Generalized Single Degree of Freedom Systems

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1 Single Degree of Freedom Dipartimento di Ingegneria Civile Ambientale e Territoriale Politecnico di Milano April, 16 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. 1. 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.

2 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.

3 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 L G x Unit mass Length L, Centre of Mass Total Mass Moment of Inertia m = constant, x G = L/, m = ml, J = m L 1 = ml 1 Rigid Rectangle y G b a Unit mass Sides γ = constant, a, b Centre of Mass x G = a/, y G = b/ Total Mass m = γab, Moment of Inertia J = m a + b 1 = γ a b + ab 1

4 Rigid Triangle y For a right triangle. G a b Unit mass Sides γ = constant, a, b Centre of Mass x G = a/, y G = b/ Total Mass m = γab/, Moment of Inertia J = m a + b 18 = γ a b + ab 6 Rigid Oval When a = b = D = R the oval is a circle. y b x Unit mass Axes a γ = constant, a, b Centre of Mass x G = y G = Total Mass m = γ πab, Moment of Inertia J = m a + b 16 trabacolo1 p(x,t) = P x/a f(t) m, J c1 k1 c k a a a a a a N The mass of the left bar is m 1 = m a and its moment of inertia is (a) J 1 = m 1 1 = a m 1 /. The maximum value of the external load is P max = P a/a = P and the resultant of triangular load is R = P a/ = 8Pa

5 Forces and Virtual Displacements J1 Z a 8Pa f(t) Z(t) J Z a N c1ż m1 Z k1z c Ż m Z kz δθ 1 = /(a) δθ = /(a) δu u = 7a a cos θ 1 a cos θ, δu = a sin θ 1 δθ 1 + a sin θ δθ δθ 1 = /(a), δθ = /(a) sin θ 1 Z/(a), sin θ Z/(a) δu = ( 1 a + ) 1 a Z = 7 1a Z Principle of Virtual Displacements 8Pa f(t) J1 Z J Z a a Z(t) N c1ż m1 Z k1z cż m Z kz δθ1 = /(a) δθ = /(a) The virtual work of the InertialDampingElasticExternal forces: Z δw I = m 1 J Z 1 a = a m ( m1 + m 9 + J 1 16a + J 9a δu Z J Z ) a Z a Ż δw D = c 1 c Z = (c + c 1 /16) Ż ( Z δw S = k 1 k Z = 9k k ) Z 9 δw Ext = 8Pa f(t) + N 7 1a 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 ( m1 + m 9 + J 1 16a + J ) Z+ 9a ( 9k1 + (c + c 1 /16) Ż k ) Z = 9 8Pa f(t) + N 7 1a Z

6 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 = 1 m m a J a J, c = 1 16 c 1 + c, k = 9 16 k k 7 1a N, p = 16 Pa. It is worth writing down the expression of k : k = 9k k 9 7 1a 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, v = v(x, t) p(x, t) N x v(x, t) 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() = 1 because Z(t) = v(, t) v(, t) = Ψ() Z(t) and

7 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 = p(x, t)ψ(x) dx = p (t) δw Inertia = m(x) vδv dx = m(x)ψ(x) ZΨ(x) dx [ ] = m(x)ψ (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) = cos φ dx = (1 cos φ) dx, substituting the well known approximation cosφ 1 φ in the above equation we have φ u(t) = dx = Ψ (x)z (t) dx hence and δu = Ψ (x)z(t) dx = Ψ (x) dx Z [ ] δw N = Ψ (x) dx N Z = k G Z

8 δw Int Approximating the internal work with the work done by bending moments, for an infinitesimal slice of beam we write dw Int = 1 Mv (x, t) dx = 1 MΨ (x)z(t) dx with M = EJ(x)v (x) δ(dw Int ) = EJ(x)Ψ (x)z(t) dx integrating [ ] 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 1 = x and Ψ = 1 cos πx are accettable shape functions for our example, as Ψ 1 () = Ψ () = and Ψ 1 () = 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 1 = constant and Ψ = π πx cos the second choice is preferable..5 f 1 =1-cos(pi*x/) f =x f 1 " f " 1.8 vi"/z(t) v/z(t) x/

9 Example Using Ψ(x) = 1 cos πx, with m = constant and EJ = constant, with a load characteristic of seismic excitation, p(t) = m v g (t), m = m k = EJ π 16 k G = N π (1 cos πx ) dx = m( π ) p g = m v g (t) cos πx π dx = sin πx dx = π 8 N EJ 1 cos πx dx = ( 1 ) m v g (t) π