Double Integral with Varying Limits
1 Ansicht (letzte 30 Tage)
Ältere Kommentare anzeigen
syms x y
t=(30:1:90);
nu=0.3;
ms=1+nu;
E=2.07*10^11;
G=8.28*10^10;
R=0.0185;
Ep=E/(1-nu^2);
k=(6*(1+nu))/(7+6*nu);
A=pi*R^2;
L=0.254;
sigma=1;
I=(1/4)*pi*R^4;
mu=R*x;
mup=2*R*(1-sigma+x);
beta=R*y;
for i=1:length(t)
F1=(sqrt(((2*mup)/(pi*mu))*tan((pi*mu)/(2*mup))))*((0.752+2.02*(mu/mup)+0.37*(1-sin((pi*mu)/(2*mup)))^3)/(cos((pi*mu)/(2*mup))));
Flll=sqrt(((2*mup)/(pi*mu))*tan((pi*mu)/(2*mup)));
x_s=(((0.5*L)/R)+y*cosd(t(i)));
f11=matlabFunction(32*(x_s.^2)*(y^2)*x*(((sind(t(i))).^6)*F1^2+ms*((sind(t(i))).^4).*((cosd(t(i))).^2)*Flll^2)+...
16*x_s*k*y*x*((sind(t(i)).^3).*sind(2*t(i))*F1^2+ms*(sind(t(i)).^2).*cosd(t(i)).*cosd(2*t(i))*Flll^2)+...
2*k^2*x*(F1^2*(sind(2*t(i)).^2)+ms*(cosd(2*t(i))).^2*Flll^2));
xmin=0;
xmax=(sqrt(1-y^2*(sind(t(i))).^2))-1+sigma;
ymax=(sqrt(1-(1-sigma)^2))/(sind(t(i)));
ymin=-ymax;
Q(i)=(L)/(k*G*A)+(L^3)/(3*E*I)+1/(Ep*pi*R)*int(int(f11,x,xmin,xmax),y,ymin,ymax); % double integral
end
plot(t,Q,'g','LineWidth',3);grid on;
title('Flexibility Coefficient');
xlabel('theta');
ylabel('f11');
The above code that I write it, but when I run for sigma=0.5 don't response.
How to plot the solution for sigma=0:0.01:1 and t=20:1:180 ?
best regards
0 Kommentare
Antworten (0)
Siehe auch
Kategorien
Mehr zu Formula Manipulation and Simplification finden Sie in Help Center und File Exchange
Community Treasure Hunt
Find the treasures in MATLAB Central and discover how the community can help you!
Start Hunting!