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User:Eml4500.f08.wiki1.ambrosio/hw7

Correction to HW 6
After HW6 was submitted, an important error was discovered in the pylon code. The -1000 lb force was intended to be applied at node 33 in the y direction. This was mistakenly entered into the R matrix as row 33. The problem is that row 33 is DOF 33, not node 33. As a result, the code has been changed so the 1000 lbs is in row 66, which is the vertical DOF on node 33.

With these new results, it is determined that the highest compressive stress is -8.6957e+006 Pa, which is located in element 55. The highest tensile stress is 9.0511e+006 Pa in element 81.

HW 7 Code
The highest bending moment is 6.287 N-m which is located on element 81. The highest shear stress is -5.48269 Pa, which is also located on element 81.

Notes 12/5 Meeting 40
At this point, a dimensional analysis of Eqn. (2) from page 39-2 is done.

$$ \begin{Bmatrix} u(\tilde{x})\\ v(\tilde{x}) \end{Bmatrix} = \begin{bmatrix} N_1&0 &0 &N_4 &0&0\\ 0& N_2 & N_3&0&N_5&N_6 \end{bmatrix} \begin{Bmatrix} \tilde{d}_1^{(e)}\\ \tilde{d}_2^{(e)}\\ \tilde{d}_3^{(e)}\\ \tilde{d}_4^{(e)}\\ \tilde{d}_5^{(e)}\\ \tilde{d}_6^{(e)} \end{Bmatrix} $$

The dimensional values of N1, N2, N4, and N5 are known to be 1, and the dimensional value of N3 and N6 are known to be L. $$\begin{bmatrix} \tilde{d}_1 \end{bmatrix}$$, $$\begin{bmatrix} \tilde{d}_2 \end{bmatrix}$$ , $$\begin{bmatrix} \tilde{d}_4 \end{bmatrix}$$ , and $$\begin{bmatrix} \tilde{d}_5 \end{bmatrix}$$ are deformations in the $$\tilde{x}$$ and $$\tilde{y}$$ axis, so their dimensional values are L. $$\begin{bmatrix} \tilde{d}_3 \end{bmatrix}$$ and $$\begin{bmatrix} \tilde{d}_6 \end{bmatrix}$$ are roational DOF's, so their deimensional values are 1.

When expanded, the the result is

$$ \begin{bmatrix} N_1 \end{bmatrix} \begin{bmatrix} \tilde{d}_1 \end{bmatrix} = L $$

$$ \begin{bmatrix} N_4 \end{bmatrix} \begin{bmatrix} \tilde{d}_4 \end{bmatrix} = L $$

$$ \begin{bmatrix} N_2 \end{bmatrix} \begin{bmatrix} \tilde{d}_2 \end{bmatrix} = L $$

$$ \begin{bmatrix} N_3 \end{bmatrix} \begin{bmatrix} \tilde{d}_3 \end{bmatrix} = L $$

$$ \begin{bmatrix} N_5 \end{bmatrix} \begin{bmatrix} \tilde{d}_5 \end{bmatrix} = L $$

$$ \begin{bmatrix} N_6 \end{bmatrix} \begin{bmatrix} \tilde{d}_6 \end{bmatrix} = L $$

title: User:Eml4500.f08.wiki1.ambrosio/hw7


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

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User:EML4500.f08.RAMROD.E/HW7

Frame Element Definition
A frame element is an element that is made up of a beam element and a truss element. Recall that for a truss element, all deformations were axial while in a beam element, the deformation was transverse to the axis. Knowing this, we can assume that for a frame element, the deformation will be some linear combination of both axial and transverse displacements at each node.

Model Problem for Frame
As with other types of elements, we will do our simple analysis using a model problem. The model here is a two-bar frame with nodes 1 and 3 simply supported and node 2 is a rigid connection between elements 1 and 2, as shown in the diagram below:

Combining the two elemental FBDs to a global FBD is shown in the picture below.

Elemental and Global Stiffness Matrices
The next step is deriving the stiffness matrix for our model problem. Since there are more dof's at each node now, the elemental stiffness matrices will each be a 6x6 matrix. We combine them very similarly as with trusses, with the overlapping dof's being at node two, such that the global stiffness matrix will be a 9x9 and be made up as shown below:

title: User:EML4500.f08.RAMROD.E/HW7


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User:Bjs

Former user Bjs
When usurping this account for SUL purposes, the former user Bjs was moved to User:Bjs (moved)

title: User:Bjs


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User:Eml4500.f08.delta 6.lopez/HW7

Monday 12/1
The applied internal forces

$$ N_{5}\left(\tilde{x}\right) $$ $$ N_{6}\left(\tilde{x}\right) $$

are shown below.

Recall the equation of the displacement matrix below, where $$ \mathbf{d}_{6x1}^{(e)} $$

is known after solving the system.

$$ \mathbf{\tilde{d}}_{6x1}^{(e)} = \mathbf{\tilde{T}}_{6x6}^{(e)}\mathbf{d}_{6x1}^{(e)} $$

Next, we will look at the computation of

$$ u\left(\tilde{x}\right) $$

and

$$ v\left(\tilde{x}\right) $$

$$ \mathbf{u}\left(\tilde{x}\right) = u(\tilde{x})\tilde{i} + v(\tilde{x})\tilde{j} = u_{x}(\tilde{x})\tilde{i} + u_{y}(\tilde{x})\tilde{j} $$

The computation of

$$ u\left(\tilde{x}\right) $$ and $$ v\left(\tilde{x}\right) $$

required using the following equations:

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

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

The computation of

$$ u_{x}\left(\tilde{x}\right) $$ and $$ u_{y}\left(\tilde{x}\right) $$ from $$ u\left(\tilde{x}\right) $$ and $$ v\left(\tilde{x}\right) $$ is shown below.

$$ \begin{Bmatrix} u_{x}\left(\tilde{x}\right)\\u_{y}\left(\tilde{x}\right) \end{Bmatrix} = \mathbf{R}^{T}\begin{Bmatrix} u\left(\tilde{x}\right)\\v\left(\tilde{x}\right) \end{Bmatrix} $$

$$ \begin{Bmatrix} u\left(\tilde{x}\right)\\v\left(\tilde{x}\right) \end{Bmatrix} = \begin{bmatrix} N_{1} & 0 & 0 & N_{4} & 0 & 0 \\ 0 & N_{2} & N_{3} & 0 & N_{5} & N_{6} \end{bmatrix}\begin{Bmatrix} \tilde{d}_{1}^{(e)} \\ \tilde{d}_{2}^{(e)} \\ \tilde{d}_{3}^{(e)} \\ \tilde{d}_{4}^{(e)} \\ \tilde{d}_{5}^{(e)} \\ \tilde{d}_{6}^{(e)} \end{Bmatrix} $$

where

$$ \begin{bmatrix} N_{1} & 0 & 0 & N_{4} & 0 & 0 \\ 0 & N_{2} & N_{3} & 0 & N_{5} & N_{6} \end{bmatrix} $$

is represented by

$$ \mathbf{N}\left(\tilde{x}\right) $$ and

$$ \begin{Bmatrix} \tilde{d}_{1}^{(e)} \\ \tilde{d}_{2}^{(e)} \\ \tilde{d}_{3}^{(e)} \\ \tilde{d}_{4}^{(e)} \\ \tilde{d}_{5}^{(e)} \\ \tilde{d}_{6}^{(e)} \end{Bmatrix} $$

is denoted

$$ \mathbf{\tilde{d}}^{(e)} $$

$$ \begin{Bmatrix} u_{x}\left(\tilde{x}\right)\\u_{y}\left(\tilde{x}\right) \end{Bmatrix} = \mathbf{R}^{T}\mathbf{N}\left(\tilde{x}\right)\mathbf{\tilde{T}}^{(e)}\mathbf{d}^{(e)} $$

Monday 12/8
$$ \tilde{k}_{23} = \frac{6EI}{L^2} = \int_{0}^{L}{\frac{d^2N_2}{dx^2}(EI)\frac{d^2N_3}{dx^2}dx} $$

In general,

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

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

Elastodynamics (trusses, 2-D frames, 3-D elasticity)
Analyzing the Modal problem using the discrete Principle of Virtual Work is as follows:

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

Above, 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 is labeled as Equation (1).

These are the complete ordinary differential equations (ODE's, 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:

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

The above is labeled as Equation (2).

Assume:

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

Where the phi matrix is not time dependent.

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 is:

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

for i = 1 through n

Now we perform step 2:

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 known as the Kronecker delta.

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

Now, applying this to Equations (1) and (2) 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,

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

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

Equation (1) can be written as:

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

The first term is equal to the second derivative of the d matrix, while the second term is equal to the d matrix itself. Also,

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

The first term in this equation is equal to the Kronecker delta, while the second term 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

Wikiversity vs. E-learning
There are many pros and cons when talking about using Wikiversity versus using E-learning for keeping up a class website.

First, I will discuss the pros and cons of using E-learning. This website it available only to students enrolled in certain classes. The layout and organization of the site is highly efficient and easy to learn, with all sections of a class website on the left hand side of the webpage, in neat tabs. The information posted to one's e-learning page is safe from inaccuracies, since only the person in charge of the account can alter the page in any way. Also, the e-learning system is frequently slow and sometimes shut down for maintenance. These are just some of the traits of the ELS website.

Now, I will evaluate the Wikiversity website. At first, the entering of lecture notes was a tiring and tedious task. However, after a few weeks, submitting my notes for the day became a simple, quick assignment. Images and videos could be entered in the middle of a lecture, creating a fluid easy to read article. These articles became a great way to study for exams, since the notes were accurate and easy to follow. A table of contents allows the user to quickly skip to a section of the notes, allowing for easy reviewing. Also, Wikiversity allows strangers to access your work, and correct or add their own thoughts, alsoallowing groups of people to work together on articles, each entering their own opinion or facts.

Overall, I enjoy using both Wikiversity and E-learning, however, for the normal class, where lectures and all other downloadable content is posted by the teacher, I prefer to use the E-learning system.

title: User:Eml4500.f08.delta 6.lopez/HW7


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User:Eml4500.f08.ateam.rivero/w7

Meeting 38: November 24,2008
Motivation: deformed shape of truss, interpertation of transverse displacements $$ v(\tilde x) $$ element.

Principle of Virtual Work: Beams $$ \int\limits_{0}^{L} w(\tilde x) \left \{ \frac{-\partial^2 }{\partial x^2}\left [(EI) \frac{\partial^2 v}{\partial x^2}\right ] + f_{t} - m \ddot v \right \} dx = 0 \qquad \qquad (1) \qquad$$ for all possible $$ w(\tilde x) $$

Integration by Parts of 1st Term: $$ \alpha = \int\limits_{0}^{L} w(\tilde x) \frac{-\partial^2 }{\partial x^2}\left [(EI) \frac{\partial^2 v}{\partial x^2}\right ] dx = 0 \qquad \qquad  \qquad$$ $$ \frac{\partial}{\partial x} \left \{\frac{\partial}{\partial x} \left [(EI) \frac{\partial^2 v}{\partial x^2}\right ] \right \} $$ $$ = \left \{w \frac{\partial}{\partial x} \left [(EI) \frac{\partial^2 v}{\partial x^2} \right ] \right \}_{0}^{L} - \int\limits_{0}^{L} \frac{\partial w}{\partial x} \frac{\partial}{\partial x} \left \{(EI) \frac{\partial^2 v}{\partial x^2}\right \} dx  $$ Note: Where $$ \left \{w \frac{\partial}{\partial x} \left [(EI) \frac{\partial^2 v}{\partial x^2} \right ] \right \}_{0}^{L} $$ will be noted as $$ \beta_{1} $$ $$ \beta_{1} - \left [\frac{dw}{dx}(EI)\frac{\partial^2 v}{\partial x^x} \right ]_{0}^{L} + \int\limits_{0}^{L} \frac{d^2w}{dx^2}(EI)\frac{\partial^2v}{\partial x^2}dx$$ Note: Where $$ \left [\frac{dw}{dx}(EI)\frac{\partial^2 v}{\partial x^x} \right ]_{0}^{L} $$ will be noted as $$ \beta_{2} $$ and $$ \int\limits_{0}^{L} \frac{d^2w}{dx^2}(EI)\frac{\partial^2v}{\partial x^2}dx $$ will be noted as $$ \gamma $$ Note: Also notice the symmetry in the expression above Therefore Equation (1) becomes: $$ -\beta_{1} + \beta_{2} - \gamma + \int\limits_{0}^{L} wf_{t}dx-\int\limits_{0}^{L}wm \ddot v dx $$, for all possible $$ w(x) $$

Now let us focus on the stiffness term $$ \gamma $$ for now to derive the beam stiffness matrix and to identify the beam shape functions.

$$ v(\tilde x) = N_{2}(\tilde x)\tilde d_{2} + N_{3}(\tilde x)\tilde d_{3} + N_{5}(\tilde x)\tilde d_{5} + N_{6}(\tilde x)\tilde d_{6} $$

Recall: $$ u(\tilde x) = N_{1}(\tilde x)\tilde d_{1} + N_{4}(\tilde x)\tilde d_{4} $$

Shape Functions: $$ N_{2}(\tilde x) = 1 - \frac{3\tilde x^3}{L^2} + \frac{2\tilde x^3}{L^3} => \tilde d_{2} $$ $$ N_{3}(\tilde x) = \tilde x - \frac{2\tilde x^2}{L} + \frac{\tilde x^3}{L^2} => \tilde d_{3} $$ $$ N_{5}(\tilde x) = \frac{3\tilde x^2}{L^2} - \frac{2\tilde x^3}{L^3} => \tilde d_{5} $$ $$ N_{6}(\tilde x) = \frac{-\tilde x^2}{L} + \frac{\tilde x^3}{L^2} => \tilde d_{6} $$

Meeting 41: December 8, 2008
$$ \tilde k_{23} = \frac {6EI}{L^2} = \int\limits_{0}^{L} \frac{d^2N_{2}}{dx^2}(EI) \frac{d^2N_{3}}{dx^2}dx$$

In general, $$ \tilde k_{ij} = \int\limits_{0}^{L} \frac{d^2N_{i}}{dx^2}(EI) \frac{d^2N_{j}}{dx^2}dx $$, where $$ i,j = 2,3,5,6 $$

Structural Dynamics
Elastodynamics (truss, frames, 2D & 3D elasticity)

Discrete Principle of Virtual Work:

Note: Bold Face implies that boundry conditins are already applied; Therefore the expression below is the reduced version.

$$ \mathbf w \cdot \left [\mathbf M \mathbf \ddot d + \mathbf K \mathbf d - \mathbf F \right ] = 0 $$, for all  $$  \mathbf w $$ Equation (1): $$ \mathbf M \mathbf \ddot d + \mathbf K \mathbf d = \mathbf F_{t} $$ $$ \mathbf d(0) = \mathbf d_{0} $$ $$ \mathbf \dot d_{0} = \mathbf v_{0} $$ Note: Complete ordinary differential equations and initial conditions governing the elastodynamics of the discretized problem.

Solving Equation (1):

1. Consider unforced vibrations problem

$$ \mathbf M \underline \ddot v + \mathbf K \underline v = \underline 0 $$

Assume:

$$ \underline v(t)_{nx1} = (sin(wt)) \phi_{nx1(not line dependent)} $$

Then:

$$ \underline \ddot v = w^2sin(wt) \underline \phi $$

Then:

$$ -w^2sin(wt) \mathbf M \phi + sin(wt) \mathbf K \phi = \underline 0 $$

Then giving us the generalized eigen value problem below:

$$ \mathbf K \phi = w^2 \mathbf M \phi $$

Which we can compare to the general form below:

$$ \underline A \underline x = \lambda \underline B \underline x $$, where $$ \lambda $$ is the eigen value

Where in your standard eigen value problem $$ \underline A \underline x = \lambda \underline B \underline x $$, where $$ ( \underline B = \underline I ) $$, where $$ \underline I $$ is your identity matrix shown below:

$$ \underline I = \begin{bmatrix} 1 &   &   & 0 \\   &. &  &   \\   &   & . &   \\ 0 &   &   & 1 \end{bmatrix} $$

Where:

$$ \lambda = w^2 $$  eigen value

and,

$$ (\lambda_{i}, \underline \phi_{i}) $$  eigen pairs, where i = 1,. . ., n

where,

Mode $$ i => animation => \underline V_{i}(t) - (sin(w_{i}t))d_{i} $$

2. Model Superposition Method

Orthogonal problem of eigen pairs:

$$ \phi_{i}^T \mathbf M \phi_{j} = \delta_{ij} = 1 $$ if $$ i=j $$

or,

$$ \phi_{i}^T \mathbf M \phi_{j} = \delta_{ij} = 0 $$ if $$ i $$ does not equal $$ j $$

Mass orthogonal of eigenvalue.

$$ \mathbf M \phi_{j} = \lambda_{j}^{-1} \mathbf K \phi_{j} $$

$$ \phi_{i}^T \mathbf M \phi_{j} = \lambda_{j}^{-1} \phi_{j}^{-1} \mathbf K \phi_{j} $$

Giving us:

=> $$ \phi_{i}^T \mathbf K \phi_{j} = \lambda_{j} \delta_{ij} $$

$$ \mathbf d(t) = \sum_{i=1}^{n} \xi_{i}(t) \phi_{i} $$

$$ \mathbf M (\sum_{j} \xi_{j} \underline \phi_{j}) + \mathbf K(\sum_{j} \xi_{j} \underline \phi_{j}) = \underline F $$

Where,

$$ \sum_{j} \xi_{j} \underline \phi_{j} = \mathbf \ddot d $$

$$ \sum_{j} \xi_{j} \underline \phi_{j} = \mathbf \dot d $$

$$ \sum_{j} \xi_{j}(\phi_{i}^{T} \mathbf M \phi_{j}) + \sum_{j} \xi_{j} \phi_{i}^{T} \mathbf K \phi_{j} = \phi_{i}^{T}F $$

Finally Giving:

$$ => \ddot \xi_{i} + \lambda_{i} \xi_{i} = \phi_{i}^T \underline F $$, where $$ i=1,....,n $$

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User:Eml4500.f08.jamama.justin/HW7

Meeting 41

$$\tilde{k}_{23}=\frac{6EI}{L^{2}}=\int_{0}^{l}{\frac{d^{2N_{2}}}{dx^{2}}(EI)\frac{d^{2N_{3}}}{dx^{2}}dx}$$

p. 36-2       P 38-2          HW7!!!! also on P 247 (second derivative of shape function

In general,  $$\tilde{k}_{ij}=\frac{6EI}{L^{2}}=\int_{0}^{l}{\frac{d^{2N_{i}}}{dx^{2}}(EI)\frac{d^{2N_{j}}}{dx^{2}}dx}$$ i,j=2,3,5,6

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

P. 31-1: Model pb

Discrete PVW: $$\mathbf{\bar{w}}\bullet [\bar{M}\bar{\ddot{d}}+\bar{K}\bar{d}-\bar{F}]=0$$ for all w

(boundary conditions already applied) EQ 1: $$\Rightarrow \bar{M}\bar{\ddot{d}}+\bar{K}\bar{d}=\bar{F}(t)$$ $$\bar{d}(0)=\bar{d_{0}}$$ $$\bar{\dot{d}}(0)=\bar{V_{0}}$$

Complete ordinary diff eqs (ODE's) (second order in time) + initial conditions governing the elastodynamic of discretized cont pb (MDOF)

Solving equation 1

1) Consider unforced vibration problem:

$$\bar{M}\ddot{v}+\bar{K}v=0$$

nxn nx1 nxn nx1  nx1 (unforced)

Assume $$v(t)=(sin\omega t)\phi $$

nx1 nx1

$$\phi $$ is not time dependent

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

Generalized igenvalue problem: In general form

$$Ax=\lambda Bx$$ where lambda is the eigen value

Standard eval problem: $$Ax=\lambda x$$

(B=I,identity matrix) $$\begin{bmatrix} 1 & & 0\\  &  ..& \\  0&  & 1 \end{bmatrix}$$

$$\lambda =\omega ^{2} $$ Eigen Value

$$(\lambda _{i},\phi _{i})$$ Eigen pairs

i=1,....,n

mode $$i\Rightarrow V_{i}(t)=(sin\omega _{i}t)\phi _{i}$$ where i=1,...,n (animation)

2) Model Superposition method: Orthogonal prop of eigenpairs:

$$\phi _{i}^{T}\bar{M}\phi _{j}=\delta _{ij}=\begin{cases} 1 & \text{ if } i=j \\ 0 & \text{ if } i\neq j \end{cases}$$

1xn nxn  nx1 (Kronecker delta)

Mass orthog. of Eigen Vector

Eq(2) p 41-2: Eq(1) p 41-3:

$$\bar{M}\phi _{j}=\lambda _{j}\bar{K}\phi _{j}$$ $$ \phi _{i}^{T}\bar{M}\phi _{j}=\lambda _{j}\phi _{i}^{T}\bar{K}\phi _{j}$$ where $$ \phi _{i}^{T}\bar{M}\phi _{j}= \delta _{ij}$$ $$\Rightarrow \phi _{i}^{T}\bar{K}\phi _{j}=\frac{d}{\lambda _{j}}\delta _{ij}$$

Eq 1 (P 41-2): $$\bar{d}(t)=\sum_{i=1}^{n}{\xi _{i}(t)\phi _{i}}$$

nx1 1x1   nx1

$$\bar{M}\sum_{j}^{}{\ddot{\xi}_{j}\phi _{j} }+\bar{K}\sum_{j}^{}{\ddot{\xi}_{j}\phi _{j} }=F$$

$$\ddot{\bar{d}}$$      $$\bar{d}$$

$$\sum_{j}^{}{\ddot{\xi }(\phi _{i}^{T}\bar{M}\phi _{j})}+\sum_{j}^{}{\xi _{j}(\phi _{i}^{T}\bar{K}\phi _{j})}=\phi _{i}^{T}F$$

$$\delta \ddot{y}$$ $$x_{i}\delta _{ij}$$

$$\Rightarrow \ddot{\xi} _{i}+\lambda _{i}\xi _{i}=\phi _{i}^{T}F $$ where i=1,....,n

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User:Eml4500.f08.delta 6.guzman/Lecture Notes 11/19

It is important to recognize what elements constitute the local stiffness matrix for each element in the system, the following diagrams show the local element degrees of freedom of element 1 and element 2

Where $$ \tilde{k}^{(e)}_{6x6}\tilde{d}^{(e)}_{6x6}=\tilde{f}^{(e)}_{6x6} \! $$ and $$\tilde{d}^{(e)}=\begin{Bmatrix} \tilde{d}_{1}^{(e)}\\ \cdot \cdot \cdot \\ \tilde{d}_{6}^{(e)} \end{Bmatrix} \! $$ and $$ \tilde{f}^{(e)}=\begin{Bmatrix} \tilde{f}_{1}^{(e)}\\ \cdot \cdot \cdot \\ \tilde{f}_{6}^{(e)}

\end{Bmatrix} \! $$

Note: $$ \tilde{f}^{(e)}_{3}=f^{(e)}_{3} \! $$ and $$ \tilde{f}^{(e)}_{6}=f^{(e)}_{6} \! $$ since these are the moments about the $$ \tilde{z} \! $$ axis.

By using the elements of the diagrams above for element 1 and element 2, the local stiffness matrix $$ \tilde{k}\! $$ would be: $$ \begin{bmatrix}\frac{EA}{L} & 0 & 0 & \frac{-EA}{L} & 0 & 0 \\ 0 & \frac{12EI}{L^{3}} & \frac{6EI}{L^{2}} & 0 & \frac{12EI}{L^{3}} & \frac{6EI}{L^{2}} \\ 0 & \frac{6EI}{L^{2}} & \frac{4EI}{L} & 0 & \frac{-6EI}{L^{2}} & \frac{2EI}{L} \\ \frac{-EA}{L} & 0 & 0 & \frac{EA}{L} & 0 & 0 \\ 0 & \frac{-12EI}{L^{3}} & \frac{-6EI}{L^{2}} & 0 & \frac{12EI}{L^{3}} & \frac{-6EI}{L^{2}} \\ 0 & \frac{6EI}{L^{2}} & \frac{2EI}{L} & 0 & \frac{-6EI}{L^{2}} & \frac{4EI}{L} \end{bmatrix} $$

Note: All the matrix coefficients would not have the same dimensions but when they are multiplied with rotations or displacements depending on location they will end up with the right units.

Dimensional Analysis
$$ \left [ \tilde {d}_1 \right ] = L = \left [ \tilde {d}_i \right ] \qquad i=1, 2, 4, 5$$

where: $$ \left [ \tilde {d}_1 \right ] $$ is a dimension of length and $$ \left [ \tilde {d}_i \right ] $$ is displacement.

Note: For the dimensional analysis the $$ [ ] \! $$ symbolizes "dimension of" instead of the $$ [ ] \! $$ commonly used for matrices.

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

where: $$ 1 $$ is dimensionless and $$ \left [ \tilde {d}_6 \right ] $$ is the rotational displacement

From the graph it can be observed that $$ \bar{AB} = R * \theta $$ where $$ \theta $$ is measured in radians and $$\bar{AB}$$ is the arc length from point A to point B. Another way to express $$ \theta $$ is $$\frac{\bar{AB}}{R}$$ where both have units of length making $$ \theta $$  dimensionless: $$ \left [ \theta \right ] = \frac {[\bar{AB}]}{[R]} = \frac {L}{L} = 1 $$

Applying the same non-dimensional analysis principle: $$ \sigma = \Epsilon \varepsilon \rightarrow \left [ \sigma \right ] = \left [ \Epsilon \right ] \left [ \varepsilon \right ] = 1 $$


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User:EAS4200C.F08.WIKI.A/Homework7

=Shear Flow-Open Section=

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.

Superposition revisited
3) Back to Superposition :

$$q_{ij} = q + q_i$$ where $$q_{ij}$$ is true shear flow $$q$$ is the closed cell constant shear flow $$q_i$$ is the open cell piecewise constant shear flow

M.1)W/o stringers

$$R^z=\sum_{i=1}^{n cells}{R^z_i}$$ where $$n=$$ the number of cells

$$R^z_1=0$$

$$R^z=0\;$$

M.2)With Stringers

$$R^z=R^{z1}+R^{z2}\;$$ where $$R^z \;$$ is the resultant of P

$$R^{z1}\;$$ is the resultant of P1 and $$R^{z2}\;$$ is the resultant of P2.

Recalling $$R^z=\sum_{i=1}^{n cells}{R^z_i}=0\;$$ provides us with the formula:

$$R^z=R^{z2}=V_z\neq 0\;$$

Equilibrium of Stringer Three
We will now look at the equilibrium of stringer three. The two and three dimensional drawings of this stringer and its spar webs are shown below.

We will first take the sum of the forces in the x-direction.

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

The first term of the integral is the taylor series expansion $$\displaystyle \frac{d\sigma_{xx}}{dx}dx $$ plus the higher order terms. the sumation of the force in the x-direction can therefore be writen as the following;

$$\displaystyle \tilde{q}_{31} - \tilde{q}_{23} +\tilde{q}_{43}+q^{(3)}= 0$$

The term $$\displaystyle q^{(3)}$$ is the contribution to the shear flow by stringer three and equals the following.

$$\displaystyle q^{(3)} = - \int_{A_3}^{}{\frac{d\sigma_{xx}}{dx}dA_3}$$

If you recall that the shear flow in the y- and z-directions are $$\displaystyle V_y = \frac{dM_z}{dx}, V_z = \frac{dM_y}{dx}$$ respectively, the contribution of the shear flow by stringer three can be rewriten as the following;

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

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

Equilibrium of Stringer 2
The process is the same as before. The results of the equilibrium of stringer two are shown below.

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

$$\displaystyle \tilde{q}_{12} = \tilde{q}_{24} = 0$$ because of cuts

$$\displaystyle Q_z^{(2)} = y_2A_2 $$ and $$\displaystyle Q_z^{(2)} = z_2A_2$$ $$(y_2,z_2)$$ are coordinates of stringer 2

$$\displaystyle q^{(2)}$$ is computed as before.

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

$$\displaystyle \tilde{q}_{24} = \tilde{q}_{41} = 0$$ because of cuts

$$\displaystyle Q_z^{(4)} = y_4A_4 $$ and $$\displaystyle Q_z^{(4)} = y_4A_4$$ $$(y_2,z_2)$$ are coordinates of stringer 4

Equilibrium of Stringer 3 Revisited
Now we can deduce the equation of $$\displaystyle \tilde{q}_{31}$$ because we now know the components of the equation.

$$\displaystyle \tilde{q}_{31} - \tilde{q}_{23} + \tilde{q}_{43}+q^{(3)}= 0$$

$$\displaystyle \tilde{q}_{31} = \tilde{q}_{23} - \tilde{q}_{43} + q^{(3)} = q^{(2)} - q^{(4)} + q^{(3)}$$

Total Shear in the Potato
With the information provided above we can no go back and use superposition in and add all the components to find the shear flow in the original potato.

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

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

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

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

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

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

We now have three unknowns $$\displaystyle q_1,q_2,q_3$$ so we need three equations. the first equation is the moment equation. You need to take the moment of $$\displaystyle V_y, V_z $$ and $$\left\{q_{12},...........q_{41} \right\}$$ about any point. this point is usually where the lines of action of $$\displaystyle V_y, V_z $$ intersect. The second and third equations are the compatibility equations which are shown below.

$$\displaystyle \theta_1 = \theta_2$$

$$\displaystyle \theta_2 = \theta_3$$

Another Method
Solving P2: For each cell

-Follow path $$S_i$$

-Find the equilibrium of each stringer on this path: There are two ways of solving

1) Complete method FBD as seen before

2) Consequence of the first method:

$$\tilde{q}_{j6}=\tilde{q}_{2j}-\tilde{q}_{j5}-\tilde{q}_{j8}+\tilde{q}^{j}$$

In this example, $$\tilde{q}_{6j}$$ is now going in the opposite direction as previous but the answer will remain the same as long as the conventions stay consistent.

$$-\tilde{q}_{6j}=\tilde{q}_{2j}-\tilde{q}_{j5}-\tilde{q}_{j8}+\tilde{q}^{j}$$

Note: What if we cut cell walls such that one stringer was isolated?

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

This equation says that $$\tilde{q}^{3}= 0$$ which is not true. Therefore you cannot isolate the stinger in this way because it results in an answer that is not valid.

Trying to make different cuts to isolate stringer 2 will result in the same false answer.

 $$\tilde{q}_{31}=\tilde{q}_{12}-\tilde{q}_{24}+\tilde{q}^{2}$$

$$\tilde{q}^{2}= 0$$

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,

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

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

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

 $$\displaystyle Q_y = \int_A zdA $$

 $$\displaystyle Q_z = \int_A ydA $$

= MATLAB Airfoil Coding =

Single Cell Airfoil
EDU>> NewAirfoil(2415,.5,[0 0 0])

Single Cell

q_0 =

2.9437e+004

q_hf =

2.9437e+004

q_fb =

-2.2512e+005

q_be =

3.2195e+004

q_eh =

2.8555e+005

Note: All values in N/m

Multi Cell Airfoil
EDU>> NewAirfoil(2415,.5,[0 0 0])

Multi Cell

q_1 =

3.2721e+004

q_2 =

2.6482e+004

q_3 =

3.7939e+004

q_hf =

3.7939e+004

q_fb =

-2.2807e+005

q_be =

3.5480e+004

q_eh =

2.8260e+005

q_w1 =

-6.2391e+003

q_w2 =

1.1456e+004

Note: All values in N/m

Also, it can be seen here that the shear flow values in the skin panels are similar between the single and multi cell cases. This verifies the validity of the two cases.

= Buckling Analysis =

$$ \displaystyle K_{2\times 2} = \left[ \begin{array}{ll} \frac{\lambda (1 + \vartheta^2)^2}{\vartheta^2} &	 \frac{4}{9} \\	 \frac{4}{9} &	 \frac{16 \lambda (1 + \vartheta^2)^2}{\vartheta^2} \end{array} \right] $$

 $$ \displaystyle \det K_{2\times 2} = \frac{16\lambda ^2(1+\vartheta ^2)^4}{\vartheta ^4}-\frac{16}{81} = 0 $$

 $$ \displaystyle \lambda ^2 = \frac{\vartheta ^4}{81(1+\vartheta ^2)^4} $$

 $$ \displaystyle \lambda = \pm \frac{1}{9}\frac{\vartheta ^2}{(1+\vartheta ^2)^2}$$

=Mediawiki vs. E-learning=

During this course, we used wikimedia to turn in our homework. Our homework consisted of writing up the class notes and doing some matlab assignments. We were asked to form teams to work on this Wikimedia homework. The format of these homework assignments forced us to attend class once every two weeks in order to get the notes for our assigned day. When doing this type of work, Wikimedia is a much more convenient and usable program to upload and format than E-learning. E-learning was not designed to do what was require for this course. Wikimedia's ability to save all archived versions of pages as well as the ability to easily interface with equation editors made it user friendly for students. It also allowed for easier grade formatting and uploading. I see E-learning as an older, less usable and less versatile version of Wikimedia.

We do believe that Wikimedia is a more versatile program than E-learning and was a much better decision for us to accomplish our homework as assigned. However, We do not believe that Wikimedia allowed me to learn Aerospace structures effectively. We would prefer to have traditional homework problems that would allow us to practice the method rather than regurgitating theory that we may or may not understand. Also, assigning Matlab to try to patch this significant problem only compounded it because most of us have never used Matlab, adding another program we needed to learn to succeed in this class, without learning any actual Aerospace Structures. We believe assigning homework in this way has hindered our learning. An argument could be made that we can always look back at these pages to find information about structures. We would respond to this argument by saying that none of our homework was graded carefully enough to catch errors or that it was written with any understanding or authority of the subject it was discussing. We would much rather turn to a validated source like a textbook than look on Wikiveristy. We feel that this style hindered our learning experience. Because we have learned near nothing in the class, we would venture to say that the homework's effectiveness on our learning experience was an utter failure.

Eas4200c.f08.WIKI.E 21:10, 9 December 2008 (UTC)

=Individual Contributions=

Christopher Pergola EAS4200C.F08.WIKI.A 17:53, 9 December 2008 (UTC)

Braden Barnes Eas4200c.f08.wiki.b 22:59, 8 December 2008 (UTC)

Eric Viale Eas4200c.f08.wiki.c 01:11, 9 December 2008 (UTC)

Michael Rolle Eas4200c.f08.wiki.d 05:26, 9 December 2008 (UTC)

Chris Fontana Eas4200c.f08.WIKI.E 20:25, 9 December 2008 (UTC)

Jeff Huenink Eas4200c.f08.wiki.f 21:32, 9 December 2008 (UTC)

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User:Eml4500.f08.jamama.justin/HW7b

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

6x1 6x6 6x1 (known after solving finite element system)

Compute $$u(\tilde{x}),v(\tilde{x})$$

$$u(\tilde{x})=u(\tilde{x})\vec{\tilde{i}}+v(\tilde{x})\vec{\tilde{j}}$$

$$=U_{x}(\tilde{x})\vec{i}+U_{y}(\tilde{x})\vec{j}$$

Compute $$U(\tilde{x}), V(\tilde{x})$$ using eq.(1) and (2) from page 38-3.

*Compute $$U_{x}(\tilde{x}), V_{y}(\tilde{x})$$ from $$U(\tilde{x}), V(\tilde{x})$$

$$\begin{Bmatrix} U_{x}(\tilde{x})\\ U_{y}(\tilde{x}) \end{Bmatrix}=R^{T}\begin{Bmatrix} U(\tilde{x})\\ V(\tilde{x}) \end{Bmatrix}$$ Where R is from P 37-3 $$\begin{Bmatrix} U(\tilde{x})\\ V(\tilde{x}) \end{Bmatrix}=\begin{bmatrix} 1 & 2 & 3 & 4 & 5 & 6 \\   N_{1} & 0 & 0 & N_{4} & 0 & 0\\ 0& N_{2} & N_{3} & 0 & N_{5} & N_{6} \end{bmatrix}\begin{Bmatrix} \tilde{d}_{1}^{(e)}\\ .\\   .\\    .\\    .\\    \tilde{d}_{6}^{(e)} \end{Bmatrix}$$ $$N(\tilde{x})$$$$\tilde{d}^{(e)}$$ $$\begin{Bmatrix} U_{x}(\tilde{x})\\ U_{y}(\tilde{x}) \end{Bmatrix}=R^{T}N(\tilde{x})\tilde{T}^{(e)}d^{(e)}$$

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User:Eas4200c.f08.aeris.guan/hw7

Matlab HW7: Determining the True Shear Flow in Different-Celled NACA Airfoils
The problem statement in this homework was to determine the true shear flows of two different cases of the NACA airfoil subject to shear forces. Part 1 was to determine these shear flows in a single-cell airfoil and part 2 was to determine these shear flows in a double-cell airfoil. The airfoils in both cases have four stringers: 2 positioned at a quarter of the chord length and 2 positioned at three quarters of the chord length. In each case, the problem was solved using superposition of the shear flows in a case analyzed without stringers and a case with stringers. The equation of moment and the compatibility equation were also used to find the true shear flows. The code in which was developed to solve this problem is shown in the collapsible boxes below. For part 1, the code starts out with finding the values ky, kz, and kyz by using the moment of inertias obtained by previous homework assignments. Then, it goes about solving the P2 problem as posed in lecture. The trailing edge is the portion of the skin that is designated to be cut. Here, Qy and Qz are calculated at each stringer node and the equilibrium equations of the shear flows at each stringer are written. This solves for all the q tilda's in P2. After these shear flows are obtained, it is necessary to find a the shear flow of the two cells (q_11 and q_21) solving for the P1 problem as posed in lecture. The indices in the notation are described as follows: the first subscript is the cell number, and the second subscript is the part of the homework the shear flow corresponds to. To do this, the moment equation is used and the moment was taken about point D. To find the moments contributed by the shear flows found in P2, the code was written to sweep from stringer to stringer that had a non-zero shear flow. At each sweep, an infinitesimal length of the skin was calculated and multiplied by the shear flow through that skin portion to obtain a force. Once that force was determined, the perpendicular distance from point D to the line of action of the force was obtained and then multiplied to the force to get an infinitesimally small moment. All the moments were added together to get the moment produced by each shear flow from P2. Along with the compatibility equation, q_11 and q_21 are obtained. Back to superposition, the shear flows of P1 and P2 are added together to get the true shear flows. For part 2, the code uses the same moments as part 1 because nothing has changed. However, the change is in the moment equation. The moment equation now has three unknowns (q_12 and q_22 and q_32) so it alone is not sufficient in solving for the P1 shear flows. This is where the compatibility equation comes in. We know that the twist angle of the first cell is equal to the twist angle of the second and third cells. The moment and compatibility equations are arranged to that it has coefficients multiplied to q_12 and q_22 and q_32. Then, the equations were arranged in matrix form and solved for in Matlab. Then, superposition was used to find the true shear flow in the three-cell case.

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User:Eml4500.f08.team.foskey.ckf/teamHW7tosubmit

Team "TEAM"  Note:

Edit: See the links for the two frame problem, and the combined frame and truss system for updates.

Jin Yu Guan Eml4500.f08.team.guan 18:24, 10 December 2008 (UTC)

HW7 Wiki Report

Lecture Notes

Nov 17 (meeting 35) - Frame Elements

Nov 19 (meeting 36) - Frame Elements & Dimensional Analysis

Nov 21 (meeting 37) - Dimensional Analysis & The Frame System Transformation Matrix T

Nov 24 (meeting 38) - Principle Of Virtual Work For Beams

Dec 1 (meeting 39) - Principle Axial & Transverse Dislocations Of A Beam (Shape Functions)

Dec 5 (meeting 40) - Derivation & Application Of Shape Functions

Dec 8 (meeting 41) - Derivation Elastodynamics (Combining FEA & Vibrations)

 * Lecture Notes are the summaries of the notes provided by Dr. Vu-Quoc in class

Additional Items

Determining if the electric pylon frame is statically indeterminate

Wikiversity vs E-Learning

Software used for Wiki homework reports

MATLAB Projects

Two bar truss with variable element properties

Two bar frame

Combined frame and truss system (1 frame element + 1 truss element)

Electric pylon as a frame system

Electric pylon frame modes

Electric pylon mode animations

Contributing group members

Cory Foskey Eml4500.f08.team.foskey.ckf 23:33, 8 December 2008 (UTC) Timothy Bengtson Eml4500.f08.team.Bengtson 07:43, 9 December 2008 (UTC) James Dwyer Eml4500.f08.team.dwyer.jd 15:25, 9 December 2008 (UTC) Joseph Figuerrez Eml4500.f08.team.figuerrez 12:16, 9 December 2008 (UTC) Jin Yu Guan Eml4500.f08.team.guan 17:38, 9 December 2008 (UTC)


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User:Eml4500.f08.a-team.melvin/Lecture 38

PVW for beams:  $$ \int_{0}^{b} w(\underbrace{x}_{\tilde{x}}) [-\frac{\delta^2}{\delta x^2} ((EI) \frac{\delta^2 v}{\delta x^2}) +ft-m \ddot{v}] \  dx = 0$$ for all possible w(x)

Integration by parts of 1st term

$$ \alpha = \int_{0}^{L} \underbrace{w(x)}_{g(x)} \frac{\delta^2}{\delta x^2}{(EI) \frac{\delta^2 v}{\delta x^2}} \ dx $$ $$ \underbrace{\frac{\delta}{\delta x} (\underbrace{\frac{\delta}{\delta x}(EI) \frac{\delta^2 v}{\delta x^2}}_{r(x)})}_{r'(x)} $$

$$ \alpha = [\underbrace{w \frac{\delta}{\delta x} {(EI) \frac{\delta^2 v}{\delta x^2}}}_{\beta_{1}}]^{2}_{0} - \int_{0}^{L} \underbrace{\frac{dw}{dx}}_{g'(x)} \underbrace{\frac{\delta}{\delta x} {(EI) \frac{\delta^2 v}{\delta x^2}}}_{r(x)} \ dx$$

$$ \alpha = \beta_{1} - [\underbrace{w \frac{\delta}{\delta x} {(EI) \frac{\delta^2 v}{\delta x^2}}}_{\beta_{2}}]^{2}_{0} + \underbrace{\int_{0}^{L} \frac{\delta^2 w}{\delta x^2} (EI) \frac{\delta^2 v}{\delta x^2} \ dx}_{\gamma}$$

Note : symmetric

Eq. (1) becomes: $$ - \beta_{1} + \beta_{2} - \gamma + \int_{0}^{L} wft \ dx - \int_{0}^{L} wm \ddot{v} \ dx = 0 $$ for all possible w(x)

Let's focus on the stiffness term $$ \gamma $$ for now to derive the beam stiffness matrix and to identify the beam shape functions.

$$ v( \tilde{x}) = N_{2} (\tilde{x})\tilde{d}_{2} + N_{3} (\tilde{x})\tilde{d}_{3} + N_{5} (\tilde{x})\tilde{d}_{5} + N_{6} (\tilde{x})\tilde{d}_{6} $$

Recall : $$ u(\tilde{x}) = N_{1} (\tilde{x})\tilde{d}_{1} + N_{4} (\tilde{x})\tilde{d}_{4} $$

p.246:

$$ N_{2} (\tilde{x}) = 1 - \frac{3 \tilde{x}^2}{L^2} + \frac{2 \tilde{x}^3}{L^3} \ \tilde{d}_{2} $$

$$ N_{3} (\tilde{x}) = \tilde{x} - \frac{2 \tilde{x}^2}{L} + \frac{\tilde{x}^3}{L^3} \ \tilde{d}_{3} $$

$$ N_{5} (\tilde{x}) = \frac{3 \tilde{X}^2}{L^2} - \frac{2 \tilde{x}^3}{L^3} \ \tilde{d}_{5} $$

$$ N_{6} (\tilde{x}) = - \frac{\tilde{x}^2}{L} + \frac{\tilde{x}^3}{L^2} $$


title: File:FIGURE2MATLAB.jpg


text:
User:KTucker/My Sandbox/Samples/Activities

This is a sample page for use while thinking about and developing templates for learning Activity pages. Ultimately, it will probably be most useful to have an Activity list with links to activity summaries which in turn link to the detail of the activities (activity sheets, facilitator's guides, assessment guides, etc.).

Latest sample page. The page has a link to add activity (summary) which adds a new section to the page with Template:Activity_preload preloaded which inserts a Template:Activity summary which in turn creates a link to the activity title in a sub-page called / of the page it is on which preloads Template:Activity Detail.

Aim: Add Activity

Adds a new section on an Activity page with Template:Activity Summary preloaded.

Template:Activity Summary includes a link to /Activity Detail/ - a sub-page named after the 'activity title' in the Activity Summary.

Template:Activity Detail consists of a template which includes various fields and sub-pages:

See the ideas below.

Template:Activity Summary Preload

Possibly a subst with // linking to the detail.

Relevant templates in preparation:

Template:Activity Detail which uses Template:Activity box themes combo, Template:Activity type icon, Template:Box Theme Attribute, ...

Result

 * Sample 1 (code as above)
 * Sample 2 (same template, a different theme)

SSE Activities Page

 * Sample Page (uses the earlier Three Box Sections).

Result
Illustrating several themes and variable left column width:

Raw Wiki Page
Alternatively, the page could be blank and the users may do what they want.

Activity 1

 * Act 1a
 * Act 1b
 * etc.

Activity 2

 * Act 2a
 * Act 2b
 * etc.

Activity 3

 * Act 3a
 * Act 3b
 * etc.


text:
User:KTucker/My Sandbox/Samples/Activities/template

{{Three Box Section L
 * left theme = 6
 * right top theme = 2
 * right bottom theme = 3
 * left image  = Isimple system icons web find.png
 * right top image  = Blue-question-mark.jpg
 * right bottom image  = Victoria 3 RR LE 080.png
 * left title  = ==Activity Set Title (web search) ==
 * right top title  = ==Hints==
 * right bottom title  = ==Solutions==
 * left subtitle = Left subtitle
 * right top subtitle = and tools, etc.
 * right bottom subtitle = find/write ...
 * left text  =

Activity 1

 * 1) Do this ...
 * 2) Do that ...
 * 3) Write up ...
 * 4) Publish ...

Activity 2

 * 1) Do this ...
 * 2) Do that ...
 * 3) Write up ...
 * 4) Publish ...
 * right top text  =
 * Hint 1 - e.g. recall ...
 * Resources
 * Community Help

 * right bottom text  =
 * Sample solutions Activity 1
 * Sample solutions Activity 2
 * Where to publish
 * etc.

}} {{Three Box Section L
 * left theme = 6
 * right top theme = 2
 * right bottom theme = 3
 * left image  = Nuvola apps bookcase.svg
 * right top image  = Global thinking.svg
 * right bottom image  = Nuvola apps bookcase.svg
 * left title  = ==Activity Set Title (reading and discussion) ==
 * right top title  = ==Hints==
 * right bottom title  = ==Further reading==
 * left subtitle = and related discussions
 * right top subtitle = and tools, etc.
 * right bottom subtitle = find/write ...
 * left text  =

Activity 1

 * 1) Do this ...
 * 2) Do that ...
 * 3) Write up ...
 * 4) Publish ...

Activity 2

 * 1) Do this ...
 * 2) Do that ...
 * 3) Write up ...
 * 4) Publish ...
 * right top text  =
 * Hint 1 - e.g. recall ...
 * Resources
 * Community Help

 * right bottom text  =
 * Sample solutions Activity 1
 * Sample solutions Activity 2
 * Where to publish
 * etc.

}} {{Three Box Section L
 * left theme = 6
 * right top theme = 2
 * right bottom theme = 3
 * left image  = Crystal 128 desktop.png
 * right top image  = Global thinking.svg
 * right bottom image  = Nuvola apps edu miscellaneous.png
 * left title  = ==Activity Set Title (writing activity ) ==
 * right top title  = ==Hints==
 * right bottom title  = ==Further reading==
 * left subtitle = and related discussions
 * right top subtitle = and tools, etc.
 * right bottom subtitle = find/write ...
 * left text  =

Activity 1

 * 1) Do this ...
 * 2) Do that ...
 * 3) Write up ...
 * 4) Publish ...

Activity 2

 * 1) Do this ...
 * 2) Do that ...
 * 3) Write up ...
 * 4) Publish ...
 * right top text  =
 * Hint 1 - e.g. recall ...
 * Resources
 * Community Help

 * right bottom text  =
 * Sample solutions Activity 1
 * Sample solutions Activity 2
 * Where to publish
 * etc.

}} {{Three Box Section L
 * left theme = 6
 * right top theme = 2
 * right bottom theme = 3
 * left image  = Battle for Wesnoth Map Editor.png
 * right top image  = Global thinking.svg
 * right bottom image  = Nuvola apps bookcase.svg
 * left title  = ==Activity Set Title (experiential learning) ==
 * right top title  = ==Hints==
 * right bottom title  = ==Further reading==
 * left subtitle = and related discussions
 * right top subtitle = and tools, etc.
 * right bottom subtitle = find/write ...
 * left text  =

Activity 1

 * 1) Do this ...
 * 2) Do that ...
 * 3) Write up ...
 * 4) Publish ...

Activity 2

 * 1) Do this ...
 * 2) Do that ...
 * 3) Write up ...
 * 4) Publish ...
 * right top text  =
 * Hint 1 - e.g. recall ...
 * Resources
 * Community Help

 * right bottom text  =
 * Sample solutions Activity 1
 * Sample solutions Activity 2
 * Where to publish
 * etc.

}} {{Three Box Section L
 * left theme = 6
 * right top theme = 2
 * right bottom theme = 3
 * left image  = Nuvola apps iconthemes.png
 * right top image  = Global thinking.svg
 * right bottom image  = Victoria 3 RR LE 080.png
 * left title  = ==Activity Set Title (assessment activities) ==
 * right top title  = ==Hints==
 * right bottom title  = ==Model solutions==
 * left subtitle = and comments
 * right top subtitle = and tools, etc.
 * right bottom subtitle = find/write ...
 * left text  =

Activity 1

 * 1) Do this ...
 * 2) Do that ...
 * 3) Write up ...
 * 4) Publish ...

Activity 2

 * 1) Do this ...
 * 2) Do that ...
 * 3) Write up ...
 * 4) Publish ...
 * right top text  =
 * Hint 1 - e.g. recall ...
 * Resources
 * Community Help

 * right bottom text  =
 * Sample solutions Activity 1
 * Sample solutions Activity 2
 * Where to publish
 * etc.

}}


text:
User:KTucker/My Sandbox/Samples/Introduction

''To make an introductory module page with links to sub-pages for Activities, Resources, Community, etc., create the page by inserting this

 *

into a page that will refer to it (or in your sandbox if you wish to work on it before releasing it):

Was: SSE Full Page Intro Now: Module Intro Page

edit it to specify the titles you prefer and save it.

Then use the resulting page. See instructions here on usage and how to make a navigator for the standard set of sub-pages.

Further tests etc.:

 * test of  using:
 * Template:Module Intro Page, and in future (possibly):
 * Template:SSE Module Page Generator Template - a form to generate an intro page.
 * temp (a temporary area for use during template development and application to existing pages)
 * test subst as a subpage - fine here under "User:..."
 * test subst in default name space - now fine.

Technical Questions
...

Solved
editpage=:/
 * It all works fine if the new page is here under User:Ktucker/My Sandbox/... or under Portal: ... but not if the page created is in the default name space?? The 'Edit' links point to " " but the editable content of the box comes from "Template ".
 * How do I get this to work for default name space pages?
 * I think the relevant code to change might be in Portal:Box-header ??
 * check for  in abs FULLPAGENAME and if exists go ahead else modify the generated links ...
 * maybe some form of redirection ...
 * Solved: needed to put in colons to indicate default name space:


text:
User:KTucker/My Sandbox/Samples/Introduction/template

try subst Module Intro Page ...


text: File:FEABBQ 11.17.08 Fig 1.JPG

title: File:FEABBQ 11.17.08 Fig 1.JPG


text: File:Areaforq.svg

title: File:Areaforq.svg


text: File:FEABBQ 11.17.08 Fig 2.JPG

title: File:FEABBQ 11.17.08 Fig 2.JPG


text:
User:Eas4200c.f08.blue.a/Lecture 39

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

title: User:Eas4200c.f08.blue.a/Lecture 39


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Template:Bcpm/prag/7

July

This template makes a link to the July key for Prague, and adds plant pages to Category:BCP/PRAGUE/7 and Category:BCP/Flora of the Czech Republic (it automatically adds ), for plants that have been recorded as flowering in June in Prague.

This template should only be used in the "Prague = " field of Bloom clock plant middle

title: Template:Bcpm/prag/7


text:
User:Eml4500.f08.wiki1.schaet/hw7

Image for section 35-2

Images from lecture 37:

Two Bar Frame using MATLAB analysis
The following depicts the two bar system to be analyzed using MATLAB. It is very similar to the 2-bar system analyzed in Homework 3 except for this system is a frame system rather than a simple truss system. Here, Element 1 is a frame element with a square cross-sectional area and Element 2 is a truss system.

Since the frame element is assumed to be sqaure, the values for the moment of inertia were calculated by using the equation $$ I=\frac{s_{1}s_{2}^{3}}{12} $$ where the s-values are the sides of the square and thus $$s_{1}=s_{2}$$. These values along with the values given in the problem are stated in the following table. P is known to be 7.

Matlab Code
Below is the matlab code used to obtain the comparison between the undeformed 2 bar truss, the 2 bar truss system (determined in homework 3), and the combined frame/truss system explained above.

 "2 Bar Truss/Frame Comparison Matlab Code"

Below is the matlab code for the m-file PlaneFrameElement.m as given in the companion for the textbook. This file obtains the K matrix and R matrix. The one given in the book is for an element with a distributed load. Plugging in zeros for these values simplifies the distributed load to the one point load we have in this problem. These lines of code could have been removed but were left in to keep it simple. PlaneFrameElement.m Matlab Code

Below is the matlab code used to obtain the results for shear, bending moment, and axial forces. It is important to note that while this file is similar to the books companion site, some changes were made. One of the changes was to increase the number of increments along the frame element from 3 points in the book (i = 0:L/2:) to 16 points in our code (i=0:L/15:L). This gives us more points along the element to analyze so that the curve is more accurate. Other changes were just simple definition changes. Instead of using the word 'inertia', it is defined as I. Similarly, the word 'modulus' was replaced with E.  This serves just to simplify the code.  Modified PlaneFrameResults.m Matlab Code

Below are the results obtained. Note too that we also have the reaction forces as defined by rf. The shear, bending moment, and axial forces are much larger matricies because they are the results at each of the points along element 1 as defined by (i = 0:L/15:L).  Results

Below is the image obtained by the code. The undeformed system is the dotted blue line. The deformed truss system is the green line. These are the same as obtained in HW3. The red line shows the deformations for the frame/truss system given in this homework. As you can see, for the frame system, the angle at node 1 does not change significantly. This is due to the change in the type of connection at that node. Instead, there is a sort of wavy motion going up to node 2. This is due to the added degree of freedom and moments at each node. There is also a close up view to see these differences.

title: User:Eml4500.f08.wiki1.schaet/hw7


text:
Template:Bcpm/prag/8

August

This template makes a link to the August key for Prague, and adds plant pages to Category:BCP/PRAGUE/8 and Category:BCP/Flora of the Czech Republic (it automatically adds ), for plants that have been recorded as flowering in June in Prague.

This template should only be used in the "Prague = " field of Bloom clock plant middle

title: Template:Bcpm/prag/8


text:
Template:Bcpm/prag/9

September

This template makes a link to the June key for Prague, and adds plant pages to Category:BCP/PRAGUE/9 and Category:BCP/Flora of the Czech Republic (it automatically adds ), for plants that have been recorded as flowering in June in Prague.

This template should only be used in the "Prague = " field of Bloom clock plant middle

title: Template:Bcpm/prag/9


text:
Template:Bcpm/prag/11

November

This template makes a link to the November key for Prague, and adds plant pages to Category:BCP/PRAGUE/11 and Category:BCP/Flora of the Czech Republic (it automatically adds ), for plants that have been recorded as flowering in October in Prague.

This template should only be used in the "Prague = " field of Bloom clock plant middle

title: Template:Bcpm/prag/11


text:
User:Junior91

THe Names Junior

1. LOVE music 2. articulate as they get 3. appreciate ACURATE knowledge

title: User:Junior91


text: Category:Bloom clock regional tags/Early Spring


text: Category:Urantia Book

title: Category:Urantia Book


title: File:Hw7arivero.jpg


text: File:Qarea.svg

title: File:Qarea.svg


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User:Eml4500.f08.ramrod.D/HW7

Functions
The functions PlaneTrussElement.m, NodalSoln.m, PlaneTrussResults.m were all called in the code for the 5 truss system. They are all necessary for the completion of the truss problem.

PlaneTrussElement.m

This function take the Young's Modulus (e), the Area of the element (A) and the coordinates of the element ends (coord) and generates the stiffness matrix for a plane truss element. The function first calculates the lengths of the the elements. Once those are calculated the stiffness matrix is calculated.

NodalSoln.m

The function takes the global coefficient matrix (K), the global right hand side vector (R), the list of degrees of freedom with specified values, and the specified values to determine the displacements and reactions at each node. The dof of the the system is first found by using the  command which finds the longest dimension of R. df then finds the difference between the dofs that have known values (a value of zero) and the dof that were found in the previous line. The displacements and the reactions are then calculated.

PlaneTrussResults.m

This function computes the plane truss element results. It takes in the modulus of elasticity (e), the area of the cross-section (A), the coordinates at the element ends (coord), and the displacements at the elemtent ends (disps) and calculates the axial strain (eps), stress(sigma) and force (force). The results are stored in the variable  and sent back to the main program.

MATLAB Code to analyze a varying area and modulus of elasticity
Below is the code for a truss system that contains a varying modulus of elasticity and varying area.

2 Bar Frame System
Below is the code for solving the 2 bar system as a frame. The displaced results are shown in the plot the system in the section below.

Plots Comparing Different 2 Bar Truss Systems
Below is the code to plot the three 2 bar truss systems that were analyzed above. The dotted line is the original 2 bar truss system, the solid blue line is the deformed original system, the red line is the deformed varying 2 bar truss system, and the green is the frame system.

Electric Pylon Frame Analysis
The following code attempts to analyze the electric pylon with the frame technique.

title: User:Eml4500.f08.ramrod.D/HW7


text: User:KTucker/My Sandbox/Samples/Introduction/temp


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