1振动模态分析的基本理论 1.2复模态理论 cm1.「k0「f (8) h Ay+ By= p 0/4 c m P 0 Shanghai Jiaotong University -Engineering Mechanics Experimental Ceneter
Shanghai Jiaotong University - Engineering Mechanics Experimental Ceneter 11 1.振动模态分析的基本理论 1.2 复模态理论 (8) 0 0 0 0 = − + f y m k y m c m Ay + By = P (9) = − = = 0 , 0 0 , 0 f P m k B m c m A
1振动模态分析的基本理论 1.2复模态理论 的y+By=P 由 x (3) y 得 y=Pes_/Y st (10) Shanghai Jiaotong University -Engineering Mechanics Experimental Ceneter 12
Shanghai Jiaotong University - Engineering Mechanics Experimental Ceneter 12 1.振动模态分析的基本理论 1.2 复模态理论 (3) st x = Xe Ay + By = P (9) (7) = x x y (10) s t s t e sX X y Ye = = 由 得
1振动模态分析的基本理论 1.2复模态理论 把式(10)代入式(9),得特征问题 (As+B)Y=0(11) 其特征方程 (ms2+cs+k)X=0(4) ms2+Cs+k=0 As+B=0(12) 由于式(11)和式(4)是解决的同一系统的自由振动,所以式(12)和式(5 )具有相同的特征根 S,和s(r=1,2,N) Shanghai Jiaotong University -Engineering Mechanics Experimental Ceneter 13
Shanghai Jiaotong University - Engineering Mechanics Experimental Ceneter 13 1.振动模态分析的基本理论 1.2 复模态理论 ( ) 0 (4) 2 ms + cs + k X = (As + B)Y = 0 (11) 把式（10）代入式（9），得特征问题 其特征方程 As + B = 0 (12) 由于式（11）和式（4）是解决的同一系统的自由振动，所以式（12）和式（5 ）具有相同的特征根 0 (5) 2 ms + cs + k = r s * r 和 s (r =1,2,...N)
1振动模态分析的基本理论 1.2复模态理论 其特征向量 (13) 式中为N阶对角阵,主元分别为 NS1,S22… DN Wyiy2-ynl Shanghai Jiaotong University -Engineering Mechanics Experimental Ceneter 14
Shanghai Jiaotong University - Engineering Mechanics Experimental Ceneter 14 1.振动模态分析的基本理论 1.2 复模态理论 = r r r r s ( ) ( ) ) ~ ( 其特征向量 式中 = r r r r s ( ) ( ) ) ~ ( * * * * (13) ` ` * * * = r r s s r ` s * ` r s 为N阶对角阵，主元分别为 N s ,s ,...,s 1 2 * * 2 * 1 , ,..., N s s s N ... = 1 2 * * 2 * 1 * ... = N
1振动模态分析的基本理论 1.2复模态理论 若m,c,k均为对称阵,这A,B也是对称阵,可证明模态向量分别对于AB具有正 交性。所以 A =diag o VB=diag 0 b (14 Shanghai Jiaotong University -Engineering Mechanics Experimental Ceneter 15
Shanghai Jiaotong University - Engineering Mechanics Experimental Ceneter 15 1.振动模态分析的基本理论 1.2 复模态理论 若m,c,k均为对称阵，这A,B也是对称阵，可证明模态向量分别对于A,B具有正 交性。所以 (14) 0 ` ` 0 ~ ~ 0 ` ` 0 ~ ~ * * = = r T r r T r b b B diag a a A diag