Root finding and plotting
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How do I find the three roots of " f "
I got the plot of " f." And because the " f " and X asis has three intersections, it would have three roots.
I would like to know how to find the three roots at the same time and plot the three roots on my original plot.
Below is my coding.
Thank you very much!!
clc
clear
format long
v=0.3;
E=209e+3;
G=E/(2*(1+v));
q=-1;
h=15;
D=(E*h^3)/(12*(1-v^2));
I=(h^3)/12;
a=600;b=2400;
n =3;
[T1, T2] = meshgrid(1:2:n);
mn = [T1(:), T2(:)]
syms x y
x_value=301;
y_value=0:1:2400;
len=length(mn);
amn=zeros(1,len);
for i=1:len
m=mn(i,1);
n=mn(i,2);
amn(i)=(16*q/(m*n*D*pi^6))*(1/((m/a)^2+(n/b)^2)^2);
end
wmn=sym(zeros(1,len));
wxx=sym(zeros(1,len));
wyy=sym(zeros(1,len));
for i=1:len
m=mn(i,1);
n=mn(i,2);
wmn(i)=amn(i).*((sin(m.*pi.*x./a)).*(sin(n.*pi.*y./b)));
wxx(i)=diff(wmn(i),x,2);
wyy(i)=diff(wmn(i),y,2);
end
combine_wxx=sum(wxx);
combine_wyy=sum(wyy);
My=-D*(combine_wyy+v*combine_wxx);
Differentiation_My=diff(My,y,1)
f=subs(Differentiation_My,x,301)
%fsolve(f,500)
Differentiation_My_value=double(subs(Differentiation_My,{x,y},{x_value,y_value}));
plot(Differentiation_My_value)
grid on
0 个评论
采纳的回答
Star Strider
2021-6-17
One way is to search for the indices nearest the zero-crossings, then interpolate (if necessary, to get more exact values) —
format long g
v=0.3;
E=209e+3;
G=E/(2*(1+v));
q=-1;
h=15;
D=(E*h^3)/(12*(1-v^2));
I=(h^3)/12;
a=600;b=2400;
n =3;
[T1, T2] = meshgrid(1:2:n);
mn = [T1(:), T2(:)]
syms x y
x_value=301;
y_value=0:1:2400;
len=length(mn);
amn=zeros(1,len);
for i=1:len
m=mn(i,1);
n=mn(i,2);
amn(i)=(16*q/(m*n*D*pi^6))*(1/((m/a)^2+(n/b)^2)^2);
end
wmn=sym(zeros(1,len));
wxx=sym(zeros(1,len));
wyy=sym(zeros(1,len));
for i=1:len
m=mn(i,1);
n=mn(i,2);
wmn(i)=amn(i).*((sin(m.*pi.*x./a)).*(sin(n.*pi.*y./b)));
wxx(i)=diff(wmn(i),x,2);
wyy(i)=diff(wmn(i),y,2);
end
combine_wxx=sum(wxx);
combine_wyy=sum(wyy);
My=-D*(combine_wyy+v*combine_wxx);
Differentiation_My=diff(My,y,1)
f=subs(Differentiation_My,x,301)
%fsolve(f,500)
Differentiation_My_value=double(subs(Differentiation_My,{x,y},{x_value,y_value}));
zxi = find(diff(sign(Differentiation_My_value))) % Zero-Crossing Indices
for k = 1:numel(zxi)
idxrng = max(zxi(k)-2,1):min(zxi(k)+2,numel(Differentiation_My_value)); % Index Range For Interpolation
zc(k) = interp1(Differentiation_My_value(idxrng),idxrng,0); % Interpolate
end
Zero_Crossings = zc
plot(Differentiation_My_value)
hold on
plot(zc, zeros(size(zc)), 'rs')
hold off
grid on
To plot it against an ‘x’ vector instead of the indices, the ‘zc’ calculation becomes:
% zc(k) = interp1(Differentiation_My_value(idxrng), x(idxrng), 0);
I commented it here because it would otherwise execute in the online Run feature, and throw an error.
.
8 个评论
Star Strider
2021-6-18
I’ve definitely been there (although by now a few decades removed).
As always, my pleasure!
更多回答(1 个)
Sulaymon Eshkabilov
2021-6-17
There are a couple of ways to find roots of this eqn:
(1) ginput() --> graphical method, e.g.:
[Roots, y] = ginput(3) % click on three crossing points
(2) fsolve()
另请参阅
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