The following code seems to be taking forever to run. I don't get the reasons behind it though. Any suggestions to make it run quicker are highly appreciated
1 Ansicht (letzte 30 Tage)
Ältere Kommentare anzeigen
Sanjit Ganti
am 12 Apr. 2018
Bearbeitet: Walter Roberson
am 13 Apr. 2018
%Prob 1.b - Distributed Loading.
n=6; %No. of mesh elements considered
L=1; %Length of total beam
l=L/n;%Length of each element.
c=0.1*L; h=0.1*L;b=1.875; zeta=0.7341;
Rho=1000; a=c*h; E=12*(10^9); I=c*(h^3)/12;k_spring = E*I/(h*L*L);
m=(Rho*a*l/420)*[156 22*l 54 -13*l; 22*l 4*l*l 13*l -3*l*l; 54 -13*l 156 -22*l; -13*l -3*l*l -22*l 4*l*l];
k=(E*I/(l^3))*[12 6*l -12 6*l; 6*l 4*l*l -6*l 2*l*l; -12 -6*l 12 -6*l; 6*l 2*l*l -6*l 4*l*l];
global A
D=0.1*eye(2*n+2);
M=zeros(2*n+2);
K=zeros(2*n+2);
for i=1:n
p=2*i-2;
for j=1:4
for s=1:4
M(p+j,p+s)=M(p+j,p+s)+m(j,s); %Global mass and stiffness matrices.
K(p+j,p+s)=K(p+j,p+s)+k(j,s);
end
end
end
M_sub=M(3:length(M),3:length(M));
K_sub=K(3:length(K),3:length(K)); %deleting the first two rows and columns to account for fixed end.
D_sub=D(3:length(D),3:length(D));
q=length(K_sub);
K_sub(q-1,q-1)=K_sub(q-1,q-1)+k_spring;%adding the spring stiffness at the boundary
A = [zeros(q) eye(q); -M_sub\K_sub -M_sub\D_sub];
tspan = linspace(0,20,21);
syms eta
si(1)=1-3*(eta^2)+2*(eta^3); %shape functions used
si(2)=(eta-2*(eta^2)+(eta^3))*l;
si(3)=3*(eta^2)-2*(eta^3);
si(4)=(-(eta^2)+(eta^3))*l;
%g=zeros(n,1); %distributed forcing load per element
for i=1:n
g(i)=(cos(b*l*(eta+i-1))-cosh(b*l*(eta+i-1))-zeta*(sin(b*l*(eta+i-1))-sinh(b*l*(eta+i-1))));
end
for i=1:4
for j=1:n
F(i,j)=int(g(j)*si(i)*l,eta,0,1);
end
end
F_single(1:2,1)=F(1:2,1);
for i=1:5
F_single(2*i+1,1)=F(3,i)+F(1,i+1);
F_single(2*i+2,1)=F(4,i)+F(2,i+1);
end
F_single(13:14,1)=F(3:4,6);
F_sub=F_single(3:14,1);
[t,Q]=ode45(@(t,Q) ganti(t,Q,M_sub,F_sub,A,n,l),tspan,zeros(24,1));
save('prob1b2r')
And the function is
function dQdt=ganti(t,Q,M_sub,F_sub,A,n,l)
omega=2363.28;
dQdt = zeros(24,1);
F_final=F_sub*sin(omega*t);
U = [zeros(12,1);M_sub\F_final];
for i=1:24
dQdt(i)=A(i,:)*[Q(1);Q(2);Q(3);Q(4);Q(5);Q(6);Q(7);Q(8);Q(9);Q(10);Q(11);Q(12);Q(13);Q(14);Q(15);Q(16);Q(17);Q(18);Q(19);Q(20);Q(21);Q(22);Q(23);Q(24)]+U(i);
end
end
0 Kommentare
Akzeptierte Antwort
Walter Roberson
am 13 Apr. 2018
You do not initialize F, so it is growing on the fly (but it is not large.) More importantly, the right hand side is int() which is for finding closed form integrals, giving exact results such as
(82126976*exp(-5/16))/390625 - (18263296*cos(5/16))/78125 + (9189504*exp(5/16))/390625 + (47337472*sin(5/16))/390625 - 2447/3125
This leads to F_sub being symbolic. You then pass F_sub into your ode function, so much of the math in the ode ends up being done in symbolic form, before ending up getting converted to numeric form.
Considering your A values are pure numeric, you cannot justify calculating with F_sub in symbolic form: you should be passing double(F_sub) to the ode. Or, more likely, instead of using int() you should be using vpaintegral() and assigning into F that has been initialized to double precision.
Note: there is no obvious reason why you made A global.
2 Kommentare
Walter Roberson
am 13 Apr. 2018
Bearbeitet: Walter Roberson
am 13 Apr. 2018
I am not sure exactly how MATLAB treats a global that is passed as a parameter. I think it becomes ordinary within the routine it is passed to, in which case performance within that routine should not be affected, but... hmmm... I'm not completely sure how it would work.
Generally speaking globals are the slowest form of variables to look up.
Okay, my tests show that if you alter a global that has also been passed as a parameter, that in the places that know it only as a parameter, the value is not changed. This means that the slowdown of using globals only occurs in the places that it is declared as global (or a shared variable.)
Weitere Antworten (0)
Siehe auch
Community Treasure Hunt
Find the treasures in MATLAB Central and discover how the community can help you!
Start Hunting!