in MDOF Systems http://intranet.dica.polimi.it/people/boffi-giacomo Dipartimento di Ingegneria Civile Ambientale e Territoriale Politecnico di Milano April 9, 2016 Outline Additional Static Condensation Today we will study the properties of structural matrices, that is the operators that relate the vector of system coordinates x and its time derivatives ẋ and ẍ to the forces acting on the system nodes, f S, f D and f I, respectively. In the end, we will see again the solution of a MDOF problem by superposition, and in general today we will revisit many of the subjects of our previous class, but you know that a bit of reiteration is really good for developing minds.
Additional We already met the mass and the stiffness matrix, M and K, and tangentially we introduced also the dampig matrix C. We have seen that these matrices express the linear relation that holds between the vector of system coordinates x and its time derivatives ẋ and ẍ to the forces acting on the system nodes, f S, f D and f I, elastic, damping and inertial force vectors. M ẍ + C ẋ + K x = pt) f I + f D + f S = pt) Also, we know that M and K are symmetric and definite positive, and that it is possible to uncouple the equation of motion expressing the system coordinates in terms of the eigenvectors, xt) = q i ψ i, where the q i are the modal coordinates and the eigenvectors ψ i are the non-trivial solutions to the equation of free vibrations, K ω 2 M ) ψ = 0 Additional Free Vibrations From the homogeneous, undamped problem introducing separation of variables M ẍ + K x = 0 xt) = ψ A sin ωt + B cos ωt) we wrote the homogeneous linear system K ω 2 M ) ψ = 0 whose non-trivial solutions ψ i for ω 2 i such that K ω 2 i M = 0 are the eigenvectors. It was demonstrated that, for each pair of distint eigenvalues ω 2 r and ω 2 s, the corresponding eigenvectors obey the ortogonality condition, Additional ψ T s M ψ r = δ rs M r, ψ T s K ψ r = δ rs ω 2 r M r.
Additional From K ψ s = ω 2 s M ψ s premultiplying by ψ T r KM 1 we have ψ T r KM 1 K ψ s = ω 2 s ψ T r K ψ s = δ rs ω 4 r M r, premultiplying the first equation by ψ T r KM 1 KM 1 ψ T r KM 1 KM 1 K ψ s = ω 2 s ψ T r KM 1 K ψ s = δ rs ω 6 r M r Additional and, generalizing, ψ T r KM 1 ) b K ψs = δ rs ω 2 r ) b+1 Mr. Additional, 2 From M ψ s = ω 2 s K ψ s premultiplying by ψ T r MK 1 we have ψ T r MK 1 M ψ s = ω 2 s ψ T M s r M ψ s = δ rs ω 2 s premultiplying the first eq. by ψ T r MK 1 ) 2 we have ψ T r MK 1 ) 2 M ψs = ω 2 s ψ T r MK 1 M s M ψ s = δ rs ω 4 s Additional and, generalizing, ψ T r MK 1 ) b M ψs = δ rs M s ω 2 b s Additional, 3 Defining X rs k) = ψ T r M M 1 K ) k ψs we have X rs 0) = ψ T ) r Mψ s = δ rs ω 2 0 s Ms X rs 1) = ψ T ) r Kψ s = δ rs ω 2 1 s Ms X rs 2) = ψ T r KM 1 ) 1 ) Kψs = δ rs ω 2 2 s Ms X rs n) = ψ T r KM 1 ) n 1 ) Kψs = δ rs ω 2 n s Ms Observing that M 1 K ) 1 = K 1 M ) 1 X rs 1) = ψ T r MK 1 ) 1 ) M ψs = δ rs ω 2 1 s Ms X rs n) = ψ T r MK 1 ) n ) M ψs = δ rs ω 2 n s Ms Additional finally X rs k) = δ rs ω 2k s M s for k =,...,.
Flexibility Given a system whose state is determined by the generalized displacements x j of a set of nodes, we define the flexibility coefficient f jk as the deflection, in direction of x j, due to the application of a unit force in correspondance of the displacement x k. The matrix F = [ f jk ] is the flexibility matrix. In general, the dynamic degrees of freedom correspond to the points where there is application of external forces and/or presence of inertial forces. Given a load vector p = { p k }, the displacementent xj is x j = f jk p k or, in vector notation, x = F p m, J a b x 2 x 1 The dynamical system The degrees of freedom 1 f 11 1 f 12 f 21 Displacements due to p 1 = 1 and due to p 2 = 1. f 22
Elastic Forces Momentarily disregarding inertial effects, each node shall be in equilibrium under the action of the external forces and the elastic forces, hence taking into accounts all the nodes, all the external forces and all the elastic forces it is possible to write the vector equation of equilibrium p = f S and, substituting in the previos vector expression of the displacements x = F f S The stiffness matrix K can be simply defined as the inverse of the flexibility matrix F, K = F 1. As an alternative definition, consider an unary vector of displacements, e i) = { δ ij }, j = 1,..., N, and the vector of nodal forces k i that, applied to the structure, produces the displacements e i) F k i = e i), i = 1,..., N. Collecting all the ordered e i) in a matrix E, it is clear that E I and we have, writing all the equations at once, F [ k i ] = [ e i) ] = E = I. Collecting the ordered force vectors in a matrix K = [ k i ] we have FK = I, K = F 1, giving a physical interpretation to the columns of the stiffness matrix. Finally, writing the nodal equilibrium, we have p = f S = K x.
The elastic strain energy V can be written in terms of displacements and external forces, p V = 1 2 pt x = 1 T F p, }{{} x 2 x } T {{ K} x. p T Because the elastic strain energy of a stable system is always greater than zero, K is a positive definite matrix. On the other hand, for an unstable system, think of a compressed beam, there are displacement patterns that are associated to zero strain energy. Two sets of loads p A and p B are applied, one after the other, to an elastic system; the work done is V AB = 1 2 pat x A + p AT x B + 1 2 pb T x B. If we revert the order of application the work is V BA = 1 2 pb T x B + p B T x A + 1 2 pat x A. The total work being independent of the order of loading, p AT x B = p B T x A., 2 Expressing the displacements in terms of F, p AT F p B = p B T Fp A, both terms are scalars so we can write p AT F p B = p B T Fp A) T = p A T F T p B. Because this equation holds for every p, we conclude that F = F T. The inverse of a symmetric matrix is symmetric, hence K = K T.
A practical consideration For the kind of structures we mostly deal with in our examples, problems, exercises and assignments, that is simple structures, it is usually convenient to compute first the flexibility matrix applying the Principle of Virtual Displacements and later the stiffness matrix, using inversion, K = F 1. On the other hand, the PVD approach cannot work in practice for real structures, because the number of degrees of freedom necessary to model the structural behaviour exceeds our ability to apply the PVD... The stiffness matrix for large, complex structures to construct different methods required are. E.g., the Finite Element Method. FEM The most common procedure to construct the matrices that describe the behaviour of a complex system is the Finite Element Method, or FEM. The procedure can be sketched in the following terms: the structure is subdivided in non-overlapping portions, the finite elements, bounded by nodes, connected by the same nodes, the state of the structure can be described in terms of a vector x of generalized nodal displacements, there is a mapping between element and structure DOF s, i el r, the element stiffness matrix, K el establishes a linear relation between an element nodal displacements and forces, for each FE, all local k ij s are contributed to the global stiffness k rs s, with i r and j s, taking in due consideration differences between local and global systems of reference. Note that in the r-th global equation of equilibrium we have internal forces caused by the nodal displacements of the FE that have nodes i el such that i el r, thus implying that global K is a banded matrix. Consider a 2-D inextensible beam element, that has 4 DOF, namely two transverse end displacements x 1, x 2 and two end rotations, x 3, x 4. The element stiffness is computed using 4 shape functions φ i, the transverse displacement being vs) = i φ is) x i, 0 s L, the different φ i are such all end displacements or rotation are zero, except the one corresponding to index i. The shape functions for a beam are s ) 2 s ) 3, s ) 2 s ) 3, φ 1 s) = 1 3 + 2 φ2 s) = 3 2 L L L L s s ) 2 s ) 3 s ) 2 s ) 3. φ 3 s) = 2 + φ 4 s) = + L) L L L L
, 2 The element stiffness coefficients can be computed using, what else, the PVD: we compute the external virtual work done by a variation δ x i by the force due to a unit displacement x j, that is k ij, δ W ext = δ x i k ij, the virtual internal work is the work done by the variation of the curvature, δ x i φ i s) by the bending moment associated with a unit x j, φ j s)ejs), δ W int = L 0 δ x i φ i s)φ j s)ejs) ds., 3 The equilibrium condition is the equivalence of the internal and external virtual works, so that simplifying δ x i we have For EJ = const, k ij = f S = EJ L 3 L 0 φ i s)φ j s)ejs) ds. 12 12 6L 6L 12 12 6L 6L 6L 6L 4L 2 2L 2 x 6L 6L 2L 2 4L 2 Blackboard Time! L EJ 4EJ 2L EJ x 2 x 3 x 1
The mass matrix maps the nodal accelerations to nodal inertial forces, and the most common assumption is to concentrate all the masses in nodal point masses, without rotational inertia, computed lumping a fraction of each element mass or a fraction of the supported mass) on all its bounding nodes. This procedure leads to a so called lumped mass matrix, a diagonal matrix with diagonal elements greater than zero for all the translational degrees of freedom and diagonal elements equal to zero for angular degrees of freedom. Consistent Mass Matrix Discussion The mass matrix is definite positive only if all the structure DOF s are translational degrees of freedom, otherwise M is semi-definite positive and the eigenvalue procedure is not directly applicable. This problem can be overcome either by using a consistent mass matrix or using the static condensation procedure. Consistent Mass Matrix Discussion Consistent A consistent mass matrix is built using the rigorous FEM procedure, computing the nodal reactions that equilibrate the distributed inertial forces that develop in the element due to a linear combination of inertial forces. Using our beam example as a reference, consider the inertial forces associated with a single nodal acceleration ẍ j, f I,j s) = ms)φ j s)ẍ j and denote with m ij ẍ j the reaction associated with the i-nth degree of freedom of the element, by the PVD δ x i m ij ẍ j = δ x i φ i s)ms)φ j s) ds ẍ j simplifying m ij = ms)φ i s)φ j s) ds. For ms) = m = const. 156 54 22L 13L f I = ml 54 156 13L 22L 420 22L 13L 4L 2 3L 2 ẍ 13L 22L 3L 2 4L 2 Consistent Mass Matrix Discussion
Consistent, 2 Pro some convergence theorem of FEM theory holds only if the mass matrix is consistent, sligtly more accurate results, no need for static condensation. Contra M is no more diagonal, heavy computational aggravation, static condensation is computationally beneficial, inasmuch it reduces the global number of degrees of freedom. Consistent Mass Matrix Discussion For each element c ij = cs)φ i s)φ j s) ds and the damping matrix C can be assembled from element contributions. However, using the FEM C = Ψ T C Ψ is not diagonal and the modal equations are no more uncoupled! The alternative is to write directly the global damping matrix, in terms of the underdetermined coefficients c b, C = b c b M M 1 K ) b. With our definition of C, C = b c b M M 1 K ) b, assuming normalized eigenvectors, we can write the individual component of C = Ψ T C Ψ cij = ψ T i C ψ j = δ ij c b ω 2b j due to the additional orthogonality relations, we recognize that now C is a diagonal matrix. Introducing the modal damping C j we have b C j = ψ T j C ψ j = b c b ω 2b j = 2ζ j ω j and we can write a system of linear equations in the c b.
We want a fixed, 5% damping ratio for the first three modes, taking note that the modal equation of motion is Using we have q i + 2ζ i ω i q i + ω 2 i q i = p i C = c 0 M + c 1 K + c 2 KM 1 K 2 0.05 ω 1 ω 2 ω 3 1 ω 2 = 1 ω 4 1 1 ω 2 2 ω 4 2 1 ω 2 3 ω 4 3 Solving for the c s and substituting above, the resulting damping matrix is orthogonal to every eigenvector of the system, for the first three modes, leads to a modal damping ratio that is equal to 5%. c 0 c 1 c 2 Computing the coefficients c 0, c 1 and c 2 to have a 5% damping at frequencies ω 1 = 2, ω 2 = 5 and ω 3 = 8 we have c 0 = 1200/9100, c 1 = 159/9100 and c 2 = 1/9100. Writing ζω) = 1 c0 ) 2 ω + c 1ω + c 2 ω 3 we can plot the above function, along with its two term equivalent c 0 = 10/70, c 1 = 1/70). Damping ratio 0.1 0.05 0-0.05 Two and three terms solutions three terms two terms -0.1 0 2 5 8 10 15 20 Circular frequency Negative damping? No, thank you: use only an even number of terms. A common assumption is based on a linear approximation, for a beam element +1 1 0 0 f G = N 1 +1 0 0 L 0 0 0 0 x 0 0 0 0 N f 1 f 2 N x 1 x 2 L f 2 = f 1 f 1 L = N x 2 x 1 ) It is possible to compute the geometrical stiffness matrix using FEM, shape functions and PVD, k G,ij = Ns)φ is)φ js) ds, for constant N K G = 36 36 3L 3L N 36 36 3L 3L 30L 3L 3L 4L 2 L 2 3L 3L L 2 4L 2
s Following the same line of reasoning that we applied to find nodal inertial forces, by the PVD and the use of shape functions we have p i t) = ps, t)φ i s) ds. For a constant, uniform load ps, t) = p = const, applied on a beam element, p = pl { 1 1 L 2 2 12 L } T 12 Static Condensation Simplified Approach Some structural parameter is approximated, only translational DOF s are retained in dynamic analysis. Consistent Approach All structural parameters are computed according to the FEM, and all DOF s are retained in dynamic analysis. If we choose a simplified approach, we must use a procedure to remove unneeded structural DOF s from the model that we use for the dynamic analysis. Enter the Static Condensation Method. Static Condensation
Static Condensation We have, from a FEM analysis, a stiffnes matrix that uses all nodal DOF s, and from the lumped mass procedure a mass matrix were only translational and maybe a few rotational) DOF s are blessed with a non zero diagonal term. In this case, we can always rearrange and partition the displacement vector x in two subvectors: a) x A, all the DOF s that are associated with inertial forces and b) x B, all the remaining DOF s not associated with inertial forces. Static Condensation x = { x A x B } T Static Condensation, 2 After rearranging the DOF s, we must rearrange also the rows equations) and the columns force contributions) in the structural matrices, and eventually partition the matrices so that { } [ ] } fi MAA M = AB {ẍa 0 M BA M BB ẍ B [ ] { } KAA K f S = AB xa K BA K BB x B Static Condensation with M BA = M T AB = 0, M BB = 0, K BA = K T AB Finally we rearrange the loadings vector and write... Static Condensation, 3... the equation of dynamic equilibrium, p A = M AA ẍ A + M AB ẍ B + K AA x A + K AB x B p B = M BA ẍ A + M BB ẍ B + K BA x A + K BB x B The terms in red are zero, so we can simplify M AA ẍ A + K AA x A + K AB x B = p A K BA x A + K BB x B = p B Static Condensation solving for x B in the 2nd equation and substituting x B = K 1 BB p B K 1 BB K BAx A p A K AB K 1 BB p B = M AA ẍ A + K AA K AB K 1 BB K BA) xa
Static Condensation, 4 Going back to the homogeneous problem, with obvious positions we can write K ω 2 M ) ψ A = 0 but the ψ A are only part of the structural eigenvectors, because in essentially every application we must consider also the other DOF s, so we write { } ψa,i ψ i =, with ψ B,i = KBB 1 K BAψ A,i ψ B,i Static Condensation L x 2 x 3 4EJ EJ 2L EJ Disregarding the factor 2EJ/L 3, [ ] K BB = L 2 6 2, K 1 2 6 BB = 1 32L 2 x 1 K = 2EJ L 3 12 3L 3L 3L 6L 2 2L 2 3L 2L 2 6L 2 [ ] 6 2, K 2 6 AB = [ 3L 3L ] Static Condensation The matrix K is K = 2EJ L 3 12 KAB K 1 BB K T AB ) = 39EJ 2L 3