Effacer les filtres
Effacer les filtres

substitute value for variable

2 vues (au cours des 30 derniers jours)
Carolyn
Carolyn le 21 Fév 2018
Commenté : Karan Gill le 21 Fév 2018
I have created a code to solve a differential in terms of one variable (all C variables and B are constants and defined previously). Now I want to plug a value in for that variable into the differential equation, but it won't let me. I tried using the 'subs' function, but that doesn't work. Help?
syms I1 I2
F1 = C1*(exp(C2*(I1-3)-1))-((C1*C2)/2)*(I2-3);
W1 = 0;
F2 = piecewise(x<1,W1,1<x<lambda,W2,x>lambda,W3);
Wo = F1 + F2;
Wo1 = diff(Wo,I1);
Wo2 = diff(Wo,I2);
clear I1 I2
I1 = trace(B);
I2 = .5*((trace(B))^2+trace(B^2));
subs(Wo1,I1,I2);
  2 commentaires
Birdman
Birdman le 21 Fév 2018
Share all variables, the entire code.
Stephen23
Stephen23 le 21 Fév 2018
Carolyn's "Answer" moved here:
Here's the whole code. It all works until the last step.
% Define coefficients for all the fiber families
syms x p Wx I1 I2
pres = [p 0 0
0 p 0
0 0 p];
theta = [2 122 241];
C1 = .00124;
C2 = 1.07;
C3 = [1.03*10^-7 1.67*10^-7 1.59*10^-7];
C4 = [26.6 35 38];
lambda = [2.78 5.6 7.74];
C5 = C3.*C4.*lambda.*(exp(C4.*(lambda-1)));
% Define displacement matrix and B matrix
F = [x 0 0
0 1/sqrt(x) 0
0 0 1/sqrt(x)];
B = F*transpose(F);
% Define F1 as given
F1 = C1*(exp(C2*(I1-3)-1))-((C1*C2)/2)*(I2-3);
W1 = 0;
% Run pipline for each fiber family to get individual strain energy
% functions
for n = 1:3
a = [cos(theta(n))
-sin(theta(n))
0];
W2 = C3(n)*exp(C4(n)*(x-1))-1;
W3 = C5(n);
F2 = piecewise(x<1,W1,1<x<lambda(n),W2,x>lambda(n),W3);
Wo(n) = F1 + F2;
Wo1(n) = diff(Wo(n),I1);
Wo2(n) = diff(Wo(n),I2);
ao(n) = transpose(a(n))*a(n);
end
% Plug in values for a matrix and invariantes
clear I1 I2
I1 = trace(B);
I2 = .5*((trace(B))^2+trace(B^2));
%Substitute in the invariant values
for n=1:3
subs(Wo1(n),I1);
subs(Wo2(n),I2);
end

Connectez-vous pour commenter.

Réponse acceptée

Karan Gill
Karan Gill le 21 Fév 2018
You shouldn't overwrite I1. Overwriting I1 will not automatically substitute for it in Wo1. Instead, remove I1 = trace(B); and use
>> I1_val = trace(B)
I1_val =
2/x + x^2
>> Wo1 = subs(Wo1,I1,I1_val)
Wo1 =
[ piecewise(x < 1 | 139/50 < x | x in Dom::Interval(1, 139/50), (3317*exp((107*I1)/100 - 421/100))/2500000), piecewise(x < 1 | 28/5 < x | x in Dom::Interval(1, 28/5), (3317*exp((107*I1)/100 - 421/100))/2500000), piecewise(x < 1 | 387/50 < x | x in Dom::Interval(1, 387/50), (3317*exp((107*I1)/100 - 421/100))/2500000)]
This way, you separate the symbolic variable from the value you want to substitute, which keeps the meaning of variables in your code clean.
  2 commentaires
Carolyn
Carolyn le 21 Fév 2018
Thank you! This worked perfectly.
Karan Gill
Karan Gill le 21 Fév 2018
Great! :) (And I understand overwriting I1 was the intuitive thing to do.)

Connectez-vous pour commenter.

Plus de réponses (0)

Catégories

En savoir plus sur Get Started with MuPAD dans Help Center et File Exchange

Community Treasure Hunt

Find the treasures in MATLAB Central and discover how the community can help you!

Start Hunting!

Translated by