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User:Eml4500.f08.RAMROD.A/MyContributionHW6

Ethics, Honesty and Imagination
It is always important to take a moment to remember the importance of being creative and imaginative. However, we must focus on our own creativity and not plagiarize the works of others. We cannot grow if we simply copy the imagination of others.

Rube Goldberg Device
Here is an impressive Rube Goldberg Device to inspire the creative mind in all. Rube Goldberg devices are impressive, complicated machines designed to do simple tasks - a show of human imagination.

Conversion of PVW to Discrete PVW
a look at Ni(x) and Ni+1(x)

Ni(x) and Ni+1(x) are linear functions and thus any linear conversion of Ni(x) to Ni+1(x) is also linear.

The following expressions illustrate the conversion:

$$ N_i(x)=\alpha_i+\beta_{i}x $$

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

we will use di to to create the following expression:

$$ N_id_i+N_{i+1}d_{i+1}=(\alpha_i+\beta_ix)d_i+(\alpha_{i+1}+\beta_{i+1}x)d_{i+1}

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

The above is clearly a linear function in x

Images of Ni and Ni+1
images of Ni(x) and right: Ni+1(x)


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User:Eml4500.f08.team.foskey.ckf/hw6b

MATLAB Assignment The Two Bar Truss With Variable Material Properties

Two Bar Truss Problem Overview
In all of the previous MATLAB Assignments, the modulus of elasticity, and the cross sectional area of an element in a truss system have been constant. In reality however, this may not always be the case. The modulus of elasticity and the cross sectional area may vary with length in an element.

In this problem, a truss system is composed of two elements. In the elements, the modulus of elasticity and cross sectional area vary linearly in the element.

This problem uses much of the same code as the original two bar truss problem.

The following data was given for the problem:

for element 1 e(1)1=10      modulus of elasticity at end 1 e(1)2=20      modulus of elasticity at end 2 A(1)1=15      cross section area at end 1 A(1)2=25      cross section area at end 2 L(1) = 4      length of element 1 for element 2 e(2)1=1       modulus of elasticity at end 1 e(2)2=2       modulus of elasticity at end 2 A(2)1=14      cross section area at end 1 A(2)2=25      cross section area at end 2 L(2) = 2      length of element 2 P = 7N    (Applied load)

MATLAB Code
The following MATLAB code is used to solve this problem:

Additional Functions
The above code uses three separate MATLAB mfiles as functions. These are PlaneTrussElement, PlaneTrussResults, and NodalSoln.

A complete description of these functions can be seen here

Comments On The Code
This problem is nearly identical to the original two bar truss problem, with the exception of the Young's moduli and cross sectional areas. Each of those properties vary along each bar. The values of those properties were given for each end of each bar. The properties were assumed to vary linearly across the length of the bars.

To be able to use the property values, and average value was found. This was accomplished by adding the properties at each end and then dividing by the length of the bar. For example: the average cross sectional area for element 1 was found by

$$ \frac{(A^{(1)}_1 +A^{(1)}_2)}{L^{(1)}}$$

where $$A^{(1)}_1$$ and $$A^{(1)}_2$$ are the cross sectional areas of element 1 at local nodes 1 and 2 respectively, and $$L^{(1)}$$ is the length of element 1.

These averages were saved and used in the same manner(but in place of) the constant properties in the original problem.

To plot the truss system, two matrices of the nodal locations were created, one for the undeformed locations, and the second for the displaced node locations. Then lines were drawn to connect the nodes for the undeformed and deformed cases.

To compare the casde of a truss with elements with variable moduli of elasticity and cross sectional areas, the original two bar truss code was added to the end of the code. The original truss code overrides the existing data, and uses the same plotting code as used earlier to plot the deformed truss with constant properties.

A legend is then created to distinguish the seperate truss systems.

Results Output
Running the above MATLAB code generates the following output:

Comments On The Results Output
The MATLAB output for this program gives the local and global element stiffness matrices (k_local and k) for elements 1 and 2, and the global stiffness matrix for the entire system (K). The output then shows the applied loads R, which was an upward force of 7 N applied at global node 2. The displacement matrix d shows the distances that the nodes are displaced under the load. In this problem, nodes 1 and 3 are constrained, so their displacements are zero in both the x and y directions. The non-zero displacements are those of node 2. Looking at the figure and the data output, it is obvious that node 2 moves upward and to the left. The reactions matrix gives the horizontal and vertical reaction forces at the constrained points. The results matrix shows the strain, stress and axial forces of each element in columns in that order, with the values for element 1 being in row 1 and the values for element 2 in row 2.

Note : all of the output for the original truss problem (with constant properties) has been suppressed. To see the results, click here

Figure
Running the above MATLAB code generates the following figure.

In the figure, the red dotted lines represent the undeformed truss, the blue solid lines show the deformed truss with the variable material properties, and the green solid lines represent the deformed truss with the constant material properties.

The node and element identifiers were created manually using Paint after running the MATLAB code.

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

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User:Eas4200c.f08.WIKI.E/hw6

=Mtg 34 Friday, November 14 =

Unsymmetrical thin walled cross sections
There will be a comparison drawn between cross sections with symmetry and cross sections of solids without symmetrical cross sections.

This is the generalized formula for a unsymmetrical cross section. (Equation 5.1)

$$ \int_{A_i}^{}{\frac{d\sigma _{xx}}{dx} dA}=-q_A$$

Case 1: Symmetrical about the Y-axis
The Stress factor is given under symmetrical cross sectional conditions as follows:

$$\sigma _{xx}=\frac{M_yZ}{I_y} $$

The shear flow factor $$q\;$$ is:

$$q(s)=\frac{-V_zQ_y}{I_y}$$

$$Q_y=\int_{A_S}^{}{ZdA}=Z_cA_s$$

Case 2:Unsymmetrical about the Y-axis
The Stress factor is given under unsymmetrical cross sectional conditions as follows:

$$\sigma _{xx}=(K_yM_z-K_{yz}M_y)y+(K_yM_y-K_{yz}M_z)z \;$$

$$K_y=\frac{I_y}{D}$$

$$K_{yz}=\frac{I_yz}{D}$$

$$K_z=\frac{I_Z}{D}$$

The stress in the x-direction can be solved for in matrix format.

$$\sigma _{xx}=\begin{vmatrix} y & z \end{vmatrix}_{1x2}\begin{vmatrix} K_y & -K_{yz}\\ -K_{yz} & K_z \end{vmatrix}\begin{Bmatrix} M_z\\M_y

\end{Bmatrix}\Leftrightarrow \sigma _{xx}=\begin{vmatrix} y & z \end{vmatrix}_{1x2}\begin{vmatrix} K_yM_Z & -K_{yz}M_y\\ -K_{yz}M_z & K_zM_y \end{vmatrix}$$

$$\sigma _{xx}=\begin{vmatrix} z & y \end{vmatrix}_{1x2}\begin{vmatrix} K_y & -K_{yz}\\ -K_{yz} & K_z \end{vmatrix}\begin{Bmatrix} M_y\\M_z

\end{Bmatrix}$$ It is more convenient for convention issues to use this version of the formula for $$\sigma _{xx}$$

Setting the moment about the z-axis to zero we are particularizing the to cross section of a symmetrical case

$$M_z=0\;$$

$$I_{yz}=0\Rightarrow D=I_yI_z$$

D being previously stated, $$K_y$$ and $$K_z$$ are as follows: $$K_y=\frac{1}{I_z}$$

$$K_y=0\;$$

$$K_z=\frac{1}{I_y}$$

The Shear flow for the particularize case is:

$$q(s)=-(K_yV_y-K_{yz}V_z)Q_z-(K_zV_z-K_{zy}V_y)Q_y\;$$

 $$Q_Z=\int_{A_s}^{}{ydA}\; \; \; \; \;

Q_y=\int_{A_s}^{}{zdA} $$

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Saint Catherine Institute for Orthodox Christian Studies

The Saint Catherine Institute for Orthodox Christian Studies is a pan-Orthodox educational organization in the Chicago area.

Goals
It aims to provide a forum for the study and research of Orthodox thought and practice as well as to promote Orthodox scholarship in fields of academic interest and with respect to issues of our age.

Through its work, this pan-Orthodox Institute aims to be a witness to Orthodox Christianity by advancing appreciation and awareness of Orthodox scholarship.

The Institute is working to realize these goals through a variety of programs, which include:
 * regular meetings with academic and research presentations for Institute members;
 * an Internet forum for communication among members;
 * public lectures, workshops, and symposia in the greater Chicago area; and
 * classes that complement existing parish adult education programs.

In the future, the Institute hopes to sponsor a biennial scholarly conference and establish an online journal.

Membership
The Institute is an independent organization whose membership is open to all Orthodox Christians with an advanced academic degree or with advanced training and background so as to actively contribute to the programs and goals of the Institute. The Institute also welcomes the participation of non-Orthodox scholars who can so contribute to the Institute.

The Institute serves, conducts activity within, and draws its members from the greater Chicago Metropolitan Area, Illinois, southern Wisconsin, and northern Indiana. There are three classes of membership within the Institute: Regular, Student, and Affiliate.

Our website
On our website, you will find a list of Forthcoming Orthodox events in the Chicago area, a list of all local Orthodox parishes, a speakers bureau listing our members available to give conferences, retreats, workshops, etc, on a variety of Orthodox topics, the last meeting minutes, and much more. Please visit us at www.saintcatherineinstitute.org

External link

 * Official site


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