User:Eml4500.f08.ateam.didomenico/hw 7

MATLAB: Two-Bar Truss System - Varying Cross-sectional Areas
The set-up and data for this two-bar truss system are seen below.



In this problem, two different methods of assembling the element stiffness matrices were explored. In one method, the varying cross-sectional areas are accounted for using the following relation general stiffness matrix formula.

$$\hat{k}^{(i)}_{gen}=\frac{1}{6\cdot L}\left(2E_{1}A_{1}+(E_{1}A_{2} + E_{2}A_1)+2E_2A_2\right)\begin{bmatrix} 1 & -1\\ -1 & 1 \end{bmatrix} $$

Additionally, the areas were averaged and assembled into the average stiffness matrix. This relation is as follows.

$$\hat{k}^{(i)}_{avg}=\frac{(E_1+E_2)\cdot (A_1 +A_2)}{4L}\begin{bmatrix} 1 & -1\\ -1 & 1 \end{bmatrix}$$

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

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

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

Plot of Deformed Shape and Deformed Shape of Two-Bar Truss System Using Average Areas
This plot depicts the deformation of the two-bar truss system using the average stiffness matrix in red and the deformation of the system using the general stiffness matrix in green. The difference is clearly shown in the plot and the table below.



MATLAB: Two-Bar Frame System
This two-bar frame system is very similar to the one analyzed in | Homework Report #2 and further in | Homework Report #5. It uses the same data as before but, for this case, Element 1 is changed to a frame element, creating a frame system. The cross-section for this system is assumed to be square. The applied load, P, has a value of 7.



This functions generates the element stiffness matrices an the element R matrix. It acts similarly to the  seen in previous problems.

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

Plot of Undeformed and Deformed Shapes
This plot depicts the undeformed and deformed shapes of the two-bar frame system. Additionally, the deformed shape of the two-bar truss system is plotted in green for comparison. This deformation is scaled by a factor of 0.5 for viewing purposes.

From the plot, the difference in frame elements and truss elements is apparent. The frame element was divided into 20 segments to show its complete deformation as a curve. The fixed boundary conditions cause an additional degree of freedom at each node, resulting in rotation or bending of the bar.



MATLAB: Electric Pylon Frame System
The structure in this problem is a electric-line pylon consisting of 91 elements. The MATLAB code for accurately assembling this system was obtained from Dr. Loc Vu-Quoc and can be found here or seen below.

The objective of this problem is to expand upon analysis performed in the previous homework report, | Homework Report #6. For this problem, the electric pylon truss system was converted to a frame system with each element having a square cross-section. The applied load P is applied vertically downward at the far right tip of the arm of the pylon. Clamped boundary conditions are applied at the base of the frame model, replacing the simply supported conditions of the previous problem. All elements are constructed of 300M steel and have identical cross-sectional areas. Properties for this material were given by Dr. Vu-Quoc and verified online here. They are displayed in the table below.



Analysis Requirements

 * Convert the electric pylon truss system to a frame system
 * Redo the static analysis using the same geometric, material and loading data as for Homework Report #6
 * Compare the nodal displacements of the node under the applied force for the frame model to those of the truss model
 * Identify the frame element or elements with the highest nodal bending moment and transverse shear force
 * Plot the undeformed shape, deformed shape of the truss model and "incomplete" deformed shape of the frame model using the same magnification factor
 * Determine if the problem is statically determinant and provide justification through argument or MATLAB programming
 * Compare the reactions obtained from the truss model to those from the frame model
 * Construct the lumped mass matrix of the electric pylon
 * Solve the generalized eigenvalue problem
 * Find the lowest three eigenpairs and, in 3 separate plots, plot the eigenvectors as deformed shapes superposed onto the undeformed shape
 * Find the three lowest vibrational periods of the electric pylon
 * Generate an animation of the three lowest vibrational periods of the electric pylon

MATLAB Analysis Code
The code below is the  file written to solve the problem as assigned. This program requires additional MATLAB functions to run. These functions are also detailed below.

This functions generates the element stiffness matrices an the element R matrix. It acts similarly to the PlaneTrussElement.m seen in previous problems.

This function returns three matrices consisting of the Axial Force Matrix,, the Bending Moment Matrix,  , and the Shear Force Matrix,. Each of these matrices lists the X and Y coordinates and the respective value for each element.

This function returns the length of an element based on the element coordinates it receives. It was independently written and cannot be found with the preceding functions.

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

This function was written to output the reduced matrices, $$\bar{M}$$ and $$\bar{K}$$. It was independently written and cannot be found with the preceding functions.

Running Analysis Code
The code below is the complete results from running the MATLAB program. Each part is detailed in later sections of this homework report.

Plot of Deformed and Undeformed Shapes
The extended MATLAB code produces the following plot of the deformation of the truss system. The frame system is detailed in red while the truss system is shown in green. Since the systems overlap, it is difficult to view. The red frame system is plotted on top of the green truss system. To scale the electric pylon truss to be 60m tall, the nodal positions were multiplied by a factor of 8.0753701211306. It is important to note that the deformation has been magnified by a factor of 20 for viewing purposes.



Statically Determinant?
The electric pylon structure is not statically determinate. While two nodes of the system are fixed, these two nodes act as the termination point of four elements. Thus, there are a total of four unknown reaction forces in a system with only three equations with which to attempt to solve them. This cannot be solved using the principles of statics.

It is important to note that, while similar in number of fixed nodes to previous systems analyzed, this system differs in that the fixed nodes contain more than one element.

Comparison of Reactions
The reactions for the truss system and the frame system are almost identical. They are displayed in the following table.

Comparison of Nodal Displacements of Truss and Frame Systems
The nodal displacements of the frame and truss systems vary by minuscule amounts. The following picture magnifies the displacements for viewing. To better understand the scale, the distance between the two nodes is approximately one ten-thousandth of a meter.



The exact displacements are presented in the table below. The values for the truss and frame systems are almost identical.

Highest Nodal Bending Moment
As shown on the plot and results from running, the greatest nodal bending moment occurs in Element 81. The value of the moment is 6.2874 N-m.

Highest Transverse Shear Force
As shown on the plot and results from running, the highest transverse shear force occurs in Element 81. The value of the force is -5.4827 N.

Lumped Mass Matrix
The lumped mass matrix was assembled as a diagonal matrix of the masses of both degrees of freedom for each node. To determine the masses, each elements length was multiplied by its cross-sectional area and its density. Once the global nodes of each end of the element were known, the half of the element's mass was assigned to both nodes. Continuing this for each element created the lumped mass matrix.

Eigenpair Analysis
The three smallest eigenvalues for the truss system were determined using the generalized eigenvalue problem shown below. In this relation $$\bar{K}$$ is the reduced stiffness matrix and $$\bar{M}$$ is the reduced lumped mass matrix.

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

By plotting the columns of the eigenvector matrix that correspond to the smallest eigenvalues, modes of the truss can be depicted. In this case, the smallest eigenvalues were 132.86, 2471.2 and 2942.6. Therefore, the columns of the eigenvector matrix, V, that correspond to these values are plotted similarly to plotting displacements. These values are simply added to the original nodal positions to give the mode shape.

The following code was appended to the original truss problem MATLAB file to plot the eigenvectors. It is important to note that the eigenvectors were multiplied by a factor of 4 for viewing purposes.

Smallest Eigenpair Plot
This plot represents the eigenvectors corresponding to the smallest eigenvalue.

Second Smallest Eigenpair Plot
This plot represents the eigenvectors corresponding to the second smallest eigenvalue.

Third Smallest Eigenpair Plot
This plot represents the eigenvectors corresponding to the third smallest eigenvalue.

Vibrational Period Analysis
To find the three lowest vibration periods of the electric pylon, the eigenvalues are set equal to the angular frequencies as shown in the equation below.

$$\lambda = \omega^2$$

The angular frequencies are converted to vibrational frequencies as follows.

$$f = \frac{\omega}{2\cdot \pi}$$

From the vibrational frequencies, the vibrational periods can be determined using the following equation.

$$T = \frac{1}{f} = \frac{2\cdot \pi}{\omega}$$

Following these relations reveals the three lowest vibrational periods, shown in the table below.

Eigenmode Analysis of the Electric Pylon Truss System
movies were created by adding the following MATLAB code segment to the code from Homework Report #6. These files were converted to  format for compatibility with Wikimedia.

Creating Videos for Upload to Wikimedia Commons
A more detailed explanation of how to generate and convert video files for upload to Wikimedia Commons can be found here or seen in the collapsible table below.

Animation of Lowest Eigenmode
This animation represents the lowest eigenmode of the electric pylon system.



Animation of Second Lowest Eigenmode
This animation represents the second lowest eigenmode of the electric pylon system.



Animation of Third Lowest Eigenmode
This animation represents the third lowest eigenmode of the electric pylon system.