Figure 4: Open single-cell section with stringers

P=P1+P2

It is known that the sum of shear forces in P is not equal to zero and that the shear flows between stringers in P are not equal but the shear flow is constant in the within each parcel. Therefore, the presence of stringers results in a non-constant shear flow.

$$R^{z}=\sum{V_{z}}\neq 0$$

$$q_{12}\neq q_{23}\neq q_{31}$$

$$q_{ij}(s)=const$$

Problem P1 is solved just as in S.1 of the miniPLAN and the resulting shear flows between stringers can be solved with the superposition of P1 and P2.

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

The method for solving the problem consists of the following steps:

1) Solve P2 for $$\tilde{q}_{12},\tilde{q}_{23},\tilde{q}_{31} $$

2) Take the moment about any point in the plane of the section

2.1) Superposition $$q_{ij}=q+\tilde{q}_{ij} $$

2.2) Selection point in plane and find moment about that point

3) Back to superposition

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

Multi Cell Shear Flow
M. Multicell

First, consider a case without stringers

For n cells we have:

$$R^{\bar{z}}+\sum_{i=1}^{n cells}{R^{\bar{z}}_{i}}$$

and

$$R^{\bar{z}}=0$$

The probelm can be broken down using superposition. the first part of the problem consists of computing the shear flow of the web. The second problem consists of computing the shear in the stringers. The cuts to the walls have been made in read int eh figure below.

The shear flow across all of the cut sections can be assumed to be zero. Once the solutions to the two parts of the problem have been solved, they can simply be added together like so:

$$R^{\bar{z}}=R^{z_{1}}+R^{z_{2}}$$

Note that $$R^{z_{1}}=0$$

and thus,

$$R^{\bar{z}}=R^{z_{2}}=V_{z}$$

Continuation of Mulit Cell Shear Flow and Euler Cut Principle
Continuation of problem using Euler cut principle Stringer 3: Refer to the image below

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

$$\int_{A_{3}}^{}{\left[\sigma_{xx}(x+dx)-\sigma_{xx}(x) \right]dA_{3}}+\left[\tilde{-q_{23}}-\tilde{q_{43}}+\tilde{q_{31}} \right]dx$$

Taylor series: $$\frac{d\sigma_{xx}dx}{dx} + h.o.t.$$

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

$$q^{3}\Rightarrow$$ stringer 3 contribution $$-\int_{A_{3}}^{}{\frac{d\sigma_{xx}}{dx}}dA_{3}$$

Recall $$v_{y}=\frac{\partial Mz}{\partial x}$$, $$v_{z}=\frac{\partial My}{\partial x}$$

$$q^{(3)}=-\left(k_{y}v_{y}-k_{yz}v_{z}\right)Q_{z}^{(3)}-\left(k_{z}v_{z}-k_{yz}v_{y} \right)Q_{y}^{(3)}$$

$$Q_{z}^{(3)}=\int_{A_{3}}^{}{y}dA_{3}$$ and  $$Q_{y}^{(3)}=\int_{A_{3}}^{}{z}dA_{3}$$

String 2

$$\tilde{q_{23}}=\tilde{q_{12}}+\tilde{q_{24}}+\tilde{q}^{(2)}$$ but because of the way you cut the cells $$\tilde{q_{12}}$$ and $$\tilde{q_{24}}$$ equal zero.

$$q^{(2)}$$ is computed the same as $$q^{(3)}$$ except $$A_{3}$$ becomes $$A_{2}$$

$$Q_{z}^{(2)}=y_{2}A_{2}$$ and $$Q_{y}^{(2)}=z_{2}A_{2}$$

Stringer 4 is $$\tilde{q}_{43}=\tilde{q}_{24}-\tilde{q}_{41}+q^{(4)}$$ but $$\tilde{q}_{24}$$ and $$\tilde{q}_{41}$$ become zero because of the cut of the cells.$$q^{(4)}$$ can be computed as above for $$q^{(2)}$$

This marks the end of problem 2

Now returning to the superposition principle

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

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

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

All of the $$\tilde{q}$$ are known

Continuing this process for $$q_{24}$$, $$q_{43}$$, and $$q_{41}$$

$$q_{24}=\tilde{q}_{24}+q_{1}$$

$$q_{43}=\tilde{q}_{43}+q_{1}$$

$$q_{41}=\tilde{q}_{41}+q_{1}$$

To solve problems, such as problem 2, you'll have a certain number of unknown shear stresses and you'll get the same amount of equations to solve for the unknowns from,

1)The moment equation: Take the moment of $$v_{y}$$, $$v_{z}$$, and $$\left\{q_{12}, ..., q_{41} \right\}$$ about any convenient point (usually where lines of action of $$v_{y}$$, $$v_{z}$$ cross)

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

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

Continuation of Stringer Shear Force
Resulting for the consequences of the first method,

$$\tilde q_{j6}=\tilde q_{2j}-\tilde q_{j5}- \tilde q_{j8}+q(j)$$

then if the $$\tilde q_{6j}$$ is reversed

then

$$-\tilde q_{j6}=\tilde q_{2j}-\tilde q_{j5}- \tilde q_{j8}+q(j)$$

The question was raised that if one stringer was isolated if the system could still be solved.

this is not true because when one stringer is isolated it would yield that the q of the actual stringer is zero which is known not to be true.

$$\tilde q_{j6}=\tilde q_{2j}-\tilde q_{j5}- \tilde q_{j8}+q(j)$$ which would yeild $$0=0+0+0+q(j)$$ and q(j) is a number that is non zero $$q(j) \neq 0$$ so this yields that the system is unsolvable whenever a stringer is isolated.

Plate Buckling Problems
Refer to the course page for more information with regards to this lecture. Equation numbers referenced in this lecture come from the aforementioned page.

Plotting buckling shapes under shear stress:

Express $$\displaystyle\{C_{22},C_{13},C_{31},C_{33} \}$$ in terms of $$\displaystyle C_{11}$$ for $$\displaystyle \vartheta = 1.5$$ 1) Find $$\displaystyle \lambda$$ for $$\displaystyle \vartheta = 1.5$$ using equation (30).
 * $$\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}$$ 2) Evaluate $$\displaystyle \mathbf K_{5 \times 5}$$ numerically in equation (26).
 * $$\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\}$$ 3) $$\displaystyle \mathbf K = [K_{ij}]$$
 * $$\displaystyle \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}{9}C_{11}\\0\\0\\0 \end{Bmatrix}$$

Solve for $$\displaystyle\{C_{22},C_{13},C_{31},C_{33} \}$$ in terms of $$\displaystyle C_{11}$$
 * $$\displaystyle \{C\}= [K]^{-1}\begin{Bmatrix} -\frac{4}{9}C_{11}\\0\\0\\0 \end{Bmatrix}$$
 * $$\displaystyle u_z = C_{11}\sin\left(\frac{\pi x}{a}\right)\sin\left(\frac{\pi y}{b}\right) + C_{22}\sin\left(\frac{2\pi x}{a}\right)\sin\left(\frac{2\pi y}{b}\right) + C_{13}\sin\left(\frac{\pi x}{a}\right)\sin\left(\frac{3\pi y}{b}\right) + C_{31}\sin\left(\frac{3\pi x}{a}\right)\sin\left(\frac{\pi y}{b}\right) + C_{33}\sin\left(\frac{3\pi x}{a}\right)\sin\left(\frac{3\pi y}{b}\right)$$

Set $$\displaystyle C_{11} = 1$$, plot $$\displaystyle u_z$$.

Results:
 * $$\displaystyle \lambda = 0.0288$$
 * $$\displaystyle \mathbf K = \begin{bmatrix}   2.1597  & -0.8000 &  -0.8000   & 1.4400\\

-0.8000 &  5.7706     &    0    &     0\\   -0.8000    &     0  &  1.6174    &     0\\    1.4400     &    0    &     0  & 10.9334\end{bmatrix}$$
 * $$\displaystyle \begin{Bmatrix}C_{22}\\C_{13}\\C_{31}\\C_{33} \end{Bmatrix}= \mathbf K^{-1}\begin{Bmatrix} -\frac{4}{9}\\0\\0\\0 \end{Bmatrix} = \begin{Bmatrix}  -0.3037\\

-0.0421\\  -0.1502\\    0.0400\end{Bmatrix}$$ Setting a = 3, b=2 (to keep the plate aspect ratio = 1.5).
 * $$\displaystyle u_z = \sin\left(\frac{\pi x}{3}\right)\sin\left(\frac{\pi y}{2}\right) -0.3037\sin\left(\frac{2\pi x}{3}\right)\sin\left(\frac{2\pi y}{2}\right) -0.0421\sin\left(\frac{\pi x}{3}\right)\sin\left(\frac{3\pi y}{2}\right) -0.1502\sin\left(\frac{3\pi x}{3}\right)\sin\left(\frac{\pi y}{2}\right) + 0.0400\sin\left(\frac{3\pi x}{3}\right)\sin\left(\frac{3\pi y}{2}\right)$$

Note: (continued from lecture on 12/1/08) Answer: in 2 parts Part 1: Equilibrium of isolated stringer Part 2: Closed cell, equilibrium of stringers Stringer 1:
 * $$\displaystyle \tilde{q_{12}} = \tilde{q_{31}} + \tilde{q_{41}} + q^{(1)}$$
 * $$\displaystyle \tilde{q_{31}} = 0$$
 * $$\displaystyle \Rightarrow \tilde{q_{12}} = \tilde{q_{41}} + q^{(1)}$$

Stringer 2:
 * $$\displaystyle \tilde{q_{24}} = \tilde{q_{12}} - \tilde{q_{23}}+q^{(2)}$$
 * $$\displaystyle \tilde{q_{23}} = 0$$
 * $$\displaystyle \Rightarrow \tilde{q_{24}} = \tilde{q_{12}} +q^{(2)}$$

Stringer 4:
 * $$\displaystyle \tilde{q_{41}} = \tilde{q_{24}} + \tilde{q_{34}} + q^{(4)} $$
 * $$\displaystyle \tilde{q_{34}} = 0$$
 * $$\displaystyle \tilde{q_{24}} = \tilde{q_{12}} +q^{(2)} = \tilde{q_{41}} + q^{(1)}+q^{(2)}$$
 * $$\displaystyle \Rightarrow \tilde{q_{41}} = \tilde{q_{41}}+q^{(1)}+q^{(2)}+q^{(4)}$$

Matlab Analysis
This Matlab assignment again analyzes the NACA 2415 airfoil. This time, it looks at the shear flow in the panels (shear flow and spar webs) when it is subjected to a transverse load. The load forces are a result of the lift and drag on the NACA 2415 airfoil.

Matlab Code
Now, here is the m-file used to calculate the Plate Buckling Characteristics:

Review of Wikiversity and Comparison to WebCT
This course relied heavily on the MediaWiki software for homework reports. As a group, we felt that this was not a very effective method of learning aerospace structures. Throughout the year, we spent more time formatting the Wiki pages and making them look good as opposed to working practice problems and working on Matlab projects. Writing equations and uploading pictures was a somewhat time consuming process that we feel took away from the learning experience. (By now we are all experts at using it, but even now it is still very time consuming)

When compared to the WebCT software that UF provides, MediaWiki is by far superior to WebCT for what this class uses it for. For online collaboration of teammates and making notes availiable to the entire class, MediaWiki is the way to go. It is definately the way of the future for online/teamwork collaboration, but we feel that it is still not user friendly enough to warrant a switch to it right now.

It was very interesting learning this software and using these new teaching methods this semester. This review is simply a collection of our thoughts about the MediaWiki software and the influence on the class.

Team Members Who Contributed to this Report
Mark Barry Eas4200c.f08.blue.a 14:13, 8 December 2008 (UTC)

Thomas Sexton EAS4200c.f08.blue.c 23:37, 9 December 2008 (UTC)

Marc Pelletier EAS4200c.f08.blue.b 17:10, 8 December 2008 (UTC)

Jeffrey Shea EAS4200c.f08.Blue.E 19:10, 9 December 2008 (UTC)

Clay Robertson eas4200c.f08.blue.f 21:48, 9 December 2008 (UTC)

Christian Garabaya

Wei-Teck Lee Eas4200c.f08.blue.g 02:10, 9 December 2008 (UTC)

title: User:Eas4200c.f08.blue.a/Homework7


text:
User:Eas4200c.f08.wiki.b/HW7

=Mtg 40 & 41 - Friday 12/5, Monday 12/8=

Cont'd from Mtg 39
Similarly, developing an equilibrium equation for stringer one yields

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

where $$\displaystyle \tilde q_{12} $$ and $$\displaystyle \tilde q_{41} $$ are unknown. $$\displaystyle \tilde q_{31} $$ becomes zero due to the cut between stringers three and one. Doing the same for stringer two yields

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

In this case, $$\displaystyle \tilde q_{23} $$ becomes zero due to the cut between stringers two and three. Again, developing the equation for stringer four yields

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

Substituting the equations for stringers one and two into this equation gives us

 $$\displaystyle \tilde q_{41} = ( \tilde q_{41} + q^{(1)} ) + q^{(2)} + \tilde q_{34} + q^{(4)} $$

Due to the cut between stringers three and four, $$\displaystyle \tilde q_{34} $$ becomes zero. In addition, $$\displaystyle \tilde q_{41} $$ cancels out, leaving

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

This, of course is not true, or even possible. The following steps explain why this is not true.

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

 $$\displaystyle 0 = \sum_{e=1}^4 q^{(e)} $$

Recall,

 $$\displaystyle q^{(e)} = n_z Q_z^{(e)} + n_y Q_y^{(e)} $$

 $$\displaystyle n_z := -(K_y V_y - K_yz V_z) $$

 $$\displaystyle n_y := -(K_z V_z - K_yz V_y) $$

Thus,

 $$\displaystyle \sum_{e=1}^4 q^{(e)} = n_z \sum_{e=1}^4 Q_z^{(e)} + n_y \sum_{e=1}^4 Q_y^{(e)} $$

Where $$\displaystyle \sum_{e=1}^4 Q_z^{(e)} $$ and $$\displaystyle \sum_{e=1}^4 Q_y^{(e)} $$ must both be equal to zero.

$$\displaystyle Q_y $$ can be expressed as

 $$\displaystyle Q_y (\hat z) = \int_{A(\hat z)} z dA $$

 $$\displaystyle Q_y = \int_A z dA = 0 = z_c \int_A dA $$

Notice that the last part of the above equation is from the mean value theorem. $$\displaystyle z_c $$ would be zero in this case, which is not possible.

Steps for Shear Buckling
The steps for plotting the buckling shape and expressing the coefficients $$\displaystyle \{ C_{11}, C_{22}, C_{13}, C_{31}, C_{33} \} $$ in terms of $$\displaystyle C_{11} $$ for the aspect ratio $$\displaystyle \vartheta = 1.5 $$ are as follows.

1) Find $$\displaystyle \lambda $$ for $$\displaystyle \vartheta = 1.5 $$ using equation 30 from Professor Vu-Quoc's page on plate buckling.

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

2) Evaluate $$\displaystyle \mathbf K _{5 \times 5} $$ numerically using equation 26 from Professor Vu-Quoc's page on plate buckling.

 $$  \displaystyle \mathbf \bar K _{5 \times 5} = \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]

$$

3) Solve $$\displaystyle \{ C_{11}, C_{22}, C_{13}, C_{31}, C_{33} \} $$ in terms of $$\displaystyle C_{11} $$.

 $$  \displaystyle \left[ \begin{array}{lllll} K_{22} &	 K_{23} &	 K_{24} &	 K_{25} \\	 K_{32} &	 K_{33} &	 K_{34} &	 K_{35} \\	 K_{42} &	 K_{43} &	 K_{44} &	 K_{45} \\	 K_{52} &	 K_{53} &	 K_{54} &	 K_{55} \end{array} \right] \left\{ \begin{array}{l} C_{22} \\	 C_{13} \\	 C_{31} \\	 C_{33} \end{array} \right\} =  \left\{ \begin{array}{l} -{4 \over 9} C_{11} \\	 0	 \\	 0	 \\	 0     \end{array} \right\} $$

 $$ \displaystyle u_z = C_{11} \sin{\pi x \over a} \sin{\pi y \over b} + C_{22} \sin{2 \pi x \over a} \sin{2 \pi y \over b} + C_{13} \sin{\pi x \over a} \sin{3 \pi y \over b} + C_{31} \sin{3 \pi x \over a} \sin{\pi y \over b} + C_{33} \sin{3 \pi x \over a} \sin{3 \pi y \over b} $$

Setting $$\displaystyle C_{11} = 1 $$, $$\displaystyle u_z(x,y) $$ can be plotted.

Proof of Stress Equation
[READ AND REPORT SEC. 4.2 (eqn 4.22a to eqn 4.28c)]

The following is a proof for the stress equation:

 $$\sigma _{xx}=\begin{bmatrix} z & y \end{bmatrix} \begin{bmatrix} K_y & -K_{yz}\\ -K_{yz} & K_z \end{bmatrix}\begin{Bmatrix} M_y\\M_z \end{Bmatrix}$$

First, the moments can be written as

 $$\displaystyle M_y = - EI_{yz} {d^2 V_o \over dx^2} - EI_y {d^2 W_o \over dx^2} $$

 $$\displaystyle M_z = - EI_{z} {d^2 V_o \over dx^2} - EI_{yz} {d^2 W_o \over dx^2} $$

Let the curvatures $$\displaystyle {d^2 V_o \over dx^2} = \chi_y $$ and $$\displaystyle {d^2 W_o \over dx^2} = \chi_z $$

In matrix form, whe have

 $$\displaystyle \begin{Bmatrix} M_y\\M_z \end{Bmatrix} = \begin{bmatrix} I_y & I_{yz}\\ I_{yz} & I_z \end{bmatrix}\begin{Bmatrix} -E \chi_z \\ -E \chi_y \end{Bmatrix} $$

 $$\displaystyle \varepsilon_{xx} = -y \chi_y - z \chi_z = \begin{bmatrix} z & y \end{bmatrix} \begin{Bmatrix} \chi_z \\ \chi_y \end{Bmatrix} $$

Thus,

 $$\displaystyle \sigma_{xx} = E \varepsilon_{xx} = E \begin{bmatrix} z & y \end{bmatrix} \begin{Bmatrix} -\chi_z \\ -\chi_y \end{Bmatrix} = E \begin{bmatrix} z & y \end{bmatrix} {1 \over E} \textbf{I}^{-1} \begin{Bmatrix} M_y \\ M_z \end{Bmatrix} $$

Notice that $$\displaystyle E $$ cancels out. Also note that

 $$\displaystyle \textbf{I}^{-1} = {1 \over D} \begin{bmatrix} I_z & -I_{yz}\\ -I_{yz} & I_y \end{bmatrix} $$

 $$\displaystyle D= I_y I_z - (I_{yz})^2 $$

Recall the expression for the shear flow

 $$\displaystyle q = -\int_A {d \sigma_{xx} \over dx} dA $$

Thus,

 $$\displaystyle {d \sigma_{xx} \over dx} = \begin{bmatrix} z & y \end{bmatrix} \textbf{I}^{-1} \begin{Bmatrix} {dM_y \over dx} \\ {dM_z \over dx} \end{Bmatrix} $$

 $$\displaystyle V_y = {dM_y \over dx} $$

 $$\displaystyle V_z = {dM_z \over dx} $$

 $$\displaystyle Q_y = \int_A zdA $$

 $$\displaystyle Q_z = \int_A ydA $$

license: Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/

title: User:Eas4200c.f08.wiki.b/HW7


text: User:Mardetanha

title: User:Mardetanha


text:
Portal:Social entrepreneurship/Curriculum/General/Introduction to Social Entrepreneurship/Navigation

=How to Navigate the Student Social Entrepreneurship Course=

Each module has a navigation template at the top of the page with links to: Introduction, Activities, Resources, Community and Links.

Example:

Usually you will start with the Introduction page and then move on to the Activities page with reference to the Resources and Community.

Some of the navigation templates also have a Home link which will either take you to the Introduction page, the main Curriculum page, or to some other page acting as the Home for the module.

Of course, you can also use the Search box in the left panel, and other features of Wikiversity.


text:
Portal:Social entrepreneurship/Curriculum/General/Introduction to Social Entrepreneurship/Educators

As an educator, you probably have lots of students/learners to guide, and most likely have to use the systems of your institution.

Use these wiki resources as an additional means of enhancing your education praxis and introducing your learners to commons-based peer production and Open Educational Resources.

To do this well requires the ability to edit the resources here, so you can improve them and be able to create special pages for your own students.

To learn how to edit wiki pages etc. see See How to get involved.

For special pages for your own students, we recommend creating sub-pages under the Community pages - e.g. browse to the relevant community page using the navigation template (see Navigation); on the Community page, click on 'edit this page' and create a link like this:

My Institution Name

This will create a link to a sub-page of the Community page which you can edit and add instructions for your students and areas for them to share resources (e.g. links to pages they have created, etc..


text:
Portal:Social entrepreneurship/Curriculum/General/Introduction to Social Entrepreneurship/Learners

Communicate with your instructor(s) if applicable and find out how they will be running the course. Their approach might be different from the approach presented here, and you may be required to use other resources in your institution.

Welcome to the world of commons-based peer production and Open Educational Resources :-)

To learn how to edit wiki pages etc. see See How to get involved.


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User:Eml4500.f08.team.foskey.ckf/hw7d

Electric Pylon Modes

Mode Problem
This is a continuation of the electric pylon frame problem. For this part, the eigenvectors are calculated and analyzed. The three lowest eigenvectors are also plotted. The three lowest vibrational periods are also found.

MATLAB Mode Code (electricframemodes.m)
The following MATLAB code was used to find the eigenvectors, and vibrational periods, and plot the figures. It is based heavily off of the earlier electric pylon problem with a truss system.

Additional Function
The function TransientPlaneFrameElement is required for the mode code to function.

This function works by receiving the element properties (modulus of elasticity, cross sectional area, moment of inertia, density, and end point coordinates). The length of the element is found from the coordinates, and the director cosines are found from the length and coordinates. The local element stiffness matrix k_ is then calculated, as is the transformation matrix T. The local elemental stiffness matrix is then converted to the global elemental stiffness matrix k by using the transformation matrix. Next, the local mass matrix m_ is calculated, and the global mass matrix mis found by again using the transformation matrix

Comments On The Mode Code
The mode code (electricpylonmodes.m) begins by running the loading code (electricframe.m) to receive the already calculated values that will be needed. These values include nodal coordinates, and element properties such as cross sectional area and modulus of elasticity. The density of the elements is then given in g/cm3, and then converted to kg/m3. The length of each element is calculated, and the mass of each element is found by multiplying the element's length by the cross sectional area and by the density.

The mass of each element is then divided by two, and each half mass is added to the global node represented by each end of the element. This is done for all of the elements until the entire mass of the truss is concentrated at the nodes.

A mass matrix is then created by using the function TrainsientPlaneFrameElement.

A reduced mass matrix M_bar, and a reduced stiffness matrix K_bar is then created using MATLAB's setdiff function. This function eliminates the values of the matrices that correspond to the constrained points. This is valid due to the principle of virtual work. The eigenvalues and eigenvectors are then found using MATLAB's eig function.

The natural frequencies $$ \omega $$, vibrational frequencies f, and the vibrational periods T are then found using the following formulas: $$ \omega = \sqrt{\lambda} $$

$$f = \frac{\omega}{2\pi}$$

$$T = \frac{1}{f} $$

where $$ \lambda $$ is an eigenvalue.

The code then proceeds to find the three lowest eigenvalues. This is accomplished by setting variables for the lowest eigenvalues (low1, low2, and low3) equal to large numbers, and then looping through all of the eigenvalues, replacing low1, low2, and low3 with any eigenvalues that are encountered that are lower than the previous values. While doing this, the code keeps track of which rows the lowest eigenvalues were in.

The code then organizes the eigenvectors. The calculated eigenvectors are contained in an 88x88 matrix that is very cumbersome. Each column of the eigenvector matrix represents one eigenvector. Since the assignment required the finding of the lowest three eigenvalue-eignevector pairs, the needed eigenvectors are found in the columns that correspond to the rows in which the lowest eigenvalues were in. The code extracts the desired eigenvectors and puts them into column matrices called modes1, modes2, and modes3.

In a similar manner as the determining of the lowest eigenvalues, the lowest three vibrational periods are found.

Since there was a number of things calculated by this code, with a majority of the values being large matrices, only a selected amount of data was deemed important for output. The next section of code outputs the important data. This data consists of the eigenvalues, the three lowest eigenvalues and their corresponding eigenvectors, and the three lowest vibrational periods.

To finish the code, three plots are created. Each plot consists of an undeformed and a deformed frame. Each plot is only an "incomplete" plot, as it does not account for the bending of elements. The undeformed frame is plotted using the same lines of code from the loading code. The deformed frame is plotted using lines of code similar to the code from the loading code, but with adjustments to the coordinates of the nodes. For this situation, the "deformed" nodal positions are created by adding the values from an eigenvector to the original nodal coordinates. This is done because the eigenvectors represent the natural displacements for the truss system. By adding the natural displacements to the original nodal coordinates, the result is the naturally deformed truss. This is how the truss system will move, even under no loads, due to the natural frequencies associated with the elements. These natural displacements, however, are very small. In order for the natural displacements to be seen on the plots, a scaling factor was used for each plot. This scaling factor was 20 for figure 1, 50 for figure 2, and 10 for figure 3.

Three plots are created from the mode code. Each of these plots represents the deformations that occur as a result of one of the three lowest eigenvectors.

Output Results
Running the mode code "electricframemodes.m" creates the following output results.

Comments On The Output
The results output shows several things. The first is eigenvalues, which is a column matrix that is a diagonal matrix of the matrix of eigenvalues. The next items displayed are the three lowest eigenvalues: low1, low2,& low3 (low1 being the lowest, low3 the highest). These eigenvalues are -0.020486, -0.020234, and, -0.0086386. The output then shows three column matrices: modes1, modes2, and modes3, which are the three lowest eigenvectors (modes1 being the lowest, modes3 the highest). The last items displayed are the three lowest vibrational periods: lowT1, lowT2, & lowT3. These lowest vibrational periods are 55.474, 57.446, and 58.344. All three are in terms of seconds.

Figures
Running the mode code will create three figures. Each figure represents the displacements from one of the three lowest eigenvectors. In each figure, the red dotted lines represent the undeformed frame system, and the solid blue lines represent the deformed frame. The x and y axis for all three figures are in terms of meters. The scaling factor was 20 for figure 1, 50 for figure 2, and 10 for figure 3. A different scaling factor was used for each figure so that the vibrational modes might be better seen for each case. Again, these frame plots are incomplete plots as they do not show the bending that each element undergoes. To simplify the problem, a straight line is made from node to node for each element.

Figure 1

Figure 2

Figure 3

Comparison Of Modes In Frame Systems And Truss System
Comparing the results from the frame system to the truss system, one notices that the frame system deforms more in natural vibration. A scaling factor of 100,000 was needed for the truss system, while a scaling factor of 10 to 30 worked for the frames. Also, the vibrational periods of the frame system are lower than those of the truss system. However, it is believed that there was an error in the coding of the truss problem, as all of the vibrational periods are very high, and two of the values are equal.

title: User:Eml4500.f08.team.foskey.ckf/hw7d


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User talk:Johnny-boy

Welcome
Hello Johnny-boy, and welcome to Wikiversity! If you need help, feel free to visit my talk page, or contact us and ask questions. After you leave a comment on a talk page, remember to sign and date; it helps everyone follow the threads of the discussion. The signature icon in the edit window makes it simple. To get started, you may
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And don't forget to explore Wikiversity with the links to your left. Be bold, and see you around Wikiversity! --Emesee 22:33, 7 December 2008 (UTC)

title: User talk:Johnny-boy


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User:Eas4200c.f08.wiki.c/homework report 7

=Meeting 35=

Mean (Average) Value Theorem (MVT)

$$\int_{A}^{}{\bar{y}dA} = \bar{y}_{c}\int_{A}^{}{dA}$$

$$=\bar{y}A$$

$$\int_{A}^{}{\bar{z}dA} = \bar{z}_{c}A$$

With $$\bar{z}_{c}$$ denoting the average z value.

$$A = \sum_{i=1}^{4}{A_{i}}$$

(Neglect skin and spar webs)

This is a similar problem as was done in homework 5.

Equation 5.5 in the book is given as:

$$q(s) = (k_{yz}Q_{z}(s) - k_{z}Q_{y}(s))V_{z}$$

Since: $$1) V_{z} 2) k_{yz},k_{z} 3) Q_{z}, Q_{y}$$ are all concentrated in the stringers and are all independent of s, this implies that the shear flow q(s) is constant between 2 stringers.

However, q(s) would increment (jump) when crossing a stringer.

Now, to find q(s), the following recipe is given:

Step 1:

Find $$(\bar{y}_{c}, \bar{z}_{c})$$ by the following formulas:

$$\bar{y} = \sum{\frac{yA}{A}}$$

$$\bar{z} = \sum{\frac{zA}{A}}$$

Step 2:

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

Step 3:

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

Step 4:

Follow the path of 's' to find $$q_{12}, q_{23}, q_{34}$$

Where the subscripts denote the shear flow from one stringer to the next.

It should be noted that:

$$Q_{z}^{23} = y_{1}A_{1} + y_{2}A_{2}$$

and so on. In other words, each Q is the sum of the Q's before it.


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User:Eml4500.f08.FEABBQ.koby/HW7

Notes for Nov. 21
Continuation of Dimensional Analysis

$$[ \varepsilon ] = \frac{du}{dx} = \frac{L}{L}=1$$

$$[ \sigma ] = [E] = \frac{F}{L^2}$$

$$\left[ \frac{EA}{L}\right] = \left[ \tilde{k_{11}}\right]=\frac{(F/L^2)(L^2)}{L}=\frac{F}{L}$$

$$\left[ \tilde{k_{11}} \tilde{d_1} \right] = [\tilde{k_{11}}][\tilde{d_1}]=\frac{[6][E][I]}{[L^2]} = \frac{1(F/L^2)L^4}{L^2} = F$$

Element F-d relation in global coordinates from element F=d relations in local coordinates, i.e. obtain

$$\mathbf{k}_{6x6}^{(e)}\mathbf{d}_{6x1}^{(e)}=\mathbf{f}_{6x1}^{(e)}$$

$$\mathbf{k}_{6x6}^{(e)}=\tilde{\mathbf{T}}_{6x6}^{(e)T}\tilde{\mathbf{k}}_{6x6}^{(e)}\tilde{\mathbf{T}}_{6x6}^{(e)}$$

Notes for Dec. 5
$$\begin{Bmatrix} u_x(\tilde{x})\\ u_y(\tilde{x}) \end{Bmatrix} = \mathbf{R}^T\begin{Bmatrix} u(\tilde{x})\\ v(\tilde{x}) \end{Bmatrix}$$  (1)

$$\begin{Bmatrix} u_x(\tilde{x})\\ u_y(\tilde{x}) \end{Bmatrix} = \begin{bmatrix} N_1 & 0 & 0 & N_4 & 0 & 0\\ 0 & N_2 & N_3 & 0 & N_5 & N_6 \end{bmatrix} \begin{Bmatrix} \tilde{d_1}^{(e)}\\ \vdots\\ \tilde{d_1}^{(e)} \end{Bmatrix}$$  (2)

Perform dimensional analysis of equation (2):

$$[u]=L$$
 * First, a look at the normal linear degrees of freedom:

$$[N_1]=[N_4] = 1$$

$$[N_1][\tilde{d_1}],[N_4][\tilde{d_4}]$$

$$1\cdot L,1\cdot L = L$$

The dimensional analysis for DOF's 2 and 5

 * Now a look at the dimensional analysis of a rotational DOF:

$$[v]=L$$

$$[\tilde{d_3}]=[1]$$

$$[N_3][\tilde{d_3}],[N_6][\tilde{d_6}]$$

$$L\cdot 1,L\cdot 1 = L$$

Derivation of Shape Functions

Recall: Governing PDE for beams. Look at governing equations without internal forces of transverse forces or distributed truss loads. This is the static case:

$$\frac{\delta ^2}{\delta x^2} \left\{(EI) \frac{\delta v^2}{\delta x^2}\right\}=0$$

If EI is constant along the beam: $$\frac{\delta ^2}{\delta x^2}v=0$$

After integrating this equation four times, you arrive at: $$\Rightarrow v(x)=C_0+C_1x^1+C_2x^2+C_3x^3$$

Now the boundary conditions can be applied to find $$C_0, C_1, C_2, C_3$$ for each shape function.

 * $$\mathbf{N_2}$$

$$v(0)=1=C_0$$

$$v(L)=0=1+C_1L+C_2L^2+C_3L^3$$

$$v'(x)=C_1+2C_2x+C_3x^2$$

$$v'(0)=0=C_1$$

$$v'(L)=0=2C_2L+3C_3L^2$$

Now solve for $$C_2, C_3$$:

$$C_3={-2C_2}/{3L}$$

plugging $$C_3$$ back into the equation for v(L): $$0=1+C_2L^2+{-2C_2}/{3L}L^3$$

$$C_2=-3/L, C_3=2/L^3$$

$$N_2(x)=1-\left( \frac{3}{L^2}\right)x+\left( \frac{2}{L^3}\right)x^3$$

 * $$\mathbf{N_3}$$

change... $$v(0)=0=C_0$$

$$v'(0)=1=C_1$$

$$v(L)=0=L+C_2L^2+C_3L^3$$

solving for $$C_3=-1/L^2-C_2/L$$

$$v'(L)=0=1+2C_2L+3C_3L^2$$

solving for $$C_3=-1/3L^2-2C_2/3L$$

Now solve for $$C_2, C_3$$ by first setting $$C_3$$ equal to itself.

$$C_3=-1/3L^2-2C_2/3L=-1/L^2-C_2/L$$

$$C_2=-2/L$$

$$C_3=1/L^2$$

$$N_2(x)=x-\left( \frac{2}{L}\right)x^2+\left( \frac{1}{L^2}\right)x^3$$

 * $$\mathbf{N_5}$$

$$v(0)=0=C_0$$

$$v'(0)=0=C_1$$

$$v(L)=1=0+C_2L^2+C_3L^3$$

$$v'(x)=C_1+2C_2x+C_3x^2$$

$$v'(L)=0=2C_2L+3C_3L^2$$

Now solve for $$C_2, C_3$$:

$$C_3={1-C_2L^2}/{L^3}$$ from v(L)

$$C_3={-2C_2}/{3L}$$ from v'(L)

Setting $$C_3$$ equal to itself: $${(1-C_2L^2)}/{L^3}={-2C_2}/{3L}$$

$$C_2={3}/{L^2}, C_3={-2}/{L^3}-2L/3$$

$$N_5(x)=\left( \frac{3}{L^2}\right)x^2-\left( \frac{2}{L^3}\right)x^3$$

 * $$\mathbf{N_6}$$

$$v(0)=0=C_0$$

$$v'(0)=0=C_1$$

$$v(L)=0=C_2L^2+C_3L^3$$

Solve for $$C_3$$... $$C_3=-C_2/L$$

$$v'(L)=1-2C_2L+3C_3L^2$$

Solving again for $$C_3$$... $$C_3 = 1/3L^2-{2C_2}/{3L}$$

Setting $$C_3$$ equal to itself yields $$C_2$$: $$C_2=-1/L$$

$$C_3=1/L^2$$

$$N_6(x)=-\left( \frac{1}{L}\right)x^2+\left( \frac{1}{L^2}\right)x^3$$

title: User:Eml4500.f08.FEABBQ.koby/HW7


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User:Eml4500.f08.team.foskey.ckf/hw7e

Animation of the modes from the elecric pylon truss

Animation Problem
In this problem, the mode plots from the original truss problem are to be animated. In order to accomplish this, the old code is reused. A new MATLAB code (shown later in this page) animates the modes using the information from the old code.

To run the Animation Code, the following MATLAB m-files are required:

electric_pylon.m (original provided code) solvingpylon1.m (loading code) solvingpylon2.m (lumped mass code) PlaneTrussElement.m (function) PlaneTrussResults.m (function) NodalSoln.m (function) TransientPlaneTrussElement.m (function)

All of the above files can be found in the Electric Pylon Truss Problem.

Matlab Code (Animation Code)
The following MATLAB code is used to create the frames that will be used to make the animations

Comments On The Animation Code
The animation code begins by running the original electric pylon truss code for finding the modes "solvingplyon2.m". This loads all of the needed data. Next the number of frames is selected, in this case "N=8". The next lines are critical to how the code works. In the lines:

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% choose=3;  %select which mode to animate (1,2 or 3) %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

the user must manually select which mode to animate. This saves time when running the code, as 16 images will be displayed for each mode. It also saved time when writing the code. The lines following the selection display the only output for this code, which is information regarding the number of frames, and which mode is being animated. The next sections of the code are essentially the same concept repeated three times, once for each mode. There are three "if" statements that check which mode is being animated, and then use the necessary data from "solvingpylon2.m".

In each mode loop, the vibrational period is divided by the number of frames to obtain increments of the period. A matrix del is created to hold the steadily increasing increments (cell 1 of del is one segment, cell 2 is two segments, and cell 8 is eight segments, or the full vibrational period).

Next, partial modes are created by looping through the number of frames and creating a column matrix of partial modes based on del. All of the partial modes are put into a single matrix, in which the next partial mode begins as the previous one ends. A scaling factor of 1,000,000,000 is also created here.

Next, the code plots the original undeformed truss. Then the deformed truss is plotted by adding the partial modes as displacements to the nodal coordinates (this will be referred to as a "positive deformation"). This is repeated for all eight frames, with each figure showing a slightly more deformed truss. Each figure is saved as a frame using MATLAB's getframe code.

This process is repeated again for eight frames, this time subtracting the partial modes from the original nodal coordinates (referred to as "negative" deformation). Doing this results in 16 frames, 8 of which show the truss moving in one direction, 8 show it moving in the opposite.

The above code process is repeated for all three modes, only one of the three will actually run when the code is run, based upon the user's input as to which mode should be animated.

To better show the animation, there will be 24 images in one mode animation cycle. The first eight image frames will be the truss moving from undeformed to a "positive deformation". The next eight are the first eight in reverse order so that the truss returns to its undeformed shape. The next eight images will show the truss undergoing "negative deformation", and the final eight images show the truss returning to its original shape.

The above is accomplished by using loops and hard coding to assign the image frames to their proper order to create a 24 frame video.

A "close all" statement is seen in the last few lines. This is commented out in the above code because the figures were needed for saving purposes (as explained later in the animations section). For a more efficient code, this line should be left in place to close the open figures.

The line "movie(H,3)" of the code will play the animation in MATLAB. This animation will loop through 3 cycles of the 24 frame animation.

Three "if" statements end the code, each creating a video file based on the mode chosen by the user. This is done using MATLAB's movie2avi code.

In order to obtain complete results, the user must run this code three times, each time changing the value of the variable "choose" to 1, 2 and 3. The user will then have three avi videos in their MATLAB directory folder entitled mode1animate.avi, mode2animate.avi, and mode3animate.avi.

Animations
The above MATLAB mode code will create three avi videos for the truss modes. Each of these can be opened with a media player, such as Windows Media Player. However, MediaWiki does not allow for the uploading of avi files.

In order to upload the animations, they must be converted to a different format, such as GIF. In order to do this, each image frame needed to be saved as its own file. This was done by manually saving each figure created by MATLAB as a tiff file. For each mode there are 16 different images, each representing one frame of the final animation.

The images were put together in their proper order and made into a single gif file by using Photoshop.

title: User:Eml4500.f08.team.foskey.ckf/hw7e


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User:Eml4500.f08.delta 6.krueger/HW Evaluation of wiki versus E-learning

 Wiki versus E-learning

Wiki: Pros: 1) The multiple variations of notes and homeworks made studying more efficient and decreased errors on homeworks that were already graded. 2) The Compilation of the best of the notes can be used by later classes. This is put in the pros, but could also be a con. If the homework was wrong and not corrected it would influence the reader negatively. 3) The wiki network always seems to be available. There hasn't been a time when I could not log on. This is not the case with E-learning. 4) Now that the code for wiki is known the students can use wiki for other projects. Cons: 1) Having homework on a continuously available website with no way of restricting access makes cheating extremely easy. 2) Learning the code added extra work to an already work intense degree program. 3) The time to grade the class notes and the homeworks was slow. The response was much longer than with regular courses. 4) The materials sometimes felt as though it was spread all over the place. I don't feel like it was a compact user friendly way of relating the course material. 5) The pages can be changed by others. This could also lead to inaccuracies in studying. 6) The errors in the homeworks were not always well defined and could lead to inaccuracies in studying.

E-learning: Pros: 1) The format is compact and easy to navigate. 2) Cheating is very difficult. Cons: 1) This format cannot be used to study. 2) E-learning is not really a learning tool. It is more of a line of information between the teacher and the students. 3) The system is often slow and many times the student is unable to log on.

 My Opinion  I feel that wiki can be a good learning tool, but this takes an extremely large amount of time and diligence by the teacher and the T.A.'s. The homeworks would need to be graded in a shorter amount of time and the errors would need to be well marked and the correct answer given. Also the open site could lead to errors in studying. The pages could be changed and this could introduce errors. Without these problems wiki would be a good tool for courses. I feel that E-learning is a better way to deliver and receive information and reduces errors. Therefore, I feel that E-learning is a better system for the classroom.


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User:Eml4500.f08.bike.viale/homework report 7

= Mtg 41 =

Use above boundary conditions to solve for $$c_0$$ through $$c_3$$

$$V(0) = 1 = C_0$$

$$V(L) = 1 + C_1L + C_2L^2 + C_3L^3$$

$$V'(x) = C_1 + 2C_2x + 3C_3x^2$$

$$V'(0) = C_1 = 0$$

(2) $$V'(L) = 2C_2L +3C_3L^2 = 0$$

(2) $$\Rightarrow C_3 = -\frac{2}{3}\frac{C_2}{L}$$

(1) $$\Rightarrow 0 = 1 + C_2L^2 + (-\frac{2}{3}\frac{C_2}{L}) - L^3$$

Where the second, third, and fourth terms combine to be:

$$C_2 = -\frac{2}{L^2}$$

This is similar to the result for $$N_2$$ found previously.

For $$N_3$$

Boundary Condition

$$V(0) = V(L) = 0$$

$$\tilde{d}_3$$ rotational

$$V'(0) = 1, V'(L) = 0$$

For $$N_5$$:

$$V(0) = 0, V(L) = 1$$

$$V'(0) = 0, V'(L) = 0$$

For $$N_2$$:

$$V(0) = V(L) = 0$$

$$V'(0) = 0, V'(L) = 1$$

$$\tilde{d}_6$$ rotational

The plots for $$N_5 and N_6$$ where previously provided.

Now we must derive the coefficients in $$\mathbf{\tilde{k}}$$ ( the element stiffness matrix)

The coefficients with EA are already complete, however the ones with EI still need to be computed.

$$\tilde{k}_{22} = \frac{12EI}{L^3} = \int_{0}^{L}{\frac{d^2N_2}{dx^2}(EI)\frac{d^2N_2}{dx^2}dx}$$

=Mtg 41=

$$\hat{k}_{23} = \frac{6EI}{L^2} = \int_{0}^{L}{\frac{d^2N_2}{dx^2}(EI)\frac{d^2N_3}{dx^2}dx}$$

In general,

$$\hat{k}_{ij} = \int_{0}^{L}{\frac{d^2N_i}{dx^2}(EI)\frac{d^2N_j}{dx^2}dx}$$

With i,j = 2,3,5,6

Elastodynamics (trusses, 2-D frames, 3-D elasticity)

Analyzing the Modal problem from lecture 31 using discrete Principle of Virtual Work is as follows:

$$ \mathbf{\bar{w}} = [\mathbf{\bar{M}}\mathbf{\ddot{\bar{d}}} + \mathbf{\bar{K}}\mathbf{\bar{d}} - \mathbf{\bar{F}}] = 0$$

In the above equation, the boundary conditions are already applied and the K matrix represents the reduced stiffness matrix.

This equation is true for all $$\mathbf{\bar{w}}$$.

$$\mathbf{\bar{M}}\mathbf{\ddot{\bar{d}}} + \mathbf{\bar{k}}\mathbf{\bar{d}} = \mathbf{\bar{F}}(t)$$

$$\mathbf{\bar{d}}(0) = \mathbf{\bar{d}}_0$$

$$\mathbf{\dot{\bar{d}}}(0) = \mathbf{\bar{V}}_0$$

The above equations will be referred to as (1) for the rest of this discussion.

These are the complete ordinary differential equations (ODEs) which are second order in time with initial conditions governing the elastodynamics of the discretized continuous problem with multiple degrees of freedom.

Now, solving the method for solving equation (1) is as follows:

1) Consider the unforced vibrations problem:

$$\mathbf{\bar{M}}_{nxn}\mathbf{\ddot{v}}_{nx1} + \mathbf{\bar{K}}_{nxn}\mathbf{v}_{nx1} = \mathbf{0}_{nx1}$$

This equation is equal to zero becuase it is unforced. This will be referred to as equation (2) from now on.

Assume:

$$\mathbf{v}(t)_{nx1} = (sinwt)\mathbf{\phi _{nx1}}$$

Where the phi matrix is not time dependent. This solution gives the oscillating motions that have been seen in the FEA animations that have been shown.

Thus:

$$\mathbf{\ddot{v}} = -\omega ^2 sin\omega t\mathbf{\phi }$$

This means equation (2) becomes:

$$-\omega ^2 sin\omega t\mathbf{\bar{M}}\mathbf{\phi } + \omega ^2 sin\omega t\mathbf{\bar{K}}\mathbf{\phi } = \mathbf{0}$$

Therefore,

$$\mathbf{\bar{k}}\mathbf{\phi } = \omega ^2\mathbf{\bar{M}}\mathbf{\phi }$$

Which is the generalized eigenvalue problem, as it is of the form:

$$\mathbf{A}\mathbf{x} = \lambda \mathbf{B}\mathbf{x}$$

Where lambda is an eigenvalue.

A standard eigenvalue problem is given when the B matrix is equal to the identity matrix and the equation becomes:

$$\mathbf{A}\mathbf{x} = \lambda \mathbf{x}$$

This means that

$$\lambda =\omega ^2$$ is an eigenvalue.

Also the eigenpairs can be represented as:

$$(\lambda _i,\mathbf{\phi _i})$$

with i = 1 through n

Now the 'ith' mode for the animation can be represented as:

$$\mathbf{v}_i(t) = (sinw_it)\mathbf{\phi _i}$$

for i = 1 through n

Now step 2 in the solution is as follows:

2) Modal superposition method:

By using the orthogonal properties of the eigenpairs we can equate:

$$\mathbf{\phi _i}^T_{1xn}\mathbf{\bar{M}}_{nxn}\mathbf{\phi }_{nx1} = \delta _{ij} = \begin{cases} & \text{1 if } i=j \\ & \text{0 if } i\neq j \end{cases}$$

This delta is referred to as the Kronecker delta.

This reduction is possible due to the mass orthogonality of the eigenvector.

Now, applying this to Eq (1) and two gives:

$$\mathbf{\bar{M}}\mathbf{\phi }_j = \lambda _j \cdot \mathbf{\bar{k}}\mathbf{\phi }_j$$

$$ \mathbf{\phi }_i^T\mathbf{\bar{M}}\mathbf{\phi }_j = \lambda _j\phi _i^T\mathbf{\bar{k}}\mathbf{\phi }_j$$

where the left hand side is equal to the Kronecker delta, therefore:

$$\phi _i^T\mathbf{\bar{k}}\mathbf{\phi }_j = \frac{1}{\lambda _j}\delta _{ij}$$

$$\mathbf{\bar{d}}_{nx1}(t) = \sum{\zeta _i(t)\mathbf{\phi }_{inx1}}$$

This means equation (1) can be written as:

$$\mathbf{\bar{M}}(\sum_{j}^{}{\ddot{\zeta }_j\mathbf{\phi }_j}) + \mathbf{\bar{K}}(\sum_{j}^{}{\zeta _j\mathbf{\phi }_j}) = \mathbf{F}$$

In the above equation, the first term in parenthesis is equal to the second derivative of the d matrix, while the second term in parenthesis is equal to the d matrix itself. We can also write:

$$\sum_{j}^{}{\ddot{\zeta }_j(\mathbf{\phi }_i^T\mathbf{\bar{M}}\mathbf{\phi }_j} + \sum_{j}^{}{\zeta _j(\mathbf{\phi }_i^T\mathbf{\bar{K}}\mathbf{\phi }_j} = \mathbf{\phi }_i^T\mathbf{F}$$

The first term in parenthesis in this equation is equal to the Kronecker delta, while the second term in parenthesis is equal to an eigenvalue times the Kronecker delta. This means that this equation can be written as an ordinary differential equation as follows:

$$\ddot{\zeta } + \lambda _i\zeta _i = \mathbf{\phi }_i^T\mathbf{F}$$

with i = 1 through n


text:
User:Eml4500.f08.Lulz.js/HW7

Frame Elements
A basic frame element consists of a truss element combined with a beam element, which gives the frame element the capacity to handle both axial and transverse deformations. Consider Figure 1, which shows a model problem consisting of a system of two frame elements. The point of connection between the two elements, where the force $$\;P$$ is applied, is a rigid connection. This means that the angle between the two elements remains constant after deformation.

The free body diagrams for both elements are shown in Figures 2 and 3. Generally speaking, the degrees of freedom for such a system correspond directly to the general forces, or $$d_i^{(e)} \Rightarrow f_i^{(e)}$$. However, for this problem, two new degrees of freedom have been introduced for both elements. These degrees of freedom correspond as follows from the force vector:

 * $$\underbrace{\left\{\begin{matrix} d_3^{(e)} \\ d_6^{(e)} \end{matrix}\right\}}_{\mbox{rotational degrees of freedom}} \Rightarrow \underbrace{\left\{\begin{matrix} f_3^{(e)} \\ f_6^{(e)} \end{matrix}\right\}}_{\mbox{bending moments}}$$

These new degrees of freedom correspond to rotations, and the associated forces are actually bending moments. Figure 4 shows the global view of the system, with the global degrees of freedom. Once element stiffness matrices $$\mathbf{k}^{(e)}\;$$ are generated, they can be assembled into the global stiffness matrix as follows:

 * $$\mathbf{K} = A\mathbf{k}^{(e)}$$

$$A\;$$ is the assembly operator. An overview of the global stiffness matrix is given in Figure 5.

In order to generate the element stiffness matrices, consider the local degrees of freedom for element 1, as shown in Figure 6. The stiffness matrices can be determined from the following familiar relationship:

 * $$\mathbf{\tilde{k}}^{(e)}\mathbf{\tilde{d}}^{(e)} = \mathbf{\tilde{f}}^{(e)}$$

Where $$\mathbf{\tilde{d}}^{(e)}$$ and $$\mathbf{\tilde{f}}^{(e)}$$ are given as:

 * $$\mathbf{\tilde{d}}^{(e)} = \begin{Bmatrix} \tilde{d}^{(e)}_1\\ \vdots \\\tilde{d}^{(e)}_6 \end{Bmatrix}$$

 * $$\mathbf{\tilde{f}}^{(e)} = \begin{Bmatrix} \tilde{f}^{(e)}_1\\ \vdots \\\tilde{f}^{(e)}_6 \end{Bmatrix}$$

For the rotational degrees of freedom, the following relationships are true:

 * $$\tilde{f}^{(e)}_3 = f^{(e)}_3$$, $$\tilde{f}^{(e)}_6 = f^{(e)}_6$$

Using these considerations, the general element stiffness matrix $$\mathbf{\tilde{k}}^{(e)}$$ can be constructed as follows:

 * $$\mathbf{\tilde{k}}^{(e)} = \begin{bmatrix} \frac{EA}{L}& 0 & 0 & \frac{-EA}{L} & 0 & 0\\ & \frac{12EI}{L^2} & \frac{6EI}{L^2} & 0 & \frac{-6EI}{L^2} & \frac{6EI}{L^2}\\  &  & \frac{4EI}{L} & 0 & \frac{-6EI}{L^2} & \frac{-2EI}{L}\\  &  &  & \frac{EA}{L} & 0 & 0\\  &  &  &  & \frac{12EI}{L^2} & \frac{-6EI}{L^2}\\ sym. &  &  &  &  & \frac{4EI}{L}\end{bmatrix}$$

Dimensional Analysis
It is useful to look at the dimensions of some of the variables involveed in the frame system. Consider first the non-rotational degrees of freedom:

 * $$[\tilde{d}_1] = L = [\tilde{d}_i]$$, for $$i=1,2,4,5\;$$

Conversely, consider the rotational degrees of freedom:

 * $$[\tilde{d}_3] = 1 = [\tilde{d}_6]$$

Figure 7 displays the arc length relationship for circular sections. This relationship is summarized as:

 * $$\widehat{AB} = R\theta \Rightarrow \theta = \frac{\widehat{AB}}{R}$$

Performing a dimensional analysis on this concept results in:

 * $$[\theta] = \frac{[\widehat{AB}]}{[R]} = \frac{L}{L} = 1$$

Furthermore, consider the stress strain relation:

 * $$\sigma = E\varepsilon \Rightarrow [\sigma] = [E]\underbrace{[\varepsilon]}_{1} = [E]$$

 * $$[\sigma] = [E] = \frac{F}{L^2}$$

Consider the axial displacement $$q\;$$:

 * $$[q] = \frac{[du]}{[dx]} = \frac{L}{L} = 1$$

For some more fundamental variables, consider the area and moment of inertia:

 * $$[A] = L^2\;$$

 * $$[I] = L^4\;$$

With the dimensions of these more basic terms defined, it is possible to analyze the dimensions of more complex variables. For example, consider the stiffness factor $$k\;$$:

 * $$\left[\frac{EA}{L}\right] = [\tilde{k}_1] = \frac{(\frac{F}{L^2})(L^2)}{L} = \frac{F}{L}$$

Analyzing the product of the global stiffness coefficient and the global degree of freedom proves that this product generates the global force components:

 * $$[\tilde{k}_{11}\tilde{d}_1] = [\tilde{k}_{11}][\tilde{d}_1] = (\frac{F}{L})(L) = F$$

 * $$[\tilde{k}_{23}\tilde{d}_3] = [\tilde{k}_{23}]\underbrace{[\tilde{d}_3]}_1 = \frac{[6][E][I]}{[L^2]} = \frac{(1)(\frac{F}{L^2})(L^4)}{L^2} = F$$

Element Displacement Relation
Returning to the local degrees of freedom, the following relation can be written to convert $$\mathbf{d}^{(e)}\;$$ to $$\mathbf{\tilde{d}}^{(e)}\;$$:

 * $$\begin{Bmatrix} \tilde{d}_1\\ \tilde{d}_2\\ \tilde{d}_3\\ \tilde{d}_4\\ \tilde{d}_5\\ \tilde{d}_6 \end{Bmatrix} = \underbrace{\begin{bmatrix} l^{(e)}& m^{(e)} & 0 & 0 & 0 & 0\\ -m^{(e)} & l^{(e)} & 0 & 0 & 0 & 0\\ 0 & 0 & 1 & 0 & 0 & 0\\ 0 & 0 & 0 & l^{(e)} & m^{(e)} & 0\\ 0 & 0 & 0 & -m^{(e)} & l^{(e)} & 0\\ 0 & 0 & 0 & 0 & 0 & 1\end{bmatrix}}_{\mathbf{\tilde{T}^{(e)}}} \begin{Bmatrix} d_1\\ d_2\\ d_3\\ d_4\\ d_5\\ d_6 \end{Bmatrix}$$

Where $$\begin{bmatrix} l^{(e)} & m^{(e)} \\ -m^{(e)} & l^{(e)}\end{bmatrix}$$ is the previously discussed $$\mathbf{R}\;$$.

Derivation of Element Stiffness Matrix from the Principle of Virtual Work
Consider the Principle of Virtual Work (PVW) for beams given as Equation 1 below:

 * $$\int_0^L w(\underbrace{x}_\tilde{x}) [\frac{-\partial^2}{\partial x^2}[(EI)\frac{\partial v^2}{\partial x^2}] + f_t(x) - m(x)\ddot{v}] = 0\;,\;\mbox{for all possible} \;w(x)\;\;(1)$$

Isolating the first term and performing an integration by parts yields the following:

 * $$\alpha := \int_0^L \underbrace{w(x)}_{s(x)} \underbrace{\frac{\partial^2}{\partial x^2}\left\{(EI)\frac{\partial v^2}{\partial x^2}\right\}}_{\underbrace{\frac{\partial}{\partial x} \underbrace{\left(\frac{\partial}{\partial x}\left\{(EI)\frac{\partial v^2}{\partial x^2}\right\}\right)dx}_{r(x)}}_{r'(x)}}$$

 * $$\alpha := \underbrace{\left[w(x)\frac{\partial}{\partial x}\left\{(EI)\frac{\partial v^2}{\partial x^2}\right\}\right]^L_0}_{\beta_1} - \int_0^L \underbrace{\frac{\partial w}{\partial x}}_{s'(x)}\underbrace{\left(\frac{\partial}{\partial x}\left\{(EI)\frac{\partial v^2}{\partial x^2}\right\}\right)dx}_{r(x)}$$

Performing another integration by parts on the second term of the above yields the following:

 * $$\alpha := \underbrace{\left[w(x)\frac{\partial}{\partial x}\left\{(EI)\frac{\partial v^2}{\partial x^2}\right\}\right]^L_0}_{\beta_1} - \int_0^L \underbrace{\frac{\partial w}{\partial x}}_{s'(x)}\underbrace{\left(\frac{\partial}{\partial x}\left\{(EI)\frac{\partial v^2}{\partial x^2}\right\}\right)dx}_{r(x)}$$

 * $$\alpha := \beta_1 - \underbrace{\left[\frac{-\partial w}{\partial x}(EI)\frac{\partial v^2}{\partial x^2}\right]^L_0}_{\beta_2} + \int_0^L \underbrace{\frac{\partial^2 w}{\partial x^2}\left\{(EI)\frac{\partial v^2}{\partial x^2}\right\}dx}_{\gamma}$$

Thus, Equation 1 can be rewritten as:

 * $$ -\beta_1 + \beta_2 - \gamma + \int^L_0 w f_t dx - \int^L_0 w m \ddot{v} dx = 0$$

Consider the beam diagram in Figure 7. Focusing on the $$\gamma\;$$ term for now, the beam stiffness matrix and shape functions can be derived as follows. First, consider the definitions for the deformation components $$v(\tilde{x})\;$$ and $$u(\tilde{x})\;$$:

 * $$ v(\tilde{x}) = N_2(\tilde{x})\tilde{d}_2 + N_3(\tilde{x})\tilde{d}_3 + N_5(\tilde{x})\tilde{d}_5 + N_6(\tilde{x})\tilde{d}_6$$

 * $$ u(\tilde{x}) = N_1(\tilde{x})\tilde{d}_1 + N_4(\tilde{x})\tilde{d}_4$$

Figure 8 shows the drawings for the various shape functions. The functions themselves are defined as:

 * $$ N_2(\tilde{x}) = 1 - \frac{3\tilde{x}^2}{L^2} - \frac{2\tilde{x}^3}{L^3}\;\;\tilde{d}_2$$

 * $$ N_3(\tilde{x}) = \tilde{x} - \frac{2\tilde{x}^2}{L} + \frac{\tilde{x}^3}{L^2}\;\;\tilde{d}_3$$

 * $$ N_5(\tilde{x}) = \frac{3\tilde{x}^2}{L^2} - \frac{2\tilde{x}^3}{L^3}\;\;\tilde{d}_5$$

 * $$ N_6(\tilde{x}) = - \frac{\tilde{x}^2}{L} - \frac{\tilde{x}^3}{L^2}\;\;\tilde{d}_6$$

Recall the general element transformation relation:

 * $$\mathbf{\tilde{d}^{(e)}} = \mathbf{\tilde{T}^{(e)}}\mathbf{d^{(e)}}$$

Consider Figure 9, the deformation components, and then the following relation for computing them:

 * $$\mathbf{u}(\tilde{x}) = u(\tilde{x}) \vec{\tilde{i}} + v(\tilde{x}) \vec{\tilde{j}} = u_x(\tilde{x}) \vec{\tilde{i}} + u_y(\tilde{x}) \vec{\tilde{j}}$$

Using the above and the previously discussed shape-function relationships, the following relationships can be derived:

 * $$\begin{Bmatrix} u_x(\tilde{x})\\u_y(\tilde{x})\end{Bmatrix} = \mathbf{R}^T \begin{Bmatrix} u(\tilde{x})\\v(\tilde{x})\end{Bmatrix}$$

 * $$\begin{Bmatrix} u(\tilde{x})\\v(\tilde{x})\end{Bmatrix} = \underbrace{\begin{bmatrix} N_1 & 0 & 0 & N_4 & 0 & 0 \\ 0 & N_2 & N_3 & 0 & N_5 & N_6 \end{bmatrix}}_{\mathbb{N}(\tilde{x})} \mathbf{\tilde{d}^{(e)}}$$

Combining the results of this section, the following overall relation can be determined:

 * $$\begin{Bmatrix} u_x(\tilde{x})\\u_y(\tilde{x})\end{Bmatrix} = \mathbf{R}^T \mathbb{N}(\tilde{x}) \mathbf{\tilde{T}^{(e)}}\mathbf{d^{(e)}}$$

Dimensional Analysis
As before, it is useful to revisit the dimensional analysis of some of the results for the previous section:

 * $$[u] = L\;$$

 * $$ [N_1] = \cdots = [N_6] = 1 $$

 * $$[N_1 \tilde{d}_1] = [N_1][\tilde{d}_1] = L$$

 * $$[N_4 \tilde{d}_4] = [N_4][\tilde{d}_4] = L$$

 * $$[v] = L\;$$

 * $$[N_2][\tilde{d}_2] = [N_5][\tilde{d}_5] = L$$

 * $$[N_3][\tilde{d}_3] = [N_6][\tilde{d}_6] = 1$$

Solving the Beam PDE
Consider again the PDE for beams:

 * $$\frac{\partial^2}{\partial x^2} \left\{(EI)\frac{\partial^2 v}{\partial x^2}\right\} = 0$$

Making the assumption that $$EI\;$$ is a constant leads to the following:

 * $$\frac{\partial v^4}{\partial x^4} = 0$$

This is an easily integrable equation which yields the following relationship and its derivative:

 * $$v(x) = C_0 + C_1x + C_2x^2 + C_3x^3\;$$

 * $$v'(x) = C_1 + 2C_2x + 3C_3x^2\;$$

In order to solve for the coefficients of integration, certain boundary conditions must be used. The following set are used to obtain $$N_2(x)\;$$, where $$\tilde{x} = x\;$$ for simplicity:

 * $$v(0) = 1\;$$

 * $$v'(0) = 1\;$$

 * $$v(L) = 1\;$$

 * $$v'(L) = 0\;$$

Application of the first two boundary condition leads directly to $$C_0\;$$ and $$C_1\;$$

 * $$v(0) = 1 = C_0\;$$

 * $$v'(0) = C_1 = 0\;$$

Evaluating the equation and its derivative at $$x = L$$ yields the following:

 * $$v(L) = \underbrace{C_0}_1 + \underbrace{C_1L}_0 + C_2L^2 + C_3L^3 = 0\;$$

 * $$v'(L) = \underbrace{C_1}_0 + 2C_2L + 3C_3L^2 = 0\;$$

Evaluating for the determined coefficients allows the above to be rewritten as:

 * $$C_2L^2 + C_3L^3 = -1\;$$

 * $$2C_2L + 3C_3L^2 = 0\;$$

These equations form a simple system of two equations, which can be easily solved to yields expressions for the remaining two coefficients of integration:

 * $$C_2 = \frac{-3}{L^2}$$

 * $$C_3 = \frac{2}{L^3}$$

Thus, the original equation can be rewritten as follows:

 * $$v(x) = 1 - \frac{3}{L^2} + \frac{2}{L^3}$$

Obtain $$N_3(x)\;$$, $$N_5(x)\;$$, and $$N_6(x)\;$$ can be achieved with an identical solution method and the following boundary conditions (the order of sets corresponds to the order of the list):

 * $$v(0) = 0\;$$

 * $$v'(0) = 1\;$$

 * $$v(L) = 0\;$$

 * $$v'(L) = 0\;$$

 * $$v(0) = 0\;$$

 * $$v'(0) = 1\;$$

 * $$v(L) = 0\;$$

 * $$v'(L) = 0\;$$

 * $$v(0) = 0\;$$

 * $$v'(0) = 0\;$$

 * $$v(L) = 0\;$$

 * $$v'(L) = 1\;$$

Derive Coefficients in Element Stiffness Matrix
A sample coefficient of the element stiffness matrix can be written as follows:

 * $$\tilde{k}_{23} = \frac{6EI}{L^2} = \int_0^L \frac{\partial^2 N_2}{\partial x^2}(EI)\frac{\partial^2 N_3}{\partial x^2} dx $$

In general, the following relationship is true:

 * $$\tilde{k}_{ij} = \int_0^L \frac{\partial^2 N_i}{\partial x^2}(EI)\frac{\partial^2 N_j}{\partial x^2} dx $$

Elastodynamics (Vibrations via FEA)
Consider the following discrete PVW equation:

 * $$\mathbf{\bar{w}} \cdot [\mathbf{\bar{M}}\mathbf{\ddot{\bar{d}}} + \mathbf{\bar{k}}\mathbf{\bar{d}} - \mathbf{\bar{F}}] = 0\; \mbox{for all}\; \mathbf{\bar{w}}$$

This can easily be rewritten as Equation 1, with the associated initial conditions:

 * $$\mathbf{\bar{M}}\mathbf{\ddot{\bar{d}}} + \mathbf{\bar{k}}\mathbf{\bar{d}} = -\mathbf{\bar{F}}(t)\;\;(1)$$

 * $$\mathbf{\bar{d}}(0) = \bar{d}_0$$

 * $$\mathbf{\dot{\bar{d}}}(0) = \bar{v}_0$$

Solving Equation 1
Consider the following "unforced" vibrations problem, given as Equation 2. The problem can be considered unforced because the right hand side is the zero vector:

 * $$\mathbf{\bar{M}}\mathbf{\ddot{v}} + \mathbf{\bar{k}}\mathbf{v} = \mathbf{0}\;\;(2)$$

Assume that $$\mathbf{v}(t) = (\sin \omega t) \boldsymbol{\phi}$$, where $$\boldsymbol{\phi}$$ is not time dependent. Applying this relationship to Equation 2 yields:

 * $$-\omega^2 (\sin \omega t)\mathbf{\bar{M}} \boldsymbol{\phi} + (\sin \omega t)\mathbf{\bar{k}}\boldsymbol{\phi} = \mathbf{0}$$

This leads to the following generalized eigenvalue problem given as Equation 3:

 * $$\mathbf{\bar{k}}\boldsymbol{\phi} = \omega^2 \mathbf{\bar{M}} \boldsymbol{\phi}\;\;(3)$$

 * {| class="collapsible collapsed"

!Standard Eigenvalue Problems
 *
 *

The general form of the eigenvalue problem is as follows:

$$\mathbf{A}\mathbf{x} = \lambda \mathbf{B}\mathbf{x}$$

This can be reduced to the standard eigenvalue problem by making the assumption that $$\mathbf{B} = \mathbf{I}\;$$, where $$\mathbf{I}\;$$ is the identity matrix:

$$\mathbf{A}\mathbf{x} = \lambda \mathbf{x}$$

$$\mathbf{I} = \left[\begin{matrix} 1\\ 0 \end{matrix}\right. \ddots \left.\begin{matrix} 0\\ 1 \end{matrix}\right]$$
 * }

 * $$\lambda = \omega^2\;$$ is defined as the eigenvalue, and the eigenpair is given as $$(\lambda_i,\phi_i)\;$$, for $$i = 1, \cdots, n\;$$. A general eigen mode is defined as:

 * $$\mathbf{v}_i(t) = (\sin \omega_i t) \boldsymbol{\phi_i}$$, for $$i = 1, \cdots, n\;$$

Now, considering the following method based on superposition. The orthogonal project of eigenpairs is given as:

 * $$\boldsymbol{\phi_i}^T \mathbf{\bar{M}}\boldsymbol{\phi_j} = \delta_{ij} = \left\{ \begin{matrix} 1, \mbox{if}\; i = j \\ 0, \mbox{if}\; i \neq j \end{matrix} \right.$$

 * $$\delta_{ij}\;$$ refers to the Kronecker delta. The mass orthogonality of eigenvectors is given as:

 * $$\mathbf{\bar{M}}\boldsymbol{\phi_j} = \lambda_j^{-1} \mathbf{\bar{k}}\boldsymbol{\phi_j}$$

 * $$\underbrace{\boldsymbol{\phi_i}^T \mathbf{\bar{M}}\boldsymbol{\phi_j}}_{\delta_{ij}} = \lambda_j^{-1} \boldsymbol{\phi_i}^T \mathbf{\bar{k}}\boldsymbol{\phi_j}$$

This leads to the following relationship:

 * $$\boldsymbol{\phi_i}^T \mathbf{\bar{k}}\boldsymbol{\phi_j} = \lambda_j \delta_{ij}$$

From Equation 1, the following can be derived:

 * $$\mathbf{\bar{d}}(t) = \Sigma \xi_i(t) \boldsymbol{\phi_i}$$

Combining these results yields the following:

 * $$\mathbf{\bar{M}} \underbrace{\left(\sum_j \ddot{\xi_j} \boldsymbol{\phi_j}\right)}_{\mathbf{\ddot{\bar{d}}}} + \mathbf{\bar{k}} \underbrace{\left(\sum_j \xi_j \boldsymbol{\phi_j}\right)}_{\mathbf{\bar{d}}} = \mathbf{\bar{F}}$$

Rearranging and multiplying by $$\boldsymbol{\phi_i}^T\;$$ yields:

 * $$\sum_j \ddot{\xi_j} \underbrace{\left(\boldsymbol{\phi_i}^T \mathbf{\bar{M}}\boldsymbol{\phi_j}\right)}_{\delta_{ij}} + \sum_j \xi_j \underbrace{\left(\boldsymbol{\phi_i}^T \mathbf{\bar{k}}\boldsymbol{\phi_j}\right)}_{\lambda_j \delta_{ij}} = \boldsymbol{\phi_i}^T \mathbf{\bar{F}}$$

Finally, the overall result can be written as:

 * $$\ddot{\xi_j} + \lambda_j\xi_j = \boldsymbol{\phi_i}^T \mathbf{\bar{F}} i = 1, \cdots, n\;$$

title: User:Eml4500.f08.Lulz.js/HW7


text:
User:Eas4200c.f08.vqcrew.E/Homework7

Mini Plan
Single-Cell Sections
 * - without stringers
 * - with stringers

Multi-Cell Sections
 * - without stringers
 * - with stringers

Without Stringers
Question: Can the setup shown in Figure # resist $$\displaystyle V_{z}$$?

Answer: No.

 * $$R^{z} = R_{AB}^{z} + R_{BA}^{z}$$

$$\displaystyle R^{z}$$ is the total resultant in the z-direction

$$R_{AB}^{z}$$ is the reslutant of q in AB

$$R_{BA}^{z}$$ is the resultant of q in BA

 * $$R_{AB}^{z} = -q\bar{A^{'}B^{'}} = R_{BA}^{z}$$

 * so, $$\displaystyle R^{z} = 0$$

Therefore, the configuration can not resist any $$\displaystyle V_{z}$$.

With Stringers

Figure # shows that q(s) is piecewise constant with respect to the curvilinear coordinate, s. (i.e. $$\displaystyle q_{ij}$$ = constant within each panel)

However, the value for q is not the same throughout all of the panels.

 * $$q_{12} \neq q_{23} \neq q_{31}$$

The presence of the stringers causes a non-constant q.

 * $$R^{z} = V_{z} \neq 0$$

Superposition can be used to solve for the values of $$\displaystyle q_{ij}$$ in each section by solving for values of q in the case without stringers, Figure #, and for the case including an imaginary cut, Figure #, in one of the sections. This is a valid use of superposition due to the linearity of the relationship.

 * $$q_{12} = q + \tilde{q}_{12}$$
 * $$q_{23} = q + \tilde{q}_{23}$$
 * $$q_{31} = q + \tilde{q}_{31}$$

Analysis Algorithm
Objectives: One unknown q ($$\displaystyle \tilde{q_{ij}}$$ are known after solving P1), need to solve one equation for one unknown.

Method: Knowing the values for $$\displaystyle V_{y}$$ and $$\displaystyle V_{z}$$

 * 1) Solve P2 for $$\displaystyle \tilde{q_{12}}$$ and $$\tilde{q_{23}}$$

 * 2) Moment Equation - take the moment about any point in the plane (y,z)

 * a) Superposition: $$\displaystyle q_{ij} = q + \tilde{q_{ij}}$$

 * b) Select point $$\displaystyle \bar{O}$$ in the plane (y,z)

Without Stringers

The same type of analysis can be performed for multi-cell sections as for single cell sections. Figure # shows the multi-cell case without stringers. The resultant of all of the shear flows in each cell can be calculated by the sum of the resultants in every cell.

 * $$\displaystyle R^{z} = \sum_{i = 1}^{N_{cells}}R_{i}^{z}$$

Similar to the single case, the resultant for each cell can be found, therefore yielding the result for the entire multi-cell section. Once again, this value is equal to zero. Without the support of the stringers, this setup can not resist $$\displaystyle V_{z}$$.

 * $$\displaystyle R_{i}^{z} = 0 \Rightarrow R^{z} = 0 $$

With Stringers

The true shear flow, $$\displaystyle q_{ij}$$, is equal to the sum of the closed cell constant shear flow, q, and the open cell piecewise constant shear flow, $$\displaystyle \tilde{q_{ij}}$$.

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

In order to solve for all of the true shear flows, imaginary cuts must be made into some of the walls. This must be done carefully to ensure that no stringers are left unattached to the rest of the assembly. Figure # shows the overall problem including no cuts and all of the stringers. Figure # is displaying the case in which the stringers are removed so that the closed cell shear flow can be calculated. Figure # represents the case in which the cuts have been made and the open cell piecewise constant shear flows can be found. The total resultant of the multi-cell case with stringers can be calculated by the sum of the resultants of P1 and P2.

 * $$\displaystyle R^{z} = R^{z_1} + R^{z_2}$$

Recall: $$\displaystyle R^{z_1} = 0$$

Therefore: $$\displaystyle R^{z} = R^{z_2} = V_{Z} \neq 0$$

title: User:Eas4200c.f08.vqcrew.E/Homework7


text:
User talk:Countrymike/Archive2008

Reading Groups
Hi Brent,

Perhaps we can arrange Illich discussions on a weekly basis, to ensure that the Illich reading group really is active?--Daanschr 10:12, 2 December 2007 (UTC)

 * Daanschr: I thought you'd quit the Illich reading group anyways? Initially I liked the idea of trying to bring some kind of 'weekly' schedule, but I then realized that I don't think its going to help and that it's kind of artificial in a project that can for all intents and purposes go over a long period of time on the wiki and have people join at different points. I think for the portal we should have some kind of designation as to whether groups are active or not, which may mean that an edit has been made at least within 6 months or so. Countrymike 20:29, 2 December 2007 (UTC)

 * Well, the reading groups are a new phenomenon. At the moment there are only 4 or 5 people occupied with it, so we can leave the organization for some time in the future. For the moment, i am confident with a text about a scheduled activity for Thucydides and The Diamond Age. I'll prefer to wait for more participants before taking new steps.--Daanschr 21:01, 2 December 2007 (UTC)

 * Yea, I think that if you wish to try scheduling for a group that is up to the primary authors/participants. I think that the idea of "promoting" more active groups on the portal page is good though. Countrymike 21:52, 2 December 2007 (UTC)

 * I agree that the primary authors, participants should deal with the schedule of group activity. What i try to prevent is that Wikiversity has only inactive groups to present. For the moment everything is swell as it is.

 * I am sorry for disliking illich. I would be pleased to be part of a reading group with you on another topic.--Daanschr 09:38, 3 December 2007 (UTC)

Talk with Cormac and Wayne
I pinged Cormac and he has limited availability on the dates you said Wayne was free. Cormac said "night of the 20th though, or the morning of the 17th" his time, and I don't know how that translates for the rest of us. However, even if it doesn't match up, we can do two shorter sessions, one with each. Ping me or drip into IRC and let me know. Historybuff 17:16, 7 December 2007 (UTC)

 * Let me know when you'd like to do an interview ... I'm flexible for most days. We should also touch base on questions and what not, so we can have something to chat about with both of them. Historybuff 06:47, 10 December 2007 (UTC)

Pinging you again on this. Things are starting to gain momentum, and to make sure we have traction, I'd like to schedule a chat with WikiEdu reps as one session, and once we are all happy that we've got the initial connections up and working, we can embark on something more ambitious. Let me know how you feel about timing and doability. Historybuff 09:55, 4 April 2008 (UTC)

custodianship
Thanks for listing yourself at Candidates for Custodianship. Decisions about custodianship are community decisions. This means you have to communicate with the community. The instructions for how to do this are at the top of that page. Please edit this page and add a short description of your wiki experience.....for example, you can say that you are sysop and bureaucrat at WikiEducator. If you have a particular choice for a mentor, say so. --JWSchmidt 16:55, 14 December 2007 (UTC)

Thanks for being willing to help at Wikiversity as a custodian. Please familiarize yourself with Category:Wikiversity administration and Category:Official policy. If you have any questions, please let me know. --JWSchmidt 14:51, 15 December 2007 (UTC)

You are now listed at Nominations for Full Custodianship. There will now be a five-day-long community discussion period. Please pay some attention to the discussion and reply to any questions that might be asked of you. I think these questions are intended to be part of the community discussion. If there are any questions that you do not understand or do not know how to answer I think it is best to say that rather than not respond at all. --JWS 15:44, 15 January 2008 (UTC)

Main page editing
I've read your concerns about main page editing. In fact in an unrelated development earlier today, I downgraded the protection on some parts of the main page so that normal registered users could edit parts of it. This move is aimed particularly at allowing ordinary users to post news items. In the longer term, it is probable that we will be looking more closely at which parts of the main page should be open to normal users to edit. In addition, I posted ideas on the colloquium about more regular main page edits, with a main page task force set up to fast-track certain kinds of change. In future sysop status should therefore not be necessary for text and hyperlink changes in certain segments of the main page. McCormack 20:59, 14 December 2007 (UTC)

Talk page and colloquium lists
Hi Brent. A long time ago I learnt somewhere on Wikimedia that it wasn't regarded as a good thing to break up other people's lists in talk page and colloquium discussions. It results in text that is not properly comprehensible for other users, because coherent numbering gets disrupted and signatures get detached from comments. I can't find a policy or etiquette page on this at the current time, but nevertheless I have sorted out your insertions into my posting on the colloquium and put them at the end of my post. I'm leaving you this message here so that you have an opportunity to check that your comments remain intact and comprehensible after the move. McCormack 09:04, 28 December 2007 (UTC)

Thanks
--Faustnh 14:27, 30 December 2007 (UTC)

Thanks Coutrymike for your kind welcome.

Thanks
Thank you for your welcome! Shmooshkums 04:48, 3 January 2008 (UTC)

Thank you for welcoming me
Thank you for welcoming me to wikiversity. I hope to help it become even better than it is already. :) Captain panda 04:08, 5 January 2008 (UTC)

Thanks!
Thanks for welcoming me too! I am around mostly at Wikipedia, and have been there for quite some time now, so I already know pretty much all there is to know about editing and all, but thanks anyway! Also if you want my attention quickly email me or leave me a message on my Wikipedia talk page! Thanks again for your kindness!--Kushan I.A.K.J 15:27, 11 January 2008 (UTC)

Yet Another Paean of Gratitude
Hey, thanks for linking my lecture to Augustine's wikipedia page. I'll take all the help I can get. Seriously, it's awesome that people care enough about this venture (and other wiki projects) to give so much of themselves. Good on ya! Mike 04:49, 15 January 2008 (UTC)

Learning about list editing
Hi Countrymike. I've noticed that you generally have difficult with editing lists. In order not to break up a numbered list, you must combine hashes and colons like this: #: rather than ::. Could you perhaps reinstate the correct numering in the interview? Thanks. --McCormack 11:33, 16 January 2008 (UTC)

Your comment on colloquium
Please see my response to one of your comments at Educational_Wikis/constructive_interactions/Colloquium_thread. I appreciate your position, and I respect your opinion, however I would ask that you respect that of others as well, and address concerns as opposed to denying their validity or that of those who raise them. (I am aware that this is an archive page.) --Draicone (talk) 11:48, 19 January 2008 (UTC)

:-)
Hi Countrymike, congrats on becoming a full custodian! --HappyCamper 00:18, 21 January 2008 (UTC)

Discussions
Hi dear wikiversitan, friend or I think I can call you Mike (if you allow).Thanks for your kind welcome , Yeh! I know it's a sweet tradition on wiki-projects.secondly, I wanna know if there are some particular sort of discussion portals to focus and exchange the knowledge of all the good guys here on any topic of knowledge .I think there should be something like this.I'm not talking about minor disscussions that is usually about the specific topic but a place where we can talk about any thing of the knowledge kingdom.Do share with me if there is something like that.Cheers,Un-predictable 18:37, 21 January 2008 (UTC)

Wikimedian Demographics update
Hello Countrymike. I'm writing because you requested a reminder about progress on Wikimedian Demographics. For the time being these will be about once per month.

Speaking of reminders, it turns out the template question isn't the most efficient way of coming up with a mailing list. If you want to recieve this message on another of your talk pages (or don't want them at all), please edit the list on Wikimedian Demographics/Reminders.

To update on progress so far, there have been somewhere around 50 respondents over the past month, with more trickling in slowly but steadily. There are 7 major surveys (or 12, depending on how you count them):
 * 1) The basic survey (about your participation in Wikimedia)
 * 2) The "geography" surveys (actually 6 of those, for each inhabited continent)
 * 3) The Operating System survey (still stubby)
 * 4) The Wikipedia survey
 * 5) The Commons survey
 * 6) The US presidential elections survey
 * 7) The sexuality and relationships survey

Only the geography surveys have had any major modifications since being originally posted (more countries were added to those).

Several new surveys have been discussed at Talk:Wikimedian Demographics, as well as other places within Wikiversity. A few that might be ready sometime in February include:
 * A survey about Women in Wikimedia
 * A survey about Wikibooks
 * A survey about Wikiversity
 * A survey about administrative issues (including cross-wiki topics such as CheckUser and "global blocking")
 * A survey about environmental issues
 * A survey about religion
 * A survey about international trade

If you have ideas for more surveys, please share them on Talk:Wikimedian Demographics. Thanks for your participation thus far, and we hope to see you soon! --SB_Johnny | talk 11:51, 5 February 2008 (UTC)

Reading
Hi Brent, thanks for adding Illich to Wikiversity_learning_model/Reading_group. (I will get around to reading it - I got it for Christmas for god's sake!!) I also added the Seely Brown group to that page asking if it should be a part of the wider group - what do you think? Cormaggio talk 17:07, 8 February 2008 (UTC)
 * Yes, I very much want to have the reading group(s) as a case study. That's how I see action research working here - doing activity (ie reading), reflecting on the activity, informing further activity... Cormaggio talk 12:59, 9 February 2008 (UTC)

I saw you added yourself to Wikiversity research. What aspects of Wikiversity are you interested in researching, and how? It'd also be great if you added your thoughts to this discussion about viable methods for building learning communities - or any other aspect of the Developing Wikiversity through action research project. Cormaggio talk 13:38, 12 February 2008 (UTC)

Personal learning environments
Hi Brent - thanks to Mchua, there are some interesting thoughts on Making Wikiversity a personal learning space if you want to take a look... Cormaggio talk 15:55, 4 March 2008 (UTC)

Licensing
hi brent, could you explain to me more about "I dual license my contributions under the CC-BY-SA license as well as the GFDL in order to make my contributions work on the widest number of OER wikis possible.". I would like to understand. I just whacked CC-BY-SA on my stuff, but get the feeling that it might not be quite that simple. Feel free to direct me to what I should read. -- Jtneill - Talk 09:04, 20 March 2008 (UTC)

Uploading Files
Hi Brent, I am working on a project located at "http://en.wikiversity.org/wiki/DFE2008_Dry_Cleaning_Detergent", being really new to wikipedia I didn't realize that I need to be an account holder for 4 days before I could upload files for use. I realize that wikimedia allows people to bypass the rule for urgent cases. My project is due at 9am UTC+4 on Wednesday, and I need to upload files urgently. Any help would be appreciated.

Thank You, Naweed
 * Yes, I got help from the irc channel and used wikimedia to host the files.

Wikimedia Radio ...
There is a proposal concering a Wikimedia Radio project n:WN:RADIO 62.56.54.110 20:27, 27 April 2008 (UTC)

IRC meeting
Hi Brent - just notifying you of a suggested time/date for an IRC meeting about 'Wikiversity learning'. Could you indicate if you would be available, or would prefer a different time/date? Cormaggio talk 14:47, 1 June 2008 (UTC)

GNUmedia
Hey Brent, How're things in the pasture?

I just migrated some content from gnumedia.org to the GNUmedia Wikiversity page. You said you were interested in updates :) Any suggestions/contributions are welcome. --Brylie 10:25, 27 June 2008 (UTC)

Your colloquium post and disruptive users
Hi Brent. Thanks for your post to the colloquium. On the subject of your custodianship, you've been a very much better custodian than your mentor. Since opposing your custodianship request a long time ago, I've slowly come to think that it was probably John's actions I should have been opposing back then, not you. Anyway, to come back to your post on the colloquium, I think you may not have been following the edit histories. I'll give you just one example of what John was doing. Go to the history of the deletion request template here: http://en.wikiversity.org/w/index.php?title=Template:Deletion_request&action=history. You'll see that someone called Countrymike made a sensible version in March (good sensible solid edit). You'll then see that on 11th August, User:ViperSnake151 made some design changes, not content changes. ViperSnake151 was doing a number of template design improvements at the time. ViperSnake151's edit drew John's attention and John made this edit which was problematic for two (smallish) reasons: firstly, the box title didn't fit the content any more, and second, the box content started to move in an anti-deletionist direction rather than being just neutral. Nobody did anything for a while. 20 days later, User:Adambro spotted the first of these problems (title misfit) and undid the change - however he left the anti-deletionist content alone. Adambro's edit was small, harmless and undoubtedly a small improvement - see. JWS promptly reverted Adambro, insisting that a "reason must be given". Actually what was going on here was that JWS wanted a reason to be entered on the template using a parameter, but the template had no parameter, and (as I slowly realised) JWS didn't know how to edit templates to use parameters. I think you do know, however! Basically, the argument was over the introduction of a parameter which neither of JWS (nor perhaps Adambro) could programme in. The argument between Adambro and JWS goes back to at least 20th August, when Adambro was active on the deletion requests pages and JWS got sarcastic with him using the comment "Adambro: you do not recognize any difference between editing a wiki and using a word processor?" (note also JWS's comment to this edit). Anyway, at this point (31st August again), I didn't like how JWS was treating Adambro and I stepped in and restored Adambro's edit as well as making a few further changes to the template - reformating, and making JWS's content edits of 11th August more neutral in tone. At the same time, I created the talk page for the template and opened a discussion of the edits. Instead of using the talk page, JWS promptly started using the template itself as a talk page with this large edit, transcluding his opinions into every page currently up for deletion. This was probably the first truly "disruptive" edit on this page, because by repurposing the template as a discussion forum, JWS disrupted the functionality of Wikiversity and forced his agenda. It's something called "disrupting a project to make a point". The template couldn't actually be left in that state, regardless of points of view. The first person to step in and correct the problem was User:Adambro, who reverted JWS's edits. I think this was right. JWS immediately reverted Adambro and this time left an even bigger essay right in the middle of the template:. The template was now at 340 words or so and about 10 times larger than a template should be. John's sarcastic comment for his edit was that the template "was not long enough". Adambro then reverted John. John then reverted Adambro again, leaving an even longer essay in the middle of the template (370 words). At this point, Adambro was on 2 reverts and John was on 3 reverts. I think if we'd been applying Wikipedia rules, JWS could have been legitimately blocked from editing Wikiversity for a short cool-off period. Watching this, I decided not to block JWS and I also decided not to revert JWS. I felt this would worsen the situation, although JWS was clearly wrong. Instead I left a note on the talk page about the edit war and backed off. At this point, another custodian (User:Darklama) took up the problem of putting the template back into order after the disruptive edits by JWS. See Darklama's edits here where he removes JWS's essay and tries to improve the template. User:Darklama is a good programmer and introduced the "reason parameter" as well. Darklama's edits were good. Adambro changed Darklama's edits and removed the reason parameter - I don't agree with that particular edit by Adambro, as the presence or absence of the parameter should have been discussed on the talk page first. I can see arguments both ways between Darklama and Adambro. I think both of their edits were good faith. Darklama then reverted Adambro. I tend to agree with Darklama. Darklama continued to prgrammatically improve the template (good edits). User:JWSchmidt then returns and starts inserting essays into the middle of the template again - this is once again disruptive editing, and the edit summaries are not exactly polite towards Darklama. Darklama reverts JWS, referring to JWS's "anger and frustration". The final word in this saga was when a sockpuppet account suddenly appeared and added a penis-icon to the template - pornographic vandalism. Without a checkuser action, we do not know who was behind this edit. The current version of the template is as left by Darklama and Hillgentleman. Anyway, Brent, I hope that as a custodian, you read all of this. If you feel that John's behaviour here was anything other than "disruptive", please state why. He really wasted a lot of time on the part of Darklama, myself and Adambro. And this is only one of his disruptive editing phases in recent days. I'd be very happy to see you change your opinion on this. If you do, many thanks and hopefully no hard feelings about your comment on the colloquium - ? --McCormack 07:42, 9 September 2008 (UTC)

title: User talk:Countrymike/Archive2008


text: File:Graph.jpg

title: File:Graph.jpg


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

=Homework 7=

 * Lecture Notes (current version) Lecture Notes (archived)
 * HW7 Contributors (current version) HW7 Contributors (archived)
 * MATLAB (current version) MATLAB (archived)

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