Multi Degrees of Freedom Systems
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1 Multi Degrees of Freedom Systems MDOF s Dipartimento di Ingegneria Civile Ambientale e Territoriale Politecnico di Milano March 9, 07 Outline, a System of Linear Differential Equations are are are a base EoM in Modal Coordinates Consider an undamped system with two masses and two degrees of freedom. p (t) p (t) m m k k k 3 x x are
2 We can separate the two masses, single out the spring forces and, using the D Alembert Principle, the inertial forces and, finally. write an equation of dynamic equilibrium for each mass. k x p m ẍ k (x x ) m ẍ + (k + k )x k x = p (t) are k (x x ) p m ẍ k 3 x m ẍ k x + (k + k 3 )x = p (t) equation of motion of a DOF system With some little rearrangement we have a system of two linear differential equations in two variables, x (t) and x (t): { m ẍ + (k + k )x k x = p (t), m ẍ k x + (k + k 3 )x = p (t). are equation of motion of a DOF system Introducing the loading vector p, the vector of inertial forces f I and the vector of elastic forces f S, { } { } { } p (t) fi, fs, p =, f p (t) I =, f S = f I, we can write a vectorial equation of equilibrium: f I + f S = p(t). f S, are
3 f S = K x It is possible to write the linear relationship between f S and the vector of displacements x = { } T x x in terms of a matrix product, introducing the so called stiffness matrix K. In our example it is [ ] k + k f S = k x = K x k k + k 3 stiffness matrix K has a number of rows equal to the number of elastic forces, i.e., one force for each DOF and a number of columns equal to the number of the DOF. stiffness matrix K is hence a square matrix K ndof ndof are f I = M ẍ Analogously, introducing the mass matrix M that, for our example, is [ ] m 0 M = 0 m we can write f I = M ẍ. Also the mass matrix M is a square matrix, with number of rows and columns equal to the number of DOF s. are Matrix Equation Finally it is possible to write the equation of motion in matrix format: M ẍ + K x = p(t). Of course it is possible to take into consideration also the damping forces, taking into account the velocity vector ẋ and introducing a damping matrix C too, so that we can eventually write M ẍ + C ẋ + K x = p(t). But today we are focused on undamped systems... are
4 K K is symmetrical. elastic force exerted on mass i due to an unit displacement of mass j, f S,i = k ij is equal to the force k ji exerted on mass j due to an unit diplacement of mass i, in virtue of Betti s theorem (also known as Maxwell-Betti reciprocal work theorem). K is a positive definite matrix. strain energy V for a discrete system is V = xt f S, and expressing f S in terms of K and x we have are V = xt K x, and because the strain energy is positive for x 0 it follows that K is definite positive. M Restricting our discussion to systems whose degrees of freedom are the displacements of a set of discrete masses, we have that the mass matrix is a diagonal matrix, with all its diagonal elements greater than zero. Such a matrix is symmetrical and definite positive. Both the mass and the stiffness matrix are symmetrical and definite positive. are Note that the kinetic energy for a discrete system can be written T = ẋt M ẋ. Generalisation of previous results findings in the previous two slides can be generalised to the structural matrices of generic structural systems, with two main exceptions.. For a general structural system, in which not all DOFs are related to a mass, M could be semi-definite positive, that is for some particular displacement vector the kinetic energy is zero.. For a general structural system subjected to axial loads, due to the presence of geometrical stiffness it is possible that for some particular displacement vector the strain energy is zero and K is semi-definite positive. are
5 problem k Graphical statement of the problem p(t) m k m x x k = k, k = k; m = m, m = m; p(t) = p 0 sin ωt. equations of motion m ẍ + k x + k (x x ) = p 0 sin ωt, m ẍ + k (x x ) = but we prefer the matrix notation... are steady state solution We prefer the matrix notation because we can find the steady-state response of a SDOF system exactly as we found the s-s solution for a SDOF system. Substituting x(t) = ξ sin ωt in the equation of motion and simplifying sin ωt, [ ] [ ] { } 3 k ξ mω 0 ξ = p dividing by k, with ω 0 = k/m, β = ω /ω 0 and st = p 0 /k the above equation can be written ([ ] [ ]) [ ] { } 3 β 0 3 β ξ = 0 β ξ = st. 0 are steady state solution determinant of the matrix of coefficients is Det = β 4 5β + but we want to write the polynomial in β in terms of its roots Det = (β /) (β ). Solving for ξ/ st in terms of the inverse of the coefficient matrix gives [ ] { } ξ β = st (β )(β ) 3 β 0 { } β = (β. )(β ) are
6 solution, graphically Normalized displacement 0 steady-state response for a dof system, harmonic load ξ /Δ st ξ /Δ st are β =ω /ω o Comment to the Steady State Solution steady state solution is x s-s = st (β )(β ) { } β sin ωt. As it s apparent in the previous slide, we have two different values of the excitation frequency for which the dynamic amplification factor goes to infinity. For an undamped SDOF system, we had a single frequency of excitation that excites a resonant response, now for a two degrees of freedom system we have two different excitation frequencies that excite a resonant response. are We know how to compute a particular integral for a MDOF system (at least for a harmonic loading), what do we miss to be able to determine the integral of motion? equation of motion To understand the behaviour of a MDOF system, we have to study the homogeneous solution. Let s start writing the homogeneous equation of motion, M ẍ + K x = 0. solution, in analogy with the SDOF case, can be written in terms of a harmonic function of unknown frequency and, using the concept of separation of variables, of a constant vector, the so called shape vector ψ: are x(t) = ψ(a sin ωt + B cos ωt). Substituting in the equation of motion, we have ( K ω M ) ψ(a sin ωt + B cos ωt) = 0
7 Eigenvalues previous equation must hold for every value of t, so it can be simplified removing the time dependency: ( K ω M ) ψ = 0. This is a homogeneous linear equation, with unknowns ψ i and the coefficients that depends on the parameter ω. Speaking of homogeneous systems, we know that there is always a trivial solution, ψ = 0, and non-trivial solutions are possible if the determinant of the matrix of coefficients is equal to zero, det ( K ω M ) = 0 are eigenvalues of the MDOF system are the values of ω for which the above equation (the equation of frequencies) is verified or, in other words, the frequencies of vibration associated with the shapes for which Kψ sin ωt = ω Mψ sin ωt. Eigenvalues, cont. For a system with N degrees of freedom the expansion of det ( K ω M ) is an algebraic polynomial of degree N in ω. A polynomial of degree N has exactly N roots, either real or complex conjugate. In Dynamics of Structures those roots ω i, i =,..., N are all real because the structural matrices are symmetric matrices. Moreover, if both K and M are positive definite matrices (a condition that is always satisfied by stable structural systems) all the roots, all the eigenvalues, are strictly positive: are ω i 0, for i =,..., N. Substituting one of the N roots ω i in the characteristic equation, ( K ω i M ) ψ i = 0 the resulting system of N linearly independent equations can be solved (except for a scale factor) for ψ i, the eigenvector corresponding to the eigenvalue ω i. are
8 scale factor being arbitrary, you have to choose (arbitrarily) the value of one of the components and compute the values of all the other N components using the N linearly indipendent equations. It is common to impose to each eigenvector a normalisation with respect to the mass matrix, so that ψ T i M ψ i =. are Please consider that, substituting different eigenvalues in the equation of free vibrations, you have different linear systems, leading to different eigenvectors. most general expression (the general integral) for the displacement of a homogeneous system is x(t) = ψ i (A i sin ω i t + B i cos ω i t). i= In the general integral there are N unknown constants of integration, that must be determined in terms of the initial conditions. are Usually the initial conditions are expressed in terms of initial displacements and initial velocities x 0 and ẋ 0, so we start deriving the expression of displacement with respect to time to obtain ẋ(t) = ψ i ω i (A i cos ω i t B i sin ω i t) i= and evaluating the displacement and velocity for t = 0 it is x(0) = ψ i B i = x 0, ẋ(0) = ψ i ω i A i = ẋ 0. i= i= are above equations are vector equations, each one corresponding to a system of N equations, so we can compute the N constants of integration solving the N equations ψ ji B i = x 0,j, i= ψ ji ω i A i = ẋ 0,j, j =,..., N. i=
9 ity - Take into consideration two distinct eigenvalues, ω r and ω s, and write the characteristic equation for each eigenvalue: K ψ r = ω rm ψ r K ψ s = ω sm ψ s premultiply each equation member by the transpose of the other eigenvector are ψ T s K ψ r = ω rψ T s M ψ r ψ T r K ψ s = ω sψ T r M ψ s ity - term ψ T s K ψ r is a scalar, hence ψ T s K ψ r = ( ψ T ) T s K ψ r = ψ T r K T ψ s but K is symmetrical, K T = K and we have ψ T s K ψ r = ψ T r K ψ s. By a similar derivation are ψ T s M ψ r = ψ T r M ψ s. ity - 3 Substituting our last identities in the previous equations, we have ψ T r K ψ s = ω rψ T r M ψ s ψ T r K ψ s = ω sψ T r M ψ s subtracting member by member we find that (ω r ω s) ψ T r M ψ s = 0 We started with the hypothesis that ω r ω s, so for every r s we have that the corresponding eigenvectors are orthogonal with respect to the mass matrix are ψ T r M ψ s = 0, for r s.
10 ity - 4 eigenvectors are orthogonal also with respect to the stiffness matrix: ψ T s K ψ r = ω rψ T s M ψ r = 0, for r s. By definition and consequently M i = ψ T i M ψ i ψ T i K ψ i = ω i M i. are M i is the modal mass associated with mode no. i while K i ω i M i is the respective modal stiffness. are a base eigenvectors are linearly independent, so for every vector x we can write x = ψ j q j. j= coefficients are readily given by premultiplication of x by ψ T i M, because N ψ T i M x = ψ T i M ψ jq j = ψ T i M ψ iq i = M i q i j= in virtue of the ortogonality of the eigenvectors with respect to the mass matrix, and the above relationship gives are a base EoM in Modal Coordinates q j = ψt j M x M j. are a base Generalising our results for the displacement vector to the acceleration vector and expliciting the time dependency, it is x(t) = x i (t) = ψ j q j (t), ẍ(t) = j= Ψ ij q j (t), ẍ i (t) = j= ψ j q j (t), j= ψ ij q j (t). j= are a base EoM in Modal Coordinates Introducing q(t), the vector of modal coordinates and Ψ, the eigenvector matrix, whose columns are the eigenvectors, we can write x(t) = Ψ q(t), ẍ(t) = Ψ q(t).
11 EoM in Modal Coordinates... Substituting the last two equations in the equation of motion, premultiplying by Ψ T M Ψ q + K Ψ q = p(t) Ψ T M Ψ q + Ψ T K Ψ q = Ψ T p(t) introducing the so called starred matrices, with p (t) = Ψ T p(t), we can finally write are a base EoM in Modal Coordinates M q + K q = p (t) vector equation above corresponds to the set of scalar equations p i = m ij q j + k ij q j, i =,..., N.... are N independent equations! We must examine the structure of the starred symbols. generic element, with indexes i and j, of the starred matrices can be expressed in terms of single eigenvectors, m ij = ψt i M ψ j = δ ij M i, k ij = ψt i K ψ j = ω i δ ijm i. where δ ij is the Kroneker symbol, are a base EoM in Modal Coordinates { i = j δ ij = 0 i j Substituting in the equation of motion, with p i = ψt i p(t) we have a set of uncoupled equations M i q i + ω i M iq i = p i (t), i =,..., N Revisited initial displacements can be written in modal coordinates, x 0 = Ψ q 0 and premultiplying both members by Ψ T M we have the following relationship: Ψ T M x 0 = Ψ T M Ψ q 0 = M q 0. Premultiplying by the inverse of M and taking into account that M is diagonal, are a base EoM in Modal Coordinates q 0 = (M ) Ψ T M x 0 q i0 = ψt i M x 0 M i and, analogously, q i0 = ψ i T M ẋ 0 M i
12 k m k m p(t) x x k = k, k = k; m = m, m = m; p(t) = p 0 sin ωt. { } { } x 0 x =, p(t) = sin ωt, x p 0 [ ] [ ] 0 3 M = m, K = k. 0 frequencies equation of frequencies is K ω M 3k ω m k = k k ω m = 0. Developing the determinant (m )ω 4 (7mk)ω + (k )ω 0 = 0 Solving the algebraic equation in ω ω = k 7 33 m 4 ω = k m ω = k m 4 ω = k m Substituting ω for ω in the first of the characteristic equations gives the ratio between the components of the first eigenvector, while substituting ω gives k ( )ψ kψ = 0 k ( )ψ kψ = 0. Solving with the arbitrary assignment ψ = ψ = gives the unnormalized eigenvectors, { } ψ =, ψ = { }
13 Normalization We compute first M and M, M = ψ T M ψ = { , } [ ] { } m m = {.6864m, m } { } =.453m M =.70346m the adimensional normalisation factors are α =.453, α = Applying the normalisation factors to the respective unnormalised eigenvectors and collecting them in a matrix, we have the matrix of normalized eigenvectors [ ] Ψ = Modal Loadings modal loading is p (t) = Ψ T p(t) [ ] { } = p 0 sin ωt { } = p 0 sin ωt Modal EoM Substituting its modal expansion for x into the equation of motion and premultiplying by Ψ T we have the uncoupled modal equation of motion { m q k q = p 0 sin ωt m q k q = p 0 sin ωt Note that all the terms are dimensionally correct. Dividing by m both equations, we have q + ω q = p 0 sin ωt m q + ω q = p 0 sin ωt m
14 Particular Integral We set ξ = C sin ωt, and substitute in the first modal EoM: solving for C C = p m with ω = K /m m = K /ω : ξ = ω C sin ωt ( C ω ω ) sin ωt = p sin ωt m ω ω C = p ω = K ω () st ω β with () st = p K =.047 p 0 k and β = ω ω of course C = () st β with () st = p K = p 0 k and β = ω ω Integrals integrals, for our loading, are thus q (t) = A sin ω t + B cos ω t + () sin ωt st β q (t) = A sin ω t + B cos ω t + () sin ωt st β for a system initially at rest q (t) = () st (sin ωt β β sin ω t) q (t) = () st (sin ωt β β sin ω t) we are interested in structural degrees of freedom, too... disregarding transient ( ) () st () ( st.096 x (t) = ψ + ψ β sin ωt = ) p0 sin ωt β β β k ( ) () st () ( st.3575 x (t) = ψ + ψ β sin ωt = ) p0 sin ωt β β β k response in modal coordinates To have a feeling of the response in modal coordinates, let s say that the frequency of the load is ω = ω 0, hence β = = and β = = q i /Δ st q (α)/δ s t α = ω o t q (α)/δ s t In the graph above, the responses are plotted against an adimensional time coordinate α with α = ω 0 t, while the ordinates are adimensionalised with respect to st = p0 k
15 response in structural coordinates Using the same normalisation factors, here are the response functions in terms of x = ψ q + ψ q and x = ψ q + ψ q : x i /Δ st x (α)/δ s t α = ω o t x (α)/δ s t
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