Solve a system of symbolic equations
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maxroucool
am 9 Dez. 2013
Beantwortet: Walter Roberson
am 9 Dez. 2013
Hi all,
I am discovering symbolic calculation in Matlab. I have to solve a system of equations that I don't want to solve by hand... :D
It is a mechanical problem based on the classical laminate theory for composite materials. The aim is to exploit data from tensile coupons tests.
% Stack-up of the coupon
% [orientation (in °), thickness (in mm)]
plies = sym([-60 1; 60 1; 60 1; -60 1]);
plies(:,1) = plies(:,1).*pi/180;
syms E1 E2 G12 nu12 nu21; % Mechanical properties of the ply
syms Q11 Q12 Q22 Q66; % Stiffness matrix in the orthogonal frame
syms Q11L Q12L Q22L Q66L; % Stiffness matric in the laminate frame
syms A11 A12 A22 A66 D11 D12 D16 D22 D26 D66; % Laminate ABD matrix
nu21 = nu12 * E2/E1;
Q11 = E1/(1-nu12*nu21);
Q12 = nu21*E1/(1-nu12*nu21);
Q22 = E2/(1-nu12*nu21);
Q66 = G12;
Qply = [ Q11 Q12 0;
Q12 Q22 0;
0 0 Q66];
ATemp = zeros(3,3);
% Generates Aij of the laminate ABD matrix
for i=1:size(plis,1)
ALam = ATemp + frameRotation(Qply, plies(i,1)).*plies(i,2);
ATemp = ALam;
end
clear ATemp
subexpr(ALam)
% Test hypothesis:
% 1st tensile test in the direction 1
% - Tensile load NL1
% - 2 deformation gages dir 1 et 2 : j11 et j21
% 2nd tensile test in the direction 2
% - Tensile load NL2
% - 2 deformation gages dir 1 et 2 : j12 et j22
% System 1 :
% { NL1 = A11 j11 + A12 j21
% { 0 = A12 j11 + A22 j21
% System 2 :
% { NL2 = A22 j12 + A12 j22
% { 0 = A12 j12 + A11 j22
% ==> 4 equations / 4 unknowns (E1, E2, nu12, G12)
NL1 = (1000);
NL2 = (500);
j11 = (0.000001);
j21 = (0.0000001);
j12 = (0.0000002);
j22 = (0.000002);
solu = solve(ALam(1,1)*j11 + ALam(1,2)*j21 - NL1, ...
ALam(1,2)*j11 + ALam(2,2)*j21, ...
ALam(2,2)*j12 + ALam(1,2)*j22 - NL2, ...
ALam(1,2)*j12 + ALam(1,1)*j22, ...
E1, E2, G12, nu12)
And the frameRotation function :
function Qlam = frameRotation(Qply, theta)
% This function computes the ply stiffness matrix Q in the laminate frame
% Qlam = Qply(theta)
Q11 = Qply(1,1);
Q12 = Qply(1,2);
Q22 = Qply(2,2);
Q66 = Qply(3,3);
t = theta;
Q11L = Q11*cos(t)^4 + Q22*sin(t)^4 + 2*(Q12+Q66)*sin(t)^2*cos(t)^2;
Q12L = (cos(t)^4+sin(t)^4)*Q12 + (Q11+Q22-2*Q66)*sin(t)^2*cos(t)^2;
Q16L = (Q11-Q12-Q66)*sqrt(2)*sin(t)*cos(t)^3 + (Q12-Q22+Q66)*sqrt(2)*cos(t)*sin(t)^3;
Q22L = Q22*cos(t)^4 + Q11*sin(t)^4 + 2*(Q12+Q66)*sin(t)^2*cos(t)^2;
Q26L = (Q11-Q12-Q66)*sqrt(2)*sin(t)^3*cos(t) + (Q12-Q22+Q66)*sqrt(2)*cos(t)^3*sin(t);
Q66L = (cos(t)^4+sin(t)^4)*Q66 + (Q11+Q22-2*Q12-Q66)*2*sin(t)^2*cos(t)^2;
Qlam = [ Q11L Q12L Q16L;
Q12L Q22L Q26L;
Q16L Q26L Q66L];
The problem is that I don't understand the results that the solve function gives me :
>> [solu.E1 solu.E2 solu.G12 solu.nu12]
ans =
[ (961*z1)/169, z1, z, -31/13]
[ 0, z1, z, 0]
[ 0, 0, z1, z]
What are these z and z1?? Do you understand where is the problem?
Thank you very much!
Max
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Akzeptierte Antwort
Walter Roberson
am 9 Dez. 2013
MuPAD has found that the set of solutions is most easily expressed by introducing some temporary variables, z and z1, instead of fully writing out the same subexpression each time.
Imagine for example that one was solving a quadratic with real-valued solutions. Instead of writing out some long expression on both parts, with the second solution just being the negative of the first, MuPAD might introduce a temporary variable z to represent the long expression, and then say that the results are [z, -z] where z is ....
Unfortunately the interface between MuPAD and MATLAB fails to return back what the temporary expression is in some cases, in some releases.
There are some alternatives to firing up the MuPAD notebook, but they mostly involve giving a sequence of MuPAD commands into an evalin() expression. For example, telling MuPAD to write the results to a file.
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