Outline of parts 1 and 2

Transcription

1 to Harmonic Loading Dipartimento di Ingegneria Civile Ambientale e Territoriale Politecnico di Milano March, 6 Outline of parts and of an Oscillator to Harmonic Load The Equation of Motion of an Oscillator from of a Oscillator to Harmonic Load The Equation of Motion for a Oscillator The Angle of Phase of Imaginary Argument Measuring Acceleration and Displacement The Accelerometre Measuring Displacements Outline of parts 3 and 4 Vibration Force Displacement Effectiveness Evaluation of damping Free vibration decay amplification Half Power Resonance Energy Loss

2 Part I of an Oscillator to Harmonic Load The Equation of Motion The SDOF equation of motion for a harmonic loading is: m ẍ + x = p sin ωt. The solution can be written, using superposition, as the free vibration solution plus a particular solution, ξ = ξ(t) EOM from x(t) = A sin ω n t + B cos ω n t + ξ(t) where ξ(t) satisfies the equation of motion, m ξ + ξ = p sin ωt. The Equation of Motion A particular solution can be written in terms of a harmonic function with the same circular frequency of the excitation, ω, whose second time derivative is ξ(t) = C sin ωt EOM from ξ(t) = ω C sin ωt. Substituting x and its derivative with ξ and simplifying the time dependency we get C ( ω m) = p.

3 Starting from our last equation, C ( ω m) = p, and introducing the frequency ratio β = ω/ω n : solving for C we get C = p ω m, collecting in the right member divisor: C = p ω m but /m = ω n, so that, with β = ω/ω n, we get: C = p β. We can now write the particular solution, with the dependencies on β singled out in the second factor: EOM from ξ(t) = p sin ωt. β The general integral for p(t) = p sin ωt is hence x(t) = A sin ω n t + B cos ω n t + p sin ωt. β Ratio and Factor Introducing the static deformation, st = p /, and the Ratio, R(t; β) the particular integral is ξ(t) = st R(t; β). The Ratio is eventually expressed in terms of the dynamic amplification factor D(β) = ( β ) as follows: EOM from R(t; β) = sin ωt = D(β) sin ωt. β 4 D(β) D(β) is stationary and almost equal to when ω << ω n (this is a quasi-static behaviour), it grows out of bound when β (resonance), it is negative for β > and goes to when ω >> ω n (high-frequency loading) β = ω/ω n 4 3 Factor, the plot D(β) EOM from β = ω/ω n

4 from Conditions Starting from rest conditions means that x() = ẋ() =. Let s start with x(t), then evaluate x() and finally equate this last expression to : x(t) = A sin ω n t + B cos ω n t + st D(β) sin ωt, x() = B =. We do as above for the velocity: EOM from ẋ(t) = ω n (A cos ω n t B sin ω n t) + st D(β) ω cos ωt, ẋ() = ω n A + ω st D(β) = A = st ω ω n D(β) = st βd(β) Substituting, A and B in x(t) above, collecting st and D(β) we have, for p(t) = p sin ωt, the response from rest: x(t) = st D(β) (sin ωt β sin ω n t). from Conditions, cont. Is it different when the load is p(t) = p cos ωt? You can easily show that, similar to the previous case, x(t) = x(t) = A sin ω n t + B cos ω n t + st D(β) cos ωt and, for a system starting from rest, the initial conditions are EOM from x() = B + st D(β) = ẋ() = A = giving A =, B = st D(β) that substituted in the general integral lead to x(t) = st D(β) (cos ωt cos ω n t). from Conditions We have seen that the response to harmonic loading with zero initial conditions is x(t; β) = st sin ωt β sin ω n t β. To determine resonant response, we compute the limit for β using the de l Hôpital rule (first, we write βω n in place of ω, finally we substitute ω n = ω as β = ): EOM from lim x(t; β) = lim (sin βω n t β sin ω n t)/ β st β β ( β )/ β = st (sin ωt ωt cos ωt). As you can see, there is a term in quadrature with the loading, whose amplitude grows linearly and without bounds.

5 , the plot x(t) / Δ st ±sqrt[(π α) +] EOM from α = ω t / π x(t) = sin ωt ωt cos ωt = sin πα πα cos πα. st note that the amplitude A of the normalized envelope, with respect to the normalized abscissa α = ωt/π, is A = + (πα) for large α πα, as the two components of the response are in quadrature. home wor Derive the expression for the resonant response with p(t) = p cos ωt, ω = ω n. EOM from Part II of the Oscillator to Harmonic Loading

6 The Equation of Motion for a Oscillator The SDOF equation of motion for a harmonic loading is: m ẍ + c ẋ + x = p sin ωt. A particular solution to this equation is a harmonic function not in phase with the input: x(t) = G sin(ωt θ); it is however equivalent and convenient to write : EOM Particular ξ(t) = G sin ωt + G cos ωt, where we have simply a different formulation, no more in terms of amplitude and phase but in terms of the amplitudes of two harmonics in quadrature, as in any case the particular integral depends on two free parameters. Substituting x(t) with ξ(t), dividing by m it is ξ(t) + ζω n ξ(t) + ω nξ(t) = p m sin ωt, Substituting the most general expressions for the particular integral and its time derivatives ξ(t) = G sin ωt + G cos ωt, ξ(t) = ω (G cosωt G sin ωt), ξ(t) = ω (G sin ωt + G cos ωt). EOM Particular in the above equation it is ω (G sin ωt + G cos ωt) + ζω n ω (G cosωt G sin ωt)+ +ω n(g sin ωt + G cos ωt) = st ω n sin ωt The particular integral, Dividing our last equation by ω n and collecting sin ωt and cos ωt we obtain ( G ( β ) G βζ ) sin ωt+ + ( G βζ + G ( β ) ) cos ωt = st sin ωt. Evaluating the eq. above for t = π ω and t = we obtain a linear system of two equations in G and G : G ( β ) G ζβ = st. G ζβ + G ( β ) =. EOM Particular The determinant of the linear system is and its solution is det = ( β ) + (ζβ) ( β ) ζβ G = + st, G = st det det.

7 ι, 3 Substituting G and G in our expression of the particular integral it is ξ(t) = st ( ( β ) sin ωt βζ cos ωt ). det The general integral for p(t) = p sin ωt is hence x(t) = exp( ζω n t) (A sinω D t + B cos ω D t) + + st ( β ) sin ωt βζ cos ωt det For p(t) = p sin sin ωt + p cos cos ωt, sin = p sin /, cos = p cos / it is EOM Particular x(t) = exp( ζω n t) (A sinω D t + B cos ω D t) + ( β ) sin ωt βζ cos ωt + sin + det ( β ) cos ωt + βζ sin ωt + cos. det Examination of the general integral x(t) = exp( ζω n t) (A sinω D t + B cos ω D t) + + st ( β ) sin ωt βζ cos ωt det shows that we have a transient response, that depends on the initial conditions and damps out for large values of the argument of the real exponential, and a so called steady-state response, corresponding to the particular integral, x s-s (t) ξ(t), that remains constant in amplitude and phase as long as the external loading is being applied. From an engineering point of view, we have a specific interest in the steady-state response, as it is the long term component of the response. EOM Particular The Angle of Phase To write the stationary response in terms of a dynamic amplification factor, it is convenient to reintroduce the amplitude and the phase difference θ and write: ξ(t) = st R(t; β, ζ), R = D(β, ζ) sin (ωt θ). Let s start analyzing the phase difference θ(β, ζ). Its expression is: θ(β, ζ) = arctan ζβ β. EOM Particular π π/ θ(β;ζ=.) θ(β;ζ=.) θ(β;ζ=.5) θ(β;ζ=.) θ(β;ζ=.7) θ(β;ζ=.).5.5 β θ(β, ζ) has a sharper variation around β = for decreasing values of ζ, but it is apparent that, in the case of slightly damped structures, the response is approximately in phase for low frequencies of excitation, and in opposition for high frequencies. It is worth mentioning that for β = we have that the response is in perfect quadrature with the load: this is very important to detect resonant response in dynamic tests of structures.

8 Ratio The dynamic magnification factor, D = D(β, ζ), is the amplitude of the stationary response normalized with respect to st : ( β D(β, ζ) = ) + (βζ) ( β ) + (βζ) = ( β ) + (βζ) D(β,ζ=.) D(β,ζ=.) D(β,ζ=.5) D(β,ζ=.) D(β,ζ=.7) D(β,ζ=.) D(β, ζ) has larger pea values for decreasing values of ζ, the approximate value of the pea, very good for a slightly damped structure, is /ζ, for larger damping, peas move toward the origin, until for ζ = the pea is in the origin, for dampings ζ > we have no peas. EOM Particular Ratio () The location of the response pea is given by the equation the 3 roots are d D(β, ζ) d β =, β 3 + β β = β i =, ± ζ. We are interested in a real, positive root, so we are restricted to < ζ. In this interval, substituting β = ζ in the expression of the response ratio, we have EOM Particular D max = ζ. ζ For ζ = there is a maximum corresponding to β =. Note that, for a relatively large damping ratio, ζ = %, the error of /ζ with respect to D max is in the order of %. Harmonic Consider the EOM for a load modulated by an exponential of imaginary argument: ẍ + ζω n ẋ + ω nx = st ω n exp(i(ωt φ)). Note that the phase can be disregarded as we can represent its effects with a constant factor, as it is exp(i(ωt φ)) = exp(iωt)/ exp(iφ). The particular solution and its derivatives are ξ = G exp(iωt), ξ = iωg exp(iωt), ξ = ω G exp(iωt), where G is a complex constant. Substituting, dividing by ω n, removing the dependency on exp(iωt) and solving for G yields [ ] [ ( β ] ) i(ζβ) G = st ( β = st ) + i(ζβ) ( β ) + (ζβ). Note how simpler it is to represent the stationary response of a damped using the complex exponential representation. EOM Particular

9 in the Complex Plane Im[G exp(i ωt)] cost*(zb) Im G exp(i ωt) cost*( b^) G ω t Re[G exp(i ωt)] Re The stationary response is ξ(t) = st ( β ) i(ζβ) ( β ) + (ζβ) exp(iωt) we plot G in the complex plane, we multiply G by exp(iωt), that is equivalent to rotate G by the angle ωt, projecting the resulting vector on the axes, we have the real and imaginary part of the response, these two vectors are rotated 9 degrees with respect to the response to the real harmonic load, p sin ωt that we have studied, what if p(t) = p cos ωt? EOM Particular Measuring Support Accelerations We have seen that in seismic analysis the loading is proportional to the ground acceleration. A simple, when properly damped, may serve the scope of measuring support accelerations. The Accelerometre Measuring Displacements Measuring Support Accelerations, With the equation of motion valid for a harmonic support acceleration: ẍ + ζβω n ẋ + ω nx = a g sin ωt, the stationary response is ξ = m a g D(β, ζ) sin(ωt θ). If the damping ratio of the is ζ.7, then the D(β) for. < β <.6! Oscillator s displacements will be proportional to the accelerations of the support for applied frequencies up to about six-tenths of the natural frequency of the instrument. As it is possible to record the displacements by means of electro-mechanical or electronic devices, it is hence possible to measure, within an almost constant scale factor, the ground accelerations component up to a frequency of the order of 6% of the natural frequency of the. The Accelerometre Measuring Displacements

10 Measuring Displacements Consider now a harmonic displacement of the support, u g (t) = u g sin ωt. The support acceleration (disregarding the sign) is a g (t) = ω u g sin ωt. With the equation of motion: ẍ + ζβω n ẋ + ω nx = ω u g sin ωt, the stationary response is ξ = u g β D(β, ζ) sin(ωt θ). Let s see a graph of the dynamic amplification factor derived above. We see that the displacement of the instrument is approximately equal to the support displacement for all the excitation frequencies greater than the natural frequency of the instrument, for a damping ratio ζ β D(β,ζ=.) β D(β,ζ=/6) β D(β,ζ=/4) β D(β,ζ=/) β D(β,ζ=.) The Accelerometre Measuring Displacements It is possible to measure the support displacement measuring the deflection of the, within an almost constant scale factor, for excitation frequencies larger than ω n. Vibration Part III Vibration Vibration Vibration isolation is a subject too broad to be treated in detail, we ll present the basic principles involved in two problems,. prevention of harmful vibrations in supporting structures due to y forces produced by operating equipment,. prevention of harmful vibrations in sensitive instruments due to vibrations of their supporting structures. Vibration Force Displacement Effectiveness

11 Force Consider a rotating machine that produces an y force p sin ωt due to unbalance in its rotating part, that has a total mass m and is mounted on a spring-damper support. Its steady-state relative displacement is given by x s-s = p D sin(ωt θ). Vibration Force Displacement Effectiveness This result depend on the assumption that the supporting structure deflections are negligible respect to the relative system motion. The steady-state spring and damper forces are f S = x ss = p D sin(ωt θ), f D = c ẋ ss = cp D ω cos(ωt θ) = ζ β p D cos(ωt θ). Transmitted force The spring and damper forces are in quadrature, so the amplitude of the steady-state reaction force is given by f max = p D + (ζβ) Vibration Force Displacement Effectiveness The ratio of the maximum transmitted force to the amplitude of the applied force is the transmissibility ratio (TR), TR = f max p = D + (ζβ) TR(β,ζ=.) TR(β,ζ=/5) TR(β,ζ=/4) TR(β,ζ=/3) TR(β,ζ=/) TR(β,ζ=.) For β <, TR is always greater than : the transmitted force is amplified.. For β >, TR is always smaller than and for the same β TR decreases with ζ. Displacement Another problem concerns the harmonic support motion u g (t) = u g exp iωt forcing a steady-state relative displacement of some supported (spring+damper) equipment of mass m (using exp notation) x ss = u g β D exp iωt, and the mass total displacement is given by ( β ) x tot = x s-s + u g (t) = u g ( β ) + i ζβ + exp iωt = u g ( + iζβ) ( β exp iωt ) + i ζβ = u g + (ζβ) D exp i (ωt ϕ). If we define the transmissibility ratio TR as the ratio of the maximum total response to the support displacement amplitude, we find that, as in the previous case, TR = D + (ζβ). Vibration Force Displacement Effectiveness

12 Effectiveness Define the isolation effectiveness, IE = TR, IE= means complete isolation, i.e., β =, while IE= is no isolation, and taes place for β =. As effective isolation requires low damping, we can approximate TR /(β ), in which case we have IE = (β )/(β ). Solving for β, we have β = ( IE)/( IE), but Vibration Force Displacement Effectiveness β = ω /ω n = ω (m/) = ω (W /g) = ω ( st /g) where W is the weight of the mass and st is the static deflection under self weight. Finally, from ω = π f we have f = g IE π st IE Effectiveness () The strange looing f = g IE π st IE 5 can be plotted f vs st for different values of IE, obtaining a design 5 chart st [cm] Knowing the frequency of excitation and the required level of vibration isolation efficiency (IE), one can determine the minimum static deflection (proportional to the spring flexibility) required to achieve the required IE. It is apparent that any isolation system must be very flexible to be effective. Input frequency [Hz] IE=. IE=.5 IE=.6 IE=.7 IE=.8 IE=.9 IE=.95 IE=.98 IE=.99 Vibration Force Displacement Effectiveness Part IV Evaluation of damping Evaluation of Viscous Damping Ratio

13 Evaluation of damping The mass and stiffness of phisycal systems of interest are usually evaluated easily, but this is not feasible for damping, as the energy is dissipated by different mechanisms, some one not fully understood... it is even possible that dissipation cannot be described in term of viscous-damping, But it generally is possible to measure an equivalent viscous-damping ratio by experimental methods: free-vibration decay method, resonant amplification method, half-power (bandwidth) method, resonance cyclic energy loss method. Evaluation of damping Free vibration decay amplification Half Power Resonance Energy Loss Free vibration decay We already have discussed the free-vibration decay method, ζ = δ m π m (ω n /ω D ) Evaluation of damping Free vibration decay amplification Half Power Resonance Energy Loss with δ m = ln x n x n+m, logarithmic decrement. The method is simple and its requirements are minimal, but some care must be taen in the interpretation of free-vibration tests, because the damping ratio decreases with decreasing amplitudes of the response, meaning that for a very small amplitude of the motion the effective values of the damping can be underestimated. amplification This method assumes that it is possible to measure the stiffness of the structure, and that damping is small. The experimenter (a) measures the steady-state response x ss of a SDOF system under a harmonic loading for a number of different excitation frequencies (eventually using a smaller frequency step when close to the resonance), (b) finds the maximum value D max = max{x ss} st of the dynamic magnification factor, (c) uses the approximate expression (good for small ζ) D max = ζ to write Evaluation of damping Free vibration decay amplification Half Power Resonance Energy Loss and finally (d) has D max = ζ = max{x ss} st ζ = st max{x ss }. The most problematic aspect here is getting a good estimate of st, if the results of a static test aren t available.

14 Half Power The adimensional frequencies where the response is / times the pea value can be computed from the equation ( β ) + (βζ) = ζ ζ Evaluation of damping Free vibration decay amplification Half Power Resonance Energy Loss squaring both sides and solving for β gives β, = ζ ζ ζ For small ζ we can use the binomial approximation and write β, = ( ζ ζ ζ ) ζ ζ ζ Half power () From the approximate expressions for the difference of the half power frequency ratios, and their sum β β = ζ ζ ζ β + β = ( ζ ) Evaluation of damping Free vibration decay amplification Half Power Resonance Energy Loss we can deduce that β β = f f ζ ζ β + β f + f ( ζ ) ζ, or ζ f f f + f where f, f are the frequencies at which the steady state amplitudes equal / times the pea value, frequencies that can be determined from a dynamic test where detailed test data is available. Resonance Cyclic Energy Loss If it is possible to determine the phase of the s-s response, it is possible to measure ζ from the amplitude ρ of the resonant response. At resonance, the deflections and accelerations are in quadrature with the excitation, so that the external force is equilibrated only by the viscous force, as both elastic and inertial forces are also in quadrature with the excitation. The equation of dynamic equilibrium is hence: Evaluation of damping Free vibration decay amplification Half Power Resonance Energy Loss Solving for ζ we obtain: p = c ẋ = ζω n m (ω n ρ). ζ = p mω nρ.