Error running for loop

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HINA
HINA 2020-1-5
评论： Cameron B 2020-1-6
Hello,
I am working with the following MATLAB code, it works properly if I want to calculate the values at some fixed points, but if I run the same code using for loop (for n=0:4), it reports an error (Error using vertcat
Dimensions of arrays being concatenated are not consistent.)
How could I resolve this issue.
I would be very grateful for your kind support.
Here's my MATLAB code:
syms P
l=100; a=50; b=20; h0=50; E=200; h1=50;%n=1.5;
for n=1:0.5:4
I0=b*h0^3/12; I1=b*h1^3/12;
L0=sqrt(P/(E*I0-P*n));%L0=simplify(L00);
L1=sqrt(P/(E*I1-P*n));%L1=simplify(L11);
v=0.3;
s=0.5;
f=1.93-3.07*s+14.53*s^2-25.11*s^3+25.8*s^4;
k=(E*I1)/(6*pi*h1*f*(1-v^2));
a11=cos(L1*l) - cos(L1*a) - L1*a*sin(L1*l) + L1*l*sin(L1*l);
a12=sin(L1*l) - sin(L1*a) + L1*a*cos(L1*l) - L1*l*cos(L1*l);
a13=cos(L0*a) - 1;
a14=sin(L0*a) - L0*a;
a21=-(L1*sin(L1*a) - L1*sin(L1*l)) - L1^2*cos(L1*a)*((P*n - E*I1)/k);
a22=(L1*cos(L1*a) - L1*cos(L1*l)) - L1^2*sin(L1*a)*((P*n - E*I1)/k);
a23=L0*sin(L0*a);
a24=(L0 - L0*cos(L0*a));
a31=-L1^2*cos(L1*a)*(P*n - E*I1);
a32=-L1^2*sin(L1*a)*(P*n - E*I1);
a33=L0^2*cos(L0*a)*(P*n - E*I0);
a34=L0^2*sin(L0*a)*(P*n - E*I0);
a41=L1^3*sin(L1*a)*(P*n - E*I1);
a42=-L1^3*cos(L1*a)*(P*n - E*I1);
a43=-L0^3*sin(L0*a)*(P*n - E*I0);
a44=L0^3*cos(L0*a)*(P*n - E*I0);
A=[a11 a12 a13 a14;a21 a22 a23 a24;a31 a32 a33 a34;a41 a42 a43 a44];
delta1=det(A);
delta2=subs(delta1);
delta=simplify(delta2);
F=matlabFunction(delta);
%FF=fplot(F)
%P=vpasolve(delta,P,1e3)
%P=feval(F,1e3 )% sol=fzero(F,1e3)
P=[];
for j = 0.01:10
PP = fzero(F, j);
P = [P; PP];
end
P_cr = min(P(P > 0))
D=plot(n,P_cr,'k.','linewidth',1.5)
%axis([0 4 0 1e6])
DD=double(D)
hold on
end

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采纳的回答

Cameron B
Cameron B 2020-1-5
if true
syms P
l=100; a=50; b=20; h0=50; E=200; h1=50;%n=1.5;
for n=1:0.02:4
I0=b*h0^3/12; I1=b*h1^3/12;
L0=sqrt(P/(E*I0-P*n));%L0=simplify(L00);
L1=sqrt(P/(E*I1-P*n));%L1=simplify(L11);
v=0.3;
s=0.5;
f=1.93-3.07*s+14.53*s^2-25.11*s^3+25.8*s^4;
k=(E*I1)/(6*pi*h1*f*(1-v^2));
a11=cos(L1*l) - cos(L1*a) - L1*a*sin(L1*l) + L1*l*sin(L1*l);
a12=sin(L1*l) - sin(L1*a) + L1*a*cos(L1*l) - L1*l*cos(L1*l);
a13=cos(L0*a) - 1;
a14=sin(L0*a) - L0*a;
a21=-(L1*sin(L1*a) - L1*sin(L1*l)) - L1^2*cos(L1*a)*((P*n - E*I1)/k);
a22=(L1*cos(L1*a) - L1*cos(L1*l)) - L1^2*sin(L1*a)*((P*n - E*I1)/k);
a23=L0*sin(L0*a);
a24=(L0 - L0*cos(L0*a));
a31=-L1^2*cos(L1*a)*(P*n - E*I1);
a32=-L1^2*sin(L1*a)*(P*n - E*I1);
a33=L0^2*cos(L0*a)*(P*n - E*I0);
a34=L0^2*sin(L0*a)*(P*n - E*I0);
a41=L1^3*sin(L1*a)*(P*n - E*I1);
a42=-L1^3*cos(L1*a)*(P*n - E*I1);
a43=-L0^3*sin(L0*a)*(P*n - E*I0);
a44=L0^3*cos(L0*a)*(P*n - E*I0);
A=[a11 a12 a13 a14;a21 a22 a23 a24;a31 a32 a33 a34;a41 a42 a43 a44];
delta1=det(A);
delta2=subs(delta1);
delta=simplify(delta2);
F=matlabFunction(delta);
%FF=fplot(F)
%P=vpasolve(delta,P,1e3)
P1=[];
for j = 0.01:10
PP = fzero(F, j);
P1 = [P1; PP];
end
P_cr = min(P1(P1 > 0));
D=plot(n,P_cr,'k.','linewidth',1.5);
%axis([0 4 0 1e6])
DD=double(D);
hold on
end
end
  3 个评论
显示 1更早的评论
HINA
HINA 2020-1-5
Yes, it works for me now. Thank you so much for your time and support. I have one more question regarding this code, that would be great if you could give any suggestion for that.
Since I am interested to get the smallest positive value of P, my question is that the syntax that i used to get smallest positive value is correct or not? as after running this file still I am not able to get the desired solutions.
I am talking about the following part:
P1=[];
for j = 0.01:10
PP = fzero(F, j);
P1 = [P1; PP];
end
P_cr = min(P1(P1 > 0));
Cameron B
Cameron B 2020-1-6
Yea I was looking at the output and it seems as if something is off. I think the way you went about finding the lowest positive value is ok, but I’m not sure about the other code. I know this is related to fracture toughness calculations for 3 point bends, but something does seem to be off a bit. However I can’t be sure. Try this:
for j = 0.001:0.001:10 PP = fzero(F, j); P1 = [P1; PP]; end
Or you can try something else for the j range.

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更多回答（1 个）

Colo
Colo 2020-1-5
Hi,
I copied your code and ran it. First loop works fine because P is 1x1sym.
in line 38 you change the size of P to 10x1: P=[P;PP]
Thus, a11 becomes 10x10 and a41 10x1 during the following iteration. So the dimensions cannot match in A.
The solution is to keep P as a 1x1sym.
You can either add a line at the beginning of the loop declaring P again as syms:
syms P
or use another variable instead of deleting P in line 36 and the following nested for loop.
I hope this helped :)
  1 个评论
HINA
HINA 2020-1-5
Thank you so much for your answer, but Can you please help me how to keep P as 1x1sym. As I am not an expert in MATLAB. I 'll be very thankful to you for that.

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