悬置位移matlab,Matlab讨论区 - 声振论坛 - 振动,动力学,声学,信号处理,故障诊断 - Powered by Discuz!...
clear
clc
%動力總成質量
m=323.98;%kg
%慣性積
Jx=25.706;Jy=13.607;Jz=23.014;Jxy=-3.029;Jyz=3.359;Jzx=-0.876;%kg*m*m
%各個懸置點坐標
Xz=72.44;Yz=-431.92;Zz=148.82;
Xy=4.34;Yy=452.98;Zy=250.22;
Xh=199.64;Yh=-7.72;Zh=-245.93;
%mm在下面將位移坐標變成m
%各個懸置的三向剛度
Kzx=380e3*1.5;Kzy=72e3*1.5;Kzz=220e3*1.5;
Kyx=380e3*1.5;Kyy=72e3*1.5;Kyz=220e3*1.5;
Khx=213e3*1.5;Khy=10e3*1.5;Khz=10e3*1.5;%N/m
%各個懸置的安裝角度
%左懸置安裝角度
ALFAzu=0;BETAzu=90*pi/180;GAMAzu=90*pi/180;
ALFAzv=90*pi/180;BETAzv=0;GAMAzv=90*pi/180;
ALFAzw=90*pi/180;BETAzw=90*pi/180;GAMAzw=0;
%右懸置安裝角度
ALFAyu=0;BETAyu=90*pi/180;GAMAyu=90*pi/180;
ALFAyv=90*pi/180;BETAyv=0;GAMAyv=90*pi/180;
ALFAyw=90*pi/180;BETAyw=90*pi/180;GAMAyw=0;
%后懸置安裝角度
ALFAhu=8*pi/180;BETAhu=90*pi/180;GAMAhu=82*pi/180;
ALFAhv=90*pi/180;BETAhv=0;GAMAhv=90*pi/180;
ALFAhw=98*pi/180;BETAhw=90*pi/180;GAMAhw=8*pi/180;
%安裝角矩陣Tj
Th=[cos(ALFAhu),cos(BETAhu),cos(GAMAhu);cos(ALFAhv),cos(BETAhv),cos(GAMAhv);cos(ALFAhw),cos(BETAhw),cos(GAMAhw)];
Tz=[cos(ALFAzu),cos(BETAzu),cos(GAMAzu);cos(ALFAzv),cos(BETAzv),cos(GAMAzv);cos(ALFAzw),cos(BETAzw),cos(GAMAzw)];
Ty=[cos(ALFAyu),cos(BETAyu),cos(GAMAyu);cos(ALFAyv),cos(BETAyv),cos(GAMAyv);cos(ALFAyw),cos(BETAyw),cos(GAMAyw)];
%參數輸入完畢
%%%%%%%%%%%====================%%%%%%%%%%
%%%%%%%%%%%---能量表達式中的Fj和kj---%%%%%%%%%%
Fh=[1e3,0,0,0,Zh,-Yh;0,1e3,0,-Zh,0,Xh;0,0,1e3,Yh,-Xh,0]*0.001;
Fz=[1e3,0,0,0,Zz,-Yz;0,1e3,0,-Zz,0,Xz;0,0,1e3,Yz,-Xz,0]*0.001;
Fy=[1e3,0,0,0,Zy,-Yy;0,1e3,0,-Zy,0,Xy;0,0,1e3,Yy,-Xy,0]*0.001;
%%%此處將位移坐標轉變成m
kh=[Khx,0,0;0,Khy,0;0,0,Khz];
kz=[Kzx,0,0;0,Kzy,0;0,0,Kzz];
ky=[Kyx,0,0;0,Kyy,0;0,0,Kyz];
%%%%%%%%%%%---振動方程中的M,D和K---%%%%%%%%%%
M=[m,0,0,0,0,0;0,m,0,0,0,0;0,0,m,0,0,0;0,0,0,Jx,-Jxy,-Jzx;0,0,0,-Jxy,Jy,-Jyz;0,0,0,-Jzx,-Jyz,Jz];
% D=[1200,0,0,0,0,0;0,1600,0,0,0,0;0,0,2000,0,0,0;0,0,0,60,0,0;0,0,0,0,80,0;0,0,0,0,0,100];
K=Fh'*Th'*kh*Th*Fh+Fz'*Tz'*kz*Tz*Fz+Fy'*Ty'*ky*Ty*Fy;
%%%%%%%%%%%%%---下面計算系統固有特性---%%%%%%%%%%%%%
[v,d]=eig(K,M);
f=diag(sqrt(d)/2/pi)'
for j=1:6
for k=1:6
for l=1:6
ENERGY(k,l)=M(k,l)*v(k,j)*v(l,j);
end
end
qq=sum(ENERGY);
qqt=sum(qq);
dig=[qq/qqt];
EGEN(:,j)=dig;
ep1=max(dig);
ep2=sum(dig);
eper(j)=ep1/ep2*100;
end
eper
EGEN=EGEN*100
總結
以上是生活随笔為你收集整理的悬置位移matlab,Matlab讨论区 - 声振论坛 - 振动,动力学,声学,信号处理,故障诊断 - Powered by Discuz!...的全部內容,希望文章能夠幫你解決所遇到的問題。
- 上一篇: 数数课堂·第四期:一个运营人的数据分析成
- 下一篇: linux 安卓打印服务器,我无法在打印