User:Eml4500.f08.qwiki.bishop/Lecture 10

 Comment: The pictures that were previously used in this homework submission have been changed. The previous images ( and ) were not originally authored by Team Qwiki, however I did not intentionally plagiarize Team Echo's work. 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. The correct images that were authored by Team Qwiki are included in this updated version of Homework 2.

A comparison between the current version of Homework 2 and our prior submission can be found here. Eric GatchEml4500.f08.qwiki.gatch 18:33, 13 November 2008 (UTC)

Elimination of Known Degree's of Freedom
If d1 = d2 = d5 = d6 = 0, then rows 1, 2, 5 and 6 can be eliminated from the global stiffness matrix. This is possible because when the stiffness matrix is multiplied by a displacement matrix containing those zero values, the respective values for the force components will equal zero. So. ..

$$	\mathbf{D} = \begin{Bmatrix} d_1=0\\d_2=0\\d_3\\d_4\\d_5=0\\d_6=0 \end{Bmatrix}$$

Respective Global Stiffness Matrix
Now. ..

$$	\mathbf{K} = \begin{vmatrix} k_{13} & k_{14} \\ k_{23} & k_{24}\\k_{33} & k_{34}\\k_{43} & k_{44}\\k_{53} & k_{54}\\k_{63} & k_{64}\end{vmatrix}$$

Then. . . by the Principle of Virtual Work (PVW), we delete the Global Stiffness Matrix rows corresponding to those rows that equal zero in the Global Stiffness Matrix.

Therefore, the Global Stiffness Matrix, K, reduces to

$$	\mathbf{K} = \begin{vmatrix} k_{33} & k_{34}\\k_{43} & k_{44}\end{vmatrix}$$

Resulting Overall Force Displacement Relationship
$$\begin{bmatrix} k_{33} & k_{34} \\ k_{43} & k_{44} \end{bmatrix} \begin{Bmatrix} d_3\\d_4\end{Bmatrix}=\begin{Bmatrix} F_3\\F_4\end{Bmatrix}=\begin{Bmatrix} 0\\P\end{Bmatrix}$$

What is the inverse of K?
Find the determinant of K:

$$det \mathbf{K}=k_{33}k_{44}-k_{34}k_{43}$$

Then the inverse of K, K-1, equals. ..

$$\mathbf{K}^{-1}=\frac{1}{det\mathbf{K}}\begin{bmatrix} k_{44} & -k_{34} \\ -k_{43} & k_{33} \end{bmatrix}$$

Verification of K-1

$$\mathbf{K}\mathbf{K}^{-1}=\mathbf{K}^{-1}\mathbf{K}=\mathbf{I}=\begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}$$

Note. ..

$$\mathbf{K}^T=\begin{bmatrix} k_{33} & k_{43} \\ k_{34} & k_{44} \end{bmatrix}$$

So. ..

$$\mathbf{K}^{-1}\ne \ \frac{1}{det(\mathbf{K})}\mathbf{K}^T$$

Now we go back to the 2 bar truss problem, for which the solution is:

$$\begin{Bmatrix} d_3 \\ d_4 \end{Bmatrix}=\mathbf{K}^{-1}\begin{Bmatrix} 0 \\ P \end{Bmatrix}=\begin{Bmatrix} 4.352 \\ 6.1271 \end{Bmatrix}$$

Computing the Reactions
There are two different methods for finding the reactions:

Method 1

Use the Element Force Displacement Relationship

For this case, the element stiffness matrix is known and the element displacements are known. Some of the element forces are known.

Element 1

$$\mathbf{k}^{(1)}\mathbf{d}^{(1)}=\mathbf{f}^{(1)}$$



$$d^{(1)}=\begin{Bmatrix} 0 \\ 0 \\ 4.352 \\ 6.1271 \end{Bmatrix}$$

Element 2

$$\mathbf{k}^{(2)}\mathbf{d}^{(2)}=\mathbf{f}^{(2)}$$



$$d^{(2)}=\begin{Bmatrix} 4.352 \\ 6.1271 \\ 0 \\ 0 \end{Bmatrix}=\begin{Bmatrix} d_1^{(2)} \\ d_2^{(2)} \\ d_3^{(2)} \\ d_4^{(2)} \end{Bmatrix}$$