声振论坛

 找回密码
 我要加入

QQ登录

只需一步,快速开始

查看: 2536|回复: 5

[其他相关] 请教转子传递矩阵

[复制链接]
发表于 2007-7-9 11:02 | 显示全部楼层 |阅读模式

马上注册,结交更多好友,享用更多功能,让你轻松玩转社区。

您需要 登录 才可以下载或查看,没有账号?我要加入

x
帮忙看一个转子动力学传递矩阵程序,一直得不到标准的剩余量曲线,还出现振荡,如果哪位前辈以前编过此程序,麻烦给传一个,解燃眉之急啊,多谢谢谢谢谢!!!!!
clear;
l=[0.25,0.5,0.5,0.25,0];
m=[20,20,20,20,20];
Jp=[0.0360,0.0360,0.0360,0.0360,0.0360];
Jd=[0.0180,0.0180,0.0180,0.0180,0.0180];
I=[3.9761e-008,3.9761e-008,3.9761e-008,3.9761e-008,3.9761e-008];
K=[15000000,0,15000000,0,15000000];
E=2.000e+011;
v=[0,0,0,0,0];
S=[0,0;0,0];
s=1;
x=[];
for n=200:800
for i=1:5
u11=[1,l(i);0,1];
u12=[l(i)*(m(i)*n^2-K(i)),(Jp(i)-Jd(i))*n^2;m(i)*n^2-K(i),0];
u21=(l(i)/(E*I(i))).*[l(i)/2,l(i)^2*(1-v(i))/6;1,l(i)/2];
u22=[1+l(i)^3*(1-v(i))*(m(i)*n^2-K(i))/(6*E*I(i)),l(i)+l(i)^2*(Jp(i)-Jd(i))*n^2/(2*E*I(i));l(i)^2*(m(i)*n^2-K(i))/(2*E*I(i)),1+l(i)*(Jp(i)-Jd(i))*n^2/(E*I(i))];
s=sign(det(u21*S+u22))*s;
S=[u11*S+u12]*inv([u21*S+u22]);
end
D=det(S)*s;
x=[x,D];
end
n=200:1:800;
grid on
plot(n,x)
回复
分享到:

使用道具 举报

发表于 2007-7-12 10:46 | 显示全部楼层
riccat 传递矩阵法 计算得到的剩余l量就是振荡的i
 楼主| 发表于 2007-7-12 11:01 | 显示全部楼层
我用此程序验算书上的算例(闻邦椿的《转子动力学》),结果相差很大,我反复检查过很多遍程序实在发现不了新的问题,如果前辈对此方面有所研究,请不吝赐教啊,很困惑,完整程序如下:
clear;
l=[1.3,1.3,1.3,1.3,1.3,1.3,1.3,1.3,1.3,1.3,1.3,1.3,0];
m=[2940,5880,5880,5880,5880,5880,5880,5880,5880,5880,5880,5880,2940];
Jp=[0,0,0,0,0,0,0,0,0,0,0,0,0];
Jd=[0,0,0,0,0,0,0,0,0,0,0,0,0];
I=[100,100,100,100,100,100,100,100,100,100,100,100,100];
E=4393;
v=[0,0,0,0,0,0,0,0,0,0,0,0,0];
k=[1.96*10^9,0,0,1.96*10^9,0,0,1.96*10^9,0,0,1.96*10^9,0,0,1.96*10^9];
kb=[2.7048*10^9,0,0,2.7048*10^9,0,0,2.7048*10^9,0,0,2.7048*10^9,0,0,2.7048*10^9];
mb=[3577,3577,3577,3577,3577,3577,3577,3577,3577,3577,3577,3577,3577];
S=[0,0;0,0];
s=1;
x=[];
for n=1864:1864
for i=1:13
K=k(i)*(kb(i)-mb(i)*n^2)/(k(i)+kb(i)-mb(i)*n^2)
%l(i)/(E*I(i))
u11=[1,l(i);0,1];
u12=[l(i)*(m(i)*n^2-K),(Jp(i)-Jd(i))*n^2;m(i)*n^2-K,0];
u21=(l(i)/(E*I(i))).*[l(i)/2,l(i)^2*(1-v(i))/6;1,l(i)/2];
u22=[1+l(i)^3*(1-v(i))*(m(i)*n^2-K)/(6*E*I(i)),l(i)+l(i)^2*(Jp(i)-Jd(i))*n^2/(2*E*I(i));l(i)^2*(m(i)*n^2-K)/(2*E*I(i)),1+l(i)*(Jp(i)-Jd(i))*n^2/(E*I(i))];
%s=sign(det(u21*S+u22))*s;
S=[u11*S+u12]*inv([u21*S+u22])
end
D=det(S);
x=[x,D];
end
n=1864:1:1864;
grid on
plot(n,x)
发表于 2007-7-14 06:36 | 显示全部楼层
加注释
发表于 2007-7-15 22:12 | 显示全部楼层
Jp=[0,0,0,0,0,0,0,0,0,0,0,0,0];
Jd=[0,0,0,0,0,0,0,0,0,0,0,0,0];
I=[100,100,100,100,100,100,100,100,100,100,100,100,100];
E=4393;
v=[0,0,0,0,0,0,0,0,0,0,0,0,0];
你能解释一下这些的含义吗??
感觉不太对!!
 楼主| 发表于 2007-7-21 10:23 | 显示全部楼层

回复 #5 wenzihui 的帖子

Jp=[0,0,0,0,0,0,0,0,0,0,0,0,0];
Jd=[0,0,0,0,0,0,0,0,0,0,0,0,0];
以上两组数据全设为0,是因为不计圆盘的转动惯量及陀螺力矩

v=[0,0,0,0,0,0,0,0,0,0,0,0,0];
令v等于零是不计剪切变形的影响

I=[100,100,100,100,100,100,100,100,100,100,100,100,100];
E=4393;
因为书上给的算例是l/(EI),而各点的E*I的积是一样的,所以就写成上面的形式
您需要登录后才可以回帖 登录 | 我要加入

本版积分规则

QQ|小黑屋|Archiver|手机版|联系我们|声振论坛

GMT+8, 2024-11-29 06:39 , Processed in 0.063049 second(s), 18 queries , Gzip On.

Powered by Discuz! X3.4

Copyright © 2001-2021, Tencent Cloud.

快速回复 返回顶部 返回列表