Rigid-body motion and modal decomposition When structure is not anchored in space, some of the vibration modes represent rigid body motion What is the frequency? What does orthogonality mean? Expanding response in terms of vibration modes What decides modal amplitudes7.9: Systems admitting rigid-body motions Not all systems are fully constrained Potential and kinetic energies( )( ) ( )2 2 21 1 2 2 3 32 21 2 1 2 3 21 1 12 1 1 2 23 2 212120 0 00 0 ,0 0 0T I I IV k kI k kM I K k k k kI k k = + + = + = = +

Ki=GJi/LiOrthogonality to rigid-body modes Eigenvalue problem The lowest eigenvalue is zero corresponding to rigid body motion For other eignevectors we get zero change in angular momentum. For a beam vibrating in space, what does orthogonolity with respect to rigid body modes implied?2K M = 0 00[1 1 1] 00TK = = =01 1 2 2 3 31 1 2 2 3 3000TMI I II I I =+ + =+ + =

7.9: Modal expansion of response Analogous to Fourier series it is advantageous to express general motion in terms of vibration modes From orthogonality conditionswe get1nr rrU c== =u c u2 ,,Ts r rsTs r s rsMK ==u uu u2,T Ts s s s s rT Tc M c KU M U K = == =u u u uc u c u7.10: Response to initial conditions by modal analysis Original problem In modal coordinates From Section 7.5 with diagonal K and Mgiving decoupled equations For initial conditions use( ) ( ) 0 (0), (0)given Mq t Kq t + = q q1( ) ( )nr rrt t ==q u0' ' 0' , 'T TMU KUM KM U MUK U KU + =+ == =

2( ) ( ) 0r rt t + =

(0) (0), (0) (0)r r r r r rM M M = = = u q u q u q

T T TSpring example Example Find response to unit initial displacement for mass #31 0 0 2 1 00 1 0 , 1 6 10 0 2 0 1 3Mm K k = = Non-dimensionalization Original equations We want to solve without choosing kand m Done by defining non-dimensional time1 12 23 31 0 0 2 1 00 1 0 1 6 1 00 0 2 0 1 3x xkx xmx x + =

22' "ktmdx k k k d x kx x x xd m m m d m == = = =1 12 23 31 0 0 " 2 1 00 1 0 " 1 6 1 00 0 2 " 0 1 3x xx xx x + = Initial modal conditions from MatlabK=[2,-1,0;-1,6,-1;0,-1,3]; M=[1,0,0;0,1,0;0,0,2]; q0=[0,0,1]; [V,D]=eig(K,M);V =0.3160 -0.9223 -0.22260.2083 -0.16140.96470.65450.2484 -0.0998D =1.3409 0 001.8250 00 06.3342V'*M*q0'1.30900.4967-0.19951 1 12 2 23 3 31.3409 0, (0) 1.3091.8250 0, (0) 0.49676.3342 0, (0) 0.1995 + = =+ = =+ = =

Modal equations:Solution1231.309cos1.1580.4967cos1.3510.1995cos2.517 === Checks?Retrieving original coordinatesV =0.3160 -0.9223 -0.22260.2083 -0.16140.96470.65450.2484 -0.09980.3160 0.92231.309 0.2083 cos1.158 0.4967 0.1614 cos1.3510.6545 0.24840.22260.1995 0.9647 cos2.5170.0998 = + xReading assignmentSections 7.11-13Source: www.library.veryhelpful.co.uk/ Page11.htm