Lecture02-Prof.Ju

Lecture02-Prof.Ju - CEE M237A / MAE M269A Lecture 2:...

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CEE M237A / MAE M269A Lecture 2: Dynamic SDOF Response to Harmonic Loading Professor J. Woody Ju
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Harmonically Forced Vibration 2 () 0 0 = frequency of the f sin si orcin n Note: g fu 1 nction cr pt P t mv cv kv P t c c k m ω ξ = ++= =< = ±± ± Loading: E.O.M.: (Damping ratio) (Natural frequency)
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Undamped System 3 () where sin cos homog. complementary solution sin particular solution c p vt A t B t G t ωω ω =+ = 3 3 () () () 0 0, 0 sin cp mv kv P c t v t ξ == = + + = ±± E.O.M.: Displ.:
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Undamped System (Cont’d) y Particular solution: 4 2 0 00 2 2 frequency ratio sin sin sin 11 1 1 where mG tk G tP t PP G m kk k ω β ωω ⇒− + = ⎛⎞ ⎜⎟ == ⎝⎠ = ±
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Undamped System (Cont’d) y Complete solution: 5 () 0 0 0 2 2 2 1 cos sin cos 1 sin cos sin 1 To find , , use I.C 1 (0) ( .'s (at the t 0) : 1 ime 0) P vt P v A tB t t k tA t B t Bv t k AB t P v A k ω ωω β ⎛⎞ =+ + ⇒= + ⎜⎟ ⎝⎠ = ⎡⎤ =− ⎢⎥ = ± ±
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Undamped System (Cont’d) y Assume that the system is quiescent (at-rest) at t=0 6 () 0 2 1 sin sin 1 P vt t t k ω βω β ⎛⎞ = 0 0 2 : static displacement 1 M.F. magnification factor 1 response ratio st st P k P Note v k Rt v == ± ± ±
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Damped System (Underdamped) 7 () () () () ( ) () 2 0 12 22 0 1 21 2 out of phase ( damping) 2 sin , 2 sin cos sin cos 2s i n + 2 cos 0 cr t cD D p P cc vt t mm c e A t B t G tG t P GG G t m G t ξω ω ξ ωω ++ = = = =+ ⎡⎤ ⇒− + ⎢⎥ ⎣⎦ −+ + = ±± ± orthogonal E.O.M.:
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Damped System (Underdamped) 8 () 2 1 2 0 12 1 22 21 2 2 0 1 2 2 0 2 2 2 2 2 2 solve for and 20 2 0 1 2 Satisfy all P GG G m G P G k P G t G k G ωω ξ ω β βξ ξβ ⎧⎫ ⎡⎤ −− + = ⎪⎪ ⎢⎥ ⎣⎦ ⎨⎬ −+ + = ⎩⎭ ⎛⎞ = ⎜⎟ ⎝⎠ =
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Lecture02-Prof.Ju - CEE M237A / MAE M269A Lecture 2:...

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