OPGAVE 1. f(t) = f 0 cos(ωt)

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1 SKRIFTLIG EKSAMEN I STRUKTUREL DYNAMIK Bygge- og Anlægskonstruktion, 7. semester Tirsdag den 3. januar 007 kl Alle hjælpemidler er tilladt OPGAVE 1 M f(t) = f 0 cos(ωt) K Figuren viser et maskinfundament med massen M påvirket af en lodret harmonisk variende kraft f(t) = f 0 cos(ωt) fra en ubalanceret roterende maskine. ω er maskinens omløbstal (vinkelfrekvens), og f 0 angiver kraftamplituden. For at dæmpe svingningspåvirkningerne på den omgivende bygning svingningsisoleres maskinfundamentet ved ophængning af dette på et system af lodret virkende, parallelle, lineært elastiske fjedre uden dæmpning. Den samlede fjederstivhed af fjedersystemet er lig K. Spørgsmål 1 (10%) Opstil bevægelsesligningen til beskrivelse af fundamentets lodrette bevægelse, og angiv et udtryk for kraften på bygningen. Spørgsmål (10%) Bestem fjederstivheden K, således at kraftamplituden på bygningen bliver lig 1 5 f 0.

2 OPGAVE m k f(t) = f 0 cos(ωt) M K Fundamentet i opgave 1 betragtes igen. Nu ønskes svingningspåvirkningerne yderligere dæmpet ved påmontering af en tunet massedæmper som vist på figuren. Dæmperen består af en punktformig masse m og en lodret virkende, lineært elastisk fjeder med fjederkonstanten k uden dæmpning, placeret lodret over fundamentets massetyngdepunkt. Spørgsmål 1 (10%) Opstil bevægelsesligningerne for masserne M og m, og angiv et udtryk for kraften på bygningen. Spørgsmål (10%) Bestem forholdet k, således at kraften på bygningen bliver lig 0. m

3 3 OPGAVE 3 m, J A µ, EI B µ, EI C l l Figuren viser en vandret, plan sammensat Bernoulli-Euler bjælke ABC, der er fast simpelt understøttet i punkterne A og C. Delbjælkerne AB og BC har begge længden l, konstant masse pr. længdeenhed µ, konstant bøjningsstivhed EI, og er uendeligt stive over for axialdeformationer. Bjælkerne er bøjningsstift forbundet i punkt B, hvor der er anordnet en udbredt masse med massen m og masseinertimomentet J om bøjningscenteret. Der betragtes kun små svingninger i konstruktionsplanen omkring den lodrette statiske ligevægtstilstand, ligesom der både ses bort fra eventuelle normalkræfters indflydelse på bøjningsdeformationerne, og eventuel dæmpning. Spørgsmål 1 (15%) Vis, at systemets egensvingningsformer enten er symmetriske eller antisymmmetriske omkring en lodret linie gennem punkt B. Opstil herved ækvivalente systemer til bestemmelse af henholdsvis symmetriske og antisymmetriske egensvingninger. Spørgsmål (15%) Opstil efter eget valg frekvensbetingelsen til bestemmelse af enten symmetriske eller antisymmetriske egensvingninger af rammen. Der fordres ingen numerisk løsning af frekvensbetingelsen.

4 4 OPGAVE 4 Konstruktionen i opgave 3 betragtes igen. Nu tilstræbes en numerisk løsning af det udæmpede egensvingningsproblem, idet bjælkerne AB og BC hver modelleres dynamisk ved hjælp et enkelt bjælkeelement. Der benyttes konsistente massematricer til diskretiseringen af den kinetiske energi. Iøvrigt benyttes de samme geometriske og fysiske antagelser som anført i opgave 3. Ved de numeriske beregninger benyttes m = µl og J = µl 3. Spørgsmål 1 (15%) Afhængig af om det symmetriske eller anti-symmetriske problem er analyseret i opgave 3, ønskes det tilsvarende diskrete generaliserede egenværdiproblem opstillet. Spørgsmål (15%) Bestem de udæmpede cirkulære egenfrekvenser og de tilhørende egensvingningsformer for det i spørgsmål 1 opstillede egenværdiproblem, og skitser egensvingningsformerne for den samlede bjælke ABC.

5 5 SOLUTIONS PROBLEM 1 Question 1: f(t) = f 0 cos(ωt) M K f b (t) x(t) Fig. 1: Sign definition of degree of freedom and load on building. Since the foundation is only moving in the vertical direction, it is described by a single degree of freedom x(t), which is selected as the vertical displacement of the mass center of gravity in the downward direction from the static equilibrium position, see Fig. 1. Then, the equation of motion reads Mẍ + Kx = f 0 cos(ωt) (1) The force on the building is denoted f b (t), with a positive sign co-directional to x(t), see Fig. 1. The reaction force on the building is caused by a compression force Kx(t) in the spring, i.e. f b (t) = Kx(t) () Question : Since the system is free of damping the stationary motion must be in phase or counter phase with the excitation. Then, the stationary solution of (1) has the form x(t) = x 0 cos(ωt) (3) Insertion of (3) into (1) provides the following result for the unknown amplitude x 0, cf. (-61), (-6)

6 6 x 0 = f 0 M(ω 0 ω ) (4) where ω 0 is the undamped circular eigenfrequency of the foundation given by (-7) ω 0 = K M (5) From (), (3), (4), (5) follows that the force on the building becomes f b (t) = f b,0 cos(ωt) (6) where K f b,0 = Kx 0 = f 0 M(ω0 ω ) = f ω0 0 ω0 (7) ω The requirement is that f b,0 = ± 1f 5 0. Hence ± 1 5 = ω 0 ω 0 ω (8) (8) only has a solution for ω > ω 0 and f b,0 = 1 5 f 0, i.e. when the displacement x(t) is in counter phase to the excitation f(t). Then ω 0 = K M = 1 6 ω K = 1 6 Mω (9) The solution (9) will only work, when the machine is running with the considered angular frequency ω.

7 7 PROBLEM Question 1: m x k d (t) f(t) = f 0 cos(ωt) M x(t) K f b (t) Fig. 1: Definition of degrees of freedom. The system has degrees of freedom x(t) and x d (t), which describes the vertical displacement of the machine foundation and the mass of the tuned mass damper. Both are selected as the vertical displacement in the downward direction of the mass centers of gravity from their static equilibrium position, see Fig. 1. The masses are cut free from the springs, and Newton s nd law of motion is applied to the free masses. Then, the equations of motion become, cf. (3-37) } Mẍ = Kx + k(x d x) + f(t) mẍ d = k(x d x) Mẍ + Kx = f(t) (1) where x(t) = [ ] x(t) x d (t), f(t) = [ ] 1 f(t), M = 0 [ ] M 0 0 m [ ] K + k k, K = k k () The force on the building is unchanged given as f b (t) = Kx(t) (3)

8 8 Question : The stationary solution of (1) reads x(t) = x 0 cos(ωt) (4) where the amplitude vector becomes, cf. (3-101), (3-10) x 0 = ( K ω M ) [ ] 1 K + k ω f(t) = 1 [ ] M k 1 k k ω m 0 f 0 x 0 = [ x0 x d,0 ] = 1 [ ][ ] k ω m k 1 D k K + k ω f M 0 0 (5) D denotes the determinant of the dynamic stiffness matrix given as D = (K + k ω M)(k ω m) k (6) According to (3) we have f b (t) 0, if x(t) 0. This is the case if the amplitude x 0 = 0. As seen from (5) this requires that k ω m = 0 k m = ω The mass dæmper with the frequency tuning (7) will only work, when the machine is running at the considered rotational speed ω. (7)

9 9 PROBLEM 3 Question 1: a) A B C A EI, µ Bm x u B x mü B mü B mü B b) A J θ B J θ B B C A EI, µ J B θ B x l J θ B l x l Fig. 1: Symmetric structural system. a) Symmetric eigenvibrations and equivalent system. b) Antisymmetric eigenvibrations and equivalent system. The system is mechanically and geometrically symmetric around a vertical line through point B. Hence, the eigenvibrations separate into symmetric and anti-symmetric eigenvibrations as shown on figs. 1a and 1b. The vertical displacement and rotation of point B are denoted u B (t) and θ B (t) with signs defined in figs. 1a and 1b. Symmetric eigenvibrations have rotation θ B (t) = 0. Hence, the d Alembert moment is J θ B (t) = 0 in these eigenvibrations. The inertial force becomes mü B (t), which is distributed equally to the beams to the left and right of point B. Hence, beam AB has zero rotation at point B and is exposed to an inertial force of magnitude mü b (t) immediately to the left of point B. Then, symmetric eigenvibrations may be analyzed by the equivalent system shown in fig. 1a to the right. Anti-symmetric eigenvibrations have displacement u B (t) = 0. Hence, the inertial force is mü b (t) = 0 in these eigenvibrations. The d Alembert moment becomes J θ B (t), which is distributed equally to the beams to the left and right of point B. Hence, beam AB has zero displacement and is exposed to a d Alembert moment of magnitude J θ B (t) immediatly to the left of point B. Then, anti-symmetric eigenvibrations may be analyzed by the equivalent system shown in fig. 1b to the right.

10 10 Question : Since the beam AB has constant bending stiffness EI and constant mass per unit length µ, the eigenmodes for both equivalent systems are given by, cf. (4-18), (4-19) ( Φ(x) = A sin λ x ) ( + B cos λ x ) ( + C sinh λ x ) ( + D cosh λ x ) a a a a λ 4 = µω l 4 EI The boundary conditions at point A (x = 0) for both equivalent systems read Φ(0) = 0 d dx Φ(0) = 0 (3) implies that D = B = 0. Then, (1) reduces to ( Φ(x) = A sin λ x ) ( + C sinh λ x ) l l (1) () (3) (4) The symmetric eigenvibration problem fulfills the following boundary conditions at point B (x = l), cf. (4-13) d dx Φ(l) = 0 ) d (EI d dx dx Φ(l) = ω mφ(l) (5) Insertion of (4) into (5) provides the following homogeneous equations for the determination of A and C [ ] [A ] cos λ cosh λ cos λ + m λ sin λ cosh λ + m λ sinh λ = µl µl C [ ] 0 0 (6) The frequency condition follows from the singularity of the coefficient matrix of (6), cf. (4-1). Then, the following frequency condition for symmetric eigenvibrations may be derived cosλcosh λ + m µl λ( cosλsinh λ sin λ cosh λ ) = 0 (7)

11 11 For m = µl the lowest three solutions to (7) become , j = 1 λ j = , j = , j = 6 (8) The anti-symmetric eigenvibration problem fulfills the following boundary conditions at point B (x = l), cf. (4-13) Φ(l) = 0 EI d dx Φ(l) = ω J d dx Φ(l) Insertion of (4) into (9) provides the following homogeneous equations for the determination of A and C (9) [ ][ ] sin λ sinh λ A sin λ J λ 3 cosλ sinh λ J λ 3 = cosh λ C ml 3 ml 3 Then, the following frequency condition for anti-symmetric eigenvibrations may be derived [ ] 0 0 (10) sin λ sinh λ + J µl 3λ3( cosλsinh λ sin λ cosh λ ) = 0 (11) For J = µl 3 the lowest three solutions to (11) become , j = λ j = , j = , j = 5 (1)

12 1 PROBLEM 4 Question 1: θ A u A EI, µ l m, J u B θ B Fig. 1: Definition of degrees of freedom of beam element. Since axial deformations are ignored, the plane beam element has four degrees of freedom. With the sign definition shown in Fig. 1 the local stiffness matrix and consistent mass matrix read, cf. (5-4), (5-43) k = EI l 3 1 6l 1 6l 6l 6l l 1 6l 1 6l 6l l 6l (1) m = µl l 54 13l l 13l 3l 54 13l 156 l 13l 3l l () For symmetric eigenvibrations we have u A = θ B = 0. Then the corresponding rows and columns can be removed from (1) and (). Further, the translational degree of freedom u B is related with a discrete mass m. Then, the global stiffness and mass matrices become K = EI l 3 [ ] 6l 6l 1 (3) M = µl [ ] 13l 40 13l m µl (4)

13 13 For anti-symmetric eigenvibrations we have u A = u B = 0. Then, the corresponding rows and columns can be removed from (1) and (). Further, the rotational degree of freedom θ B is related with a discrete mass moment of inertia J. Then, the global stiffness and mass matrices become K = EI l 3 [ ] l l (5) M = µl [ ] 3l 40 3l + 40 J µl With the use of the relevant system matrices the two lowest eigenfrequencies and eigenmodes of symmetric or antisymmetric eigenvibrations are determined from the eigenvalue problem, cf. (3-4) (6) ( K ω j M ) Φ (j) = 0, j = 1, (7) Question : For symmetric eigenvibrations with m = µl the eigenvalue problem (7) may be written as [[ ] 6l 6l 1 λ4 40 [ ]][ ] [ ] 13l Φ1 0 = 13l Φ (8) where [Φ, Φ ] denotes the components of the eigenvector Φ, and λ 4 is the same frequency parameter as defined in the continuous case in problem, given as λ 4 = µω l 4 EI The characteristic equation may be written, cf. (3-43) (9) 135s 508s + 1 = 0, s = λ4 40 s = { , j = , j = λ = { , j = , j = (10)

14 14 As expected (10) represent upper-bounds to the analytical solutions given by Eq. (8), Problem 3. Especially, the first symmetric eigenfrequency is accurately determined by the numerical solution. Using standard methods, the corresponding eigenmodes become [ ] Φ (1) 1 = [ ], Φ () 1 = (11) For anti-symmetric eigenvibrations with J = µl 3 the corresponding generalized eigenvalue problem (6) becomes [[ ] l l λ4 40 [ ]] [ ] [ ] 3l Φ1 0 = 3l 4 0 Φ (1) The characteristic equation may be written 1687s 174s + 1 = 0, s = λ4 40 s = { , j = , j = λ = { , j = , j = (13) The corresponding eigenmodes become [ ] Φ (1) 1 = [ ], Φ () 1 = (14) a) λ 1 = λ 4 = θ A = 1 u B = θ A = 1 u B = b) λ = λ 3 = θ A = 1 θ B = θ A = 1 θ B = Fig. : Undamped eigenmodes. a) 1st and nd symmetric eigenmode. b) 1st and nd anti-symmetric eigenmode. The eigenmode corresponding to the solutions (11) and (14) have been sketched on Fig.. Notice that the indicated displacements and rotations lack dimensional accuracy.