Generalized Single Degree of Freedom Systems

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Single Degree of Freedom http://intranet.dica.polimi.it/people/boffi-giacomo Dipartimento di Ingegneria Civile Ambientale e Territoriale Politecnico di Milano March, 8 Outline Until now our were described as composed by a single mass connected to a fixed reference by means of a spring and a damper. While the mass-spring is a useful representation, many different, more complex systems can be studied as SDOF systems, either exactly or under some simplifying assumption.. SDOF rigid body assemblages, where the flexibility is concentrated in a number of springs and dampers, can be studied, e.g., using the Principle of Virtual Displacements and the D Alembert Principle.. simple structural systems can be studied, in an approximate manner, assuming a fixed pattern of displacements, whose amplitude (the single degree of freedom) varies with time.

Further on Rigid Assemblages Today we restrict our consideration to plane, -D systems. In rigid body assemblages the limitation to a single shape of displacement is a consequence of the configuration of the system, i.e., the disposition of supports and internal hinges. When the equation of motion is written in terms of a single parameter and its time derivatives, the terms that figure as coefficients in the equation of motion can be regarded as the generalised properties of the assemblage: generalised mass, damping and stiffness on left hand, generalised loading on right hand. m ẍ + c ẋ + k x = p (t) Further on systems have an infinite variety of deformation patterns. By restricting the deformation to a single shape of varying amplitude, we introduce an infinity of internal contstraints that limit the infinite variety of deformation patterns, but under this assumption the system configuration is mathematically described by a single parameter, so that our model can be analysed in exactly the same way as a strict SDOF system, we can compute the generalised mass, damping, stiffness properties of the SDOF model of the continuous system. Final on Generalised SDOF From the previous comments, it should be apparent that everything we have seen regarding the behaviour and the integration of the equation of motion of proper SDOF systems applies to rigid body assemblages and to SDOF models of flexible systems, provided that we have the means for determining the generalised properties of the dynamical systems under investigation.

Assemblages of planar, or bidimensional, rigid bodies, constrained to move in a plane, the flexibility is concentrated in discrete elements, springs and dampers, rigid bodies are connected to a fixed reference and to each other by means of springs, dampers and smooth, bilateral constraints (read hinges, double pendulums and rollers), inertial forces are distributed forces, acting on each material point of each rigid body, their resultant can be described by a force applied to the centre of mass of the body, proportional to acceleration vector (of the centre of mass itself) and total mass M = dm a couple, proportional to angular acceleration and the moment of inertia J of the rigid body, J = (x + y )dm. Rigid Bar Unit mass Centre of Mass Length L, Total Mass Moment of Inertia L G m = constant, x G = L/, m = ml, J = m L = ml x Rigid Rectangle Unit mass Sides y γ = constant, a, b Centre of Mass x G = a/, y G = b/ Total Mass m = γab, Moment of Inertia J = m a + b a G = γ a b + ab b

Rigid Triangle For a right triangle. Unit mass Sides y γ = constant, a, b Centre of Mass x G = a/, y G = b/ Total Mass m = γab/, Moment of Inertia J = m a + b 8 G a = γ a b + ab 6 b Rigid Oval When a = b = D = R the oval is a circle. y Unit mass Axes b a γ = constant, a, b Centre of Mass x G = y G = Total Mass m = γ πab, Moment of Inertia J = m a + b 6 x trabacolo p(x,t) = P x/a f(t) c k c k m, J a a a a a a The mass of the left bar is m = m a and its moment of inertia is (a) J = m = a m /. The maximum value of the external load is P max = P a/a = P and the resultant of triangular load is R = P a/ = 8P a N

Forces and Virtual Displacements J Z a 8P a f(t) Z(t) J Z a N cż m Z δθ = /(a) kz c Ż m Z δθ = /(a) kz δu u = 7a a cos θ a cos θ, δu = a sin θ δθ + a sin θ δθ δθ = /(a), δθ = /(a) sin θ Z/(a), sin θ Z/(a) δu = ( a + ) a Z = 7 a Z Principle of Virtual Displacements cż J Z a m Z 8P a f(t) δθ = /(a) kz cż Z(t) J Z a m Z kz δθ = /(a) N The virtual work of the InertialDampingElasticExternal forces: Z δw I = m J Z a = a m Z ) ( m + m 9 + J 6a + J 9a Ż δw D = c δw S = k Z δu J Z a Z a c Z = (c + c /6) Ż ( k Z = 9k 6 + k ) Z 9 δw Ext = 8P a f(t) + N 7 a Z _ Principle of Virtual Displacements For a rigid body in condition of equilibrium the total virtual work must be equal to zero δw I + δw D + δw S + δw Ext = Substituting our expressions of the virtual work contributions and simplifying, the equation of equilibrium is ( m + m 9 + J 6a + J ) Z+ 9a ( + (c + c /6) Ż + 9k 6 + k 9 ) Z = 8P a f(t) + N 7 a Z

Principle of Virtual Displacements Collecting Z and its time derivatives give us m Z + c Ż + k Z = p f(t) introducing the so called generalised properties, in our example it is m = m + 9 9m + 6a J + 9a J, c = 6 c + c, k = 9 6 k + 9 k 7 a N, p = 6 P a. It is worth writing down the expression of k : k = 9k 6 + k 9 7 a N Geometrical stiffness Let s start with an example... Consider a cantilever, with varying properties m and EJ, subjected to a load that is function of both time t and position x, p = p(x, t). The transverse displacements v will be function of time and position, x H v = v(x, t) v(x, t) p(x, t) N EJ = EJ(x) m = m(x)... and an hypothesis To study the previous problem, we introduce an approximate model by the following hypothesis, v(x, t) = Ψ(x) Z(t), that is, the hypothesis of separation of variables Note that Ψ(x), the shape function, is adimensional, while Z(t) is dimensionally a generalised displacement, usually chosen to characterise the structural behaviour. In our example we can use the displacement of the tip of the chimney, thus implying that Ψ(H) = because Z(t) = v(h, t) and v(h, t) = Ψ(H) Z(t)

Principle of Virtual Displacements For a flexible system, the PoVD states that, at equilibrium, δw E = δw I. The virtual work of external forces can be easily computed, the virtual work of internal forces is usually approximated by the virtual work done by bending moments, that is δw I M δχ where χ is the curvature and δχ the virtual increment of curvature. δw E The external forces are p(x, t), N and the forces of inertia f I ; we have, by separation of variables, that δv = Ψ(x) and we can write δw p = [ p(x, t)δv dx = δw Inertia = m(x) vδv dx = [ ] = m(x)ψ (x) dx ] p(x, t)ψ(x) dx = p (t) m(x)ψ(x) ZΨ(x) dx Z(t) = m Z. The virtual work done by the axial force deserves a separate treatment... δw N The virtual work of N is δw N = Nδu where δu is the variation of the vertical displacement of the top of the chimney. We start computing the vertical displacement of the top of the chimney in terms of the rotation of the axis line, φ Ψ (x)z(t), u(t) = H cos φ dx = ( cos φ) dx, substituting the well known approximation cosφ φ we have φ u(t) = dx = Ψ (x)z (t) dx hence and δu = [ δw N = Ψ (x)z(t) dx = in the above equation Ψ (x) dx Z ] Ψ (x) dx N Z = kg Z

δw Int Approximating the internal work with the work done by bending moments, for an infinitesimal slice of beam we write dw Int = Mv (x, t) dx = MΨ (x)z(t) dx with M = EJ(x)v (x) integrating δ(dw Int ) = EJ(x)Ψ (x)z(t) dx [ δw Int = ] EJ(x)Ψ (x) dx Z = k Z the shape function must respect the geometrical boundary conditions of the problem, i.e., both Ψ = x and Ψ = cos πx H are accettable shape functions for our example, as Ψ () = Ψ () = and Ψ () = Ψ () = better results are obtained when the second derivative of the shape function at least resembles the typical distribution of bending moments in our problem, so that between Ψ = constant and Ψ = π πx cos H H the second choice is preferable..5 f =-cos(pi*x/) f =x f " f ".8 vi"/z(t).5.6. v/z(t).5....6.8 x/h

Example Using Ψ(x) = cos πx H, with m = constant and EJ = constant, with a load characteristic of seismic excitation, p(t) = m v g (t), m = m ( cos πx H ) dx = m( π )H π k = EJ 6H k G = N π H p g = m v g (t) cos πx π EJ dx = H H sin πx H dx = π 8H N cos πx H dx = ( ) mh v g (t) π Analysis The process of estimating the vibration characteristics of a complex system is known as vibration analysis. We can use our previous results for flexible systems, based on the SDOF model, to give an estimate of the natural frequency ω = k /m A different approach, proposed by Lord Rayleigh, starts from different premises to give the same results but the Quotient method is important because it offers a better understanding of the vibrational behaviour, eventually leading to successive refinements of the first estimate of ω. Quotient Our focus will be on the free vibration of a flexible, undamped system. inspired by the free vibrations of a proper SDOF we write Z(t) = Z sin ωt v(x, t) = Z Ψ(x) sin ωt, and v(x, t) = ω Z Ψ(x) cos ωt. the displacement and the velocity are in quadrature: when v is at its maximum v =, hence V = V max, T = and when v is at its maximum it is v =, hence V =, T = T max, disregarding damping, the energy of the system is constant during free vibrations, V max + = + T max

Rayleigh s Quotient Now we write the expressions for V max and T max, V max = Z EJ(x)Ψ (x) dx, S T max = ω Z m(x)ψ (x) dx, equating the two expressions and solving for ω we have ω S = EJ(x)Ψ (x) dx S m(x)ψ (x) dx. S Recognizing the expressions we found for k and m we could question the utility of Quotient... Quotient in method we know the specific time dependency of the inertial forces f I = m(x) v = m(x)ω Z Ψ(x) sin ωt f I has the same shape we use for displacements. if Ψ were the real shape assumed by the structure in free vibrations, the displacements v due to a loading f I = ω m(x)ψ(x)z should be proportional to Ψ(x) through a constant factor, with equilibrium respected in every point of the structure during free vibrations. starting from a shape function Ψ (x), a new shape function Ψ can be determined normalizing the displacements due to the inertial forces associated with Ψ (x), f I = m(x)ψ (x), we are going to demonstrate that the new shape function is a better approximation of the true mode shape mode shapes Given different shape functions Ψ i and considering the true shape of free vibration Ψ, in the former cases equilibrium is not respected by the structure itself. To keep inertia induced deformation proportional to Ψ i we must consider the presence of additional elastic constraints. This leads to the following considerations the frequency of vibration of a structure with additional constraints is higher than the true natural frequency, the criterium to discriminate between different shape functions is: better shape functions give lower estimates of the natural frequency, the true natural frequency being a lower bound of all estimates.

mode shapes In general the selection of trial shapes goes through two steps,. the analyst considers the flexibilities of different parts of the structure and the presence of symmetries to devise an approximate shape,. the structure is loaded with constant loads directed as the assumed displacements, the displacements are computed and used as the shape function, of course a little practice helps a lot in the the choice of a proper pattern of loading... mode shapes p = m(x) P = M p = m(x) (a) (b) p = m(x) (c) (d) p = m(x) Refinement R Choose a trial function Ψ () (x) and write our first estimate R of ω is v () = Ψ () (x)z () sin ωt V max = Z() EJΨ () dx T max = ω Z () mψ () dx ω = EJΨ () dx mψ () dx.

Refinement R We try to give a better estimate of V max computing the external work done by the inertial forces, the deflections due to p () are p () = ω m(x)v () = Z () ω Ψ () (x) v () = ω v() ω = ω () Z() Ψ ω = ω Ψ () Z(), where we write Z () because we need to keep the unknown ω in evidence. The maximum strain energy is V max = p () v () dx = ω () Z Z() m(x)ψ () Ψ () dx Equating to our previus estimate of T max we find the R estimate ω = Z() Z () m(x)ψ () Ψ () dx m(x)ψ () Ψ () dx Refinement R With little additional effort it is possible to compute T max from v () : T max = ω m(x)v () dx = ω6 Z() m(x)ψ () dx equating to our last approximation for V max we have the R approximation to the frequency of vibration, ω = Z() Z () m(x)ψ () Ψ () dx m(x)ψ () Ψ () dx. Of course the procedure can be extended to compute better and better estimates of ω but usually the refinements are not extended beyond R, because it would be contradictory with the quick estimate nature of the Quotient method and also because R estimates are usually very good ones. Refinement Example m.5m m k k k.5 /5 6/5 Ψ () p() ω m Ψ () T = ω.5 m Z V = k Z ω = k 9/ m = k m v () = 5 Z () = 5 m k ω Ψ () m k V () = m 5 m k ω ( + / + /5) = m 5 m 87 ω k ω = 9 m m 87 8 m k = k 9 m =.8 k m