User:Eml4500.f08.wiki1.brannon

HW1
Bar Element 1: Global Free-Body Diagram

Section 13 Text Strings, Error Messages, Input
Text strings can be entered into MATLAB using single quotation marks as follows:

string='text string'

Using the disp function, text strings can be displayed in the following way:

disp('displays text string')

Error messages can be displayed with the error function:

error('error message')

which is useful with M-files, as an M-file execution will be aborted.

Within M-files, the user may be prompted to input data using the input function:

variable=input('Enter number: ')

The variable is assigned the input value and execution continues.

Section 14 Managing M-files
Creating or editing an M-file from within MATLAB can be accomplished using the !-feature. This is useful because all variables would be reset upon exiting MATLAB. System sommands such as copying, editing, and printing can be executed using this feature. As an example, !ed can be used to access the system editor:

>> !ed fileforediting.m

After editing the file, the user will return to MATLAB as it was before entering the editor.

In addition, the editor and MATLAB may be active simultaneously on systems permitting multiple processes, suspending one while working in another.

Debugging tools can be seen using the help dbtype command.

The present working directory can be viewed with the pwd command and cd can be used to change directories. The dir and ls commands list the contents of the directory and what lists the M-files within the directory only. The delete command deletes a diskfile. The type command prints a M-file to the screen.

M-files must be stored in a MATLAB-accesible directory, such as the present working directory. Typically, these files are stored in a subdirectory of the home directory titled matlab.

Section 15 Comparing Efficiency of Algorithms: Flops, Tic and Toc
Efficiency can be measured by counting the floating point operations (flops)performed or the elapsed time.

The flops count is reset using flops(0). Using the command flops again immediately after the operation gives the flops count.

flops(0), var=2+2; flops

The time elapsed for any given command can be measured using the tic command immediately before the command and toc just after it.

tic, command, toc

The processing speed available for the operation will also affect these measurements.

Section 16 Output Format
Computations within MATLAB are performed using double precision, though the format with which the results are shown can be changed with the following commands:

format short          fixed point with four decimal places (default) format long           fixed point with fourteen decimal places format short e        scientific notation with four decimal places format long e         scientific notation with fifteen decimal places format rat            approximation by ratio of small integers format hex            hexadecimal format format bank           fixed dollars and cents format +              +, -, blank

Changes to format remain until changed again.

Section 17 Hardcopy
Hardcopy is obtained using the diary command. The command below causes text that appears subsequently on the screen to be written to the named diskfile (default: diary):

diary named diskfile

The command diary on will write text to the file until the command diary off is called.

Conceptual Step of Assembly: Topology of K




\begin{bmatrix} K_{11} & K_{12} & K_{13} & K_{14} & K_{15} & K_{16}\\ K_{21} & K_{22} & K_{23} & K_{24} & K_{25} & K_{26}\\ K_{31} & K_{32} & K_{33} & K_{34} & K_{35} & K_{36}\\ K_{41} & K_{42} & K_{43} & K_{44} & K_{45} & K_{46}\\ K_{51} & K_{52} & K_{53} & K_{54} & K_{55} & K_{56}\\ K_{61} & K_{62} & K_{63} & K_{64} & K_{65} & K_{66} \end{bmatrix} = \begin{bmatrix} k_{11}^{(1)} & k_{12}^{(1)} & k_{13}^{(1)} & k_{14}^{(1)} & 0 & 0\\ k_{21}^{(1)} & k_{22}^{(1)} & k_{23}^{(1)} & k_{24}^{(1)} & 0 & 0\\ k_{31}^{(1)} & k_{32}^{(1)} & k_{33}^{(1)}+k_{11}^{(2)} & k_{34}^{(1)}+k_{12}^{(2)} & k_{13}^{(2)} & k_{14}^{(2)}\\ k_{41}^{(1)} & k_{42}^{(1)} & k_{43}^{(1)}+k_{21}^{(2)} & k_{44}^{(1)}+k_{22}^{(2)} & k_{23}^{(2)} & k_{24}^{(2)}\\ 0 & 0 & k_{31}^{(2)} & k_{32}^{(2)} & k_{33}^{(2)} & k_{34}^{(2)}\\ 0 & 0 & k_{41}^{(2)} & k_{42}^{(2)} & k_{43}^{(2)} & k_{44}^{(2)} \end{bmatrix} $$

K_{11}=k_{11}^{(1)}= {9 \over 16}$$

K_{12}=k_{12}^{(1)}= {3 \sqrt{3} \over 16}$$

K_{13}=k_{13}^{(1)}= -{9 \over 16}$$

K_{14}=k_{14}^{(1)}= -{3 \sqrt{3} \over 16}$$

K_{21}=k_{21}^{(1)}= {3 \sqrt{3} \over 16}$$

K_{22}=k_{22}^{(1)}= {3 \over 16}$$

K_{23}=k_{23}^{(1)}= -{3 \sqrt{3} \over 16}$$

K_{24}=k_{24}^{(1)}= -{3 \over 16}$$

K_{31}=k_{31}^{(1)}= -{9 \over 16}$$

K_{32}=k_{32}^{(1)}= -{3 \sqrt{3} \over 16}$$

K_{33}=k_{33}^{(1)}+k_{11}^{(2)}= {9 \over 16}+{5 \over 2}=3.0625$$

K_{34}=k_{34}^{(1)}+k_{12}^{(2)}= {3 \sqrt{3} \over 16}+-{5 \over 2}=-2.1752$$

K_{43}=k_{43}^{(1)}+k_{21}^{(2)}= {3 \sqrt{3} \over 16}+-{5 \over 2}=-2.1752$$

K_{44}=k_{44}^{(1)}+k_{22}^{(2)}= {3 \over 16}+{5 \over 2}=2.6875$$ ...

The final global stiffness matrix, K, becomes:

\begin{bmatrix} {9 \over 16} & {3 \sqrt{3} \over 16} & -{9 \over 16} & -{3 \sqrt{3} \over 16} & 0 & 0\\ {3 \sqrt{3} \over 16} & {3 \over 16} & -{3 \sqrt{3} \over 16} & -{3 \over 16} & 0 & 0\\ -{9 \over 16} & -{3 sqrt{3} \over 16} & 3.0625 & -2.1752 & -{5 \over 2} & {5 \over 2}\\ -{3 \sqrt{3} \over 16} & -{3 \over 16} & -2.1752 & 2.6875 & {5 \over 2} & -{5 \over 2}\\ 0 & 0 & -{5 \over 2} & {5 \over 2} & -{5 \over 2} & -{5 \over 2}\\ 0 & 0 & {5 \over 2} & -{5 \over 2} & -{5 \over 2} & {5 \over 2} \end{bmatrix} $$

Eml4500.f08.wiki1.brannon 20:44, 26 September 2008 (UTC)

Infinitesimal displacement (related to virtual displacement)
Global Free-Body Diagram showing Infinitesimal Displacements

$$ \overline{AC}= {|P_2^{(1)}| \over k^{(1)}}={5.1243 \over 3/4}=6.8324 $$

$$ \overline{AB}= {|P_1^{(1)}| \over k^{(2)}}={6.2762 \over 5}=1.2552 $$

Find (x,y) coordinates of B and C:

Point B:

$$ x= \overline{AB} cos{135^\circ}=1.2552*-{\sqrt 2 \over 2}=-0.8876 $$

$$ y= \overline{AB} sin{135^\circ}=1.2552*{\sqrt 2 \over 2}=0.8876 $$

Point C:

$$ x= \overline{AC} cos{30^\circ}=6.8324*{\sqrt 3 \over 2}=5.917 $$

$$ y= \overline{AC} sin{30^\circ}=6.8324*{1 \over 2}=3.4162 $$

We now have two unknowns, (XD,YD)

We need the equations for line $$\overline{AB}$$ and $$\overline{BC}$$

Method for the Determination of the Slope between Two Arbitrary Points
PQ Line Segment Diagram

$$ \overline{PQ}=(PQ) \tilde{i}=(PQ)[cos \theta \vec i +sin \theta \vec j]=(x-x_P)\vec i +(y-y_P)\vec j $$

$$ x-x_P=(PQ)cos \theta$$

$$ y-y_P=(PQ)sin \theta$$

$$ {y-y_P \over x-x_P}=tan \theta$$

$$ y-y_P=tan \theta(x-x_P)$$

Equation for line perpendicular to PQ passing through P:

$$ y-y_P=tan( \theta + {\pi \over 2})(x-x_P)$$

Determination of (XD,YD)
Line Perpendicular to Point B:

$$y-0.8876=tan{225^\circ}(x+0.8876)$$

$$y=x+1.7752$$

Line Perpendicular to Point C:

$$y-3.4162=tan{120^\circ}(x-5.917)$$

$$y=-1.732x + 13.665$$

Intersection at Point (XD,YD):

$$y_D+1.732(y_D-1.7752)=13.665$$

$$y_D=6.127$$

$$(x_D+1.7752)=-1.732x_D + 13.665$$

$$x_D=4.352$$

$$\vec {AD} = (x_D-x_A) \vec i + (y_D-y_A) \vec j$$

Simplification: $$x_A=0, y_A=0$$

$$\vec {AD} = x_D \vec i + y_D \vec j$$

$$\vec {AD} = 4.352 \vec i + 6.127 \vec j$$

$$ \vec {AD} = d_3 \vec i + d_4 \vec j$$

$$ d_3=4.352, d_4=6.127$$

3-bar Truss System
3-bar Truss System

$$ E^{(1)}=2, A^{(1)}=3, L^{(1)}=5$$

$$E^{(2)}=4, A^{(2)}=1, L^{(2)}=5$$

$$E^{(3)}=3, A^{(3)}=2, L^{(3)}=10$$

$$\theta^{(1)}=30^\circ, \theta^{(2)}=-30^\circ , \theta^{(3)}=45^\circ$$

$$P=30$$

Convenient Local Node Numbering:

3-bar Truss System Local Node Numbering

Eml4500.f08.wiki1.brannon 01:29, 8 October 2008 (UTC)

Eigenvalues and Eigenvectors of K
For the global stiffness matrix K computed previously, the eigenvectors and eigenvalues are displayed below as matrix V and D, respectively. EDU>> k

k =

0.5625   0.3248   -0.5625   -0.3248         0         0    0.3248    0.1875   -0.3248   -0.1875         0         0   -0.5625   -0.3248    3.0625   -2.1752   -2.5000    2.5000   -0.3248   -0.1875   -2.1752    2.6875    2.5000   -2.5000         0         0   -2.5000    2.5000    2.5000   -2.5000         0         0    2.5000   -2.5000   -2.5000    2.5000

EDU>> [V,D]=eig(k)

V = Mode Shapes =

0.1201  -0.5000   -0.0021    0.5951    0.6174   -0.0139    0.0693    0.8660    0.0036    0.3436    0.3565   -0.0080    0.4636    0.0000    0.0000    0.4796   -0.5409    0.5123   -0.5257    0.0000    0.0000    0.5438   -0.4330   -0.4904    0.4947    0.0030   -0.7071   -0.0321   -0.0765   -0.4984   -0.4947    0.0030   -0.7071    0.0321    0.0765    0.4984

D = Diag[Eigenvalues] =

-0.0000        0         0         0         0         0         0    0.0000         0         0         0         0         0         0    0.0000         0         0         0         0         0         0    0.0000         0         0         0         0         0         0    1.4706         0         0         0         0         0         0   10.0294

EDU>> frequency=sqrt(D)

frequency =

0 + 0.0044i       0                  0                  0                  0                  0 0            0.0000                  0                  0                  0                  0                  0                  0             0.0000                  0                  0                  0                  0                  0                  0             0.0046                  0                  0                  0                  0                  0                  0             1.2127                  0                  0                  0                  0                  0                  0             3.1669

EDU>> modeShapes=V

modeShapes =

0.1201  -0.5000   -0.0021    0.5951    0.6174   -0.0139    0.0693    0.8660    0.0036    0.3436    0.3565   -0.0080    0.4636    0.0000    0.0000    0.4796   -0.5409    0.5123   -0.5257    0.0000    0.0000    0.5438   -0.4330   -0.4904    0.4947    0.0030   -0.7071   -0.0321   -0.0765   -0.4984   -0.4947    0.0030   -0.7071    0.0321    0.0765    0.4984

  Truss System Eigenvectors Plot

These mode shapes (corresponding to the zero eigenvalues) may come out as a linear combination of the pure mode shapes. The mode shapes are in the form of pure rigid body motion or translation in the x, y, and z directions, as well as one representing the mechanism motion, the pin joint joining the two elements at the point of applied force, allowing independent motion of each element.

Eigenvectors and Eigenvalues of Modified K for Case A
$$\mathbf{\overline{k}v} = \lambda \mathbf{v}$$

$$\Rightarrow \mathbf{\overline{k}}$$ is stiffness matrix for constrained system

  Truss System for Case A

  Truss System for Case A (Local Node Numbering)

Matlab code: Modify 2-bar truss code

$$conn= \begin{bmatrix} 1&2\\ 2&3\\ 3&4 \end{bmatrix}$$

$$lmm= \begin{bmatrix} 1&2&3&4\\ 3&4&5&6\\ 5&6&7&8 \end{bmatrix}$$

$$ a = b = 1m$$

$$E = 2$$

$$A = 3$$

$$ k^{(1)}= \begin{bmatrix} 0&0&0&0\\ 0&6&0&-6\\ 0&0&0&0\\ 0&-6&0&6 \end{bmatrix},

k^{(2)}= \begin{bmatrix} 6&0&-6&0\\ 0&0&0&0\\ -6&0&6&0\\ 0&0&0&0 \end{bmatrix},

k^{(3)}= \begin{bmatrix} 0&0&0&0\\ 0&-6&0&6\\ 0&0&0&0\\ 0&6&0&-6\\ \end{bmatrix} $$



K= \begin{bmatrix} 0&0&0&0&0&0&0&0\\ 0&6&0&-6&0&0&0&0\\ 0&0&6&0&-6&0&0&0\\ 0&-6&0&6&0&0&0&0\\ 0&0&-6&0&6&0&0&0\\ 0&0&0&0&0&-6&0&6\\ 0&0&0&0&0&0&0&0\\ 0&0&0&0&0&6&0&-6\\ \end{bmatrix} $$

ka =

0    0     0     0     0     0     0     0     0     6     0    -6     0     0     0     0     0     0     6     0    -6     0     0     0     0    -6     0     6     0     0     0     0     0     0    -6     0     6     0     0     0     0     0     0     0     0    -6     0     6     0     0     0     0     0     0     0     0     0     0     0     0     0     6     0    -6

EDU>> [Va,Da]=eig(ka)

Va = Mode Shapes =

0        0         0         0    1.0000         0         0         0         0   -0.7071         0         0         0         0   -0.7071         0         0         0   -0.7071         0         0         0         0   -0.7071         0   -0.7071         0         0         0         0    0.7071         0         0         0   -0.7071         0         0         0         0    0.7071   -0.7071         0         0         0         0    0.7071         0         0         0         0         0   -1.0000         0         0         0         0    0.7071         0         0         0         0    0.7071         0         0

Da = Diag[Eigenvalues]

-12    0     0     0     0     0     0     0     0     0     0     0     0     0     0     0     0     0     0     0     0     0     0     0     0     0     0     0     0     0     0     0     0     0     0     0     0     0     0     0     0     0     0     0     0     0     0     0     0     0     0     0     0     0    12     0     0     0     0     0     0     0     0    12

  Case A Truss System Eigenvectors Plot

The global stiffness matrix K for case A shows 5 zero eigenvalues, representing 3 rigid body motions and 2 mechanism motions for the two pin joints in the truss system

Eml4500.f08.wiki1.brannon 01:56, 24 October 2008 (UTC)

Justification for Eliminating Rows 1,2,5, and 6 to Obtain $$\overline{K}$$ in 2-bar Truss
Force-Displacement Relationship: K*d=F K*d-F=0 $$\rightarrow Equation(1)$$ By the Principle of Virtual Work:W•(Kd-F)=0 for all W $$\rightarrow Equation(2)$$ $$Equation(1)\Leftrightarrow Equation(2)$$ Proof: $$(1) \rightarrow (2)$$: Trivial $$(2) \rightarrow (1)$$: Not Trivial We want to show $$(2) \rightarrow (1)$$. Choice 1: Select W such that W1=1, W2=0, W3=0, W4=0, W5=0, W6=0 W=$$ \begin{Bmatrix} 1\\ 0\\ 0\\ 0\\ 0\\ 0\end{Bmatrix}$$ W•(Kd-F)= 1.) $$[\sum_{j=1}^6 K_{1j}d_j-F_1]+0\cdot [\sum_{j=1}^6 K_{2j}d_j-F_2]+0\cdot [\sum_{j=1}^6 K_{3j}d_j-F_3]+0\cdot [\sum_{j=1}^6 K_{4j}d_j-F_4]$$$$+0\cdot [\sum_{j=1}^6 K_{5j}d_j-F_5]+0\cdot [\sum_{j=1}^6 K_{6j}d_j-F_6]$$ $$\sum_{j=1}^6 K_{1j}d_j=F_1$$   $$\rightarrow (1st Equation)$$ Choice 2: Select W such that W1=0, W2=1, W3=0, W4=0, W5=0, W6=0  W=$$ \begin{Bmatrix} 0\\1\\0\\0\\0\\0\end{Bmatrix}$$  2.) $$0\cdot [\sum_{j=1}^6 K_{1j}d_j-F_1]+[\sum_{j=1}^6 K_{2j}d_j-F_2]+0\cdot [\sum_{j=1}^6 K_{3j}d_j-F_3]+0\cdot [\sum_{j=1}^6 K_{4j}d_j-F_4]$$$$+0\cdot [\sum_{j=1}^6 K_{5j}d_j-F_5]+0\cdot [\sum_{j=1}^6 K_{6j}d_j-F_6]$$ $$\sum_{j=1}^6 K_{2j}d_j=F_2 \rightarrow (2nd Equation)$$ Choice 3: Select W such that W1=0, W2=0, W3=1, W4=0, W5=0, W6=0 W=$$ \begin{Bmatrix} 0\\0\\1\\0\\0\\0\end{Bmatrix}$$ 2.) $$0\cdot [\sum_{j=1}^6 K_{1j}d_j-F_1]+0\cdot[\sum_{j=1}^6 K_{2j}d_j-F_2]+[\sum_{j=1}^6 K_{3j}d_j-F_3]+0\cdot [\sum_{j=1}^6 K_{4j}d_j-F_4]$$$$+0\cdot [\sum_{j=1}^6 K_{5j}d_j-F_5]+0\cdot [\sum_{j=1}^6 K_{6j}d_j-F_6]$$ $$\sum_{j=1}^6 K_{3j}d_j=F_3 \rightarrow (3rd Equation)$$ Choice 4: Select W such that W1=0, W2=0, W3=0, W4=1, W5=0, W6=0 W=$$ \begin{Bmatrix} 0\\0\\0\\1\\0\\0\end{Bmatrix}$$  2.) $$0\cdot [\sum_{j=1}^6 K_{1j}d_j-F_1]+0\cdot[\sum_{j=1}^6 K_{2j}d_j-F_2]+0\cdot[\sum_{j=1}^6 K_{3j}d_j-F_3]+[\sum_{j=1}^6 K_{4j}d_j-F_4]$$$$+0\cdot [\sum_{j=1}^6 K_{5j}d_j-F_5]+0\cdot [\sum_{j=1}^6 K_{6j}d_j-F_6]$$ $$\sum_{j=1}^6 K_{4j}d_j=F_4 \rightarrow (4th Equation)$$ Choice 5: Select W such that W1=0, W2=0, W3=0, W4=0, W5=1, W6=0 W=$$ \begin{Bmatrix} 0\\0\\0\\0\\1\\0\end{Bmatrix}$$ 2.) $$0\cdot [\sum_{j=1}^6 K_{1j}d_j-F_1]+0\cdot[\sum_{j=1}^6 K_{2j}d_j-F_2]+0\cdot[\sum_{j=1}^6 K_{3j}d_j-F_3]+0\cdot [\sum_{j=1}^6 K_{4j}d_j-F_4]$$$$+[\sum_{j=1}^6 K_{5j}d_j-F_5]+0\cdot [\sum_{j=1}^6 K_{6j}d_j-F_6]$$ $$\sum_{j=1}^6 K_{5j}d_j=F_5 \rightarrow (5th Equation)$$ Choice 6: Select W such that W1=0, W2=0, W3=0, W4=0, W5=0, W6=1 W=$$ \begin{Bmatrix} 0\\0\\0\\0\\0\\1\end{Bmatrix}$$  2.) $$0\cdot [\sum_{j=1}^6 K_{1j}d_j-F_1]+0\cdot[\sum_{j=1}^6 K_{2j}d_j-F_2]+0\cdot[\sum_{j=1}^6 K_{3j}d_j-F_3]+0\cdot [\sum_{j=1}^6 K_{4j}d_j-F_4]$$$$+0\cdot [\sum_{j=1}^6 K_{5j}d_j-F_5]+[\sum_{j=1}^6 K_{6j}d_j-F_6]$$ $$\sum_{j=1}^6 K_{6j}d_j=F_6 \rightarrow (6th Equation)$$ $$\begin{bmatrix} K_{11}&\cdots&K_{16}\\ \vdots&\ddots&\vdots\\ K_{61}&\cdots&K_{66}\end{bmatrix}* \begin{Bmatrix} d_1\\d_2\\d_3\\d_4\\d_5\\d_6\end{Bmatrix}= \begin{Bmatrix} F_1\\F_2\\F_3\\F_4\\F_5\\F_6\end{Bmatrix}$$ Hence, Kd=F (or Equation (1) from above)

Principle of Virtual Work, Relating to W
The Principle of Virtual Work accounts for the boundary conditions.

For the two-bar truss system, d1=d2=d5=d6=0 Weighting coefficients must be "kinematically admissible". In other words, the weighting coefficients cannot violate the boundary conditions.

Weighting Coefficients$$\equiv$$Virtual Displacements $$W_1=W_2=W_5=W_6=0$$ W•(Kd-F) $$\begin{Bmatrix}W_3\\W_4\end{Bmatrix}\cdot (\bar{\mathbf{K}} \bar{\mathbf{d}} -\bar{\mathbf{F}})=0$$ for all $$\begin{Bmatrix}W_3\\W_4\end{Bmatrix}$$ (3) $$\bar{\mathbf{K}}=\begin{bmatrix}K_{33}&K_{34}\\K_{43}&K_{44}\end{bmatrix}\qquad \bar{\mathbf{d}}=\begin{Bmatrix}d_3\\d_4\end{Bmatrix}\qquad \bar{\mathbf{F}}=\begin{Bmatrix}F_3\\F_4\end{Bmatrix}$$

$$\begin{Bmatrix}W_3=n\\W_4=n\end{Bmatrix}\cdot (\begin{bmatrix}K_{33}&K_{34}\\K_{43}&K_{44}\end{bmatrix} \begin{Bmatrix}d_3\\d_4\end{Bmatrix} -\begin{Bmatrix}F_3\\F_4\end{Bmatrix})=0$$

Because the $$\begin{bmatrix}K_{33}&K_{34}\\K_{43}&K_{44}\end{bmatrix} \begin{Bmatrix}d_3\\d_4\end{Bmatrix}$$ term is equal to $$\begin{Bmatrix}F_3\\F_4\end{Bmatrix}$$, $$\begin{bmatrix}K_{33}&K_{34}\\K_{43}&K_{44}\end{bmatrix} \begin{Bmatrix}d_3\\d_4\end{Bmatrix} -\begin{Bmatrix}F_3\\F_4\end{Bmatrix}=\begin{Bmatrix}0\\0\end{Bmatrix}$$ $$\begin{bmatrix}K_{33}&K_{34}\\K_{43}&K_{44}\end{bmatrix} \begin{Bmatrix}d_3\\d_4\end{Bmatrix}=\begin{Bmatrix}F_3\\F_4\end{Bmatrix}$$(4)

Part 4
The conversion from axial forces and element forces in global coordinates is as follows: $$\mathbf{f}^{(e)}=\begin{pmatrix}f_{1}^{(e)}\\f_{2}^{(e)}\\f_{3}^{(e)}\\f_{4}^{(e)}\\f_{5}^{(e)}\\f_{6}^{(e)}\end{pmatrix}$$ in global coordinates $$\mathbf{\hat{k}^{(e)}q^{(e)}=p^{(e)}}={EA \over L} \begin{bmatrix}1&-1\\-1&1\end{bmatrix}\begin{pmatrix} q_1^{(e)}\\q_2^{(e)} \end{pmatrix}= \begin{pmatrix}p_1^{(e)}\\p_2^{(e)}\end{pmatrix}$$ in axial coordinates $$\mathbf{q^{(e)}=T^{(e)}d^{(e)}}$$ $$\mathbf{\hat{k}^{(e)}T^{(e)}d^{(e)}=p^{(e)}}$$ $${EA \over L} \begin{bmatrix}1&-1\\-1&1\end{bmatrix}\begin{pmatrix} l^{(e)} & m^{(e)} & n^{(e)}  & 0 & 0 & 0\\ 0 & 0 &0 & l^{(e)}  & m^{(e)} & n^{(e)} \end{pmatrix}\begin{pmatrix} d_1^{(e)}\\ d_2^{(e)}\\ d_3^{(e)}\\ d_4^{(e)}\\ d_5^{(e)}\\ d_6^{(e)} \end{pmatrix}=\begin{pmatrix}p_1^{(e)}\\p_2^{(e)}\end{pmatrix}$$ Multiplying both sides by T(e)T as completed in the 2D case, $$\mathbf{T^{(e)T}\hat{k}^{(e)}T^{(e)}d^{(e)}=T^{(e)T}p^{(e)}}$$ $$\begin{pmatrix} l^{(e)} & 0\\m^{(e)} & 0\\n^{(e)}  & 0\\ 0 &l^{(e)}\\0&m^{(e)}\\ 0&n^{(e)} \end{pmatrix} {EA \over L} \begin{bmatrix}1&-1\\-1&1\end{bmatrix} \begin{pmatrix} l^{(e)} & m^{(e)} & n^{(e)}  & 0 & 0 & 0\\ 0 & 0 &0 & l^{(e)}  & m^{(e)} & n^{(e)} \end{pmatrix} \begin{pmatrix} d_1^{(e)}\\ d_2^{(e)}\\ d_3^{(e)}\\ d_4^{(e)}\\ d_5^{(e)}\\ d_6^{(e)} \end{pmatrix}=\begin{pmatrix} l^{(e)} & 0\\m^{(e)} & 0\\n^{(e)}  & 0\\ 0 &l^{(e)}\\0&m^{(e)}\\ 0&n^{(e)} \end{pmatrix} \begin{pmatrix}p_1^{(e)}\\p_2^{(e)}\end{pmatrix} $$ and with $$\mathbf{k^{(e)}=T^{(e)T}\hat{k}^{(e)}T^{(e)}},\qquad\mathbf{f^{(e)}=T^{(e)T}p^{(e)}}$$ We have $$\mathbf{(T^{(e)T}\hat{k}^{(e)}T^{(e)})d^{(e)}=(T^{(e)T}p^{(e)})}$$ or $$\mathbf{k^{(e)}d^{(e)}=f^{(e)}}$$ $$\begin{pmatrix}f_{1}^{(e)}\\f_{2}^{(e)}\\f_{3}^{(e)}\\f_{4}^{(e)}\\f_{5}^{(e)}\\f_{6}^{(e)}\end{pmatrix}=\begin{pmatrix} l^{(e)} & 0\\m^{(e)} & 0\\n^{(e)}  & 0\\ 0 &l^{(e)}\\0&m^{(e)}\\ 0&n^{(e)} \end{pmatrix}\begin{pmatrix}p_1^{(e)}\\p_2^{(e)}\end{pmatrix}$$

Eml4500.f08.wiki1.brannon 18:40, 7 November 2008 (UTC)

HW6
Bar Broken Up Into Elements

<p style="text-align:center;">Single Element Function Interpolation

Assume $$u(x) \,$$ for $$x_i \le x \le x_{i+1} \,$$

Motivation for Linear Interpolation of $$u(x) \,$$
2-bar Truss

<p style="text-align:center;">2-Bar Truss Deformation

Rube Goldberg Machine
[Rube Goldberg Machine]

Honesty, Imagination, and Ethics
[Honesty],[Imagination], and [Ethics]

[The University of Florida Honor Code] We, the members of the University of Florida community, pledge to hold ourselves and our peers to the highest standards of honesty and integrity.

The University of Florida Honor Pledge

"On my honor, I have neither given nor received unauthorized aid in doing this assignment.”

Continuous Principle of Virtual Work to Discrete Case (continued)
Lagrangian Interpolation

Motivation for Form of $$N_i(x) \,$$ and $$N_{i+1}(x) \,$$:

$$N_i(x) \,$$ and $$N_{i+1}(x) \,$$ are linear (straight lines), thus any linear combination of $$N_i(x) \,$$ and $$N_{i+1} \,$$ is also linear, and in particular, the expression for $$u(x) \,$$.

$$N_i(x)=\alpha_i + \beta_i x \,$$, with $$\alpha_i \,$$, $$\beta_i \,$$ numbers

$$N_{i+1}(x)=\alpha_{i+1}+\beta_{i+1}x \,$$, with $$\alpha_{i+1} \,$$, $$\beta_{i+1} \,$$ numbers

Linear Combination of $$N_i \,$$ and $$N_{i+1} \,$$:

$$N_id_i+N_{i+1}d_{i+1}=(\alpha_i+\beta_ix)d_i+(\alpha_{i+1}+\beta_{i+1}x)d_{i+1} \,$$ $$\quad=(\alpha_id_i+\alpha_{i+1}d_{i+1})+(\beta_id_i+\beta_{i+1}d_{i+1})x \,$$ is clearly linear.

Recall equation for $$u(x) \,$$ (interpolation of $$u(x) \,$$):

$$u(x_i)=N_i(x_i)d_i+N_{i+1}(x_i)d_{i+1} \,$$ $$u(x_i)={x_{i+1}-x_i \over x_{i+1}-x_i}d_i+{x_i-x_i \over x_{i+1}-x_i}d_{i+1} \,$$ $$N_i(x_i)=1 \,$$ $$N_{i+1}(x_i)=0 \,$$ $$u(x_i)=d_i \,$$

Using the same logic:

$$u(x_{i+1})=d_{i+1} \,$$

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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)<p style="text-align:right;">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



$$v(\tilde{x})=N_2(\tilde{x})\tilde{d}_2+N_3(\tilde{x})\tilde{d}_3+N_5(\tilde{x})\tilde{d}_5+N_6(\tilde{x})\tilde{d}_6$$<p style="text-align:right;">Equation (2)

Recall:$$u(\tilde{x})=N_1(\tilde{x})\tilde{d}_1+N_4(\tilde{x})\tilde{d}_4$$<p style="text-align:right;">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)}