User:Eml4500.f08.Ateam.philbin/HW7

Beam Shape Functions
The beam shape functions for the six forces is as follows.







Note the similarities between the corresponding horizontal and vertical forces as well as the moments. The vertical sides of the first four shapes have a magnitude of 1. The initial slope of mode 3 and the final slope of mode 6 both have a magnitude of 1. The final displacement vector can be written in the following form.

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

$$u(\tilde{x}) = N_2(\tilde{x})\hat{d}_1 + N_2(\tilde{x})\hat{d}_4 $$

In order to compute these components, the relationship developed in a previous lecture will be used.

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

This relationship is used to reorient the axial coordinate system to the global coordinate system.



Computing the displacement vector
The displacement vector can be written in terms of its components, both in the axial and global coordinate systems.

$$\mathbf{u}(\tilde{x})=u(\tilde{x})\vec{\tilde{i}}+v(\tilde{x})\vec{\tilde{j}}\quad (axial\ coordinates)$$

$$\mathbf{u}(\tilde{x})=u_x(\tilde{x})\vec{i}+v_y(\tilde{x})\vec{j}\quad (global\ coordinates) $$

In order to find $$u_x(\tilde{x})\ and\ v_y(\tilde{x})$$ a transformation matrix $$R^T$$ will be used.

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

Now the equation for $$\begin{Bmatrix} u(\tilde{x}) \\ v(\tilde{x})  \end{Bmatrix}$$ must be found. It is related to the displacements and the equation derived from the mode shapes.

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

Using these equations, they can be rewritten to give the final form of the global displacement vector.

$$\begin{Bmatrix} u_x(\tilde{x})\\v_y(\tilde{x})\end{Bmatrix} = R^T \begin{bmatrix} N_1 &0&0&N_4&0&0\\0&N_2&N_3&0&N_5&N_6\end{bmatrix} \tilde{T}^{(e)} d^{(e)}$$

Team View of Mediawiki
Mediawiki is an interesting medium for our class to use to complete our homework. Its learning curve is hard at first but once the proper syntax is learned, very professional work can be created. The math function in particular can create very precise and systematic functions that are easy to understand. Since all members of the group are using the same editors, a cohesive presentation can be achieved throughout each homework assignment. Since there are multiple groups working on each assignment, once turned in, each student has access to many different views and approaches to each problem.

Suggested software for Mediawiki
Most of the aspects of each homework assignment are similar and flow throughout the assignment. However, it was noticed that the programs that each student used for creating images were not the same. Therefore, some images were more plain that the others. In order to create a more uniform look, each group should choose an image software program at the beginning of the class and stick with it throughout the course.

Besides the Paint program that comes on every Windows PC, there are several programs available online that have enhanced features. Some such as Paint.NET and e-Paint are available for free and offers more advanced tools that the basic paint program. Another program that can easily be found online is GIMP (GNU Image Manipulation Program) which is more designed to modify images. Inkscape is a vector based graphics editor which can create Scalable Vector Graphics (SVG) file format. This is different that the usual scalar approach and allows the size of images to be easily changed with little loss of quality.

Anyone of these programs would be effective at creating the images that were needed for this class. In future courses, if a group were to decide on one, it would increase the overall continuity between individual contributions.

12/08/08
$$\tilde{k}_{23}=\frac{6EI}{L^2}= \int_{0}^{L} \frac{d^2N_2}{dx^2}(EI)\frac{d^2N_3}{dx^2}\, dx$$

Or to write this in a more generalized form:

$$\tilde{k}_{ij}= \int_{0}^{L} \frac{d^2N_i}{dx^2}(EI)\frac{d^2N_j}{dx^2}\, dx\quad i,j=2,3,5,6$$

Discrete Principal of Virtual Work
$$\bar{w}\cdot[\bar{M}\ddot{\bar{d}}+\bar{k}\bar{d}-\bar{F}]=0 \quad\ for\ all\ \bar{w}$$

The bar notation denotes that the boundary conditions have already been applied. The ordinary differential equations (second order in time) and the initial conditions governing the elastodynamics of the discrete continuity problem are as follows.

$$\bar{M}\ddot{\bar{d}}+\bar{k}\bar{d}=\bar{F}(t)\quad\ (1)$$

$$\bar{d}(0)=\bar{d}_0\quad \quad (initial\ position)$$

$$\dot{\bar{d}}(o)=\bar{v}_0 \quad \quad (initial\ velocity)$$

Eigenvalue Solution
1) Consider the unforced vibration problem:

$$\bar{M}\ddot{v}+ \bar{k} v = 0 \quad \ (2)$$

Assume $$v(t)=(sin {wt})\mathbf{\phi}$$. Here $$\phi$$ is a column matrix of size (nx1) that is not time dependent.

$$\ddot{v}=-w^2sin {wt}\phi$$

Plug the previous equation back into equation (1):

$$-w^2 sin{wt} \bar{M}\phi + sin {wt} \bar{k} \phi = 0 $$

This can be simplified to the generalized eigenvalue problem:

$$\bar{k}\phi = w^2 \bar{M} \phi$$

An even more generalized version can be written in terms of matrices A and B.

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

Once evaluated, the following relations are made.

$$\mathbf{\lambda} = w^2$$

$$Mode\ i=\begin{cases} V_i(t)=(sin {w_it})\phi_i \\ i=1,...,n\end{cases}$$

2) Model superposition method:

Orthogonal property of eigenvalues is shown in the following equation:

$$\phi_{i} \bar{M} \phi_{j} = \delta _{i j} = \begin{cases} 1, & \mbox{if }\ {i=j} \\ 0, & \mbox{if }\ {i \ne j} \end{cases}$$

Here $$ \mathbf{\delta}_{ij}$$ is the Kronecker delta which has a variety of uses including digital processing.

$$\bar{M}\phi_j=\lambda^{-1}_j\phi^T_i\bar{k}\phi_j$$

$$\phi_i^T\bar{M}\phi_j=\delta_{ij}\lambda^{-1}_j\phi^T_i\bar{k}\phi_j$$

$$\phi^T_i\bar{k}\phi_j=\lambda_j \delta_{ij}$$

Next look at:

$$\bar{M}\ddot{\bar{d}}+\bar{k}\bar{d}=\bar{F}\quad \ (3)$$

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

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

then

$$\bar{d}(t)=\sum_{i=1}^n \xi_i(t)\phi_i $$

Now plug the previous equation into equation (3):

$$\bar{M}(\sum_j \xi_j\phi_j)+\bar{k}(\sum_j\xi_j\phi_j)=\bar{F}$$

where

$$\ddot{\bar{d}}=\sum_j \xi_j\mathbf{\phi_j}\quad and\ \bar{d}=sum_j\xi_j\mathbf{\phi_j}$$

Finally,

$$\sum_j \ddot{\xi}_j(\mathbf{\phi_i^T}\mathbf{\bar{M}}\mathbf{\phi_j})+ \sum_j \ddot{\xi}_j(\mathbf{\phi_i^T}\mathbf{\bar{k}}\mathbf{\phi_j})=\mathbf{\phi^T_i} \mathbf{F}$$

$$\ddot{\xi}_i+\lambda_i\xi_i=\phi^T_i\mathbf{F}$$