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.