1 MDOF Systems Introduction: Why MDOF? General Form of Equation of Motion Undamped Free Vibrations Modal Frequencies and Mode Shapes Dynamic Response of MDOF Systems Normal Coordinates and Mode Superposition method Modal Participation; EQ Response; Vibration Mitigation 2006/2007 CE4258: Structural Stability magnitudes of the two masses are determined simply by statics For uniform mass distribution, each node will take half the element mass This leads to (a) diagonal mass matrix and (b) zero masses for rotational DOFs memd mf mcmbma ed f cba 10 2006/2007 CE4258: Structural Stability s i n)( 21 p(t) L/2 x 2m L/4 L/4 constant, EIm 2006/2007 CE4258: Structural Stability only the amplitude varies with time) and is a phase angle 22 2006/2007 CE4258: Structural Stability the mode with the next higher frequency is the 2nd mode, etc. 2006/2007 CE4258: Structural Stability 0 ; 0 ; 0 ; ; 0 ; 0 ; 0 ; ; T T T T T T T T T k K k k k k K k k k k K For example, a 3-DOF system with 3 mode shapes 1, 2, 3, where each is a 3 x 1 column vector. Then, 112 1 2 3 2 33 00 00 00 K kK K Alternatively, we write 28 2006/2007 CE4258: Structural Stability thus When the first orthogonality condition is applied to the RHS, it is clear that which shows that vibrating shapes are orthogonal with respect to the stiffness and mass matrices 2nnnku mu 2 TTnm n m nu k u u m u 0 T mnmnu k u 0 T mnmnu m u if Orthogonality Conditions thus 0 0 TTm n m nm k m n 30 2006/2007 CE4258: Structural Stability equations uncoupled 2006/2007 CE4258: Structural Stability equations uncoupled 2006/2007 CE4258: Structural Stability 2 4 . 2 r a d /s 0 . 9 1 3 0 . 4 0 2 0 . 0 6 8 7 7 . 7 u2u1 u3 m m m1 2 3 L 2L 2L 2006/2007 CE4258: Structural Stability thus higher modes of structure have very little damping. (So what?) Natural frequency, n Rayleigh damping Stiffness proportional damping (a0 = 0) Mass proportional damping (a1 = 0)Da mp ing rat io, n 1 12iia 02 i ia When damping is proportional to stiffness matrix, the damping ratio is directly proportional to frequency of vibration; thus higher modes will be very heavily damped. 2006/2007 CE4258: Structural Stability General response expression given by Duhamel integral for each mode is 01 s i nnnt tn n D nn D nY t P e t dM 6. MODAL FREE VIBRATIONS If initial conditions are not zero, the following free vibration response must be added to Duhamel integral for each mode: 2006/2007 CE4258: Structural Stability Other secondary response parameters such as stress or forces developed can be determined once the primary response has been found 2006/2007 CE4258: Structural Stability & Dynamics 92 Elastic forces fS which resist deformation of structure are given directly by Alternatively, ()Sf t k u t k Y t 1 2 31 2 3()Sf t k Y t k Y t k Y t 2 2 21 1 2 2 3 31 2 3()Sf t m Y t m Y t m Y t ()Sf t m Z t 211 222 Yt Z t Y t Since each modal contribution is multiplied by the square of modal frequency, it is evident that higher modes are of greater significance in evaluating forces than displacements. Consequently, it is necessary to include more modal components to define forces 46 2006/2007 CE4258: Structural Stability & Dynamics 93 Example Determine the dynamic response of a 3-storey shear building (previously discussed) subjected to a sine- pulse loading (kN). Neglect damping effects. 0 0.100.05-0.05-0.10 12 3 50 100 100 pt p t f t pt cos / pf t t t 0 .0 2 / 2 / 2p p pt s t t t 1000 kg 1500 kg 2000 kg 600 kN/m 1200 kN/m 1800 kN/m 3u 2u 1u 1pt 2pt 3pt 2006/2007 CE4258: Structural Stability & Dynamics 94 Solution Stiffness Matrix: k33 = 3000 k11 = 600 u1=1 k21 = -600 k31 = 0 k12 = 0 u3=1 k23 = -1200 k12 = -600 u2=1 k22 = 1800 k32 = -1200 5 1 1 0 1 3 2 6 1 0 / 0 2 5 k N m 3 100 0 1 .5 0 1 0 0 0 2 m 2 1 1 0 1 3 1 .5 2 0 2 5 2 km 2600 47 2006/2007 CE4258: Structural Stability & Dynamics 95 325 .5 7 .5 2 0 1 2 31 4 . 5 2 2 , 3 1 . 0 4 8 , 4 6 . 1 0 0 r a d / s Modal Periods (Descending Order): 1 2 30 . 4 3 3 s , 0 . 2 0 2 s , 0 . 1 3 6 sT T T Roots of Frequency Equation and Modal Frequencies: 1 2 30 . 3 5 1 5 , 1 . 6 0 6 6 , 3 . 5 4 2 0 Frequency Equation: 2 0km First Mode Shape (n=1): 1 0.3515 21 1 2 1 1 1 3 1 e q u a ti o n n o t u s e d s e t to 1 0 1 3 1 . 5 2 0 0 2 5 2 0 km 2 1 3 10 . 6 4 8 6 0 . 3 0 1 8 2006/2007 CE4258: Structural Stability & Dynamics 96 1.0000 0.6486 0.3018 2.4382 Mode 1 1=14.522 rad/s -0.6790 -0.6066 1.0000 -2.5405 1.0000 Mode 2 2=31.048 rad/s Mode 3 3=46.100 rad/s 1 1 1 0 .6 4 8 6 0 .6 0 6 6 2 .5 4 0 5 0 .3 0 1 8 0 .6 7 9 0 2 .4 3 8 2 Modal Mass (kg): Tn nnMm 2 2 31 1 1 . 5 0 . 6 4 8 6 2 0 . 3 0 1 8 1 0 1 8 1 3M 2 2 32 1 1 . 5 0 . 6 0 6 6 2 0 . 6 7 9 0 1 0 2 4 7 4M 2 2 33 1 1 . 5 2 . 5 4 0 5 2 2 . 4 3 8 2 1 0 2 2 5 7 1M 48 2006/2007 CE4258: Structural Stability & Dynamics 97 Modal Stiffness (kN/m) and Load (kN): Tn nnKk 1 2 33 8 2 . 4 , 2 3 8 5 , 4 8 0 2 0 K K K 12 3 50 100 100 pt p t p t f t k N pt Tn nP p t 1 145.0Pf 2 78.56Pf 3 39.78Pf Modal Equation of Motion (damping Cn = 0): 1 , 2 , 3n n n n nM Y K Y P t n Since pulse duration is short (tp/Tn < 0.25), free vibration solution can be used as an approximation. cos / pf t t t 0 .0 2 / 2 / 2p p pt s t t t 13/ 0 . 0 2 / 0 . 4 3 3 0 . 0 4 6 8 , / 0 . 0 2 / 0 . 1 3 6 0 . 1 4 7 ppt T t T 2006/2007 CE4258: Structural Stability & Dynamics 98 The initial velocity due to the pulse can be obtained from If the initial displacement = 0, then Normal coordinate solutions (m): 0.01 1 0.02 0.01 0.01 2 0.02 0.01 0.01 3 0.02 0.01 () ( 0) 14 5 , 00 0 ( 0) c os 1. 01 8 1813 78560 ( 0) c os 0. 40 42 2474 39780 ( 0) c os 0. 02 25 22571 n n n P t dtim pu ls e Y m ass M Y t dt Y t dt Y t dt 0 s in 0 c o sYY t t Y t 1 2 2 1.018 sin 14. 522 0.0 701 2 sin 14. 522 14.522 0.4042 sin 31. 048 0.0 130 2 sin 31. 048 31.048 0.0225 sin 46. 100 0.0 004 9 sin 46. 1 46.1 Y t t t Y t t t Y t t t 49 2006/2007 CE4258: Structural Stability & Dynamics 99 Physical solutions (m): 1 2 31 2 3u t Y Y Y 1 1 2 30 . 0 7 0 1 2 s i n 0 . 0 1 3 0 2 s i n 0 . 0 0 0 4 9 s i nu t t t 2 1 2 30 . 0 4 5 4 8 s i n 0 . 0 0 7 9 0 s i n 0 . 0 0 1 2 4 s i nu t t t 3 1 2 30 . 0 2 1 1 6 s i n 0 . 0 0 8 8 4 s i n 0 . 0 0 1 1 9 s i nu t t t 1 2 3 0 .0 7 0 1 2 sin 1 4 .5 2 2 0 .0 1 3 0 2 sin 3 1 .0 4 8 0 .0 0 0 4 9 sin 4 6 .1 0 0 Y t t Y t t Y t t 1 2 3 1 1 1 ( ) 0 . 6 4 8 6 ( ) 0 . 6 0 6 6 ( ) 2 . 5 4 0 5 ( ) 0 . 3 0 1 8 0 . 6 7 9 0 2 . 4 3 8 2 u t Y t Y t Y t 1 1 1 0 .6 4 8 6 0 .6 0 6 6 2 .5 4 0 5 0 .3 0 1 8 0 .6 7 9 0 2 .4 3 8 2 1 2 3 0 . 0 7 0 1 2 0 . 0 1 3 0 2 0 . 0 0 0 4 9 ( ) 0 . 0 4 5 4 8 s in 0 . 0 0 7 9 0 s in 0 . 0 0 1 2 4 s in 0 . 0 2 1 1 6 0 . 0 0 8 8 4 0 . 0 0 1 1 9 u t t t t 2006/2007 CE4258: Structural Stability & Dynamics 100 D i s p l a c e m e n t ( m ) - - A l l m o d e s i n c l u d e d -0 . 1 -0 . 0 5 0 0 . 0 5 0 . 1 0 0 . 5 1 1 . 5 2 Ti m e ( s ) u1 u2 u3 D i s p l a c e m e n t ( m ) - - F i r s t m o d e o n l y -0 . 1 -0 . 0 5 0 0 . 0 5 0 . 1 0 0 . 5 1 1 . 5 2 Ti m e ( s ) Displacement (m) first mode onlyu 1 u 2 u3 50 2006/2007 CE4258: Structural Stability & Dynamics 101 Question 4-3 2006/2007 CE4258: Structural Stability & Dynamics 102 Question 4-3 gu gu Fixe d Re fer enc e Figure Q3(a) Area = 1.6 m/s 200 ms Figure Q3(b) t 1u 2u 3u 51 2006/2007 CE4258: Structural Stability & Dynamics 103 Example For the same 3-storey shear building, an explicit damping matrix is to be defined such that the damping ratio in the first and third modes will be 5% critical. Assume Rayleigh damping. 1000 kg 1500 kg 2000 kg 600 kN/m 1200 kN/m 1800 kN/m 3u 2u 1u 2006/2007 CE4258: Structural Stability & Dynamics 104 Solution 1 bb b bbc m a m k c 212 bn b nn b a 1 11 0 3 13 3 1 1 12 a a For Rayleigh damping, let b = 0 and 1 in above equation: 0 1 1 1 4 .5 2 2 0 .0 5 1 1 4 .5 2 2 0 .0 5 12 4 6 .1 4 6 .1 a a 0 1 1 .1 0 40 .0 0 1 6 5aa 01 1 . 1 0 4 0 . 0 0 1 6 5 c a m a k m k 5 1 1 0 1 3 2 6 1 0 / 0 2 5 k N m 3 100 0 1 .5 0 1 0 0 0 2 m kg 1 2 31 4 . 5 2 2 , 3 1 . 0 4 8 , 4 6 . 1 0 0 r a d / s 52 2006/2007 CE4258: Structural Stability & Dynamics 105 2 0 9 4 9 9 0 0 9 9 0 4 6 2 6 1 9 8 0 / 0 1 9 8 0 7 1 5 8 c N s m Now, it is of interest to determine what damping ratio this matrix will yield in the second mode 022 12 1 . 1 0 41 1 1 1 3 1 . 0 4 8 0 . 0 0 1 6 52 2 3 1 . 0 4 8aa 2 0 .0 4 3 4 4 .3 4 % 1 1 0 4 0 0 9 9 0 9 9 0 0 0 1 6 5 6 0 9 9 0 2 9 7 0 1 9 8 0 0 0 2 2 0 8 0 1 9 8 0 4 9 5 0 c 2006/2007 CE4258: Structural Stability & Dynamics 106 Solution Natural frequency, n Rayleigh damping Da mp ing rat io, n 1 2 3 0.05 0.0434 53 2006/2007 CE4258: Structural Stability & Dynamics 107 2DOF structure under base acceleration: 11 1 1 2 2 1 1 2 2 2 2 2 22 00 100 1 g g g mum u k k k u m u m u k k u mmu e ff gm u c u k u p t m I u t Modal Participation Factors n-DOF structure: uY T T T T gm Y k Y p t m I u Giving n uncoupled eqns: ith equation of motion: T gM Y KY m I u ()01 ( ) s i n t Tii g iii iY t m I u t dMith modal response: () () 2 i Ti ii i ii i g Ti i i g i g ii M Y K Y m I u mI Y Y u uM 2006/2007 CE4258: Structural Stability & Dynamics 108 e ff gm u c u k u p t m I u t Modal Participation Factors, i n-DOF structure: ith modal response: ith equation of motion: ()2 i Ti i i g i gii mIY Y u u M () 0 1( ) ( ) sin Ti i ii t i i g i i mI M Y t u t d Total response: ( ) ( ) 11 0 0 () 1 0 1 ( ) ( ) ( ) sin 1 m a x ( ) m a x ( ) si 1 m n a x ( ) sin t g tnn ii i i g i iii n i i i t gi i i i u t Y t u t d u t d is e stim ate u t u t d re spo n se spe cd f rom tru m ith modal participation factor: 54 2006/2007 CE4258: Structural Stability & Dynamics 109 SDOF system under earthquake excitation ()01 ( ) s in ( ) t tgDDu t u e t d m u 2k 2k c 2006/2007 CE4258: Structural Stability & Dynamics 110 2 ( ) ( ) 11 0 m a x ( ) 1m a x ( ) sin nnii i i i i ii t i g i i u t o r u t d ()01 ( ) s in ( ) t tgDDu t u e t d Earthquake displacement response spectrumGet max u(t) for each SDOF system to plot response spectrum For MDOF system, estimate max u(t) from max of each mode using some combination rule, e.g. 55 2006/2007 CE4258: Structural Stability & Dynamics 111 Course Overview Overview and Basic Concepts SDOF Systems Free-Vibration Response Forced-Vibration Response MDOF Systems Introduction to Advanced Topics CE4258 Structural Stability and Dynamics End of Lectures