README.md

# me3255_finalproject_group29
## Thomas Ressler, Charlotte Veitner
### *The following is an overview of the processes used in this project.  For more depth, refer to the documentation within the corresponding ".m" file.*
### ------------------------------ 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.

![Analytical Solution (1)](IMG_1275.png)
![Analytical Solution (1)](IMG_1276.png)

##### 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.

![Beam Deflection vs. Horizontal Beam Distance](plot_1a_1.png)

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.

![Distributed Beam Loading vs. Maximum Deflection](plot_1a_2.png)

#### 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.

![Problem 2 setup](problem2_setup.png)

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:


![6-Segment maximum deflection](plot_2_6seg.png)

This is the deflection of the beam using 6 segments and 5 different values for the distributed load.


![10-Segment maximum deflection](plot_2_10seg.png)

This is the deflection of the beam using 10 segments and 5 different values for the distributed load.


![20-Segment maximum deflection](plot_2_seg20.png)

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.

![Problem 3 setup](problem3_setup.png)

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.

![6-Segment maximum deflection](plot_3_6seg.png)

This the deflection depending on different values for P  using 6 segments and 5 different values for the distributed load.


![10-Segment maximum deflection](plot_3_10seg.png)

This the deflection depending on different values for P  using 6 segments and 5 different values for the distributed load.


![20-Segment maximum deflection](plot_3_20seg.png)

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.

![Handwritten Eigenvalue Problem (1)](eigen_setup.png)

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:

![Handwritten Eigenvalue Problem (2)](eigen.png)

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.

![6-Segment Natural Frequencies](plot_5_e_6seg.png)

![10-Segment Natural Frequencies](plot_5_e_10seg.png)

![20-Segment Natural Frequencies](plot_5_e_20seg.png)
	# me3255_finalproject_group29
	## Thomas Ressler, Charlotte Veitner
	### The following is an overview of the processes used in this project. For more depth, refer to the documentation within the corresponding ".m" file.
	### ------------------------------ 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.

	![Analytical Solution (1)](IMG_1275.png)
	![Analytical Solution (1)](IMG_1276.png)

	##### 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.

	![Beam Deflection vs. Horizontal Beam Distance](plot_1a_1.png)

	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.

	![Distributed Beam Loading vs. Maximum Deflection](plot_1a_2.png)

	#### 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.

	![Problem 2 setup](problem2_setup.png)

	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:


	![6-Segment maximum deflection](plot_2_6seg.png)

	This is the deflection of the beam using 6 segments and 5 different values for the distributed load.


	![10-Segment maximum deflection](plot_2_10seg.png)

	This is the deflection of the beam using 10 segments and 5 different values for the distributed load.


	![20-Segment maximum deflection](plot_2_seg20.png)

	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.

	![Problem 3 setup](problem3_setup.png)

	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.

	![6-Segment maximum deflection](plot_3_6seg.png)

	This the deflection depending on different values for P using 6 segments and 5 different values for the distributed load.


	![10-Segment maximum deflection](plot_3_10seg.png)

	This the deflection depending on different values for P using 6 segments and 5 different values for the distributed load.


	![20-Segment maximum deflection](plot_3_20seg.png)

	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.

	![Handwritten Eigenvalue Problem (1)](eigen_setup.png)

	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:

	![Handwritten Eigenvalue Problem (2)](eigen.png)

	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.

	![6-Segment Natural Frequencies](plot_5_e_6seg.png)

	![10-Segment Natural Frequencies](plot_5_e_10seg.png)

	![20-Segment Natural Frequencies](plot_5_e_20seg.png)