User:Eml4500.f08.qwiki.bishop/HW3

 Comment: The pictures that were previously used in this homework submission have been changed. The previous images ( and ) were based on images that were not originally authored by Team Qwiki, however I did not intentionally plagiarize Team Echo's work. As Lee Harris, fellow Team Qwiki member, also stated in Homework 4, while working on Homework 2, both Team Qwiki and Team Echo created two images, with the same name: and . Due to the nature of MediaWiki, if two different teams create an image with the same name, only the latest version is displayed when called upon in an article. Therefore, even though Team Qwiki uploaded two of our own images (Element1.jpg and Element2.jpg) during homework 2, Team Echo's images were displayed because their upload was the most recent. In the future, we will make sure to upload files that have a unique name. For Homework 3, I created and  on October 5 because it was the two bar truss used in Homework 2, and to keep the images showing the same two bar truss with identical angles and lengths, I based the mentioned images on them. I was under the impression that Element1.jpg was the work of Team Qwiki. The problem has been resolved and Team Qwiki files are now displayed here. In the future Team Qwiki and myself will be sure to: A comparison between the current version of Homework 3 and our prior submission can be found here. Andrea Booher EML4500.f08.qwiki.booher 00:17, 13 November 2008 (UTC)
 * Create unique names when uploading files
 * Check the file history of all images before I right click and save to my computer

=Lecture 12 - Monday, September 22, 2008=

For a more thorough understanding of the Finite Element Method, it is wise to derive the element force displacement with respect to the global coordinate system.

Recall from Page 6-1, $$k^{(e)}d^{(e)}=f^{(e)}$$ (Equation 1)



Therefore, the element force displacement matrix can be written as follows:

$$\mathbf{k^{(e)}}\begin{bmatrix} 1 & -1 \\ -1 & 1\\ \end{bmatrix}\begin{pmatrix} q_{1}^{(e)}\\ q_{2}^{(e)}\\ \end{pmatrix}=\begin{pmatrix} P_{1}^{(e)}\\ P_{2}^{(e)}\\ \end{pmatrix} $$

$$q^{(e)}_{i}$$=axial displacement of element e at local node $$i$$

$$P^{(e)}_{i}$$=axial force of element e at local node $$i$$

The overall goal is to derive equation 1 from equation 2 (already derived in Meeting 4). We want to find the relationship between: The relationships can be expressed in the form: $$q^{(e)}_{2x1}=T^{(e)}_{2x4}d^{(e)}_{4x1}$$
 * $$q^{(e)}_{2x1}$$ and $$d^{(e)}_{4x1}$$
 * $$P^{(e)}_{2x1}$$ and $$f^{(e)}_{4x1}$$

Consider the displacement vector of local node i, denoted by $$d^{(e)}_{i}$$:




 * $$\vec{d^{(e)}_{[i]}}=d^{(e)}_{1}\vec{i}+d^{(e)}_{2}\vec{j}$$


 * $$q^{(e)}_{i}$$=axial displacement of node [1] is the orthogonal projection of the displacement vector $$\vec{d_{[1]}^{(e)}}$$ of node [1] on the $$\tilde{x}$$ axis of element e

Therefore, for node [1], $$q_1^{(e)}$$ can be derived using the following steps.

$$q_{1}^{(e)}=d_{[1]}^{(e)}*\vec{\tilde{i}}$$ $$q_{1}^{(e)}=(d_{1}^{(e)}\vec{i}+d_{2}^{(e)}\vec{j})\vec{\tilde{i}}$$ $$q_{1}^{(e)}=d_{1}^{(e)}(\vec{i}*\vec{\tilde{i}})+d_{2}^{(e)}(\vec{j}*\vec{\tilde{i}})$$ $$q_{1}^{(e)}=l^{(e)}d_{1}^{(e)}+m^{(e)}d_{2}^{(e)}$$ $$q_{1}^{(e)}=\begin{bmatrix} l^{(e)} & m^{(e)}\\ \end{bmatrix}\begin{pmatrix} d_{1}^{(e)}\\ d_{2}^{(e)}\\ \end{pmatrix}$$

The values for $$q_2^{(e)}$$ and $$q_2^{(e)}$$ can now be substituted into the following matrix:

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

=Lecture 13 - Wednesday, September 24, 2008=

Test day, no lecture notes given. Exam 1 Solutions

=Lecture 14 - Friday, September 26, 2008=

Similar to our objective in Meeting 12, today we would like to transform from a 4x2 matrix to a 2x2. The transformation relationship

\begin{bmatrix} P_1^{(e)}\\ P_2^{(e)}\\ \end{bmatrix} =T_{(2x4)}^{(e)} \begin{bmatrix} f_1^{(e)}\\ f_2^{(e)}\\ f_3^{(e)}\\ f_4^{(e)}\\ \end{bmatrix} $$ derived in Meeting 12 can also be expressed in a more general form:
 * $$\mathbf{P}^{(e)}_i=\mathbf{T}^{(e)}\mathbf{f}^{(e)}_i$$, where


 * $$\mathbf{P}^{(e)}_{i}$$ is the axial force of element e at local node $$i$$
 * $$\mathbf{T}^{(e)}_{(2x4)}$$ is the matrix used to transform force components to an axial force
 * $$\mathbf{f}^{(e)}_{i}$$ is the $$i$$th force component of element e

The overall goal is to find the relationship between: Let us consider the force transformation relationship:
 * $$\mathbf{q}^{(e)}_{(2x2)}$$ and $$\mathbf{d}^{(e)}_{(4x2)}$$, (axial displacements of each element node and their respective displacement component)
 * $$\mathbf{P}^{(e)}_{(2x2)}$$ and $$\mathbf{f}^{(e)}_{(4x2)}$$, (axial forces at each element node and their respective force components)
 * $$\mathbf{P}^{(e)}_{(2x2)}=\mathbf{T}^{(e)}_{(2x4)}\mathbf{f}^{(e)}_{(4x2)}$$

Recall that the element axial FD relationship is given by:
 * $$\mathbf{\hat{k}}^{(e)}_{(2x2)}\mathbf{q}^{(e)}_{(2x2)}=\mathbf{P}_{(2x2)}^{(e)}$$

By introducing a transformation matrix into the FD relationship, we obtain the following relationship:
 * $$\mathbf{\hat{k}}^{(e)}\mathbf{T}^{(e)}\mathbf{d}^{(e)}=\mathbf{T}^{(e)}\mathbf{f}^{(e)}$$

The ultimate goal is to express the above equation in the form $$\mathbf{k}^{(e)}\mathbf{d}^{(e)}=\mathbf{f}^{(e)}$$. If $$\mathbf{T}^{(e)}$$ were an nxn matrix, then one could simply "move" $$\mathbf{T}^{(e)}$$ from the right hand side to the left hand side by premultiplying the entire equation by the inverse $$\mathbf{T^{(e)}}^{-1}$$. Unfortunately, $$\mathbf{T}^{(e)}$$ is a rectangular 2x4 matrix, which means it is non-invertible.

Since we cannot solve for the inverse of the transformation ($$\mathbf{T}^{(e)}$$ is not an nxn matrix), we must instead make use of the transpose:
 * $$\mathbf{T}^{(e)T}_{(4x2)}\hat{k}^{(e)}_{(2x2)}\mathbf{T}^{(e)}_{(2x4)}\mathbf{d}^{(e)}_{(4x1)}=\mathbf{f}^{(e)}_{(4x1)}$$

Note that $$\mathbf{k}^{(e)}=\mathbf{T}^{(e)T}_{(4x2)}\mathbf{\hat{k}}^{(e)}_{(2x2)}\mathbf{T}^{(e)}_{(2x4)}$$ Therefore, $$\mathbf{k}^{(e)}\mathbf{d}^{(e)}=\mathbf{f}^{(e)}$$

The relationship $$\mathbf{k}^{(e)}=\mathbf{T}^{(e)T}\mathbf{\hat{k}}^{(e)}\mathbf{T}^{(e)}$$ can be justified by the principle of virtual work, which was first introduced in Meeting 10. Additionally, applying the principle of virtual work to the relationship $$K_{(6x6)}d_{(6x1)}=F_{(6x1)}$$ reduces it to $$K_{(2x2)}d_{(2x1)}=F_{(2x1)}$$.

One may wonder why we cannot simply solve $$d=K^{(-1)}F$$ Answer : The equation cannot be solved in the current form due to the singularity of $$K$$, i.e, $$detK=0$$ and thus $$K$$ is not invertible. (See above for definition of interbile matrix) Recall that in order to find $$K^{(-1)}$$ we must first compute $$1/{(detK)}$$. If a matrix is singular then $$1/{(detK)}$$ is undefined and cannot be solved. For our case of an unconstrained structural system, there are three possible rigid body motions in 2-dimensions (2 translations and 1 rotation) for each node.

References



The blue solid line represents the undeformed truss system, while the green dashed line represents the deformed truss system.

=Contributing Team Members= The following students contributed to this report
 * Michael Berry Eml4500.f08.qwiki.berry 10:41, 8 October 2008 (UTC)
 * Andrea Booher Eml4500.f08.qwiki.booher 11:32, 8 October 2008 (UTC)
 * Chris Bishop Eml4500.f08.qwiki.bishop 13:02, 8 October 2008 (UTC)
 * David Nobles Eml4500.f08.qwiki.nobles 15:29, 8 October 2008 (UTC)
 * Eric Gatch Eml4500.f08.qwiki.gatch 12:15, 8 October 2008 (UTC)
 * Lee HarrisEml4500.f08.qwiki.harris 17:16, 8 October 2008 (UTC)