Matlab Solve a system of equations for a set of data

Hello,

So i want to solve this strain, stress problem by knowing 2 out of 6 unknowns and the 4 equations i need. I have a set of data for ex28 and ey28 of 9 values. I am noob to matlab and I started to write a code to solve it as follows:

E=73*10^9; v=0.33;

ey28=-1 *10^-6;
ex28=-3*10^-6;
syms sigmax28 sigmay28 ez28 sigmaz28;

eqn1=sigmax28==v*E*(ex28+ey28+ez28)/((1+v)*(1-2*v))+E/(1+v)*ex28;
eqn2=sigmay28==v*E*(ex28+ey28+ez28)/((1+v)*(1-2*v))+E/(1+v)*ey28;
eqn3=sigmaz28==v*E*(ex28+ey28+ez28)/((1+v)*(1-2*v))+E/(1+v)*ez28;
eqn4=ez28==sigmaz28/E-v/E*(sigmay28+sigmax28);

sol28=solve([eqn1, eqn2,eqn3,eqn4],[sigmax28,sigmay28,sigmaz28,ez28])

When solved, I get a 'struct' with very large numbers, so I have the feeling I am doing something wrong. I also could not figure out how to get the values from the struct without doubleclicking on it on the variable tree

I am also finding this process of going value by value very tedious, is there a way I could make the whole system of equations at once?

Thanks for throwing some light on it :)

Answer 1

[John D'Errico]

You get huge numbers, because you are trying to use symbolic tools here, but with double precision numbers. See what happens here:

x = (sqrt(5) + 1)/2
x =
          1.61803398874989
>> sym(x)
ans =
910872158600853/562949953421312

At that point, x is just a number. You and I happen to know where it came from, but MATLAB does not. So in order to do exact arithmetic on it, the symbolic toolbox converts the floating point number to a ratio of integers. You can always convert a result to a floating point number using double, or to a floating point high precision number using vpa.

double(sym(x))
ans =
          1.61803398874989
          
vpa(sym(x))
ans =
1.6180339887498949025257388711907

Don't trust those extra digits that vpa yields here, because there were only roughly 16 significant digits to start with.

double(sol28.sigmax28)
ans =
         -190572.704674325

[Manuel Soler]

Thank you so much

I was able to solve it and make it more automate too :)

Answer 2

[David Hill]

You could just solve Ax=b (A\b):

E=73*10^9; 
v=0.33;
ey28=-1*10^-6;
ex28=-3*10^-6;
b=[v*E*(ex28+ey28)/(1+v)/(1-2*v)+E*ex28/(1+v);v*E*(ex28+ey28)/(1+v)/(1-2*v)+E*ey28/(1+v);v*E*(ex28+ey28)/(1+v)/(1-2*v);0];
A=[1 0 0 -v*E/(1+v)/(1-2*v);0 1 0 -v*E/(1+v)/(1-2*v);0 0 1 -v*E/(1+v)/(1-2*v)-E/(1+v);v/E v/E -1/E 1];
x=A\b;