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Help with a Matrix for different angles

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Eddy Ramirez
Eddy Ramirez 2021 年 3 月 28 日
コメント済み: Walter Roberson 2021 年 3 月 31 日
Greetings,
I am running the following code and I want to find out the value of sigma1, sigma2, and tau6 at each angle starting from 0. I was thinking of doing a for-lopp, but the problem is that I am not able to do negative thetas or 0.
I think the problem is that MatLAb is taking ALL the angles and creating one big solution based on all angles? i might be wrong here, but any help to achieve my goal would be greatly appreciated
theta=0:5:90;
m=cosd(theta);
n=sind(theta);
sigma1=sym('sigma_1');
sigma2=sym('sigma_2');
tau6=sym('tau_6');
stress=[sigma1; sigma2; tau6];
sigmax=0;
sigmay=0;
tau5=sym('taus');
stress_rotation=[sigmax; sigmay; tau5];
T=[m.^2 n.^2 2.*m.*n;
n.^2 m.^2 -2.*m.*n;
-m.*n m.*n m.^2-n.^2];
equation=stress==T.*stress_rotation;
solution=solve(equation, stress);
sigma1_f=vpa(solution.sigma_1);
sigma2_f=vpa(solution.sigma_2);
tau6_f=vpa(solution.tau_6);
  9 件のコメント
Eddy Ramirez
Eddy Ramirez 2021 年 3 月 31 日
yeah still no luck, i might have to recode it i just find it weird that a for loop doesnt do the trick. I wrote it as shown below but it does not work at all
theta=0:5:90;
m=cosd(theta);
n=sind(theta);
m2=m.^2;
n2=n.^2;
for i=1:length(theta)
T=[m2(i) n2(i) 2.*m(i).*n(i);
n2(i) m2(i) -2.*m(i).*n(i);
-m(i)*n(i) m(i)*n(i) m2(i)-n2(i)];
sigma1=sym('sigma_1');
sigma2=sym('sigma_2');
tau6=sym('tau_6');
stress=[sigma1; sigma2; tau6];
sigmax=0;
sigmay=0;
tau5=sym('taus');
stress_rotation=[sigmax; sigmay; tau5];
equation=stress(i)==T.*stress_rotation(i);
solution=solve(equation, stress);
sigma1_f=vpa(solution.sigma_1);
sigma2_f=vpa(solution.sigma_2);
tau6_f=vpa(solution.tau_6);
end
Walter Roberson
Walter Roberson 2021 年 3 月 31 日
stress=[sigma1; sigma2; tau6];
That is a vector of length 3
equation=stress(i)==T.*stress_rotation(i);
but you index it at i where i can be up to length(theta)
Also your stress_rotation vector is length 3 as well.

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