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

Global Picture
The first step in the recipe to solve problems using the finite element method is to create a global picture of the problem itself. This global picture must contain the Global Degrees of Freedom (DOFs), also referred to as displacement DOFs and Global Forces. In order to determine what needs to be solved for, the displacement DOFs are split up between those that are known, such as constraints, and those that are unknown. The unknown parts must be solved for using the Finite Element Method.

Similarly, the Global Forces are split up into a known part, which is comprised of the applied forces to the system, and an unknown part, which are the reactions of the system.

Element Picture
Once the Global Picture has been completed, it is necessary to create an Element Picture for each of the elements of the global system. These element pictures must label all of the Element DOFs and Element Forces.

Force Displacement Relations
The first step in creating the Global Force Displacement (FD) Relation is to populate the Element Stiffness and Force Matrices. Once the Element Stiffness Matrices have been created, the Global Stiffness Matrix can be assembled. This Global Stiffness Matrix can then be inserted into the Global FD Relation.

Elimination of Known Displacement DOFs
In order to solve the Global FD Relation created in the last step, it is necessary to eliminate the known displacement DOFs in the system. As stated above, these known DOFs usually come in the form of constraints on the system. Once these constraints have been imposed on the system, it is possible to fully determine the values of the Global Displacement Matrix using the Global FD Relation.

Compute Element Forces
Once the Global Displacement Matrix has been fully determined, it is necessary to relate these global displacements with the corresponding element displacements. The fully populated Element Displacement Matrices can then be used in the Element FD Relations to solve for the forces in each element.

Compute Reactions
Because the Element Force Matrices are composed of internal element forces and reactions, computing the reactions is as simple as determining the correspondence between the Global Force Matrix and Element Force Matrices.

Global Picture
The image below shows the Global Picture of the two-member truss system.

The main reason for the creation of the Global Picture is to label each node in the two-member truss system and introduce the Global DOFs and Global Forces.

$$

\textbf{K}_{n x n}=\left[ \begin{array}{c} K_{ij} \end{array} \right]_{n x n}=\textrm{Global\ Stiffness\ Matrix}

$$

$$

\textbf{d}_{n x 1}=\begin{Bmatrix} d_{j} \end{Bmatrix}_{n x 1}=\textrm{Global\ Displacement\ Matrix}

$$

$$

\textbf{F}_{n x 1}=\begin{Bmatrix} F_{i} \end{Bmatrix}_{n x 1}=\textrm{Global\ Force\ Matrix}

$$

Numbering the Global Displacement DOFs

 * Follow the order of the global node numbers
 * For each node, follow the order of the global coordinate axes and number the displacement definitions for that node

Numbering the Global Forces

 * Follow the order of the global node numbers
 * For each node, follow the order of the global coordinate axes and number the displacement definitions for that node

Element Picture
$$

\textbf{k}_{4 x 4}^{(e)}\textbf{d}_{4 x 1}^{(e)}=\textbf{f}_{4 x 1}^{(e)}

$$

$$

\textbf{k}_{4 x 4}^{(e)}=\left[ \begin{array}{c} k_{ij}^{(e)} \end{array} \right]_{4 x 4}=\textrm{Element\ Stiffness\ Matrix}

$$

$$

\textbf{d}_{4 x 1}^{(e)}=\begin{Bmatrix} d_{j}^{(e)} \end{Bmatrix}_{4 x 1}=\textrm{Element\ Displacement\ Matrix}

$$

$$

\textbf{f}_{4 x 1}^{(e)}=\begin{Bmatrix} f_{i}^{(e)} \end{Bmatrix}_{4 x 1}=\textrm{Element\ Force\ Matrix}

$$

The Element Picture for Element 1 (e=1) is shown below.

The Element Picture for Element 2 (e=2) is shown below.

Element Force Displacement Relations
The condensed version of the Element Stiffness Matrix is shown below for an element with four different DOFs.

$$

\left[ \begin{array}{c} \textbf{k}^{(e)} \end{array} \right]_{4 x 4}\begin{Bmatrix} \textbf{d}^{(e)} \end{Bmatrix}_{4 x 1}=\begin{Bmatrix} \textbf{f}^{(e)} \end{Bmatrix}_{4 x 1}

$$

 * At the element level, the element force matrix is equal to the element stiffness matrix for element e multiplied by the element displacement matrix of element e.

From page 225 in Finite Element Analysis with Mathematica and MATLAB Computations, The second equation from the bottom is the element equation when the matrix multiplication is carried out. This equation is shown below.

$$

k^{(e)} \left[ \begin{array}{cccc} (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{array} \right] \left[\begin{array}{c} u_{1} \\ v_{1} \\ u_{2} \\ v_{2}\end{array}\right] = \left[\begin{array}{c} F_{1x} \\ F_{1y} \\ F_{2x} \\ F_{2y} \end{array}\right]

$$

Where the axial stiffness of the bar element "e" is

$$k^{(e)} = \frac{E^{(e)}A^{(e)}}{L^{(e)}} $$

When a problem with any plane assembly of bars needs to be solved, the above equation can be used. The individual elements, however, are assumed to resist axial loads.

$$

l^{(e)}\textrm{,\ }m^{(e)}=\textrm{director\ of\ cosines\ of\ the\ }\tilde {x} \textrm{\ axis\ (goes\ from\ the\ local\ node\ numbers\ 1\ to\ 2)\ with\ respect\ to\ global\ (x,y)\ coordinates.}

$$

The director of cosines are shown below as trigonometric functions.

$$ l^{(e)}=\mathbf{\vec{i}}\cdot \vec{i}=\vec{i}\cdot (\cos \theta ^{(e)}\vec{i +\sin \theta ^{(e)}}\vec{j)}=\cos \theta ^{(e)}\vec{i}\cdot \vec{i}+\sin \theta ^{(e)}\vec{j}\cdot \vec{i}=\cos \theta ^{(e)}\times 1 +\sin \theta ^{(e)}\times 0 $$

$$ l^{(e)}=\cos \theta ^{(e)} $$

In general, $$k_{ij}^{(e)} = k_{ji}^{(e)}$$ or $$\textbf{k}^{(e)T} = \textbf{k}^{(e)}.$$ The lower triangle of the $$\textbf{k}^{(e)}$$ matrix is symmetric. The upper triangle part is the remaining part shown below. $$ \textbf{k}^{(e)}= \left[ \begin{array}{cccc} k_{11}^{(e)} & k_{12}^{(e)} & k_{13}^{(e)} & k_{14}^{(e)} \\ & k_{22}^{(e)} & k_{23}^{(e)} & k_{24}^{(e)} \\ & & k_{33}^{(e)} & k_{34}^{(e)} \\ & & & k_{44}^{(e)} \end{array} \right]$$

This is useful for finding the elemental stiffness matrices for any element. This allows for the reduction of the number of calculations required in order to fill in the matrices.

To assemble the elemental stiffness matrix for element 1, the director cosines and axial stiffness of the bar must be determined as shown below. The axial stiffness $$k^{(1)}$$ is found using the Elastic Modulus E, the cross-sectional area of the element A, and the length of the element L.

$$\theta^{(1)} = 30^{\circ}$$ $$l^{(1)} = \cos(\theta^{(1)}) = \cos(30^{\circ}) = \frac{\sqrt{3}}{2} $$ $$m^{(1)} = \sin(\theta^{(1)}) = \cos(30^{\circ}) = \frac{1}{2} $$ $$k^{(1)} = \frac{E^{(1)}A^{(1)}}{L^{(1)}} = \frac{(3)(1)}{4} = \frac{3}{4}$$

Once these have been determined they can be used to create the elemental stiffness matrix shown below in coefficient form.

$$ \textbf{k}^{(1)}= k^{(1)}\left[ \begin{array}{cccc} k_{11}^{(1)} & k_{12}^{(1)} & k_{13}^{(1)} & k_{14}^{(1)} \\ k_{21}^{(1)} & k_{22}^{(1)} & k_{23}^{(1)} & k_{24}^{(1)} \\ k_{31}^{(1)} & k_{32}^{(1)} & k_{33}^{(1)} & k_{34}^{(1)} \\ k_{41}^{(1)} & k_{42}^{(1)} & k_{43}^{(1)} & k_{44}^{(1)} \end{array} \right]$$ $$ = \left[ \begin{array}{c} k_{ij}^{(1)} \end{array} \right]_{4x4} $$ where $$i = 1, 2, 3, 4 $$ and $$ j = 1, 2, 3, 4$$

The coefficients can be determined using the previously found director cosines and the axial stiffness of the bar. Three sample calculations are shown below.

$$k_{11}^{(1)} = k^{(1)}(l^{(1)})^2 = (\frac{3}{4})(\frac{\sqrt{3}}{2})^2 = \frac{9}{16}$$ $$k_{12}^{(1)} = k^{(1)}(l^{(1)})(m^{(1)}) = (\frac{3}{4})(\frac{\sqrt{3}}{2})(\frac{1}{2}) = \frac{3\sqrt{3}}{16}$$ $$k_{22}^{(1)} = k^{(1)}(m^{(1)})^2 = (\frac{3}{4})(\frac{1}{2})^2 = \frac{3}{16}$$ Observations:

1) It is only necessary to compute 3 numbers: $$k_{11}^{(1)}, k_{12}^{(1)}, k_{22}^{(1)}$$. The other coefficients have the same absolute value but differ depending on whether they are positive or negative.  2) Matrix $$\textbf{k}^{(1)}$$ is symmetric; therefore, $$k_{ij}^{(1)} = k_{ji}^{(1)}$$. For example: $$k_{13}^{(1)} = k_{31}^{(1)}$$ You only need to interchange the row and column indices.

The director cosines and axial stiffness of the bar calculations have to be repeated for element 2. These calculations are shown below. $$\theta^{(2)} = -45^{\circ}$$ $$l^{(2)} = \cos(\theta^{(2)}) = \cos(-45^{\circ}) = \frac{\sqrt{2}}{2} $$ $$m^{(2)} = \sin(\theta^{(2)}) = \cos(-45^{\circ}) = \frac{-\sqrt{2}}{2} $$ $$k^{(2)} = \frac{E^{(2)}A^{(2)}}{L^{(2)}} = \frac{(5)(2)}{5} = 5$$

These are determined to find the coefficients of the stiffness matrix for element 2. That matrix is shown below in coefficient form.

$$ \textbf{k}^{(2)}= k^{(2)}\left[ \begin{array}{cccc} k_{11}^{(2)} & k_{12}^{(2)} & k_{13}^{(2)} & k_{14}^{(2)} \\ k_{21}^{(2)} & k_{22}^{(2)} & k_{23}^{(2)} & k_{24}^{(2)} \\ k_{31}^{(2)} & k_{32}^{(2)} & k_{33}^{(2)} & k_{34}^{(2)} \\ k_{41}^{(2)} & k_{42}^{(2)} & k_{43}^{(2)} & k_{44}^{(2)} \end{array} \right]$$ $$ = \left[ \begin{array}{c} k_{ij}^{(2)} \end{array} \right]_{4x4} $$ where $$i = 1, 2, 3, 4 $$ and $$ j = 1, 2, 3, 4$$

The only required calculation to determine the value of the matrix above is shown below.

$$k_{11}^{(2)} = k^{(2)}(l^{(2)})^2 = (5)(\frac{\sqrt{2}}{2})^2 = \frac{5}{2}$$ Observations: 1) The absolute value of all the coefficients in the $$\textbf{k}_{ij}^{(2)}$$ where (i,j) = 1, 2, 3, and 4 are the same. Compute one of the coefficients and all the other coefficients will have the same value but will differ depending on whether they are positive or negative.  2) It is also known that $$\textbf{k}^{(2)T} = \textbf{k}^{(2)}$$. This leads to the conclusion that $$\textbf{k}^{(2)}$$ is symmetric.

The element level displacement degrees of freedom are shown below.

$$

\textrm{Element\ 1:\ }\begin{Bmatrix} d_{1}^{(1)} & d_{2}^{(1)} & d_{3}^{(1)} & d_{4}^{(1)} \end{Bmatrix}

$$

$$

\textrm{Element\ 2:\ }\begin{Bmatrix} d_{1}^{(2)} & d_{2}^{(2)} & d_{3}^{(2)} & d_{4}^{(2)} \end{Bmatrix}

$$

Global Force Displacement Relation
The generalized Global Force Displacement Relation for a "Free-Free Structure" is shown below.

$$

\textbf{K}_{n x n}\textbf{d}_{n x 1}=\textbf{F}_{n x 1}

$$

Where $$\textbf{K}_{n x n}$$ is the Global Stiffness Matrix for the two-member truss system described above, $$\textbf{d}_{n x 1}$$ is the Global Displacement Matrix, and $$\textbf{F}_{n x 1}$$ is the Global Force Matrix.

The expanded matrix form of the Global Force-Displacement Relation for the "Free-Free Structure" above is shown below.

$$

\left[ \begin{array}{cccccc} 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{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}

$$

The global level displacement degrees of freedom are shown below.

$$

\begin{Bmatrix} d_{1} & d_{2} & d_{3} & d_{4} & d_{5} & d_{6} \end{Bmatrix}

$$

In order to go from the Element Matrices (Stiffness, Displacement, and Force) to the Global Matrices, it is necessary to go through an assembly process. This assembly process starts with identifying the correspondence between element displacement degrees of freedom and global displacement degrees of freedom as shown below.

$$

\textrm{Node\ 1:\ }d_{1}=d_{1}^{(1)}\textrm{,\ }d_{2}=d_{2}^{(1)}

$$

$$

\textrm{Node\ 2:\ }d_{3}=d_{3}^{(1)}=d_{1}^{(2)}\textrm{,\ }d_{4}=d_{4}^{(1)}=d_{2}^{(2)}

$$

$$

\textrm{Node\ 3:\ }d_{5}=d_{3}^{(2)}\textrm{,\ }d_{6}=d_{4}^{(2)}

$$

The conceptual step of the assembly process is shown below.

The section in the above matrix where the red shaded section and blue shaded section overlap is caused by the fact that $$d_{3}$$ and $$d_{4}$$ are common to both elements of the two-member truss system.

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.

Elimination of Known Displacement DOFs
This matrix in compact notation is shown below.

$$

\left[ \begin{array}{c} K_{ij} \end{array} \right]_{6 x 6}\begin{Bmatrix} d_{j} \end{Bmatrix}_{6 x 1}=\begin{Bmatrix} F_{i} \end{Bmatrix}_{6 x 1}

$$

The summation notation form of the above compact matrix notation is shown below.

$$

\displaystyle\sum_{j=1}^6 K_{ij}d_{j}=F_{i}, i=1,...,6

$$

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 Virtual 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}

$$

Step 4 in the finite element analysis process begins with the reduction of the global force-displacement relations. This reduction is done by first eliminating the known degrees of freedom(DOF’s). $$ d_{1}=d_{2}=d_{5}=d_{6}=0 $$

Deleting the corresponding columns in the global stiffness matrix and the force matrix results in the following relationship.

$$ \left[ \begin{array}{cc} k_{33} & k_{34}\\ k_{43}& k_{44} \end{array} \right]_{2x2} $$ $$\begin{Bmatrix} d_{3}\\d_{4} \end{Bmatrix}_{2x1}=\begin{Bmatrix}F_{3}\\F_{4} \end{Bmatrix}_{2x1}$$

In order to find the inverse of the stiffness matrix one must first calculate the determinant as follows. det $$k = k_{33}k_{44}-k_{34}k_{43} $$

The inverse of the stiffness matrix can then be calculated as shown below.

$$k^{-1} = \frac {1}{\text{det(k)}}$$$$ \left[ \begin{array}{cc} k_{44} & -k_{34}\\ -k_{43}& k_{33} \end{array} \right] $$

If done correctly the stiffness matrix multiplied by its inverse will yield the identity matrix, I. $$I= \left[ \begin{array}{cc} 1 & 0\\ 0& 1 \end{array} \right] $$

For the two beam truss system shown below, the displacement matrix is calculated by multiplying the inverse matrix by $$\begin{Bmatrix}{0} \\ {P} \end{Bmatrix}$$, where P=7.

$$\begin{Bmatrix}d_{3} \\ d_{4} \end{Bmatrix}=k^{-1}\begin{Bmatrix}{0}\\{7}\end{Bmatrix}$$

det $$k=(3.0625)(2.6875)-(-2.175)(-2.175)=3.50$$

Therefore, $$ k^{-1}=\frac {1}{3.50}$$$$ \left[ \begin{array}{cc}2.6875&2.175\\ 2.175&3.0625\end{array} \right]=\left[ \begin{array}{cc}.7678&.6214\\ .6214&.8750\end{array} \right]$$

so $$\begin{Bmatrix}d_{3} \\ d_{4} \end{Bmatrix}=\left[ \begin{array}{cc}.7678&.6214\\ .6214&.8750\end{array} \right]\begin{Bmatrix}{0}\\{7}\end{Bmatrix}=\begin{Bmatrix}{4.352}\\{6.1271}\end{Bmatrix}$$

Step 5 consists of computing the reactions, which can be done by two different methods. Method 1 is to use the element force-displacement relations and method 2 is the Euler cut principle. The force-displacement relation corresponding to the two beam truss system is shown below.

$$k_{4x4}^{(e)}d_{4x1}^{(e)}=f_{4x1}^{(e)}$$ where $$e=1,2$$

$$d^{(1)}=\begin{Bmatrix}{0}\\{0}\\{4.352}\\{6.1271}\end{Bmatrix}$$ and $$f^{(1)}=\begin{Bmatrix}{f_{1}^{(1)}}\\{f_{2}^{(1)}}\\{f_{3}^{(1)}}\\{f_{4}^{(1)}}\end{Bmatrix}$$

$$d^{(2)}=\begin{Bmatrix}{4.352}\\{6.1271}\\{0}\\{0}\end{Bmatrix}$$ and $$f^{(2)}=\begin{Bmatrix}{f_{1}^{(2)}}\\{f_{2}^{(2)}}\\{f_{3}^{(2)}}\\{f_{4}^{(2)}}\end{Bmatrix}$$

Compute Element Forces
The resulting element displacement and stiffness matrices can be used to determine the element forces. The forces on Element 1 are calculated below.

Element 1 :

The element stiffness matrix multiplied by the element displacement matrix is k (1) d (1)

$$

=\left[ \begin{array}{cccc} K_{11} & K_{12} & K_{13} & K_{14} \\ K_{21} & K_{22} & K_{23} & K_{24} \\ K_{31} & K_{32} & K_{33} & K_{34} \\ K_{41} & K_{42} & K_{43} & K_{44} \end{array} \right]\begin{Bmatrix} 0 \\ 0 \\ 4.352 \\ 6.127  \end{Bmatrix}

$$

The matrices can be reduced due to the zeros in the displacement matrix and set equal to the element force matrix k (1) d (1) = f (1)

$$

=\left[ \begin{array}{cc} K_{13} & K_{14} \\ K_{23} & K_{24} \\ K_{33} & K_{34} \\ K_{43} & K_{44} \end{array} \right]\begin{Bmatrix} 4.352 \\ 6.127  \end{Bmatrix} = \left[ \begin{array}{c} -4.4378 \\ -2.5622 \\ 4.4378 \\ 2.5622 \end{array} \right]  =  \left[ \begin{array}{c} f_{1}^{(1)} \\ f_{2}^{(1)} \\ f_{3}^{(1)} \\ f_{4}^{(1)} \end{array} \right]

$$

Observation:

Element 1 is in equilibrium therefore:

(1) $$\sum F_x = f_1^{(1)} + f_3^{(1)} = 0 $$

(2) $$\sum F_y = f_2^{(1)} + f_4^{(1)} = 0 $$

(3) $$\sum M_{any pt} = 0 $$

Element 2 :

The figure below shows element 2 labeled with axial forces P.

Method 2 to solve for a 2-bar truss system using statics method:

P can be brought back into the big picture by performing an equilibrium balance of node 2.

Compute Reactions
The final step in solving for the reactions is to relate the element forces found in the previous step with the global forces that correspond to the reactions of the system. The reactions of the system are shown below.

$$

\begin{Bmatrix} f_{1} \\ f_{2} \\ f_{5} \\ f_{6} \end{Bmatrix}=\begin{Bmatrix} -4.438 \\ -2.562 \\ 4.438 \\ -4.438 \end{Bmatrix}

$$

Analysis and Verification of MATLAB Code for Two-Member Truss System
In this section, the two-member truss problem covered during class will be solved manually using the Finite Element Method and compared with the execution of the Finite Element Method by means of a MATLAB code written by Professor Loc Vu Quoc from the University of Florida Department of Mechanical and Aerospace Engineering.

Finite Element Method Manual Calculations

The next step to solving the two-member truss system shown above is to populate the Element Stiffness Matrices, $$\textbf{k}^{(1)}$$ and $$\textbf{k}^{(2)}$$. This step is shown below using the equation for the Element Stiffness Matrices given in the course notes above.

$$

\textbf{k}^{(1)}=k^{(1)} \left[ \begin{array}{cccc} (l^{(1)})^{2} & l^{(1)}m^{(1)} & -(l^{(1)})^{2} & -l^{(1)}m^{(1)} \\ l^{(1)}m^{(1)} & (m^{(1)})^{2} & -l^{(1)}m^{(1)} & -(m^{(1)})^{2} \\ -(l^{(1)})^{2} & -l^{(1)}m^{(1)} & (l^{(1)})^{2} & l^{(1)}m^{(1)} \\ -l^{(1)}m^{(1)} & -(m^{(1)})^{2} & l^{(1)}m^{(1)} & (m^{(1)})^{2} \end{array}\right]=\left[\begin{array}{cccc} 0.5625 & 0.3248 & -0.5625 & -0.3248 \\ 0.3248 & 0.1875 & -0.3248 & -0.1875 \\ -0.5625 & -0.3248 & 0.5625 & 0.3248 \\ -0.3248 & -0.1875 & 0.3248 & 0.1875 \end{array} \right]

$$

$$

\textbf{k}^{(2)}=k^{(2)} \left[ \begin{array}{cccc} (l^{(2)})^{2} & l^{(2)}m^{(2)} & -(l^{(2)})^{2} & -l^{(2)}m^{(2)} \\ l^{(2)}m^{(2)} & (m^{(2)})^{2} & -l^{(2)}m^{(2)} & -(m^{(2)})^{2} \\ -(l^{(2)})^{2} & -l^{(2)}m^{(2)} & (l^{(2)})^{2} & l^{(2)}m^{(2)} \\ -l^{(2)}m^{(2)} & -(m^{(2)})^{2} & l^{(2)}m^{(2)} & (m^{(2)})^{2} \end{array}\right]=\left[\begin{array}{cccc} 2.5 & -2.5 & -2.5 & 2.5 \\ -2.5 & 2.5 & 2.5 & -2.5 \\ -2.5 & 2.5 & 2.5 & -2.5 \\ 2.5 & -2.5 & -2.5 & 2.5 \end{array} \right]

$$

Once both Element Stiffness Matrices have been populated, it is necessary to assemble the Global Stiffness Matrix using an assembly process involving both $$\textbf{k}^{(1)}$$ and $$\textbf{k}^{(2)}$$. This process is shown below.

$$

\textbf{K}=\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]=\left[ \begin{array}{cccccc} 0.5625 & 0.3248 & -0.5625 & -0.3248 & 0 & 0 \\ 0.3248 & 0.1875 & -0.3248 & -0.1875 & 0 & 0 \\ -0.5625 & -0.3248 & 3.0625 & -2.1752 & -2.5 & 2.5 \\ -0.3248 & -0.1875 & -2.1752 & 2.6875 & 2.5 & -2.5 \\ 0 & 0 & -2.5 & 2.5 & 2.5 & -2.5 \\ 0 & 0 & 2.5 & -2.5 & -2.5 & 2.5 \end{array} \right]

$$

Next, the constraints imposed on the Global Displacement Matrix due to the pinned nodes are taken into account. These constraints are: $$d_{1}=0$$, $$d_{2}=0$$, $$d_{5}=0$$, $$d_{6}=0$$.

$$

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

$$

Because the first and last two rows of the Global Displacement Matrix are all 0, the corresponding rows and columns can be deleted from the Global Stiffness Matrix and Global Force Matrix in order to reduce the Global Force Displacement Relation to a 2 x 2 linear system. This system is shown below.

$$

\left[ \begin{array}{cc} K_{33} & K_{34} \\ K_{43} & K_{44}\end{array} \right]\begin{Bmatrix} d_{3} \\ d_{4}\end{Bmatrix}=\begin{Bmatrix} F_{3} \\ F_{4}\end{Bmatrix}

$$

The process of solving this linear system for the global displacements $$d_{3}$$ and $$d_{4}$$ is shown below.

$$

\left[ \begin{array}{cc} 3.0625 & -2.1752 \\ -2.7152 & 2.6875\end{array} \right]\begin{Bmatrix} d_{3} \\ d_{4}\end{Bmatrix}=\begin{Bmatrix} 0 \\ 7\end{Bmatrix}

$$

$$

\begin{Bmatrix}d_{3} \\ d_{4} \end{Bmatrix}=\textbf{K}^{-1}\begin{Bmatrix}{0}\\{7}\end{Bmatrix}

$$

$$

\textbf{K}^{-1}=\dfrac{1}{\text{det} \textbf{K}}\left[ \begin{array}{cc} K_{44} & -K_{34} \\ -K_{43} & K_{33}\end{array} \right]=\dfrac{1}{(3.0625)(2.6875)-(-2.1752)(-2.1752)}\left[ \begin{array}{cc} 2.6875 & 2.1752 \\ 2.1752 & 3.0625\end{array} \right]=\left[ \begin{array}{cc} 0.7681 & 0.6217 \\ 0.6217 & 0.8753\end{array} \right]

$$

$$

\begin{Bmatrix}d_{3} \\ d_{4} \end{Bmatrix}=\left[ \begin{array}{cc} 0.7681 & 0.6217 \\ 0.6217 & 0.8753\end{array} \right]\begin{Bmatrix}{0}\\{7}\end{Bmatrix}

$$

$$

\begin{Bmatrix}d_{3} \\ d_{4} \end{Bmatrix}=\begin{Bmatrix}4.3519 \\ 6.1271 \end{Bmatrix}

$$

In order determine the internal forces and reactions in both members, it is necessary to use the newly determined values from the Global Displacement Matrix in the Element Force Displacement Relation. The process of solving for the internal forces and reactions for Element 1 is shown below.

$$

\textbf{k}^{(1)}\textbf{d}^{(1)}=\textbf{f}^{(1)}

$$

$$

\textbf{d}^{(1)}=\begin{Bmatrix} 0 \\ 0 \\ 4.3519 \\ 6.1271 \end{Bmatrix}

$$

$$

\left[\begin{array}{cccc} 0.5625 & 0.3248 & -0.5625 & -0.3248 \\ 0.3248 & 0.1875 & -0.3248 & -0.1875 \\ -0.5625 & -0.3248 & 0.5625 & 0.3248 \\ -0.3248 & -0.1875 & 0.3248 & 0.1875 \end{array} \right]\begin{Bmatrix} 0 \\ 0 \\ 4.3519 \\ 6.1271 \end{Bmatrix}=\begin{Bmatrix} f_{1}^{(1)} \\ f_{2}^{(1)} \\ f_{3}^{(1)} \\ f_{4}^{(1)} \end{Bmatrix}

$$

$$

\begin{Bmatrix} f_{1}^{(1)} \\ f_{2}^{(1)} \\ f_{3}^{(1)} \\ f_{4}^{(1)} \end{Bmatrix}=\begin{Bmatrix} -4.438 \\ -2.562 \\ 4.438 \\ 2.562 \end{Bmatrix}

$$

In the above Element Force Matrix for Element 1, $$f_{1}^{(1)}$$ and $$f_{2}^{(1)}$$ represent the reactions at Global Node 1, whereas $$f_{3}^{(1)}$$ and $$f_{4}^{(1)}$$ represent the internal forces in Element 1.

The process shown above is repeated below in order to solve for the internal forces and reactions of Element 2.

$$

\textbf{k}^{(2)}\textbf{d}^{(2)}=\textbf{f}^{(2)}

$$

$$

\textbf{d}^{(2)}=\begin{Bmatrix} 4.3519 \\ 6.1271 \\ 0 \\ 0 \end{Bmatrix}

$$

$$

\left[\begin{array}{cccc} 2.5 & -2.5 & -2.5 & 2.5 \\ -2.5 & 2.5 & 2.5 & -2.5 \\ -2.5 & 2.5 & 2.5 & -2.5 \\ 2.5 & -2.5 & -2.5 & 2.5 \end{array} \right]\begin{Bmatrix} 4.3519 \\ 6.1271 \\ 0 \\ 0 \end{Bmatrix}=\begin{Bmatrix} f_{1}^{(2)} \\ f_{2}^{(2)} \\ f_{3}^{(2)} \\ f_{4}^{(2)} \end{Bmatrix}

$$

$$

\begin{Bmatrix} f_{1}^{(2)} \\ f_{2}^{(2)} \\ f_{3}^{(2)} \\ f_{4}^{(2)} \end{Bmatrix}=\begin{Bmatrix} -4.438 \\ 4.438 \\ 4.438 \\ -4.438 \end{Bmatrix}

$$

In the above Element Force Matrix for Element 2, $$f_{1}^{(2)}$$ and $$f_{2}^{(2)}$$ represent the internal forces in Element 2, whereas $$f_{3}^{(2)}$$ and $$f_{4}^{(2)}$$ represent the reactions at Global Node 3.

Relating the element forces solved for with the global forces, an array possessing only the reactions at Global Node 1 and Global Node 3 is shown below.

$$

\begin{Bmatrix} f_{1} \\ f_{2} \\ f_{5} \\ f_{6} \end{Bmatrix}=\begin{Bmatrix} -4.438 \\ -2.562 \\ 4.438 \\ -4.438 \end{Bmatrix}

$$

MATLAB Code
% Two bar truss example clear all; e = [3 5]; A = [1 2]; P = 7; L=[4 2]; alpha = pi/3; beta = pi/4; nodes = [0, 0; L(1)*cos(pi/2-alpha), L(1)*sin(pi/2-alpha); L(1)*cos(pi/2-alpha)+L(2)*sin(beta),L(1)*sin(pi/2-alpha)-L(2)*cos(beta)]; dof=2*length(nodes); conn=[1,2; 2,3]; lmm = [1, 2, 3, 4; 3, 4, 5, 6]; elems=size(lmm,1); K=zeros(dof); R = zeros(dof,1); debc = [1, 2, 5, 6]; ebcVals = zeros(length(debc),1); %load vector R = zeros(dof,1); R(4) = P; % Assemble global stiffness matrix K=zeros(dof); for i=1:elems lm=lmm(i,:); con=conn(i,:); k_local=e(i)*A(i)/L(i)*[1 -1; -1 1] k=PlaneTrussElement(e(i), A(i), nodes(con,:)) K(lm, lm) = K(lm, lm) + k; end K R % Nodal solution and reactions [d, reactions] = NodalSoln(K, R, debc, ebcVals) results=[]; for i=1:elems results = [results; PlaneTrussResults(e, A, ... nodes(conn(i,:),:), d(lmm(i,:)))]; end format short g results k_local = 0.75       -0.75      -0.75         0.75 k = 0.5625     0.32476      -0.5625     -0.32476      0.32476       0.1875     -0.32476      -0.1875      -0.5625     -0.32476       0.5625      0.32476     -0.32476      -0.1875      0.32476       0.1875 k_local = 5   -5      -5     5 k = 2.5        -2.5         -2.5          2.5      -2.5          2.5          2.5         -2.5      -2.5          2.5          2.5         -2.5       2.5         -2.5         -2.5          2.5 K = 0.5625     0.32476      -0.5625     -0.32476            0            0      0.32476       0.1875     -0.32476      -0.1875            0            0      -0.5625     -0.32476       3.0625      -2.1752         -2.5          2.5     -0.32476      -0.1875      -2.1752       2.6875          2.5         -2.5            0            0         -2.5          2.5          2.5         -2.5            0            0          2.5         -2.5         -2.5          2.5 R = 0         0          0          7          0          0 d = 0         0      4.352     6.1271          0          0 reactions = -4.4378      -2.5622        4.4378       -4.4378 results = 1.7081      5.1244       8.5406       5.1244       17.081        0.6276       1.8828        3.138       1.8828        6.276

Conclusions
As shown in the previous sections, the results of the manual calculations match the results determined from the MATLAB Code written by Professor Loc Vu Quoc. This shows that the proper methodology was taken and accurate calculations were made throughout the process.

Contributing Team Members
Andrew McDonald - Eml4500.f08.bike.mcdonald 22:01, 25 September 2008 (UTC) Garrett Pataky - Eml4500.f08.bike.pataky 22:02, 25 September 2008 (UTC) Samuel Bernal - Eml4500.f08.bike.bernal 13:07, 26 September 2008 (UTC) Bobby Sweeting - Eml4500.f08.bike.sweeting 18:33, 25 September 2008 (UTC) Shawn Gravois - Eml4500.f08.bike.gravois 18:22, 25 September 2008 (UTC) Eric Viale - Eml4500.f08.bike.viale 21:41, 25 September 2008 (UTC)


text:
User:Eml4500.f08.bike.mcdonald/homework report 3

Element Force-Displacement Relation With Respect To The Global Coordinate System
The element force displacement (FD) relation can be determined with respect to the global coordinate system. The following equation shows the general force displacement relation for an element member.

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

The local free body diagram for the element member is pictured in the figure below.

The local axial force displacement diagram for the element member is pictured in the figure below.

The following force displacement equation can be constructed from the above axial force displacement diagram.

$$\mathbf{k}^{(e)}\begin{bmatrix}1 & -1\\-1 & 1\end{bmatrix}\begin{Bmatrix}q_1^{(e)}\\q_2^{(e)}\end{Bmatrix}=\begin{Bmatrix}p_1^{(e)}\\ p_2^{(e)}\end{Bmatrix}$$
 * $$ \mathbf{\hat{\; k}}^{(e)}_{2x2}\,\;\;\;\;\;\;\;\; \mathbf{q}^{(e)}_{2x1} \;\; = \;\;\;\; \mathbf{p}^{(e)}_{2x1}$$

Where:
 * qi(e) = axial displacement of element e at local node i.
 * pi(e) = axial force of element e at local node i.

The goal is to find the relations between $$\begin{cases} \mathbf{q}^{(e)}_{2x1} & \text{ and}\;\;\;\;\;   \mathbf{d}^{(e)}_{4x1}  \\ \mathbf{p}^{(e)}_{2x1} & \text{ and}\;\;\;\;\;  \mathbf{f}^{(e)}_{4x1} \end{cases}$$

One of the relations can be expressed in the form below.

$$\mathbf{q}^{(e)}_{2x1} = \mathbf{T}^{(e)}_{2x4}\, \mathbf{d}^{(e)}_{4x1}$$

The figure below depicts the displacement vector of local node 1 as well as a portion of the element member e.

In the figure above, the displacements $$d^{(e)}_1$$ and $$d^{(e)}_2$$ in the $$\vec{i}\,$$ and $$\,\vec{j}$$ directions are shown as components of $$\mathbf{\hat{\; d}}\,$$1(e). The equation of the $$\vec{i}\,$$ and $$\,\vec{j}\,$$ components of $$\mathbf{\hat{\; d}}\,$$1(e) can be seen in the following equation.

$$\bar{d}^{(e)}_1=d^{(e)}_1\vec{i}+d^{(e)}_2\vec{j}$$

q1(e) = axial displacement of node 1 is the orthagonal projection of the displacement vector $$\bar{d}^{(e)}_1$$ of node 1 on the $$\tilde{x}$$ axis of element e. This is represented in the equation below.

$$q^{(e)}_1=\vec{d}^{(e)}_1\cdot \vec{\tilde{i}}$$

The displacement vector of node 1 can be broken down into components. $$q^{(e)}_1= (d^{(e)}_1\vec{i}+d^{(e)}_2\vec{j}) \cdot \vec{\tilde{i}}$$

Which is equivalent to. $$q^{(e)}_1= d^{(e)}_1(\vec{i}\cdot \vec{\tilde{i}})+d^{(e)}_2(\vec{j} \cdot \vec{\tilde{i}})$$
 * $$ \cos \theta _e \;\;\;\;\; \sin \theta _e$$
 * $$ l^{(e)} \;\;\;\;\;\;\;\;\;\; m^{(e)}$$

q1(e) is a 1x1 scalar which can be calculated with the displacements of node 1 and the director cosines as shown in the following equations.

$$q_1^{(e)}=l^{(e)}d_1^{(e)}+m^{(e)}d_2^{(e)}$$

$$q_1^{(e)}=\begin{bmatrix}l^{(e)} & m^{(e)} \end{bmatrix}\begin{Bmatrix}d_1^{(e)}\\ d_2^{(e)}\end{Bmatrix}$$ $$1x1 \;\;\;\;\;\;\;\;\; 1x2 \;\;\;\;\;\;\;\;\; 2x1$$

The matrices that result from combining the displacement equations for nodes 1 and 2 are listed below. $$\begin{Bmatrix}q_1^{(e)}\\q_2^{(e)}\end{Bmatrix}=\begin{bmatrix} l^{(e)} & m^{(e)} & 0 & 0\\ 0 & 0 & l^{(e)} & m^{(e)}\end{bmatrix}\begin{Bmatrix} d_1^{(e)}\\ d_2^{(e)}\\ d_3^{(e)}\\ d_4^{(e)}\end{Bmatrix}$$ $$\mathbf{q}^{(e)}=\mathbf{T}^{(e)}\mathbf{d}^{(e)}$$

Now, the equation below can be used to relate the axial forces to the internal forces and the transfer function. $$\textbf{P}_{2x1}^{(e)}=\textbf{T}^{(e)}\textbf{f}_{4x1}^{(e)}$$

Expanding the matrices out, it is seen that $$\textbf{P}^{(e)}$$ is comprised of the two axial forces while $$\textbf{f}^{(e)}$$ is comprised of the four internal forces.

$$\begin{Bmatrix}P_1^{(e)}\\P_2^{(e)}\end{Bmatrix}=\textbf{T}^{(e)}\begin{Bmatrix}f_1^{(e)}\\ f_2^{(e)} \\ f_3^{(e)} \\f_4^{(e)} \end{Bmatrix}$$

It can be observed that in order for the equation to be computable, the transfer matrix must be a 2 x 4 rectangular matrix.

Recalling the elemental axial force-displacement relation, the transfer matrix can be used to create an equation that replaces the axial force and displacement matrices with the elemental displacement and force matrices.

$$\hat{\textbf{k}}^{(e)}\textbf{q}^{(e)}=\textbf{p}^{(e)}\ = \hat{\textbf{k}}^{(e)}(\textbf{T}^{(e)}\textbf{d}^{(e)})\ = (\textbf{T}^{(e)}\textbf{f}^{(e)})$$

In order to eliminate all axial forces, the transfer matrix must be moved from the right hand side of the equation to the left hand side. In order to do this, the inverse of the transfer matrix must be determined.

Unfortunately, the transfer matrix was previously determined to be a $$2 x 4$$ rectangular matrix and thus an inverse cannot be taken. To fix this problem, the transpose will be used instead of the inverse to find the elemental forces.

$$[\textbf{T}^{(e)T}\hat{\textbf{k}^{(e)}}\textbf{T}^{(e)}]\textbf{d}^{(e)}=\textbf{f}^{(e)}$$

The principle of virtual work can also be used to reduce this matrix down to one that would be easier to compute. The principle is based on the reduction of the global force-displacement relation and can be reduced to what is shown below by assuming that $$d_1 = d_2 = d_5 = d_6 = 0$$.

$$\textbf{k}_{6x6}\textbf{d}_{6x1}=\textbf{F}_{6x1}$$ --> $$\bar{\textbf{k}}_{2x2}\bar{\textbf{d}}_{2x1}=\bar{\textbf{F}}_{2x1}$$

The matrix cannot be solved by simply moving the stiffness matrix over because the stiffness matrix is singular. Singular matrices are defined as having a determinant which equals zero. Since the inverse of a matrix requires one over the determinant, this creates an invertible situation.

In an unconstrained structural system, there are three possible rigid body movements: two translations and one rotation.

Using dynamics and vibrations, the eigenvalues found can be used in the equation shown below where $$\textbf{k}$$ is the stiffness matrix, $$\lambda$$ is the eigenvalue, and $$\textbf{M}$$ is the mass matrix. This equation relates the stiffness and the mass matrices.

The four zero eigenvalues correspond to the system having no stored elastic energy. This becomes a rigid body mode. Modes can also be referred to as the eigenvectors.

Alternative Method For Determining Reactions of Two Member Truss System
When solving for the reactions of the two member truss system, the only method described so far was to use the Element FD Relation in order to relate the Element Displacement Matrices with the Element Force Matrices. Another method that can be used to solve for the reactions of the system is to use the Global FD Relation.

The use of the Global FD for this purpose was assigned as a homework assignment, and is illustrated below.

Closing the Loop Between the Finite Element Method and Statics
The key to closing the loop between the Finite Element Method and Statics is to apply the Principle of Virtual Work. This principle has been used in Homework Report 2 when attempting to reduce the Global Stiffness Matrix.

The image below shows a Free Body Diagram of a two-force member along with the relationship between the axial forces in the member.

Due to the axial forces that exist in the two force members in the figure below, a certain amount of axial deformation, $$q$$ will occur. The figure below shows the two member truss shape before the given load, $$P$$ is applied to the system causing axial forces and displacements.

Because the reactions in both two-force members in the two member truss system are now known, the methods of statics allow for the axial member forces $$P_{1}^{(1)}$$ and $$P_{2}^{(2)}$$ to be determined. Knowing $$P_{1}^{(1)}$$ and $$P_{2}^{(2)}$$, the axial displacement DOFs (the amount of extension of each member) can be determined as shown below.

$$

q_{1}^{(1)}=0,\text{ fixed node 1}

$$

$$

q_{2}^{(1)}=\dfrac{P_{1}^{(1)}}{k^{(1)}}=\dfrac{P_{2}^{(1)}}{k^{(1)}}

$$

$$

q_{1}^{(2)}=\dfrac{-P_{2}^{(2)}}{k^{(2)}}

$$

$$

q_{2}^{(2)}=0,\text{ fixed node 2}

$$

Knowing the axial displacements in each member, the overall displacement DOFs of Node 2 can be determined using a graphical process shown in the figure below.

In the above figure, the red dotted line is the deformed shape of the two member truss system, caused by application of the load, $$P$$. The geometric length  corresponds to the value of axial displacement in Element 1, $$q_{2}^{(1)}$$, whereas the geometric length   corresponds to the value of axial displacement in Element 2, $$ q_{1}^{(2)}$$.

NOTE: The reason that segments  and   are perpendicular to   and , respectively, is due to the Principle of Virtual Work and the use of extremely small angles of deflection in the principle. The below figure shows the derivation of this principle by means of the small angle theorem.

The Small Angle Theorem states that for small angles, $$\alpha$$, $$\sin {\alpha}=\alpha$$ and $$\cos {\alpha}=1$$. Applying this theorem to the relations for $$u_{x}$$ and $$u_{y}$$ shown in the above figure, the following values are determined.

$$

u_{y}=R\sin {\alpha }\cong R\alpha

$$

$$

u_{x}=R(1-\cos \alpha )\cong 0

$$

The method of infinitesimal displacement, which is related to virtual displacement, is used to analyze the two bar truss system.

Some sample calculations for the lengths of AC and AB were performed as follows:

$$AC = \frac{|P_{2}^{(1)}|}{k^{(1)}} = \frac{5.1243}{.75} = 6.8324$$

$$ AB = \frac{|P_{1}^{(2)}|}{k^{(2)}} = \frac{6.276}{5} = 1.255$$

The (x,y) coordinates $$(x_{B}, y_{B})$$ and $$(x_{C}, y_{C})$$ must then be found. This can be done using simple trigonometry and the values of the magnitude of the line and the values of theta obtained in HW 2. This method is further explained using the figure on the left, along with the intermediate calculation steps shown below.

2 unknowns $$(x_{D}, y_{D})$$

The equations for lines AB and BC are then needed.

$$ \vec{PQ} = (PQ)\vec{\tilde{i}} = (PQ)[cos\Theta \vec{i} + sin\theta \vec{j}]$$

$$= (x - x_{p})\vec{i} + (y - y_{p})\vec{j}$$

$$\Rightarrow x - x_{p} = (PQ)cos\Theta$$

and $$y - y_{p} = (PQ)sin\Theta$$

$$\Rightarrow \frac{y - y_{p}}{x - x_{p}} = tan\Theta$$

$$y - y_{p} = (tan\Theta )(x - x_{p})$$

Using this equation along with $$\Theta^{1} = 30^{o}$$, it can be found that $$(x_{C}, y_{C})$$ = $$(5.92, 3.416)$$

Also, the equation for a line perpendicular to the above line, passing P is as follows:

$$y - y_{p} = (tan(\Theta + \frac{\pi}{2}))(x - x_{p})$$

This is useful in finding the value of $$(x_{B}, y_{B})$$, and using the value of $$\Theta^{2} = 135^{o}$$ it can be found that $$(x_{B}, y_{B})$$ = $$(-.877, .877)$$

To find the value of $$(x_{D}, y_{D})$$ one can use the following equation:

$$ \vec{AD} = (x_{D} - x_{A})\vec{i} + (y_{D} - y_{A})\vec{j}$$

Where the value of $$(x_{A}, y_{A})$$ are zero due to choosing the axis' origin to be point A.

By definition:

$$\vec{AD} = d_{3}\vec{i} + d_{4}\vec{j}$$

This means that, using the values obtained in HW 2, the line AD can be expressed as:

$$\vec{AD} = 4.35\vec{i} + 6.127\vec{j}$$

Which is the displacement vector of A from FEM. This ends closing the loop between statics and the Finite Element Method. The next topic will cover the analysis of a three-bar truss system.

Analysis of the Three-Bar Truss System
The three bar truss system shown above will now be analyzed using the values below as parameters.

$$ E^{(1)} = 2$$ $$ E^{(2)} = 4$$ $$ E^{(3)} = 3$$

$$ A^{(1)} = 3$$ $$ A^{(2)} = 1$$ $$ A^{(3)} = 2$$

$$ L^{(1)} = 5$$ $$ L^{(2)} = 5$$ $$ L^{(3)} = 10$$

$$P = 30$$ $$\Theta ^{(1)} = 30^{o}$$ $$\Theta ^{(2)} = 30^{o}$$ $$\Theta ^{(3)} = 45^{o}$$

Care should be taken when numbering the local nodes in order to make the assembly of the $$K$$ matrix convenient.

The 3-bar truss system is shown with a known force P and three unknown axial reaction forces $$R_{1}, $$$$R_{2}, $$$$ R_{3}$$.

In this system, the statics method yields the following equations. $$

\sum{F_{X}=0} $$ $$ \sum{F_{Y}=0}$$ $$ \sum{M_{A}=0}$$

The moment equation about point A does not solve for any of the unknown force values because all the forces' lines of action go through point A. This means that only the first two equations, the sum of the forces in the X and Y directions can be used. In order to solve this system statically, a third equation must be found. A moment equation around another point could have been the answer, but as shown in the following 3-D explanation, the moment equation is proven to be unusable because it is also equal to zero.

Analyzing the system shown on the left, the following equations are formulated. $$ \sum{M_{B}}=\vec{BA}\times \vec{F}=\vec{BA'}\times \vec{F}$$

$$A'$$ is assumed to be on the line of action of $$\vec{F}$$. The vector $$BA'$$ can be rewritten as shown. $$ \vec{BA'}=\vec{BA}\times \vec{AA'}$$ Therefore, the moment around point B can be rewritten as shown. $$ \sum{M_{B}}=(\vec{BA}+ \vec{AA'})\times \vec{F}= \vec{AA'}\times \vec{F}+\vec{BA}\times \vec{F} $$

$$\vec{AA'}\times \vec{F}$$ becomes zero because $$\vec{AA'} $$ and $$\vec{F}$$ are in the same direction.

Now the moment equation around B is as shown: $$\sum{M_{B}}=\vec{BA}\times \vec{F} $$

Now, back to analyzing a specific 3-bar truss system, the diagram is shown below. Node A in the 3-bar truss system is in equilibrium. $$\sum_{i=0}^{3}{F_{i}=\vec{0}}$$ The moment equation around point B, which was derived above, is rewritten as shown. $$\sum_{i}^{}{M_{B,i}=\sum_{i}^{}{\vec{BA'_{i}}}}\times \vec{F_{i}}$$ $$A'_{i}$$= any point on the line of action of $$\vec{F_{i}}$$. The moment equation around point B can be rewritten as shown. $$\sum_{i}^{}{M_{B,i}=\sum_{i}^{}{\vec{BA}}}\times \vec{F_{i}}$$ $$\sum_{i}^{}{M_{B,i}=\vec{BA}}\times \sum{\vec{F_{i}}}$$ As stated above, the sum of the forces is equal to zero due to point A being in equilibrium. The moment around B is proved to be zero.

$$\sum_{i}^{}{M_{B,i}=\vec{0}}$$

With this conclusion, it was determined that the system is statically indeterminate. The Finite Element Analysis Method must be used to solve the 3-bar truss system.

To begin the Finite Element Method, the stiffness matrix is created. This matrix is shown below.

Plot of Undeformed and Deformed Two-Bar Truss System Using MATLAB
The plot below shows the deformed and undeformed states of the two-bar truss system analyzed during class. This plot was created through the use of a MATLAB script developed by Andrew J. McDonald, using the MATLAB script, developed by X.G. Tan and hosted on Professor Loc Vu Quoc's University of Florida website.

The blue dotted lines represent the undeformed state of the two-member truss system, and the solid red lines represent the deformed state of the two-member truss system.

The code of the MATLAB script developed to produce this plot is given below.

% Matlab Script %********************************************************************* % filename: two_bar_truss_def_and_und.m  % % PURPOSE: %   Plot the two-bar truss system for both deformed %   and undeformed cases % % AUTHOR: Andrew J. McDonald % % Modified on: ______________________________ % Created on : Sat, 04 Oct 2008, 11:12:54 EST % % DEPENDENCIES: %  call: % % REMARKS: % %*********************************************************************

clear; close;

% model with 2-D beam elements with 2 dofs per node dof = 2;        %  dof per node

% % input the nodal coordinates %    n_node = 3;              % total number of nodes n_elem = 2;             % total number of elements total_dof = 6 * n_node; % total dofs of system

% two-bar truss system data %    L_1 = 4; L_2 = 2; theta_1 = 30*(pi/180); theta_2 = -45*(pi/180);

position(:, 1) = [ 0; 0]; position(:, 2) = [ L_1*cos(theta_1); L_1*sin(theta_1)]; position(:, 3) = [ L_1*cos(theta_2)+L_2*cos(theta_2); L_1*sin(theta_1)+L_2*sin(theta_2)];

% set up the nodal coordinate arrays x, y    for i = 1 : n_node x(i) = position(1,i); y(i) = position(2,i); end

% set up the element connectivity array node_connect, defined as % follows: % node_connect(local node number, element number) = global node number

node_connect(1, 1) = 1;  % element 1 node_connect(2, 1) = 2;

node_connect(1, 2) = 2;  % element 2 node_connect(2, 2) = 3;

% plot the whole truss system % loop over the elements for i = 1 : n_elem % for element i, do the following: % node_1 = global node number corresponding to the local node 1 node_1 = node_connect(1,i); % node_2 = global node number corresponding to the local node 1 node_2 = node_connect(2,i); % xx : 1x2 array containing x coordinates of node_1 and node_2 xx = [x(node_1),x(node_2)]; % yy : 1x2 array containing y coordinates of node_1 and node_2 yy = [y(node_1),y(node_2)]; % plot the element i in 2-D using command plot % use axis command to avoid having the plot window too tight axis([0 5 0 5]) % solid line plot(xx,yy,'--') % hold the current figure for the next element hold on    end

text(x(node_connect(1,1)),y(node_connect(1,1)),'Global Node 1',...      'HorizontalAlignment', 'center') text(x(node_connect(2,1)),y(node_connect(2,1)),'Global Node 2',...      'HorizontalAlignment', 'center') text(x(node_connect(2,2)),y(node_connect(2,2)),'Global Node 3',...      'HorizontalAlignment', 'center') text(x(node_connect(2,1))/2,y(node_connect(2,1))/2,'Global Element 1',...     'HorizontalAlignment', 'center') text( x(node_connect(2,1)) + (x(node_connect(2,2))-x(node_connect(2,1)))/2, y(node_connect(2,1)) +(y(node_connect(2,2))-y(node_connect(2,1)))/2 ,'Global Element 2',...     'HorizontalAlignment', 'center')

hold on

d_3 = 4.352; d_4 = 6.1271;

position_d(:, 1) = [ 0; 0]; position_d(:, 2) = [ L_1*cos(theta_1)+ d_3; L_1*sin(theta_1)+ d_4 ]; position_d(:, 3) = [ L_1*cos(theta_2)+L_2*cos(theta_2); L_1*sin(theta_1)+L_2*sin(theta_2)];

% set up the nodal coordinate arrays x, y    for i = 1 : n_node x(i) = position_d(1,i); y(i) = position_d(2,i); end

% set up the element connectivity array node_connect, defined as % follows: % node_connect(local node number, element number) = global node number

node_connect_d(1, 1) = 1;  % element 1 node_connect_d(2, 1) = 2;

node_connect_d(1, 2) = 2;  % element 2 node_connect_d(2, 2) = 3;

% plot the whole truss system % loop over the elements for i = 1 : n_elem % for element i, do the following: % node_1 = global node number corresponding to the local node 1 node_1 = node_connect_d(1,i); % node_2 = global node number corresponding to the local node 1 node_2 = node_connect_d(2,i); % xx : 1x2 array containing x coordinates of node_1 and node_2 xx = [x(node_1),x(node_2)]; % yy : 1x2 array containing y coordinates of node_1 and node_2 yy = [y(node_1),y(node_2)]; % plot the element i in 2-D using command plot % use axis command to avoid having the plot window too tight axis([0 8 0 9]) % solid line plot(xx,yy,'-r') % hold the current figure for the next element hold on    end

text(x(node_connect_d(1,1)),y(node_connect_d(1,1)),'Global Node 1',...      'HorizontalAlignment', 'center') text(x(node_connect_d(2,1)),y(node_connect_d(2,1)),'Global Node 2',...      'HorizontalAlignment', 'center') text(x(node_connect_d(2,2)),y(node_connect_d(2,2)),'Global Node 3',...      'HorizontalAlignment', 'center') text(x(node_connect_d(2,1))/2,y(node_connect_d(2,1))/2,'Global Element 1',...      'HorizontalAlignment', 'center') text( x(node_connect_d(2,1)) + (x(node_connect_d(2,2))-x(node_connect_d(2,1)))/2, y(node_connect_d(2,1)) + (y(node_connect_d(2,2))-y(node_connect_d(2,1)))/2 ,'Global Element 2',...     'HorizontalAlignment', 'center')

% put the title on the figure title('Deformed and Undeformed States of Two-Bar Truss System') % label x axis xlabel('x') % label y axis ylabel('y')

Contributing Team Members
Andrew McDonald - Eml4500.f08.bike.mcdonald 01:31, 5 October 2008 (UTC) Garrett Pataky - (Eml4500.f08.bike.pataky 00:55, 7 October 2008 (UTC)) Sam Bernal - Eml4500.f08.bike.bernal 15:11, 7 October 2008 (UTC) Bobby Sweeting - Eml4500.f08.bike.sweeting 22:51, 6 October 2008 (UTC) Shawn Gravois - Eml4500.f08.bike.gravois 00:26, 30 September 2008 (UTC) Eric Viale - Eml4500.f08.bike.viale 02:50, 7 October 2008 (UTC)

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


text:
University of Florida/Eml4500/f08.qwiki/Lecture 11

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.

Meeting 12

Recall from Page 6-1, $$k^{(e)}d^{(e)}=f^{(e)}$$ (Equation 1) Note to self; make sure these are 4x4, 4x1, 4x1

Note to self: insert diagrams (2) and the matrices for kq=P

$$q^{(e)}_{i}$$=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 of local node i, denoted by $$d^{(e)}_{i}$$: Note to self: make sure the i is enclosed by a square

Insert figure 12-3

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


text:
User:Eas4200c.f08.aero6.villalba/circle second moment of area

Derivation of second moment of area of a circle

Given a circle with $$r=a\,\!$$ as shown, we start from the definition of second moment area:

$$I_y=\iint\limits_A \,z^2 dy\,dz$$

which can also be written as,

$$I_y=\int\,z^2\,dA$$

For a circular cross section, a transformation to polar coordinates is needed. Therefore,

$$z=r\sin \theta\,\!$$

$$y=r\cos \theta\,\!$$

Next, we look at our differential area in polar coordinates. This is obtained from the area of a sector.

The area of the sector is $$A=\frac{1}{2} r^2 d\theta$$ and differentiating we obtain $$dA=r dr d\theta\,\!$$, where this differential area is shown below.

Thus, our integral becomes

$$I_y=\int\limits_0^{2\pi}\int\limits_0^a (r\sin \theta)^2 r dr d\theta$$

Next, we solve the integral

$$I_y=\int\limits_0^{2\pi}\int\limits_0^a r^3 \sin^2 \theta dr d\theta = \int\limits_0^{2\pi}\sin^2 \theta \left [ \int\limits_0^a r^3 dr \right ] d\theta = \int\limits_0^{2\pi}\sin^2 \theta \left [ \frac{r^4}{4} \right ]_0^a d\theta$$

$$= \frac {a^4}{4} \int\limits_0^{2\pi}\sin^2 \theta d\theta = \frac {a^4}{4} \int\limits_0^{2\pi} \frac{1-\cos (2\theta)}{2} d\theta = \frac {a^4}{4} \int\limits_0^{2\pi} \frac{1}{2} - \frac{1}{2} \cos(2\theta) d\theta$$

$$= \frac {a^4}{4} \left [ \frac{1}{2}\theta - \frac{1}{4} \sin(2\theta) \right ]_0^{2\pi} = \frac {a^4}{4} \left [ \frac{1}{2}(2\pi) - \frac{1}{4}\sin(4\pi) - \frac{1}{2}(0) + \frac{1}{4}\sin(0) \right ]$$

$$= \frac {a^4}{4} \left ( \pi - 0 - 0 + 0 \right ) = \frac{\pi a^4}{4}$$

Therefore, for any circular cross-section with a radius of $$r\,\!$$, the second moment of area about the y-axis, $$I_y\,\!$$ is equals to $$\frac{\pi r^4}{4}$$.

Note: Through the same means and a similar derivation, it can also be shown the same results would be obtained for $$I_z\,\!$$.


text:
User:Emesee/Ologies

-logy

A

 * Acarology, the study of ticks and mites
 * Actinobiology, the study of the effects of radiation| upon living organisms
 * Actinology, the study of the effect of light on chemicals
 * Aerobiology, a branch of biology that studies organic particles, such as bacteria, fungal spores, very small insects and pollen, which are passively transported by the air.
 * Aerology, the study of the atmosphere
 * Aetiology, the medical study of the causation of disease
 * Agnoiology, the study of things of which we are by nature ignorant, or of things which cannot be known
 * Agrobiology, the study of plant nutrition and growth in relation to soil conditions
 * Agrology
 * Canada: the art and science of agriculture
 * Non-Canada: the study of soils
 * Agrostology, the study of grasses
 * Algology, the study of algae
 * Allergology, the study of the causes and treatment of allergies|; a branch of medicine
 * Andrology, the study of male health and disease
 * Anesthesiology, the study of anesthesia and anesthetics; a branch of medicine
 * Angelology, the study of angels
 * Angiology, the study of the anatomy of blood and lymph vascular systems
 * Anthropology, the study of humans
 * Apiology, the study of bees (apiaries)
 * Arachnology, the study of spiders and their kind
 * Archaeology, the study of past cultures through the analysis of material remains
 * Archaeozoology, the study of relationships between humans and animals over time through examination of animal remains at archaeological sites (also see Zooarchaeology)
 * Areology, the study of Mars
 * Assyriology, the study of the Assyrians
 * Astacology, the study of crayfish
 * Asteroseismology, the study of the internal structures of stars as revealed by their oscillations (see also Helioseismology).
 * Astrobiology, the study of origin of life
 * Astrogeology, the study of geology of celestial bodies (e.g., planets, asteroids, comets)
 * Astrology, the study of the purported influence(s) of celestial bodies on earthly affairs
 * Astrometeorology: using astrology to forecast the weather
 * Atmology, the study of the laws and phenomena of aqueous vapors
 * Audiology, the study of hearing; a branch of medicine
 * Autecology|, the study of the ecology of any individual species
 * Autology, the attribute (of a word) of being self descriptive. The word "polysyllabic" is autological because it is polysyllabic. (compare heterology)
 * Auxology, the study of human growth.
 * Axiology, the study of the nature of values and value judgements

B

 * Bacteriology, the study of bacteria|
 * Balneology, the scientific study of baths, bathing and of their application to disease
 * Batology, the study of blackberries
 * Bibliology, the study of books, printing, and publishing; also called Bibliography
 * Bioclimatology, the study of the effects of climate on living organisms
 * Biogeomorphology, or ecogeomorphology, the study of interactions between organisms and geomorphological processes|
 * Bioecology, the study of the relationship of organisms to each other and to their environment|
 * Biology, the study of life
 * Biometeorology, the study of the effects of atmospheric conditions on living organisms
 * Biotechnology industrial use of living organisms or their components to improve human health and food reduction
 * Boxology is the study of schematic layouts, for example of corporate org chart|s.
 * Bryology, the study of bryophytes
 * Bryozoology|, the study of the phylum Bryozoa
 * Buddhology, the study of the nature of Buddha

C

 * Campanology, the study and the art of bell ringing
 * Cardiology, the study of the heart
 * Cariology, the study of cell|s
 * Cereology, the study of crop circles
 * Cetology, the study of cetaceans - whales, dolphins, and porpoise
 * Characterology, the study of character|
 * Christology, the theological| study of redemption (christos > christein: to sacrament), not only speaking of Christianity but also extended to religion in general
 * Chorology, the study of the relationship of biological or other phenomena to their locations
 * Chronology the study of things in order of time or the study of time
 * Climatology, the study of the climate
 * Codicology, the study of the preparation of books and manuscripts (ink, paper, vellum etc)
 * Coleopterology, the study of beetles
 * Coniology, the study of dust in the atmosphere and its effects on plants and animals, also spelled "koniology"
 * Conchology, the study of shells| and of molluscs
 * Contrology|, a neologism for the Pilates method, a system of physical exercise
 * Cosmology, the study of the cosmos| or our place in it
 * Cosmetology, the study of cosmetics and their use
 * Craniology, the study of the characteristics of the skull
 * Criminology, the scientific study of crime
 * Cryology, the study of very low temperatures and related phenomena.
 * Cryptology, the study of how to encrypt and decrypt secret messages
 * Cryptozoology, the study of animals that may or may not be mythical
 * Cynology, the study of dogs
 * Cytomorphology, the study of the structure of cells
 * Cytology, the study of cell|s

D

 * Deltiology, the study of, but more often the collecting of picture postcards
 * Demonology, the study of demons
 * Dendrochronology, the study of the age of trees and the records in their rings
 * Dendrology, the study of trees
 * Deontology, in ethics, the study of duty.
 * Dermatology, the field of medicine that deals with the skin
 * Dermatopathology, the field of dermatological| anatomical pathology
 * Desmology, the study of ligaments
 * Dialectology, the study of dialects
 * Dipterology, the study of flies

E

 * Ecclesiology, the study of Church architecture and decoration, or separately the study of the Christian Church
 * Ecogeomorphology, or biogeomorphology, the study of interactions between organisms and geomorphological processes|
 * Ecohydrology, the study of interactions among organisms and the water cycle.
 * Ecology, the study of the interrelationships between living organisms and their environment. Sometimes spelled "œcology".
 * Economic geology is concerned with earth materials that can be utilized for economic and/or industrial purposes.
 * Ecophysiology, the study of the interrelationship between an organism's physical functioning and its environment
 * Edaphology, a branch of soil science that studies the influence of soil on life.
 * Egyptology, the study of the ancient Egyptians
 * Electrophysiology, the study of the relationship between electric phenomena and bodily processes
 * Embryology, the study of embryos
 * Emetology, the study or knowledge of emetics
 * Endocrinology, the study of internal secretory glands
 * Enigmatology, the study of puzzles
 * Enology (or Oenology), the study of wine and winemaking
 * Entomology, the study of insects
 * Enzymology, the study of enzymes
 * Epidemiology, the study of the origin and spread of diseases
 * Epistemology, the study of the nature and origins of knowledge
 * Escapology, the practice of escaping from restraints| or other traps.
 * Eschatology, a branch of theology concerned with the final events in the history of the world or of mankind
 * Ethnology, the study of race
 * Ethnomusicology, the study of music in society, usually non-western music
 * Ethology, the study of animal behaviour
 * Etiology, same as Aetiology
 * Etymology, the study of word origins
 * Eulogy, the speech of praise
 * Evolutionary biology, the study of the process of biological evolution
 * Evolutionary psychology, the study of the process of (usually) human psychological evolution.
 * Exobiology, the study of life in outer space

F

 * Felinology, the study of cats
 * Fetology, the study of the fetus, especially when within the uterus. Sometimes spelled foetology.
 * Formicology, the study of ants
 * Fungology, fungus + -ology, see mycology.
 * Futurology, the study of the future

G

 * Garbology - study of refuse and trash
 * Gastrology or Gastroenterology - diseases of stomach and intestines
 * Gemmology| or Gemology, the study of gemstones and ornamental materials
 * Genealogy (commonly misspelled and mispronounced as "geneology"), the study of relationships within families particularly with a view to constructing family trees
 * Genecology, the study of genetic differences in relation to the environment
 * Geobiology, the study of the biosphere and its relations to the lithosphere and atmosphere.
 * Geochronology, the study of the age of the Earth
 * Geology, the study of the Earth
 * Geomorphology, the study of present-day landforms, traditionally on Earth but with increasing frequency on nearby planetary objects
 * Gerontology, the study of old age
 * Glaciology, the study of glaciers
 * Grammatology, the study of writing systems. Also the study of deconstructive literary criticism, popularized in the 1960s by Jacques Derrida
 * Graphology, the study of handwriting for the purpose of analysing the character of the writer
 * Grossology, the study of gross and disgusting things. Fictional; book title
 * Gynaecology or Gynecology, the study of medicine relating to women, or of women in general

H

 * Hematology|/Haematology, the study of blood
 * Heliology|, the study of the Sun
 * Helioseismology, the study of vibrations and oscillations in the Sun
 * Helminthology, the study of parasitic worms. See Vermeology.
 * Hepatology, the study of the liver; a branch of medicine
 * Herbology, the study of the therapeutic use of plants
 * Herpetology, the study of reptiles and amphibians
 * Heterology, the attribute (of a word) of not being self descriptive. The word "monosyllabic" is heterological because it is not monosyllabic. (compare autology)
 * Heteroptology, the study of true bugs
 * Hieroglyphology, Egyptian hieroglyphics, an ancient writing system
 * Hierographology, the study of sacred texts
 * Hierology, a more inclusive term for theology
 * Hippology, study of horses
 * Histology, the study of living tissues
 * Histopathology, the study of the (microscopic) structure of diseased tissue|s
 * Historiology, the study of the writings and practices of historians
 * Homology, concept in anthropology, biology, algebraic topology, and sociology, meaning "likeness in structure"
 * Horology, the study of making timepieces, measuring time and timekeeping (also the study of time)
 * Hydrogeology, the study of underground water
 * Hydrology, the study of water
 * Hypnology, the study of sleep (also: somnology)

I

 * Ichnology, the study of fossil footprints, tracks and burrows
 * Ichthyology, the study of fish
 * Ideology, the study of ideas throughout society both socially,politically,historically and literary
 * Immunology, the study of the immune system
 * Indology, the study of the history, languages, and cultures of the Indian subcontinent
 * Iranology, the study of history, literature, art and culture of Iran (Persia)
 * Islamology, the study of the Islam

J

 * Japanology, the study of Japanese people

K

 * Karyology, the study of karyotypes (a branch of cytology)
 * Killology, the study of human beings killing other human beings (Grossman|'s theory)
 * Kinesiology, the study of movement in relation to human anatomy; a branch of medicine
 * Kremlinology, the study of the Soviet Union
 * Kymatology, the study of waves or wave motions

L

 * Laryngology, the study of the larynx, or voice box; a branch of medicine
 * Lepidopterology, the study of butterflies and moths
 * Lexicology, the study of the signification and application of words
 * Limnology, the study of fresh water environments, particularly lakes
 * Lithology, the study of rocks|
 * Ludology, the study of video games
 * Lymphology, the study of the lymph system and glands

M

 * Malacology, the study of mollusks
 * Mammalogy, the study of mammals
 * Mereology, a branch of logic focusing on the study of part-whole relationships|
 * Meteorology, the study of weather
 * Methodology, (properly) the study of methods
 * Metrology, the study of measurement
 * Microbiology, the study of microorganisms.
 * Micrology, the science of preparing and handling microscopic objects for study.
 * Mineralogy, the study of minerals
 * Mixology, the study or skill of preparing mixed drinks.
 * Molinology, the study of windmills, watermills and animal engines.
 * Monadology, a book by Leibniz on the study of his theory of monads.
 * Morphology, the study of forms that is used in several other fields (biology, linguistics, astronomy, etc.)
 * Museology, the study of museum management.
 * Musicology, the study of music
 * Mycology, the study of fungi|
 * Myology the scientific study of muscles
 * Myrmecology, the study of ants
 * Mythology, the study of myths

N

 * Nanotechnology, the study and design of machines at the molecular level
 * Nanotribology, the study of friction at very small (atomic) scale
 * Nematology, the study of nematodes
 * Neonatology, the study of diseases and the care of newborn infants; a branch of pediatrics/paediatrics
 * Nephology, the study of clouds
 * Nephrology, the study of the kidneys and their diseases, a branch of medicine
 * Neurology, the study of nerves
 * Neuropathology, the study of neural diseases
 * Neurophysiology, the study of the functions of the nervous system
 * Nosology, the study of disease classification
 * Numerology, the study of numbers (often in a non-mathematical sense)
 * Nutriology, the study of nutrition

O

 * Oceanology, the study of oceans
 * Odonatology, the study of dragonflies| and damselflies|
 * Odontology, the study of the structure, development, and abnormalities of the teeth
 * Oenology, the study of wine
 * Omnology, the study of everything
 * Oncology, the study of cancer
 * Oneirology, the study of dreams
 * Onomatology|, the study of proper names
 * Ontology, the study of existence
 * Oology, the study of egg|s
 * Ophthalmology, the study of the eyes
 * Organology, the study of musical instruments (not just organ|s); alternatively, the study of anatomical| organ|s
 * Ornithology, the study of birds
 * Orology, the study of mountains and their mapping
 * Orthopterology, the study of grasshoppers and crickets|
 * Osteology, the study of bones
 * Otolaryngology, the study of the ear and throat; a branch of medicine
 * Otology, the study of the structure, function, and pathology of the ear
 * Otorhinolaryngology, the same as otolaryngology

P

 * Paleoanthropology, the study of prehistoric people and human origins
 * Paleobiology, the study of prehistoric life
 * Paleobotany, the study of prehistoric metaphyte|s (i.e., multicellular plants)
 * Paleoclimatology, the study of prehistoric climates
 * Paleoecology, the study of prehistoric environments by analyzing fossils and rock| strata|
 * Paleontology, the study of fossils of ancient life
 * Paleophytology (see "paleobotany"), the study of ancient multicelled plants
 * Paleozoology, the study of prehistoric metazoans (i.e., multicellular animals)
 * Palynology, the study of pollen
 * Parapsychology, the study of paranormal or psychic phenomenon that defy conventional scientific explanations
 * Parasitology, the study of parasites
 * Pathology, the study of illness
 * Pedology, one of two main branches of soil science
 * Pedology
 * Penology, the study of prison management and criminal rehabilitation|.
 * Personology, a variant of physiognomy
 * Petrology, the study of rocks and the conditions by which they form
 * Pharmacology, the study of drugs|
 * Phenology, the study of periodic biological| phenomena such as flowering, migration, breeding, etc.
 * Phenomenology, the study and science of phenomena as distinct from the science of actual existence or being; also a movement founded by Husserl which studies conscious experience without its metaphysical concerns
 * Philology, the study of a language together with its literature
 * Phlebology, a branch of medicine that deals with the venous system|
 * Phonology, the study of vocal sounds
 * Phrenology, the derivation of a persons character traits, by studying the shape of their skull
 * Phycology, the study of algae
 * Physiology, the study of mechanical, physical, and biochemical functions of living organisms
 * Phytology, the study of plants; botany
 * Phytopathology ("Plant pathology"), the study of plant diseases
 * Phytosociology, the study of the ecology of plant communities
 * Piphilology, the mnemonic practice of writing sentences such that the numbers of letters in the words correspond to the consecutive digits of pi
 * Planetology, the study of planets and solar systems; more commonly called Planetary science
 * Planktology, the study of plankton
 * Pneumology, a synonym of Pulmonology
 * Pomology, the scientific study and cultivation of fruits
 * Posology, the study of drug dosage
 * Primatology, the study of primates
 * Proctology, the study of the hind regions (rectum, anus, colon| and pelvic floor) of the body
 * Psephology, the study, especially statistical, of elections.
 * Psychobiology, the study and psychology of organisms with regard to their functions and structures
 * Psychology, the study of mental processes in living creatures
 * Psychopathology, the study of the mental processes within mental illness or disorders
 * Psychopharmacology, the study of psychotropic or psychiatric drugs
 * Psychophysiology, the study of the physiological bases of psychological processes
 * Pulmonology, the specialty in medicine that deals with diseases of the lungs and the respiratory tract

R

 * Radiology, the study of rays, usually ionising radiation|
 * Redology, the study of the classical Chinese literary masterpiece Dream of the Red Chamber
 * Reflexology, originally the study of reflexes or of reflex responses; but see also non-study list
 * Rheology, the study of flow
 * Rheumatology, the study of rheumatic diseases, a branch of medicine
 * Rhinology, the study of the nose and its diseases
 * Ripperology, the study of Jack the Ripper and the crimes associated with him

S

 * Sarcology, a subsection of anatomy that studies the soft parts. It includes myology, angiology, neurology, and splanchnology
 * Scatology, the study of feces
 * Scientology, the applied religious philosophy created by American writer L. Ron Hubbard
 * Sedimentology, a branch of geology that studies sediments.
 * Seismology, the study of earthquakes
 * Selenology, the study of the moon
 * Semiology, the study of sign|s
 * Semitology, the study of Semitic cultures
 * Serology, the study of blood serum|
 * Serpentology, the study of snakes
 * Sexology, the study of sex
 * Sindonology, the study of the Shroud of Turin
 * Sinology, the study of China
 * Sitiology, dietetics, the study of diet. Also, a treatise on diet. Also spelled "sitology"
 * Sociology, the study of society
 * Sociobiology, the study of the effect of evolution on ethology
 * Somnology, the study of sleep, also: hypnology
 * Somatology, study of human characteristics
 * Soteriology the study of the doctrine of salvation, especially as related to Christianity
 * Sovietology, the study of communist Soviet Union
 * Speleology, the study or exploration of caves
 * Sporalogy, a parody of astrology
 * Stemmatology, a discipline that attempts to reconstruct the transmission of a text (especially a text in manuscript form) on the basis of relations between the various surviving manuscripts
 * Stomatology, study of the mouth and its diseases.
 * Sumerology, the study of the Sumerians.
 * Symbology, the study and interpretation of symbols, or a set of symbols. Also used for types of barcodes
 * Symptomatology, the study of symptoms
 * Synecology|, the study of the ecological interrelationships among communities of organisms
 * Syphilology, the study of syphilis

T

 * Taxology, a synonym for taxonomy
 * Technology, the study of the practical arts (but see above)
 * Teleology, the study of ends or final causes
 * Teratology, the study of wonders, or monsters
 * Terminology, the usage and study of terms.
 * Thanatology, the study of physical, psychological and social problems associated with dying.
 * Thermology, the study of heat.
 * Theology, the study in religion
 * Tibetology, the study of Tibet
 * Tocology, the study of childbirth, sometimes spelled "tokology".
 * Tonology, the study of intonation in speech
 * Topology, the mathematical study of closeness and connectedness
 * Toxicology, the study of poisons
 * Traumatology, the study of wounds and injuries| caused by accidents or assaults and their surgical treatment and repair; a branch of medicine
 * Tribology, the study of friction and lubrication
 * Trichology, the study of hair and scalp
 * Typology, the study of classification

U

 * Urology, the study and treatment of diseases of the urogenital tract, a branch of medicine. Sometimes spelled "ourology".
 * Ufology, the study of Unidentified flying object (UFO) phenomena.

V

 * Vaccinology, the study of vaccines
 * Venereology, the study of venereal diseases
 * Vexillology, the study of flags
 * Victimology, the study of victims of crime, often applied to characterizing the criminal
 * Virology, the study of viruses
 * Volcanology (also spelled vulcanology), the study of volcanoes and related phenomena (traps)

X

 * Xenobiology, the study of non-terrestrial life
 * Xylology, the study of wood

Z

 * Zooarchaeology, the study and analysis of animal remains at archaeological sites to reconstruct relationships between people, animals, and their environment (also see Archaeozoology)
 * Zoology, the study of animals
 * Zoopathology ("Animal pathology"), the study of animal diseases
 * Zymology, the study of fermentation

Ologies that are not fields of study
Words ending in -ology that are not fields of study, and thus not "ologies" in the sense of this article, are:
 * Anthology, a collection of literary pieces (such as poems).
 * Apology
 * a statement of regret.
 * an explanation for or justification of beliefs.
 * Arcology - enormous habitat (hyperstructure) of extremely high human population density.
 * Chronology is the arrangement or setting out of past events in order of occurrence; the recording of historical events in date sequence.
 * Cosmetology, the art and career of using cosmetics to improve beauty.
 * Dilogy is
 * ambiguous or equivocal speech, or
 * a work composed in two parts (see trilogy)
 * Docimology, a treatise on the art of testing, e.g. in assaying metals.
 * Doxology, a spoken or sung end of a prayer.
 * Eulogy, though not an -ology, is a commemoration of a person's life at his/her funeral.
 * Hagiology is literature dealing with the life of a saints|saint or, indeed, any revered person, a biography of an individual, rather than a study of saints, sainthood or saintliness in general.
 * Heterology, a dissimilarity of parts often attributable to a difference in origin.
 * Homology (biology)|Homology, a similarity often atributable to common origin
 * Ideology, sometimes spelled idealogy, is a system of ideas about politics, human life or culture.
 * Kibology, joke religion worshiping Kibo.
 * Menology, a register of months, or a calendar of the lives of the saints for each day of the year.
 * Necrology, a list of people who have died, especially in the recent past or during a specific period.
 * Philology, the historical study of languages. This is not a ology in the strict sense, because it is not the study (-ologia) of love (philo-), but the love (philo-) of literature (logia).
 * Phraseology is the way words are put together, therefore the style guide|style being used in a Sentence (linguistics)|sentence, or the set of phrases or the choice of words used by any particular group of people, or a type of register (linguistics)|register that reflects the form of language used in a certain social situation in which particular subjects are being discussed.
 * Piphilology seems to be a borderline case with some aspects of a field of study, but not a scientific discipline.
 * Reflexology, alternative method of massage, manipulative therapy|therapy or pressure on certain points of the sole of the feet as a means of relieving nervous tension.
 * Scientology, the belief system/cult religion founded by L. Ron Hubbard, self-described as a study of knowledge.
 * Tautology (logic)|Tautology, a self-affirming truth.
 * Terminology, a set of words and/or phrases, usually in relation to some particular canon or field of study e.g. 'mathematical terminology'.
 * Tetralogy of Fallot, an abnormality of the heart consisting of four deformities that often occur together.
 * Trilogy (although not strictly an -ology) is a body of writing in three parts, as tetralogy is that in four parts. Other words such as pentalogy, hexalogy, and heptalogy or septology cover larger series.
 * Tropology, the use of tropes in speech or writing.
 * A Zumology is a treatise on the fermentation of liquors.

title: User:Emesee/Ologies


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title: Template:Smwplink


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Dissertations

External dissertations

 * Computational Asteroseismology

title: Dissertations


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User:Eml4500.f08.ramrod.c/HW

=Trusses, Beams, and Frames=

Diagram of Truss
Below is a diagram of a truss with 2 elastic, or deformable, bars. There is a known load P applied where the two bars join. The two bases of the truss are both fixed, or constrained, in both the x and y axes. There will be no displacement at these points. The goal is to find all of the unknown forces and displacements.

Global Free Body Diagram
Below is the Global (the whole structure) Free Body Diagram of the same 2-bar truss. Notice that the constraints have been replaced with reaction forces.

Currently, there are 4 unknown reaction forces (R1x, R1y, R3x, and R3y). However, there are only three equations that can be used (ΣFx=0, ΣFy=0, and ΣM=0). In other words, we have 4 unknowns but only 3 equations to solve for them, therefore this system is statically indeterminant.

Free Body Diagrams of the 2 bar elements
To continue, we will separate the two bars and look at them independently. A new local free body diagram is drawn for each bar element. Forces and the resulting displacements are again drawn and the nodes are renumbered accordingly.

Bar Element 1 Bar Element 2

Notes
 * fdegree of freedomundefined

 * fi(e) = ith eternal force of element e with:

 * i=1,2,3,4

 * e=1,2

 * for this example

There are two possible cases in this situation:

 * Case 1: The observer sits on node 1

 * In this case:

 * f2 = k(d2 -d1)

 * Case 2: The observer sits on node 2 (or equivalently f1+f2=0)

 * In this case

 * f1= -f2= -k(d2 -d1)

 * f1=k(d1 -d2)

Alternate Node Numbering Method
Below is another way to number the local nodes.

Force Displacement Relation
An elastic bar can be considered a spring with a spring constant of EI/L (E is Young's Modulus, I is the moment of inertia, and L is the length of the bar or spring). Recall the two following spring systems:

 * Below is a 1-dimensional spring with two free ends. A force f is applied and a displacement d results. The force-displacement relation is f=kd.

 * Another system is a 1-D (one dimensional) spring with 2 free ends. A force f is applied and a displacement d results. The matrix method (on the right) is used to find the force-displacement relationship. (The 2x2 matrix with the k's in it is known as the stiffness matrix).

title: User:Eml4500.f08.ramrod.c/HW


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User:Eml4500.f08.ramrod.c/9-12-08

Lecture Notes for 9/12/08
In compact notation:

 * [kij] 6x6 {dj} 6x1 = {Fi} 6x1
 * (more generally, the matrices are nxn)

$$\sum_{j=1}^{6}{ K_{ij}d_j }\! = F_i$$,    i = 1, 2, 3, ...., 6

The three matrices used are shown below:

 * K nxn = [Kij]nxn = global stiffness matrix

 * d nxn = {dj}nxn = global displacement matrix

 * F nxn = {Fi}nxn = global force matrix

Recall the element force-displacement relation
 * k (e)4x4 d (e)4x1= f (e)4x1

 * k n4x4 = [kij(e)] = element stiffness matrix

 * d nxn = {dj(e)} = element displacement matrix

 * F nxn = {fi(e)} = elementforce matrix

title: User:Eml4500.f08.ramrod.c/9-12-08


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User:Eml4500.f08.qwiki/Lecture 7

 WARNING: Please move your work to the user namespace, and request a deletion of all of your wiki page under the main namespace by inserting the delete or delete my page template at the top of the page so wikiversity custodian could delete them. Remember to log in when you perform these operations so you are not mistaken to be vandals. Failure to do the above immediately will result in a lower HW grade. Eml4500.f08 09:28, 25 September 2008 (UTC)

To continue with modeling the 2-bar truss system, element 1 is further examined to find the stiffness matrix k (1). Element 1 is located at an angle $$\theta ^{(1)}={\color{red}30^{\circ}}$$

Therefore inputting this angle into the following equations of the director cosines yields:

$$l^{(1)}=cos\theta ^{(1)}=cos30^{\circ} =\frac{\sqrt{3}}{2}=0.866$$

$$m^{(1)}=sin\theta ^{(1)}=sin30^{\circ} =\frac{1}{2}=0.50$$

Next, to find the stiffness coefficient the below equation is used.

$$k^{(1)}=\frac{E^{(1)} A^{(1)}}{L^{(1)}}=\frac{3*1}{4}=0.75$$

Element Stiffness Matrix:

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

Knowing the above element stiffness matrix, is the result of:

$$\mathbf{k^{(e)}}=k^{(e)}*\begin{bmatrix} (l^{(e)})^{2} & l^{(e)}m^{(e)} & -(l^{(e)})^{2} & -l^{(e)}m^{(e)}\\ l^{(e)}m^{(e)} & (m^{(e)})^{2} & -l^{(e)}m^{(e)} & -(m^{(e)})^{2}\\ -(l^{(e)})^{2} & -l^{(e)}m^{(e)} & (l^{(e)})^{2} & l^{(e)}m^{(e)}\\ -l^{(e)}m^{(e)} & -(m^{(e)})^{2} & l^{(e)}m^{(e)} & (m^{(e)})^{2} \end{bmatrix}$$

Where (e) is the element number. An example of finding out what each element of the stiffness matrix is shown below.

$$\mathbf{k_{11}}=k^{(1)}*(l^{(1)})^{2}=\frac{3}{4}*(\frac{\sqrt{3}}{2})^{2}=\frac{9}{16}=0.5625 $$

Doing this for all rows and columns yields an element 1 stiffness matrix of:

$$\mathbf{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}

=\begin{bmatrix} 0.563 & 0.325 & -0.563 & -0.325\\ 0.325 & 0.188 & -0.325 & -0.188\\ -0.563 & -0.325 & 0.563 & 0.325\\ -0.325 & -0.188 & 0.325 & 0.188 \end{bmatrix}$$

Some observations made from seeing the computed element stiffness matrix are as follows,

 * 1) Only 3 numbers need to be computed, the other values only differ by a sign change.
 * 2) Matrix k(1) is symmetric. Thus

$$k_{ij}=k_{ji} \rightarrow   k_{13}=k_{31}  \rightarrow   -0.563=-0.563$$

Observation 2 also means that the transpose of k is the same as k.

title: User:Eml4500.f08.qwiki/Lecture 7


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User:Devourer09/Project documentation

Graphing program
I used Cairo and I may have also used Pango. I think the graph executable in the directory is the one that outputs the graph PNG image. You may want to check out gnuplot if you find my graphing program too troublesome to modify and use. I didn't realize that in the end I would just be using a shellscript to get this all working. The graph.h and graph.c files are where the graphing functions are defined I guess, and then main4.c is where the entry point is.

Usage of graph
My graph program has a lot of arguments, so I'll explain them here. They are all required and in this order.

 * width is the width of the graph. I forget why it's not the same as the image width but it is different.
 * height is basically the same as the width except for height.
 * xAxisBegin is what value the X axis will start at.
 * xAxisEnd is the value at which the X axis ends.
 * xAxisScale is, I think, what value the X axis will count by starting from xAxisBegin until it reaches xAxisEnd.
 * yAxisBegin same as the X axis version.
 * yAxisEnd same as the X axis version.
 * yAxisScale same as the X axis version.
 * graphTitle is the title of the whole graph.
 * xAxisLabel is the label of the X axis, so you can name it.
 * yAxisLabel is the label of the Y axis, so you can name it.
 * dataFileName is the text file with the values to be plotted.
 * imageWidth is the width of the actual PNG image.
 * imageHeight is the height of the image.
 * pngName is the name of the PNG image you want to be generated.

Important Links

 * Cairo's website has the documentation and such.
 * Cairo API reference
 * Good Cairo tutorial
 * Make a PNG with Cairo
 * How to compile the program
 * Pango API manual

USB driver
So the create.sh bash script actually builds the module from the source file test.c, I believe, into test.ko and then unloads the previous test module in the kernel that hopefully exists, and then will load the new module, and output will be displayed to the console for feedback in case of errors and such. Makefile is the makefile that actually builds the module using KBuild and then the rest of create.sh is just unloading and loading modules. Of course test.c is derived from usb-skeleton.c and edited to our liking. When you write to the device, /dev/spoon, it will make the URB and send it on its way. A read will of course read from the device.

Data app
This program is the foo executable, and it basically talks to the driver by reading and writing to it. I got it to a point where I could actually type in the hex commands on the command line and it would actually send that command and return the data. I was working on getting it so I could put the name of the text file I wanted the data to be stored in, and I could send a command and it would store the data in that file. Once the data was in the file, the graphing program could read the data and create the graphs.

Usage
This usage works.

But I was going to have it look something like this:

And it would end up creating the file if it didn't exist or add to the already existing one.

title: User:Devourer09/Project documentation


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User talk:Bodymap0

 Hello Bodymap0, and welcome to Wikiversity! If you need help, feel free to visit my talk page, or contact us and ask questions. After you leave a comment on a talk page, remember to sign and date; it helps everyone follow the threads of the discussion. The signature icon in the edit window makes it simple. To get started, you may

 * Take a guided tour and learn to edit.
 * Visit a (kind of) random random project.
 * Browse everything on Wikiversity, or visit a portal corresponding to your educational level: pre-school, primary, secondary, tertiary, non-formal education.
 * Find out about research activities on Wikiversity.

 * Read an introduction for teachers and find out how to write an educational resource for Wikiversity.
 * Give feedback about your initial observations
 * Discuss Wikiversity issues or ask questions at the colloquium.
 * Chat with other Wikiversitans on #wikiversity-en.

And don't forget to explore Wikiversity with the links to your left. Be bold to contribute and to experiment with the sandbox or your userpage, and see you around Wikiversity! Jtneill - Talk - c 22:36, 22 September 2008 (UTC)

 Woof! My name is Jack(Russell). I am a dog and a Wikiversity mascot. I am pretty new around Wikiversity. Perhaps we can learn together! Tail wag...

title: User talk:Bodymap0


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User talk:Matty (temp)

 Hello Matty (temp), and welcome to Wikiversity! If you need help, feel free to visit my talk page, or contact us and ask questions. After you leave a comment on a talk page, remember to sign and date; it helps everyone follow the threads of the discussion. The signature icon in the edit window makes it simple. To get started, you may

 * Take a guided tour and learn to edit.
 * Visit a (kind of) random random project.
 * Browse everything on Wikiversity, or visit a portal corresponding to your educational level: pre-school, primary, secondary, tertiary, non-formal education.
 * Find out about research activities on Wikiversity.

 * Read an introduction for teachers and find out how to write an educational resource for Wikiversity.
 * Give feedback about your initial observations
 * Discuss Wikiversity issues or ask questions at the colloquium.
 * Chat with other Wikiversitans on #wikiversity-en.

And don't forget to explore Wikiversity with the links to your left. Be bold to contribute and to experiment with the sandbox or your userpage, and see you around Wikiversity! --mikeu talk 23:25, 22 September 2008 (UTC)

I was your note about usernames. Please see Changing username. --mikeu talk 23:25, 22 September 2008 (UTC)

title: User talk:Matty (temp)


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Nutrition facts labels

Nutrition Facts Labels

“The Federal Food Drug and Cosmetic Act and its amendments are responsible for the requirement of food labels. Food labeling is required for most prepared foods, such as breads, cereals, canned and frozen foods, snacks, desserts, drinks, etc. Nutrition labeling for raw produce (fruits and vegetables) and fish is voluntary.” (US Food and Drug Association, August 2008)

There are eight major sections to note when reading a nutrition label: serving size; total calories and calories from fat; percent daily values; nutrients to limit, such as total fat, sodium, and cholesterol; nutrients to get plenty of, such as vitamins, minerals, and fiber; additional nutrients; the ingredients list; and the footnotes. The daily values on nutritional labels use Recommended Daily Intakes. Optimum Daily Intakes are an updated finding on nutritional needs.

Labels are to be read with discretion. One of the first things to look at on a nutrition facts label is the number of servings. This is important because the rest of the label’s information will be based off of one serving size, which may or may not be what the consumer believes is one serving. If one were to eat more or less than the listed serving size, the total amount of nutrients would also be more or less.

Total calories and calories from fat tell how much energy the food contains. Taking note of the calories is important and helpful when managing weight gain, loss, or maintaining current weight.

Percent daily values are also to be considered. These values are given based on a 2,000 calorie/day diet. Some individuals require a higher calorie intake and some require less, therefore it is important to adapt certain diets to adhere to specific needs. Individuals are different so it key to use the percent daily values as a starting block or more general estimate.

Nutrients to limit are listed right below the total calories and calories from fat. They typically consist of fat (including Trans fats and saturated fats), cholesterol, and sodium. These nutrients tend to increase risk of heart disease, high blood pressure, and cancer. It is recommended to aim for low intakes of these nutrients.

Some nutrients that should be focused on include fiber, Vitamins A and C, calcium, and iron. Higher percentages of these in food products are usually better for most diets due to lack of proper nutrient intake.

Additional nutrients include those such as carbohydrates and sugars. Healthy sources of these nutrients include whole grains, fruits, and vegetables. Starchy foods, such as white breads and enriched carbohydrates, should be limited to achieve adequate nourishment.

Ingredients lists show what products are in the food. Items are listed in descending order by weight. The list also indicates any allergens present in the food product.

Footnotes remind individuals that all percentages are based on diets of 2,000 calories per day, so it is important to keep in mind personal daily caloric needs may be more or less than what is listed.

References: American Dietetics Association; Mayo Clinic; US Food and Drug Association; Department of Health and Human Services

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User:Mu301/Learning blog/2008-09

September 2008

 * At the bottom of every page there is text that reads: "All text is available under the terms of the GNU Free Documentation License (see Copyrights for details)." The page that gives details, however, is a proposed policy.  We might want to change the link to a WMF copyright page until we make our own policy official.  It doesn't look good to link to a copyright page that basically says "maybe this is our policy."  --mikeu talk 00:04, 23 September 2008 (UTC)

 * Thought of the day: consensus and the zeroth law of robotics --mikeu talk 13:48, 24 September 2008 (UTC)

 * Thought for another day... --mikeu talk 21:19, 25 September 2008 (UTC)

 * Philosophical Question #3: Is the individual greater than the community?
 * That depends on the nature of the community. In a combat unit fighting for material gains, it is customary to sacrifice a chess piece for the sake of winning the game. In a spiritual community, saving the individual is an overarching goal. In an educational community, the dilemma is comparable to the one faced by Mrs. Zajac in Tracy Kidder's book, Among Schoolchildren. In that story, the teacher spent an enormous amount of time and energy trying to save two problem children (Clarence and Robert). In the end, Mrs. Zajac failed to reach them, and so everyone lost, including the other children in the class who were neglected while Mrs. Zajac spent so much time on a futile effort to reach Clarence and Robert.

 * Reminder to self: check the links in Special:Statistics and find the MediaWiki: page to edit it. --mikeu talk 23:39, 26 September 2008 (UTC)

 * Handy wikimedia tools:  --mikeu talk 17:11, 27 September 2008 (UTC)

 * User:Sj's thoughts. --mikeu talk 16:06, 29 September 2008 (UTC)

 * Reminder to check up on students who appear to be creating a project at wikipedia. --mikeu talk 17:36, 29 September 2008 (UTC)

 * The Association of American Colleges and Universities will be holding a conference in Providence called Engaging Science, Advancing Learning: General Education, Majors, and the New Global Century.  --mikeu talk 21:17, 29 September 2008 (UTC)

title: User:Mu301/Learning blog/2008-09


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