4-1 图4-1所示系统中，各个质量只能沿铅垂方向运动，假设m1?m2?m3?m，

k1?k2?k3?k4?k5?k6?k。试求系统的固有频率及振型矩阵

图4-1

解：如图选择广义坐标。求质量矩阵及利用刚度影响系数法求刚度矩阵为

?m00??3k?K???kM??0m0??????00m??，??k由频率方程

K?p2M?0?k3k?k?k??k??3k??

，得

?k3k?mp2?k?k?k3k?mp2?03k?mp2?k?k

解出频率为

p1?kkkp2?2p3?2m，m，m

?(3k?mp2)2?k2?????k2?k(3k?mp2)??k2?k(3k?mp2)???

2由特征矩阵B?K?pM的伴随矩阵的第一列，

adjB(1)

将

p1?km代入得系统的第一阶主振型为

A(1)??111?

TA(2)满足如下关系：

2(A(1))TMA(2)?0，(K?p2M)A(2)?0

(2)(2)(2)(2)(2)(2)展开以上二式得，A1?A2?A3?0。取A2?0，A1??1，可得到A3?1。

即有

A(2)???101?

TA(3)满足如下关系：

2(3)(A(1))TMA(3)?0，(A(2))TMA(3)?0(K?p3M)A?0

(3)(3)(3)(3)(3)(3)(3)A?A?A?0?A?A?0A?A1231313展开以上二式得，，，联立得。取

(3)(3)??2。即得 A1(3)?1，A3?1，可得到A2主振型矩阵为

A(3)??1?21?

T

?1?11??A??10?2???1??11?

图4-2

4-2 试计算图4-2所示系统对初始条件x0??0000?和x0??v00v?的响应。

解：在习题4-6中已求得系统的主振型矩阵和质量矩阵分别为

TTAp?A??1?A?2?A?3?A?4???11?1??1?11?2?11?2??A???12?1?1?1?2???1111???1?1???1???1?1 1 1?2?1?(1?2)?1 1 1?? 1?2??(1?2)??? 1??m000??0m00??M???00m0???000m?? ?1主质量振型为

000? ?4.000?0??0.41400T? MP?AMA?m??004.0000???00013.657??

i)A(N?1MiA(i),由此得到正则振型振型为

?0.65730.5000?0.2706??0.2706?0.50000.6533??0.2706?0.5000?0.6533??0.65330.50000.2706?

000??0??0?????100???0???0?010??0??0??????001??0??0?正则振型的第i列为

?0.5000?1?0.5000AN?m?0.5000??0.5000正则坐标初始条件为

?0.5000??0.6533TxN(0)?ANMx0?m??0.5000???0.27060.50000.50000.5000??1??0.27060.27060.6533???0?0.5000?0.50000.5000??0??0.6533?0.65330.2706??00.5000??1?0.6533???00.5000??0??0.2706??00.50000.5000?0.5000??0.6533?0.27060.2706TxN(0)?ANMx0?m??0.5000?0.5000?0.5000???0.27060.6533?0.6533000??v??1?????0?100??0??mv???1?010??0??????001??v??0? m?v0v0?

T?N(0)xN(0)?ATx0= 0，xNM??ATNMx0= xN3?正则坐标的响应为xN1?mvt，xN2?0，为p3?2km。 vmsinp3tp3，xN4?0其中频率

(2)(3)(1)(4)Ax?Ax?x?Ax?ANN2NN3NN1NxN4，展开得到 最终得到响应,由?x1??1??1??x?????v??1??2?vt?1??cosp3t???????x1?122p3?3???????x1???1??4? 1?v?(t?sinpt)3??2p3??1?v?(t?sinpt)3??2p3?1??2??3??4??x?ANxN1?ANxN2?ANxN3?ANxN4???v?1(t?sinpt)3??2p3???v?1(t?sinpt)?3?2p3??

解：从6—6中可得主频率和主振型矩阵为

p1?0 , p2?k?2?2?m,p3?2kk , p4?(2?2)mm

?m000???0m00?M???00m0???000m??，可求出主质量矩阵 由质量矩阵

000??1??0??02?20Mp?APTMAP?4m?0010????0002?2???

则正则振刑矩阵为

??(2?2)2?2?11??22????22?1?1??1?22?AN?2m?2?2??1?1?22???2?22?2?1?1??22?

1111????22?2???(2?2)?2m?2222?AN?1???1?1?11?2??2?22?22?2????2222?

T?1X0?AX?0000????N0于是 N

于是得

XN?0??ANX0?vm?1?0vm0?T

XN1?XN1?0?t?vtm

XN2?XN2?0?sinp2t?0p2 XN3?XN4?1XN3?0?vmsinp3t?sinp3tp3p3 XN4?0?sinp4t?0p4

234 所以响应为

X?AN??XN1?AN??XN2?AN??XN3?AN??XN4，