ode23 , matrix equation , 2nd order DE
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How I will solve this matrix in Matlab ?
w=10.02;
p= 7850; %%%Kg/m^3;
g=9.81;
G=77*10^9; %%% N/m^2
E=206*10^9; %%% N/m^2
L=9.6; %%%m
D=0.15 ; %%m
m=pi*(D/2)^2*L*p;
A=pi*D^2/4 ; %%m2
J=m*D^2/8; %%Kg*m^2
Ip=pi*D^4/32 ; %% m^4
I=pi*D^4/64 ;
cx=0.05*m*w;
cy=0.05*m*w;
ct=0.08*J*w;
kx=3*E*I/L;
ky=3*E*I/L;
kt=G*Ip/L;

採用された回答
その他の回答 (1 件)
Torsten
2019 年 6 月 14 日
0 投票
Convert the system to a first order system and write it as
M*y' = f(t,y)
Then use the mass-matrix option of the ODE solvers to supply M, define f in a function file and use ODE45, ODE15S, ... to solve.
6 件のコメント
Gloria
2019 年 6 月 14 日
Torsten
2019 年 6 月 14 日
s1' = s4
s2' = s5
s3' = s6
m*s4' - m*e*sin(w*t)*s6' = -c_x*s4 - k_x*s1 + m*e*(w+s6)^2*cos(w*t) + F_x
m*s5' + m*e*cos(w*t)*s6' = -c_y*s5 -k_y*s2 - m*g + m*e*(w+s6)^2*sin(w*t) + F_y
-m*e*sin(w*t)*s4' + m*e*cos(w*t)*s5' + (J+m*e^2)*s6' = -c_theta*s6 -k_theta*s3 -m*e*g*cos(w*t) + M_theta
where
s1 = x, s4 = x', s2 = y, s5 = y', s3 = theta, s6 = theta'
Now write the system as
M*s' = f(s,t)
and you are nearly done.
Gloria
2019 年 6 月 14 日
Jan
2019 年 6 月 14 日
@Gloria: You can't. s6' is not Matlab code, but a mathematical expression.
Gloria
2019 年 6 月 16 日
Jan
2019 年 6 月 16 日
You can write s(6)', because ' is the Matlab operator for the complex conjugate transposition:
a = [1, 1i; ...
2, 2i]
a'
You provide real scalars. Then the ' operaotr does not change the value. So it is valid, but simply a confusing waste of time.
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