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User:Eml4500.f08.bike.viale/homework report 1

By using Matlab, a user can produce plots of planar curves, 3-D curves, 3-D mesh surfaces, and 3-D faceted surfaces. This section deals with each of these different plots, using the commands 'plot', 'plot3', 'mesh', and 'surf', respectively. Rather than uploading all graphs produced in this section, the reader is encouraged to try the provided Matlab code for themselves to see the plots.

The 2-D Plot
First, the 'plot' command is demonstrated by graphing a sine function over the interval -4 to 4 as follows: This will produce a sine curve.

For another example, the graph y = e- x 2  is produced over the interval -1.5 to 1.5 as follows: The period precedes the ^ in order to perform entry wise operations on the x vector.

The Figure Command
The command 'figure' can be used to produce an empty figure if the user desires. The most recently created figure will be considered the current figure by Matlab, unless the user reselects a different figure. This is done by typing the command 'figure(1)', with the 1 designating the first created figure to be the one made current. The command 'gcf' is used to return the number of the current figure.

The Fplot Command
If an M-file of a function has already been created, the command 'fplot' can be used to quickly plot the graph of that function. As an example, one can create the M-File 'expnormal', with the contents of that function as follows: Once this is saved, then typing: In the workspace will produce the same results as obtained in the 'The Plot Function' section above.

Parametrically Defined Curves
Plots of parametrically defined curves can be made as follows:

Graph Labels
The following is a list of commands which take a string as their argument and the effect the command has on the current plot.

An example is:

>> title('Best Least Squares Fit')

Which will give the title 'Best Least Squares Fit' to the current plot.

To produce grid lines on the graph, the command 'grid' is used'.

Changing Scales on Axes
While Matlab auto-scales the axes, sometimes it is desired to customize the scales. This is done by the following commands:

These commands are only valid after a plot is produced.

Producing Multiple Graphs on One Plot
One way of putting multiple graphs on one plot is as follows:

>> x = 0:.01:2*pi; y1 = sin(x); y2 = sin(2*x); y3 = sin(4*x); plot(x,y1,x,y2,x,y3)

This produces three sine waves on one plot. Another way to produce the same result is:

>> x = 0:.01:2*pi; Y = [sin(x)', sin(2*x)', sin(4*x)']; plot(x,Y)

Also, typing the command 'hold' will freeze the current graph so that another one can be entered on top of it, typing 'hold off' will release the hold.

Line Types and Colors
There are many different line types and colors that can be used in Matlab. They are as follows:

Line types: solid (-), dashed (--), dotted, dashdot(-.) Mark types: point (.), plus (+), star(*), circle(o), x-mark(x)

Colors are chosen as follows:

Colors: yellow (y), magenta (m), cyan (c), red (r), green (g), blue (b), white (w), black (k)

Thus, if a user wants a plot with a red dashed line, the command:

>> plot(x,y,'r--')

Should be used.

Specialized Commands
The 'subplot' command will partition the plot screen into several small plots so that they can be seen all at once in the same figure, but on different plots. Other less used 2-D plotting functions are: 'polar', 'bar', 'quiver', 'compass', 'feather', 'rose', 'stairs', 'fill'.

The 3-D Plot
The command 'plot3' is used to produce 3-D curves in Matlab. It works in basically the same way that the 'plot' command works for 2-D graphs.

A quick example of producing a 3-D plot follows:

>> t = .01:.01:20*pi; x = cos(t); y = sin(t); z = t.^3; plot3(x,y,z)

This produces a helix whose points are all the elements of x,y, and z. The vectors passed into the 'plot3' command must all be the same length to work properly.

3-D Mesh and Surface Plots
3-D wire meshes are produced with the command 'mesh(z)' with z being a 2 dimensional matrix. An example is as follows:

>> mesh(eye(10))

The mesh surface is defined by the elements in the 2 dimensional matrix equaling the z coordinates of the mesh.

Surface plots are similarly created with the command 'surf' as follows:

>> surf(eye(10))

To draw a mesh over a rectangle such as a graph of z = e- x 2 - y 2 over the square [-2,2] x [-2,2], the following procedure is used:

>> xx = -2:.2:2; >> yy = xx; >> [x,y] = meshgrid(xx,yy); >> z = exp(-x.^2 - y.^2); >> mesh(z)

All 2-D formatting commands work for 3-D plots, so please reference that section.

Color and Shading
Color shading of 3-D plots is done using the 'shading' command. These are set by typing either:

shading faceted, shading interp, or shading flat

Into the workspace, with a shading of faceted, interpolated, or flat being the result. The user is encouraged to experiment with all three in order to find which fits the graph best.

The color of the surface can be set with the 'colormap' command, with the input argument being one of:

hsv (default), hot, cool, jet, pink, copper, flag, gray, bone

Thus, the command 'colormap(pink)' will result in a pink colormap.

Parametrically Defined Surfaces
The M-File below gives an example of how to plot a parametrically defined 3-D function:

Sparse Matrix Computations
Matlab has many functions that can speed up computation time if they are used when a matrix is sparse, in other words, many of the entries are zero. To demonstrate the matrix:

>> F = floor(10*rand(6)); F = triu(tril(F,1),-1);

is created. Next the statement:

>> S = sparse(F)

is used to convert F into sparse mode. This statement also returns the nonzero entries of F since the output is not suppressed.

>> F = full(S)

restores S to full storage, and the function issparse(F) can be used to check whether a matrix is sparse or full.

However, sparse matrices are usually generated directly rather than through the sparse function. For example, a sparse tridiagonal matrix is created as shown below:

>> m = 6; n = 6; e = ones(n,1); d = -2*e; >> T = spdiags([e,d,e],[-1,0,1],m,n)

T =

(1,1)      -2       (2,1)        1       (1,2)        1       (2,2)       -2       (3,2)        1       (2,3)        1       (3,3)       -2       (4,3)        1       (3,4)        1       (4,4)       -2       (5,4)        1       (4,5)        1       (5,5)       -2       (6,5)        1       (5,6)        1       (6,6)       -2

Sparse Analogs
The sparse analogs of full matrix functions are as follows:

Operations With Sparse and Full Matrices
Since Matlab's default storage is full matrix, if a sparse and full matrix are operated on together (such as in S*F), Matlab will return a full matrix.

If the only matrices involved in the operation are sparse, the resulting matrix will be stored as sparse.


text: File:Wiki Fig 8.JPG

license: Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/

title: File:Wiki Fig 8.JPG


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User:Eml4500.f08.bike.viale/homework report 2

The above figure from the previous lecture shows the combination of the two local 4x4 k matrices into a 6x6 global k matrix. The top right corner is the k matrix corresponding to the first element and the bottom left corner corresponds to the k matrix of element 2. The 4x4 section in the middle of this global matrix is a combination of the 2 local matrices. The global matrix in terms of the local matrix elements is as follows:

$$

\left[ \begin{array}{cccccc} K^{(1)}_{11} & K^{(1)}_{12} & K^{(1)}_{13} & K^{(1)}_{14} & 0 & 0 \\ K^{(1)}_{21} & K^{(1)}_{22} & K^{(1)}_{23} & K^{(1)}_{24} & 0 & 0 \\ K^{(1)}_{31} & K^{(1)}_{32} & (K^{(1)}_{33} + K^{(2)}_{11} & (K^{(1)}_{34} + K^{(2)}_{12})& K^{(2)}_{13} & K^{(2)}_{14} \\ K^{(1)}_{41} & K^{(1)}_{42} & (K^{(1)}_{43} + K^{(2)}_{21}) & (K^{(1)}_{44} + K^{(2)}_{22}) & K^{(2)}_{23} & K^{(2)}_{24} \\ 0 & 0 & K^{(2)}_{31} & K^{(2)}_{32} & K^{(2)}_{33} & K^{(2)}_{34} \\ 0 & 0 & K^{(2)}_{41} & K^{(2)}_{42} & K^{(2)}_{43} & K^{(2)}_{44} \end{array} \right]\begin{Bmatrix} d_{1} \\ d_{2} \\ d_{3} \\ d_{4} \\ d_{5} \\ d_{6} \end{Bmatrix}=\begin{Bmatrix} F_{1} \\ F_{2} \\ F_{3} \\ F_{4} \\ F_{5} \\ F_{6} \end{Bmatrix}

$$

Using the K values given earlier in class, some sample element calculations are as follows, with the rest able to be found using the symmetry of the matrix:

$$ K_{11} = \frac{9}{16}$$

$$ K_{12} = \frac{3\sqrt{3}}{16}$$

$$ K_{33} = \frac{49}{16}$$

$$ K_{34} = \textbf{-2.175}$$

$$ K_{43} = \textbf{-2.175}$$

$$ K_{44} = \textbf{2.6875}$$

The notation used above is for the global elements, rather than the local elements.

Also, due to the symmetry of the matrix, $$ \textbf{K}_{34} = \textbf{K}_{43}$$, therefore only one needs to be calculated when computing the values of the elements.

The green shaded top right and bottom left corners of the figure correspond to the $$ \textbf{0}$$ elements of the matrix.

After constructing this global matrix, the next step is to reduce the equation by utilizing the boundary conditions. In this particular problem, global nodes 1 and 2 are fixed, thus there is no displacement involved, and:

$$ \textbf{d} = \begin{Bmatrix} 0 \\ 0 \\ d_{3} \\ d_{4} \\ 0 \\ 0 \end{Bmatrix} $$

Utilizing the Principle of Virtural Work allows us to ignore the first, second, fifth and sixth column of the $$ \textbf{k}$$ matrix, as all $$ \textbf{F}$$ elements that it would produce would not be used in the calculation. Thus:

$$

\left[ \begin{array}{cc} 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{array} \right]\begin{Bmatrix} d_{3} \\ d_{4}\end{Bmatrix}=\begin{Bmatrix} F_{3} \\ F_{4}\end{Bmatrix}

$$


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User:Eml4500.f08.qwiki.bishop/Lecture 12

=Lecture 12 - Monday, September 22, 2008=

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

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

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

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

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

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

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

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

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

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

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

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

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

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


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User:Eml4500.f08.FEABBQ.Jayma/notes2

Class Notes: September 17, 2008
Certain degrees of freedom in our example problem are already known; the boundary conditions set these displacements to equal zero, while the displacements of global node 2 are to be determined. Setting $$ d_1 = d_2 = d_5 = d_6 =0 $$ is the same as deleting columns 1, 2, 5 and 6 in the stiffness matrix. Shown below are columns 3 and 4 and the simplified displacement vector..

$$\underline{K}\left[ \begin{array}{l} d_1=0 \\ d_2=0 \\ d_3=? \\ d_4=? \\ d_5=0 \\ d_6=0 \end{array} \right]= \left[ \begin{array}{ll} K_{13} & K_{14} \\ K_{23} & K_{24} \\ K_{33} & K_{34} \\ K_{43} & K_{44} \\ K_{53} & K_{54} \\ K_{63} & K_{64} \end{array} \right] \left[ \begin{array}{ll} d_3 \\ d_4 \end{array} \right] = \underline{F} $$

Since the stiffness matrix reduces down to a 6-by-2 size and the displacement reduces to a 2-by-1 matrix, the resulting force matrix will have dimensions 6-by-1.

Further reduction of this system of matrices can be performed by the Principal of Virtual Work, which allows us to delete the same rows from the stiffness matrix as we did the columns (1, 2, 5 and 6) as shown below.

$$\left[ \begin{array}{ll} K_{33} & K_{34} \\ K_{43} & K_{44} \end{array} \right] \left[ \begin{array}{l} d_3 \\ d_4 \end{array} \right] = \left[ \begin{array}{c} F_3=0 \\ F_4=P=7 \end{array} \right] $$

Since the goal now is to determine the displacements $$d_3$$ and $$d_4$$, we can rearrange the equation above to solve for these displacements.

$$\left[ \begin{array}{l} d_3 \\ d_4 \end{array} \right] = \underline{\overline{K}}^{-1} \left[ \begin{array}{c} 0 \\ P \end{array} \right] $$

We now have an inverse matrix. As a review...

$$\underline{\overline{K}}^{(-1)} = \dfrac{1}{det(\underline{\overline{K}})} \left[ \begin{array}{cc} K_{44} & -K_{34} \\ -K_{43} & K_{33} \end{array} \right]; det(\underline{\overline{K}}) = K_{33}K_{44}-K_{34}K_{43}$$

With the values $$K_{34} = K_{43} = -2.175$$, $$K_{33} = 3.0625$$ and $$ K_{44} = 2.6875 $$, we have

$$det(\underline{\overline{K}}) = 3.0625(2.6875)-(-2.175)(-2.175) = 3.4998$$

$$\underline{\overline{K}}^{(-1)} = \left[ \begin{array}{cc} 2.6875/3.4998 & 2.175/3.4998 \\ 2.175/3.4998 & 3.0625/3.4998 \end{array} \right] = \left[ \begin{array}{ll} 0.76790 & 0.62146 \\ 0.62146 & 0.87505 \end{array} \right] $$

$$\left[ \begin{array}{l} d_3 \\ d_4 \end{array} \right] = \left[ \begin{array}{ll} 0.76790 & 0.62146 \\ 0.62146 & 0.87505 \end{array} \right] \left[ \begin{array}{c} 0 \\ 7 \end{array} \right] = \left[ \begin{array}{cc} 4.35022 \\ 6.12535 \end{array} \right]$$

Since finding the inverse stiffness matrix was the prominent calculation, it is worthwhile to check the solution, especially given such a simple check as shown below.

$$\underline{\overline{K}}(\underline{\overline{K}}^{-1})=\underline{\overline{K}}^{-1}(\underline{\overline{K}})=\underline{I}=\left[ \begin{array}{cc} 1 & 0 \\ 0 & 1 \end{array} \right]$$

The $$\underline{I}$$ above is called the Identity Matrix and results when a matrix is multiplied by it's own inverse.

 * A special note about the transpose of a matrix...

$$\underline{\overline{K}}^{T} = \left[ \begin{array}{cc} K_{33} & K_{43} \\ K_{34} & K_{44} \end{array} \right] $$

The transpose of the matrix $$\underline{\overline{K}}$$ is not the same as the matrix appearing in the $$\underline{\overline{K}}^{-1}$$ equation.

In order to compare to Statics, we compute the reaction forces.

5) Compute Reactions
We use two methods:

5.1) Use the element Force-Displacement relationships; $$\underline{k}^{(e)}\underline{d}^{(e)}=\underline{f}^{(e)}$$, where $$e=1,2$$.

Element 1 (e=1)

$$\underline{k}^{(1)}\underline{d}^{(1)}=\underline{f}^{(1)}$$. The contents of the stiffness matrix were previously calculated, and the displacement matrix is complete from our calculations above for the global displacements, while the force (reaction) matrix is unknown.

$$\underline{d}^{(1)}=\left[ \begin{array}{c} 0\\0\\4.352\\6.1271 \end{array} \right]$$

Element 2 (e=2)

Similarly, for $$\underline{d}^{(2)}$$, the stiffness matrix is already known and the following equation can be used to solve for the reactions of element 2.

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

$$\underline{d}^{(2)}=\left[ \begin{array}{c} 4.352\\6.1271\\0\\0 \end{array} \right]$$

Class Notes: September 19, 2008
Revisiting element 1 and the known $$\underline{k}^{(1)}$$ and $$\underline{d}^{(1)}$$ matrices, we can solve for the reactions.

$$\left[ \begin{array}{cccc} 0.5625 & 0.32476 & ... & ...\\ ...\\...\\...\end{array} \right] \left[ \begin{array}{c} 0\\ 0\\ 4.352\\ 6.1271 \end{array} \right] = \left[ \begin{array}{c} \underline{f}^{(1)}_{1}\\ \underline{f}^{(1)}_{2}\\ \underline{f}^{(1)}_{3}\\ \underline{f}^{(1)}_{4} \end{array} \right]$$

Matrix multiplication results in the following values for the reactions of element 1...

$$ \left[ \begin{array}{c} -4.4378\\-2.5622\\4.4378\\2.5622 \end{array} \right] $$ where $$\underline{f}^{(1)}_{1}$$ and $$\underline{f}^{(1)}_{2}$$ are the reactions at the constrained node and $$\underline{f}^{(1)}_{3}$$ and $$\underline{f}^{(1)}_{4}$$ are the internal forces.

Similarly, for the stiffness and displacement matrices of element 2

$$\left[ \begin{array}{cccc} 2.5 & -2.5 & ... & ...\\ ...\\...\\...\end{array} \right] \left[ \begin{array}{c} 4.352\\ 6.1271\\ 0\\ 0 \end{array} \right] = \left[ \begin{array}{c} \underline{f}^{(2)}_{1}\\ \underline{f}^{(2)}_{2}\\ \underline{f}^{(2)}_{3}\\ \underline{f}^{(2)}_{4} \end{array} \right] = \left[ \begin{array}{c} -4.4377\\4.4377\\4.4377\\-4.4377 \end{array} \right]$$

Through the sum of the forces we can see that $$\Sigma F_{x} = \underline{f}^{(1)}_{1}+\underline{f}^{(1)}_{3}=-4.4378+4.4378=0$$ and $$\Sigma F_{y} = \underline{f}^{(2)}_{1}+\underline{f}^{(1)}_{4}=-2.5622+2.5622=0$$. We also note that $$ \Sigma M_{any point}=0$$. These three conditions satisfy the criteria for self-equilibrium.

HW for September 19, 2008
For element 1, if we call the reactions at local node 1, $$ -P^{(1)}_{1} $$, that is

$$ -P^{(1)}_{1}=\underline{f}^{(1)}_{1}+\underline{f}^{(1)}_{2}=-4.4378x-2.5622y$$

and if we also call the internal forces at node 2, $$ P^{(1)}_{2} $$ and $$ P^{(1)}_{2}=\underline{f}^{(3)}_{1}+\underline{f}^{(4)}_{2}=4.4378x+2.5622y$$

then $$ P^{(1)}_{2}=-(-P^{(1)}_{1})=P^{(1)}_{1} $$, further demonstrating equilibrium of element 1.

Similarly, for element 2, if we call local node 2, $$P^{(2)}_{2}$$, that is

$$ P^{(2)}_{2}=\underline{f}^{(2)}_{3}+\underline{f}^{(2)}_{4}=4.4377x-4.4377y$$

then $$ P^{(2)}_{1}= -P^{(2)}_{2} = \underline{f}^{(2)}_{3}+\underline{f}^{(2)}_{4}=-4.4377x+4.4377y$$

5.2) The second method for solving the two-bar truss problem involves using Statics. Here we use the Cut Principal by Euler.

 * Image of 2 different Euler Cut methods go here.

To bring the Finite Element Solution into the picture, we verify the equilibrium of global node 2.

HW for September 19, 2008

 * Image of global node 2 with three P forces goes here.

title: User:Eml4500.f08.FEABBQ.Jayma/notes2


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User:Eml4500.f08.FEABBQ.koby/HW2

Sept 12
Expressing the global force-matrix function can be simplified by using the compact notation:

$$\left [ K_ij \right ] \left \{ d_j \right \} = \left \{ F_i \right \}$$

$$\sum_{j=1}^6 K_ij d_j = F_i,\qquad i=1...6$$

How do you go from several local element matrices to one global matrix? An assembly plan! First you must identify the correspondence between the various elements and their degree's of freedom:

http://en.wikipedia.org/wiki/Help:Formula

title: User:Eml4500.f08.FEABBQ.koby/HW2


text:
User:Eml4500.f08.delta 6.ramirez/nodeequilibrium

Verification of Equilibrium of element 1:
Note: The reactions for node 1 were calculated to be

F1(1) = -4.4278

F2(1) = -2.5622

F3(1) = 4.4278

F4(1) = 2.5622

Note: Negative values signify that the direction of the forces in element 1 were assumed to be in the wrong direction.

Redrawing the free body diagram with all positive forces shows more clearly how the resultant forces P1 and P2 cancel each other out.

Now,

F1(1) = 4.4278

F2(1) = 2.5622

F3(1) = 4.4278

F4(1) = 2.5622

Determining Resultants:

P1= [(f1(1))2 + (f2(1))2]1/2= [(4.4278)2+ (2.5622)2]1/2= 5.12

P2= [ (f3(1))2 + (f4(1))2]1/2= [(4.4278)2+ (2.5622)2]1/2= 5.12

Note: P1 = P2, thus element 1 is in equilibrium.


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User:Eml4500.f08.delta 6.lopez/HW2

***WEDNESDAY***
The following displays the F matrix calculated two different ways.

$$ \textbf{K}\begin{bmatrix} d_{1}=0\\ d_{2}=0\\ d_{3}\\ d_{4}\\ d_{5}=0\\ d_{6}=0 \end{bmatrix} =\begin{bmatrix} K_{13} & K_{14}\\ K_{23} & K_{24}\\ K_{33} & K_{34}\\ K_{43} & K_{44}\\ K_{53} & K_{54}\\ K_{63} & K_{64} \end{bmatrix} \begin{bmatrix} d_{3}\\ d_{4} \end{bmatrix} = \textbf{F} $$

Here, we delete the corresponding columns in the global stiffness matix, K. Due to the Principle of Virtual Work (PVW), we must also delete the corresponding rows (rows 1, 2, 5, 6).

Below, we have the Resulting Force Displacement relation:

$$ \begin{bmatrix} K_{33} &K_{34} \\ K_{43} & K_{44} \end{bmatrix} \begin{bmatrix} d_{3}\\ d_{4} \end{bmatrix} = \begin{bmatrix} F_{3}\\ F_{4} \end{bmatrix} $$ where the F matrix is found below.

$$ \textbf{F} = \begin{bmatrix} F_{3}\\ F_{4} \end{bmatrix} = \begin{bmatrix} 0\\ P \end{bmatrix} $$ Before we continue, we will find the determinant and inverse of the K matrix. The determinant is simply equal to:

$$ det\; \textbf{K} = K_{33}K_{44}-K_{34}K_{43} $$

Now that we have the determinant, we can calculate the inverse using the formula below.

$$ \textbf{K}^{-1} = \frac{1}{det\; \textbf{K}} \begin{bmatrix} K_{44} & -K_{34}\\ -K_{43} & K_{33} \end{bmatrix} $$

*NOTE* The inverse of the K matrix cannot be found by using the transpose of the matrix, illustrated below.

$$ \textbf{K}^{T} = \begin{bmatrix} K_{33} & K_{34}\\ K_{43} & K_{44} \end{bmatrix} , where\; \textbf{K}^{-1}\neq \frac{1}{det\; \textbf{K}} \textbf{K}^{T} $$

With the inverse of the K matrix known, we can now find the unknown forces from the d matrix.

$$ \begin{bmatrix} d_{3}\\ d_{4} \end{bmatrix} = \textbf{K}^{-1}\begin{bmatrix} 0\\ P \end{bmatrix} = \begin{bmatrix} 4.352\\ 6.127 \end{bmatrix} $$

Step 5: Compute Reactions
There are two main methods for solving this 2-bar truss system.

*Method 1*
For the first method, we will use the element force displacement relationship: $$ \textbf{k}^{e}\textbf{d}^{e}=\textbf{f}^{e}, where\; e=1,2 $$

We can now separate the truss into each individual bar, and compute the calculations for each bar alone.

Element 1:
$$where\; \textbf{d}^{(1)}= \begin{bmatrix} 0\\ 0\\ 4.352\\ 6.127 \end{bmatrix} = \begin{bmatrix} d_{1}^{(1)}\\ d_{2}^{(1)}\\ d_{3}^{(1)}\\ d_{4}^{(1)} \end{bmatrix} $$

Element 2:
$$ where\; \textbf{d}^{(2)}= \begin{bmatrix} 4.352\\ 6.127\\ 0\\ 0 \end{bmatrix}= \begin{bmatrix} d_{1}^{(2)}\\ d_{2}^{(2)}\\ d_{3}^{(2)}\\ d_{4}^{(2)} \end{bmatrix} since\; d_{3}=d_{1}^{(2)}, d_{4}=d_{2}^{(2)} $$

title: User:Eml4500.f08.delta 6.lopez/HW2


text:
User:Eas4200c.f08.carbon.w

=Aircraft Structures=

Course Information

Website - www.mae.ufl.edu/~vql → "teaching" → "my course website" → "EML 4500 Finite Element Anslysis and Design" EML 4500 Course website Password: fall08.wiki

Course Content

Finite Element Analysis (FEM) makes solving partial differential equations (PDEs) convenient. This is integral to the study of Aerospace Structures, or its structural components.

The emphasis of the course will be to understand the mechanics, formulate problems, judge solution correctness, and avoiding ad-hoc structural analysis methods.

The plan of the course is to maintain confidentiality while using large-scale peer-to-peer domains including Wikiversity and MIT's OpenCourseWare in addition to the textbook. The course will examine the method of work instead of other approaches and such as e-learning and the old approach to EML4500 (10% HW and 30% x 3 for each exam). This format allows students to be introduced to Media Wiki and stimulate collaboration between students with other students, teams, the professor, and the class as a whole.

There are several relevant time zones important to us (UTC, EDT, and EST). HW will be turned in according to the UTC time to maintain consistency with the Wikipedia time stamp format.

HW shortcuts should be created by attaching the URL from the archived version by going to "History" → click on the date to open archived version → copy this address to submit

 * {| class="toccolours collapsible collapsed" width="60%" style="text-align:left"

!Contribution Metals Stress-Strain Curve, from team Carbon
 * The graph depicts the general graph of the relationship between stresses and strains in metals.
 * The graph depicts the general graph of the relationship between stresses and strains in metals.

 * }

--

Steel alloys have very high stiffness and strength. Titanium alloys have less high stiffness and strength, and Aluminum alloys still less. Modern aluminum lithium alloys are 10% stiffer and 10% lighter than conventional aluminum alloys.

-Aluminum alloys are resistant to fatigue and cyclic tensile stress, and thus are used on the fuselage and the lower wing skins. Aluminum has historically played a major roll as material used in aircraft structures because it is cheap to buy, manufacture and is abundant, as opposed to Titanium which is expensive.

-Titanium is heavier than aluminum and lighter than steel, and double the yield and ultimate stress of aluminum. It is popular for its corrosion resistance and ability to withstand extreme temperatures up to 1000°F.

-Steel Alloys have extremely high stiffness and strength, but is expensive and has poor resistance to corrosion. Thus, despite its unique material property, is general reserved for landing gear and fittings with high loads.

Reference Sun [2006], pg. 14-15

Because of the very low toughness of glass, it has a very limited use on aircraft. Because of the extreme forces acting on the fuselage in-flight, windows are designed to be very small and sturdy. Recent discoveries of new glass-like materials allow for larger windows. Also, the study of metallic glass has become increasingly significant and could prove to have major importance on aircraft in the future. The article below describes some of the issues and discoveries:

Metallic Glass

Fiber-reinforced composites allow for complex loads by introducing fibers embedded in a matrix (polymers, metals, ceramics, etc.). They are stiff, strong, and light, and are ideal to withstand high loads experienced during flight while still minimizing weight to increase the aerodynamic properties. The composite laminates "have excellent fatigue life, damage tolerance, and corrosion resistance". Fiber composites are becoming very popular for their unique material properties and compose 50% of the new Boeing 787. More facts on the 787 can be found at:

Boeing 787 Fact Sheet

Reference Sun [2006], pg. 15-16

=Pb 1.1=

HW PROBLEM 1.1
BELOW IS THE SOLUTION FROM CLASS.

Consider a square tube (seen to the right)

 * wall thickness << height (a)
 * wall thichness << width (b)

ASSUMPTION 1: $$\displaystyle L = 2 * (a + b)$$

Q: Find optimal b/a ratio to maximize load bearing capability of beam assuming:

ASSUMPTION 2: $$\displaystyle M = T$$ ASSUMPTION 3: $$\displaystyle \sigma$$ allowable $$\displaystyle = 2*\tau$$ allowable

In other words, find optimal cross section carrying max bending moment and max torque:

$$\displaystyle \tau = T /(2 * a * b * t)$$

Since cross section walls are very thin, we can assume shear stress distribution to be uniform along the wall. See the wikipedia article on shear flow to learn more. Shear Flow

Proof:

$$\displaystyle T = T$$ab $$\displaystyle + T$$bc $$\displaystyle + T$$cd $$\displaystyle + T$$da $$\displaystyle$$

$$\displaystyle T$$ab $$\displaystyle= b/2 * (\tau t * a ) = \tau *a*b*t/2$$

$$\displaystyle T$$bc $$\displaystyle= a/2 * (\tau * t * b) = \tau *a*b*t/2$$

$$\displaystyle T$$cd $$\displaystyle= b/2 * (\tau * t * a) = \tau *a*b*t/2$$

$$\displaystyle T$$da $$\displaystyle= a/2 * (\tau * t * b) = \tau *a*b*t/2$$

Recalling that $$\displaystyle T = T$$ab $$\displaystyle + T$$bc $$\displaystyle + T$$cd $$\displaystyle + T$$da $$\displaystyle$$, this proves that $$\displaystyle T = 2 \tau a b t$$ and  $$\displaystyle  \tau = T/(2 a b t)$$.

$$\displaystyle $$

Case:1
Assume stress (bending normal stress) reaches stress allowable, first. What to verify that $$\displaystyle \tau << \tau$$allowable

Recall: $$\displaystyle Stress = \sigma = (M *z)/I$$

M = bending moment, Z = ordinate of a point on axis perpendicular to neutral bending axis

I = 2nd area moment of inertia

Assume $$\sigma_{max} = \sigma_{allowable}$$, that the maximum bending normal stress increases to the allowable stress first. The goal is to find the maximum allowable bending moment which corresponds to the max bending stress at z=b/2.

Recall: $$\sigma = \frac {Mz} I, ~ Z = \frac b 2$$

=> $$M = \frac {2 I \sigma_{max}} b \frac {2 I \sigma_{allowable}} b$$

To maximize the bending moment, simply maximize the ratio I/b. First, find the moment of inertia with respect to 'b',

Recall from assumption 1:

=> $$a = \frac L 2 - b$$

=> (I/b) fuction of b

Maximize M : Max = max, from assumption 2 (T=M)

According to assumption 3, the allowable shear stress is equal to half of the allowable normal stress. Therefore, if the max shear stress is less than the allowable shear stress, then the optimal ratio is acceptable. If not, then it is not acceptable.

Assume that the maximum bending normal stress reaches the allowable stress first. Then the moment of inertia of the cross section is the sum of the moments fo inertia of the cross section is the sum of the moments of inertia of the four segments of the cross section. If you use the parallel axis theorem, it is as follows:

t^3 << t, hence the approximation

Plot f(b) we get inverted parabola, 0 at 0 and 3L/4. max at 3L/8

We can see that this case can NOT exist! We have to violate a known equation to use this cross-section!

Case 2:
Assume torsional shear stress reaches the allowable shear stress first. Torsional shear stress is constant along the thin wall because the cross section is negligibly thin.

$$\tau^{(1)}_{max} < \tau_{allowable}$$

$$\tau_{max} = \tau_{allowable} = \tau = \frac T {2~a~b~t}$$

Maximize T by maximizing the product of a and b. For a constant perimeter L, the cross section of the box beam is square. The product of a and b is the area of the cross section and a=b, therefore L/4 = a = b. We need to check to make sure that the maximum bending stress for the cross section is below the allowable bending stress. If this is true, then the optimal ratio b/a = 1 is acceptable, otherwise it is not acceptable.

$$\frac {a^{(2)}} {b^{(2)}} = 1$$

$$T^{(2)}_{max} = (2 + \tau_{all}) * (\frac L 4) = \frac 1 8 + L_{tall} = M^{(2)}_{max}$$

$$M_{max} = 2 \sigma_{all} (\frac I b) = \frac {3 + L^2} {32} \sigma_{all}$$

$$\tau_{max} = \frac {M_{max}} {2abt} = \frac {M_{max}} {2 (\frac L 8) (3 \frac L 8) t} = \frac {32M_{max}} {3tL^2} = \sigma_{all}$$

Shear stress due to maximum Torsion(1)=Tmax(1)/(2a(1)b(1)t) = (by assumption 2) Mmax(1)/[2(L/8)(3L/8)t] = σallowable

τmax(1) = σallowable = (by assumption 3) 2τallowable > τallowable

Because τmax(1) > τallowable, therefore this case is not acceptable.

=HW3=

uy = -(PP')sinβ. Because PP'=(OP)α

→ uy = -α(OP)sinβ, where (OP)sinβ=zp

=Warping=

Torsion relates the rate of twist of an object and the displacement of any point along a non-centerline axis to an applied torque.

θ=α/x=rate of twist.

The following are mathematical definitions for the warping along the x,y, and z-axes.

1: y-displacement of a point on the cross section, horizontal

→ uy=-rαsinβ=-αy=-θxz

2: z-displacement of a point on the cross section, vertical

→ uz=v=rαcosβ=+(PP')cosβ=+αyp=+θxy, where PP'=OPα, α=θx, and yp=y because the variable is general for all coordinates

3: x-displacement, along centerline

→ ux=θψ(y,z), where this value is proportional to the rate of twist (θ)

These three equations compose the kinematic assumptions of warping.

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!Contribution Physical Representation of Torsion, from team Carbon
 * Torsion relates the rate of twist of an object and the displacement of any point along a non-centerline axis to an applied torque.
 * Torsion relates the rate of twist of an object and the displacement of any point along a non-centerline axis to an applied torque.

Contribution by carbon.w
 * }

Roadmap for Torsional Analysis of Aircraft Wing
Below is a picture of a cross section of a wing with a two-cell stringer-skin-web section with a thin, variable wall.

A) Kinematic Assumptions described above with the three equations.

B) Strain-displacement relationship:

εxx = εyy = εzz = γxy = 0

σxx = σyy = σzz = τxy = 0

$$\gamma _{xz} = \frac{\partial w}{\partial x} + \frac{\partial u}{\partial z}$$

$$\gamma _{yz} = \frac{\partial w}{\partial y} + \frac{\partial v}{\partial z}$$

C) Equilibrium Equation for Stresses:

$$\frac{\partial \tau _{xz}}{\partial x} + \frac{\partial \tau_{yz}}{\partial y} = 0$$

$$\frac{\partial \sigma _{xx}}{\partial x} + \frac{\partial \tau _{yx}}{\partial y} + \frac{\partial \tau _{zx}}{\partial z} = 0$$

$$\frac{\partial \tau _{xy}}{\partial x} + \frac{\partial \sigma _{yy}}{\partial y} + \frac{\partial \tau _{zy}}{\partial z} = 0 $$

$$\frac{\partial \tau _{xz}}{\partial x} + \frac{\partial \tau _{yz}}{\partial y} + \frac{\partial \sigma _{zz}}{\partial z} = 0 $$

$$\tau _{xy} = \tau _{yx}$$

$$\tau _{yz} = \tau _{zy}$$

$$\tau _{xz} = \tau _{zx}$$

D) Prandtl Stress Funtion, $$\phi $$

$$\tau _{xz} = \frac{\partial \phi }{\partial y}$$

$$\tau _{yz} = \frac{\partial \phi }{\partial x}$$

E) Strain Compatibility

$$\frac{\partial \gamma _{yz}}{\partial x} - \frac{\partial \gamma _{xz}}{\partial y} = 2\theta $$

F) Equation for $$\phi $$

$$\frac{\partial ^{2}\phi }{\partial x^{2}} = -2G\theta$$

G) Boundary Conditions for $$\phi$$

$$\frac{\partial \phi }{\partial s} = 0$$ or $$\phi $$ = constant (on the lateral surface)

If cross section is solid and thus has only a single contour boundary, then the constant can be chosen arbitrarily as zero: $$\phi$$ = 0 on lateral surface of the bar

H) Torque, $$T = 2\int \int_{A}^{}{\phi dA}$$

→ $$T = GJ\theta $$, where using the above equation and the equation for $$\phi $$ we obtain $$J = -\frac{4}{\bigtriangledown^{2}\phi }\int \int_{A}^{}{\phi dxdy}$$ and $$(\phi dxdy = dA)$$

I) Thin Walled Cross Section

$$\tau = \frac{T}{2abt}$$ Ad-hoc assumption of shear flow

$$T = \oint_{}^{}{\rho qds} = \int \int_{\bar{A}}^{}{2qdA} = 2q\bar{A}$$

J) Twist Angle, θ: Method 1

\theta = \frac{1}{2G\bar{A}}\oint_{}^{}{\tau ds} = \frac{1}{2G\bar{A}}\oint_{}^{}{\frac{q}{t}ds}

=Multicell Airfoil=

To consider an example of an airfoil separated by spars (creating multicells), one must first consider the simplified case using right angles. In the figure below:

t1 = .3cm, t2 = .5cm, t3 = .4cm

Looking back to the Roadmap Section K, and using the equation $$\theta_i = \frac{1}{2G_i\bar{A_i}} \oint \frac{q_i}{t_i} ds$$ we can find θ as a function of T and J (torsional constant).

(1) $$T = T_{1} + T_{2} = 2q_{1}\bar{A_{1}} + 2q_{2}\bar{A_{2}}$$, where $$\bar{A_{1}} = ac$$ and $$\bar{A_{2}} = bc$$

(3) $$\theta_1 = \frac{1}{2G\bar{A_1}}[\frac{2q_{1}a}{t_{1}} + \frac{q_{1}c}{t_{1}} + \frac{(q_{1} - q_{2})c}{t_{12}}]$$

(4) $$\theta_2 = \frac{1}{2G\bar{A_2}}[\frac{2q_{2}a}{t_{2}} + \frac{q_{2}c}{t_{2}} + \frac{(q_{2} - q_{1})c}{t_{12}}]$$

Because the two cells are connected the angle of each cell must rotate together. In other words, C1 and C2 have the same rate of twist angle, where q1 and q2 are two unknowns.

(2) $$\theta_1 = \theta_2$$ Equations (1) and (2) can be used to solve for an expression of q1 and q2 in terms of the resultant of T.

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!HW Torsion of Multicell Airfoil, from team Carbon
 * $$q_{1} = \frac{T_{1}}{2\bar{A_{1}}}$$, where $$\beta _{1} = \frac{1}{2\bar{A_{1}}} = \frac{1}{2*a*c} \approx 4.167m$$
 * $$q_{1} = \frac{T_{1}}{2\bar{A_{1}}}$$, where $$\beta _{1} = \frac{1}{2\bar{A_{1}}} = \frac{1}{2*a*c} \approx 4.167m$$

→ $$q_{1} = 4.167T_{1}$$

$$q_{2} = \frac{T_{2}}{2\bar{A_{2}}}$$, where $$\beta _{2} = \frac{1}{2\bar{A_{2}}} = \frac{1}{2*b*c} \approx 2.083m$$

→ $$q_{2} = 2.083T_{2}$$

From (3): $$\theta = \frac{1}{2G\bar{A_1}}[\frac{2q_{1}a}{t_{1}} + \frac{q_{1}c}{t_{1}} + \frac{(q_{1} - q_{2})c}{t_{12}}]$$, where $$\theta _1 = \theta _2 = \theta$$

→ $$\theta = \frac{1}{2Gac}[\frac{24.167T_{1}a}{t_{1}}  \frac{4.167T_{1}c}{t_{1}}   \frac{(4.167T_{1} - 2.083T_{2})c}{t_{12}}]$$

→ $$\theta = \frac{1}{.24G}[T_{1}(833.4 + 333.36) + 416.7T_{1} - 208.3T_{2}]$$

→ $$\theta = \frac{1}{.24G}(1583.46T_{1} - 208.3T_{2})$$

Using $$T = GJ \theta $$ and $$\theta _1 = \theta _2 = \theta$$, plug into the equation for $$\theta$$

→ $$\theta = \frac{1}{.24G}(1583.46GJ \theta - 208.3GJ \theta)$$

→ $$\theta = 5729.83J \theta$$

→ $$J = 1.745e-4$$

Contribution by carbon.w
 * }

Use $$T = T_{1} + T_{2}$$ to find the first equation:

$$\bar{A}_{1} = a*c = (.3m)(.4m) = .12m^{2}$$ and $$\bar{A}_{2} = b*c = (.6m)(.4m) = .24m^{2}$$

→ $$T = 2q_{1}\bar{A}_{1} + 2q_{2}\bar{A}_{2} = .24q_{1} + .48q_{2}$$ (1)

Now solve for $$\theta _{1}$$ and $$\theta _{2}$$:

 * $$\theta _{1} = \frac{1}{2G\bar{A_1}}[\frac{2q_{1}a}{t_{1}} + \frac{q_{1}c}{t_{1}} + \frac{(q_{1} - q_{2})c}{t_{12}}]$$, where $$G_{1} = G_{2} = G$$

$$\theta _{1} = \frac{1}{.24G}[200q_{1} + 133.33q_{1} + 100(q_{1} - q_{2})]$$

$$\theta _{1} = \frac{1}{.24G}(433.33q_{1} - 100q_{2})$$

 * $$\theta _{2} = \frac{1}{2G\bar{A_2}}[\frac{2q_{2}a}{t_{2}} + \frac{q_{2}c}{t_{2}} + \frac{(q_{2} - q_{1})c}{t_{12}}]$$

$$\theta _{2} = \frac{1}{.48G}[240q_{2} + 80q_{2} + 100(q_{2} - q_{1})]$$

$$\theta _{2} = \frac{1}{.48G}(420q_{2} - 100q_{1})$$

 * $$\theta _{1} = \theta _{2}$$

$$\frac{1}{.24}(433.33q_{1} - 100q_{2}) = \frac{1}{.48}(420q_{2} - 100q_{1})$$

$$2013.875q_{1} = 1291.67q_{2}$$, where G cancels. (2)

 * Solving the system of equations of (1) and (2):

from (1): $$q_{1} = 4.167T - 2q_{2}$$

substitute into (2): $$(2013.87)(4.167T - 2q_{2}) = 1291.67q_{2}$$

→ $$q_{2} = .000783T$$

 * Substitute into (2):

→ $$q_{1} = .0005024T$$

Since $$\theta _{1} = \theta _{2} = \theta $$, we can plug into either $$\theta _{1}$$ or $$\theta _{2}$$

$$\theta _{2} = \theta = \frac{1}{.48G}(420q_{2} - 100q_{1})$$

$$\theta = \frac{.58}{G}T$$

 * Because $$\theta = \frac{T}{2GJ}$$, therefore $$\frac{.58}{G}T = \frac{T}{2GJ}$$

→ → $$J = .862m^{4}$$

$$\theta = \frac{1}{2G\bar{A_1}}[\frac{2q_{1}a}{t_{1}} + \frac{q_{1}c}{t_{1}} + \frac{(q_{1} - q_{2})c}{t_{12}}]$$

Lecture 22
$$\boldsymbol{\varepsilon } = \begin{bmatrix} \varepsilon _{11} & \varepsilon _{12} & \varepsilon _{13}\\ \varepsilon _{21} & \varepsilon _{22} & \varepsilon _{23}\\ \varepsilon _{31} & \varepsilon _{32} & \varepsilon _{33} \end{bmatrix} = [\varepsilon _{ij}]$$ (tensorial notation), where x → 1, y → 2, z → 3, i = row, j = column, and i,j = 1,2,3

By using the tensorial notation, it allows you to only need to memorize one equation for $$\boldsymbol{\varepsilon }$$ rather than three:

$$\varepsilon _{ij} = \frac{1}{2}(\frac{\partial u_{i}}{\partial x_{j}}  \frac{\partial u_{j}}{\partial x_{i}})$$

where x → x1, y → x2, z → x3

$$\varepsilon _{11} = \varepsilon _{xx} = \frac{1}{2}(\frac{\partial u_{1}}{\partial x_{1}}  \frac{\partial u_{1}}{\partial x_{1}}) = \frac{\partial u_{1}}{\partial x_{1}} = \frac{\partial u_{x}}{\partial x}$$

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!HW Strain Symmetry, from team Carbon
 * By symmetry, $$\varepsilon _{ij} = \varepsilon _{ji}$$. If we make this assumption, then:
 * By symmetry, $$\varepsilon _{ij} = \varepsilon _{ji}$$. If we make this assumption, then:

$$\varepsilon _{ij} = \frac{1}{2}(\frac{\partial u_{i}}{\partial x_{j}} + \frac{\partial u_{j}}{\partial x_{i}}) = \varepsilon _{ji} = \frac{1}{2}(\frac{\partial u_{j}}{\partial x_{i}} +  \frac{\partial u_{i}}{\partial x_{j}})$$

Contribution by carbon.w
 * }

Use $$T = T_{1} + T_{2}$$ to find the first equation:

$$\bar{A}_{1} = a*c = (.3m)(.4m) = .12m^{2}$$ and $$\bar{A}_{2} = b*c = (.6m)(.4m) = .24m^{2} $$

→ $$T = 2q_{1}\bar{A}_{1} + 2q_{2}\bar{A}_{2} = .24q_{1} + .48q_{2}$$ (1)

Now solve for $$\theta _{1}$$ and $$\theta _{2}$$:

=Equation of Equilibrium=

Using dimensional analysis, we can find the dimension, which is not the same as the unit. All measures have units, but some are dimensionless.

To find the dimension of $$\begin{bmatrix} \frac{f}{A} \end{bmatrix}$$, let's analyze each individual component:

$$[f] = \frac{F}{L}$$ and $$[A] = L^2$$

→ $$\begin{bmatrix} \frac{f}{A} \end{bmatrix} = \frac{F}{L^3}$$

This same methodology can be applied to any measurement such as stress and strain. The ad-hoc proof below show that you can approach the same dimension no matter what form of a measure you use:

$$[\sigma ] = \frac{F}{L^{2}}$$ and $$[dx] = L \Rightarrow \begin{bmatrix} \frac{d\sigma }{dx} \end{bmatrix} = \frac{[d\sigma ]}{[dx]} = \frac{F/L^{2}}{L} = \frac{F}{L^{3}} = [f]$$

Below is the dimensional analysis for strain, and shows that strain is non-dimensional (but still has a unit).

Unit: $$\varepsilon = \frac{du}{dx} = \frac{\Delta L}{L}$$

Dimension: $$[\varepsilon ] = \frac{[du]}{[dx]} = \frac{L}{L} = 1$$ (non-dimensional)

Also, $$\nu = -\frac{\varepsilon _{yy}}{\varepsilon _{xx}} \Rightarrow [\nu ] = \frac{[\varepsilon _{yy}]}{[\varepsilon _{xx}]} = 1$$

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!Contribution Eigenvalues and Buckling, from team Carbon
 * Eigenvalues are very important in determining material capabilities and making judgments for better designs. A study at Rice University using eigenvalue analysis showed that a column would be strongest if it were widest at the top, middle, and bottom of the column and skinny 1/4 distance from the top and 1/4 from the bottom.
 * Eigenvalues are very important in determining material capabilities and making judgments for better designs. A study at Rice University using eigenvalue analysis showed that a column would be strongest if it were widest at the top, middle, and bottom of the column and skinny 1/4 distance from the top and 1/4 from the bottom.

This study showed that this is the most material-conscious and strength-conscious way to build a column. Eigenvalues showed that naturally a column, regardless of its shape buckles at the top and lower quarter location. To conserve material, the thinning of these sections would not make a contribution to the buckling.

Contribution by carbon.w
 * }

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!HW Particularize symmetrical cross section, from team Carbon
 * For the second moment of inertia about the y-axis then the z-axis, the symmetry would be at the centroid if the cross section is symmetrical. Because there is symmetry about the centroid, integration on either side would cancel out and equal zero.
 * For the second moment of inertia about the y-axis then the z-axis, the symmetry would be at the centroid if the cross section is symmetrical. Because there is symmetry about the centroid, integration on either side would cancel out and equal zero.

Considering if $$M_{z} = 0$$ and that $$I_{yz} = 0$$. From previous lectures, it was shown that by definition

$$D:= I_{y}I_{z} - (I_{yz})^{2}$$. Therefore, $$D = I_{y}I_{z}$$

$$\Rightarrow k_{y} = \frac{I_{y}}{D} = \frac{I_{y}}{I_{y}I_{z}} = \frac{1}{I_{z}}$$

$$k_{yz} = 0$$

$$k_{z} = \frac{I_{z}}{I_{y}I_{z}} = \frac{1}{I_{y}}$$

$$\sigma _{xx} = \begin{bmatrix} z & 0 \end{bmatrix}\begin{bmatrix} \frac{1}{I_{y}} & 0\\ 0 & \frac{1}{I_{z}} \end{bmatrix}\begin{Bmatrix} M_{y}\\ 0 \end{Bmatrix}$$

$$\sigma _{xx} = \begin{bmatrix} z & 0 \end{bmatrix} \begin{Bmatrix} \frac{M_{y}}{I_{y}}\\ 0 \end{Bmatrix}$$

$$\Rightarrow \sigma _{xx} = \frac{zM_{y}}{I_{y}}$$

Contribution by carbon.w
 * }

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!HW Matrix Form of q(s), from team Carbon
 * The following equation can can be written in matrix form $$q(s) = -(k_{y}V_{y} - k_{yz}V_{z})Q_{z} - (k_{z}V_{z} - k_{yz}V_{y})Q_{y}$$
 * The following equation can can be written in matrix form $$q(s) = -(k_{y}V_{y} - k_{yz}V_{z})Q_{z} - (k_{z}V_{z} - k_{yz}V_{y})Q_{y}$$

$$q(s) = \begin{bmatrix} -Q_{z} & -Q_{y} \end{bmatrix}\begin{bmatrix} k_{y} & -k_{yz}\\ -k_{yz} & k_{z} \end{bmatrix}\begin{Bmatrix} V_{y}\\ V_{z} \end{Bmatrix}$$

To arrive at the particular case where $$q(s) = -\frac{V_{z}Q_{y}}{I_{y}}$$, make the assumption that $$Q_{z} = 0$$ and $$I_{yz} = 0$$

Therefore $$D = I_{y}I_{z}$$ and this can be plugged into the k-values:

$$\Rightarrow k_{y} = \frac{I_{y}}{D} = \frac{I_{y}}{I_{y}I_{z}} = \frac{1}{I_{z}}$$

$$k_{yz} = 0$$

$$k_{z} = \frac{I_{z}}{I_{y}I_{z}} = \frac{1}{I_{y}}$$

The matrix then becomes

$$q(s) = \begin{bmatrix} 0 & -Q_{y} \end{bmatrix}\begin{bmatrix} \frac{1}{I_{z}} & 0\\ 0 & \frac{1}{I_{y}} \end{bmatrix}\begin{Bmatrix} V_{y}\\ V_{z} \end{Bmatrix} = q(s) = -\frac{V_{z}Q_{y}}{I_{y}}$$

Contribution by carbon.w
 * }

Lecture 38
$$Q_{y}^{(3)} = \int_{A_{3}}^{}{zdA_{3}}$$

$$Q_{y}^{(2)} = z_{2}A_{2}$$

Resuperposition of Problem 2
The basic formula for shear flow along the skin:

$$q_{ij} = \tilde{q}_{ij} + q_{k}$$

Following this formula, several equations can be found:

 * $$q_{12} = \tilde{q}_{12} + q_{1}$$, where $$\tilde{q}_{12}$$ is known as zero and $$q_{1}$$ is unknown

 * $$q_{23} = \tilde{q}_{23} + q_{1} - q_{2}$$

 * $$q_{31} = \tilde{q}_{31} + q_{1} - q_{3}$$

 * $$q_{24} = \tilde{q}_{24} + q_{2}$$

 * $$q_{43} = \tilde{q}_{43} + q_{3} - q_{4}$$

 * $$q_{41} = \tilde{q}_{41} + q_{4}$$

Looking at all these equations, there are 3 unknowns, and thus we need 3 equations.

1) The first equation would be one equation; finding the moments of Vy and Vz, and $${q_{12}, q_{23}, q_{31}, q_{24}, q_{43}, q_{41}}$$ about any convenient point.  The point about which the moment is taken should be where the lines of action of $$V_{y}$$ and $$V_{z}$$ intersect.

Because each of the three sections are combined, if one section twists by a certain angle, then the other two must also twist by the same angle. Therefore,

2) $$\theta _{1} = \theta _{2}$$ 3) $$\theta _{2} = \theta _{3}$$

Lecture 39
To solve problem 2, follow each step for each cell:

 * Follow the path "$$s_{i}$$"
 * Write the equilibrium of each stringer on the path "$$s_{i}$$"

There are two ways to write the equilibrium:

1) Method shown in previous section, using FBD 2) The consequence of the first method is shown below:

The equilibrium equation is: $$\tilde{q}_{j6} = \tilde{q}_{3j} - \tilde{q}_{j5} - \tilde{q}_{j7}  q^{j} $$

The equilibrium equation is: $$-\tilde{q}_{6j} = \tilde{q}_{3j} - \tilde{q}_{j5} - \tilde{q}_{j7}  q^{j} $$

Note:

Consider what would happen if the cell walls were cut such that one stringer was isolated. Theoretically, it cannot be solved.

$$\tilde{q}_{23} = \tilde{q}_{31} = \tilde{q}_{34} = 0$$

$$\tilde{q}_{31} = \tilde{q}_{23} - \tilde{q}_{34} + q^{(3)}$$

This means that $$q^{(3)} = 0$$, but this is not true!

A second example is shown below:

$$\tilde{q}_{24} = \tilde{q}_{23} = \tilde{q}_{12} = 0$$

$$\tilde{q}_{23} = \tilde{q}_{12} - \tilde{q}_{24} + q^{(2)}$$

This shows that $$q^{(2)} = 0$$ which also is not true!

Using the peace sign, which was popular in the '60s:

 * $$\tilde{q}_{15} = \tilde{q}_{54} = \tilde{q}_{42} = \tilde{q}_{21} = 0$$

$$\tilde{q}_{15} = \tilde{q}_{31} + \tilde{q}_{21} + q^{1}$$

 * $$\Rightarrow \tilde{q}_{31} = -q^{(1)}$$

 * $$\tilde{q}_{53} = q^{(5)}$$

 * $$\tilde{q}_{43} = q^{(4)}$$

 * $$\tilde{q}_{23} = q^{(2)}$$

 * $$\tilde{q}_{15} = \tilde{q}_{31} + \tilde{q}_{21} + q^{1}$$

$$\tilde{q}_{31} = \tilde{q}_{53} + \tilde{q}_{43} + \tilde{q}_{23} + q^{(3)}$$

$$\Rightarrow \tilde{q}_{31} = q^{(5)} + q^{(4)} + q^{(2)} - q^{(1)}$$

There are four unknowns, so there will need to be four equations described in the previous section:

1) Moment equation

2) $$\theta _{1} = \theta _{2}$$

3) $$\theta _{2} = \theta _{3}$$

4) $$\theta _{3} = \theta _{4}$$

$$\tilde{q}_{53} = \tilde{q}_{43} = \tilde{q}_{23} = \tilde{q}_{31} = 0$$

$$\tilde{q}_{31} = \tilde{q}_{53} + \tilde{q}_{43} + \tilde{q}_{23} + q^{(3)}$$

This leads to $$q^{(3)} = 0$$ which is not true.

title: User:Eas4200c.f08.carbon.w


text:
User:Eml4500.f08.group.w

=Finite Element Analysis=

Course Information

Website - www.mae.ufl.edu/~vql → "teaching" → "my course website" → "EML 4500 Finite Element Anslysis and Design" EML 4500 Course website Password: fall08.wiki

MWF Period 8 (3:00-3:50PM), FLG 270 Office: NEB135

Course Content

Finite Element Analysis (FEM) makes solving partial differential equations (PDEs) convenient. This is integral to the study of Aerospace Structures, or its structural components.

The emphasis of the course will be to understand the mechanics, formulate problems, judge solution correctness, and avoiding ad-hoc structural analysis methods.

The plan of the course is to maintain confidentiality while using large-scale peer-to-peer domains including Wikiversity and MIT's OpenCourseWare in addition to the textbook. The course will examine the method of work instead of other approaches and such as e-learning and the old approach to EML4500 (10% HW and 30% x 3 for each exam). This format allows students to be introduced to Media Wiki and stimulate collaboration between students with other students, teams, the professor, and the class as a whole.

There are several relevant time zones important to us (UTC, EDT, and EST). HW will be turned in according to the UTC time to maintain consistency with the Wikipedia time stamp format.

HW shortcuts should be created by attaching the URL from the archived version by going to "History" → click on the date to open archived version → copy this address to submit

=Chapter 4: Trusses, Beams, and Frames=

Trusses
The image below is a two-truss system with elastic (deformable) bars.

Global node numbers are denoted by circles while element numbers are denoted by triangles. The elements are separable members of a system which can be individually analyzed. Nodes are points on a system on which forces appear. The x-y coordinate system can be located in any rotation about the z-axis, however in this case the y-axis is chosen to be parallel to the applied force P. Furthermore, the triangles beneath nodes 1 and 2 denote that the node displacement is fixed to zero in both the x- and y-direction.

Forces acting on the system are shown below with a systematic example of how to name the forces on the member. The names given to each of forces are arbitrary, but must be systematic and consistent.

The R1 and R3 forces can be replaced with two forces instead of four that act along the line of the trusses. Taking the moment about node 2 equal to zero, gives no conclusion because both R1 and R3 forces run through the node. The figure then has two unknown reactions, and only two equations of equilibrium (the sum of forces about the x- and y-axes. This means that the system is statically indeterminate (equations of statics cannot be used to solve R1 and R3.

To examine the forces, the system can be split into two independent bar elements.

Bar element 1:

This picture is another example of how to name the forces, nodes and elements in a systematic manner, rather than a rigid, consistent manner. The element number is denoted by a triangle again. The local nodes (nodes relative only to the specific FBD section) are denoted by squares as opposed to the global nodes (nodes relative to the entire structure) which are circled. The subscript of the forces are the degrees of freedom (dof) or internal force of the element. Each force is a separate degree of freedom and should have their own number. Also they are numbered in a consistent order: x-, y-, x-, y-axis. The superscript represent the element number in parentheses. For this global system, the dof's are 1, 2, 3, and 4 and the element numbers are 1, 2, which come to eight total numbers in both elements.

Bar element 2:

Force Displacement (FD) Relation

The diagram above shows the relationship between force and displacement in a spring with one end fixed. It both ends are free, the case becomes as follows.

Because there are now two independent variables, the spring must be analyzed with a matrix form of F = kd. Matrices are names by "row x column"

\begin{pmatrix} f_1 \\ f_2 \end{pmatrix}_{2x1} = \begin{bmatrix} k & -k \\ -k & k \end{bmatrix}_{2x2} \begin{pmatrix} d_1 \\ d_2 \end{pmatrix}_{2x1} $$

There are 2 cases which can be examined:

Case 1 is the result with respect to node 1:

$$f_2 = k (d_2 - d_1)$$

Case 2 is the result with respect to node 2:

$$f_1 + f_2 = 0$$

→ $$f_1 = -f_2 = -k (d_2 - d_1) = k (d_1 - d_2)$$

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!Contribution title, from team group
 * Text
 * }
 * }

=MATLAB Primer=

Hard copy
Command:  filename

This command pulls up the hardcopy and copies the content excluding graphics to that file, and will go to a default file if no filename is given. It continues to copy to the file until you command it to  or.

Graphics
MATLAB plots curves, 3D plots of curves , #D mesh surface plots , and 3D faceted surface plots. allows you to preview these features.

Planar plots
The command  creates an x-y plot as shown below the format for the range is 'lower rang:meshsize: upper range'.

Figures can be suppressed or exposed as the "current" figure by the command according to its figure number, and command   will give you the number of the current figure.

To make graphs, first put a function in an M-file then command it open with a domain:

Parametric graphs can also be defined as shown:

To enter text in the graph, follow these commands:

title of graph x-axis label y-axis label text on graph (placement defined by mouse) text position at specific coordinates places grid on the graph axis limits makes standard axis for all graphs auto-scaling shows current scaling for vector v puts same scale on both axes puts same scale and tic marks on both axes turns off both axis scaling and tic marks turns on both axis scaling and tic marks

Put the  commands after the   command.

To make multiple plots on a graph the format may be one of the two as follows

You can also command  to freeze the current plot and from there superimpose another plot on the same graph. This can be turned off with. To make line types use the codes: solid(-), dashed (--), dotted, dashdot (-.). To make marktypes, use the codes: point(.), plus (+), star (*), circle (o), x-mark (x). To make colors, use the codes: yellow(y), magent(m), cyan(c), red(r), green(g), blue(b), white(w), black(k). An example is a green dotted line:.

Several other features are:

Graphics hardcopy
To send the graphics to the printer, command, while   filename saves the figure to the specified format or to a default format if not specified.

3-D line plots
An example format of a 3-D line plot is as follows:

A 3-D graph has all the same commands plus.

3-D mesh and surface plots
3-D mesh plots shows a plots of matrix z with the command. A surface plot is controlled by. To obtain a mesh graph of a function over a rectangle, an example is as follows:

The shading can be  and are entered after the   command. Color profiles may be:.

Handle Graphics
To see a more complete list of the options to control a graph enter  and.

Sparse matrix computations
=HW3=

If you use the same argument, $$\begin{Bmatrix} p^{(e)}_{1}\\ p^{(e)}_{2} \end{Bmatrix}_{2x1} = \bar{T}^{(e)}_{2x4} \begin{Bmatrix} f^{(e)}_{1}\\ f^{(e)}_{2}\\ f^{(e)}_{3}\\ f^{(e)}_{4} \end{Bmatrix}_{4x1}$$

This yields the symbolic simplification:

P (e) = T (e) f (e)

For each node, the relationship can be expressed as:

q (e)2x1 = T (e)2x4 d (e)4x1

Substituting variables into the element axial Force-Distance Relationship:

$$\hat{k}^{(e)}_{2x2} q^{(e)}_{2x1}= p^{(e)}_{2x1}$$ → $$\hat{k}^{(e)} (T^{(e)}d^{(e)})= (T^{(e)}f^{(e)})$$

The goal is to have k (e) d (e) = f (e). Therefore, we want to move the T (e) from the right hand side to the left hand side by multiplying both sides by T (e)-1 (or the inverse of T). But because T is not a square matrix, it cannot be inverted.

k (e) d (e)= f (e) → $$[T^{(e)T}_{4x2} (transpose) \hat{k}^{(e)}_{2x2}T^{(e)}_{2x4}]_{(4x4)}d^{(e)}_{4x1} = f^{(e)}_{4x1}$$

HW: Verify that $$k^{(e)} = T^{(e)T}\hat{k}^{(e)}T^{(e)}$$ $$\bar{T}^{(e)T}k^{(e)}\begin{bmatrix} 1 & -1\\ -1 & 1 \end{bmatrix}\begin{bmatrix} l^{(e)} & m^{(e)} & 0 & 0\\ 0 & 0 & l^{(e)} & m^{(e)} \end{bmatrix} = \begin{bmatrix} l^{(e)} & 0\\ m^{(e)} & 0\\ 0 & l^{(e)}\\ 0 & m^{(e)} \end{bmatrix}k^{(e)}\begin{bmatrix} l^{(e)} & m^{(e)} & -l^{(e)} & -m^{(e)}\\ -l^{(e)} & -m^{(e)} & l^{(e)} & m^{(e)} \end{bmatrix} = k^{(e)}\begin{bmatrix} (l^{(e)})^{2} & l^{(e)}m^{(e)} & -(l^{(e)})^{2} & -(l^{(e)}m^{(e)})\\ l^{(e)}m^{(e)} & (m^{(2)})^2 & -(l^{(e)}m^{(e)}) & -(m^{(2)})^2\\ -(l^{(e)})^{2} & -(l^{(e)}m^{(e)}) & (l^{(e)})^{2} & l^{(e)}m^{(e)}\\ -(l^{(e)}m^{(e)}) & -(m^{(2)})^2 & l^{(e)}m^{(e)} & (m^{(2)})^2 \end{bmatrix} = \bar{k}^{(e)}$$

Principle of Virtual Work: K 6x6 d 6x1 = F 6x1 → K' 2x2 d' 2x1 = F' 2x1

We cannot solve for d as d = K -1 F because of the singularity of K -1 where the det[ K ]=0 and therefore we cannot invert K.

In 2D motion, there are three possible rigid body motions: two in translation and one rotation

HW: To find the eigenvalues: $$det\begin{bmatrix} k_{11}^{(1)} -\lambda & k_{12}^{(1)} & k_{13}^{(1)} & k_{14}^{(1)} & 0 & 0\\ k_{21}^{(1)} & k_{22}^{(1)} - \lambda & k_{23}^{(1)} & k_{24}^{(1)} & 0 & 0\\ k_{31}^{(1)} & k_{32}^{(1)} & (k_{33}^{(1)}  k_{11}^{(2)}) - \lambda & (k_{34}^{(1)}   k_{12}^{(2)}) & k_{13}^{(2)} & k_{14}^{(2)}\\ k_{41}^{(1)} & k_{42}^{(1)} & (k_{43}^{(1)}  k_{21}^{(2)}) & (k_{44}^{(1)}   k_{22}^{(2)}) - \lambda  & k_{23}^{(2)} & k_{24}^{(2)}\\ 0 & 0 & k_{31}^{(2)} & k_{32}^{(2)} & k_{33}^{(2)} - \lambda & k_{34}^{(2)}\\ 0 & 0 & k_{41}^{(2)} & k_{42}^{(2)} & k_{43}^{(2)} & k_{44}^{(2)} - \lambda \end{bmatrix} = 0$$ This can be reduced to (because of the Principle of Virtual Work) $$det\begin{bmatrix} (k_{33}^{(1)} + k_{11}^{(2)}) - \lambda & (k_{34}^{(1)} + k_{12}^{(2)})\\ (k_{43}^{(1)} + k_{21}^{(2)}) & (k_{44}^{(1)} + k_{22}^{(2)}) - \lambda \end{bmatrix} = 0$$ Solving becomes $$[(k_{33}^{(1)} + k_{11}^{(2)}) - \lambda]*[(k_{44}^{(1)} + k_{22}^{(2)}) - \lambda] - [(k_{34}^{(1)} + k_{12}^{(2)})(k_{43}^{(1)} + k_{21}^{(2)})] = 0$$ But because K43=K34, this can be written as $$\lambda ^{2}- \lambda(k_{33}^{(1)} + k_{11}^{(2)} + k_{44}^{(1)} + k_{22}^{(2)}) + [(k_{33}^{(1)} + k_{11}^{(2)})(k_{44}^{(1)} + k_{22}^{(2)}) - (k_{34}^{(1)} + k_{12}^{(2)})^{2}] = 0 $$ Therefore, the eigenvalues are λ = 0,$$(k_{33}^{(1)} + k_{11}^{(2)})(k_{44}^{(1)} + k_{22}^{(2)}) - (k_{34}^{(1)} + k_{12}^{(2)})^{2}$$, and $$k_{33}^{(1)} + k_{11}^{(2)} + k_{44}^{(1)} + k_{22}^{(2)}$$ There is a zero eigenvalue because there is no rotational component of the system. This means that there is no elastic energy stored in the structure in the rotational mode. → Dynamic eigenvalue as realted to vibration frequency: K v =λ M  v, where K is the stiffness matrix, λ is the eigenvalue, and M is the mass matrix.

=HW 4=

Lecture 19
$$\mathbf{\tilde{d}}^{(e)}_{4x1} = \mathbf{\tilde{T}}^{(e)}_{4x4}\textbf{d}^{(e)}_{4x1} \Rightarrow \begin{Bmatrix} \tilde{d^{(e)}_{1}}\\ \tilde{d^{(e)}_{2}}\\ \tilde{d^{(e)}_{3}}\\ \tilde{d^{(e)}_{4}} \end{Bmatrix} = \begin{bmatrix} \textbf{R}^{(e)}_{2x2} & \textbf{0}_{2x2}\\ \textbf{0}_{2x2} & \textbf{R}^{(e)}_{2x2} \end{bmatrix}\begin{Bmatrix} d^{(e)}_{1}\\ d^{(e)}_{2}\\ d^{(e)}_{3}\\ d^{(e)}_{4} \end{Bmatrix}$$

$$\mathbf{\tilde{f}}^{(e)} = k^{(e)}\begin{bmatrix} 1 & 0 & -1 & 0\\ 0 & 0 & 0 & 0\\ -1 & 0 & 1 & 0\\ 0 & 0 & 0 & 0 \end{bmatrix}\mathbf{\tilde{d}}^{(e)}$$

$$\mathbf{\tilde{f}}^{(e)}_{4x1} = \mathbf{\tilde{k}}^{(e)}_{4x4}\mathbf{\tilde{d}}^{(e)}_{4x1}$$

Considering the case where $$\mathbf{\tilde{d}}^{(e)}_{4} \neq 0 \Rightarrow \mathbf{\tilde{d}}^{(e)}_{1} = \mathbf{\tilde{d}}^{(e)}_{2} = \mathbf{\tilde{d}}^{(e)}_{3} = 0$$

$$\mathbf{\tilde{f}}^{(e)}_{4x1} = \mathbf{\tilde{k}}^{(e)}_{4x4}\mathbf{\tilde{d}}^{(e)}_{4x1} = \textbf{0}_{4x1}$$, the 4th column of the stiffness matrix.

Interpretation of Transverse Degrees of Freedom

 * {| class="toccolours collapsible collapsed" width="60%" style="text-align:left"

!HW Force relationship, from team group
 * Using the relationship between distance and T, we can find the relationship between force and T: $$\mathbf{\tilde{d}}^{(e)}_{4x1} = \mathbf{\tilde{T}}^{(e)}_{4x4}\textbf{d}^{(e)}_{4x1}$$:
 * Using the relationship between distance and T, we can find the relationship between force and T: $$\mathbf{\tilde{d}}^{(e)}_{4x1} = \mathbf{\tilde{T}}^{(e)}_{4x4}\textbf{d}^{(e)}_{4x1}$$:

$$\mathbf{\tilde{f}}^{(e)}_{4x1} = \mathbf{\tilde{T}}^{(e)}_{4x4}\textbf{f}^{(e)}_{4x1} \Rightarrow \begin{Bmatrix} \tilde{f^{(e)}_{1}}\\ \tilde{f^{(e)}_{2}}\\ \tilde{f^{(e)}_{3}}\\ \tilde{f^{(e)}_{4}} \end{Bmatrix} = \begin{bmatrix} \textbf{R}^{(e)}_{2x2} & \textbf{0}_{2x2}\\ \textbf{0}_{2x2} & \textbf{R}^{(e)}_{2x2} \end{bmatrix}\begin{Bmatrix} f^{(e)}_{1}\\ f^{(e)}_{2}\\ f^{(e)}_{3}\\ f^{(e)}_{4} \end{Bmatrix}$$

contribution by carbon.w
 * }

In addition, using the force-distance relationship:

$$\mathbf{\tilde{k}}^{(e)}\mathbf{\tilde{d}}^{(e)} = \mathbf{\tilde{f}}^{(e)} \Rightarrow \mathbf{\tilde{k}}^{(e)}\mathbf{\tilde{T}}_{(e)}\mathbf{\tilde{d}}^{(e)} = \mathbf{\tilde{T}}_{(e)}\mathbf{\tilde{f}}^{(e)}$$

If $$\mathbf{\tilde{T}}_{(e)}$$ is invertible, then the relationship becomes:

$$[\mathbf{\tilde{T}}^{(e)-1}\mathbf{\tilde{k}}^{(e)}\mathbf{\tilde{T}}^{(e)}]\mathbf{\tilde{d}}^{(e)} = \mathbf{\tilde{f}}^{(e)}$$

$$\mathbf{\tilde{T}}_{(e)}$$ is a block diagonal matrix. The general form of a block diagonal matrix is as follows:

$$\textbf{A} = \begin{bmatrix} \textbf{D}_{1} & & \textbf{0}\\ & ... & \\ \textbf{0} & & \mathbf{D_{s}} \end{bmatrix}$$

A simpler example of the diagonal matrix to display more clearly the format:

$$\textbf{B} = \begin{bmatrix} d_{11} & & & \textbf{0}\\ & d_{22} & & \\ & & ... & \\ \textbf{0} & & & d_{nn} \end{bmatrix} = Diag[d_{11}, d_{22},...,d_{nn}]$$

$$\mathbf{B^{-1}} = Diag[\frac{1}{d_{11}}, \frac{1}{d_{22}},...,\frac{1}{d_{nn}}]$$

If we assume that dii does not equal zero for i=1,...,n and a block diagonal matrix would be:

$$\mathbf{A} = Diag[\mathbf{D_{1}},...,\mathbf{D_{s}}]$$

$$\mathbf{A^{-1}} = Diag[\mathbf{D^{-1}_{1}},...,\mathbf{D^{-1}_{s}}]$$

$$\mathbf{T^{(e)-1}} = Diag[\mathbf{R^{(e)-1}},\mathbf{R^{(e)-1}}]$$

$$\mathbf{R^{(e)T}} = \begin{bmatrix} l^{(e)} & -m^{(e)}\\ m^{(e)} & l^{(e)} \end{bmatrix} * \begin{bmatrix} l^{(e)} & m(e)\\ -m^{(e)} & l^{(e)} \end{bmatrix} = [(l^{(e)})^{2}  (m^{(e)})^{2}][(-l^{(e)}m^{(e)})   (m^{(e)}l^{(e)}]$$

$$\mathbf{R^{(e)T}}*\mathbf{R^{(e)}} = \begin{bmatrix} 1 & 0\\ 0 & 1 \end{bmatrix}_{2x2} = \mathbf{\bar{I}_{2x2}}$$ (the identity matrix)

$$\Rightarrow \mathbf{R^{(e)-1}} = \mathbf{R^{(e)T}}$$

$$\mathbf{\tilde{T}^{(e)-1}} = Diag[\mathbf{R^{(e)T}}, \mathbf{R^{(e)T}}] = (Diag[\mathbf{R^{(e)}}, \mathbf{R^{(e)}}])^{T}$$

$$\Rightarrow \mathbf{\tilde{T}^{(e)-1}} = \mathbf{\tilde{T}^{(e)T}}$$

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!HW Force relationship, from team group
 * Using the relationship shown above,
 * Using the relationship shown above,

$$[\mathbf{\tilde{T}}^{(e)-1}\mathbf{\tilde{k}}^{(e)}\mathbf{\tilde{T}}^{(e)}]\mathbf{\tilde{d}}^{(e)} = \mathbf{\tilde{T}}_{(e)}\mathbf{\tilde{f}}^{(e)}$$,

and plugging in $$\mathbf{\tilde{T}^{(e)-1}} = \mathbf{\tilde{T}^{(e)T}}$$, we arrive at:

$$[\mathbf{\tilde{T}}^{(e)T}\mathbf{\tilde{k}}^{(e)}\mathbf{\tilde{T}}^{(e)}]\mathbf{\tilde{d}}^{(e)} = \mathbf{\tilde{f}}^{(e)}$$

$$\mathbf{k}^{(e)} = \mathbf{\tilde{T}}^{(e)-1}\mathbf{\tilde{k}}^{(e)}\mathbf{\tilde{T}}^{(e)} $$

contribution by carbon.w
 * }

=HW6=

Stiffness term:

In the figure above, assuming the displacement u(x) for $$x_{i} \leq x \leq x_{i+1}$$ (for example, $$x\in [x_{i}, x_{i+1}]$$, where the symbol $$\in $$ means "belongs to". The figure shows the interpolation of u(x) where there is only axial displacement (or, no transverse displacement)

The motivation for linear interpolation of u(x)
In a 2-bar truss the deformation can be shown as:

The deformed shape is a straight line, meaning that there is an implicit asumption of linear interpolation of displacement between two nodes.

u(x) can be expressed in terms of $$d_{i} = u(x_{i})$$ and $$d_{i+1} = u(x_{i+1})$$ as a linear function in x (for example, linear interpolation) as follows:

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

where $$N_{i}(x)$$ and $$N_{i+1}$$ are linear functions in x.

$$N_{i} = \frac {x-x_{i+1}}{x_{i} - x_{i+1}}$$

$$N_{i+1} = \frac {x - x_{i}}{x_{x+1} - x_{i}}$$

Therefore, $$u(x) = \frac {x-x_{i+1}}{x_{i} - x_{i+1}}d_{i} + \frac {x - x_{i}}{x_{x+1} - x_{i}}d_{i+1}$$

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

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!HW Expression for $$\mathbf{k^{(i)}}$$, from team group
 * The expression for $$\mathbf{k^{(i)}}$$ can be solved as such:
 * The expression for $$\mathbf{k^{(i)}}$$ can be solved as such:

$$\mathbf{k^{(i)}} = \int_{x=0}^{\tilde{x} = L^{(i)}}{\mathbf{B^{T}}(\tilde{x})(EA)(\tilde{x})\mathbf{B}(\tilde{x})d\tilde{x}} = \frac {EA}{\tilde{x}}\begin{bmatrix} 1 & -1\\ -1 & 1 \end{bmatrix}^{L^{(i)}}_{0} = \frac {EA}{L^{(i)}}\begin{bmatrix} 1 & -1\\ -1 & 1 \end{bmatrix}$$

The equation of the area is: $$A(\tilde{x}) = N_1^{(i)}(\tilde{x})A_1 + N_2^{(i)}(\tilde{x})A_2$$

The equation of the modulus of elasticity is: $$E(\tilde{x}) = N_1^{(i)}(\tilde{x})E_1 + N_2^{(i)}(\tilde{x})E_2$$

Substituting the equations above into the stiffness matrix, is as follows:

$$\mathbf{k^{(i)}} = \frac {[N_1^{(i)}(\tilde{x})E_1 + N_2^{(i)}(\tilde{x})E_2][N_1^{(i)}(\tilde{x})A_1 + N_2^{(i)}(\tilde{x})A_2]}{L^{(i)}}\begin{bmatrix} 1 & -1\\ -1 & 1 \end{bmatrix}$$

contribution by carbon.w
 * }

=HW6 MATLAB=

The truss system is statically indeterminate. It requires the use of the Principle of Virtual Work or Mechanics of Materials to solve.

Consider isolating an arbitrary node located within the truss system. Also, determine each reaction at a node as a single transverse reaction along the line of action of a beam rather than two reactions along the vertical and horizontal 2-D coordinate system.

Now begin with the node on the far right tip of the arm of the pylon as shown below:

As you can see there are two unknowns and three possible statics equations. Taking a moment however at the node in consideration reveals no conclusion and cannot be considered a possible statics equation in this example.

Using $$\sum{F_{y}} = 0$$, it is easily found that the y-component of R2 = 1000N.

Using $$\sum{F_{x}} = 0$$, it is also easily found that $$R_{1} = R_{2}cos\theta $$

From here, the final reactions cannot be computed and equations of Mechanics must be introduced (if not using FEM). The node shown above is the simplest FBD in the system.

Consider the node at (25, 30):

From this node, it is clear to see why six reactions cannot be computed with only two equations of statics.

With each node there are two, three, four, five or six unknown reactions. There are six nodes with two unknown reactions at the extreme locations of the truss system which cannot be solved as shown above. The nodes with four, five or six unknown reactions can't be computed using only two equations of statics.

The reason we cannot use the moment equation as a third equation of statics is because if the moment is taken at any point the moment equals zero. At each node, each isolated truss intersects. If the moment is taken at the node, it is clear why the moment equals zero, however at any other point it does as well (as shown in previous homework assignments). Each member is a two-force body and there is no resulting moment arm and no moment. This shows that the nodes with three unknown reactions cannot be solve as well.

Lecture 36
Below is a deformable truss with horizontal, vertical and rotational displacements:

The matrix formulas are as follows:

$$\mathbf{\tilde{k}}^{(e)}_{6x6}\mathbf{\tilde{d}}^{(e)}_{6x6} = \mathbf{\tilde{f}}^{(e)}_{6x6}$$, where

$$\mathbf{\tilde{d}}^{(e)} = \begin{Bmatrix} \tilde{d}^{(e)}_{1}\\ ...\\ \tilde{d}^{(e)}_{6} \end{Bmatrix}$$ and $$\mathbf{\tilde{f}}^{(e)} = \begin{Bmatrix} \tilde{f}^{(e)}_{1}\\ ...\\ \tilde{f}^{(e)}_{6} \end{Bmatrix}$$

Note: $$\tilde{f}_{3}^{(e)} = f_{3}^{(e)}$$ and $$\tilde{f}_{6}^{(e)} = f_{6}^{(e)}$$

Taking the moments about $$\tilde{z}$$ (where $$\tilde{z} = z$$), the displacement matrix is as follows:

.......$$\tilde{d}_{1}$$  .........   $$\tilde{d}_{2}$$   .........  $$\tilde{d}_{3}$$   .........   $$\tilde{d}_{4}$$  .........   $$\tilde{d}_{5}$$    ......... $$\tilde{d}_{6}$$

$$\begin{Bmatrix} \frac{EA}{L} & 0 & 0 & -\frac{EA}{L} & 0 & 0\\ 0 & \frac{12EI}{L^{3}} & \frac{6EI}{L^{2}} & 0 & -\frac{12EI}{L^{3}} & \frac{6EI}{L^{2}}\\ 0 & \frac{6EI}{L^{2}} & \frac{4EI}{L} & 0 & -\frac{6EI}{L^{2}} & \frac{2EI}{L}\\ \frac{-EA}{L} & 0 & 0 & \frac{EA}{L} & 0 & 0\\ 0 & -\frac{12EI}{L^{3}} & -\frac{6EI}{L^{2}} & 0 & \frac{12EI}{L^{2}} & -\frac{6EI}{L^{2}}\\ 0 & \frac{6EI}{L^{2}} & \frac{2EI}{L} & 0 & -\frac{6EI}{L^{2}} & \frac{4EI}{L} \end{Bmatrix}\begin{matrix} \tilde{d}_{1}\\ \tilde{d}_{2}\\ \tilde{d}_{3}\\ \tilde{d}_{4}\\ \tilde{d}_{5}\\ \tilde{d}_{6} \end{matrix}$$

The dimensional analysis is as follows:

$$[\tilde{d}_{1}] =L=[\tilde{d}_{i}]$$, where i = 1,2,3,4,5

dimensions of rotation = $$[\tilde{d}_{3} = 1 \Rightarrow$$ (no dimension, because the unit is degree)

The length of the arc AB is equal to $$R\theta \Rightarrow \theta = \frac{arcAB}{R}$$

$$[\theta ] = \frac{[arcAB]}{[R]} = \frac{1}{1} = 1$$

To note, the unit is not the same as the dimension. $$\theta$$ of course does not have a unit of 1!

$$\sigma = E\varepsilon \Rightarrow [\sigma ] = [E][\varepsilon ]$$

title: User:Eml4500.f08.group.w


text:
User:Eml4500.f08.delta 6.ramirez/classnotes

Matrix 2-Bar Truss System
(continued)  Element 1:

θ(1) = 30° l(1) = cos θ(1) = cos30° = $$\frac{\sqrt{3}}{2}$$ m(1) = sin θ(1) = sin30° = $$\frac{1}{2}$$ k(1) = E(1)·A(1)/ L(1) = $$\frac{3}{4}$$

$$k^{(1)} = k_{(1)} \begin{bmatrix} k_{11}^{(1)} & k_{12}^{(1)} & k_{13}^{(1)} & k_{14}^{(1)}\\ k_{21}^{(1)} & k_{22}^{(1)} & k_{23}^{(1)} & k_{24}^{(1)}\\ k_{31}^{(1)} & k_{32}^{(1)} & k_{33}^{(1)} & k_{34}^{(1)}\\ k_{41}^{(1)} & k_{42}^{(1)} & k_{43}^{(1)} & k_{44}^{(1)} \end{bmatrix}$$

k11(1) = k(1) · ( l(1) )2 = $$\frac{3}{4}$$ · $$(\frac{\sqrt{3}}{2})^{2}$$ = $$\frac{9}{16}$$

k12(1) = k(1) · l(1) · m(1) = $$\frac{3}{4}$$ · $$\frac{\sqrt{3}}{2}$$· $$\frac{1}{2}$$ = $$\frac{\sqrt{3}}{16}$$

k42(1) = - k(1) · ( m(1) )2 = - $$\frac{3}{4}$$ · $$(\frac{2})^{2}$$ = - $$\frac{3}{16}$$

Observation:

1) Only three numbers need to be computed. Other coefficients have the same absolute value just differ by sign (+ or -) 2) Matrix k(1) is symmetrical i.e., kij(1) = kji(1); just interchange the row and column indices

For example: k13(1) = k31(1)

In general,

kij(e) = kji(e) or k ij(1)Т = k (1)

where Т represents the transpose of the matrix.

$$k^{(e)} = \begin{bmatrix} k_{11}^{(1)} & k_{12}^{(1)} & k_{13}^{(1)} & k_{14}^{(1)}\\ & k_{22}^{(1)} & k_{23}^{(1)} & k_{24}^{(1)}\\ & & k_{33}^{(1)} & k_{34}^{(1)}\\ & &  & k_{44}^{(1)} \end{bmatrix}$$

Since the matrix is symmetrical only the upper right triangle part needs to be calculated.

$$k^{(1)} = \begin{bmatrix} \frac{9}{16} & \frac{\sqrt{3}}{16} & -\frac{9}{16} & -\frac{\sqrt{3}}{16}\\ \frac{\sqrt{3}}{16} & \frac{3}{16} & -\frac{\sqrt{3}}{16} & -\frac{3}{16}\\ -\frac{9}{16} & -\frac{\sqrt{3}}{16} & \frac{9}{16} & \frac{\sqrt{3}}{16}\\ -\frac{\sqrt{3}}{16} & -\frac{3}{16} & \frac{\sqrt{3}}{16} & \frac{3}{16} \end{bmatrix}$$

Element 2:

k(2) = E(2)·A(2)/ L(2) = 5 θ(2) = $$-\frac{\Pi }{4}$$ l(2) = cos θ(2) = $$cos(-\frac{\Pi }{4}) = \frac{\sqrt{2}}{2}$$ m(2) = sin θ(2) = $$\sin (-\frac{\Pi }{4}) = -\frac{\sqrt{2}}{2}$$

k(2) = [kji(2)]4x4

Observation:

1) The absolute value of all coefficients kji(e), e=2 (i,j) = 1,...,4 are the same (|l| = |m|) → Only four coefficients need to be computed; all other coefficients add + or -. 2) k ij(2)Т = k (2)

$$k^{(2)} = \begin{bmatrix} \frac{5}{2} & -\frac{5}{2} & -\frac{5}{2} & \frac{5}{2}\\ -\frac{5}{2} & \frac{5}{2} & \frac{5}{2} & -\frac{5}{2}\\ -\frac{5}{2} & \frac{5}{2} & \frac{5}{2} & -\frac{5}{2}\\ \frac{5}{2} & -\frac{5}{2} & -\frac{5}{2} & \frac{5}{2} \end{bmatrix}$$


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User:Eas4200c.f08.wiki.c/homework report 1

Reasons for Using Wikiversity for Homework Submission
The Aerospace Structures class will be using Wikiversity in order to submit homework due to the versatility of the mediawiki software. Instead of having to meet in one place to collaborate, the group will be able to just post their part in their user web space and everyone else in the group will be able to view and edit in any additional information they have on the subject. Also, any other group will be able to see the other groups pages and share information.

The structure will be even better if it is carried out in future years as any subsequent classes will have the previous years work to look at.

Also, since the student is already on Wikiversity, it will be easier for him or her to investigate the subject further by looking at any related Wikipedia pages.

Dangers of Using Wikiversity
A downside to using Wikiversity is that any one will be able to see your work and change it. This includes people who are not in the class, and the student is asked to keep an eye on their page at all times to make sure no vandalism has been done to their work. Also, any other group in the class could steal another groups work and post it as their own since it's freely available on line. One way around this is for TA's to cross check the work of one group against every other group, but this won't guard against a group rewording the work of another group and doing the same thing. The suggested way around this is to wait until right before the homework is due to post it on line.


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User:Eas4200c.f08.wiki.c/homework report 2

= Sept 12th Notes =

A box beam is a rectangular structure that is important to study due to the fact that many aerospace structures follow this same basic design. A box beam is a form of shell beam, and structures such as an airplane fuselage are also shell beams, the only difference being that their cross sections are circular rather than a rectangle. Likewise, the wings on an airplane are also shell beams with a slightly more complex cross section.

Now that there is a motivation for solving the problem, Case 1 from last lecture is continued as follows:

$$\tau ^{(1)}_{max}$$ is the shear stress corresponding to the maximum torque $$T^{(1)}_{max}$$

Continuing the solution, replace $$T^{(1)}_{max}$$ in the shear equation by $$M^{(1)}_{max}$$. This is done due to the 1st assumption in the problem statement. Secondly, assumption 2 allows for the substitution of $$\tau ^{(1)}_{max}$$ with $$\sigma _{allow}$$

$$\tau ^{(1)}_{max} = \frac{T^{(1)}_{max}}{2a^{(1)}b^{(1)}t}$$

$$\tau ^{(1)}_{max} = \frac{M^{(1)}_{max}}{2(\frac{L}{8})\frac{3L}{8}t} = \sigma _{allow}$$

$$\tau _{max}^{(1)} = \sigma _{allow} = 2\tau _{allow} > \tau _{allow}$$

After these substitutions, it is easy to see that $$\tau ^{(1)}_{max}$$ is equal to $$\sigma _{allow}$$, which is twice as big as $$\tau _{allow}$$. Therefore this answer is not acceptable, and the solver must move on to case 2.

Pictured to the right is a graphical representation of normal stress. The stress is distributed through the length of the bar with and is equal to the force, F, divided by the cross sectional area A.

A graphical representation of this result can be shown using a Mohr's circle. As shown on the figure to the right, as long as the shear and normal stress stay enclosed within the circle, the structure will not fail. However, since in this case the maximum shear encountered is greater than half the allowable normal stress, this result is unacceptable

Case 2

Assume that $$\tau _{max}$$ reaches $$\tau _{allow}$$ first.

Therefore:

$$\tau = \frac{T}{2abt}$$

$$T = (2(t)\tau_{allow})(ab)$$

$$T_{max} = (2(t)\tau_{allow})(ab)_{max}$$

Everything on the right hand side of the above equation is constant, therefore to maximize the torque, the cross sectional area must be maximized. With the constraint of keeping the perimeter constant, it is easy to see that the largest cross sectional area will be achieved by making a equal to b, and thus having a square cross section as follows:

$$a^{(2)} = b^{(2)} = \frac{L}{4}$$


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User:Eas4200c.f08.gator.edwards/Additional Questions

Circle Moment of Inertia
$$I_y=\iint\limits_A z^2\,dA=\frac\pi4 r^2$$

In polar coordinates, Iy for a circle is calculated below:

$$J_o = \int_A{\rho^2 dA} = \int_0^r{\rho^2 (2\pi\rho d\rho)} = 2\pi \int_0^r{\rho^3 d\rho} = \frac{1}{2}\pi r^4$$

$$J_o = I_z + I_y = 2I_y \Rightarrow I_y = {1 \over 4} \pi r^4$$

.

Channel Cross-Section Moment of Inertia

To compute the moment of inertia of a channel cross section we first recall formula for the moment of inertia of a rectangle

$$I_{y}=\frac{1}{12}bh^{3}$$

and the parallel axis theorem

$$I_{y}=\bar{I}_{y'}+Ad^{2}$$

We also remember that the moment of inertia of a composite area is equal to the sum of the moments of inertia of its parts. Using the first equation we find the moments of inertia of the vertical sections:

$$I_{y,vert}=2(\frac{1}{12}tb^{3})=\frac{1}{6}tb^{3}$$

Next we find the moment of inertia of the horizontal piece

$$I_{y,horiz}=\frac{1}{12}(a-2t)b^{3}$$

and use the parallel axis theorem to combine the components:

$$I_{y}=I_{y,vert}+(I_{y,horiz}+Ad^{2})=\frac{1}{6}tb^{3}+[\frac{1}{12}(a-2t)t^{3}+(a-2t)(t)(\frac{b-t}{2})]$$

This procedure can be simplified with a few parameters: 1) a = b   2)  t = a / 10 3) A = 3at   4)  t << a so that a is the distance from the base to the middle of the horizontal section and the area of the horizontal section is equal to at

Applying these simplifications and reassembling the equation we get

$$I_{y}=I_{y,vert}+(I_{y,horiz}+Ad^{2})=\frac{1}{6}tb^{3}+\frac{1}{12}at^{3}+3at\left(\frac{b}{2} \right)^{2}$$

$$=\frac{1}{6}\left(\frac{a}{10} \right)a^{3}+\frac{1}{12}a\left(\frac{a}{10} \right)^{3}+3a\left(\frac{a}{10} \right)\left(\frac{a}{2} \right)^{2}=\frac{a^{4}}{60}+\frac{a^{4}}{120}+\frac{3a^{4}}{20}=\frac{7a^{4}}{40}$$

.

Problem 1.7 with Modification

The objective of this problem is to compare the moments of inertia of a solid circle (case 0) and a composite section consisting of two circles connected with a rectangle (case 1, shown in figure at right). The radius of the solid circle is R0, the radius of the smaller circles on the composite section is R1, the length of the rectangular portion is 2R0, and the thickness of the rectangular portion is t.

R0 = 10 cm t = 1 cm

R1 is to be found so that the cross sectional areas are the same for both sections.

Starting with the areas

$$A_{0}=\pi R_{0}^{2}=(10 cm)^{2}\pi = 314 cm^{2}$$

$$A_{1}=2(\pi R_{1}^{2})+2R_{0}t=A_{0}=\pi R_{0}^{2}$$

Solve for R1

$$R_{1}^{2}=\frac{\pi R_{0}^{2}-2R_{0}t}{2\pi }$$

Plugging in the known values we get R1

$$R_{1}=\sqrt{\frac{\pi \cdot10^{2}-2\cdot 10\cdot 1}{2\pi }}= 6.84 cm^{2}$$

From the above section the moment of inertia of a circle and rectangle are

Circle: $$I_{y}=\frac{1}{4}\pi r^{4}$$

Rectangle: $$I_{y}=\frac{1}{12}bh^{3}$$

Using these the moment of inertia for Case 0 is

$$I_{y,0}=\frac{1}{4}\pi R_{0}^{4}=\frac{1}{4}\pi \cdot 10^{4}= 7854 cm^{4}$$

Combining with the parallel axis theorem stated above the moment of inertia for Case 1 is

$$I_{y,1}=\frac{1}{12}bh^{3}+\frac{1}{4}\pi R_{1}^{4}+(\pi R_{1}^{2})(R_{0}+R_{1})^{2}=\frac{1}{12}\cdot 1\cdot (2\cdot 10)^{3}+\frac{1}{4}\cdot \pi \cdot 6.84^{4}+(\pi \cdot 6.84^{2})(10+6.84)^{2}= 44,068 cm^{4}$$

From the results it is clear that the cross section of Case 1 yields a much higher moment of inertia and therefore can withstand higher forces than the section of Case 0.

Mohr's Circle

Mohr's circle is a graphical representation of any 2-D stress state proposed in 1892 by Christian Otto Mohr. It can be applied to many engineering quantities such as stresses, strains, and moments of area.

Mohr's circle may also be applied to three-dimensional stress. In this case, the diagram has three circles, two within a third.

You can use Mohr's circle to find the planes of maximum normal, principle and shear stresses, as well as the stresses on known weak planes. For example, if the material is brittle, the engineer might use Mohr's circle to find the maximum component of normal stress (tension or compression); and for ductile materials, the engineer might look for the maximum shear stress.

Looking at the stress figure on the right, we see that with only a force in the x direction, in the enlarged section the stress in the x-direction $$(\sigma _{xx})$$ is equal to $$\sigma$$. The stress in the y-direction $$(\sigma _{yy})$$ is equal to 0.

Now if they are plotted on a stress-strain graph as a Mohr's Circle, we see that by plotting $$(\sigma _{xx})$$ and $$(\sigma _{yy})$$ and connecting them with a circle, $$\tau _{max} = \frac{\sigma _{allow}}{2}$$.


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User:Eml4500.f08.bottle.ranto/HW 2 Images

 * $$\textbf{k}^{(e)}\textbf{d}^{(e)}=\textbf{f}^{(e)}$$

 * $$ \textbf{k}^{(1)}= \left[ \begin{array}{cccc} k_{11}^{(1)} & k_{12}^{(1)} & k_{13}^{(1)} & k_{14}^{(1)} \\ k_{21}^{(1)} & k_{22}^{(1)} & k_{23}^{(1)} & k_{24}^{(1)} \\ k_{31}^{(1)} & k_{32}^{(1)} & k_{33}^{(1)} & k_{34}^{(1)} \\ k_{41}^{(1)} & k_{42}^{(1)} & k_{43}^{(1)} & k_{44}^{(1)} \end{array} \right]$$

$$ \textbf{k}^{(1)}= \left[ \begin{array}{cccc} l_{s}^{2} & l_{s}m_{s} & -l_{s}^{2} & -l_{s}m_{s} \\ l_{s}m_{s} & m_{s}^{2} & l_{s}m_{s} & -m_{s}^{2} \\ -l_{s}^{2} & -l_{s}m_{s} & l_{s}^{2} & l_{s}m_{s} \\ -l_{s}m_{s} & -m_{s}^{2} & l_{s}m_{s} & m_{s}^{2} \end{array} \right]$$


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User:Eml4500.f08.gravy.jad/Notes

FEA Notes and Homework 2

Truss Problem description
Take a specific example to see how the recipe works.

Data:

Element Length

$$\ L^{(1)}=4 \,$$

$$\ L^{(2)}=2 \,$$

Young's Modules

$$\ E^{(1)}=3 \,$$

$$\ E^{(2)}=5 \,$$

Cross-sectional Area

$$\ A^{(1)}=1 \,$$

$$\ A^{(2)}=2 \,$$

Inclination angle

 * $$\ \theta^{(1)} = 30^o \,$$

 * $$\ \theta^{(2)} = -45^o \,$$

Solving a Truss Problem with Statics

First take the sum of the moments about global node 3 and solve for $$R_3$$ which is the reaction at global node 3 in the y-direction

$$ \sum {M_1}= P(4cos(30)) +R_3(4cos(30) + 2sin(45))$$

$$R_3= \frac{-P(4cos(30))} {4cos(30) + 2cos(45)} = -4.4378$$

Then use this information to take the sum of the forces in the y-direction and solve for $$R_1$$ which is the reaction at global node 1 in the y-direction.

$$\sum{F_y}= -4.4378 + P+R_1$$

$$\ R_1= -2.5622 \,$$

By observation it can be determined that there are no forces in the x-direction so $$R_1$$ and $$R_3$$ are the only reaction forces

Solving a Truss Problem with Finite Element Analysis
1)Global Picture

$$\begin{Bmatrix} f_1 \\\vdots & \\ f_6 \end{Bmatrix}_{1X6}=\begin{bmatrix} & &  \\  & k & \\ & &   \end{bmatrix}_{6X6}*    \begin{Bmatrix} d_1 \\\vdots & \\ d_6 \end{Bmatrix}_{1X6}$$

$$\begin{matrix} global& &global& &global \\ force & &stiffness& &displacement \\  column &   &matrix  & &column \\ matrix & & & &matrix \end{matrix}$$

2) Element Picture

3) Global FD at element level:

$$\ k^{(e)} d^{(e)}=f^{(e)} \,$$

$$\ k^{(e)}\,$$ = the element stiffness matrix

$$\ d^{(e)}\,$$= the element displacement matrix

$$\ f^{(e)}\,$$= the element force matrix

$$k^{(e)}\begin{bmatrix} (l^{(e)})^2 & l^{(e)}m^{(e)} & -(l^{(e)})^2 & -(l^{(e)}m^{(e)})  \\  l^{(e)}m^{(e)} & (m^{(e)})^2 & -(l^{(e)}m^{(e)}) & -(m^{(e)})^2  \\ -(l^{(e)})^2 & -(l^{(e)}m^{(e)}) & (l^{(e)})^2 & l^{(e)}m^{(e)}\\ -(l^{(e)}m^{(e)}) & -(m^{(e)})^2 & l^{(e)}m^{(e)} & (m^{(e)})^2    \end{bmatrix}_{4X4} = \underline{k^{(e)}}$$

$$\ k^{(e)}= \,$$ $$\frac{E^{(e)}A^{(e)}}{L^{(e)}}$$ axial stiffness of bar element "e" where e=1,2

$$\ l^{(e)},m^{(e)} = \,$$ director cosines of x axis (goes from 1 to 2) with respect to global (x,y coordinates)

$$\ l^{(e)}= \vec \tilde{i} \cdot \vec i= cos\theta^{(e)}\,$$

$$\ m^{(e)}= \vec i \cdot \vec j= cos\left (\frac{\pi}{2}- \theta^{(e)} \right )= sin\left (\theta^{(e)} \right )\,$$

$$\ \vec \tilde{i}=cos\left(\theta^{(e)} \right)\vec i + sin\left( \theta^{(e)}  \right )\vec j \,$$

$$\ \vec \tilde{i} \cdot \vec i = \vec \tilde{i} \left( cos\left(\theta^{(e)} \right)\vec i + sin\left( \theta^{(e)}  \right )\vec j \right) i= cos\left(\theta^{(e)}  \right) \vec i \cdot \vec i + sin\left( \theta^{(e)}  \right ) \vec i  \,$$

$$\ \vec \tilde{i} \cdot \vec j= HW \,$$

Element 1:
$$\ \theta^{(1)} = 30^o \,$$

$$\ l^{(1)}=cos\theta^{(1)}= cos 30^o= \frac{\sqrt{3}}{2} \,$$

$$\ m^{(1)}=sin\theta^{(1)}= sin 30^o= \frac{1}{2} \,$$

$$\ k^{(1)}= \frac{E^{(1)}A^1}{L^1} \frac{\left(3\right ) \left(1\right )}{4}= \frac {3}{4}\,$$

$$\ \underline{k^{(1)}}=\begin{bmatrix} k_{11}^{(1)} & k_{12}^{(1)} & k_{13}^{(1)} & k_{14}^{(1)}  \\  k_{21}^{(1)} & k_{22}^{(1)} & k_{23}^{(1)} & k_{24}^{(1)}  \\ k_{31}^{(1)} & k_{32}^{(1)} & k_{33}^{(1)} & k_{34}^{(1)}\\ k_{41}^{(1)} & k_{42}^{(1)} & k_{43}^{(1)} & k_{44}^{(1)}    \end{bmatrix}_{4X4}= \left [k_{ij}^1 \right ]_{4X4} \,$$

i=row number =1,2....4

j=column number =1,2....4

$$\ k_{11}^{(1)}= k^{(1)}\left( l^{(1)} \right)^2=\frac{3}{4}*\frac{\sqrt{3}}{2}= \frac{9}{16} \,$$

$$\ k_{12}^{(1)}= k^{(1)}\left( l^{(1)}m^{(1)} \right)=\frac{3}{4}*\frac{1}{2}= \frac{3\sqrt{3}}{16} \,$$

$$\ k_{42}^{(1)}=k_{24}^{(1)}= -k^{(1)}\left( m^{(1)} \right)^2=\frac{-3}{4}*\frac{1}{2}= \frac{-3}{16} \,$$

Observation:

1) Only need to compute 3 numbers. Other coefficients have same absolute value, just differ by (+ -)

2) Matrix $$ \underline{k^{(1)}}$$ is synmetrix, $$k_{ij}^{(1)}=k_{ji}^{(1)}$$

$$k_{13}^{(e)}=k_{31}^{(e)}$$ (just interchange the row and column index)

$$\ \underline{k^{(1)}}=\begin{bmatrix} \frac{3}{4}(\frac{\sqrt{3}}{2})^2 & \frac{3}{4}(\frac{\sqrt{3}}{2})(\frac{1}{2}) & -\frac{3}{4}(\frac{\sqrt{3}}{2})^2 & -\frac{3}{4}(\frac{\sqrt{3}}{2})(\frac{1}{2})  \\  \frac{3}{4}(\frac{\sqrt{3}}{2})(\frac{1}{2}) & \frac{3}{4}(\frac{1}{2})^2 & -\frac{3}{4}(\frac{\sqrt{3}}{2})(\frac{1}{2}) & -\frac{3}{4}(\frac{1}{2})^2  \\ -\frac{3}{4}(\frac{\sqrt{3}}{2})^2 & -\frac{3}{4}(\frac{\sqrt{3}}{2})(\frac{1}{2}) & \frac{3}{4}(\frac{\sqrt{3}}{2})^2 & \frac{3}{4}(\frac{\sqrt{3}}{2})(\frac{1}{2})\\-\frac{3}{4}(\frac{\sqrt{3}}{2})(\frac{1}{2}) & -\frac{3}{4}(\frac{1}{2})^2 & \frac{3}{4}(\frac{\sqrt{3}}{2})(\frac{1}{2}) & \frac{3}{4}(\frac{1}{2})^2    \end{bmatrix}_{4X4}=  \underline{k^{(1)}} _{4X4} \,$$

$$\ \underline{k^{(1)}}=\begin{bmatrix} \frac{9}{16}& \frac{3\sqrt{3}}{16} & -\frac{9}{16} & -\frac{3\sqrt{3}}{16}  \\  \frac{3\sqrt{3}}{16} & \frac{3}{16} & -\frac{3\sqrt{3}}{16} & -\frac{3}{16}  \\ -\frac{9}{16} & -\frac{3\sqrt{3}}{16} & \frac{9}{16} & \frac{3\sqrt{3}}{16}\\ -\frac{3\sqrt{3}}{16} & -\frac{3}{16} & \frac{3\sqrt{3}}{16} & \frac{3}{16}    \end{bmatrix}_{4X4}=  \underline{k^{(1)}} _{4X4} \,$$

In general, $$k_{ij}^{(e)}=k_{ji}^{(e)} or k^{(e) T} =k^{(e)}$$

The transpose of $$ k^{(e)}$$ is equal to $$ k^{(e)}$$

$$\ \underline{k^{1}}=\begin{bmatrix} k_{11}^{(e)} & k_{12}^{(e)} & k_{13}^{(e)} & k_{14}^{(e)}  \\   & k_{22}^{(e)} & k_{23}^{(e)} & k_{24}^{(e)}  \\  &  & k_{33}^{(e)} & k_{34}^{(e)}\\ sym. & &  & k_{44}^{(e)}    \end{bmatrix}_{4X4}= \left [k_{ij}^1 \right ]_{4X4} \,$$

Element 2:
$$\ k^{(2)}= \frac{E^{(2)}A^{(2)}}{L^{(2)}} \frac{\left(5\right ) \left(2\right )}{2}= 5\,$$

$$\ \theta^{(2)} = \frac{-\pi}{4}= -45^o \,$$

$$\ l^{(2)}=cos\theta^{(2)}= cos (\frac{-\pi}{4})= \frac{\sqrt{2}}{2} \,$$

$$\ m^{(2)}=sin\theta^{(2)}= sin (\frac{-\pi}{4})= \frac{-\sqrt{2}}{2} \,$$

$$\ k^{(2)} = \left [k_{ij}^{(2)} \right ]_{4X4} \,$$

$$\ k_{11}^{(2)} = k^{(2)}(l^{(2)})^2 = 5\frac{2}{4} = 2.5 \,$$

Observation:

1) Absolute value of all coefficients $$\ k_{ij}^{(e)} \,$$, $$\ e=2 \,$$ , $$\ (i,j)=1, ..., 4 \,$$ are the same => comp 1 coefficient. For other coefficients, and (+) or (-)

2) $$\ k ^{(2)T} = k^{(2)} \,$$ ie. $$\ k^{(2)} \,$$ sym

$$\ \underline{k^{(2)}}=\begin{bmatrix} \frac{5}{2}& -\frac{5}{2} & -\frac{5}{2} & \frac{5}{2}  \\  -\frac{5}{2} & \frac{5}{2} & \frac{5}{2} & -\frac{5}{2}  \\ -\frac{5}{2} & \frac{5}{2} & \frac{5}{2} & -\frac{5}{2}\\ \frac{5}{2} & -\frac{5}{2} & -\frac{5}{2} & \frac{5}{2}    \end{bmatrix}_{4X4}=  \underline{k^{(2)}} _{4X4} \,$$

Elem FD rel: $$\ k^{(e)}d^{(e)}= f^{(e)} \,$$
 * e=1,2

$$\begin{Bmatrix} d_1 \\\vdots & \\ d_6 \end{Bmatrix}_{1X6}, \begin{Bmatrix} f_1 \\\vdots & \\ f_6 \end{Bmatrix}_{1X6} $$

Global FD relationship:

$$\ \underline{K}_{nxn} \underline{d}_{nx1} = \underline{F}_{nx1} \,$$

there, n=6

$$\ \begin{bmatrix} k_{11} & k_{12} & k_{13} & k_{14} & k_{15} & k_{16}  \\ k_{21}& k_{22} & k_{23} & k_{24} & k_{25} & k_{26} \\ k_{31}& k_{32} & k_{33} & k_{34} & k_{35} & k_{36} \\ k_{41}& k_{42} & k_{43} & k_{44} & k_{45} & k_{46} \\ k_{51}& k_{52} & k_{53} & k_{54} & k_{55} & k_{56} \\ k_{61}& k_{62} & k_{63} & k_{64} & k_{65} & k_{66}  \end{bmatrix}  \begin{Bmatrix}  d_{1} \\ d_{2} \\ d_{3} \\ d_{4} \\ d_{5} \\ d_{6} \end{Bmatrix} = \begin{Bmatrix}  F_{1} \\ F_{2} \\ F_{3} \\ F_{4} \\ F_{5} \\ F_{6} \end{Bmatrix}\,$$

Global F-D Relationship
Noting the fact that the equations for the elemental $$\underline{d^{(e)}}$$ and global $$\underline{d}$$ overlap, as seen in the below picture, the following equations can be derived.

$$d_1 = d_1^{(1)}$$ $$d_2 = d_2^{(1)}$$ $$d_3 = d_3^{(1)} = d_1^{(2)}$$ $$d_4 = d_4^{(1)} = d_2^{(2)}$$ $$d_5 = d_3^{(2)}$$ $$d_6 = d_4^{(2)}$$

Below is a representation of the global $$\underline{K}$$ matrix, which is the stiffness matrix for the system. The two colored boxes represent the contributions of the elemental $$\underline{k^{(e)}}$$. The $$\underline{k^{(1)}}$$ matrix is shown in blue, while the red box represents $$\underline{k^{(2)}}$$.

$$\ \begin{bmatrix} K_{11}^{(1)} & K_{12}^{(1)} & K_{13}^{(1)} & K_{14}^{(1)} & 0 & 0  \\ K_{21}^{(1)}& K_{22}^{(1)} & K_{23}^{(1)} & K_{24}^{(1)} & 0 & 0 \\ K_{31}^{(1)} & K_{32}^{(1)} & (K_{33}^{(1)} + K_{11}^{(2)}) & (K_{34}^{(1)} + K_{12}^{(2)}) & K_{13}^{(2)} & K_{14}^{(2)} \\ K_{41}^{(1)}& K_{42}^{(1)} & (K_{43}^{(1)} + K_{21}^{(2)}) & (K_{44}^{(1)} + K_{22}^{(2)}) & K_{23}^{(2)} & K_{24}^{(2)} \\ 0 & 0 & K_{31}^{(2)} & K_{32}^{(2)} & K_{33}^{(2)} & K_{34}^{(2)} \\ 0 & 0 & K_{41}^{(2)} & K_{42}^{(2)} & K_{43}^{(2)} & K_{44}^{(2)}  \end{bmatrix}\,$$

EX:

$$\ K_{11} = k_{11}^{(1)} \,$$

$$\ K_{33} = k_{33}^{(1)} + k_{11}^{(2)} \,$$

$$\ K_{43} = K_{34} = k_{34}^{(1)} + k_{12}^{(2)} = k_{43}^{(1)} + k_{21}^{(2)} \,$$

$$\ k_{11}^{(1)} = \frac{9}{16} \,$$

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

$$ K_{33} = \frac{9}{16} + \frac{5}{2} = 3.0625 $$

$$ K_{43} = \frac{3\sqrt{3}}{16} + \frac{-5}{2} = 2.1752 $$

$$ K_{44} = k_{44}^{(1)} + k_{22}^{(2)} = \frac{3}{16} + \frac{5}{2} = 2.6875 $$ Mbr> $$\ \ = \begin{bmatrix} \frac{9}{16}& \frac{3\sqrt{3}}{16} & -\frac{9}{16} & -\frac{3\sqrt{3}}{16} & 0 & 0 \\  \frac{3\sqrt{3}}{16} & \frac{3}{16} & -\frac{3\sqrt{3}}{16} & -\frac{3}{16} & 0 & 0 \\ -\frac{9}{16} & -\frac{3\sqrt{3}}{16} & (\frac{9}{16} + \frac{5}{2}) & (\frac{3\sqrt{3}}{16} - \frac{5}{2}) & -\frac{5}{2} & \frac{5}{2} \\ -\frac{3\sqrt{3}}{16} & -\frac{3}{16} & (\frac{3\sqrt{3}}{16} - \frac{5}{2}) & (\frac{3}{16} + \frac{5}{2}) & \frac{5}{2} & -\frac{5}{2} \\  0 & 0 & -\frac{5}{2} & \frac{5}{2} & \frac{5}{2} & -\frac{5}{2} \\ 0 & 0 & \frac{5}{2} & -\frac{5}{2} & -\frac{5}{2} & \frac{5}{2}  \end{bmatrix}_{6X6}\,$$ $$\ \ = \begin{bmatrix} 0.5625 & 0.3248 & -0.5625 & -0.3428 & 0 & 0 \\  0.3248 & 0.1875 & -0.3248 & -0.1875 & 0 & 0 \\ -0.5625 & -0.3248 & (0.5625 + 2.5) & (0.3248 - 2.5) & -2.5 & 2.5 \\ -0.3248 & -0.1875 & (0.3248 - 2.5) & (0.1875 + 2.5) & 2.5 & -2.5 \\  0 & 0 & -2.5 & 2.5 & 2.5 & -2.5 \\ 0 & 0 & 2.5 & -2.5 & -2.5 & 2.5  \end{bmatrix}_{6X6}\,$$

Elimination of Known Degrees of Freedom
Knowing that global nodes 1 and 3 are actually fixed points, it is seen that there is no possibility of them moving. Therefore, the displacement at those points is zero.

$$\ d_1 = d_2 = d_5 = d_5 = d_6 = 0 \,$$

$$\ \underline{d}=\begin{Bmatrix} d_1 \\ d_2  \\  d_3 \\ d_4 \\ d_5 \\ d_6 \end{Bmatrix}= \begin{Bmatrix}  0 \\ 0 \\  d_3 \\ d_4 \\ 0 \\ 0 \end{Bmatrix}$$

Since 4 of the values for $$\underline{d}$$ are zero, we can reduce the system. Turning $$\underline{d}$$ into a 2x2 matrix, would mean that you would be able to reduce $$\underline{K}$$ to a 6x2 matrix. This is down by removing the columns that are associated with the rows within $$\underline{d}$$ that are zero. The new $$\underline{K}$$ is shown below.

$$\ \underline{K}=\begin{bmatrix} K_{13} & K_{14} \\ K_{23} & K_{24} \\  K_{33} & K_{34} \\ K_{43} & K_{44} \\ K_{53} & K_{54} \\ K_{63} & K_{64} \end{bmatrix}$$

This makes the F-D equation: $$\ \underline{K}_{6x2} \underline{d}_{2x1} = \underline{F}_{6x1} \,$$

The next step is to see that since global nodes 1 and 3 are fixed, we can able the concept of "Virtual Work" which will allow us to ignore their effects for the moment and let us turn $$\underline{F}$$ into a 2x1 by removing the components from those fixed nodes. However, by doing this, we need to adjust $$\underline{K}$$ to keep the equation true. We can further reduce $$\underline{K}$$ down to a 2x2, and the following equation results:

$$\ \underline{K}_{2x2} * \underline{d}_{2x1} = \underline{F}_{2x1} \,$$

Where $$\underline{K}$$ is equal to:

$$\ \underline{K}=\begin{bmatrix} K_{33} & K_{34} \\ K_{43} & K_{44} \end{bmatrix}$$ Since $$\underline{F}$$ is now only made up of the forces from global node 2, we now know the value of $$\underline{F}$$. $$\ \underline{F} = \begin{Bmatrix} F_3 \\ F_4 \end{Bmatrix}= \begin{Bmatrix}  0 \\ P \end{Bmatrix}$$ We also can solve for $$\underline{K}$$: $$\ \underline{K}=\begin{bmatrix} 3.0625 & 2.1752 \\ 2.1752 & 2.6875 \end{bmatrix}$$

At this point, the only term not known in the global F-D equation is $$\underline{d}$$. In order to solve the system, we must first solve for $$\underline{d}$$, which gives the following equation: $$\ \underline{d}_{2x1} = \underline{K}^{-1}_{2x2} * \underline{F}_{2x1} \,$$

Where $$\underline{K^{-1}}$$ is equal to: $$\ \underline{K^{-1}} = \frac{1}{det(\underline{K})}*\begin{bmatrix} K_{44} & -K_{43} \\ -K_{34} & K_{33} \end{bmatrix}$$ $$\ det(\underline{K}) = K_{33} * K_{34} - K_{34} * K_{43} $$

$$\ \underline{K^{-1}} = \begin{bmatrix} 0.7681 & -0.6217 \\ -0.6217 & 0.8753 \end{bmatrix}$$ Plugging everything in and solving for $$\underline{d}$$ gives:

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

$$\begin{Bmatrix} d_3 \\ d_4 \end{Bmatrix} = \begin{bmatrix} 0.7681 & -0.6217 \\ -0.6217 & 0.8753 \end{bmatrix} * \begin{Bmatrix} 0 \\ P \end{Bmatrix} = \begin{Bmatrix} 4.352 \\ 6.1271 \end{Bmatrix} $$

$$\ \underline{d} = \begin{Bmatrix} 4.352 \\ 6.1271 \end{Bmatrix}$$

Computing Reactions
$$ Elem 1: \underline{k}^{(1)} \underline{d}^{(1)} = \underline{f}^{(1)} $$

$$\ d^{(1)} = \begin{bmatrix} 0 \\ 0 \\ 4.352 \\ 6.1271 \end{bmatrix}_{4X1}\,$$

$$\ = \begin{bmatrix} 0.5625 & 0.32476 & ... & ... \\ & & & \\ & & & \\ & & & \end{bmatrix}_{4X4} \begin{Bmatrix} 0 \\ 0 \\ 4.352 \\ 6.1271 \end{Bmatrix}_{4X1} \,$$

$$\ = \begin{bmatrix} -0.5625 & -0.32476 \\ ... & ... \\ & \\ & \end{bmatrix}_{4X2} \begin{Bmatrix} 4.352 \\ 6.1271 \end{Bmatrix}_{4X1} \,$$

$$\ = \begin{bmatrix} -4.4378 \\ -2.5622 \\ +4.4378 \\ +2.5622 \end{bmatrix}_{4X1}\,$$

$$\ = \begin{bmatrix} f_{1}^{(1)} \\ f_{2}^{(1)} \\ f_{3}^{(1)} \\ f_{4}^{(1)} \end{bmatrix}_{4X1}\,$$

 * f1(1) and f2(1) = Reactions

 * f3(1) and f4(1) = Internal Forces

 * Observation: Element 1 is in equilibrium

 * ΣFx = f1(1) + f3(1)

 * ΣFy = f2(1) and f4(1)

ΣMat any point = 0

$$P_{1}^{(1)} = [(f_{1}^{(1)})^2 + (f_{2}^{(1)})^2]^ \frac{1}{2}$$

$$ Elem 2: \underline{k}^{(2)} \underline{d}^{(2)} = \underline{f}^{(2)} $$

 * The same method to solve for element one can also be used to solve for element 2

HW: Verify equilibrium of element 1
Let $$P_{1}^{(1)} = f_{1}^{(1)} + f_{2}^{(1)} = -4.4378 i\hat{} -2.5622j\hat{}$$ and let $$P_{2}^{(1)} = f_{3}^{(1)} + f_{4}^{(1)} = 4.4378 i\hat{} + 2.5622j\hat{}$$

Note that $$P_{1}^{(1)} + P_{2}^{(1)} = 0$$ which illustrates that $$\sum{}F=0$$.

Also, note that $$P_{1}^{(1)} \times P_{2}^{(1)} = \left( -4.4378 i\hat{} -2.5622j\hat{} \right) \times \left( 4.4378 i\hat{} +2.5622j\hat{} \right) = \left( \left( -4.4378 \times 2.5622 \right) - \left( 4.4378 \times -2.5622 \right) \right) k\hat{} = 0$$.

This shows that $$P_{1}^{(1)}$$ and $$P_{2}^{(1)}$$ are in line with each other. Since all forces are in line with each other, $$\sum{}M_{anypoint} = 0$$.

HW: Verify equilibrium of node 2
Note that $$\sum{}F = P_{1}^{(1)} + P_{2}^{(2)} + P = \left( -4.4378 i\hat{} -2.5622j\hat{} \right) + \left( 4.4378 i\hat{} -4.4378j\hat{} \right) + 7j\hat{} = 0$$.

Also, since all forces are acting on the same point, $$\sum{}M_{2} = 0$$.

Contributing Team Members
The following students contributed to this report:

Jessica Dyer Eml4500.f08.gravy.jad 22:11, 25 September 2008 (UTC)

Mikal McFarlane Eml4500.f08.gravy.mmm 00:52, 26 September 2008 (UTC)

Patrick Clayton Eml4500.f08.gravy.psc 16:33, 26 September 2008 (UTC)

Brian Eggeman Eml4500.f08.gravy.bje 17:32, 26 September 2008 (UTC)

Justin Tennant Eml4500.f08.gravy.JMT 16:31, 26 September 2008 (UTC)

Sean Sullivan Eml4500.f08.gravy.sms 17:20, 26 September 2008 (UTC)

title: User:Eml4500.f08.gravy.jad/Notes