Part E

README.md

@@ -1,12 +1,23 @@
 # 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);` The
+output `w` should be a vector, but the solution represents a 2D data set w(x,y).
+#### Approach
 Central finite difference approximation for gradient of 3x3 interior nodes of w.
 ```matlab
 [w] = membrane_solution3(T,P);
 ```
 
 ## 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
@@ -16,13 +27,26 @@ P = 0.001 MPa
 
 
 ## 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);` The
+output `w` should be a vector, but the solution represents a 2D data set w(x,y).
+#### Approach
+
 General central finite difference approximation for gradient
 of nxn interior nodes of w.
 ```matlab
 [w] = membrane_solution(T,P,n)
 ```
 
 ## 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
@@ -32,12 +56,76 @@ of nxn interior nodes of w.
 ![Fig. 2](./figures/PartD.png)
 
 ## Part E
-Function that calculates the difference in the strain energy and the work done by pressure for n-by-n elements.
+#### 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
-[pw_se,w] = SE_diff(T,P,n)
+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 nodes
+% 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]
@@ -54,4 +142,24 @@ Error
 ```
 
 ## 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

SE_diff.m

@@ -1,33 +1,51 @@
 function [pw_se,w] = SE_diff(T,P,n)
-  E = 1; % 1 TPa ~= 10^6 MPa
-  v = 0.31;
-  t = 3*10^-4;
-  h = 10/(n+1);
+% 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 nodes
+% Output:
+% pw_se = Absolute value of difference between strain energy and work done by pressure
+% w = displacement vector for interior nodes
 
-  w = membrane_solution(T,P,n);
+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
 
-  z = zeros(n + 2);
-  z(2:end-1,2:end-1) = reshape(w,[n n]);
+% 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]);
 
-  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
+% 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
 
-  pw = sum(sum(wavg.*h^2.))
-  dwdx = zeros(num);
-  dwdy = zeros(num);
+% final work done by pressure
+pw = sum(sum(wavg.*h^2.*P))
 
-  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
+% 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
 
-  se = (E*t*h^2)/(2*(1-v^2)) * sum(sum((1/4).*dwdx.^4+(1/4).*dwdy.^4+(1/4).*(dwdx.*dwdy).^2));
+% 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));
 
-  pw_se = pw - se;
+% Final value of difference between strain energy and work done by pressure
+pw_se = pw - se;