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User:Eml4500.f08.ateam.nobrega/Homework Report Four

Consider the two bar truss from meeting 5 page 6:

The connectivity array,"conn" is featured in the figure above. Conn(e,j) is the global node number of local node j of element e. The rows in the connectivity array refer to the element number. For example, row 1 refers to element 1 and row 2 refers to element 2. However, the numbers in the matrix correspond to the global degree of freedom numbers. Each column refelects a local node number.

The location matrix master array (lmm)is featured in the figure to the right. lmm(i,j) is the equivalent number (global dof number) for the element of stiffness coefficient which corresponds to the jth local dof number. The rows in the mtarix correspond to the element numbers; while, the colums identify the local degree of freedom number. The numbers in the array are the global degree of freedom numbers NOT the element degree of freedom numbers.

Homework: Modify the MATLAB code for the 2-Bar truss system to solve the 3-Bar truss system. The solution to this homework assignment can be found in the section titled Three Bar Truss Deformation, of this homework report.

Transform a system with four dofs into a system with also four dofs (instead of only two dofs) so that the transform matric is now a 4x4 matrix and (hopefully) invertible.

Defining the Transform Matrix
The goal is to find $$ \tilde{T}^{(e)}$$, a 4x4 matrix that transforms the set of local element degrees of freedom $$ d^{(e)} $$ a 4x1 matrix to another set of local, element, degrees of freedom $$ \tilde{d}^{(e)}$$ such that $$ \tilde{T}^{(e)}$$ is invertible.

$$ \tilde{d}^{(e)}$$ is the rotated degrees of freedom matrix corresponding to the $$\tilde{x}$$ coordinate system. The relationship is described below,

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

The components that make up the $$\tilde{d}^{(e)}$$ matrix are as follows,

 * $$ \tilde{d_{1}}^{(e)} =

\begin{bmatrix} l^{(e)} & m^{(e)} \end{bmatrix} \begin{bmatrix} d_{1}^{(e)}\\ d_{2}^{(e)} \end{bmatrix} $$

Here $$\tilde{d_{1}}^{(e)}$$ directly corresponds to the axial displacement $$q_{1}^{(e)}$$. And,

 * $$ \tilde{d_{2}}^{(e)} = \overrightarrow{d_{[1]}}^{(e)} \cdot \overrightarrow{\tilde{j}}$$

where $$ \tilde{d_{2}}^{(e)}$$ is a component of $$\overrightarrow{d_{[1]}}^{(e)}$$ along the unit vector $$\overrightarrow{\tilde{j}}$$, i.e., along the $$\tilde{y}$$-axis.

Here,

 * $$ \overrightarrow{i} \cdot \overrightarrow{\tilde{j}} = \cos{(\theta + \frac{\pi}{2})} = -\sin\theta$$

 * $$ \overrightarrow{j} \cdot \overrightarrow{\tilde{j}} = \cos\theta $$

This becomes,

 * $$\tilde{d_{2}}^{(e)} = -d_{1}^{(e)}\sin\theta + d_{2}^{(e)}\cos\theta$$

Where,

 * $$-\sin\theta^{(e)} = -m^{(e)}$$
 * $$ \cos\theta^{(e)} = l^{(e)}$$

So,

\tilde{d_{2}}^{(e)} = \begin{bmatrix} -m^{(e)} & l^{(e)} \end{bmatrix} \begin{bmatrix} d_{1}^{(e)}\\ d_{2}^{(e)} \end{bmatrix} $$

Combining these two equations equates to

\begin{bmatrix} \tilde{d_{1}}^{(e)}\\ \tilde{d_{2}}^{(e)} \end{bmatrix} = \begin{bmatrix} l^{(e)} & m^{(e)}\\ -m^{(e)} & l^{(e)} \end{bmatrix} \begin{bmatrix} d_{1}^{(e)}\\ d_{2}^{(e)} \end{bmatrix} $$

Here,

\begin{bmatrix} l^{(e)} & m^{(e)}\\ -m^{(e)} & l^{(e)} \end{bmatrix} = R^{(e)} $$

Looking at the entire element e and combining the local degrees of freedom one gets,

\begin{bmatrix} \tilde{d_{1}}^{(e)}\\ \tilde{d_{2}}^{(e)}\\ \tilde{d_{3}}^{(e)}\\ \tilde{d_{4}}^{(e)} \end{bmatrix} = \begin{bmatrix} l^{(e)} & m^{(e)} & 0 & 0\\ -m^{(e)} & l^{(e)} & 0 & 0\\ 0 & 0 & l^{(e)} & m^{(e)}\\ 0 & 0 & -m^{(e)} & l^{(e)} \end{bmatrix} \begin{bmatrix} d_{1}^{(e)}\\ d_{2}^{(e)}\\ d_{3}^{(e)}\\ d_{1}^{(e)} \end{bmatrix} $$

The first matrix is the local element degrees of freedom corresponding to the $$\tilde{x}$$ coordinate system, the second is the transformation matrix, and the third is the matrix of the local element degrees of freedom.

The transformation matrix can be broken down as follows,

\begin{bmatrix} l^{(e)} & m^{(e)} & 0 & 0\\ -m^{(e)} & l^{(e)} & 0 & 0\\ 0 & 0 & l^{(e)} & m^{(e)}\\ 0 & 0 & -m^{(e)} & l^{(e)} \end{bmatrix} = \begin{bmatrix} R^{(e)} & 0\\ 0 & R^{(e)} \end{bmatrix} $$

If the transverse displacement is taken away, the force-displacement relationship will look the same as in previous lectures.

 * $$ k^{(e)}

\begin{bmatrix} 1 & -1\\     -1 & 1 \end{bmatrix} \begin{bmatrix} q_{1}^{(e)}\\ q_{2}^{(e)} \end{bmatrix} = \begin{bmatrix} p_{1}^{(e)}\\ p_{2}^{(e)} \end{bmatrix} $$

Looking at the force-displacement relationship,

\tilde{f}^{(e)} = k^{(e)} \begin{bmatrix} 1 & 0 & -1 & 0\\     0 & 0 & 0 & 0\\      -1 & 0 & 1 & 0\\      0 & 0 & 0 & 0 \end{bmatrix} \tilde{d}^{(e)} $$

Here the axial displacements are present and the transverse displacements are taken to be zero because if one acts along the transverse the axial displacements are zero and the spring has no effect.

This now becomes,

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

In this equation force is represented by a 4x1 matrix, the spring stiffness is represented by a 4x4 matrix, and the displacement is a 4x1 matrix.

Note: Consider the case where d4(e) ≠ 0, where d1(e)=d2(e)=d3(e)=0

Multiplying out the equation out results in a 0 matrix as listed below:

f4(e) = k4(e) d4(e) = 0 where f is a 4x1 matrix, k is a 4x4 matrix, d is a 4x1 matrix which creates a 4x1 0 matrix.

The 0 matrix represents the 4th column of the k(e) which is the interpretation of the transverse dof's.

From meeting 14 page 3: d4(e) = T4(e) d4(e)

Similarly using the same argument f4(e) = T4(e) d4(e)

Also k4(e) * d4(e) = f4(e)
 * which leads to k4(e) T4(e) d4(e) = T4(e) f4(e)

In the equation above the T4(e) is a block diagonal matrix.

\begin{bmatrix} R_{2x2}^{(e)} & 0_{2x2}\\ 0_{2x2} & R_{2x2}^{(e)}\\ \end{bmatrix} $$

If T(e) is invertible, then [T(e)-1 k(e) T(e)]d(e) = f(e).

Consider a general block diagonal matrix with:

A= \begin{bmatrix} D_1& & &  & 0\\ &  .& &  & &\\      &   &  .& & &\\      &   &   & .&&\\      0&   &  &  & D_5\\ \end{bmatrix} $$

Question: What is the inverse of A (A-1? Simple example: Diagonal Matrix

B= \begin{bmatrix} d_{11}& & &  & 0\\ &  d_{12}& &  & &\\ &  &  .& & &\\      &   &   & .&&\\      0&   &  &  & d_{mn}\\ \end{bmatrix} = diag[d_{11}, d_{22}, .... , d_{mn}] $$

Thus,

B^{-1} = Diag \begin{bmatrix} \tfrac{1}{d_{11}},& \tfrac{1}{d_{22}},& .&  .& .& .,&\tfrac{1}{d_{mn}}\\ \end{bmatrix} $$

Assuming that all values of the diagonal coefficients are not equal to 0, i.e. dii ≠ 0 for all values of i=1......n.

For a block diagonal matrix A,

A = Diag \begin{bmatrix} D_1,& .& .& .&, D_s \end{bmatrix} $$ and

A^{-1} = Diag \begin{bmatrix} D_1^{-1}, & .& .& .& .&, D_s^{-1}\\ \end{bmatrix} $$ where D1-1, is the inverse of the matrix D1.

The transform matrix is defined to be
 * T(e)-1=Diag[R(e)-1, R(e)-1]

On meeting 19 page 2 R(e) is defined as

R^{(e)} = \begin{bmatrix} l^{(e)} & m^{(e)}\\ -m^{(e)} & l^{(e)} \end{bmatrix} $$

R^{(e)T} = \begin{bmatrix} l^{(e)} & -m^{(e)}\\ m^{(e)} & l^{(e)} \end{bmatrix} $$ Thus multiplying R(e) by its transpose will result in

R^{(e)T}R^{(e)} = \begin{bmatrix} l^{(e)} & -m^{(e)}\\ m^{(e)} & l^{(e)} \end{bmatrix} \begin{bmatrix} l^{(e)} & m^{(e)}\\ -m^{(e)} & l^{(e)} \end{bmatrix} = \begin{bmatrix} 1 & 0\\     0 & l \end{bmatrix} $$ The resulting matrix represents a 2x2 I or identity matrix.

Thus,
 * R(e)-1=R(e)T
 * T(e)-1=Diag[R(e)T, R(e)T]=(Diag[R(e), R(e)])T

Also,
 * T(e)-1=T(e)T

Recall Equations:

f_{3}^{(1)} + f_{1}^{(2)} = 0 $$
 * Equation 1: $$

$$ [k_{31}^{(1)}*d_{1}^{(1)} + k_{32}^{(1)}*d_{2}^{(1)} + k_{33}^{(1)}*d_{3}^{(1)} + k_{34}^{(1)}*d_{4}^{(1)}] +         [k_{11}^{(1)}*d_{1}^{(2)} +            k_{12}^{(1)}*d_{2}^{(2)} + k_{13}^{(1)}*d_{3}^{(2)} + k_{14}^{(1)}*d_{4}^{(2)}] = 0 $$
 * Equation 1: Expanded

$$ [k_{31}^{(1)}*d_{1} + k_{32}^{(1)}*d_{2} + k_{33}^{(1)}*d_{3} + k_{34}^{(1)}*d_{4}] + [k_{11}^{(1)}*d_{3} + k_{12}^{(1)}*d_{4} + k_{13}^{(1)}*d_{5} + k_{14}^{(1)}*d_{6}] = 0 $$
 * Equation 1: Expanded using global degrees of freedom

f_{4}^{(1)} + f_{2}^{(2)} = P $$
 * Equation 2: $$

$$ [k_{41}^{(1)}*d_{1}^{(1)} + k_{42}^{(1)}*d_{2}^{(1)} + k_{43}^{(1)}*d_{3}^{(1)} + k_{44}^{(1)}*d_{4}^{(1)}] + [k_{21}^{(1)}*d_{1}^{(2)} +           k_{22}^{(1)}*d_{2}^{(2)} + k_{23}^{(1)}*d_{3}^{(2)} + k_{24}^{(1)}*d_{4}^{(2)}] = P $$
 * Equation 2: Expanded

$$ [k_{41}^{(1)}*d_{1} + k_{42}^{(1)}*d_{2} + k_{43}^{(1)}*d_{3} + k_{44}^{(1)}*d_{4}] + [k_{21}^{(1)}*d_{3} + k_{22}^{(1)}*d_{4} + k_{23}^{(1)}*d_{5} + k_{24}^{(1)}*d_{6}] = P $$
 * Equation 2: Expanded using global degrees of freedom

Assembly of   $$  \displaystyle \mathbf     k^{(e)} $$, e = 1,. . ., nel (number of elements) into the global stiffness matrix

$$\displaystyle \mathbf K = \overset{e=nel}{\underset{e=1}{\mathbb A}} \mathbf k^{(e)}$$

$$\displaystyle \overset{e=nel}{\underset{e=1}{\mathbb A}} $$ = Assembly Operator

n : total number of global degrees of freedom before eliminating the boundry conditions

ned : number of element degrees of freedom

ned << n : ned is very small when compared to n

Further Evaluation and Understanding of the Zero Eigen Values
Meeting 21 on Monday October 13th 2008 was a review and further understanding of a previously assigned homework problem dealing with the Eigen values of the Global K Matrix for a two bar truss system. The solution to the Eigen value problem actually yields four zero Eigen values which corresponds to rigid body motion. The four zero values correspond to the global dof constraints: two translations, one rotation, and one mechanism which relates the idea of zero stored energy in the truss system if you remove both supports from the two bar truss system.

The next section is a plot of the Eigen vectors which correspond to the zero Eigen values of the two bar truss system.

K =

0.5625   0.3248   -0.5625   -0.3248         0         0    0.3248    0.1875   -0.3248   -0.1875         0         0   -0.5625   -0.3248    3.0625   -2.1752   -2.5000    2.5000   -0.3248   -0.1875   -2.1752    2.6875    2.5000   -2.5000         0         0   -2.5000    2.5000    2.5000   -2.5000         0         0    2.5000   -2.5000   -2.5000    2.5000

The corresponding Eigen values in a column vector are as follows:

eig(K)

ans =

-0.0000  -0.0000    0.0000    0.0000    1.4706   10.0294

The command [U,D]=eig(K) produces a matrix U whose columns are the eigenvectors of K and a diagonal matrix D with the eigenvalues of K on its diagonal.

[U,D]=eig(K)

U =

-0.1118   0.5043   -0.0000   -0.5931    0.6174   -0.0139   -0.0814   -0.8634    0.0000   -0.3476    0.3565   -0.0080   -0.4628    0.0089    0.0000   -0.4803   -0.5409    0.5123    0.5266   -0.0053   -0.0000   -0.5429   -0.4330   -0.4904   -0.4947    0.0071   -0.7071    0.0313   -0.0765   -0.4984    0.4947   -0.0071   -0.7071   -0.0313    0.0765    0.4984

D =

-0.0000        0         0         0         0         0         0   -0.0000         0         0         0         0         0         0    0.0000         0         0         0         0         0         0    0.0000         0         0         0         0         0         0    1.4706         0         0         0         0         0         0   10.0294

From these matrices (U and D) the four Eigenvectors corresponding to the zero Eigen values can be pulled from matrix U and are as follows:

E1 =

-0.1118  -0.0814   -0.4628    0.5266   -0.4947    0.4947

E2 =

0.5043  -0.8634    0.0089   -0.0053    0.0071   -0.0071

E3 =

0        0         0         0   -0.7071   -0.7071

E4 =

-0.5931  -0.3476   -0.4803   -0.5429    0.0313   -0.0313

The results of the plots are the Mode Shapes which are a linear combination of the pure mode shapes (pure rigid body motion two translations, one rotation, and one pure mechanism). The results explained above can also be represented mathematically with the following arguments.

$$Kv = \lambda v$$

Let $$\begin{Bmatrix} U_1 & U_2 & U_3 & U_4 \end{Bmatrix}$$ be the pure eigenvetors corresponding to the four zero eigenvalues. $$K*U_i = 0*U_i$$ for i = 1,...,4.

To get W for each zero eigenvalue, we take the sum of the four contributors, each multiplied by a constant $$\alpha _{1}$$, which gives us the following relation:

$$ W = \alpha_1 u_1 + \alpha_2 u_2 + \alpha_3 u_3 + \alpha_4 u_4$$

W consists of the combination of eigenvectors for zero eigenvalues, therefore W also represents an eigenvector corresponding to a zero eigenvalue.

Eigenvalue Analysis - Continuation
The links below illustrate the application of Eigenvalue Modal Analysis to real world examples:

 * Frequency analysis of shell structures with contact
 * Frequency Solutions with Contact

The first link demonstrates various modes of the oscillations in a shell structure. The second link is an example of rigid body modes with zero eigenvalues.

In order to generate the plots of modal analysis, the Eigenvalue is simply multiplied by a sinusoidal function. For example:

$$d(t)=\lambda cos(\omega t)$$

where

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

f is the frequency of vibration.undefined

Assembly of Element Stiffness Matrix into Global Stiffness Matrix
Consider the previous example of the two bar truss and recall the element force-displacement relationship:

$$k^{e}d^{e}=f^{e}$$

We have previously proved the equilibrium of Node 2 from the free body diagrams of Element 1 and Element 2.

$$d_{3}=d_3^{(1)}=d_1^{(2)}$$

$$d_{4}=d_4^{(1)}=d_2^{(2)}$$

For Node 2, identify the global degree of freedom to element degree of freedom for both Element 1 and 2.

Equilibrium of Node 2

$$ \sum F_x=0=-f_{3}^{(1)}-f_{1}^{(2)} $$

$$ \sum F_y=0=P-f_{4}^{(1)}-f_{2}^{(2)} $$

Next replace each f above with the individual element kd relationship.

$$ \left(k^{(1)}_{31}d^{(1)}_{1}+k^{(1)}_{32}d^{(1)}_{2}+k^{(1)}_{33}d^{(1)}_{3}+k^{(1)}_{34}d^{(1)}_{4} \right)+\left(k^{(2)}_{11}d^{(2)}_{1}+k^{(2)}_{12}d^{(2)}_{2}+k^{(2)}_{13}d^{(2)}_{3}+k^{(2)}_{14}d^{(2)}_{4} \right)=0$$

$$ \left(k^{(1)}_{41}d^{(1)}_{1}+k^{(1)}_{42}d^{(1)}_{2}+k^{(1)}_{43}d^{(1)}_{3}+k^{(1)}_{44}d^{(1)}_{4} \right)+\left(k^{(2)}_{21}d^{(2)}_{1}+k^{(2)}_{22}d^{(2)}_{2}+k^{(2)}_{23}d^{(2)}_{3}+k^{(2)}_{24}d^{(2)}_{4} \right)=P $$

Principle of Virtual Work
Motivation:

 * The principle of virtual work allows us to eliminate the corresponding rows to the boundry conditions to obtain the global stiffness matrix K. Reducing the global stiffness matrix from a (6x2) matrix to a (2x2) matrix.

q^{(e)} = T^{(e)}\cdot d^{(e)} $$

k^{(e)} = T^{(e)T}\cdot k^{(e)}\cdot T^{(e)} $$

 * The ultimate goal is that the principal of virtual work allows us to derive the Finite Element Method for Partial Differential Equations which are heavily used in the world of Heat Transfer, Vibrations, Finite Element Analysis Commercial Codes and many more.

Force Displacement Relation for a bar or spring is expressed in the form $$ kd=F $$, which implies

$$ kd-F=0 $$ Equation (3)

$$ w(kd-F)=0 $$ for all w Equation (4)

Proof:

(3) => (4) Trivial

(4) => (3) Not Trivial

The reason why going from equation 3 to 4 is trivial is because since equation 4 is valid for all w, select w = 1 and then equation 4 becomes....

$$ 1(kd-F) = 0 $$ which is equal to equation 3

Evaluating the Five Bar Truss
The extra functions that are needed to finish the code include the functions PlaneTrussElement.m, NodalSoln.m, and PlaneTrussResults.m. They are as follows,

PlaneTrussElement.m NodalSoln.m PlaneTrussResults.m Once the extra functions have been created this code gives,

K =

1.0e+005 *

0.3260   0.7607   -0.3260   -0.7607         0         0         0         0    0.7607    2.9749   -0.7607   -1.7749         0   -1.2000         0         0   -0.3260   -0.7607    2.4309    1.1914   -0.3300    0.3300   -1.7749   -0.7607   -0.7607   -1.7749    1.1914    2.4309    0.3300   -0.3300   -0.7607   -0.3260         0         0   -0.3300    0.3300    1.5300   -0.3300   -1.2000         0         0   -1.2000    0.3300   -0.3300   -0.3300    1.5300         0         0         0         0   -1.7749   -0.7607   -1.2000         0    2.9749    0.7607         0         0   -0.7607   -0.3260         0         0    0.7607    0.3260

R =

0          0           0     -150000           0           0           0           0

d =

0        0    0.5390   -0.9531    0.2647   -0.2647         0         0

reactions =

1.0e+005 *

0.5493   1.5993   -0.5493   -0.0993

results =

-0.0001743     -34.859 -1.3944e+005 -3.15e-005     -6.2999       -25200 -5.2941e-005     -10.588       -31764 -5.2941e-005     -10.588       -31764 0.00032087      22.461        44922

>>

The Connectivity Array

The connectivity array, con, uses the local node numbers of each element to locate the global node numbers. The matrix shows which global node numbers are shared by what element. For the five-bar truss the connectivity matrix has the dimensions 5x2 based on the five elements of the truss and the two nodes per each element.

The Master Location Matrix

The master location matrix, lmm, uses the local degrees of freedom numbers of each element to place the global degrees of freedom numbers of each element. The matrix illustrates which global dof numbers are shared by what element. For the five-bar truss the master location matrix has the dimensions 5x4 based on the five elements of the truss system with four degrees of freedom per each element.

The Element Stiffness Coefficient
The element stiffness coefficient for the five-bar truss is computed using the function PlaneTrussElement.m. This function brings in the element modulus of elasticity, the element area, and the element node coordinate system. From this coordinate system the function calculates the cosine directors l and m which are defined as follows,

l^{(e)} = \cos{\theta^{(e)}} = \frac{x_{2} - x_{1}}{length} $$

m^{(e)} = \sin{\theta^{(e)}} = \frac{y_{2} - y_{1}}{length} $$

The stiffness coefficient matrix is then calculate based on,

K^{(e)}= k^{(e)} \cdot \begin{bmatrix} (l^{(e)})^2 & l^{(e)}m^{(e)} & -(l^{(e)})^2 & -l^{(e)}m^{(e)}\\ l^{(e)}m^{(e)} & (m^{(e)})^2 & -l^{(e)}m^{(e)} & -(m^{(e)})^2\\ -(l^{(e)})^2 & -l^{(e)}m^{(e)} & (l^{(e)})^2 & l^{(e)}m^{(e)}\\ -l^{(e)}m^{(e)} & -(m^{(e)})^2 & l^{(e)}m^{(e)} & (m^{(e)})^2 \end{bmatrix} $$

Where,

k^{(e)} = \frac{E^{(e)}A^{(e)}}{L^{(e)}} $$

Plot of Five Bar Truss Deformation
Based on the above computation and referencing the two-bar truss deformation plot the five-bar truss can now be plotted as both un-deformed and deformed. To make the deformation noticeable a magnifying factor of 1000 was used. The code is as follows,

Contributing Writers
Eml4500.f08.ateam.boggs.t 17:02, 24 October 2008 (UTC)

Eml4500.f08.ateam.carr 17:16, 24 October 2008 (UTC)

Eml4500.f08.ateam.shah 17:36, 24 October 2008 (UTC)

Eml4500.f08.ateam.nobrega 18:10, 24 October 2008 (UTC)

Eml4500.f08.ateam.rivero 18:10, 24 October 2008 (UTC)

Eml4500.f08.ateam.mcnally 18:15, 24 October 2008 (UTC


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User talk:KSch

 Hello KSch, 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 to contribute and to experiment with the sandbox or your userpage, and see you around Wikiversity! --mikeu talk 22:12, 11 October 2008 (UTC)

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

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Talk:Arrays, strings and records

This page contains many obvious errors. Most of them are fatal mistakes. I'll be working on it.

title: Talk:Arrays, strings and records


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User:Jj137

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title: User:Jj137


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User talk:Jj137

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Wikiversity:Think tank

The Think Tank is intended to serve a similar function to the "WikiProjects" on Wikipedia, as a place to organize efforts to create new community learning initiatives.

Choosing topics
Topics can come from any source, from current events that people might want to learn about to topics of interest brought up in another context. Focus could also be given to current events within the wikimedia community. The idea is to rapidly create base structures for learning initiatives in order to attract and serve contributors who may be interested or concerned about something going on in the world or their community.

Possible topics

 * 2008 global economic crisis
 * Consensus in a wiki environment
 * Creating imagemaps for use on wikis
 * Digital media and democracy -- maybe needs a better name, but the use of digital media in the 2008 U.S. election could be a very interesting topic.
 * Topic:International Year of Astronomy - 2009

Requested topics

 * article 31 of the Convention of the rights of a child, which states that every child has the right to (1)rest and leisure, to engage in (2)play and recreational activities and to participate in (3)cultural life and the arts. What are scientific and psychological arguments that can under build this article?
 * A more organized curriculum and more comprehensive course materials in neuroscience. I'm involved with the course materials in this area.  --AFriedman 22:00, 5 January 2009 (UTC)
 * /History of Science/ -- Per emailed request, incubating. --SB_Johnny talk 11:22, 17 December 2008 (UTC)
 * See also Category:History of Science. The Jade Knight (d'viser) 08:46, 24 December 2008 (UTC)
 * Possible cures for influenza virus, which is currently killing people in mexico and may become a global pandemic.
 * A more in-depth film program
 * The possibility of a total rebuild of the educational system using modern technology and ethics.

Finding projects
Many new contributors come to Wikiversity and begin creating content, but all too often receive very little assistance. Think tank participants can help identify and reach out to these users by watching the recent changes or new pages.

Participants interested in watching for new projects

 * --SB_Johnny talk 12:14, 15 December 2008 (UTC)
 * The Jade Knight (d'viser) 23:21, 15 December 2008 (UTC)
 * AFriedman 22:00, 5 January 2009 (UTC)
 * Bobby 16:07, 24 Feburary 2009 (UTC)
 * Debarko De 03:28, 16 May 2009 (UTC)
 * User:Chandra Janakiraman 11:30, 17 Jun 2009 (IST)

To-do

 * 1) Create a template for the user talk pages of people who begin new resources.

Designing projects
The skeletal structure of the initiatives should be set out in a quick and efficient way. Examples:
 * Will it use surveys or personal reflection?
 * Can participants collect data?
 * What Wikimedia resources are available? If it's a current events topic, can Wikinewsies or Commoners help provide documentation?
 * Choosing a "logo" for initiatives?
 * Navigation templates and general structure of the initiative.

Participants interested in helping develop new projects

 * --SB_Johnny talk 12:14, 15 December 2008 (UTC)
 * AFriedman 22:01, 5 January 2009 (UTC)
 * Djd011 12:24, 14 February 2009 (GMT-6)
 * Tejas V Hoskatti 18:40, 4 April 2009 (GMT-5:30)
 * Debarko De 03:29, 16 May 2009 (UTC)
 * (you?)

To-do

 * 1) Create a list of templates and other materials that can easily be adapted to new projects
 * 2) * Start a Wikiversity Project (tutorial)

Launching initiatives
Once a skeletal structure has been created, the initiative should be "advertised". In general, this could be done via regular announcements on the Colloquium and its analogs on the other Wikimedia projects. To avoid becoming a "pest", we should probably do this on a schedule announcing all of the new projects every 3 weeks or so, rather than announcing each project when it comes up (i.e, we don't want to be accused of "spamming" the other projects).

title: Wikiversity:Think tank


text:
User:Mbartley57

Welcome to MIKEB'S WIKI USERPAGE !!
Calculus prerequisite answers:

1. tan(t) = sin(t) /cos(t)

2. csc(t) = 1/x therefore x = sin(t)

3. tan2(x) + 1= sec2(x)--> (sin2(x) + cos2(x))/cos2(x) --> 1/cos2(x) --> sec2(x)

4. COS(A+B)= COSACOSB - SINASINB Proof1--> COS(A+A)= COS(2A) = COS2A - SIN2A -->2COS2A - 1 Proof2--> previous results used for half angle by substituting A for 2A and A/2 for A. Proof3--> COS2A = (COS2A+1)/2 Proof4--> COS3A =

title: User:Mbartley57


text:
User:Stevenarntson/Self Assessment

Self Assessment
A self assessment is a critique of yourself. Let it reflect your experience, and not your perception of an anyone's expectations of you.

Possible questions

 * "What did I learn in this class, and what do I have left to learn?"
 * "What did I do well in this class, and what could I have done better?"
 * "How do I feel prepared, or unprepared, to continue my studies at this point?"
 * "How did I contribute to the experiences of other students in this class?"
 * You are also encouraged to consider any other questions relevant to your experiences.

Responses should be at least 200 words, posted to your Wikiversity userspace.

"To reflect is to look back over what has been done so as to extract the net meanings which are the capital stock for intelligent dealing with further experiences. It is the heart of intellectual organization and of the disciplined mind." "Perhaps the greatest of all pedagogical fallacies is the notion that a person learns only the particular thing they are studying at the time."
 * -John Dewey

title: User:Stevenarntson/Self Assessment


text: Template:Wikipedia short

title: Template:Wikipedia short


text:
User talk:Mbartley57

 Hello Mbartley57, 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

 * Take a guided tour and learn to edit.
 * Visit a (kind of) random random project.
 * Browse everything on Wikiversity, or visit a portal corresponding to your educational level: pre-school, primary, secondary, tertiary, non-formal education.
 * Find out about research activities on Wikiversity.

 * Read an introduction for teachers and find out how to write an educational resource for Wikiversity.
 * Give feedback about your initial observations
 * Discuss Wikiversity issues or ask questions at the colloquium.
 * Chat with other Wikiversitans on #wikiversity-en.

And don't forget to explore Wikiversity with the links to your left. Be bold to contribute and to experiment with the sandbox or your userpage, and see you around Wikiversity! Jtneill - Talk - c 02:10, 12 October 2008 (UTC)

 Woof! My name is Jack(Russell). I am a dog and a Wikiversity mascot. I am pretty new around Wikiversity. Perhaps we can learn together! Tail wag...

Calculus
Hi there! In response to your answers to introduction to calculus: The others look OK. See you around Wikiversity! --HappyCamper 21:51, 12 October 2008 (UTC)
 * You can still convert the cosine to a function of sine in question 1.
 * In proof 4, you can do cos(3A) = cos (2A + A) and then expand everything.

title: User talk:Mbartley57


text:
User talk:Iliana Mirtschewa

 Hello Iliana Mirtschewa, 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

 * Take a guided tour and learn to edit.
 * Visit a (kind of) random random project.
 * Browse everything on Wikiversity, or visit a portal corresponding to your educational level: pre-school, primary, secondary, tertiary, non-formal education.
 * Find out about research activities on Wikiversity.

 * Read an introduction for teachers and find out how to write an educational resource for Wikiversity.
 * Give feedback about your initial observations
 * Discuss Wikiversity issues or ask questions at the colloquium.
 * Chat with other Wikiversitans on #wikiversity-en.

And don't forget to explore Wikiversity with the links to your left. Be bold to contribute and to experiment with the sandbox or your userpage, and see you around Wikiversity! Jtneill - Talk - c 03:30, 12 October 2008 (UTC)

 Woof! My name is Jack(Russell). I am a dog and a Wikiversity mascot. I am pretty new around Wikiversity. Perhaps we can learn together! Tail wag...

title: User talk:Iliana Mirtschewa


text:
Autism spectrum/Asperger syndrome

Asperger's Syndrome was a diagnosis created by Hans Asperger. It was introduced in the DSM-IV as a pervasive developmental disorder, with almost the exact same characteristics as autism, with the only change being that there should not be a speech delay or intellectual impairment. In the DSM-V however, all pervasive developmental disorders were combined into the broad diagnosis of autism spectrum disorder.

title: Autism spectrum/Asperger syndrome


text:
User:Stevenarntson/Seattle/Organizations

 * Home Alive
 * the Sightline Institute
 * the Seattle Farmers Market Alliance
 * Solid Ground
 * the Vera Project
 * Seattle Center for Book Arts
 * the Low-Income Housing Institute
 * Nickelsville Seattle
 * Seattle DIY
 * Seattle Free School


text:
User:Stevenarntson/Seattle/Public Space

 * Seattle Parks and Gardens
 * Seattle Public Library
 * the Ballard Locks
 * the Pike Place Market
 * Seattle Community Centers
 * the Olympic Sculpture Park

title: User:Stevenarntson/Seattle/Public Space


text:
User:Stevenarntson/The Commons

This class examines contemporary and historical art issues through the lens of "The Commons," in terms of physical spaces, materials, and ideas.

assignments and policies

schedule

 * 1) the commons
 * 2) art walk
 * 3) guerilla art
 * 4) film, TBA
 * 5) midterm. presentations due in class.
 * 6) collage
 * 7) collage, continued
 * 8) /surrealism/
 * 9) field trip, TBA
 * 10) final projects due
 * 11) final projects due. self assessment due

title: User:Stevenarntson/The Commons


text:
Business/Wikibooks

 * Accounting
 * Investing
 * Marketing
 * Corporate Social Responsibility
 * Macroeconomics
 * Microeconomics
 * Enterprise Resource Planning Internals
 * Accountancy

 * Asset Allocation
 * Asset Pricing
 * Business English
 * Consultant's Manual
 * Financial Derivatives
 * Personal finance
 * Investing
 * Investment Business Plan
 * Management for IT Professionals

 * Managerial Economics
 * Managing Groups and Teams
 * Meeting Basics
 * Management Strategy
 * Operations Strategy
 * Integration Management
 * Project Management
 * Introduction to Business/Overview

 * Technical analysis
 * Workplace Communication

title: Business/Wikibooks


text:
School:Business/Related news archives

 * April 2007 Russian plans tunnel to Alaska...
 * April 2007 Car manufacturer plans $2,500 car.
 * March 2007 Electric car maker to open five dealerships in America by Fall 2007.
 * March 2007 Worked expected to start on tunnel linking Europe to Africa in 2008
 * March 2007 More business use Wikis
 * March 2007 A company claims it will unveil radioactive waste processing technology.
 * March 2007 Delphi to try and raise 2 billion by offering stock holders 56,700,000 common shares at $35 each.
 * February 2007 Google to launch Google Apps Primere tomorrow. It should directly compete with Microsoft.
 * February 2007 National Geographic a world-wide prepaid mobile phone. It works in over 100 countries with the same number.
 * February 2007 Skype asks FCC to force cell carriers to allow their services to operate so more competition can be introduced into the market.

title: School:Business/Related news archives


text:
Business books on Google books

 * How Small Business Trades Worldwide: Your Guide to Starting Or Expanding a Small Business
 * Getting Down to Business: An Agenda for Corporate Social Innovation
 * The China Business Handbook 2005
 * The Internet Business Resources Kit: Everything You Need to Make Money Online
 * Asia and Africa in the Global Economy
 * The Rise of Open Source Licensing
 * Commercial Space Launch Industry: Congressional Hearing
 * Another book over at Google

title: Business books on Google books


text:
Template:Denotes wikipedia

(wp)

title: Template:Denotes wikipedia


text:
School:Theology/Participants

The histories of Wikiversity pages indicate who the active participants are. If you are an active participant in this school, you can list your name here (this can help small schools grow and the participants communicate better; for large schools it is not needed).

Active

 * 1) SotanScholar (discuss • contribs) as of Aug 3, 2022.
 * 2) AlistairReece 14:51, 28 August 2007 (UTC)
 * 3) Since 24 November 2013 with contributions to Theology --Marshallsumter (discuss • contribs) 14:52, 17 April 2018 (UTC)
 * 4) DErnestWachter, email:[mailto:dewnada@icloud.com dewnada@icloud.com]
 * 5) Mikeduke324
 * 1) Mikeduke324

Inactive

 * 1)  Hoopydink 12:13, 11 October 2006 (UTC)
 * 2)   Dionysios  (talk), founder of the former Department of Orthodox Christian Studies, now reorganized as the Center for Eastern Orthodox Studies, Date: 2007-10-28 (October 28, 2007) Time:  1920 UTC
 * 3) Chocoman 13:38, 2 September 2007 (UTC)
 * 4) StephenDeloney 20:43, 7 February 2007 (UTC)
 * 5) muzammilpalla
 * 6) Christve
 * 7) Andy33 13:49, 09 December 2010 (GMT)
 * 8) WendyB 20:12, 13 December 2013 (EST)
 * 9) The dreaming student 8:31 11 October 2013 (UTC)
 * 10) Magosgruss 20:33, 29 February 2008 (UTC)
 * 11) promulgation December 2007
 * 12) Guðsþegn 00:37, 23 October 2007 (UTC)
 * 13) Sjohnson 16:39, 23 August 2007 (UTC)
 * 14) Paul Danon
 * 15) JRasmussen
 * 16) davymast
 * 1) davymast

title: School:Theology/Participants


text:
Economics/Workspace

a collective workspace for learning about economics

readings@Wikipedia
lists of occupations

title: Economics/Workspace


text: Category:Otolaryngology

title: Category:Otolaryngology


text: Category:High school mathematics

title: Category:High school mathematics


text:
Astronomy college course/Saros cycle

Organization
This page will begin with a History section that links the Ecliptic, the Saros cycle and the Antikythera mechanism.

pages that need to be rewritten ASAP
The following subpages to Orbital mechanics exist and will be organized in a few days
 * Astronomy college course/Saros cycle/Saros/Quiz_1
 * Astronomy college course/Saros cycle/Saros

Resources that are already linked to wikipedia and should be moved with great care
Ecliptic Antikythera

Topics for quiz 1
223 synodic months    = 223  ×  29.530589  ≈ 6585.32  days (lunar phases)

242  draconic  months  = 242  ×  27.212221  ≈ 6585.35  days (line of nodes)

 * The synodic month is the time between two consecutive new moons, i.e., when the Moon passes the Sun in during it's monthly transit through all the constellations of the zodiac.
 * The draconic month marks two passages of the Moon through the line of nodes, i.e., when the Moon is in the plane defined by Earth's orbit around the Sun.

An eclipse can only occur

 * The anomalistic month marks successive perihelion's of Moon's elliptical orbit.  At perigelion, the Earth-Moon distance is at a minimum, and the Moon's orbital speed is at a maximum.  (But due to the rotation of the Earth, the Moon's apparent motion in the sky (from east to west) is at minimum on that day).

Topics for quiz 2
There seems no end to the complexities and patterns regarding cycles of eclipses. For reference, we note a few of them.

Anomalistic and sidereal months
From Orbit_of_the_Moon:
 * 223 synodic months    = 223  ×  29.530589  ≈ 6585.321  days (lunar phases)
 * 242  draconic  months  = 242  ×  27.212221  ≈ 6585.357  days (line of nodes)
 * 239  anomalistic  months   = 239  ×  27.55455  ≈ 6585.537  days (closest to Earth)
 * 241 sidereal months  = 241  ×  27.321662  ≈ 6584.521  days (orbit relative to the stars)

The anomalistic month marks the duration between consecutive perihelions, i.e., when the Moon is at it's closest approach. The Moon's proper motion is also governed by cycle, and the Antikythera mechanism contained a pin gear for the Moon's motion to model this effect. The matching of the Saros cycle to nearly 239 anomalistic months causes successive eclipses of the Saros cycle to endure for similar lengths of time. We also see that 241 sidereal months misses the cycle by one day, meaning that the stars are somewhat close to their original position with respect to the Moon on the next Saros. Finally, the fact that the Saros is within about 11 days of a year ensures that consecutive Saros eclipses will occur in roughly the same season.

Inex cycle
An even more precise cycle is the Inex cycle cycle:
 * 358 synodic months =358 × 29.530589 =10571.951 days
 * 388.5 draconic months = 388.5 × 27.212221 =10571.948 days

title: Astronomy college course/Saros cycle


text:
User:Harryinitiative

knowledge is for everyone

title: User:Harryinitiative


text:
User:Mrdboyell

Welcome to Mrdboyell User page

title: User:Mrdboyell


text:
User:Eas4200c.f08.nine.F/Homework 1

 Comment: This page is identical to the previous one, except that the page was moved from Wikipedia to Wikiversity. There are no other comparisons. Eas4200c.f08.Nine.F 08:18, 7 November 2008 (UTC)

Group nine - Week 1

Aircraft Design
Airplanes are composed of many different parts, the main structural ones being the wing and the fuselage. The wing and fuselage are made up of many basic structural elements, and those elements act like beams in a torsion member. One main difference between beams and aircraft structural elements is that aircraft loads are in the form of air pressure and concentrated loads such as in the landing gear and passenger seats. As an engineer, we have to decide if these local loads are not excessive enough to cause major deflections.

Aircraft Structure
The main structural purpose of the wing is to transfer the load to the fuselage. The airfoil shape of the wing is based on aerodynamics considerations and will not be discussed in this article, which focuses on the structural components of the wing. The wing acts like a beam and torsion member. The wing is composed mainly of spars and ribs. "The spar is a heavy beam running from spanwise to take transverse shear loads and spanwise bending. It is usually composed of a thin shear panel, or web, with a heavy cap or flange at the top and bottom to take bending." There may be more than one spar in a wing, but in general one carries the majority of the forces on it, and is called the main spar. The wing ribs are flat structures that are used to take in-plane loads and reduce the effective buckling length of the stringers and as a result it increases their compressive load capability. Ribs are placed chord wise and are supported by the spars.

A combination of using spars with or without stringers, thin or thick wing skin, thin or regular airfoils and stringers being manufactured or not as an integral part of the skin depends on the design characteristics of the wing. It varies according to need and performance.

The skin of the fuselage bears the shear stresses due to the torques and traverse forces. It also carries the hoop stresses due to internal pressures of the aircraft.

The stabilator of an aircraft acts as a horizontal stabilizer and elevator. Stabilators are found on the back portion of the wings and allows the pilot to have a higher pitching moment.

Fuselage Designs
The fuselage is the center of an airplane, it is the main portion that holds pay loads and passengers. The fuselage has relatively small loads applied to it. Most of the forces acting on the fuselage are forces from the wing and landing gear reactions, and the pay loads.

Monocoque is a single shell aircraft and vehicle design. The word Monocoque is french where 'mono' means single and 'coque' means shell. This design uses the shell to support all of structural loading. This design technique is not used very often, because the shell has to be fairly thick to support the loading which makes the aircraft heavy. A preferred design method is the Semi-Monocoque.

Semi-Monocoque is similar to the Monocoque design having a single shell. However in addition it has stiffening braces. The purpose of the braces are help support the necessary loading. The advantage of this design is the weight. The shell does not have to be as thick therefore reducing the weight. Most Semi-monocoque designs have a thin shell supported by braces running axially called stringers and longerons. These stringers and then supported by transverse braces called frames or rings.

Landing Gear
The landing gear of an aircraft is what makes it possible for an aircraft to take off and land. The landing gear commonly consists of a high strength steel alloy because of the extreme stresses that are experienced during take off and landing. In general, wheels are the main component in the landing gear, but other gear such as skis or floats may be used.

Aircraft Materials
Aircraft materials must be of high strength and stiffness while remaining lightweight. The typical types of materials that are used tend to be steel alloys, titanium, aluminum, and fiber-reinforced composites. Aluminum is a very attractive metal for use as an aircraft material because it is very light weight and although not as strong as steel and titanium alloys, it is sufficient to withstand the shearing loads on the fuselage of an aircraft. The 2024-T3 aluminum alloy is a common material used for the fuselage and lower wing skins. These areas are prone to fatigue and cyclic tensile stresses and that is why a stronger alloy used. The upper wing, however, uses a 7075-T6 alloy which has a higher strength and lower fracture toughness. The upper wing experiences less fatigue and that is why it is used. Carbon fiber is a non-metallic material that is used in aircraft design due to its high strength and extremely low weight. "Fiber composites are stiff, strong, and light and are thus most suitable for aircraft structures." Steel and titanium alloys see their use on the control surfaces because they are very strong metals.

Finite Element Method
The finite element method (FEM) is "a numerical technique for finding approximate solutions of partial differential equations (PDE) as well as of integral equations. The solution approach is based either on eliminating the differential equation completely (steady state problems), or rendering the PDE into an approximating system of ordinary differential equations, which are then solved using standard techniques such as Euler's method, Runge-Kutta, etc."

FEM is used often in most fields of Engineering. Partial differential equations are used in the designing of aerospace structures for example, and as a result the finite element method is used to simplify these functions for Aerospace Engineers. FEM is widely used for stiffness and strength visualizations, as well as reducing weight, material, and costs for certain projects. It can also be used to pin point where structures may bend or twist, and determine the distribution of stresses and displacements.

Stress and Strain
Stress and Strain are engineering terms that refer to a force per unit area, and the amount of deformation on an object caused by that stress respectively. Stress and Strain are proportional by a factor known as the "Young's Modulus". This quantity relates directly to an objects stiffness; the higher the modulus, the higher the stiffness. The Equation $\sigma = E \epsilon $ is read as "Stress is equal to Young's Modulus times Strain. Stress and Young's Modulus both have the units of Force per Area. Strain is a dimensionless value that denotes how much a material has deformed by a ratio of deformed length over the original length.

In the figure to the right represents a stress-strain curve. The point on the graph denoted by the number 1, signifies $$\sigma_u$$, the ultimate stress, this term represents the maximum stress possible for any ductile material. Point 2, represents $$\sigma_y$$ which represents the yield stress. This stress is the maximum stress that a material can withstand without being plastically deformed and can return to its original shape. Point 3 represents the point on the stress-strain curve where the material will fail or rupture. Region 4 represents the strain hardening region, and as discussed in the lecture is also known as the plastic hardening. Region 5 denoted the area where the material is softening or necking. To the left of point 2 known as the elastic region, the slope of that straight line is very important and is known as the Young's Modulus referred to earlier.

title: User:Eas4200c.f08.nine.F/Homework 1


text: Category:Orbital mechanics

title: Category:Orbital mechanics


text:
User:Eas4200c.f08.nine.F/Homework 2

 Comment: This page is identical to the previous one, except that the page was moved from Wikipedia to Wikiversity. There are no other comparisons. Eas4200c.f08.Nine.F 08:18, 7 November 2008 (UTC)

Lecture 1

Group nine - Week 2

___________________________________________________________________________________________________________________________________________________________

Second Moment of Inertia

"The second moment of area, also known as the area moment of inertia or second moment of inertia, is a property of a shape that is used to predict its resistance to bending and deflection which are directly proportional. This is why beams with higher area moments of inertia, such as I-beams, are so often seen in building construction as opposed to other beams with the same area."

A good example of second moment of inertia would be a persons attempt to prevent a force from turning a large lever. The ferther the person puts his or her hand from the pivot the easier it is to keep the lever from moveing. The persons hand can be thought of as the sum of all of the small parts makeing up the object. The position of the persons hand is proportional to the squar of the distance from the pivot. Each part of the object adds its own contribution depending on its shape, size and position. Each part can be cut into smaller parts and then be summed up until the infinitesimal size is reached and the result is accurate. The basic equation for the second moment of inertia is as follows.

Ix=$$ \int\ $$y2dA

The figure to the right shows different types of stringers. The moment of inertia will be calculated assuming the areas of a,b,c and d are the same.

Mohr's Circle
Mohr's Circle (FELIX)

Mohr’s circle is a graphical tool that can express a plane stresses as a function of the angle theta.

The creation of a Mohr’s circle takes few simple steps.
 * 1) Create a coordinate axis in which the horizontal axis represents the normal stress and the vertical axis will represent the shear stress.
 * 2) In an x-y coordinate system plot $$\sigma _{y}$$ and $$\tau_{xy}$$ a on the numerical position on the normal stress axis on the y-face of the object.
 * 3) The next point is the x-y coordinate of $$\sigma_x$$ and $$\tau_{xy}$$ noting that tension and clockwise shear on the x-face.
 * 4) Mark the midpoint of the line between the points described in steps 2 & 3. Refer to Figure 2 for a reference
 * 5) Create a circle as shown in Figure 2.
 * 6) With the circle now constructed each point on the circle represents a normal/shear stress combination. The line that was drawn in Figure 2 represents the principal axis of the stress. In Figure 3 the horizontal axis of the Mohr's circle represents the two principal stresses in which all he stress on an object represents a normal stress.

In reading a Mohr's circle the primary consideration comes from the angle $$\theta$$ which is measured off the principal axis. a rotation of 2 $$\theta$$ represents a rotation of $$\theta$$ in the element.

Brief History of Mohr's Circle
In 1885 Christian Otto Mohr, a German civil Engineer proposed a graphical representation of two and three-dimensional stresses. Mohr was born on October 8, 1835 and died October 2, 1918 in the town of Dresden.

___________________________________________________________________________________________________________________________________________________________

___________________________________________________________________________________________________________________________________________________________

___________________________________________________________________________________________________________________________________________________________

___________________________________________________________________________________________________________________________________________________________

Contributing Team Members
The following students contributed to this report:

David Phillips Eas4200C.f08.nine.d (talk) 18:34, 26 September 2008 (UTC)

Oliver Watmough Eas4200c.f08.nine.o 10:07, 26 September 2008 (UTC)

Stephen Featherman Eas4200c.f08.nine.s 11:48, 26 September 2008 (UTC)

Ricardo Albuquerque Eas4200c.f08.nine.r 4:30, 26 September 2008 (UTC)

Felix Izquierdo Eas4200c.f08.nine.F 4:34, 26 September 2008 (UTC)

title: User:Eas4200c.f08.nine.F/Homework 2


text:
User:Eas4200c.f08.nine.F/Homework 3

 Comment: This page is identical to the , except that the page was moved from Wikipedia to Wikiversity. There are no other comparisons. Eas4200c.f08.nine.F 08:18, 7 November 2008 (UTC)

Homework 1

Homework 2

Group nine - Homework 3

Introduction
Reasons to use open thin wall cross section for stringers

1) manufacturing: stamping of thin flat sheets

2) construction of aircraft: riveting

Why do the stringer have angled walls compared to vertical walls?

One possible reason is for storage purposes. Having angled walls allows the stringers to be stacked while saving space.

Shear Panels

$u \equiv u_{x}$

$u \equiv u_{y}$

$\gamma $ = change in the right angle due to shear deformation.

$\epsilon_{xy} $ = torsional shear strain = $ \frac{1}{2}\gamma_{xy}$

Curved Panels
$q = \tau t$  (Shear flow)

$d\vec{F} = qd\vec{l} = q(dl_{y}\hat{j} + dl_{z}\hat{k})$

$ = q(dlcos\theta \hat{j} + dlsin\theta \hat{k})$

where $dlcos\theta \hat{j}$  =  $dy$  and  $dlsin\theta \hat{k}$  =  $dz$

Resultant shear force vector

$\vec{F} = \int_{A}^{B} d\vec{F} = q(\int_{A}^{B}dy \hat{j} + \int_{A}^{B}dz \hat{k})$

$\vec{F} = q(a\hat{j} + b\hat{k})$

$\vec{F} = F_{y}\hat{j} + F_{z}\hat{k}$

$\Rightarrow F_{y} = qa$

$\Rightarrow F_{z} = qb$

$\frac{F_{y}}{F_{z}} = \frac{a}{b}$|undefined

Resultant Magnitude  $

$

$\Rightarrow R = q\sqrt{a^{2} + b^{2}}$|undefined

where $d$ = length of straight line $AB$.

$\Rightarrow R = qd$

We can relate $R = to $T = 2q\bar{A}$      (3.48)

Closed thin wall cross section:

$\vec{T} = T\hat{i}$

$d\vec{T} = \vec{r} \times d\vec{F}$

$\Rightarrow dT = \rho dF$

$T = \oint_{}^{}dT = q\oint_{}^{}\rho dl$

where $\rho dl = 2dA$

$\Rightarrow T = 2q \int_{\bar{A}}^{} dA$|undefined

$\Rightarrow T = 2q\bar{A}$

Open thin wall cross section:

$T = 2q\bar{A}$

$Re = T = 2q\bar{A}$

NACA 4-digit air foil series

ns = number of segments to the y-axis

$d\bar{A} = \frac{1}{2}\vec{r} \times \vec{PQ} =

$

Twist and Warping
Torsion of uniform, non circular bars (leads to warping of cross section)

Warping = axial displacement along x-axis of a part on the deformed cross section.

$\theta = \frac{\alpha}{x} = $   rate of twist.

$(1)\, \, u_{y} = -\theta_{xy} $ (3.11)

$(2)\, \, u_{z} = +\theta_{xy} $ (3.12)

Warping displacement along x-axis

$(3)\, \, u_{x} = \theta \psi(y,z)$

Road Map for torsional Analysis of aircraft wing

A) Kinematic assumptions (Section 3.2)

B) Strain - displacement relationship (Section 3.2) C) Equilibrium Equations ( Stresses)  (Ch.2; Section 3.2) D) Pranal Stress Function $\phi$(Section 3.2)

E) Strain compatibility  Eqn.(3.15)(3.17)

F) Eqn. for$\phi$ (3.19)

G) B conditions for$\phi$(3.24)

H)$T = 2\int\int_{A}^{} \phi dA,$ (3.25)

$T = GJ\theta$

$J = \frac{-4}{\bigtriangledown^{2} \phi} \int\int \phi dA$

I) Thin walled cross section

Formal Derivation ( Section 3.5)

$T = 2q\bar{A}$ (3.48)

J) Twist angle $\theta$ : Method 1

$\theta = \frac{1}{2G\bar{A}}\oint\frac{q}{t}ds$|undefined (3.56)

K) Multicell Section (Section 3.6)

Multicell Thin-Walled Sections
Multicell Thin-Walled Sections refers to wing sections that are composed of airfoil skin supported by thin vertical supports called spar webs that form multicell constructions. There are often stiffeners incorporated in the construction of multicell sections which are usually located above and below the spar webs themselfs. These stiffeners are very effective when used to carry bending loads, but they do little to help counteract torsional loads and are often neglected in the consideration of torsional rigidity.

For a two-cell section there are three boundary contours which we will label S0, S1, and S2. This will give us

$$\phi$$(S0)=C0

$$\phi$$(S1)=C1

$$\phi$$(s2)=C2

Where C0, C1, and C2 are all constants and $$\phi$$ is the stress function. The shear flow between two boundary contours is equal to the difference between the values of $$\phi$$ along these contours. The shear flow for each cell is considered positive if it forms a counterclockwise torque about the cell and its value is equal to the value of $$\phi$$ on the inside contour minus that on the outside contour.

q1=C1-C0

q2=C2-C0

q12=C1-C2

The Torque contributed by each cell can be calculated by using T=2Aq. The total torque of a two-cell section is

T=2A1q1+2A2q2

where A1 and A2 are the areas enclosed by the shear flows q1 and q2 respectively.

The equations for the twist angles $$\theta$$1 and $$\theta$$2 are

$$\theta$$1=$$\frac{1}{2\bar{A_{1}}G}\int_{cell 1}{\frac{qds}{t}}$$

$$\theta$$2=$$\frac{1}{2\bar{A_{2}}G}\int_{cell 2}{\frac{qds}{t}}$$

$$\theta_1$$=$$\theta_2$$=$$\theta$$

Homework Problems & Matlab Code
{| class="toccolours collapsible collapsed" style="width:100%" ! Homework Problems

Homework Problem 1

Prove: $$ A_{blue} = \frac {b h} {2} $$

$$ A_{red} = \frac {a h} {2} $$

$$ A_{blue} = A_{total} - A_{red} = \frac {(a+b) h} {2} - \frac {a h} {2} $$

so we have: $$ A_{blue} = \frac {b h} {2} + \frac {a h} {2} - \frac {a h} {2} = \frac {b h} {2} $$

Provig that: $$ A_{blue} = \frac {b h} {2} $$ ___________________________________________________________________________________________________________________________________________________________

Homework Problem 2

Show that: $$ J_{o} = \frac {\pi a^4}{2} $$

$$ J_{o}= \int_{A} {r^2 dA} = \int_{A} {r^2\times r\times dr\times d\theta} $$

(polar coordinates for dA)

Since we're integrating from (0 to a) around the whole circle, which means

$$ d\theta = 2\times \pi $$

we have: $$ 2\times \pi \times \int_{0}^{a}{r^3 dr} = \frac {2 \pi a^4} {4} = \frac {\pi a^4}{2} = J_{o} $$

and thus: $$ J_{o}= \int_{A} {r^2 dA} = \int_{A} {r^2\times r\times dr\times d\theta} $$

__________________________________________________________________________________________________________________________________________________________

Homework Problem 3

Show: $$ b^2 \cong \bar{r}^2 $$ and $$ a^2 \cong \bar{r}^2 $$

we know that: $$ \bar{r} = a + \frac {t}{2} \Rightarrow \bar{r}^2 = a^2 + \frac {t^2} {4} $$

since: $$ t \ll a, $$ we can neglect this term, leaving $$ a^2 \cong \bar{r}^2 $$

the same holds true for b since: $$ \bar{r} = b t \frac {t}{2} \Rightarrow \bar{r}^2 = b^2 - \frac {t^2} {4} $$

since: $$ t \ll b, $$ we can neglect this term, leaving $$ b^2 \cong \bar{r}^2 $$

___________________________________________________________________________________________________________________________________________________________

Homework Problem 4

Given: $$ t_{1} = 0.008 m \rightarrow t_{2} = 0.01 m = t_{3} \rightarrow b = 2 m \rightarrow a = 4 m $$

Find: $$ T_{all} $$

$$ \bar{A} = \frac {\pi}{2} \times \frac {b^2}{4} + \frac {b a}{2} $$

$$ \bar{A} = \frac {\pi}{2} \times \frac {4}{4} + \frac {8}{2} = 2\pi [m^2]$$

$$ \theta = \frac {1}{2 G \bar{A}} \times q\times [\frac{0.5 b \pi}{t_{1}} + \frac{a}{t_{2}} + \frac{\sqrt{a^2 + b^2}}{t_{3}}] $$

since: $$ q = \frac {T} {2 \bar{A}} = \frac {T}{4 \pi} [N/m] $$

$$ \theta = \frac {1}{4 G \pi} \times \frac {T}{4 \pi}\times [\frac{\pi}{0.008} + \frac{4}{0.01} + \frac{\sqrt{16 + 4}}{0.01}] = 98.669 [rad/m] $$

since: $$ T_{all} = 2\times \bar{A} \times \tau_{all} \times min_{thickness} $$

and (given): $$ \tau_{all} = 100 [GPa] $$

$$ T_{all} = 2\times 2\times \pi\times 100\times 0.008 = 10 \times 10^9 [Nm] $$

___________________________________________________________________________________________________________________________________________________________

Homework Problem 5

Compare solid circular cross section to hollow thin wall cross section:

a) solid circular cross section: $$ r_{o} = 1 cm $$

b) hollow circular cross section: $$ r_{o} = 5.1 cm \rightarrow r_{i} = 5 cm \rightarrow t = 0.1 cm $$

$$ A_{a} = \pi \times r^2 \rightarrow r_{o} = 1 cm \rightarrow A_{a}= \pi [cm^2] $$

$$ A_{b} = \pi \times r_{o}^2-r_{i}^2 \rightarrow r_{o} = 5.1 cm \rightarrow r_{i} = 5 cm \rightarrow

A_{b}= 1.0 \times \pi [cm^2] $$

$$ A_{a} \approx A_{b} \rightarrow 1.00 \approx 1.01 $$

$$ J_{a} = \frac {\pi r^4}{2} \rightarrow r = 1cm \rightarrow J_{a} = \frac {\pi}{2} [cm^4] $$

$$ J_{b} = \frac {\pi (r_{o}^4 -r_{i}^4)}{2} \rightarrow r_{o} = 5.1 cm \rightarrow r_{i} = 5 cm \rightarrow J_{b} = 80.9276 [cm^4] $$

$$ \frac {J_{a}}{J_{b}} = \frac {1.57}{80.9276} = 0.0194 $$

Find $$ r_{i}(c)$$ with $$ t=0.02 \times r_{i}(c) $$ such that $$ J_{c} = J_{a} $$ and find $$ \frac {A_{a}}{A_{c}} $$

$$ r_{i} + t = r_{o} \rightarrow r_{i} + 0.02 \times r_{i} = r_{o} $$

$$ J_{c} = J_{a} \rightarrow 0.5\times \pi\times (r_{o}^4-r_{i}^4) = \frac {\pi}{2} $$

$$ (r_{i} + 0.02 \times r_{i})^4 - r_{i}^4 = 1 \rightarrow r_{i} = 1.866 cm \rightarrow r_{o} = 1.9033 cm \rightarrow t = 0.037 cm $$

$$ \frac {A_{a}}{A_{c}} = \frac {\pi} {\pi (r_{o}^2-r_{i}^2)} = \frac {1}{(1.9^2-1.87^2)} \approx 7.16 $$

Sample Run
NACA Airfoil calculation program

Enter first digit of airfoil: 2 Enter second digit: 4 Enter the third and fourth digits: 15 Enter Py: 0 Enter Pz: 0 Enter number of segments: 60

The average area is: 0.103 The minumum number segments required to have the average area accurate within 1 percent is: 24.000

Figure 1 shows the cross-section of the NACA airfoil and Figure 2 shows the centroid line

Contributing Team Members
The following students contributed to this report:

David Phillips Eas4200C.f08.nine.d 3:10, 8 October 2008 (UTC)

Oliver Watmough Eas4200c.f08.nine.o 10:07, 8 October 2008 (UTC)

Stephen Featherman Eas4200c.f08.nine.s 1:07, 8 October 2008 (UTC)

Ricardo Albuquerque Eas4200c.f08.nine.r 4:30, 26 October 2008 (UTC)

Felix Izquierdo Eas4200c.f08.nine.F 18:34, 8 October 2008 (UTC)

title: User:Eas4200c.f08.nine.F/Homework 3


text:
User:Eml4500.f08.bottle.vitello/HW 4

Lecture Notes
Before we continue with our analysis of the 3-bar truss system we will make a quick comment on modifying a 2-bar truss program in MATLAB to solve a 3-bar truss system.

First, we define the connection array as "conn". Consider the 2-bar truss system:

$$ conn = \begin{bmatrix} 1 & 2\\ 2& 3 \end{bmatrix}$$ where column 1 corresponds to local node $$ 1 \! $$ and column 2 corresponds to local node $$ 2 \! $$. Similarly, row 1 represents element $$ 1 \! $$ and row 2 represents element $$ 2 \! $$. This array can then be represented by conn(e,j) = global node # of local node $$ j \! $$ of element $$ e \! $$

We denote the location master array by "lmm". Therefore

$$ lmm = \begin{bmatrix} 1 & 2 & 3 & 4\\ 3& 4 & 5 & 6 \end{bmatrix} $$

where row 1 represents element $$ 1 \! $$ and row 2 represents element $$ 2 \! $$, each column represents a local degree of freedom number with each number in the matrix representing the global degree of freedom number (or equation number) in $$ K \! $$.

This can be rewritten as limm(i,j)= equation number (global degree of freedom) for the element stiffness matrix coefficient corresponding to the $$jth \! $$ local degree of freedom number

Returning to our analysis of FEA, our goal is to find $$ \tilde{T}^{(e)}_{4X4} $$ that transforms the set of local element dof's $$ d^{(e)}_{4X1} $$ to another set of local element $$ \tilde{d}^{(e)}_{4X1}$$ such that $$ \tilde{T}^{(e)} $$ is invertible.

From previous lectures we know that:

$$ \underbrace {\tilde{d}_{1}^{(e)}}_{q_{1}^{(e)}} = \begin{vmatrix} l^{(e)} & m^{(e)} \end{vmatrix}\begin{Bmatrix} d_{1}^{(e)}\\ d_{2}^{(e)} \end{Bmatrix} $$

and $$ \tilde{d}_{2}^{(e)}= \vec{d}^{(e)}_{1}\circ \vec{\tilde{j}} $$ $$= -\sin \theta ^{(e)}d_{1}^{(e)}+\cos \theta ^{(e)}d_{2}^{(e)}$$

therefore:

$$ \tilde{d}_{2}^{(e)} = \begin{vmatrix} -m^{(e)} & l^{(e)} \end{vmatrix}\begin{Bmatrix} d^{(e)}_{1}\\ d_{2}^{(e)} \end{Bmatrix} $$

If we combine the two we get the following equation in matrix form: $$ \begin{Bmatrix} \tilde{d}_{1}^{(e)}\\ \tilde{d}_{2}^{(e)} \end{Bmatrix} = \begin{bmatrix} l^{(e)} & m^{(e)}\\ -m^{(e)} & l^{(e)} \end{bmatrix}\begin{Bmatrix} d_{1}^{(e)}\\ d_{2}^{(e)} \end{Bmatrix} $$

where $$ \begin{bmatrix} l^{(e)} & m^{(e)}\\ -m^{(e)} & l^{(e)} \end{bmatrix} = R^{(e)} $$

This can be expanded to : $$\underbrace{\begin{Bmatrix} \tilde{d}^{(e)}_{1}\\ \tilde{d}^{(e)}_{2}\\ \tilde{d}^{(e)}_{3}\\ \tilde{d}^{(e)}_{4} \end{Bmatrix}}_{\tilde{d}^{(e)}_{4X1}}= \underbrace{\begin{bmatrix} R^{(e)} & 0\\ 0 & R^{(e)} \end{bmatrix}}_{\tilde{T}^{(e)}_{4X4}}\underbrace{\begin{Bmatrix} d^{(e)}_{1}\\ d^{(e)}_{2}\\ d^{(e)}_{3}\\ d^{(e)}_{4} \end{Bmatrix}}_{d^{(e)}_{4X1}}$$

Examining the element below

we find $$ \tilde{f}^{(e)}= k^{(e)}\begin{bmatrix} 1 &0 & -1 & 0\\ 0&  0& 0 & 0\\  -1&  0&  1& 0\\  0& 0 & 0 &0  \end{bmatrix}\tilde{d}^{(e)}$$

where $$ \tilde{f}^{(e)}_{4X1}=\tilde{k}^{(e)}_{4X4}\tilde{d}^{(e)}_{4X1} $$

To find the values of K, we break down the problem by using interpolation of the transverse degrees of freedom using: $$ \tilde{d}^{(e)}=\tilde{T}^{(e)}d^{(e)}$$ and $$ \tilde{f}^{(e)}=\tilde{T}^{(e)}f^{(e)}$$ Using the simple force equation derived earlier $$ \tilde{K}^{(e)}\tilde{d}^{(e)}=\tilde{f}^{(e)}$$ we can then plug in the DOF equations as before and and get the result: $$ \tilde{K}^{(e)}\tilde{T}^{(e)}d^{(e)}=\tilde{T}^{(e)}f^{(e)} $$

And if we assume that T is invertable, we derive this result for the force equation:

$$ \left[\tilde{T}^{(e)^{-1}} \tilde{K}^{(e)} \tilde{T}^{(e)} \right]d^{(e)}=f^{(e)}$$

A way to prove that this inversion for T works is to consider a general block diagram matrix:

$$ B=\begin{bmatrix} d_{11} & 0 & 0 & 0\\ 0 & d_{22} & 0 & 0\\ 0 & 0 & ... & 0\\ 0 & 0 & 0 & d_{nn} \end{bmatrix}$$

By applying the inverse on this matrix gives this result:

$$ B^{-1}=\begin{bmatrix} \frac{1}{d_{11}} & 0 & 0 & 0\\ 0 & \frac{1}{d_{22}} & 0 & 0\\ 0 & 0 & ... & 0\\ 0 & 0 & 0 & \frac{1}{d_{nn}} \end{bmatrix}$$

Using the same method applied above for $$ R^{(e)}=\begin{bmatrix} l^{(e)} & m^{(e)}\\ -m^{(e)} & l^{(e)} \end{bmatrix} $$ gives the following result:

$$ R^{(e)^{T}}=\begin{bmatrix} l^{(e)} & -m^{(e)}\\ m^{(e)} & l^{(e)} \end{bmatrix} $$

Plugging the inverse of R will also give the inverse of T.

$$ \tilde{T}^{(e)^{-1}}=diag\left[ R^{(e)^{T}},R^{(e)^{T}} \right] $$

in which the value can be plugged back into the derivation

$$ \left[ \tilde{T}^{(e)^{-1}}\tilde{K}^{(e)}\tilde{T}^{(e)} \right]d^{(e)}=f^{(e)} $$

to give the results for K:

$$ k^{(e)}=\tilde{T}^{(e)^{T}}\tilde{k}^{(e)}\tilde{T}^{(e)} $$

In previous meetings we found the eigenvalues associated with a 2-bar truss system. Four eigenvalues were found, three of which were due to rigid body motion and the forth resulting from mechanisms. To understand the concept of a mechanism, we imagine that the support on a 2-bar truss system is cut and the structure can now move freely. The two bars can now move in a "scissor" motion about the pin joint. This is possible because there is no stored energy in the pin.

For the eigenvalue problem we have: $$ Kv=\lambda v \! $$, where $$ Kv $$ is the mass matrix and $$ \lambda v $$ is the identity matrix.

Let $$ \begin{Bmatrix} (u_{1})_{6X1} & (u_{2})_{6X1} & (u_{3})_{6X1} & (u_{4})_{6X1} \end{Bmatrix} $$ be pure eigenvalues corresponding to the four eigenvalues.

$$ K_{6X6}(u_{i})_{6X1}= \underbrace {0 u_{i}}_{0_{6X1}}, i = 1, ... ,4 $$

We can write the linear combination of $$ \begin{Bmatrix} u_{i}, i = 1, ... , 4 \end{Bmatrix}$$ as $$ \Sigma (\alpha _{i})_{1X1}(u_{i})_{6X1} = W_{6X1} \!$$

where $$\alpha _{i} = real numbers \!$$

Physically this means that the linear combination of movement in the  x \! and $$ y \! $$ directions and rotation.

$$ W \! $$ is also an eigenvalue that corresponds to a zero eigenvalue $$ KW=K(\sum_{i = 1}^{4} \alpha_{i}u_{i} ) $$ $$= \sum_{i = 1}^{4}{\alpha _{i}\underbrace{(Ku_{i})}_{0_{6X1}}} = 0 $$ $$ = 0_{1x1} W_{6X1} \!$$ We will now begin the justification of the assembly of the elemental stiffness matrix $$ k^{(e)} \!$$ into the global stiffness matrix.

Consider the example of the 2-bar truss and recall the following: 1. The elemental FD relationship can be written as $$ k^{(e)}d^{(2)}=f^{(e)} \!$$ 2. We used the Euler cut principle to find the equivalent of global node $$ 2 \! $$ 3. We drew the FBD's of element $$ 1 \! $$ and $$ 2 \! $$ to find the elemental degrees of freedom $$ d^{(e)}_{4X1} $$ 4. We identified the global degrees of freedom to elemental degrees of freedom for node $$ 2 \! $$  for element  $$ 1 \! $$ and $$ 2 \! $$

We can draw the following FBD to represent the equilibrium for node $$ 2 \! $$

We can now see that the addition of factors in the global stiffness matrix is because of the equilibrium position of node $$ 2 \! $$. The equations of equilibrium are $$ \sum{F_{x}}= 0 = -f^{(1)}_{3}-f^{(2)}_{1}$$ $$ \sum{F_{y}}= 0 = P-f^{(1)}_{4}-f^{(2)}_{2} $$

Leading to the summation of, $$ f^{(1)}_{3}+f^{(2)}_{1}=0$$ $$ f^{(1)}_{4}+f^{(2)}_{2}=P $$

The Kd=F matrices can be plugged into the summation of forces to get the overall big picture of the small deflections in the two bar truss and ultimately, obtain the values for the K matrix.

$$ \left[k^{(1)}_{31}d^{(1)}_{1}+k^{(1)}_{32}d^{(1)}_{2}+k^{(1)}_{33}d^{(1)}_{3}+k^{(1)}_{34}d^{(1)}_{4} \right]+\left[k^{(2)}_{11}d^{(2)}_{1}+k^{(2)}_{12}d^{(2)}_{2}+k^{(2)}_{13}d^{(2)}_{3}+k^{(2)}_{14}d^{(2)}_{4} \right]=0$$

$$ \left[k^{(1)}_{41}d^{(1)}_{1}+k^{(1)}_{42}d^{(1)}_{2}+k^{(1)}_{43}d^{(1)}_{3}+k^{(1)}_{44}d^{(1)}_{4} \right]+\left[k^{(2)}_{21}d^{(2)}_{1}+k^{(2)}_{22}d^{(2)}_{2}+k^{(2)}_{23}d^{(2)}_{3}+k^{(2)}_{24}d^{(2)}_{4} \right]=P $$

Prove $$ \textbf{k}^{(e)} = \textbf{T}^{(e)^{T}} \hat{\textbf{k}}^{(e)} \textbf{T}^{(e)} $$
Where

$$ \textbf{k}^{(e)}= \left[ \begin{array}{cccc} l_{s}^{2} & l_{s}m_{s} & -l_{s}^{2} & -l_{s}m_{s} \\ l_{s}m_{s} & m_{s}^{2} & l_{s}m_{s} & -m_{s}^{2} \\ -l_{s}^{2} & -l_{s}m_{s} & l_{s}^{2} & l_{s}m_{s} \\ -l_{s}m_{s} & -m_{s}^{2} & l_{s}m_{s} & m_{s}^{2} \end{array} \right]$$

As given in the notes above

$$ \textbf{T}^{(e)^{T}} = \left [ \begin{array}{cccc} l_{s}^{(e)} & 0 \\ m_{s}^{(e)} & 0 \\ 0 & l_{s}^{(e)} \\ 0 & m_{s}^{(e)} \end{array} \right] $$

and

$$ \hat\textbf{k}^{(e)} = \left [ \begin{array}{cccc} 1 & -1 \\ -1 & 1 \end{array} \right] $$

Therefore

$$ \hat\textbf{k}^{(e)}\textbf{T}^{(e)} = \left [ \begin{array}{cccc} l_{s}^{(e)} & m_{s}^{(e)} & -l_{s}^{(e)} & -m_{s}^{(e)} \\ -l_{s}^{(e)} & -m_{s}^{(e)} & l_{s}^{(e)} & m_{s}^{(e)} \end{array} \right] $$

Finally, multiplying by $$ T^{(e)^{T}} $$ gives

$$ \left[ \begin{array}{cccc} l_{s}^{2} & l_{s}m_{s} & -l_{s}^{2} & -l_{s}m_{s} \\ l_{s}m_{s} & m_{s}^{2} & l_{s}m_{s} & -m_{s}^{2} \\ -l_{s}^{2} & -l_{s}m_{s} & l_{s}^{2} & l_{s}m_{s} \\ -l_{s}m_{s} & -m_{s}^{2} & l_{s}m_{s} & m_{s}^{2} \end{array} \right]$$

5-Bar Truss System
The execution of this code results in:

EDU>> TrussAssemblyEx

K =

1.0e+005 *

0.3260   0.7607   -0.3260   -0.7607         0         0         0         0    0.7607    1.7749   -0.7607   -1.7749         0         0         0         0   -0.3260   -0.7607    0.3260    0.7607         0         0         0         0   -0.7607   -1.7749    0.7607    1.7749         0         0         0         0         0         0         0         0         0         0         0         0         0         0         0         0         0         0         0         0         0         0         0         0         0         0         0         0         0         0         0         0         0         0         0         0

K =

1.0e+005 *

0.3260   0.7607   -0.3260   -0.7607         0         0         0         0    0.7607    1.7749   -0.7607   -1.7749         0         0         0         0   -0.3260   -0.7607    2.1009    1.5213         0         0   -1.7749   -0.7607   -0.7607   -1.7749    1.5213    2.1009         0         0   -0.7607   -0.3260         0         0         0         0         0         0         0         0         0         0         0         0         0         0         0         0         0         0   -1.7749   -0.7607         0         0    1.7749    0.7607         0         0   -0.7607   -0.3260         0         0    0.7607    0.3260

K =

1.0e+005 *

0.3260   0.7607   -0.3260   -0.7607         0         0         0         0    0.7607    2.9749   -0.7607   -1.7749         0   -1.2000         0         0   -0.3260   -0.7607    2.1009    1.5213         0         0   -1.7749   -0.7607   -0.7607   -1.7749    1.5213    2.1009         0         0   -0.7607   -0.3260         0         0         0         0         0         0         0         0         0   -1.2000         0         0         0    1.2000         0         0         0         0   -1.7749   -0.7607         0         0    1.7749    0.7607         0         0   -0.7607   -0.3260         0         0    0.7607    0.3260

K =

1.0e+005 *

0.3260   0.7607   -0.3260   -0.7607         0         0         0         0    0.7607    2.9749   -0.7607   -1.7749         0   -1.2000         0         0   -0.3260   -0.7607    2.1009    1.5213         0         0   -1.7749   -0.7607   -0.7607   -1.7749    1.5213    2.1009         0         0   -0.7607   -0.3260         0         0         0         0    1.2000         0   -1.2000         0         0   -1.2000         0         0         0    1.2000         0         0         0         0   -1.7749   -0.7607   -1.2000         0    2.9749    0.7607         0         0   -0.7607   -0.3260         0         0    0.7607    0.3260

K =

1.0e+005 *

0.3260   0.7607   -0.3260   -0.7607         0         0         0         0    0.7607    2.9749   -0.7607   -1.7749         0   -1.2000         0         0   -0.3260   -0.7607    2.4309    1.1914   -0.3300    0.3300   -1.7749   -0.7607   -0.7607   -1.7749    1.1914    2.4309    0.3300   -0.3300   -0.7607   -0.3260         0         0   -0.3300    0.3300    1.5300   -0.3300   -1.2000         0         0   -1.2000    0.3300   -0.3300   -0.3300    1.5300         0         0         0         0   -1.7749   -0.7607   -1.2000         0    2.9749    0.7607         0         0   -0.7607   -0.3260         0         0    0.7607    0.3260

R =

0          0           0     -150000           0           0           0           0

Using this data, it is possible to create MATLAB code to graphically represent a scaled version of the deformed truss system.

Three-bar truss system
The code used to solve the two-bar truss system was altered to solve the three-bar truss system. The altered code is displayed below, followed by the functions required to solve the system, along with a diagram near the end portraying the deformation in the system. % Three bar truss example taken from 2 bar truss example % modified for three bars instead clear all; % Matrices e, A, and L represent Young's Modulus, cross sectional area, and % length of the three elements e = [2 4 3]; A = [3 1 2]; L=[5 5 10]; % Verical force exerted on node 2 P = 30; % Angles between each element and the horizontal alpha = pi/6; beta = pi/6; ceta = pi/4;  % note: "ceta" is not actually a greek letter % Coordinates of each of the 4 nodes calculated with trigonometry nodes = [0, 0; L(1)*cos(alpha), L(1)*sin(alpha); 2*L(1)*cos(alpha), 0; L(1)*cos(alpha)-L(3)*cos(ceta),L(1)*sin(alpha)-L(3)*sin(ceta)]; % Since the problem contains 4 nodes, there are 8 global degrees of freedom dof=2*length(nodes); % The connectivity matrix describes which nodes are connected to one % another by listing each row as an element and the global nodes it % connects conn=[1,2; 2,3; 2,4]; % The location matrix master array details which global degrees of freedom % are related to matching local degrees of freedom. lmm = [1, 2, 3, 4; 3, 4, 5, 6; 3, 4, 7, 8]; % The number of elements is determined by the size of the lmm matrix elems=size(lmm,1); % The K (square) and R (columnar) matrices are initialized with zeros K=zeros(dof); R = zeros(dof,1); debc = [1, 2, 5, 6, 7, 8]; ebcVals = zeros(length(debc),1); % Load vector R = zeros(dof,1); R(4) = P; % Assemble global stiffness matrix K=zeros(dof); for i=1:elems lm=lmm(i,:); con=conn(i,:); k_local=e(i)*A(i)/L(i)*[1 -1; -1 1] k=PlaneTrussElement(e(i), A(i), nodes(con,:)) K(lm, lm) = K(lm, lm) + k; end K R % Nodal solution and reactions [d, reactions] = NodalSoln(K, R, debc, ebcVals) results=[]; for i=1:elems results = [results; PlaneTrussResults(e, A, ...           nodes(conn(i,:),:), d(lmm(i,:)))]; end format short g results The above code uses the following functions obtained from the website highlighted in the Fundamental Finite Element Analysis and Applications textbook by Bhatti. function k = PlaneTrussElement(e, A, coord) % PlaneTrussElement(e, A, coord) % Generates stiffness matrix for a plane truss element % e = modulus of elasticity % A = area of cross-section % coord = coordinates at the element ends x1=coord(1,1); y1=coord(1,2); x2=coord(2,1); y2=coord(2,2); L=sqrt((x2-x1)^2+(y2-y1)^2); ls=(x2-x1)/L; ms=(y2-y1)/L; k = e*A/L*[ls^2, ls*ms,-ls^2,-ls*ms; ls*ms, ms^2,-ls*ms,-ms^2; -ls^2,-ls*ms,ls^2,ls*ms; -ls*ms,-ms^2,ls*ms,ms^2]; function [d, rf] = NodalSoln(K, R, debc, ebcVals) % [nd, rf] = NodalSoln(K, R, debc, ebcVals) % Computes nodal solution and reactions % K = global coefficient matrix % R = global right hand side vector % debc = list of degrees of freedom with specified values % ebcVals = specified values dof = length(R); df = setdiff(1:dof, debc); Kf = K(df, df); Rf = R(df) - K(df, debc)*ebcVals; dfVals = Kf\Rf; d = zeros(dof,1); d(debc) = ebcVals; d(df) = dfVals; rf = K(debc,:)*d - R(debc); function [eps, sigma, force] = PlaneTrussResults(e, A, coord, disps) % [eps, sigma, force] = PlaneTrussResults(e, A, coord, disps) % Compute plane truss element results % e = modulus of elasticity % A = Area of cross-section % coord = coordinates at the element ends % disps = displacements at element ends % The output quantities are eps = axial strain % sigma = axial stress and force = axial force. x1=coord(1,1); y1=coord(1,2); x2=coord(2,1); y2=coord(2,2); L=sqrt((x2-x1)^2+(y2-y1)^2); ls=(x2-x1)/L; ms=(y2-y1)/L; T=[ls,ms,0,0; 0,0,ls,ms]; d = T*disps; eps= (d(2)-d(1))/L; sigma = e.*eps; force = sigma.*A; Results of the solved 3-bar truss problem are as follows: k_local = 1.2        -1.2         -1.2          1.2 k = 0.9     0.51962         -0.9     -0.51962      0.51962          0.3     -0.51962         -0.3         -0.9     -0.51962          0.9      0.51962     -0.51962         -0.3      0.51962          0.3 k_local = 0.8        -0.8         -0.8          0.8 k = 0.6    -0.34641         -0.6      0.34641     -0.34641          0.2      0.34641         -0.2         -0.6      0.34641          0.6     -0.34641      0.34641         -0.2     -0.34641          0.2 k_local = 0.6        -0.6         -0.6          0.6 k = 0.3         0.3         -0.3         -0.3          0.3          0.3         -0.3         -0.3         -0.3         -0.3          0.3          0.3         -0.3         -0.3          0.3          0.3 K = 0.9     0.51962         -0.9     -0.51962            0            0            0            0      0.51962          0.3     -0.51962         -0.3            0            0            0            0         -0.9     -0.51962          1.8      0.47321         -0.6      0.34641         -0.3         -0.3     -0.51962         -0.3      0.47321          0.8      0.34641         -0.2         -0.3         -0.3            0            0         -0.6      0.34641          0.6     -0.34641            0            0            0            0      0.34641         -0.2     -0.34641          0.2            0            0            0            0         -0.3         -0.3            0            0          0.3          0.3            0            0         -0.3         -0.3            0            0          0.3          0.3 R = 0    0     0    30     0     0     0     0 d = 0           0      -11.674       44.405            0            0            0            0 reactions = -12.567     -7.2557       22.387      -12.925      -9.8194      -9.8194 results = 2.4186      6.4625       2.3145

Displaying the three-bar truss problem can be done with the code written below. clear; close; % The 3-bar truss problem has 8 connection points between lines that will % be displayed n_node = 8; % there are 9 dotted lines to be displayed, showing the 3-bar truss system % before deformation and showing the lines AB, AC, AD n_elem = 9; % P values calculated from the results above based on the equation % P = sqrt[(f1_1)^2 + (f2_1)^2] P2_1 = 14.512; P1_2 = 25.849; P1_3 = 13.88676; k_1 = 1.2; k_2 = 0.8; k_3 = 0.6; % Deformation in each bar calculated by P(e)/k(e) AB = P2_1/k_1; AC = P1_2/k_2; AD = P1_3/k_3; % Positions of all 8 nodes based on the origin at global node 4 position(:,1) = [10*cos(pi/4)-5*cos(pi/6); 10*sin(pi/4)-5*sin(pi/6); 0];  % global node 1 position(:,2) = [10*cos(pi/4); 10*sin(pi/4); 0];    % Point A, global node 2 position(:,3) = [10*cos(pi/4)+5*cos(pi/6); 10*sin(pi/4)-5*sin(pi/6); 0];     % global node 3 position(:,4) = [0; 0; 0]; % global node 4 position(:,5) = [10*cos(pi/4)+AB*cos(pi/6); 10*sin(pi/4)+AB*sin(pi/6); 0]; % point B position(:,6) = [10*cos(pi/4)-AC*cos(pi/6); 10*sin(pi/4)+AC*sin(pi/6); 0];  % point C position(:,7) = [10*cos(pi/4)+AD*cos(pi/4); 10*sin(pi/4)+AD*sin(pi/4); 0];  % point D position(:,8) = [-4.60; 51.48; 0];  % point E, or global node 2 after deformation % Creates x,y,z coordinates for the position of each node above for i = 1 : n_node; x(i) = position(1,i); y(i) = position(2,i); z(i) = position(3,i); end % node_connect(local node number, element number) = global node number node_connect(1,1) = 1; node_connect(2,1) = 2; node_connect(1,2) = 2; node_connect(2,2) = 3; node_connect(1,3) = 2; node_connect(2,3) = 4; node_connect(1,4) = 2; node_connect(2,4) = 5; node_connect(1,5) = 2; node_connect(2,5) = 6; node_connect(1,6) = 2; node_connect(2,6) = 7; node_connect(1,7) = 5; node_connect(2,7) = 8; node_connect(1,8) = 6; node_connect(2,8) = 8; node_connect(1,9) = 7; node_connect(2,9) = 8; % Plotting a dotted line between nodes for i = 1 : n_elem; node_1 = node_connect(1,i); node_2 = node_connect(2,i); xx = [x(node_1),x(node_2)]; yy = [y(node_1),y(node_2)]; zz = [z(node_1),z(node_2)]; plot3(xx,yy,zz,':') hold on end % Plotting a solid line to represent the deformed 3-bar truss system % Element 1 node_1 = 1; node_2 = 8; xx = [x(node_1),x(node_2)]; yy = [y(node_1),y(node_2)]; zz = [z(node_1),z(node_2)]; % Element 2 node_1 = 3; node_2 = 8; xxx = [x(node_1),x(node_2)]; yyy = [y(node_1),y(node_2)]; zzz = [z(node_1),z(node_2)]; % Element 3 node_1 = 4; node_2 = 8; xxxx = [x(node_1),x(node_2)]; yyyy = [y(node_1),y(node_2)]; zzzz = [z(node_1),z(node_2)]; plot3(xx,yy,zz,'-') plot3(xxx,yyy,zzz,'-') plot3(xxxx,yyyy,zzzz,'-') title('Deformed and undeformed three-bar truss system') xlabel('x') ylabel('y') zlabel('z') view([0 0 1])     % xy plane view

''Note: this MATLAB code was created using the following codes as examples:

- The MATLAB script for analyzing the Two Bar Truss System was obtained from Professor Vu-Quoc's website http://clesm.mae.ufl.edu/~vql/courses/fead/2008.fall/codes/twoBarTrussEx.txt

- The MATLAB script for the plot of a crab-leg structure created by X.G. Tan (2003) and obtained from Professor Vu-Quoc's website http://apollo.mae.ufl.edu/~vql/courses/fead/2006.fall/codes/matlab_plot_example2.m

- The MATLAB script for the plot of an electric pylon (undeformed five-bar truss structure) created by Team MeshWorks (2003) and obtained from Professor Vu-Quoc's website http://apollo.mae.ufl.edu/~vql/courses/fead/2006.fall/code.web.pages/electric_pylon.m''

From the above 2-bar truss system, expand the relationship F = k d for the sum of the vertical forces  In Local DOFs:

$$P = [k_{41}^{(1)}d_1^{(1)} + k_{42}^{(1)}d_2^{(1)} + k_{43}^{(1)}d_3^{(1)} + k_{44}^{(1)}d_4^{(1)}] + [k_{21}^{(2)}d_1^{(2)} + k_{22}^{(2)}d_2^{(2)} + k_{23}^{(2)}d_3^{(2)} + k_{24}^{(2)}d_4^{(2)}]$$

In Global DOFs:

$$P = [k_{41}^{(1)}d_1 + k_{42}^{(1)}d_2 + k_{43}^{(1)}d_3 +k_{44}^{(1)}d_4] + [k_{21}^{(2)}d_3  + k_{22}^{(2)}d_4 + k_{23}^{(2)}d_5 + k_{24}^{(2)}d_6]$$

Where

$$d_1^{(1)} = d_1$$ $$d_2^{(1)} = d_2$$ $$d_3^{(1)} = d_2^{(2)}=d_3$$ $$d_4^{(1)} = d_1^{(2)}=d_4$$ $$d_3^{(2)} = d_5$$ $$d_4^{(2)} = d_6$$

are defined as the conversions between local DOFs to global DOFs.

2-Bar Truss Eigenvectors
The eigenvalues of the global stiffness matrix are as follows:

$$\textbf{K}= \begin{bmatrix} 0.5625 & 0.3248 & -0.5625 & -0.3248 & 0 & 0 \\ 0.3248 & 0.1875 & -0.3248 & -0.1875 & 0 & 0 \\ -0.5625 & -0.3248 & 3.0625 & -2.1752 & -2.5 & 2.5 \\ -0.3248 & -0.1875 & -2.1752 & 2.6875 & 2.5 & -2.5 \\ 0 & 0 & -2.5 & 2.5 & 2.5 & -2.5 \\ 0 & 0 & 2.5 & -2.5 & -2.5 & 2.5  \end{bmatrix}\,$$

To find the eigenvectors corresponding to the eigenvalues of the global stiffness matrix, first the matrix of eigenvalues is defined in MATLAB:

k =

0.5625   0.3248   -0.5625   -0.3248         0         0    0.3248    0.1875   -0.3248   -0.1875         0         0   -0.5625   -0.3248    3.0625   -2.1752   -2.5000    2.5000   -0.3248   -0.1875   -2.1752    2.6875    2.5000   -2.5000         0         0   -2.5000    2.5000   -2.5000    2.5000         0         0    2.5000   -2.5000    2.5000   -2.5000

Then, the following command is implemented:

EDU>> E = eig(k)

E =

-7.0650  -0.0000   -0.0000    0.0000    1.4401    7.1249

Plotting these values in yields the following graph:

Plotting eigenvectors corresponding to zero eigenvectors for a 3 bar truss system
Assume the following case of a 3 bar truss system:

Where a=b=1, and the three bar properties consist of; E=2 and A=3. This means for all elements, $$\ k=\frac{(EA)}{L}=\frac{(2*3)}{1}=6 $$

The system can be described as $$\ \mathbf{K} * \mathbf{v}= \mathbf{\lambda\ }* \mathbf{v} $$

Where $$\ \mathbf{K} $$ is the 6x6 stiffness matrix for the contrained system.

Individual Contributions
Eml4500.f08.bottle.vitello 17:49, 12 October 2008 (UTC) Eml4500.f08.bottle.butler 21:01, 16 October 2008 (UTC) Eml4500.f08.bottle.hipps 20:32, 19 October 2008 (UTC) Eml4500.f08.bottle.brockmiller 20:21, 23 October 2008 (UTC) Eml4500.f08.bottle.loschak 04:49, 24 October 2008 (UTC) Eml4500.f08.bottle.barnes 15:04, 24 October 2008 (UTC) Eml4500.f08.bottle.ranto 15:48, 24 October 2008 (UTC)

title: User:Eml4500.f08.bottle.vitello/HW 4


text:
User:Stevenarntson/The Commons/The Commons

We'll discuss the Commons this quarter in terms of materials, places, and ideas.

 * The Tragedy of the Commons
 * The Tragedy of the Anticommons

Materials

 * Rivers and Tides, a film by Thomas Riedelsheimer about artist Andy Goldsworthy.
 * Kurt Schwitters.

Places

 * the Umbrellas, by Christo and Jeanne Claude
 * A video about Add Art

Ideas

 * The Amen Break, by Nate Harrison.