README.md

# ME 3255 Final Project

## Part A
#### Problem Statement
Create a central finite difference approximation of the gradient with 3-by-3 interior nodes of w
for the given membrane solution in terms of P and T. `[w]=membrane_solution3(T,P);`

#### Approach
```matlab
function [w] = membrane_solution3(T,P)
  % Central finite difference approximation of gradient
  % with 3x3 (um^2) interior nodes in terms of P & T.
  % Input:
  % T = tension per unit length (uN/um)
  % P = pressure (MPa)
  % Output:
  % w = displacement vector for interior nodes

  od = ones(8,1);
  od(3:3:end) = 0;
  k = -4 * diag(ones((3^2),1)) + diag(ones((3^2)-3,1),3) + diag(ones((3^2)-3,1),-3) + diag(od,1) + diag(od,-1);

  y = -(10/4)^2*(P/T)*ones(9,1);
  w = k\y;

  % Find displacement vector w
  % which represents 2D data set w(x,y)
  [x,y] = meshgrid(0:10/4:10,0:10/4:10);
  z = zeros(size(x));
  z(2:end-1,2:end-1) = reshape(w,[3 3]);

  % Plot gradient of membrane displacement
  surf(x,y,z)
  title('Gradient of Membrane Displacement')
  zlabel('Displacement [um] (micrometers)')
end

```

## Part B
#### Problem Statement
Solve for w given a pressure, P=0.001 MPa and tension, T=0.006 uN/um. Plot the result with
`surf(X,Y,W)` where X, Y, and W are the x-, y-, and z-coordinates of each point on the
membrane from 0-10um.
#### Approach
T = 0.006 uN/um
P = 0.001 MPa
```matlab
[w] = membrane_solution3(0.006,0.001);
```
![Fig. 1](./figures/PartB.png)


## Part C
#### Problem Statement
Create a general central finite difference approximation of the gradient with
n-by-n interior nodes of w
for the given membrane solution in terms of P and T. `[w]=membrane_solution(T,P,n);`

#### Approach
```matlab
function [w] = membrane_solution(T,P,n)
  % General Central finite difference approximation of
  % gradient with n-by-n interior nodes in terms of P & T.
  % Input:
  % T = tension per unit length (uN/um)
  % P = pressure (MPa)
  % n = # of interior node rows/columns
  % Output:
  % w = displacement vector for interior nodes

  od = ones(n^2-1,1);
  od(n:n:end) = 0;
  k = -4 * diag(ones(n^2,1)) + diag(ones((n^2)-n,1),n) + diag(ones((n^2)-n,1),-n) + diag(od,1) + diag(od,-1);

  y = -(10/(n+1))^2*(P/T)*ones(n^2,1);
  w = k\y;

  % Find displacement vector w
  % which represents 2D data set w(x,y)
  [x,y] = meshgrid(0:10/(n+1):10,0:10/(n+1):10);
  z = zeros(size(x));
  z(2:end-1,2:end-1) = reshape(w,[n n]);

  % Plot gradient of membrane displacement
  surf(x,y,z)
  title('Gradient of Membrane Displacement')
  zlabel('Displacement [um] (micrometers)')
end

```

## Part D
#### Problem Statement
Solve for w given a pressure, P=0.001 MPa and tension, T=0.006 uN/um with 10 interior nodes. Plot the result with `surf(X,Y,W)` where X, Y, and W are the x-, y-, and z-coordinates of each point on the membrane from 0-10um.

#### Approach
- T = 0.006 uN/um
- P = 0.001 MPa
- n = 10 nodes
```matlab
[w] = membrane_solution(0.006,0.001,10)
```
![Fig. 2](./figures/PartD.png)

## Part E
#### Problem Statement
Create a function `SE_diff` that calculates the difference in strain energy (right hand side Eq.
4) and work done by pressure (left hand side Eq. 4) for n-by-n elements.

`[pw_se,w]=SE_diff(T,P,n)`

Use the solution from part **c** to calculate w, then do a numerical integral over the
elements to calculate work done and strain energy.
#### Approach
```matlab
function [pw_se,w] = SE_diff(T,P,n)
  % function that calculates the difference between
  % strain energy and work done by pressure on the membrane.
  % Input:
  % T = tension per unit length (uN/um)
  % P = pressure (MPa)
  % n = # of interior node rows/columns
  % Output:
  % pw_se = Absolute value of difference between strain energy and work done by pressure
  % w = displacement vector for interior nodes

  E = 1e6;        % 1 TPa ~= 10^6 MPa
  t = 3*10^-4;    % thickness [um]
  h = 10/(n+1);   % height [um]
  v = 0.31;       % Poisson's Ratio

  % Displacement vector w found using Part C
  w = membrane_solution(T,P,n);
  z = zeros(n + 2);
  z(2:end-1,2:end-1) = reshape(w,[n n]);

  % Calculate average displacement, wavg, for each element by taking the displacement at each
  % corner and then average the found values.
  num = n + 1;
  wavg = zeros(num);
  for i = 1:num
    for j = 1:num
      wavg(i,j) = mean([z(i,j),z(i+1,j),z(i,j+1),z(i+1,j+1)]);
    end
  end

  % final work done by pressure
  pw = sum(sum(wavg.*h^2.*P))

  % to find= change in displacement, find the change in displacement on
  % the x-axis, dwdx, and the change in displacement on the y-axis, dwdy, and
  % average the found values.
  dwdx = zeros(num);
  dwdy = zeros(num);
  for i = 1:num
      for j = 1:num
          dwdx(i,j) = mean([z(i+1,j)-z(i,j),z(i+1,j+1)-z(i,j+1)]);
          dwdy(i,j) = mean([z(i,j+1)-z(i,j),z(i+1,j+1)-z(i+1,j)]);
      end
  end

% Using dwdx and dwdy, calculate the strain energy, se.
se = (E*t*h^2)/(2*(1-v^2)) * sum(sum((1/4).*dwdx.^4+(1/4).*dwdy.^4+(1/4).*(dwdx.*dwdy).^2));

% Final value of difference between strain energy and work done by pressure, pw_se.
pw_se = pw - se;
```

## Part F
#### Problem Statement
Use a root-finding method to calculate the tension in the membrane given a pressure, P=0.001 MPa, and n=[20:5:40] interior nodes.

Show that the error in tension is decreasing with a table:
#### Approach
Root-finding Method used to find tension in the Membrane, given:
- P = 0.001 MPa
- n = [20:5:40]

Error
```
|number of nodes |Tension (uN/um) |rel. error |
|---|---|---|
|3 |0.059 |n/a|
|20|0.0618|4.5%|
|25|0.0616|0.3%|
|30|0.0614|0.3%|
|40|0.0611|0.3%|
```

## Part G
#### Problem Statement
Plot the Pressure vs maximum deflection (P (y-axis) vs max(w) (x-axis)) for
P=linspace(0.001,0.01,10). Use a root-finding method to determine tension, T, at each
pressure. Use a cubic best-fit to find A, where, P(x)=A*dw^3. State how
many interior nodes were used for the graph.
#### Approach

Pressure v. Maximum deflection (MPa v. um)
## Part H
#### Problem Statement
Show that the constant A is converging as the number of nodes is
increased (Similar table to **f**).
#### Approach


## Part I
#### Problem Statement
If the square membrane sides are always equal, but have a tolerance of 0.1\%, what
should the depth of the sensor be if 2.5% of the sensors won't hit the bottom given a
maximum pressure of 0.01 MPa.
#### Approach