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Mechanical Vibrations v i b r a t i o n s . v i b r a t i o n s . Single degree of freedom system with viscous damping: System free body diagram C is the damping constant or coefficient v Mathematical model (governing equation of motion): .. . i m x c x kx 0 b . r c x Fd is thedampingforce. a It is proportion al to velocity t i Solution: o Assume x(t) = B est, substitute into the equation of motion: 2 n c c s1, 2 s m s2 cs k 0 4m k 2m . s1, 2 c 2m 2 k c m 2m v i b r a t i o n s . DefineC 2mn , then therootsof thesolutioncan be writtenas : s1, 2 2 1 n , where is known as thedam ping ratio. C , where Cc is thecritical dam ping coe fficient Cc Cc 2mn . In theface of the value of thesquare rootshown above threepossible solutionsare exist.T hesesolutionsare as follows: v i b r a t i o n s Solution # 1: under-damped vibration: When ζ<1 The roots S1,2 of the characteristic equation can now be written as: s i 1 2 1, 2 n T hesolut ionfor t hiscase becomes: xt e nt B1 cos 1 2 nt B2 sin 1 2 nt OR . xt Xent cos 1 2 nt Defined 1 2 nt as t hedam ped nat ural frequency, t hen xt Ae nt cosd t v i b r a t i o n s The solution (above) for the under-damped vibration case (<1 consists of a harmonic motion of frequency d and an amplitude n t Ae Note that: B2 A B B and t an B1 is t hephase angle. A and can be det erminedfrom t heinat ial 2 1 2 2 1 condit ionsof t hemotion. . T heamplit defor t hiscase is exponant ially decaying wit h t imeas shown. v i b r a t i o n s . Frequency of damped vibration: d 1 2 n Under-damped vibration v i b r a t i o n s . Critical damping (cc) The critical damping cc is defined as the value of the damping constant c for which the radical in s-equation becomes zero: 2 k k cc 0 cc 2m 2mn m m 2m v i b r a t i o n s . Solution # 2: critically damped vibration (ζ=1): For this case the roots of the characteristic equation become: s1, 2 n T herefor he t solutioncan be writtenas : xt e nt B1 B2t B1 and B2 are constantsthatcan be determinedfrominatial conditions. T hemotionis no longer harmonicas shown in the figure. v i b r a t i o n s . -The system returns to the equilibrium position in short time -The shape of the curve depends on initial conditions as shown -The moving parts of many electrical meters and instruments are critically damped to avoid overshoot and oscillations . v i b r a t i o n s Solution # 3: over-damped vibration (ζ>1) For this case the roots of the characteristic equation become: s1, 2 2 1 n T herefore,thesolutioncan be writtenas, xt B1e 2 1 n t B2 e 2 1 n t B1 and B2 are constantsto be determinedfrom knowing theinitialconditionsof themotion. . v i b r a t i o n s . Graphical representation of the motions of the damped systems v i b r a t i o n s . Logarithmic decrement The logarithmic decrement represents the rate at which the amplitude of a free-damped vibration decreases. It is defined as the natural logarithm of the ratio of any two successive amplitudes. v i b r a t i o n s . Logarithmic decrement: x1 (t ) Aent1 cosd t1 nt2 x2 (t ) Ae cosd t2 So t2 t1 d d But x1 (t ) ent1 n t1 d en d x2 (t ) e d cosd t2 cos2 d t1 cosd t1 Assume: δ is the logarithmic decrement x1 (t ) 2 2 ln n d n 2 x2 (t ) 1 n 1 2 Logarithmic decrement: is dimensionless 2 v i b r a t i o n s . Logarithmic decrement 2 2 2 For small damping ; ζ <<1 2 v i b r a t i o n s . Generally, when the amplitude after a number of cycles “n” is known, logarithmic decrement can be obtained as follows: 1 n ln x1 (t ) 2 xn (t ) 1 2 where x1 (t)is thefirst known amplitude,xn (t)is theotherknown amplitudeaftern cyclesof thedecayedmotion. v i b r a t i o n s . Example 2.6 An under-damped shock absorber is to be designed for a motorcycle of mass 200 kg (Fig.(a)).When the shock absorber is subjected to an initial vertical velocity due to a road bump, the resulting displacement-time curve is to be as indicated in Fig.(b). Requirements: 1. Find the necessary stiffness and damping constants of the shock absorber if the damped period of vibration is to be 2s and the amplitude x1 is to be reduced to one-fourth in one half cycle (i.e. x1.5 = x1/4 ). v i b r a t i o n s . Example 2.6 Note that this system is under-damped system v i b r a t i o n s Solution: Finding k and c Here, n 0.5 and 1 n ln x1 1 2 ln4 2.7726 x n 0 .5 1 2 0.4037 d 2 . x1 4, t hen t helogarit hmic decrementis : xn 2 d 2 n 1 2 n 2 n 1 0.4037 2 3.4338rad / s v The critical damping can be found as: i b cc 2mn 22003.4338 1373.54 N.sec/m r a t Damping coefficient C and stiffness K can be found as: i o c c 0.40371373.54 554.4981N.sec/m.......(Ans.) c n s k mn2 2003.4338 2358.2652N/m........(Ans.) 2 . S t a t i c s . End of chapter