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# 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] | |
```matlab | |
% Part F script | |
% Take the results of 'SE_diff.m' and use the 'bisect.m' | |
% root-finding method to | |
% solve for the | |
% tension T, | |
% given a pressure P and | |
% a range of interior nodes [n] | |
[pw_se,w] = SE_diff(T,P,n) | |
[root,fx,ea,iter] = bisect(@(T) SE_diff(T, 0.001,40),0.001,1,0.1)ff(T,0.001,20),0.001,1,0.1) | |
% Run a total of 5 times, varying the value of n by 5 | |
% from 20 to 40 | |
``` | |
##### Relative Error | |
| number of nodes | Tension(uN/um) | rel. error | | |
| --- | --- | --- | | |
| 3 | 0.0489 | n/a | | |
| 20 | 0.0599 | 22.3% | | |
| 25 | 0.0601 | 0.27% | | |
| 30 | 0.0602 | 0.17% | | |
| 35 | 0.0603 | 0.09% | | |
| 40 | 0.0603 | 0.06% | | |
## 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 | |
```matlab | |
% Create plot for wmax v pressure | |
clear | |
% range of pressures to be used to find tension T | |
P = linspace(0.001,0.01,10); | |
% find the tension using the 'bisect.m' method | |
for i = 1:length(P) | |
func = @(T) SE_diff(T,P(i),10); | |
[root(i),fx,ea,iter] = bisect(func,0.001,1,.1); | |
end | |
% Calculate each w using every root and pressure value | |
for i = 1:length(root) | |
w = membrane_solution(root(i),P(i),10); | |
w1(:,i) = w; % displays w values as column | |
end | |
% Find final wmax by taking the maximum w value from each | |
% column in the w vector | |
wmax = max(w1); | |
coefficients = polyfit(wmax,P,3); | |
Y = polyval(coefficients,wmax); | |
% plot the w_max vs. the Pressure and include a cubic best fit curve. | |
plot(wmax,P,wmax,Y,'or') | |
xlabel('Max Deflection (um)') | |
ylabel('Pressure (MPa)') | |
title('Pressure vs. Maximum Deflection') | |
``` |