User:Eml4500.f08.wiki1.aguilar/TeamHW7

=Team Homework 7=

Wikiversity vs. E-Learning - Wiki1 Group Verdict
While both the Wikiversity and the E-Learning systems are very useful, each was designed for a specific purpose in mind. While E-Learning was designed to facilitate the communication between instructors and students, wikis are more designed to allow groups and organizations to easily post, edit, and share information. Because they are used with different purposes in mind, they are difficult to compare.

While E-Learning is in no way perfect, it does facilitate communication between professors and students when used correctly. By using E-learning as the standard and trying to familiarize all the professors with it, it bypasses the need for teachers to create a website to post information that needs to be accessible to all the students. We are not familiar with the back end of E-Learning, but we are sure that using E-learning to post files and information is easier than teaching professors how to build their own websites with attached files and documents. E-Learning does allows easy methods for professors to do a variety of different things. For example, E-Learning allows professors and TA's to upload grades and then for the various assignments and exams, students can view their grades at any time throughout the semester. Professors can also allow it to provide statistics on the assignment or exam like the average, the high and low scores, as well as histograms showing the grade distributions. E-Learning allows the professors to easily post announcements and documents as well, like homework solution PDFs and course syllabi. There is a feature for a discussion board which could easily be used to allow students to ask other students and TAs various questions, and there is also a calendar to allow various important dates to be posted. This are just some of the features provided for by the E-Learning system, many of which cannot be adequately provided for through Wikiversity.

It becomes more of a question what is it that we are trying to do? If, for example, the end goal is to provide a large, multimedia report that is to be worked on by an entire group, then Wikiversity does indeed provide a very good solution for this. But E-Learning is currently not meant to serve this function. For the purposes of this class, "E-Learning" is more of an indirect method for group collaboration. If you had to use E-Learning to do this, the process would vary group by group. Groups could come up with various ways to create these reports. Some groups might use sharing features of Google documents to write up the report and allow all members of the group to contribute text, images and various other elements to the report. This could then be saved as either a PDF or a Word document which could then be submitted on E-Learning. The advantage to this method would be that there isn't any "coding" to learn so that would minimize some of the learning curve, and it would also protect reports from being viewed by others outside the group. The downside to this would be that reports from group to group would be much less streamlined, and having to save equations as images could become quite a hassle. Another alternative would be to use Word to allow groups to create the report. Word 2007 does contain a much improved equation editor and almost everyone uses Microsoft Word already. E-Learning could be set up to allow groups members to upload their parts, and then someone at the end would combine into a final report. This method would have similar pros and cons to the Google Shared documents method.

Clearly, for a group to put together an entire report like what was required of us, Wikiversity does seem to provide a very good solution, keeping in mind that no solution will be perfect. Wikiversity allows for students to be introduced to wikis, something that is becoming much more widespread throughout the "real world." These skills are much more likely to becomes valuable later on down the road. Wikiversity is major project and it's performance will not be affected by the activity of our groups even under peak conditions. The downside is that the Wikiversity approach does, however, lend itself to the possibility of being vandalized or copied by members from other groups.

As a group, we think that for our class purposes, the Wikiversity method could be improved. This could be done by cutting out much of the tedious busy work that tends to be more "filler" within our reports. While each homework report usually looked impressive when it was turned in, by essentially having each of the groups rewrite all the notes from the course lectures was of little added benefit to each student. Much time was spent tediously formatting the lengthy equations that time was taken away from more beneficial areas of the homework. Each student should be responsible for taking his or her own notes thus emphasizing the importance of class attendance. The reports should focus on actual homework with each group's solutions to various problems and MATLAB assignments with included images, videos, and coding from all the members of the group. In our opinion, this approach to the Wikiversity homework reports would be more beneficial to the students, and should be used in conjunction with many of the features of E-Learning, such as grade statistics and feedback, as well as others.

Course Notes
This consists of course notes from 11/17 - 12/8.

Frame Analysis
A frame element is a combination of a truss (bar element) with a beam element.

The differences between a truss and a beam is that because of it's pin connections, a truss only incurs only axial deformation while the fixed nature of the nodes of a beam experience transverse deformation.

2-D Frame Local DOFs


In general, $$d_{(i)}^{(e)}\Rightarrow f_{(i)}^{(e)}$$, where $$f_{(i)}^{(e)}$$ are the generalized forces. Here, e=1,2 and i=1,...,6

$$\left.\begin{matrix} d_{(3)}^{(e)}\\d_{(6)}^{(e)} \end{matrix}\right\}$$ rotational degrees of freedom

$$\left.\begin{matrix} f_{(3)}^{(e)}\\f_{(6)}^{(e)} \end{matrix}\right\}$$ bending moments

2-D Frame Global DOFs


Therefore, there are two element stiffness matrices, $$k_{6\times6}^{(e)}$$ where e=1,2.

This means the global stiffness matrix is $$K_{9\times9}^{(e)}=Ak_{6\times6}^{(e)}$$.

This is assembled by combining the two element stiffness matrices as shown below.



Transformed Coordinate System


$$\mathbf{\tilde{k}}_{(6\times 6)}^{(e)} \mathbf{\tilde{d}}_{(6\times 1)}^{(e)}=\mathbf{\tilde{f}}_{(6\times 1)}^{(e)}$$, where each is defined below.

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

Note: The rotational displacements and moments remain unchanged. $$\tilde{d}_{3}^{(e)}={d}_{3}^{(e)}$$ $$\tilde{d}_{6}^{(e)}={d}_{6}^{(e)}$$ $$\tilde{f}_{3}^{(e)}={f}_{3}^{(e)}$$ $$\tilde{f}_{6}^{(e)}={f}_{6}^{(e)}$$

and

$$ \mathbf{\tilde{k}}_{(6\times 6)}^{(e)}= \overbrace{

\begin{bmatrix} \frac{EI}{L} & 0 & 0 & -\frac{EI}{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{4EI}{L} & 0 & -\frac{6EI}{L^2} & \frac{4EI}{L} \end{bmatrix}}

^{\begin{matrix} \;\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{matrix}}

\left.\begin{matrix} \\ \\ \\ \\ \\ \end{matrix}\right\}

\begin{matrix} \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{matrix} $$

Dimensional Analysis
The square bracket notation around a variable means "dimension of".

$$ \begin{bmatrix} \tilde{d}_1 \end{bmatrix}=L = \begin{bmatrix} \tilde{d}_i \end{bmatrix} $$, where i=1,2,4,5 and L signifies that it has a dimension of length.

Note: There is a difference between 'units' and 'dimensions'. Displacement has a dimension of length Mass, length, time, charge and temperature are all fundamental dimensions. They cannot be broken down any further, while a dimension such as force can be broken down even further as shown below.

$$Force=mass \times acceleration = \frac {mass \times distance}{{time}^2}=\frac{m\times L}{t^2}$$

$$\begin{bmatrix} \tilde{d}_3 \end{bmatrix}=1$$, meaning it has no dimension.

$$\sigma = E\varepsilon \Rightarrow \begin{bmatrix}\sigma \end{bmatrix}= \begin{bmatrix}E\end{bmatrix} \begin{bmatrix}\varepsilon   \end{bmatrix} =1$$

$$\left[\varepsilon \right]={[du] \over [dx]}={L \over L}=1$$

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

$$[A]=L^2, [I]=L^4 \,$$

$$\left[ {EA \over L} \right]=[\tilde{k_{11}}]={{F \over L^2}L^2 \over L}={F \over L}$$

$$[\tilde{k}_{11} \tilde{d}_1]=[\tilde{k}_{11}][\tilde{d}_1]=F$$

$$[\tilde{k}_{14} \tilde{d}_4]=[\tilde{k}_{14}][\tilde{d}_4]=\left[ {-EA \over L} \right][L]={-{F \over L^2}L^2 \over L}L=-F$$

$$[\tilde{k}_{22} \tilde{d}_2]=[\tilde{k}_{22}][\tilde{d}_2]={12EI \over L^3}L={{F \over L^2}L^4 \over L^3}L=F$$

$$[\tilde{k}_{23} \tilde{d}_3]=[\tilde{k}_{23}][\tilde{d}_3]={[6][E][I] \over [L^2]}={1 {F \over L^2}L^4 \over L^2}=F$$

$$[\tilde{k}_{25} \tilde{d}_5]=[\tilde{k}_{25}][\tilde{d}_5]=-[\tilde{k}_{22} \tilde{d}_2]=-F$$

$$[\tilde{k}_{26} \tilde{d}_6]=[\tilde{k}_{26}][\tilde{d}_6]=[\tilde{k}_{23} \tilde{d}_3]=F$$

$$[\tilde{k}_{33} \tilde{d}_3]=[\tilde{k}_{33}][\tilde{d}_3]={[4][E][I] \over [L]}={1{F \over L^2}L^4 \over L}=F*L$$

$$[\tilde{k}_{35} \tilde{d}_5]=[\tilde{k}_{35}][\tilde{d}_5]=-[\tilde{k}_{23} \tilde{d}_3]=-F$$

$$[\tilde{k}_{36} \tilde{d}_6]=[\tilde{k}_{36}][\tilde{d}_6]={[2][E][I] \over [L]}={1{F \over L^2}L^4 \over L}=F*L$$

$$[\tilde{k}_{44} \tilde{d}_4]=[\tilde{k}_{44}][\tilde{d}_4]={[E][A] \over [L]}[L]={{F \over L^2}L^2 \over L}L=F$$

$$[\tilde{k}_{55} \tilde{d}_5]=[\tilde{k}_{55}][\tilde{d}_5]=[\tilde{k}_{22} \tilde{d}_2]=F$$

$$[\tilde{k}_{56} \tilde{d}_6]=[\tilde{k}_{56}][\tilde{d}_6]=-[\tilde{k}_{23} \tilde{d}_3]=-F$$

$$[\tilde{k}_{66} \tilde{d}_6]=[\tilde{k}_{66}][\tilde{d}_6]=[\tilde{k}_{33} \tilde{d}_3]=F*L$$

Element Force-Displacement Relationship in Global Coordinates from the Force-Displacement Relationship in Local Coordinates
$$\mathbf{k}^{(e)}_{6x6} \mathbf{d}^{(e)}_{6x1}=\mathbf{f}^{(e)}_{6x1}$$

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

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

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

$$\tilde{d}_3=d_3$$

$$\tilde{d}_6=d_6$$

Derivation of $$\tilde{\mathbf{k}}^{(e)}$$ from the Principle of Virtual Work, focusing only on the bending effect.

$$-{\partial^2 \over \partial x^2}((EI){\partial^2 v \over \partial x^2}) + f_t(x)=m(x)\ddot v$$

$${\partial \over \partial x}((EA){\partial u \over \partial x}) + f_a(x,t)=m(x) \ddot u$$

Motivation: deformed shape of truss element interpolation of transverse displacement $$v(s)=v(\tilde{x})$$

The Principle of Virtual Work for Beams
$$\int_0^L W(\tilde{x})[-{\partial^2 \over \partial x^2}(EI){\partial^2 v \over \partial x^2}+f_t-m \ddot v] dx=0$$ for all W(x)Equation (1)

Integration by Parts of the 1st term:

$$\alpha =\int_0^L W(x) {\partial^2 \over \partial x^2}((EI){\partial^2 v \over \partial x^2})dx$$

$$W(x){\partial \over \partial x}({\partial \over \partial x}((EI){\partial^2 v \over \partial x^2}))$$

$$=[W{\partial \over \partial x}((EI){\partial^2 v \over \partial x^2})]_0^L-\int_0^L {dW \over dx}{\partial \over \partial x}((EI){\partial^2 v \over \partial x^2})dx$$

$$=[W{\partial \over \partial x}((EI){\partial^2 v \over \partial x^2})]_0^L-[{dW \over dx}(EI){\partial^2 v \over \partial x^2}]_0^L+\int_0^L{d^2W \over dx^2}(EI){\partial^2 v \over \partial x^2}dx$$

Thus, Equation (1) becomes:

$$[W{\partial \over \partial x}((EI){\partial^2 v \over \partial x^2})]_0^L-[{dW \over dx}(EI){\partial^2 v \over \partial x^2}]_0^L+\int_0^L{d^2W \over dx^2}(EI){\partial^2 v \over \partial x^2}dx+\int_0^L W f_t dx-\int_0^L W m \ddot v dx =0 for all possible W(x)$$

Derivation of the Stiffness Matrix and Shape Functions for Beams
Focus on stiffness term, $$\int_0^L{d^2W \over dx^2}(EI){\partial^2 v \over \partial x^2}dx$$, to derive the beam stiffness matrix and identify shape functions for beams



$$v(\tilde{x})=N_2(\tilde{x})\tilde{d}_2+N_3(\tilde{x})\tilde{d}_3+N_5(\tilde{x})\tilde{d}_5+N_6(\tilde{x})\tilde{d}_6$$Equation (2)

Recall:$$u(\tilde{x})=N_1(\tilde{x})\tilde{d}_1+N_4(\tilde{x})\tilde{d}_4$$Equation (1)



$$N_2(\tilde{x})=1-{3\tilde{x}^2 \over L^2}+{2\tilde{x}^3 \over L^3}; N_3(\tilde{x})=\tilde{x}-{2\tilde{x}^2 \over L} + {\tilde{x}^3 \over L^2}; N_5(\tilde{x})={3 \tilde{x}^2 \over L^2} - {2\tilde{x}^3 \over L^3}; N_6(\tilde{x})=-{\tilde{x}^2 \over L} + {\tilde{x}^3 \over L^2}$$

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

Computation of $$u(\tilde{x})$$ and $$v(\tilde{x})$$:



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

$$=u_x(\tilde{x})\vec{i}+u_y(\tilde{x})\vec{j}$$

Computation of $$u(\tilde{x})$$ and $$v(\tilde{x})$$ using Equations (1) and (2):

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

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

$$\mathbb{N}(\tilde{x})=\begin{bmatrix}N_1 &0&0&N_4&0&0\\0&N_2&N_3&0&N_5&N_6\end{bmatrix}$$

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

 Note:  Eq. (2) (p. 39-2, Dimensional Analysis)

[u] = L p. 31-4: [N1] = [N4] = 1

Eq. (2) p. 39-2: $$\left[N_{1} \right]\left[\tilde{d}_{1} \right]+\left[N_{4} \right]\left[\tilde{d}_{4} \right]$$

$$\left[N_{1} \right]=1$$ $$\left[\tilde{d}_{1} \right]=L$$ $$\left[N_{4} \right]=1$$ $$\left[\tilde{d}_{4} \right]=L$$

[v] = L $$\left[N_{2} \right]\left[\tilde{d}_{2} \right]=L$$ $$\left[N_{2} \right]=1$$ $$\left[\tilde{d}_{2} \right]=L$$

$$\left[N_{3} \right]\left[\tilde{d}_{3} \right]=L$$ $$\left[N_{3} \right]=L$$ $$\left[\tilde{d}_{3} \right]=1$$

$$\left[N_{5} \right]\left[\tilde{d}_{5} \right]=L$$ $$\left[N_{5} \right]=1$$ $$\left[\tilde{d}_{5} \right]=L$$

$$\left[N_{6} \right]\left[\tilde{d}_{6} \right]=L$$ $$\left[N_{6} \right]=L$$ $$\left[\tilde{d}_{6} \right]=1$$

$$N_{2}, N_{3}, N_{5}, N_{6}$$ (p. 38-4) p. 38-3 plots Recall: Governing PDE for beams p. 37-4, Eq. (1), without f2 (dist. transverse load) and without inertia force $$m\ddot{v}$$ (static case):

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

Further, consider constant EI : $$\frac{\partial^4 }{\partial x^4}v = 0$$

Integrate 4 times 4 constants (c0, c1, c2, c3) $$\Rightarrow v(x)=c_{0}+c_{1}x^{1}+c_{2}x^{2}+c_{3}x^{3}$$

To obtain N2(x) ($$\tilde{x}=x$$for simplicity)

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

Use above boundary conditions to solve for c0, c1, c2, c3

v(0)=1=c0 $$(1) \; \; v(L)=1+c_{1}L+c_{2}L^2+c_{3}L^3=0$$ $$v'(x)=c_{1}+2c_{2x}+3c_{3}x^2$$ v'(0)=c_(1)=0 $$(2)\; \; v'(L)=2c_{2}L+3c_{3}L^2=0$$

$$c_{0}=1$$ $$c_{1}=0$$ $$c_{2}=\frac{-3}{L^2}$$ $$c_{3}=\frac{-2}{3}\frac{1}{L}\frac{-3}{L^2}=\frac{2}{L^3}$$ Compute with expression for N2 on p. 38-4

 For N3:  (Boundary Conditions) v(0)=v(L)=0 v'(0)=1 v'(L)=0

 For N5:  ($$\tilde{d_{5}}$$ rotation) v(0)=0 v(L)=1 v'(0)=0 v'(L)=0

 For N6:  ($$\tilde{d_{6}}$$ rotation) v(0)=0 v(L)=0 v'(0)=0 v'(L)=1

See p. 39-1 for plots of N5, N6

From p. 36-2: $$\tilde{k}_{22}=\frac{12EI}{L^3}$$

From p. 38-2: $$\tilde{k}_{22}=\int_{0}^{L}{\frac{d^2N_2}{dx^2}(EI)}\frac{d^2N_2}{dx^2}dx$$

In general:

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

Therefore, knowing that the stiffness matrix $$\tilde{K}_{6x6}$$ is symmetric about the diagonal:

From p. 36-2: $$\tilde{k}_{23}=\tilde{k}_{32}=\frac{6EI}{L^2}$$

From p. 38-2: $$\tilde{k}_{23}=\int_{0}^{L}{\frac{d^2N_2}{dx^2}(EI)}\frac{d^2N_3}{dx^2}dx=\tilde{k}_{32}=\int_{0}^{L}{\frac{d^2N_3}{dx^2}(EI)}\frac{d^2N_2}{dx^2}dx$$

Similarly: $$\tilde{k}_{25}=\tilde{k}_{52}=\frac{-12EI}{L^3}$$

$$\tilde{k}_{25}=\int_{0}^{L}{\frac{d^2N_2}{dx^2}(EI)}\frac{d^2N_5}{dx^2}dx=\tilde{k}_{52}=\int_{0}^{L}{\frac{d^2N_5}{dx^2}(EI)}\frac{d^2N_2}{dx^2}dx$$

$$\tilde{k}_{26}=\tilde{k}_{62}=\frac{6EI}{L^2}$$

$$\tilde{k}_{26}=\int_{0}^{L}{\frac{d^2N_2}{dx^2}(EI)}\frac{d^2N_6}{dx^2}dx=\tilde{k}_{62}=\int_{0}^{L}{\frac{d^2N_6}{dx^2}(EI)}\frac{d^2N_2}{dx^2}dx$$

$$\tilde{k}_{33}=\frac{4EI}{L}$$

$$\tilde{k}_{33}=\int_{0}^{L}{\frac{d^2N_3}{dx^2}(EI)}\frac{d^2N_3}{dx^2}dx$$

$$\tilde{k}_{35}=\tilde{k}_{53}=\frac{-6EI}{L^2}$$

$$\tilde{k}_{35}=\int_{0}^{L}{\frac{d^2N_3}{dx^2}(EI)}\frac{d^2N_5}{dx^2}dx=\tilde{k}_{53}=\int_{0}^{L}{\frac{d^2N_5}{dx^2}(EI)}\frac{d^2N_3}{dx^2}dx$$

$$\tilde{k}_{36}=\tilde{k}_{63}=\frac{2EI}{L}$$

$$\tilde{k}_{36}=\int_{0}^{L}{\frac{d^2N_3}{dx^2}(EI)}\frac{d^2N_6}{dx^2}dx=\tilde{k}_{63}=\int_{0}^{L}{\frac{d^2N_6}{dx^2}(EI)}\frac{d^2N_3}{dx^2}dx$$

$$\tilde{k}_{55}=\frac{12EI}{L^3}$$

$$\tilde{k}_{55}=\int_{0}^{L}{\frac{d^2N_5}{dx^2}(EI)}\frac{d^2N_5}{dx^2}dx$$

$$\tilde{k}_{56}=\tilde{k}_{65}=\frac{-6EI}{L^2}$$

$$\tilde{k}_{56}=\int_{0}^{L}{\frac{d^2N_5}{dx^2}(EI)}\frac{d^2N_6}{dx^2}dx=\tilde{k}_{65}=\int_{0}^{L}{\frac{d^2N_6}{dx^2}(EI)}\frac{d^2N_5}{dx^2}dx$$

$$\tilde{k}_{66}=\frac{4EI}{L}$$

$$\tilde{k}_{66}=\int_{0}^{L}{\frac{d^2N_6}{dx^2}(EI)}\frac{d^2N_6}{dx^2}dx$$

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

p. 31-1: Model problem

Discrete PVW
(Boundary Conditions already applied) $$\mathbf{\bar{w}\bullet \left[\bar{M}\ddot{\bar{d}}+\bar{K}\bar{d}-\bar{F} \right]}=0$$ for all $$\mathbf{\bar{w}}$$ where $$\mathbf{\bar{K}}$$ is the reduced stiffness matrix

Complete Ordinary Differential Equations (ODE's)
(2nd order in time) and initial conditions governing the elastodynamics of discretized continuity problem (MDOF) $$(1)\Rightarrow \mathbf{\bar{M}\ddot{\bar{d}}+\bar{K}\bar{d}=\bar{F}(t)}$$ $$\; \; \; \; \mathbf{\bar{d}(0)=\bar{d_{0}}}$$ $$\; \; \; \; \mathbf{\dot{\bar{d}}(0)=\bar{v}_{0}}$$

'''Solving Eq. (1):'''

Method 1: Considering the unforced vibration problem
$$\mathbf{\bar{M}_{nxn}\ddot{v}_{nx1}+\bar{K}_{nxn}v_{nx1}=0_{nx1}}$$ Assume $$\mathbf{v(t)_{nx1}=(sin\omega t)\phi _{nx1}}$$ $$\mathbf{\ddot{v}=-\omega ^2sin\omega t\phi}$$ $$\mathbf{-\omega ^2sin\omega t\bar{M}\phi+sin\omega t\bar{K}\phi=0}$$ Generalized eigenvalue problem $$\Rightarrow \mathbf{\bar{K}\phi =\omega ^2\bar{M}\phi}\; \; \; (1)$$ General form: $$\mathbf{Ax=\lambda Bx}$$, where $$\lambda$$ is an eigenvalue Standard eigenvalue problem: $$\mathbf{Ax=\lambda x}$$,where $$\mathbf{(B=I)}$$ The identity matrix $$\mathbf{I}=\begin{bmatrix}1 & & 0\\ & ... & \\ 0 & & 1\end{bmatrix}$$ $$\lambda =\omega ^2$$ $$(\lambda _i,\mathbf{\phi _i)}$$ are eigenpairs and i = 1,...,n Mode i $$\Rightarrow \mathbf{v_i(t)=(sin\omega _it)\phi _i}$$ where i = 1,...,n

Method 2: Model Superposition method
Orthogonal properties of eigenpairs:

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

Mass Orthogonality of eigenvectors: Eq. (2) p. 41-2: Eq. (1) p. 41-3:

$$\mathbf{\bar{M}\phi _j=\lambda _j^{-1}\bar{K}\phi _j}$$

$$\mathbf{\phi _i^T\bar{M}\phi _j=\lambda _j^{-1}\phi _i^T\bar{K}\phi _j}$$ where $$\mathbf{\phi _i^T\bar{M}\phi _j=\delta _{ij}}$$

$$\Rightarrow \mathbf{\phi _i^T\bar{K}\phi _j=\lambda _j\delta _{ij}}$$

Eq. (1) p. 41-2: $$\mathbf{\bar{d}(t)_{nx1}=\sum_{i=1}^{n}{\xi _i(t)_{1x1}}\phi _{nx1}}$$

$$\mathbf{M\left(\sum_{j}^{}{}\ddot{\xi _j}\phi _j\right)+\bar{K}\left(\sum_{j}^{}{}\xi _j\phi _j \right)=F}$$ where $$\sum_{j}^{}{}\ddot{\xi _j}\phi _j=\mathbf{\ddot{\bar{d}}}$$ and $$\sum_{j}^{}{}\xi _j\phi _j=\mathbf{\bar{d}}$$

$$\mathbf{\sum_{j}^{}{}\ddot{\xi _j}\left(\phi _i^T\bar{M\phi _j}\right)+\sum_{j}^{}{}\xi _j\left(\phi _i^T\bar{K}\phi _j \right)=\phi _i^TF}$$ <p style="text-align:center;">where $$\phi _i^T\bar{K}\phi _j=\delta _{ij}$$ and $$\phi _i^T\bar{K}\phi _j=\lambda _j\delta _{ij}$$

<p style="text-align:center;">$$\Rightarrow \mathbf{\ddot{\xi _i+\lambda _i\xi _i=\phi _i^TF}}$$ where i = 1,...,n

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.



Electric Pylon as Frame Elements
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.



Problem Statement and Given Information
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.

<p style="text-align:center;"> "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. <p style="text-align:center;">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. <p style="text-align:center;"> Modified PlaneFrameResults.m Matlab Code

Results Obtained
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). <p style="text-align:center;"> Results

Images
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.



Contributing Members
Eml4500.f08.wiki1.aguilar 20:06, 9 December 2008 (UTC) Eml4500.f08.wiki1.ambrosio 19:46, 9 December 2008 (UTC) Eml4500.f08.wiki1.oatley 16:26, 9 December 2008 (UTC) Eml4500.f08.wiki1.schaet 19:34, 9 December 2008 (UTC) Eml4500.f08.wiki1.brannon 17:42, 9 December 2008 (UTC)