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text: File:6 bar truss system setup.jpg

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title: File:6 bar truss system setup.jpg


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User:Eml4500.f08.Ateam.philbin/HW5

Homework 5

Mtg. 28 10-31-08

Example of a Composite Material
Some composite materials are examples of how the modulus of elasticity can be a function of x. One example is reinforced concrete walls. The modulus of the concrete is significantly different than that of the reinforcing steel. In order to accurately model this material the modulus would change as one moved through each material.

source: Rebar

Another material that has a modulus that changes with x is an aluminum composite panel which consists of two aluminum sheets sandwiched around a non-aluminum material.

source: (Aluminium_composite_panel)

Partial Differential Method

A very small slice of the beam shall be considered of width dx. The force balance in the x direction gives the following result.

$$ \sum F_x=0= -N(x,t) + N(x+dx,t) +f(x,t)dx - m(x)\ddot{u}dx$$

$$=\frac{\partial {N}}{\partial{x}}(x,t)dx + higher\ order\ terms + f(x,t)dx-m(x)\ddot{u}dx\quad (Eqn. 1)$$

These higher order terms come from the Taylor Series expansion, shown in the general form here.

$$f(x+dx)=f(x)+\frac{df(x)}{dx}dx+\frac{1}{2}\frac{d^2f(x)}{dx^2}dx^2...$$

Where all the $$dx^2...$$ and higher terms are neglected due to their small effect on the result. After those simplifications are made, Eqn. 1 can be written in the following form.

$$\frac{\partial{N}}{\partial{x}}+f(x,t)=m(x)\ddot{u}\quad (Eqn. 2)$$

In this form, it is easy to see that this is the equation of motion. In order to get to a more usable form, N(x,t) needs to be rewritten into its constituent components.

$$N(x,t)=A(x)\ \sigma(x,t)$$

$$\sigma(x,t)=E(x)\ \epsilon(x,t)$$

$$\epsilon(x,t)=\frac{\partial{u}}{\partial{x}}(x,t)$$

Using these relationships, N(x,t) can be written as follows.

$$N(x,t)=[A(x)\ E(x)\ \frac{\partial{u}}{\partial{x}}(x,t)]$$

Now Eqn. 1 can be rewritten in the final form.

$$\frac{\partial{ }}{\partial{x}}[A(x)\ E(x)\ \frac{\partial{u}}{\partial{x}}(x,t)]+f(x,t)=m(x)\ddot{u}$$

This is the final form of the equation of motion that will be used to solve finite element problems with the partial derivative method.

Initial and Boundary Conditions
In order to solve these problems, certain aspects of the conditions must be known. For the following case, two boundary conditions are known.

 $$u(0,t)=0\ and\ u(L,t)=0$$

In another example, two initial conditions are known.

$$u(0,t)=0\ and\ N(L,t)=A(L)\sigma(L,t)=F(t)$$

3-bar Space Truss Statics

The three bar space truss was found to be statically determinate. There are three unknowns and three equations to use, summing the forces in x, y, and z directions. By using these three equations, the following relationships are made.

$$\sum{F_{x}}= F_{1}l^{(1)} + F_{2}l^{(2)} = 0$$

$$\sum{F_{y}}= F_{1}m^{(1)} + F_{2}m^{(2)} = -P$$

$$\sum{F_{z}}= F_{1}n^{(1)} + F_{2}n^{(2)} + F_{3}n^{(3)} = 0$$

Plugging in P = 20,000 N, gives the following forces along each element.

$$F_{1} = 20375\ N$$

$$F_{2} = 13124\ N$$

$$F_{3} = -23149\ N$$

In order to solve for the axial stress and strain, the following equations will be used.

$$\sigma_{n} = \frac{F_{n}}{A_{n}}$$

$$\epsilon_{n} = \frac{\sigma_{n}}{E_{n}}$$

Plugging in the given values for the Young's Modulus and area for each element yields these values. (Sigma is expressed in MPa and epsilon is expressed in mm.)

$$\sigma_{1} = \frac{20375 N}{200\ mm^2}= 101.875\qquad \qquad \epsilon_{1} = \frac{\sigma_{1}}{200\ GPa}= 0.5094$$

$$\sigma_{2} = \frac{13214 N}{200\ mm^2}= 66.07\qquad \qquad \ \ \ \epsilon_{2} = \frac{\sigma_{2}}{200\ GPa}= 0.33035$$

$$\sigma_{3} = \frac{-23149 N}{600\ mm^2}= -38.58\qquad \qquad \epsilon_{3} = \frac{\sigma_{3}}{200\ GPa}= -0.1929$$

These values for the displacement of each element matches the results from the MATLAB confirming the method.

title: User:Eml4500.f08.Ateam.philbin/HW5


title: File:2bcs.JPG


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User:Eml4500.f08.ateam.didomenico/hw 5

Essential MATLAB Functions
Similar to the five-bar truss system, the following plane truss system MATLAB programs require additional functions to run. These functions are found at the course textbook website, Fundamental Finite Element Analysis and Applications. They include,  , and. The details of these functions are outlined below.

This function generates the element stiffness matrix for each element. When the function is called it calculates the X and Y coordinate positions of the element ends from the coordinates passed in. It produces the element length from this information. Next it calculates the direction cosines of the element. Using the Young's modulus and cross-sectional area of the element, the element stiffness matrix, k, is assembled. The stiffness matrix is passed back to the.

This function solves the force-displacement relationship for each node, resulting in the displacements and reactions at each node. This information is passed back to the.

This function outputs the axial strain, axial stress, and axial force of each element.

Two-Bar Truss System: MATLAB Code Debugging
In Homework Report #2, the team analyzed the MATLAB code for a two-bar truss system. This code was provided by Dr. Vu-Quoc and can be found here. After the homework report submission deadline, a bug was found in the provided code. This bug occurs during the building of the  array. Instead of passing the proper Young's modulus,, and cross-sectional area,  , indicies into the   function, the program passes in the entire   and   arrays. This is sufficient for cases in which all truss elements have an identical cross-sectional area and Young's modulus, but fails when these properties vary among elements.

Note: Applied load P = 7

To fix the bug, a simple modification was made to the following program segment:

Where the  array is assembled the variables   and   are replaced by the array indices   and   respectively.

was modified to be

The resulting MATLAB program code is as follows:

Running
Running this code produces the following output: >> twobar_debug k_local = 0.75       -0.75         -0.75         0.75 k = 0.5625     0.32476      -0.5625     -0.32476       0.32476       0.1875     -0.32476      -0.1875        -0.5625     -0.32476       0.5625      0.32476      -0.32476      -0.1875      0.32476       0.1875 k_local = 5   -5     -5     5 k = 2.5        -2.5         -2.5          2.5          -2.5          2.5          2.5         -2.5          -2.5          2.5          2.5         -2.5           2.5         -2.5         -2.5          2.5 K = 0.5625     0.32476      -0.5625     -0.32476            0            0       0.32476       0.1875     -0.32476      -0.1875            0            0       -0.5625     -0.32476       3.0625      -2.1752         -2.5          2.5      -0.32476      -0.1875      -2.1752       2.6875          2.5         -2.5             0            0         -2.5          2.5          2.5         -2.5             0            0          2.5         -2.5         -2.5          2.5 R = 0     0      0      7      0      0 d = 0            0         4.352        6.1271             0             0 reactions = -4.4378      -2.5622        4.4378       -4.4378 results = 1.7081      5.1244       5.1244        0.6276        3.138        6.276

Compared to the previously obtained  array shown below, this new   array makes much more sense. The previous array contained additional meaningless values along with the desired values.

Previous  array:

results = 1.7081      5.1244       8.5406       5.1244       17.081        0.6276       1.8828        3.138       1.8828        6.276

Verifying Array Accuracy
The new  array values correspond to the table shown below.

The  array for finite element analysis of the truss system is verified using principles of statics by simply summing the forces in the X and Y directions and setting these equations equal to zero. This produces two equations with two unknown values from which the axial forces of each element are derived. From here, using the element properties, the element stresses and strains can be determined. This is shown below.

$$\sum{F_{y}} = 7 - F_1\cdot sin(30^{\circ} ) - f_2\cdot sin(45^{\circ}) = 0$$

$$\sum{F_{x}} = F_1\cdot cos(30^{\circ} ) - F_2\cdot cos(45^{\circ}) = 0$$

Where $$F_1$$ is the axial force of Element 1 and $$F_2$$ is the axial force of Element 2.

Solving these equations simultaneously yields the element axial forces.

$$F_1 = 5.1244$$

$$F_2 = 6.2760$$

Now the axial stress for each element can be computed using the axial force and cross-sectional area for each element.

$$\sigma_1 = \frac{F_1}{A_1} = \frac{5.1244}{1.00} = 5.1244$$

$$\sigma_2 = \frac{F_2}{A_2} = \frac{6.2760}{2.00} = 3.1380$$

The values for axial stress can now be used along with the Young's modulus to determine the axial strain for each element.

$$\epsilon_1 = \frac{\sigma_1}{E_1} = \frac{5.1244}{3.00} = 1.7081$$

$$\epsilon_2 = \frac{\sigma_2}{E_2} = \frac{3.1380}{5.00} = 0.6276$$

These values are almost identical to the values obtained through finite element analysis. The MATLAB results have been verified to be accurate. For this case, principles of statics proved to be a much more efficient method to solve the problem. This is not the case as mesh complexity increases.

Six-Bar Truss System: MATLAB Finite Element Analysis
The problem to be evaluated in this section is found as Example 4.1 on pg.226 of the course textbook. The MATLAB code used to evaluate the problem is from the  file found at the Student Companion Site for the class textbook.

Additionally, a modified version of the problem is analyzed accordingly in this section.

The load $$P = 20 kN$$ acts at an angle $$\alpha = 30^\circ$$ for each version of the problem.

Original Problem: Example 4.1
Analysis of the original six-bar truss system is important before comparing to the modified six-bar truss system. The following code is from the  file found at the website for the class textbook.

Running
Running this code produces the following results.

>> SixBarTrussEx K = Columns 1 through 6 85355       35355       -50000            0            0            0         35355        35355            0            0            0            0        -50000            0        85355       -35355            0            0             0            0       -35355  1.0202e+005            0            0 0           0            0            0        71554       -35777             0            0            0            0       -35777        17889             0            0            0            0            0            0             0            0            0       -66667            0            0        -35355       -35355       -35355        35355       -71554        35777        -35355       -35355        35355       -35355        35777       -17889   Columns 7 through 10 0           0       -35355       -35355             0            0       -35355       -35355             0            0       -35355        35355             0       -66667        35355       -35355             0            0       -71554        35777             0            0        35777       -17889         71554        35777       -71554       -35777         35777        84555       -35777       -17889        -71554       -35777  2.1382e+005            0 -35777      -17889            0  1.0649e+005 R = 0            0         10000         17321             0             0             0             0             0             0 d  = 0            0       0.21311       0.24998             0             0             0             0    -0.0060971      0.012242 reactions = -10873      -217.27        874.27       -437.13       -1.7279        -16666 results = 5.3276e-005      10.655        10655 -4.6334e-006    -0.92669      -926.69 -4.8873e-006    -0.97746      -977.46 -8.3326e-005     -16.665       -16665 1.5363e-006     0.30727       307.27 -9.659e-009  -0.0019318      -1.9318

The  array values correspond to the table shown below.

The original displacements are shown in the following matrix. It is important to note that the nodal positions were magnified by a factor of 1000 in the original  file.

$$\begin{Bmatrix} d_1 \\ d_2 \\ d_3 \\ d_4 \\ d_5 \\ d_6 \\ d_7 \\ d_8 \\ d_9 \\ d_{10} \\ \end{Bmatrix}$$ = $$\begin{Bmatrix} 0 \\           0 \\   0.21311 \\   0.24998 \\            0 \\            0 \\            0 \\            0 \\   -0.0060971 \\     0.012242 \\ \end{Bmatrix}$$

Plotting Example 4.1
To plot the undeformed and deformed shapes, the following code was appended to the original  file. For viewing, the nodal magnification factor in the original file was removed and the resulting nodal displacements were multiplied by a factor of 2500.

Modified Problem: Varying Young's Modulus
This modification of Example 4.1 involves varying the Young's modulus, E, of the elements as shown in the table below. This requires re-writing the  code to allow for an array of E values. The subsequent MATLAB file is called.

To allow for the passing of array indices, the MATLAB code was modified as shown. Much like fixing the bug in the two-bar truss system code,  replaces   inside the   loop that builds the stiffness matrices and the results array.

Running
The results obtained by running the code are as follows. Again, it is important to note that the nodal positions are multiplied by a factor of 1000. >> sixbarmod K = Columns 1 through 6 76391       38891       -37500            0            0            0         38891        38891            0            0            0            0        -37500            0        69320       -31820            0            0             0            0       -31820        98486            0            0             0            0            0            0        71554       -35777             0            0            0            0       -35777        17889             0            0            0            0            0            0             0            0            0       -66667            0            0        -38891       -38891       -31820        31820       -71554        35777        -38891       -38891        31820       -31820        35777       -17889   Columns 7 through 10 0           0       -38891       -38891             0            0       -38891       -38891             0            0       -31820        31820             0       -66667        31820       -31820             0            0       -71554        35777             0            0        35777       -17889         89443        44721       -89443       -44721         44721        89027       -44721       -22361        -89443       -44721  2.3171e+005        16015 -44721      -22361        16015  1.1096e+005 R = 0            0         10000         17321             0             0             0             0             0             0 d = 0            0       0.26485       0.26083             0             0             0             0    0.00063864     -0.001246 reactions = -9908.3       23.619       -90.274        45.137       -1.4008        -17389 results = 6.6213e-005      9.9319       9931.9 5.347e-007    0.096245       96.245 5.0465e-007     0.10093       100.93 -8.6943e-005     -17.389       -17389 -1.5183e-007   -0.033402      -33.402 -6.2645e-009  -0.0015661      -1.5661

The  array values correspond to the table shown below.

The original displacements are shown in the following matrix. It is important to note that the nodal positions were magnified by a factor of 1000 in the  file.

$$\begin{Bmatrix} d_1 \\ d_2 \\ d_3 \\ d_4 \\ d_5 \\ d_6 \\ d_7 \\ d_8 \\ d_9 \\ d_{10} \\ \end{Bmatrix}$$ = $$\begin{Bmatrix} 0 \\           0 \\      0.26485 \\      0.26083 \\            0 \\            0 \\            0 \\            0 \\   0.00063864 \\    -0.001246 \\ \end{Bmatrix}$$

Plotting Example 4.1 and the Modified Six-Bar Truss System
To plot the undeformed and deformed shapes, the following code was appended to the original  file. The subsequent file was plotted over the existing plot for Example 4.1 to show how varying Young's modulus throughout the truss system can affect deformation. Subsequent deformation clearly differed from that of elements with identical Young's modulus. However, deformation did not vary greatly.

The undeformed truss system is represented by the blue dashed line. The deformed truss system of Example 4.1 is represented by the solid red line. The deformed modified truss system is represented by the green dotted line. For viewing, the nodal magnification factor in the original files was removed and the resulting nodal displacements were multiplied by a factor of 2500.

Analysis of Three-Bar Space Truss System: Example 4.2
The truss system in Example 4.2 is a pin-jointed structure with three elements. Much like problems solved earlier, the element properties vary in this truss system and are displayed below. This system adds another dimension to the FEA and results in a total of 12 degrees of freedom. Nodal displacements for this problem are denoted by,    and   in the  ,    and   directions, respectively.

Before implementing FEA to solve the problem, it is important to properly set up the  and   arrays. $$= \begin{bmatrix} 1&4\\ 2&4\\ 3&4 \end{bmatrix}$$

$$= \begin{bmatrix} 1 & 2 & 3 & 10 & 11 & 12\\ 4 & 5 & 6 & 10 & 11 & 12\\ 7 & 8 & 9 & 10 & 11 & 12 \end{bmatrix} $$

The  file for this problem was found on the Student Companion Site for the course textbook. It is as follows.

Additional MATLAB Functions
These functions were found on the Student Companion Site for the course textbook. They behave similarly to their corresponding functions for plane trusses detailed in the previous section.

Note: The program also uses, detailed in the previous section.

Running
Running this code produces the following results. Note that the nodal positions are multiplied by a factor of 1000 for simplicity. >> ThreeBarSpaceTrussEx K = Columns 1 through 6 1459.7      2919.3      -3040.9            0            0            0        2919.3       5838.6      -6081.9            0            0            0       -3040.9      -6081.9       6335.3            0            0            0             0            0            0       3566.7      -3566.7       4953.8             0            0            0      -3566.7       3566.7      -4953.8             0            0            0       4953.8      -4953.8       6880.3             0            0            0            0            0            0             0            0            0            0            0            0             0            0            0            0            0            0       -1459.7      -2919.3       3040.9      -3566.7       3566.7      -4953.8       -2919.3      -5838.6       6081.9       3566.7      -3566.7       4953.8        3040.9       6081.9      -6335.3      -4953.8       4953.8      -6880.3   Columns 7 through 12 0           0            0      -1459.7      -2919.3       3040.9             0            0            0      -2919.3      -5838.6       6081.9             0            0            0       3040.9       6081.9      -6335.3             0            0            0      -3566.7       3566.7      -4953.8             0            0            0       3566.7      -3566.7       4953.8             0            0            0      -4953.8       4953.8      -6880.3             0            0            0            0            0            0             0            0            0            0            0            0             0            0        60000            0            0       -60000             0            0            0       5026.4      -647.45       1912.9             0            0            0      -647.45       9405.4       -11036             0            0       -60000       1912.9       -11036        73216 d  = 0            0             0             0             0             0             0             0             0      -0.18705        -2.592       -0.3858 reactions = 6666.7        13333        -13889       -6666.7        6666.7       -9259.3             0             0         23148 results = 0.00050936      101.87        20375    0.00033036       66.072        13214    -0.0001929       -38.58       -23148

The  array values correspond to the table shown below.

The displacement matrix is as follows.

$$\begin{Bmatrix} d_1 \\ d_2 \\ d_3 \\ d_4 \\ d_5 \\ d_6 \\ d_7 \\ d_8 \\ d_9 \\ d_{10} \\ d_{11} \\ d_{12} \\ \end{Bmatrix}$$ = $$\begin{Bmatrix} 0 \\        0 \\         0 \\         0 \\         0 \\         0 \\         0 \\         0 \\         0 \\   -0.1871 \\   -2.5920 \\   -0.3858 \\ \end{Bmatrix}$$

The reactions are as follows.

$$\begin{Bmatrix} R_1 \\ R_2 \\ R_3 \\ R_4 \\ R_5 \\ R_6 \\ R_7 \\ R_8 \\ R_9 \\ \end{Bmatrix}$$ = $$\begin{Bmatrix} 6667 \\  13333 \\  -13889 \\   -6667 \\    6667 \\   -9259 \\       0 \\       0 \\   23148 \\ \end{Bmatrix}$$

MATLAB Plots of Example 4.2
To plot the undeformed and deformed states of the three-bar space truss system the following code was appended to the original  file. Each plotting view was manipulated within the MATLAB plot editor.

Perspective View: (1, 1, 1)
To make the deformation more clear the following perspective view was included.

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User talk:Suneelreddyvanga

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Organic chemistry/Hybridization and Molecular Orbital Theory

Hybridization and Molecular Orbital (MO) Theory are ways of describing atomic bonding in atoms and molecules. Hybridization, developed by Linus Pauling, explains bonding through electron orbitals combining to form hybrid orbitals. MO theory focuses on the bonds between two or more atoms, and explains how the characteristics of electron orbital overlap.


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Computer Architecture Lab/WS2008/Group2/Documentation

Members of Group 2, Sweet16

 * Franz-Josef Katzdobler, 0425195
 * Daniel Reichhard, 0025792
 * Stefan Resch, 0425306
 * Matthias Wenzl, 0425388

Assembler
The man page:

S16ASM(1)                       User Manuals                        S16ASM(1) NAME s16asm - sweet 16 assembler SYNOPSIS s16asm [-b | h | l | d | r] [-n][-o outfile ] [ infile ...] DESCRIPTION s16asm Transforms the given assembler source files to the selected des- tination format in the given order. If no infile is given, stdin  will be used. The default destination format is raw. OPTIONS -b    Output bytes in ASCII as binary number. -h    Output bytes in ASCII as hex number in format of hexdump. -l    Output as listing. -r    Output as vhdl rom description. -d    Output in format for bootloader. -n    Supress warnings. -o outfile Place output to file outfile. Otherwise use stdout. BUGS Currently none know. Please send a mail when you find one. AUTHOR Stefan Resch Matthias Wenzl Daniel Reichhard Franz Katzdobler Sweet16                         NOVEMBER 2008                       S16ASM(1)

Assembler program for illustration:

Pipelining
The pipeline itself consists of 4 Stages:
 * Fetch
 * Decode
 * Execute
 * Memory Stage

You can see the functional description [here].

The interaction with the buttons, leds and all kinds of memories is managed by a memory interface unit. Following memory mapping is applied:


text:
User talk:Eml4500.f08.a-team.robinson

pagemove

 * To see what is in your userspace click here

Here are two wikitext scripts that you can use to keep track of subpages in your userspace.

creates a page called My subpages

creates a list of all subpages in your userspace.

Keep in mind that these two scripts here on this talk page, and on your user page User:. They won't work on most Wikiversity pages.--Guy vandegrift (discuss • contribs) 14:49, 1 March 2024 (UTC)

title: User talk:Eml4500.f08.a-team.robinson


text:
Category talk:BCP/SEPA/11

 notcategory=BCP/SEPA/11 category=BCP/SEPA/10

 category=BCP/SEPA/11 category=BCP/Using old color field

title: Category talk:BCP/SEPA/11


text:
User:Eml4500.f08.Lulz.js/HW5

Justification of Reduction of Stiffness Matrix for 2-bar Truss Problem
From the general force displacement relationship, the following relation can be derived:

 * $$\mathbf{kd}= \mathbf{F}$$ => $$\mathbf{kd - F}= \mathbf{0}\;\;(1)$$

Using this relation, the Principle of Virtual Work can be expressed as follows:

 * $$\mathbf{W}\cdot\mathbf{(kd - F)} = \mathbf{0}$$ for all $$\mathbf{W}\;\;(2)$$

Where $$\mathbf{W}\;$$ refers to the weighting matrix. It is useful to show the equivalency of the above two expressions. While is is trivial to show how equation 2 comes from equation 1, it is more challenging to show how equation 1 comes from equation 2. In order to prove that this is the case, consider the following weighting coefficients:

 * $$W_{1} = 1\;$$, $$\;W_{2}=...=W_{6} = 0$$

Which can be expressed in transposed form as:

 * $$\mathbf{W^{T}} = \begin{bmatrix} 1&0&0&0&0&0\end{bmatrix}$$

Applying this choice for $$\mathbf{W}$$ to equation 2 yields the following result:

 * $$\mathbf{W}\cdot\mathbf{(kd - F)} = [1\cdot\sum_{j=1}^{6} (k_{1j}d_{j} - F_{1}) + 0\cdot\sum_{j=1}^{6} (k_{2j}d_{j} - F_{2}) + ... + 0\cdot\sum_{j=1}^{6} (k_{6j}d_{j} - F_{6})] = 0$$

It is easily shown that the result of the above expression is:

 * $$\sum_{j=1}^{6} k_{1j}d_{j} = F_{1}$$

Consider another set of weighting coefficients and resulting transposed form:

 * $$W_{1} = 0\;$$, $$W_{2} = 0\;$$ , $$\;W_{3}=...=W_{6} = 0$$

 * $$\mathbf{W^{T}} = \begin{bmatrix} 0&1&0&0&0&0\end{bmatrix}$$

As before, applying this matrix to equation 2 yields a result of $$\sum_{j=1}^{6} k_{2j}d_{j} - F_{2} = 0$$. Thus, it is apparent that for any choice of weighting matrix $$\mathbf{W}$$, the result is equivalent to $$\mathbf{kd - F}= \mathbf{0}$$, which is equation 1.

Returning to the specific 2-bar truss problem, accounting for the boundary conditions allows the removal of the following global displacements:

 * $$d_{1}=d_{2}=d_{5}=d_{6}=0\;$$

The weighting coefficients must be "kinematically admissable" - this means that it must not violate the boundary conditions. Thus, the following is a result of the above expression:

 * $$W_{1}=W_{2}=W_{5}=W_{6}=0\;$$

This leads to the conclusion that the weighting coefficients are essentially virtual displacements. As a result of the previous 2 relations, equation 2 can now be rewritten as:

 * $$\begin{Bmatrix}W_{3}\\W_{4}\end{Bmatrix}\cdot\mathbf{(\bar{k}\bar{d} - \bar{F})} = 0$$ for all $$\begin{Bmatrix}W_{3}\\W_{4}\end{Bmatrix}$$

Where $$\mathbf{\bar{k}}\;$$, $$\mathbf{\bar{d}}\;$$, and $$\mathbf{\bar{F}}$$ are defined as follows:

 * $$\mathbf{\bar{k}} = \begin{bmatrix}k_{33}&k_{34}\\k_{43}&k_{44}\end{bmatrix}$$

 * $$\mathbf{\bar{d}} = \begin{Bmatrix}d_{3}\\d_{4}\end{Bmatrix}$$

 * $$\mathbf{\bar{F}} = \begin{Bmatrix}F_{3}\\F_{4}\end{Bmatrix}$$

An example of weighting is the calculation of a student's final grade in this course, which is given as follows: $$course\;grade = \alpha_{0} + HW Grade + \Sigma(\alpha Exam Grade)$$

Derivation of $$\mathbf{k^{(e)}} = \mathbf{T^{(e)^{T}}}\mathbf{\hat{k}^{(e)}}\mathbf{T^{(e)}}$$
Recall the force-displacement relationship adjusted for axial degrees of freedom $$\mathbf{q^{(e)}}$$:

 * $$\mathbf{\hat{k}^{(e)}}\mathbf{q^{(e)}} = \mathbf{P^{(e)}}$$

This relationship can be rearranged to the following:

 * $$\mathbf{\hat{k}^{(e)}}\mathbf{q^{(e)}} - \mathbf{P^{(e)}} = \mathbf{0}\;\;(1)$$

Applying the Principle of Virtual Work results in the following:

 * $$\mathbf{\hat{W}}\cdot(\mathbf{\hat{k}^{(e)}}\mathbf{q^{(e)}} - \mathbf{P^{(e)}}) = \mathbf{0}$$ for all $$\mathbf{\hat{W}}$$

It has previously been shown that equations 1 and 2 are equivalent. Recalling the following two relations:

 * $$\mathbf{q^{(e)}} = \mathbf{T^{(e)}}\mathbf{d^{(e)}}\;\;(3)$$

 * $$\mathbf{\hat{W}^{(e)}} = \mathbf{T^{(e)}}\mathbf{W^{(e)}}\;\;(4)$$

$$\mathbf{\hat{W}}$$ refers to virtual axis displacements, corresponding to $$\mathbf{q^{(e)}}$$.

$$\mathbf{W}$$ refers to virtual displacements in global coordinate system, corresponding to $$\mathbf{d^{(e)}}$$.

Placing equations 3 and 4 into equation 2 yields the following:

 * $$(\mathbf{T^{(e)}}\mathbf{W})\cdot[\mathbf{\hat{k}^{(e)}}(\mathbf{T^{(e)}}\mathbf{d^{(e)}}) - \mathbf{P^{(e)}}] = \mathbf{0}$$ for all $$\mathbf{W}\;\;(5)$$

Recall the following two vector identities:

 * $$(\mathbf{A}\cdot\mathbf{B})^{T} = \mathbf{B}^{T}\mathbf{A}^{T}\;\;(6)$$

 * $$\mathbf{a}\cdot\mathbf{b} = \mathbf{a}^{T}\mathbf{b}\;\;(7)$$

Applying equations 7 and 8 into equation 5 yields the following derivation:

$$(\mathbf{T^{(e)}}\mathbf{W})^{T}[\mathbf{\hat{k}^{(e)}}(\mathbf{T^{(e)}}\mathbf{d^{(e)}}) - \mathbf{P^{(e)}}] = \mathbf{0}$$ for all $$\mathbf{W}$$

$$\mathbf{T^{(e)}}^{T}\mathbf{W}^{T}[\mathbf{\hat{k}^{(e)}}(\mathbf{T^{(e)}}\mathbf{d^{(e)}}) - \mathbf{P^{(e)}}] = \mathbf{0}$$ for all $$\mathbf{W}$$

$$\mathbf{W}^{T}[\left(\mathbf{T^{(e)}}^{T}\mathbf{\hat{k}^{(e)}}\mathbf{T^{(e)}}\right)\mathbf{d^{(e)}} - \left\{\mathbf{T^{(e)}}^{T}\mathbf{P^{(e)}}\right\}] = \mathbf{0}$$

In the last expression, the term inside the parentheses is equivalent to $$\mathbf{k^{(e)}}$$, and the term inside the curly brackets is equivalent to $$\mathbf{f^{(e)}}$$. Making these substitutions yields the following result:

$$\mathbf{W^{T}}[\mathbf{k^{(e)}}\mathbf{d^{(e)}} - \mathbf{f^{(e)}}] = \mathbf{0}$$ for all $$\mathbf{W}$$.

Continuous Case - Motivational Model Problem

Consider the elastic bar shown in Figure ##. The bar has a varying cross-section and Young's modulus, and is subject to a time-dependent loading condition consisting of a distributed, varying axial load as well as a concentrated load, signified by $$f(x,t)\;$$ and $$P(x)\;$$ respectively. A section of the bar is shown in Figure ##.

The product $$A(x)\rho (x)\;$$ can be expressed as $$m(x)\cdot dx\;$$, where $$m(x)\;$$ is the mass per unit length. The term $$\frac{\partial{^{2}u}}{\partial{x^{2}}}$$ refers to the acceleration of the deforming bar, and can be represented as $$\ddot{u}$$.

The sum of forces on the section can be written as follows:

 * $$\Sigma F_{x} = 0 = -N{x,t} + N(x+dx,t) +f(x,t)dx - m(x)\ddot{u}dx$$

 * $$\Sigma F_{x} = 0 = \frac{\partial{N(x,t)}}{\partial x}dx + f(x,t)dx - m(x)\ddot{u}dx + h.o.t\;\;(1)$$

"h.o.t" refers to the higher order terms.

Recall the general definition of the Taylor Series expansion, $$f(x+dx) = f(x) + \frac{\partial{f(x)}}{\partial x}dx + \frac{1}{2}\frac{\partial{^{2}x}}{\partial{x}^{2}} + ...$$, where $$\frac{1}{2}\frac{\partial{^{2}x}}{\partial{x}^{2}} + ...$$ are the higher order terms and are typically neglected. Dividing out the $$dx\;$$ terms and applying this concept to equation 1 yields equation 2, the equation of motion for the elastic bar:

 * $$\frac{\partial{N}}{\partial x} + f = m\ddot{u}\;\;(2)$$

The constitutive relation for the bar can be given as equation 3:

 * $$N(x,t) = A(x)\sigma (x,t)\;\;(3)$$

The $$\sigma (x,t)\;$$ term can be broken down as follows:

 * $$\sigma (x,t)= E(x)\epsilon (x,t)\;$$

 * $$\epsilon (x,t)= \frac{\partial{u}}{\partial{x}}(x,t)\;$$

The partial differential equation of motion for the elastic bar can be given as equation 4, which is the result of substituting equation 3 and resulting breakdown of terms into equation 2:

 * $$\frac{\partial}{\partial x}[A(x)E(x)\frac{\partial{u}}{\partial{x}}] + f(X,t) = m(x)\ddot{u}\;\;(4)$$

In order to manipulate this equation, boundary and initial conditions are required. Since the equation is of 2nd order in both x and t, two boundary and two initial conditions are necessary. The initial conditions consist of an initial displacement and velocity, given as $$[init. vel., init. disp]\;$$. The boundary conditions depend on the type of system being analyzed. Figure ## shows a bar of length L with both ends constrained. The boundary conditions for this system are:

 * $$u(0,t) = 0 = u(L,t)\;$$

Figure ## again shows a bar of length L, but with one end free and subject to force F. The first boundary condition, as with the constrained case, is:

 * $$u(0,t) = 0\;\;(bc\;1)$$

The second boundary condition can be derived using the following line of thought, similar to that used to derive equation 4:

 * $$N(L,t) = F(t)\;$$

 * $$N(L,t) = A(L)\sigma (L,t)\;$$

 * $$\sigma (L,t) = E(L)\epsilon (L,t)\;$$

 * $$\epsilon (L,t) = \frac{\partial u}{\partial x} (L,t)$$

Using the above relations, the second boundary condition for this case can be written as follows:

 * $$\frac{\partial u(L,t)}{\partial x} = \frac{F(t)}{A(L)E(L)}\;\;(bc\;2)$$

title: User:Eml4500.f08.Lulz.js/HW5


text: File:ContinuousBar.jpg

title: File:ContinuousBar.jpg


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User:Eml4500.f08.ateam.rivero/w5

Principle of Virtual Work Continued
The following is analogy between the weighing factor used in the principle of virtual work and how the course grade for this course, eml 4500 is calculated.

$$Course Grade = \alpha_0 + HW Grade + \sum_{i=1}^3 \alpha_i * Exam 1 $$

Deriving :$$ \mathbf k^{(e)} = \mathbf T^{(e)T} * \mathbf \hat k^{(e)} * \mathbf T^{(e)} $$ Recall Force Displacement relation with axial displacement degrees of freedom $$ \mathbf q^{(e)}$$ $$ \mathbf \hat k_{(2x2)}^{(e)} * \mathbf q_{(2x1)}^{(e)} = \mathbf p_{(2x1)}^{(e)} $$ $$ \rightarrow \mathbf \hat k^{(e)} * \mathbf q^{(e)} - \mathbf p^{(e)} = \mathbf 0 $$ : $$ (Eq.1) $$ $$ \mathbf \hat w_{(2x1)} * (\mathbf \hat k^{(e)} * \mathbf q^{(e)} - \mathbf p^{(e)})_{(2x1)} = 0 $$ for all $$ \mathbf \hat w_{(2x1)} $$ : $$ (Eq.2) $$ Recall: $$ \mathbf q_{(2x1)}^{(e)} = \mathbf T_{(2x4)}^{(e)} * \mathbf d_{(4x1)}^{(e)} $$ : $$ (Eq.3) $$ Similarly: $$ \mathbf \hat w_{(2x1)} = \mathbf T_{(2x4)}^{(e)} * \mathbf w_{(4x1)} $$ : $$ (Eq.4) $$ Note: $$ \mathbf \hat w_{(2x1)} $$ is the equivalent of the axial virtual displacement

title: User:Eml4500.f08.ateam.rivero/w5


text:
User:Eml4500.f08.wiki1.ambrosio/hw5

Part #3
It has already been shown that $$\mathbf{\tilde{d}^{(e)}} =\mathbf{\tilde{T}}^{(e)} \mathbf$$. We have already derived the transformation matrix $$\mathbf{\tilde{T}}^{(e)}$$, and we also know $$\mathbf$$ is the element DOF is the global coordinate. Therefore the axial displacement $$\mathbf{\tilde{d}^{(e)}} $$ can be written as

$$ \begin{align} \mathbf{\tilde{d}^{(e)}} & =\mathbf{\tilde{T}}^{(e)} \mathbf \\ \mathbf{q^{(e)}} & =\mathbf{\tilde{T}}^{(e)} \mathbf \\

\begin{pmatrix} q_1^{(e)} \\ q_2^{(e)} \end{pmatrix} & = \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} \end{align} $$

Part #5
We have already shown that $$\mathbf{k^{(e)}} = \mathbf{\tilde{T}}^{(e)T} \mathbf{{\tilde{k}}^{(e)}} \mathbf{{\tilde{T}}^{(e)}} $$ .By using the transformation matrix we get the following.

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

\begin{pmatrix} k_1^{(e)} \\ k_2^{(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}

\mathbf{{\tilde{k}}^{(e)}}

\begin{pmatrix} l^{(e)} & m^{(e)} & n^{(e)}  & 0 & 0 & 0\\ 0 & 0 &0 & l^{(e)} & m^{(e)}  & n^{(e)} \end{pmatrix}

\end{align} $$

Two Bar Truss
The following code was used to re-run the two bar truss system.

The following codes were used in the two bar truss code.

When the MATLAB code is ran, the following results are obtained.

Row 1 of the "results" represents element 1, while row 2 represents element 2. Column 1 is the strain, column 2 the stress, and column 3 is the axial force for each element.

Six Bar Truss Example
When the MATLAB code is ran, the following results are obtained.

As with the "results" matrix from the two bar truss code, the row indicates the element while column 1 is the strain, cloumn 2 is the stress, and column 3 is the axial force.

The figure below shows the undeformed truss (dashed blue line) and deformed truss (solid red line) with a weighting factor of 10000.

Six Bar Truss With Different Young's Modulus
When the MATLAB code is ran, the following results are obtained.

As with the "results" matrix from the two bar and six bar truss code, the row indicates the element while column 1 is the strain, cloumn 2 is the stress, and column 3 is the axial force.

The figure below shows the undeformed truss (dashed blue line), deformed truss where all elements have the same Young's Modulus (solid red line), and deformed truss where all elements have a different Young's Modulus (solid green line) with a weighting factor of 10000.

title: User:Eml4500.f08.wiki1.ambrosio/hw5


text:
User:Eas4200c.f08.WIKI.E/hw5

Mtg. 24 Monday 20 October 08
kinematic assumption drop down

4 zero strain components

5 zero stress components

$$\varepsilon -\sigma $$ relation in two forms (Tensorial 23-2 or Engineering 23-1b)

Rewrite $$\varepsilon -\sigma $$ relation: Full (not diagonal)

$$\begin{Bmatrix} \varepsilon _{i,j} \end{Bmatrix}_{6x1}= \begin{vmatrix} \mathbf{A}_{3x3} & \mathbf{0}_{3x3}\\ \mathbf{0}_{3x3} & \mathbf{B}_{3x3} \end{vmatrix} \begin{Bmatrix} \sigma _{i,j} \end{Bmatrix}_{6x1}$$

$$\varepsilon -\sigma $$ relation

$$\begin{Bmatrix} \sigma _{i,j} \end{Bmatrix}= \begin{vmatrix} \mathbf{A}_{3x3}^{-1} & \mathbf{0}_{3x3}\\ \mathbf{0}_{3x3} & \mathbf{B}_{3x3}^{-1} \end{vmatrix} \begin{Bmatrix} \varepsilon _{i,j} \end{Bmatrix}$$

$$\mathbf{C}_{6x6}= \begin{vmatrix} \mathbf{A}_{3x3} & \mathbf{0}_{3x3}\\ \mathbf{0}_{3x3} & \mathbf{B}_{3x3} \end{vmatrix}$$

$$ \mathbf{C}_{6x6}\cdot \mathbf{C}^{-1}_{6x6}=\mathbf{I}_{6x6}$$

$$\mathbf{C}_{6x6}\cdot \mathbf{C}^{-1}_{6x6}=\begin{vmatrix} \mathbf{A_{3x3}A}_{3x3}^{-1} & \mathbf{0}_{3x3}\\ \mathbf{0}_{3x3} & \mathbf{B_{3x3}B}_{3x3}^{-1} \end{vmatrix}=\mathbf{I_{6x6}} $$

We are trying to prove that $$ \sigma _{xx}=\sigma _{yy}=\sigma _{zz}=\gamma _{yz}=0$$

$$\begin{Bmatrix} \sigma _{i,j} \end{Bmatrix}=

\begin{vmatrix} \mathbf{A}_{3x3}^{-1} & \mathbf{0}_{3x3}\\ \mathbf{0}_{3x3} & \mathbf{B}_{3x3}^{-1} \end{vmatrix}\begin{Bmatrix} 0\\ 0\\ 0\\ 0\\ \varepsilon _{31}\\ \varepsilon _{12} \end{Bmatrix}$$ $$\begin{matrix} \sigma _{11}=0\\ \sigma _{22}=0\\ \sigma _{33}=0\\ \sigma _{23}=2G\varepsilon _{23}\\ \sigma _{31}=2G\varepsilon _{31}\\ \sigma _{21}=2G\varepsilon _{12} \end{matrix}$$

Consider the 1D stress model

title: User:Eas4200c.f08.WIKI.E/hw5


text:
User:Eas4200c.f08.wiki.b/HW5

=Mtg. 26 - Monday 10/27=

Bidirectional Bending Recipe
The following is a general recipe of equations used for solving a bidirectional bending problem.

The moment in the y-direction is

 $$\displaystyle M_y = \int_A z \sigma_{xx} dA $$

Similarly, the moment in the z-direction is

 $$\displaystyle M_z = \int_A y \sigma_{xx} dA $$

Where $$\displaystyle dA = dydz $$

The moment of intertia in the y-direction is

 $$\displaystyle I_y = \int_A z^2 dA $$

Similarly, the moment of intertia in the z-direction is

 $$\displaystyle I_z = \int_A y^2 dA $$

 $$\displaystyle I_{yz} = \int_A yz dA $$

The stress in the x-direction is

 $$\displaystyle \sigma_{xx} = E \epsilon_{xx} = {I_y M_z - I_{yz} M_y \over I_y I_z - (I_{yz})^2} y + {I_z M_y - I_{yz} M_z \over I_y I_z - (I_{yz})^2} z $$

Recall that

 $$\displaystyle I = \begin{bmatrix} I_{11} & I_{12} & I_{13} \\ I_{21} & I_{22} & I_{23} \\ I_{31} & I_{32} & I_{33} \end{bmatrix}$$

Note that the determinat is defined as

 $$\displaystyle D := I_y I_z - (I_{yz})^2 = I_{22} I_{33} - (I_{23})^2 $$

for $$\displaystyle \begin{bmatrix} I_{22} & I_{23} \\ I_{32} & I_{33} \end{bmatrix}$$

 $$\displaystyle I_{32} = I_{zy} = \int_A zy dA = I_{yz} = I_{23} $$

At the neutral axis, or where $$\displaystyle \sigma_{xx} = 0 $$

 $$\displaystyle \sigma_{xx} = m_y y + m_z z $$

Thus,

 $$\displaystyle z = (- {m_y \over m_z})y = (\tan \beta) y $$

=Mtg. 28 - Friday 10/31=

Taking the previous equation

 $$\displaystyle 0 = - \sigma (x)A + \sigma (x+dx)A +f(x)dx $$

The equation can be rewritten as

 $$\displaystyle 0 = A[ \sigma (x+dx) - \sigma (x) ] +f(x)dx $$

Notice that

 $$\displaystyle [ \sigma(x+dx) - \sigma(x) ] = {d \sigma (x) \over dx} $$

Also recall that

 $$\displaystyle f(x+dx) = f(x) + {df(x) \over dx} dx + {1 \over 2} {d^2 f(x) \over dx^2} (x)(dx)^2 + ... $$

If the higher orter terms are neglected, we obtain

 $$\displaystyle {d \sigma (x) \over dx} + {f(x) \over A} = 0 $$

where $$\displaystyle {f(x) \over A} $$ is the applied load or body force, which is equal to the force divided by the volume.

As seen in the above image, the sum of the forces in the x-direction can be expressed as

 $$\displaystyle \sum F_x = 0 = dydz [- \sigma_{xx}(x,y,z) + \sigma_{xx}(x+dx,y,z)] + dzdx [- \sigma_{yx}(x,y,z) + \sigma_{yx}(x,y+dy,z)] $$
 $$\displaystyle + dxdy [- \sigma_{zx}(x,y,z) + \sigma_{zx}(x,y,z+dz)] $$

Simplifying yields

 $$\displaystyle 0 = dxdydz [ {\partial \sigma_{xx} \over \partial x} + {\partial \sigma_{yx} \over \partial y} + {\partial \sigma_{zx} \over \partial z} ] $$

title: User:Eas4200c.f08.wiki.b/HW5


text:
User:Eml4500.f08.ramrod.c/11-20-08

Justification of Eliminated rows 1, 2, 5, and 6 to obtain K 2-bar truss:
Force Displacement Relation: K6x6d6x1 = F6x1

Moving F to the other side of the equation gives us Equation (1):

 Equation (1)

$$\mathbf{Kd}-\mathbf{F}=\mathbf{0}_{6x1} $$

Using the Principle of Virtual Work, we can arrive at Equation (2):

 Equation (2)

$$\mathbf{w}_{6x1}\bullet(\mathbf{Kd}-\mathbf{F})=\mathbf{0}_{1x1} $$
 * This is true for all w6x1
 * w is known as the "weighting mat."
 * Note that the dot product is used in this equation.

Equation (1)↔ Equation (2)

Using Principle of Virtual Work
Taking account for the boundary conditions of the 2 bar truss:
 * d1=d2=d5=d6=0

Weighting coefficient must be "kinematically admissible," ie cannot violate the boundary conditions
 * w1=w2=w5=w6=0

Weighting coefficients = virtual displacement

 * (calculus of variations)

From before:

$$\begin{Bmatrix} w_3\\ w_4\end{Bmatrix}\bullet(\mathbf{Kd}-\mathbf{F})=0$$

$$\mathbf{w}\bullet(\mathbf{Kd}-\mathbf{F})$$

$$\begin{Bmatrix} w_3\\ w_4 \end{Bmatrix}\bullet[\begin{bmatrix} K_{13} & K_{14} \\ K_{23} & K_{24} \\ K_{33} & K_{34}\\ K_{43} & K_{44}\\ K_{53} & K_{54}\\ K_{63} & K_{64} \end{bmatrix} \begin{Bmatrix} d_3\\ d_4 \end{Bmatrix} - \begin{Bmatrix} F_1\\ F_2\\ F_3\\ F_4\\ F_5\\ F_6 \end{Bmatrix}] = 0$$
 * This is true for all $$\begin{Bmatrix}

w_3\\ w_4\end{Bmatrix}$$

$$\mathbf{\overline{K}} = \begin{bmatrix} K_{33} & K_{34} \\ K_{43} & K_{44} \end{bmatrix}$$

$$\mathbf{\overline{d}} = \begin{Bmatrix} d_3\\ d_4 \end{Bmatrix}$$

$$\mathbf{\overline{F}} = \begin{Bmatrix} F_3\\ F_4 \end{Bmatrix} $$

Comparing Equations (3) and (4)
Equation (3) : $$(kd-F) = 0$$

Equation (4) : $$w(kd-F) = 0$$

Equation (3) → Equation (4) is trivial

Equation (4) → Equation (3)
 * Since Equation (4) is valid for all w, select w=1

 * $$1(kd-F) = 0$$ Thus proving Equation (3) is true

title: User:Eml4500.f08.ramrod.c/11-20-08


text:
User:Eml4500.f08.bike.mcdonald/homework report 6

 See my comments below. Please don't remove these comment boxes; just add your comment to these comment boxes in case you fix the problems. Eml4500.f08 19:52, 26 November 2008 (UTC)

Initial Conditions and the Continuous Principle of Virtual Work
Figure 1 shows the loaded elastic bar system in question.

The initial conditions for the system in Figure 1 at t = 0 are

$$ u\left(x,t=0\right)=\overline{u}(x) $$ for the known displacement function

and

$$ \frac{\partial u}{\partial t}(x,t=0)={\ddot{u}}(x,t=0)=\overline{V}(x) $$ for the known velocity function.

The partial differential equation for the continuous principle of virtual work(PVW) of dynamics for the elastic bar in Figure 1 can be seen below in Equation 1.

$$ PDE: \frac{\partial}{\partial x}\left[(EA)\frac{\partial u}{\partial x}\right]+f=m{\ddot{u}} $$ ⇨ $$ Equation \left(1\right)$$

The discrete equation of motion can then be shown by Equation 2, where $$ \underline{M} $$ is the mass matrix. Equation 2 represents a multiple defree of freedom system(MDOF). Note that a single degree of freedom system(SDOF) has a very similar equation of motion, shown be Equation 3.

MDOF System ⇨ $$ \underline{M}\underline{\ddot{d}}+\underline{k}\underline{d}=\underline{F} $$ ⇨ $$ Equation \left(2\right)$$

SDOF System ⇨ $$ m\ddot{d}+kd=F $$ ⇨ $$ Equation \left(3\right)$$

The integral shown in Equation 4 was used to derive Equation 2 from Equation 1. The term $$ w(x) $$ represents the weighting function.

$$ \int_{x=0}^{x=L} w(x)\left({\frac{\partial}{\partial x}\left[(EA)\frac{\partial u}{\partial x}\right]+f-m\ddot{u}}\right)dx=0 $$ for all possible $$ w(x) $$ ⇨ $$ Equation \left(4\right)$$

To go from Equation 1 to Equation 4 is a trivial task, but going from Equation 4 to Equation 1 is not trivial. This process is shown below.

For simplicity, let $$ \left({\frac{\partial}{\partial x}\left[(EA)\frac{\partial u}{\partial x}\right]+f-m\ddot{u}}\right)=g(x) $$

Equation 4 then becomes $$ \int_{x=0}^{x=L} w(x)g(x)dx=0 $$ for all $$ w\left(x\right) $$

Since Equation 4 holds for all $$ w\left(x\right) $$, select $$ w\left(x\right)=g\left(x\right) $$,

then Equation 4 becomes $$ \int_{x=0}^{x=L}g^2dx=0 $$ ⇨ $$ g\left(x\right)=0 $$

Integration by Parts and the Discrete Principle of Virtual Work
Recall the steps for and integration by parts:

$$ \left(rs\right)^{'}=r^{'}s+rs^{'} $$

$$ r^{'}=\frac{dr}{dx} $$ and $$ s^{'}=\frac{ds}{dx} $$

$$ \int \left(rs\right)^{'}=\int r^{'}s + \int rs^{'} $$

Which yields a final result of $$ \int r^{'}s=rs-\int rs^{'} $$

Now Recall the Cont. Principle of Virtual Work(PVW) shown below.

$$ \int_{x=0}^{x=L} w(x)\left({\frac{\partial}{\partial x}\left[(EA)\frac{\partial u}{\partial x}\right]+f-m\ddot{u}}\right)dx=0 $$ for all possible $$ w(x) $$

In the first term, let $$ s\left(x\right)=w(x) $$ and $$ r\left(x\right)=(EA)\frac{\partial u}{\partial x} $$

Solving using integration by parts yields

$$ \int_{x=0}^{x=L} w\left(x\right)\frac{\partial}{\partial x}\left[\left(EA\right)\frac{\partial u}{\partial x}\right]dx = \left[w(EA)\frac{\partial u}{\partial x}\right]_{x=0}^{x=L} - \int_{x=0}^{x=L}\frac{\partial w}{\partial x}(EA)\frac{\partial u}{\partial x}dx $$

$$ = w\left(L\right)(EA)(L)\frac{\partial u}{\partial x}(L,t)-w(0)(EA)(0)\frac{\partial u}{\partial x}(0,t)-\int_{0}^{L}\frac{\partial w}{\partial x}(EA)\frac{\partial u}{\partial x}dx $$

where $$ \left(EA\right)(L)\frac{\partial u}{\partial x}(L,t) = N(L,t)=F $$

and $$ (EA)(0)\frac{\partial u}{\partial x}(0,t)=N(0,t) $$

Now lets consider the model problem with 2 boundary conditions shown below.

At $$ x=0 $$ select $$ w(x) $$ such that $$ w(0)=0 $$. This means that the system is kinematically admissible.

Motivation: discrete PVW applied to the equation shown below.

$$ \underline{W}_{6x1}\left(\left[ \underline K \right]_{6x2}\begin{Bmatrix} d_{3} \\ d_{4}\end{Bmatrix}_{2x1}-\underline{F}_{6x1}\right)=0_{1x1} $$ for all $$ \underline{W} $$

where $$ \underline{F}^{T}=\left[F_{1}F_{2}F_{3}F_{4}F_{5}F_{6}\right] $$

$$ F_{3}, F_{4} $$ are known reactions and $$ F_{1}, F_{2}, F_{5}, F_{6} $$ are unknown reactions.

Since $$ \underline{w} $$ can be selected arbitrarily, select $$ \underline{w} $$ such that $$ w_{1}=w_{2}=w_{5}=w_{6}=0 $$ so to eliminate equations involving unknown reactions. Simply put, rows 1, 2, 5, and 6 are eliminated. The result is shown below in Equation 1.

$$ \underline{W}\left(\underline{K}\underline{d}-\underline{F}\right)=0 $$ for all $$ \underline{W} $$

$$ \underline{K}_{2x2}\underline{d}_{2x1}=\underline{F}_{2x1} $$ ⇨ $$ Equation \left(1\right) $$

Back to the principle of virtual work, the unknown reaction is

$$ N(0,t)=(EA)(0)\frac{\partial u}{\partial x}(0,t) $$

The equation then becomes

$$ W\left(L\right)F(t)-\int_{0}^{L}\frac{\partial w}{\partial x}(EA)\frac{\partial u}{\partial x}dx+\int w(x)\left[f-m\ddot{u}\right]dx=0 $$ for all $$ w\left(x\right) $$ such that $$ w\left(0\right)=0 $$

The final equation can be written as

$$ \int_{0}^{L}w\left(m\ddot{u}\right)dx+\int_{0}^{L} \frac{\partial w}{\partial x}(EA)\frac{\partial u}{\partial x}dx=W(L)F(t)+\int_{0}^{L}wfdx $$ for all $$ w\left(x\right) $$ such that $$ w\left(0\right)=0 $$

Motivation for the Linear Interpolation of u(x)
Elastic Bar Interpolation Example

It is assumed that the displacement u(x) is for $$x_i\leq x\leq x_{i+1}$$.

The following is an example of linear interpolation of u(x) for a 2-bar truss.

The deformed shape is a straight line, which means there was an assumption of linear interpolation of displacement between the two nodes. Below is a case where there is only axial displacement, meaning zero transverse displacement.

u(x) is expressed in terms of di=u(xi) and di+1=u(xi+1) as a linear function in x.

$$\mathbf{u(x)=N_i(x)d_i+N_{i+1}(x)d_{i+1}}$$ where $$N_i(x) and N_{i+1}(x)$$ are linear functions in x.

 Good essay. Eml4500.f08 20:08, 26 November 2008 (UTC)

Now, motivation is shown for the form of $$ N_{i}(x)+N_{i+1}(x)$$.

$$ N_{i}(x)$$ and $$N_{i+1}(x)$$ are considered to be linear, meaning they are represented as straight lines. Therefore, any linear combination of $$ N_{i}(x)$$ and $$N_{i+1}(x)$$ is also linear. The expressions below show $$ N_{i}(x)$$ and $$N_{i+1}(x)$$ written in linear form.

$$ N_{i}(x)=\alpha_{i}+\beta_{i}x$$

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

Where $$\alpha_{i}$$ and $$\beta_{i}$$ both represent real numbers.

Next, the linear combination of $$ N_{i}(x)$$ and $$N_{i+1}(x)$$ can be shown as follows.

$$ N_{i}d_{i}+N_{i+1}d_{i+1}=(\alpha_{i}+\beta_{i}x)d_{i}+(\alpha_{i+1}+\beta_{i+1}x)d_{i+1}$$

The above equation can be rearranged and written as shown below.

$$ U_{x}=(\alpha_{i}d_{i}+\alpha_{i+1}d_{i+1})+(\beta_{i}d_{i}+\beta_{i+1}d_{i+1})x$$

This final equation shows that the linear combination is clearly a linear function in x.

A second motivation for $$ N_{i}(x)+N_{i+1}(x)$$ is shown below.

Recall the linear interpolation for u(x).

$$N_i(x)=\frac{x-x_{i+1}}{x_{i}-x_{i+1}}$$ $$N_{i+1}(x)=\frac{x-x_i}{x_{i+1}-x_i}$$

Now, plugging in $$ x_{i}$$ for x, it can be rewritten as follows:

$$N_i(x_{i})=\frac{x_{i}-x_{i+1}}{x_{i}-x_{i+1}}=1$$

$$N_{i+1}(x_{i})=\frac{x_{i}-x_i}{x_{i+1}-x_i}=0$$

Rewriting $$ u(x_{i})$$,

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

$$ u(x_{i})= d_{i}$$

This shows that $$ u(x_{i}) $$ is equal to the weighting coefficient $$d_{i}$$

Elemental Stiffness Matrix for Element i
The elemental stiffness matrix for element i is found using the integral below.

$$\int_{x_i}^{x_{i+1}}[N'_i(x)W_i + N'_{i+1}W_{i+1}](EA)[N'_id_i + N'_{i+1}d_{i+1}]dx$$

where $$ N'_i = \frac{dN_i(x)}{dx}$$ and $$ N'_{i+1} = \frac{dN_{i+1}(x)}{dx}$$

Note:

$$u(x) = \underbrace{\begin{bmatrix}N'_i(x) & N'_{i+1}(x)\end{bmatrix}}_{\textbf{N}(x)_{1x2}}\begin{Bmatrix}d_i \\ d_{i+1}\end{Bmatrix}_{2x1}$$

$$\frac{du(x)}{dx} = \underbrace{\begin{bmatrix}N_i(x) & N_{i+1}(x)\end{bmatrix}}_{\textbf{B}(x)_{1x2}}\begin{Bmatrix}d_i \\ d_{i+1}\end{Bmatrix}_{2x1}$$

Similarly, the $$W(x)$$ matrix can be described as:

$$W(x) = \textbf{N}(x)\begin{Bmatrix}W_i \\ W_{i+1}\end{Bmatrix}$$

$$\frac{dW(x)}{dx} = \textbf{B}(x)\begin{Bmatrix}W_i \\ W_{i+1}\end{Bmatrix}$$

Recall the elemental degrees of freedom used with this element.

$$\begin{Bmatrix} d_i \\ d_{i+1} \end{Bmatrix} = \begin{Bmatrix} d_1^{(i)} \\ d_2^{(i)} \end{Bmatrix} = \textbf{d}^{(i)} $$

$$\begin{Bmatrix} W_i \\ W_{i+1} \end{Bmatrix} = \begin{Bmatrix} W_1^{(i)} \\ W_2^{(i)} \end{Bmatrix} = \textbf{W}^{(i)} $$

Using these matrices, the following equation can be used to find $$\beta$$

$$\beta = \int_{x^i}^{x^{i+1}}(\textbf{B}\textbf{W}^{(i)})_{1x1}(EA)_{1x1}(\textbf{B}\textbf{d}^{(i)})_{1x1}dx = \textbf{W}^i \cdot (\textbf{k}^{(i)} \textbf{d}^{(i)}) $$

Since each of these matrices becomes a scalar, they can be reorganized and multiplied in any order. The substitutions shown can also be made.

$$\beta = \int_{x^i}^{x^{i+1}}(EA)(\textbf{B}\textbf{W}^{(i)})_{1x1} \cdot (\textbf{B}\textbf{d}^{(i)})_{1x1}dx = (\textbf{BW}^{(i)})^T(\textbf{Bd}^{(i)}) = \textbf{W}^{(i)T}\textbf{B}^T(\textbf{Bd}^{(i)}) = \textbf{W}^{(i)}\cdot\textbf{B}(\textbf{Bd}^{(i)})$$

Ultimately, the $$\beta$$ matrix can be finalized to the equation below.

$$\beta = \textbf{W}^{(i)} \cdot (\int\textbf{B}^T(EA)\textbf{B}dx)d^{(i)}$$

Using B and the transpose of B, the global stiffness matrix K can be determined like the stiffness matrices were determined previously with the transfer matrix and its transpose.

$$\textbf{k}^{(i)}_{2x2} = \int_{x_i}^{x_{i+1}} \textbf{B}^T(x)_{2x1}(EA)_{1x1} \textbf{B}(x)_{1x2}dx $$

Considering EA as a constant, the K then becomes the equation shown below.

$$\textbf{k}^{(i)} = \frac{EA}{L^{(i)}}\begin{bmatrix} 1 & -1 \\ -1 & 1 \end{bmatrix}$$

Transfer of Variable Coordinates from $$x$$ to $$\tilde{x}$$
$$\tilde{x} := x - x_i $$

$$d\tilde{x} = dx$$

$$\textbf{k}^{(i)} = \int_{\tilde{x} = 0}^{\tilde{x} = L^{(i)}}\textbf{B}^T(\tilde{x})(EA)(\tilde{x})\textbf{B}(\tilde{x})d\tilde{x}$$

When EA and the length of the element are considered constant, the K matrix becomes the equation shown below.

$$\textbf{k}^{(i)} = \frac{EA}{L^{(i)}}\begin{bmatrix} 1 & -1 \\ -1 & 1 \end{bmatrix}$$

An element with varying areas and moduli of elasticity can now be solved. To prove this, consider the element shown below.

The area and modulus of elasticity equations are shown below.

$$A(\tilde{x}) = N_1^{(i)}(\tilde{x})A_1 + N_2^{(i)}(\tilde{x})A_2$$

$$E(\tilde{x}) = N_1^{(i)}(\tilde{x})E_1 + N_2^{(i)}(\tilde{x})E_2$$

The elemental stiffness matrix can then be solved as a function of $$\begin{Bmatrix} A_1, & A_2, & E_1, & E_2, & L^{(i)}\end{Bmatrix}$$

$$k_{2x2}(\tilde{x})^{(i)} = \frac{\begin{bmatrix} N_1^{(i)}(\tilde{x})A_1 + N_2^{(i)}(\tilde{x})A_2\end{bmatrix}\begin{bmatrix}N_1^{(i)}(\tilde{x})E_1 + N_2^{(i)}(\tilde{x})E_2\end{bmatrix}}{L^{(i)}}\begin{bmatrix} 1 & -1 \\ -1 & 1 \end{bmatrix}$$

 Error: The integral was missing; the above expression should be integrated with respect to the variable $$\displaystyle \tilde x$$ over the length of the element, since otherwise the stiffness matrix would be a function of $$\displaystyle \tilde x$$ as shown, and that would not make sense. This error is the same as in Team ATeam(Sean) (with annotations). Eml4500.f08 20:08, 26 November 2008 (UTC)

For the sixth homework, the class was asked to evaluate the function:

$$N_i(\tilde{x})=$$

Which equals:

$$N_i(\tilde{x})=\frac{\tilde{x}-\tilde{x}_{i+1}}{\tilde{x}_{i}-\tilde{x}_{i+1}}$$

Where:

$$\tilde{x} = x - x_{i}$$

Using the linear interpolation method described in an earlier lecture and defining $$\tilde{x}$$ as the length from the beginning of the interpolation to the value that is to be found.

A description of the shape function $$N_{2}^{(i)}$$ using the $$\tilde{x}$$ notation is as follows:

$$N_{2}^{(i)}(\tilde{x}) = \frac{\tilde{x}}{L^{(i)}} = \begin{cases} & \text{0 if } \tilde{x}= 0\\ & \text{1 if } \tilde{x}= L \end{cases}$$

Where the boundary conditions are provided, at $$\tilde{x} = 0$$ and $$\tilde{x} = L$$ respectively.

The class was also asked to complete the example provided in the book on page 159 while setting $$E_{1} = E_{2} = E$$ and to let $$A(\tilde{x})$$ be linear as it was described in the previous lecture. $$\mathbf{k}^{(i)}$$ is found from the problem given in the previous lecture and compared to the expression given in the book, which is:

$$\frac{E(A_{1}+A_{2})}{L^{(i)}2}\begin{bmatrix} 1 & -1\\ -1 & 1 \end{bmatrix} = \mathbf{k}^{(i)}$$

As seen from the previous lecture, this result for the stiffness matrix is identically the same.

Next, the general $$\mathbf{k}^{(i)}$$ matrix from the previous lecture is to be compared with the matrix obtained by using $$\frac{(A_{1}+A_{2})}{2}$$ and $$\frac{(E_{1}+E_{2})}{2}$$ while noting that in this case $$E_{1}\neq E_{2}$$. This produces the stiffness matrix:

$$\frac{(E_{1}+E_{2})(A_{1}+A_{2})}{4L^{(i)}}\begin{bmatrix} 1 & -1\\ -1 & 1 \end{bmatrix} = \mathbf{k}_{ave}^{(i)}$$

The average stiffness matrix is then subtracted from the general stiffness matrix to find that:

$$\frac{(2E - E_{1}+E_{2})(A_{1}+A_{2})}{4L^{(i)}}\begin{bmatrix} 1 & -1\\ -1 & 1 \end{bmatrix} = \mathbf{k}^{(i)} - \mathbf{k}_{ave}^{(i)}$$

In order to interpret this, it is useful to review the Mean Value Theorem from calculus.

Redo of Eigenvector Plots of Rectangular Truss Systems
In Homework Report 4, Professor Vu Quoc assigned each team in EML 4500 the task of plotting the eigenvectors of a rectangular 3-bar truss system. No teams properly completed this assignment, thus Professor Vu Quoc asked for each team to redo the task. The redo of this assignment is shown below for the basic 3-bar rectangular truss system as well as the rectangular truss system with a cross-member added for structural rigidity.

3-Bar Rectangular Truss System
The setup for the 3-bar rectangular truss system presented by Professor Loc Vu Quoc is shown below.

The global stiffness matrix of the above rectangular system was determined in Homework Report 4 and is shown below.

K =

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

The eigenvectors (V) and eigenvalues (D) of the above global stiffness matrix were determined using MATLAB and are shown below.

V =

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

D =

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     0     0     0     0     0     0     0     0    12

 Note: As explained in class, you should apply the boundary conditions to obtain the reduced stiffness matrix (dimension 4x4), then in that case you would only have one zero eigenpair corresponding to one mechanism, instead of 5 zero eigenpairs, which are 5 mechanisms. If you had applied the boundary conditions (blocking 4 dofs), only the 3rd eigenpair (swaying motion) would be the only zero eigenpair. See below. Eml4500.f08 20:08, 26 November 2008 (UTC)

Plot of the Eigenvectors
The plots of the eigenvectors corresponding to the zero eigenvectors of the 3-bar rectangular truss system are shown below. The solid red lines correspond to the deformed shape of the 3-bar rectangular truss system and the dotted blue lines correspond to the undeformed shape of the 3-bar rectangular truss system.

MATLAB Code Used to Produce Eigenvector Plots
The following MATLAB code was used to produce the above plots.

Addition of Cross-Member to 3-Bar Rectangular Truss System

The global stiffness matrix of the above modified rectangular truss system was also determined in Homework Report 4 and is shown below.

K =

2.1213   2.1213         0         0   -2.1213   -2.1213         0         0    2.1213    8.1213         0   -6.0000   -2.1213   -2.1213         0         0         0         0    6.0000         0   -6.0000         0         0         0         0   -6.0000         0    6.0000         0         0         0         0   -2.1213   -2.1213   -6.0000         0    8.1213    2.1213         0         0   -2.1213   -2.1213         0         0    2.1213    8.1213         0   -6.0000         0         0         0         0         0         0         0         0         0         0         0         0         0   -6.0000         0    6.0000

The eigenvectors (V) and eigenvalues (D) of the above global stiffness matrix were determined using MATLAB and are shown below.

V =

0.2372   0.1828   -0.6190   -0.3565    0.5948   -0.0000    0.0000    0.2150    0.2512   -0.5293    0.1627    0.1742    0.1586    0.2656    0.5127    0.4914    0.5834    0.0231    0.0449   -0.2390   -0.4362   -0.5767   -0.0263    0.2764    0.2512   -0.5293    0.1627    0.1742    0.4362   -0.2656   -0.5127   -0.2764    0.5834    0.0231    0.0449   -0.2390   -0.1586    0.5767    0.0263   -0.4914   -0.0950   -0.3695   -0.5012    0.0566   -0.1586   -0.3112    0.4863   -0.4914    0.3447    0.3636   -0.2394    0.8317    0.0000    0.0000    0.0000    0.0000   -0.0950   -0.3695   -0.5012    0.0566   -0.4362    0.3112   -0.4863    0.2764

D =

-0.0000        0         0         0         0         0         0         0         0   -0.0000         0         0         0         0         0         0         0         0   -0.0000         0         0         0         0         0         0         0         0    0.0000         0         0         0         0         0         0         0         0    3.8183         0         0         0         0         0         0         0         0   12.0000         0         0         0         0         0         0         0         0   12.0000         0         0         0         0         0         0         0         0   16.6670

 Note: These 4 zero eigenpairs are linear combinations of 4 pure eigenmodes (3 rigid body rotations and 1 mechanism). Similar to the above remark, if you had applied the boundary conditions to obtain the reduced stiffness matrix (dimension 4x4), then in that case you would only have one zero eigenpair corresponding to one mechanism. Eml4500.f08 20:08, 26 November 2008 (UTC)

MATLAB Code Used to Produce Eigenvector Plots
The following MATLAB code was used to produce the above plots.

Analysis of 91-Element Electric Pylon Structure
The setup of the 91-element electric pylon structure is shown in the figure below. This system is to be analyzed using the Finite Element Method.

Each of the 91-elements that make up the electric pylon structure were made of 300 M steel with Young's Modulus of 200 GPa and had a cross-sectional area of 4 cm2.

The following topics will be covered in this analysis:

 * 1) Develop a MATLAB code to solve the electric pylon structure
 * 2) Plot the undeformed and deformed shape
 * 3) Compute the axial stress in each element of the electric pylon structure
 * 4) Determine which elements have the highest tensile and compressive stress in the structure
 * 5) Insert arrows on the undeformed and deformed plot to show the highest tensile and compressive stress elements
 * 6) Determine if the problem is statically determinate
 * 7) Construct the lumped mass matrix of the structure by assuming that the mass of each element is divided by two and lumped at each of the two nodes of the element
 * 8) Solve the generalized eigenvalue problem
 * 9) Find the lowest three eigenpairs
 * 10) Plot the eigenvectors corresponding to the lowest three eigenvalues as deformed shapes superposed onto the undeformed plot of the electric pylon structure
 * 11) Find the lowest three vibrational periods of the electric pylon

MATLAB Code to Solve the Electric Pylon Problem Using the Finite Element Method
The MATLAB code developed by Team Bike to solve the electric pylon structure using the Finite Element Method is shown below.

Additional Functions Required
In order to properly run the above MATLAB code, the following additional functions must be saved in the working directory of MATLAB.

MATLAB Command Window Printout for Electric Pylon Structure and Results
Running the MATLAB code developed by Team Bike, shown above, produced the following results in the MATLAB command window.

d =

0           0            0            0  -0.00019023   0.00017465   -0.0001277 -1.4074e-006 -0.00016006  -0.0002967  -0.00012671   0.00032602  -0.00012798    -0.000128  -0.00012792  -0.00054327   0.00044932   0.00030901   0.00046381  -0.00061057     0.001017   0.00064583    0.0010325  -0.00019427    0.0010502  -0.00097027    0.0017167    0.0010338    0.0017162   0.00023259    0.0016833  -0.00064959     0.001683   -0.0013293    0.0023452   0.00064079    0.0021863   -0.0010297    0.0047122    0.0023902    0.0039576   -0.0024165    0.0063605    0.0057328    0.0063605    0.0037877    0.0063605    0.0029803    0.0063614     0.002384    0.0063657    0.0016463    0.0063711   0.00086194    0.0063182  -0.00024268    0.0062289   -0.0013182    0.0061363   -0.0023672    0.0060413   -0.0033231    0.0058839   -0.0051278    0.0055152    -0.012587    0.0067826    0.0042648     0.006996   -0.0063359     0.007012    0.0034574    0.0074006   -0.0043591    0.0071505    0.0029772    0.0071478    0.0020508    0.0071421    0.0012278     0.007182   0.00032026    0.0072816  -0.00073249    0.0074073   -0.0018633    0.0075643   -0.0033872    0.0090005    0.0029772     0.010455   -0.0033872

reactions =

149.64     -501.65      -149.64       1501.7

results =

1.0781e-005 2.1562e+006       862.48 -6.5356e-006 -1.3071e+006     -522.85 6.7339e-006 1.3468e+006       538.71 6.4556e-006 1.2911e+006       516.45 -3.5152e-006 -7.0304e+005     -281.22 -2.3798e-005 -4.7596e+006     -1903.8 -1.2325e-005 -2.465e+006      -986.01 7.1182e-006 1.4236e+006       569.46 1.9269e-005 3.8538e+006       1541.5 -2.8828e-005 -5.7655e+006     -2306.2 -1.8941e-007      -37882      -15.153 8.778e-009      1755.6      0.70224 1.9328e-005 3.8655e+006       1546.2 1.4698e-005 2.9395e+006       1175.8 -1.8494e-005 -3.6988e+006     -1479.5 -2.883e-005 -5.7661e+006     -2306.4 1.7948e-006 3.5896e+005       143.58 2.621e-005  5.242e+006       2096.8 9.8986e-006 1.9797e+006       791.89 -1.8874e-005 -3.7749e+006       -1510 -3.258e-005 -6.5159e+006     -2606.4 2.7892e-006 5.5784e+005       223.14 3.2194e-006 6.4388e+005       257.55 2.9828e-005 5.9656e+006       2386.3 -4.0852e-006 -8.1704e+005     -326.82 -1.9808e-005 -3.9616e+006     -1584.6 1.16e-005   2.32e+006       928.02 -4.0653e-006 -8.1307e+005     -325.23 -2.8778e-005 -5.7556e+006     -2302.2 -1.4237e-007      -28474       -11.39 -4.9677e-006 -9.9353e+005     -397.41 -7.6072e-008      -15214      -6.0857 2.9608e-005 5.9216e+006       2368.6 2.7907e-007       55813       22.325 -2.4358e-005 -4.8717e+006     -1948.7 6.9573e-006 1.3915e+006       556.59 -1.5116e-008     -3023.3      -1.2093 -2.8794e-005 -5.7588e+006     -2303.5 -2.427e-005 -4.854e+006      -1941.6 6.9522e-006 1.3904e+006       556.17 -9.4154e-007 -1.8831e+005     -75.323 1.129e-005  2.258e+006       903.22 -3.16e-006  -6.32e+005       -252.8 -2.0051e-005 -4.0102e+006     -1604.1 7.093e-019 1.4186e-007  5.6744e-011 2.4411e-019 4.8822e-008  1.9529e-011 3.2073e-007       64147       25.659 1.3501e-006 2.7001e+005       108.01 1.3575e-006 2.7149e+005        108.6 -1.0389e-005 -2.0778e+006     -831.13 -2.2136e-005 -4.4271e+006     -1770.9 -2.9388e-005 -5.8777e+006     -2351.1 -3.6774e-005 -7.3549e+006     -2941.9 -4.331e-005 -8.662e+006      -3464.8 -4.3478e-005 -8.6957e+006     -3478.3 1.2903e-019 2.5807e-008  1.0323e-011 -1.5471e-019 -3.0942e-008 -1.2377e-011 2.6985e-019  5.397e-008  2.1588e-011 -6.1024e-019 -1.2205e-007 -4.8819e-011 -8.8522e-007 -1.7704e+005     -70.818 1.1034e-006 2.2069e+005       88.276 -9.5965e-007 -1.9193e+005     -76.772 9.9436e-007 1.9887e+005       79.549 1.0732e-005 2.1464e+006       858.56 -1.0415e-005 -2.083e+006      -833.19 1.1448e-005 2.2895e+006       915.81 -1.1321e-005 -2.2643e+006     -905.71 1.0517e-005 2.1035e+006       841.38 -1.0732e-005 -2.1464e+006     -858.56 7.0689e-006 1.4138e+006       565.51 -6.8859e-006 -1.3772e+006     -550.87 7.9177e-006 1.5835e+006       633.42 -1.8464e-005 -3.6928e+006     -1477.1 -4.8696e-007      -97393      -38.957 -1.1845e-007      -23690      -9.4762 1.5604e-007       31208       12.483 4.5239e-005 9.0479e+006       3619.2 -4.6964e-019 -9.3927e-008 -3.7571e-011 4.5158e-005 9.0317e+006       3612.7 -1.5877e-018 -3.1754e-007 -1.2702e-010 4.5256e-005 9.0511e+006       3620.5 -6.5877e-007 -1.3175e+005     -52.702 -1.4822e-006 -2.9645e+005     -118.58 8.9829e-006 1.7966e+006       718.63 2.2438e-005 4.4876e+006         1795 3.3117e-005 6.6233e+006       2649.3 3.8878e-005 7.7755e+006       3110.2 1.6111e-019 3.2222e-008  1.2889e-011 -1.441e-019 -2.8821e-008 -1.1528e-011 -9.6069e-020 -1.9214e-008 -7.6855e-012 0           0            0

Bar_Stress =

2.1562e+006 -1.3071e+006 1.3468e+006 1.2911e+006 -7.0304e+005 -4.7596e+006 -2.465e+006 1.4236e+006 3.8538e+006 -5.7655e+006 -37882      1755.6  3.8655e+006 2.9395e+006 -3.6988e+006 -5.7661e+006 3.5896e+005 5.242e+006 1.9797e+006 -3.7749e+006 -6.5159e+006 5.5784e+005 6.4388e+005 5.9656e+006 -8.1704e+005 -3.9616e+006 2.32e+006 -8.1307e+005 -5.7556e+006 -28474 -9.9353e+005 -15214 5.9216e+006 55813 -4.8717e+006 1.3915e+006 -3023.3 -5.7588e+006 -4.854e+006 1.3904e+006 -1.8831e+005 2.258e+006 -6.32e+005 -4.0102e+006 1.4186e-007 4.8822e-008 64147 2.7001e+005 2.7149e+005 -2.0778e+006 -4.4271e+006 -5.8777e+006 -7.3549e+006 -8.662e+006 -8.6957e+006 2.5807e-008 -3.0942e-008 5.397e-008 -1.2205e-007 -1.7704e+005 2.2069e+005 -1.9193e+005 1.9887e+005 2.1464e+006 -2.083e+006 2.2895e+006 -2.2643e+006 2.1035e+006 -2.1464e+006 1.4138e+006 -1.3772e+006 1.5835e+006 -3.6928e+006 -97393      -23690        31208  9.0479e+006 -9.3927e-008 9.0317e+006 -3.1754e-007 9.0511e+006 -1.3175e+005 -2.9645e+005 1.7966e+006 4.4876e+006 6.6233e+006 7.7755e+006 3.2222e-008 -2.8821e-008 -1.9214e-008 0

Comp_Stress =

-8.6957e+006

Comp_Stress_Element_Number =

55

Tens_Stress =

9.0511e+006

Tens_Stress_Element_Number =

81

lambda =

132.16      2468.4       2940.3

column =

79   85    86

T =

0.54654     0.12647      0.11587

Axial Stress in Each Bar
The axial stress in each bar are shown in the above MATLAB command window printout as the variable. These values are replicated in the table below for ease of viewing.

Highest Tensile Stress Element in the Electric Pylon Structure
As shown in the MATLAB command window printout above, the highest tensile stress in the electric pylon structure exists in element number 81, and has a value of 9.0511 x 106 Pa.

Highest Compressive Stress Element in the Electric Pylon Structure
As shown in the MATLAB command window printout above, the highest compressive stress in the electric pylon structure exists in element number 55, and has a value of -8.6957 x 106 Pa.

Plot of the Undeformed and Deformed Shapes of Electric-Pylon Structure
The figure shown below is a plot of the undeformed and deformed shapes of the 91-element electric pylon structure. This plot was produced using the MATLAB code shown below. This MATLAB code solved the 91-element electric pylon structure by use of the Finite Element Method. The solid red lines in the plot below represent the deformed shape of the 91-element electric pylon structure, whereas the dotted blue lines represent the undeformed shape of the 91-element electric pylon structure. In order to more clearly show the deformation experienced by the electric pylon structure, the deformations were multiplied by a magnification factor of 120.

The arrows in the above plot show the bar elements of the electric pylon structure with the highest compressive and highest tensile stress.

Is The Electric Pylon Structure Statically Determinate?
No, the electric pylon structure is not statically determinate. While this system is similar to the 2-bar truss system analyzed in previous Homework Reports in that there are only two nodes that have reactions, it is dissimilar to that problem due to the fact that each of these nodes has two members meeting at them. Because of this, each of the nodes has two unknown forces along the direction of each of those members. This means that the system has four unknown reaction forces, with only two equations to attempt to solve them with. These two equations are developed by summing the forces in the positive X direction and summing the forces in the positive Y direction.

 Note: Then you should compute the reactions at the base of the pylon using the finite element method; let's do this computation in HW7. Eml4500.f08 19:52, 26 November 2008 (UTC)

Lowest Eigenvalues of the Electric Pylon Structure
The eigenvalues of the electric pylon structure were determined from the generalized eigenvalue problem shown below, involving the reduced global stiffness matrix and the reduced lumped mass matrix.

$$ \bar{K}\nu =\lambda \bar{M}\nu $$

Lowest Eigenvalue
The lowest eigenvalue of the electric pylon structure was determined to be 132.16. This value was shown above in the MATLAB command window printout as well.

2nd Lowest Eigenvalue
The 2nd lowest eigenvalue of the electric pylon structure was determined to be 2468.4. This value was shown above in the MATLAB command window printout as well.

3rd Lowest Eigenvalue
The 3rd lowest eigenvalue of the electric pylon structure was determined to be 2940.3. This value was shown above in the MATLAB command window printout as well.

Lowest Eigenpair
The following plot of the lowest eigenpair as the deformed shape of the electric pylon structure is shown below. The deformed structure was multiplied by a magnification factor of 50 to allow for visualization of the deformation.

2nd Lowest Eigenpair
The following plot of the 2nd lowest eigenpair as the deformed shape of the electric pylon structure is shown below. The deformed structure was multiplied by a magnification factor of 20 to allow for visualization of the deformation.

3rd Lowest Eigenpair
The following plot of the 3rd lowest eigenpair as the deformed shape of the electric pylon structure is shown below. The deformed structure was multiplied by a magnification factor of 30 to allow for visualization of the deformation.

Three Lowest Vibrational Periods of the Electric Pylon
The three lowest vibrational periods of the electric pylon structure were determined from the lowest eigenvalues determined above as follows.

$$ \lambda =\omega ^2 $$

Taking the square root of both sides of this equation gives the following equation, which allows for the circular frequency of vibration of the system to be determined.

$$ \omega= \sqrt{\lambda } $$

Substituting the above value for the circular frequency of vibration into the following relation gave the vibrational period of the electric pylon structure for each of the lowest eigenvalues.

$$ T=\frac{2\pi }{\omega } $$

Lowest Vibrational Period
The lowest vibrational period of the electric pylon structure was determined to be 0.11587 s. This value was also shown in the MATLAB command window printout above.

2nd Lowest Vibrational Period
The 2nd lowest vibrational period of the electric pylon structure was determined to be 0.12647 s. This value was also shown in the MATLAB command window printout above.

3rd Lowest Vibrational Period
The 3rd lowest vibrational period of the electric pylon structure was determined to be 0.54654 s. This value was also shown in the MATLAB command window printout above.

 Note: The mode with longest period is called the first fundamental vibration mode, then the second mode is the one with the second longest period, etc. So your "3rd lowest vibrational mode" is actually the first mode with the longest period. Eml4500.f08 20:08, 26 November 2008 (UTC)

Contributing Team Members
Andrew McDonald - Eml4500.f08.bike.mcdonald 18:36, 19 November 2008 (UTC) Garrett Pataky - Eml4500.f08.bike.pataky 06:17, 19 November 2008 (UTC) Sam Bernal - Eml4500.f08.bike.bernal 01:36, 21 November 2008 (UTC) Bobby Sweeting - Eml4500.f08.bike.sweeting 13:01, 19 November 2008 (UTC) Shawn Gravois - Eml4500.f08.bike.gravois 22:16, 19 November 2008 (UTC) Eric Viale - Eml4500.f08.bike.viale 20:10, 20 November 2008 (UTC)


text:
User:Eas4200c.f08.vqcrew.f/hw4

Homework 4

Elastic Theory

For proper understanding of the elastic theory it can help to go back to the Kinematic Assumptions within the road map laid out in previous lectures and stated in the HW 3 report. Recall that in these assumptions that $$\displaystyle \theta$$ is considered to remain constant with respect to 'x' for a uniform bar.

A) Kinematic Assumptions

 * (1) $$u_x (y,z) = \displaystyle \theta \displaystyle \psi (y,z)$$

 * (2) $$u_y (x,z) = -\displaystyle \theta x z$$

 * (3) $$u_z (x,y) = +\displaystyle \theta x y$$

These equations vary slightly from the equations used in the textbook by C.T. Sun, this is due to the use of a different coordinated system. To transform the textbook equations to the uniform notation equations used above it is advised to use cyclic permutation, as seen in figure ###. The basic relationship can be followed by the figure, but the arrows are used to represent the conversion between the x-axis of C.T. Sun becomes the y-axis in uniform notation and likewise for the other axis.

Other important relations can be generated from use of the Kinematic Assumptions and that is the relationship to certain strains and strain rate, giving the following results:

 * $$\displaystyle \epsilon_{xx} = \displaystyle \epsilon_{yy} = \displaystyle \epsilon_{zz} = \displaystyle \gamma_{yz} = 0$$

Proof

These results can easily be proven using the partial derivatives of equations (1), (2), (3) above:

 * $$\displaystyle \epsilon_{xx} = \frac{\partial u_{x} (y,z)}{\partial x} = \frac{\partial (\displaystyle \theta \displaystyle \psi (y,z))}{\partial x} = 0$$

 * $$\displaystyle \epsilon_{yy} = \frac{\partial u_y (x,y)}{\partial y} = \frac{\partial (-\displaystyle \theta x z)}{\partial y} = 0$$

 * $$\displaystyle \epsilon_{yy} = \frac{\partial u_y (x,y)}{\partial y} = \frac{\partial (+\displaystyle \theta x y)}{\partial z} = 0$$

 * $$\displaystyle \gamma_{yz} = \frac{\partial u_y (x,z)}{\partial z} + \frac{\partial u_z (x,y)}{\partial y} = -\theta x + \theta x = 0$$

Formal Derivation

Now looking at the stains in 3-D. There are a total of six strain components for a 3 dimensional object due to symmetry.

 * $$\mathbf{\displaystyle \epsilon} = \begin{bmatrix}

\epsilon_{xx} & \epsilon_{xy} & \epsilon_{xz} \\ \epsilon_{yx} & \epsilon_{yy} & \epsilon_{yz} \\ \epsilon_{zx} & \epsilon_{zy} & \epsilon_{zz} \end{bmatrix}$$

Now that we have a 3x3 matrix with 9 coefficients, we need to change the notation to that of indicial notation.

 * $$ x \Leftrightarrow 1, y \Leftrightarrow 2 , z \Leftrightarrow 3 $$

Now we can write the matrix in indicial notation.

 * $$\mathbf{\displaystyle \epsilon} = \begin{bmatrix}

\epsilon_{11} & \epsilon_{12} & \epsilon_{13} \\ \epsilon_{21} & \epsilon_{22} & \epsilon_{23} \\ \epsilon_{31} & \epsilon_{32} & \epsilon_{33} \end{bmatrix}$$ $$ = \displaystyle \epsilon_{ij}$$

$$\displaystyle \epsilon_{ij}$$ is known at tensorial notation. For this matrix i,j range from 1 to 3, where i is the row number in the matrix and j in the column designation.

Now that the matrix for $$\displaystyle \epsilon_{ij}$$ is formed, the symmetry of the matrix can now be calculated.

 * $$\displaystyle \epsilon_{ij} = \frac{1}{2} (\frac{\partial u_i}{\partial x_j} + \frac{\partial u_j}{\partial x_i})$$

The coordinates x,y,z can now be rewritten: $$ x \Leftrightarrow x_1, y \Leftrightarrow x_2 , z \Leftrightarrow x_3 $$

 * $$\epsilon_{11} = \displaystyle \epsilon_{xx} = \frac{1}{2} (\frac{\partial u_1}{\partial x_1} + \frac{\partial u_1}{\partial x_1})$$

 * $$\displaystyle \epsilon_{xx} = \frac{\partial u_1}{\partial x_1} \Rightarrow \frac{\partial u_x}{\partial x}$$

HW Symmetry $$\displaystyle \epsilon_{ij} = \epsilon_{ji}$$

The symmetry of the matrix can be seen by examining the relationship between $$\displaystyle \epsilon_{21}$$ and $$\displaystyle \epsilon_{21}$$

 * $$\epsilon_{12} = \displaystyle \epsilon_{xy} = \frac{1}{2} (\frac{\partial u_1}{\partial x_2} + \frac{\partial u_2}{\partial x_1})$$

Now look at $$\displaystyle \epsilon_{21}$$

 * $$\epsilon_{21} = \displaystyle \epsilon_{xy} = \frac{1}{2} (\frac{\partial u_2}{\partial x_1} + \frac{\partial u_1}{\partial x_2})$$

It can now clearly be seen that $$\displaystyle \epsilon_{12} = \displaystyle \epsilon_{21}$$

From the Symmetrical proof it can be the following relation can be seen:

 * $$\displaystyle \epsilon_{12} = \displaystyle \epsilon_{21} \Leftrightarrow \epsilon_{xy} = \epsilon_{yx}$$

This relationship provides the result that simplifies the $$\displaystyle \epsilon_{ij}$$ matrix from 9 varialbe to only 6 individual components.

Stress Tensor

The stress tensor can be looked at much like the the strain relations as a 3x3 matrix. Also by symmetry there will be only 6 independent components of $$\displaystyle \sigma$$ in 3-D.

 * $$\displaystyle \sigma = [\sigma]_{3x3}$$

Based of the relations between the stress and strain, several stresses and a shear strain can be found by kinematic assumptions.

 * $$\displaystyle \sigma_{xx} = \displaystyle \sigma_{yy} = \displaystyle \sigma_{zz} = \displaystyle \tau_{yz} = 0$$

title: User:Eas4200c.f08.vqcrew.f/hw4


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User:ShakespeareFan00/Lord Creator From this day

NB The following is considered dual licensed as GFDL and CC-By-SA, I've no objections to it being used and adapted, Some of the verses are from the original Eternal_Father,_Strong_to_Save (a work now in the public domain)

The contributions of User:AdamCuerden in the creation of both lyrical arrangments below is greatly appreciated.

The 2008 Wiki arrangment
Eternal Father, strong to save, Whose arm hath bound the restless wave, Who bidd'st the mighty ocean deep Its own appointed limits keep; Oh, hear us when we cry to Thee, For those in peril on the sea!

O Christ who on the earth did live And saw the evils men did give, Taught wisdom which men's pride denied And liveth on though He had died, Please heed this call, protection deign To all those that shall cross the plain O spirit who in heaven dwells, And swarming darkness ever quells, From vengeful judgement us protect, And to mercies compassion us direct. We ask thy blessings Thou disperse On those whom the blue skies traverse

Lord God creator, from this day The sin of wrath in all us stay And tie us not to cause of birth, May ev'ry man get judged at worth! Thy guided wisdom, offer thee, For all those who shall now be free

Inaugural arrangment (lyrics)
Verses Arrangement Written from (Nov 5th 2008.) The intent was to create a nominally 'Presidential' version here. (Teachers/Educators) O Father, whom to mankind gave Respected elders, prophets brave. Through word divine, deed set apart That we should learn inside our heart And Wisdom may Thou always grant To those in whom a seed we plant.

(Lawyers/men of Justice) Christ, Son redeemer, worthy judge Let not the tempter morals budge False witness none of us let bear When charge be brought, make judgement fair Let innocent ne'er trials fear When we are called to judgement here.

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User:Eml4500.f08.team.guan/spacetrusses

Space Truss Matrix Relations
Knowledge of the 2D trusses can be projected into 3 dimensional trusses as well. The following is the derivation of some matrix relations of trusses in the 3 dimensional space.

Axial Force Displacement Relationship
Here, the axial force displacement relationship is the exact same as the 2D truss. This is because the no matter if the truss is in 2D or 3D space, there can only be two axial displacements so the size of the matrix doesn't change. Hence, the following relation. $$ \frac{EA}{L} \begin{bmatrix} 1 & -1\\ -1 & 1\\ \end{bmatrix}

\begin{bmatrix} q_1\\ q_2\\ \end{bmatrix}= \begin{bmatrix} P_1\\ P_2\\ \end{bmatrix} $$ Where q is the axial displacement and P is the force. As seen in the bottom figure, the truss in 3D space is subject to 6 displacement degrees of freedom and equivalently 2 axial displacements.

Element Dofs and Element Forces in Global (x,y,z) Coordinates
3D trusses introduce 2 extra degrees of freedom in each element. This results in a 6 by 1 elemental displacement matrix and consequently a 6 by 1 elemental force matrix. This is evident from the above figure. These matrices are listed below. $$ d= \begin{bmatrix} d_1^{(e)}\\ d_2^{(e)}\\ d_3^{(e)}\\ d_4^{(e)}\\ d_5^{(e)}\\ d_6^{(e)}\\ \end{bmatrix}$$    $$ f= \begin{bmatrix} f_1^{(e)}\\ f_2^{(e)}\\ f_3^{(e)}\\ f_4^{(e)}\\ f_5^{(e)}\\ f_6^{(e)}\\ \end{bmatrix} $$

Transformation Matrix
In order to change from local displacement degrees of freedom in global coordinates to axial displacements in local coordinates, a new transformation matrix must be implemented. A new direction cosine must be added. The figure below shows how the new direction cosine can be calculated. P_1 is the location of the node 1 and P_2 is the location of node 2. The new direction cosine is similar to the other two direction cosines and describes the displacement of a local node in a the z direction. The equations for the direction cosines can also be found below.  $$l_t=\frac{x_2-x_1}{L}$$ ,   $$m_t=\frac{y_2-y_1}{L}$$   ,   $$n_t=\frac{z_2-z_1}{L}$$ The transformation matrix derivation is as follows. Consider the axial displacement of node 1 on the below diagram.  This displacement is essentially the orthogonal projection of the displacement vector $$d_{[1]}^{(e)}$$ of node 1 on the axis of the element e described by the unit vector $$\tilde {w}$$. In order to describe this relationship mathematically, the dot product is taken between the $$d_{[1]}^{(e)}$$ displacement vector and the unit vector $$\tilde {w}$$. $$ q_1^{(e)}=\vec {d_{[1]}^{(e)}} \cdot \tilde {w}= (d_1^{(e)} \vec{i} +d_2^{(e)} \vec{j} +d_3^{(e)} \vec{k} )\cdot \tilde{w}= d_1^{(e)}\cdot \tilde{w}+d_2^{(e)}\cdot \tilde{w}+d_3^{(e)}\cdot \tilde{w} $$ Node 1 is located at the origin (0,0,0) and node 2 is located at (x_2,y_2,z_2). $$\tilde {w}=\frac{x_2 \vec{i}+y_2 \vec{j}+z_2 \vec{k}}{\sqrt{x_2^2+y_2^2+z_2^2}}=\frac{x_2 \vec{i}+y_2 \vec{j}+z_2 \vec{k}}{L}$$ $$q_1^{(e)}=\frac{x_2}{L}d_1^{(e)}+\frac{y_2}{L}d_2^{(e)}+\frac{z_2}{L}d_3^{(e)}=l_td_1^{(e)}+m_td_2^{(e)}+n_td_3^{(e)}$$ In matrix form, the above  equation yields to the following. $$q_1^{(e)}= \begin{bmatrix} l_t & m_t & n_t\\ \end{bmatrix} \begin{bmatrix} d_1^{(e)}\\ d_2^{(e)}\\ d_3^{(e)}\\ \end{bmatrix} $$ $$q_2^{(e)}= \begin{bmatrix} l_t & m_t & n_t\\ \end{bmatrix} \begin{bmatrix} d_4^{(e)}\\ d_5^{(e)}\\ d_6^{(e)}\\ \end{bmatrix} $$ Combining these two equations yields the following relationship. $$ \begin{bmatrix} q_1^{(e)}\\ q_2^{(e)}\\ \end{bmatrix}= \begin{bmatrix} l_t & m_t & n_t &0 &0 &0\\ 0 &0 &0 &l_t & m_t & n_t\\ \end{bmatrix} \begin{bmatrix} d_1^{(e)}\\ d_2^{(e)}\\ d_3^{(e)}\\ d_4^{(e)}\\ d_5^{(e)}\\ d_6^{(e)}\\ \end{bmatrix} $$ Where the transformation matrix is $$T= \begin{bmatrix} l_t & m_t & n_t &0 &0 &0\\ 0 &0 &0 &l_t & m_t & n_t\\ \end{bmatrix}$$

The Principle of Virtual Work
The principle of virtual work also applied in the same way in 3 dimensional space and is pervasive in the these matrix relationships as well. The force displacement relationship in global coordinates is derived below using the principle of virtual work. $$\displaystyle Kd=F$$ Unlike the elemental stiffness, displacement, and force matrices, the size of the global force displacement matrices depend on the number of truss elements there are. The sizes are not fixed like the elemental case. Give this relation, we can subtract the force matrix from both sides and apply the PVW. Then, we have $$w \cdot (Kd-F)=0$$ for all $$w$$. To prove this, let us select some choices for the weighting function matrix. Choice 1: We select the weighting function such that, $$\displaystyle w_1=1, w_2=0, w_3=0,...,w_n=0$$ where "n" is the number of global displacement degrees of freedom we have. $$w^T= \begin{bmatrix} 1 & 0& 0& ... &...&...\\ \end{bmatrix} $$

Plugging this column matrix into equation 2 will yield the following. $$w \cdot (Kd-F)= 1[\sum_{j=1}^n{K_{1j}d_j-F_1}]+0[\sum_{j=1}^n{K_{2j}d_j-F_2}]+0[\sum_{j=1}^n{K_{3j}d_j-F_3}]+...+0[\sum_{j=1}^n{K_{nj}d_j-F_n}]=0 $$ Multiplying it all out gives $$\sum_{j=1}^n{K_{1j}d_j}=F_1 $$ We continue to choose choices for the weighting matrix strategically for "n" amount of times. Putting all the resulting equations in matrix form will give you $$\displaystyle Kd=F$$

From the 2D case, we derived a square transformation matrix that is invertible. Recall that it equals: $$

T^{(e)}= \begin{bmatrix} R^{(e)}_{2x2} & 0_{2x2} \\ 0_{2x2} & R^{(e)}_{2x2} \\ \end{bmatrix} = \begin{bmatrix} l^{(e)} & m^{(e)} & 0 & 0 \\ -m^{(e)} & l^{(e)} & 0 & 0 \\ 0 & 0 & l^{(e)} & m^{(e)} \\ 0 & 0 & -m^{(e)} & l^{(e)} \\ \end{bmatrix}

$$ In the 3D case, we must account for the extra directional cosine (n_t). It is necessary to derive the transformation matrix in 3D to derive the elemental matrix stiffness matrix in global coordinates in terms of the transformation matrix. Now, lets consider the bottom figure to assist in derive this square, invertible, 3D transformation matrix.  The top truss indicates the displacements in an arbitrary coordinate system whereas the bottom truss indicates the displacements in a specific coordinate system. This coordinate system has x tilda as the axis of the element and the y tilda and z tilda axes are mutually orthogonal with x tilda and each other. In other words, the element is normal to the y tilda and z tilda plane. Thus, the following equation is true.  $$ \begin{bmatrix} q_1\\ q_2\\ \end{bmatrix}= \begin{bmatrix} \tilde {d_1}\\ \tilde {d_4}\\ \end{bmatrix} $$ These are the axial displacements and the rest of the displacements are transverse displacements. The force displacement relationship in this form is as follows. $$

\tilde{f}^{(e)}_{6x1}=\tilde{k}^{(e)}_{6x6}\tilde{d}^{(e)}_{6x1}=

\begin{bmatrix} \tilde{f}^{(e)}_{1}\\ \tilde{f}^{(e)}_{2}\\ \tilde{f}^{(e)}_{3}\\ \tilde{f}^{(e)}_{4}\\ \tilde{f}^{(e)}_{5}\\ \tilde{f}^{(e)}_{6}\\ \end{bmatrix} =k^{(e)} \begin{bmatrix} 1 &0&0&-1&0&0\\ 0&0&0&0&0&0\\ 0 &0&0&0&0&0\\ -1 &0&0&1&0&0\\ 0&0&0&0&0&0\\ 0 &0&0&0&0&0\\ \end{bmatrix} \begin{bmatrix} \tilde{d}^{(e)}_{1}\\ \tilde{d}^{(e)}_{2}\\ \tilde{d}^{(e)}_{3}\\ \tilde{d}^{(e)}_{4}\\ \tilde{d}^{(e)}_{5}\\ \tilde{d}^{(e)}_{6}\\ \end{bmatrix}$$ The new transformation matrix is given below. $$H= \begin{bmatrix} l_t&m_t&n_t\\ l_s&m_s&n_s\\ l_r&m_r&n_r\\ \end{bmatrix} $$     ,       $$ \tilde{T}= \begin{bmatrix} H&0\\ 0&H\\ \end{bmatrix}= \begin{bmatrix} l_t&m_t&n_t&0&0&0\\ l_s&m_s&n_s&0&0&0\\ l_r&m_r&n_r&0&0&0\\ 0&0&0&l_t&m_t&n_t\\ 0&0&0&l_s&m_s&n_s\\ 0&0&0&l_r&m_r&n_r\\ \end{bmatrix}$$ Where l_s is the cosine of the angle between the s and y axes and l_r is the cosine of the angle between the r and z axes. t is the axis defined by the unit vector w and r and s are the axes mutually orthogonal to t. This gives:
$$ k^{(e)}=\tilde{T}^{(e)T}\tilde{k}^{(e)}\tilde{T}^{(e)}...(1) $$ $$ f=kd=[\tilde{T}^{(e)T}\tilde{k}^{(e)}\tilde{T}^{(e)}][\tilde{T}^T\tilde{d}$$] The derivation of the transformation matrix relationship 1 is as follows.
 (5) $$(undefined^{(e)}) \begin{bmatrix} \tilde^{(e)}(undefined^{(e)}^{(e)})-^{(e)} \end{bmatrix} =0$$ (6)  $$\displaystyle (_{pxq}_{qxr})^{T}= ^{T}^{T} $$ (7)  $$_{nx1}_{nx1}= ^{T}_{1xn}_{nx1} $$  Note:  The right side of equation (7) is scalar: $$==(Scalar)

$$

When you apply equations (7) & (6) into equation (5) it yields:

$$(undefined^{(e)})^{T} \begin{bmatrix} \tilde^{(e)}(undefined^{(e)}^{(e)})-^{(e)} \end{bmatrix} =o $$  for all $$_{6x1}$$

 Note:  $$(undefined^{(e)})^{T}$$ comes from equation (7). Applying the transpose rule for matrix multiplication from equation (6) yields:

$$(^{(T)}undefined)^{(e)T} \begin{bmatrix}\tilde^{(e)}(undefined^{(e)}^{(e)})-^{(e)} \end{bmatrix} =o $$

When you replace the transpose with the dot product it yields:

$$ \begin{bmatrix} (undefined^{(e)}\tilde^{(e)}undefined^{(e)})^{(e)}-(undefined^{(e)T}^{(e)} \end{bmatrix} =0 $$ Now we can choose the choices for the weighting function that would strategically yield the desired result.

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