# tgr12001/me3255_finalproject_group29

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Failed to load latest commit information. IMG_1275.png IMG_1276.png README.md eigen.png eigen_setup.png grade.md montecarlo.m plot_1a_1.png plot_1a_2.png plot_2_10seg.png plot_2_6seg.png plot_2_seg20.png plot_3_10seg.png plot_3_20seg.png plot_3_6seg.png plot_5_e_10seg.png plot_5_e_20seg.png plot_5_e_6seg.png plotter_1_a.m problem2_setup.png problem3_setup.png problem_2.m problem_3.m problem_4.m setdefaults.m shape_simple_support.m

# me3255_finalproject_group29

## Thomas Ressler, Charlotte Veitner

### ------------------------------ Problem 1 ------------------------------

#### Part A

##### Finding the Beam Deflection Equation (w=?)

The first steps in problem 1 include analytically solving for the deflection of the beam. The process is illustrated in the two photographs below.

First, the differential equation is integrated 4 times in a row in order to obtain the equation for beam deflection, w, using the conditions that omega and pressure are both equal to zero.

Once the beam deflection equation is derived, there are 4 unknown constants present. These constants are found through using the boundary conditions defined in the problem statement.

These constants can then be substituted back into the beam deflection equation.

##### Writing a Function to output Beam Deflection

The beam deflection equation is then pasted into the MATLAB function "w = shape_simple_support(x,q)". This function takes an input position, x, and distributed vertical load across the beam. The output of the function comes directly from the derived beam deflection equation.

##### Plotting Distributed Load vs. Maximum Deflection of the Beam

The script "plotter_1_a.m" is used to plot q against the maximum beam deflection. This script creates two graphs. The first graph is the beam deflection against the horizontal beam distance. This graph is solely used as a way to verify that the location of the maximum deflection is halfway across the beam. Viewing the plot, it is clear that this is the case. Now that it is known that the maximum deflection is 0.5*L, we are able to plot various values of q on the beam. We chose to start at 1 N/m and go up to 10000 N/m. The graph shows that the maximum deflection follows a linear slope, becoming larger as the distributed loading applied increases. #### Part B

Monte Carlo methods rely on repeated random sampling to obtain numerical results. In general, they help shed light on statistics to generate a useful solution.

This part of the project uses the Monte Carlo model to find the mean and standard deviation for the maximum deflection given that the width and height of the beam are each normally distributed random variables with 0.1% standard deviations at a distributed loading of 50 N/m.

The results are the following:

Mean of Maximum Deflection Standard Deviation of Maximum Deflection
0.0011 4.0919e-04

### ------------------------------ Problem 2 ------------------------------

This probelm requires to find the deflection of the beam by using the central difference method. The system of equation can be found by simplfying the main differential equation. P=0 and w=0 which allows to elliminate the second and third term of the differential equation. Leaving us with just the first term to determine the deflection. This process and the setup of the system of equation is shown in the photograph below. The process is repeated for 10 segments and 20 segments respectively, increasing the dimensions of the matrix. The construction of the system matrices can be seen in the file problem_2.

In order to solve for the deflections the inverse of the matrices is being multiplied by the vector representing the distributed load. Due to q=constant it can be taken as a constant and the vector is a constant vector with all of the entries being equal.

The deflection vector is being found for a total of 5 different applied loads. For every load the function returns a vector giving the deflection at the different segment points. Due to wanting to find the maximum deflection, only the maximum value of each vector is being taken into consideration. The final return of the function are 3 different graphs, for the three different number of segments. Each graph shows the applied load vs. maximum deflection meaning there are a total of 5 different points that are connected by a straight line.

The three different graphs are shown below: This is the deflection of the beam using 6 segments and 5 different values for the distributed load. This is the deflection of the beam using 10 segments and 5 different values for the distributed load. This is the deflection of the beam using 20 segments and 5 different values for the distributed load.

The graphs show that the maximum deflection yields the same value independet from the number of segments, which makes sense as the distributed load is equal and has the same effect on the beam for every chosen number of segments. The graph for the 20 segments is to be taken as the most accurate as the analysis takes more beam locations into consideration.

### ------------------------------ Problem 3 ------------------------------

In order to find the system of equations the main differential equation needs to be simpliefied. Due to the frequency being =0 and the distributed load being constant the third term of the differential is being eliminated and the right hand side is constant. As there is still an applied load the first term of the differential remains the same as in problem 2. To simplify the setup the matrices that were produced in problem 2 can be used again. In addition to that matrix another matrix that is using the force P and the second derivative of the deflection is being introduced. The setup of that matrix can be seen in the photograph below. Due to the matrixes having the same dimensions they can be subtracted from one another. In order to find the deflection vector the inverse of the matric is being multplied by the right hand side of the equation, just like in problem 2.

The function returns 4 values for every applied load and segment leading to a total of 20 vectors. They are being grouped under the number of segments and the deflection is ones again being plot against the distributed load. Only this time there are 4 lines representing the four different values of force P. There are a total of three different graphs for the three different number of segments. Each graph includes 4 lines based on the force P showing the maximum deflection for the different distributed loads connected with a straight line. This the deflection depending on different values for P using 6 segments and 5 different values for the distributed load. This the deflection depending on different values for P using 6 segments and 5 different values for the distributed load. This the deflection depending on different values for P using 6 segments and 5 different values for the distributed load.

### ------------------------------ Problem 4 ------------------------------

#### Parts A, B, and C

Before finding the system equations, the main differential equation needs to be simplified. This is done by inputting the specified conditions of P=0 and q=0. The process for deriving the system of equations into matrix form is detailed in the photograph below. A similar process is applied to the 10 segment and 20 segment formats, but is scaled up to incorporate the increased amount of segments. Details regarding the construction of the system matrices in MATLAB can be seen in the file "problem_4.m".

#### Part D

To solve for the natural frequencies of the beam for each segment length, the MATLAB function "eig" was used: This function takes an input A matrix and outputs the eigenvector matrix and corresponding eigenvalues. The eigenvalue is equal to the following:

``````Eigenvalue = (Density * Area * NatFrequency^2 * StepSize^4) / (E * I);
``````

Therefore, the natural frequency can be solved for through algebraic rearrangements.

A system with “n” degrees of freedom has “n” eigenvalues, and each eigenvalue corresponds to a natural frequency. Since the 6 segment configuration is a matrix of size 5x5, it will have 5 eigenvalues and 5 natural frequencies. Similarly, the 10 and 20 segment configurations will have 9 and 19 eigenvalues and natural frequencies, respectively.

The results for the Natural Frequencies are:

6 Segment:

Natural Frequency # Value [Hz]
1st 22.5658449501167
2nd 84.2168798695585
3rd 168.433759739117
4th 252.650639608675
5th 314.301674528117

10 Segment:

Natural Frequency # Value [Hz]
1st 22.8992638198219
2nd 89.3555157880681
3rd 192.863554934140
4th 323.291293203508
5th 467.871554830881
6th 612.451816458252
7th 742.879554727621
8th 846.387593873693
9th 912.843845841935

20 Segment:

Natural Frequency # Value [Hz]
1st 23.0411009132069
2nd 91.5970552793083
3rd 203.979787977640
4th 357.422063152273
5th 548.145622742683
6th 771.454219736559
7th 1021.84925535716
8th 1293.16517281403
9th 1578.72127377348
10th 1871.48621932352
11th 2164.25116487356
12th 2449.80726583301
13th 2721.12318328988
14th 2971.51821891048
15th 3194.82681590436
16th 3385.55037549477
17th 3538.99265066940
18th 3651.37538336774
19th 3719.93133773384

#### Part E

Corresponding to each eigenvalue and natural frequency is an eigenvector. These eigenvectors are plotted against horizontal beam distance for the first three natural frequencies at the three different segment counts. The results are shown in the three graphs below.

Looking at these graphs, an interesting trend appears. All three of them share the same general mode shapes, but the curves become more defined as the segment count increases. This is analogous to a low bit rate (digitally compressed) “.mp3” compared with a much higher bit rate (analog) “.wav” file. For example, listening to the .wav file is a richer and more satisfying experience because the audio has greater detail than the more compressed .mp3. This is because the .wav file includes much more information. In our case, the low bit rate .mp3 is the 6-segment setup and the .wav is the 20-segment. The larger segment count captures a higher range of beam characteristics, resulting in cleaner mode shape curves.   You can’t perform that action at this time.