SKRIFTLIG EKSAMEN I SVINGNINGSTEORI Bygge- og Anlægskonstruktion, 8. semester Onsdag, den 19. juni 2002 kl Alle hjælpemidler er tilladt

Transkript

1 SKRIFTLIG EKSAMEN I SVINGNINGSTEORI Bygge- og Anlægskonstruktion, 8. semester Onsdg, den 19. juni 2002 kl Alle hjælpemidler er tilldt OPGAVE 1 f t)=f 0cos ùt) A B C c 0 D Figuren viser en horizontl pln msseløs bjælke ABC f længden 2. Bjælken, der ntges uendelig stiv og fri for dæmpning, er simpelt understøttet i punkt A. I punktet C er bjælken understøttet f en fjedrende understøtning, der består f 2 lineært elstiske serieforbundne fjedre med fjederkonstnterne, der er prllelforbundet med et lineært viskost dæmperelement med dæmpningskonstnten c 0. Mellem fjedrene er plceret en punktformig msse f størrelsen. Fjedersystemet er fst understøttet ved punktet D. Fjedre og dæmpere er lle nordnet i lodret retning, og mssen ntges fstholdt således, t denne kun kn foretge bevægelser i lodret retning. I øvrigt betrgtes kun små bevægelser f bjælken omkring en vndret sttisk ligevægtstilstnd. Bjælken belstes i midtpunktet B f en lodret hrmonisk vrierende krft ft) = f 0 cosωt) med mplituden f 0 og den cirkulære frekvens ω. Spørgsmål 1 25%) Bestem den sttionære bevægelse f punkt C og f mssen,når krften ft) hr virket så længe, t responset fr eventuelle begyndelsesbetingelser er klinget ud. Spørgsmål 2 10%) Bestem den cirkulære frekvens for hvilken bjælken ABC forbliver i ro, og bestem mplituden f mssen i denne tilstnd.

2 2 OPGAVE 2 A B C D Figuren viser en pln rmmekonstruktion, bestående f de vndrette delbjælker AB og BC og den lodrette bjælke BD. Delbjælkerne er lle msseløse Bernoulli-Eulerbjælker med konstnt bøjningsstivhed og længden, der er bøjningsstift forbundet i punkt B. Konstruktionen er fst indspændt i punkt D, og lle bjælker ntges uendeligt stive overfor xildeformtioner. I de frie bjælkeender A og C er vedhæftet 2 punktformige msser f størrelsen m. Kun små svingninger omkring den sttiske ligevægtstilstnd betrgtes, og der ses bort fr indflydelsen f eventuelle normlkræfter på det dynmiske respons. Spørgsmål 1 20%) Bestem konstruktionens udæmpede cirkulære egenfrekvenser og egensvingningsformer. Hjælp: En lodret neddrettet enhedskrft i punkt C medfører en neddrettet flytning f størrelsen i punkt C, en opdrettet flytning f størrelsen i punkt A, ogen vndret flytning mod højre f størrelsen f punkt C. En vndret enhedskrft rettet mod højre i punkt C medfører en flytning i smme retning f dette punkt på

3 3 OPGAVE 3 A,ì C D 2 Systemet i opgve 1 betrgtes igen. Blot ntges nu, t den vndrette bjælke AC f længden 2 er en pln Bernoulli-Eulerbjælke med konstnt bøjningsstivhed og msse pr. længdeenhed µ. Endvidere ses der bort fr dæmpningen i det understøttende fjedersystem, svrende til c 0 = 0. Der ses endvidere bort fr indflydelsen f en eventuel normlkrft i bjælke AC på det dynmiske respons, og kun små svingninger omkring den vndrette sttiske ligevægtstilstnd betrgtes. Spørgsmål 1 20%) Opstil frekvensbetingelsen til bestemmelse f udæmpede cirkulære egenfrekvenser f konstruktionen. Der kræves ingen numerisk løsning f frekvensbetingelsen.

4 4 OPGAVE 4 mg m k V,ì l Et køretøj, der modelleres vh. et enkelt frihedsgrders system bestående f en msse m og en lineært elstisk fjeder med fjederstivhed k, bevæger sig med den konstnte hstighed V lngs en vndret flde. Til tiden t = 0 bevæger køretøjet sig ind på en vndret bro med det frie spænd l, der modelleres vh. en simpelt understøttet Bernoulli- Eulerbjælke med konstnt bøjningsstivhed og konstnt msse pr. længdeenhed µ. Broen ntges fri for dæmpning, og køretøjet ntges t belste broen, således t denne kun udfører små bevægelser i plnen omkring en vndret ligevægtstilstnd. Der ses bort fr indflydelsen f eventuelle tryknormlkræfter i broens længdekse på dennes dynmiske respons. Spørgsmål 1 25%) Idet broens respons modelleres vh. en enkelt modl frihedsgrd, og køretøjets dynmiske bevægelse ntges kun t foregå i lodret retning, skl mn opstille bevægelsesligningerne for broen og køretøjet. Køretøjets egenvægt mg, hvorg er tyngdeccelertionen, tges med i nlysen. Såvel broen som køretøjets msse ntges i ro til tiden t =0.

5 5 SOLUTIONS PROBLEM 1 Question 1: ft)=f cos ùt) A B C 0 x 1. x -x )-c x c 0 x -x ) 2 1 x 2 D x 2 Fig. 1: Forces on free bem nd free mss. The verticl displcement x 1 of point C nd the verticl displcement x 2 of the mss re introduced s degrees of freedom with signs s defined in fig. 1. Then the extension of the upper spring is x 2 x 1, nd the compression of the lower spring is x 2. The velocity of the dmper element is ẋ 1. Hence, the interction force between the bem nd the supporting spring system t point C becomes x 2 x 1 ) c 0 ẋ 1. The bem is cut free from the supporting spring system nd the interction force is pplied s n externl force with sign defined in fig. 1. Similrly, the mss is cut free from the springs, nd the spring forces x 2 x 1 )nd x 2 re pplied s externl forces with signs s defined in fig. 1. Next, moment equilibrium is formulted for the bem round point A, nd Newton s 2nd lw of motion is pplied for the free mss, leding to the following equtions of motion ft) + ) } x 2 x 1 ) c 0 ẋ 1 2 =0 ẍ 2 = x 2 x 1 ) x 2 Mẍ + Cẋ + Kx = ft) 1) where [ ] x1 t) xt) = x 2 t) [ ] 0 0 M = 0 [ ] f0, ft) =F cos ωt, F = 0 [ ] [ ] 2c0 0 2k0 2k, C =, K = )

6 6 Becuse the dmper is cting indirectly on the mss vi the upper spring,the system is not genuine single-degree-of-freedom system. Insted, the indicted twodegrees-of-freedom formultion becomes necessry. The following complex formultion is introduced for the force ft) =Re Fe iωt) The sttionry hrmonic response then becomes, cf ), 3-101) xt) =Re Xe iωt) 3) X = K + iωc ω 2 M ) 1 F = [ ] 1 [ ] 2k0 +2c 0 ωi 2 f0 2 ω 2 = 1 [ ] 2k0 ω 2 f 0 D 4) 0 D =2 ω 2 )2 +2c 0 ωi) 2k c 0 ω 3 i 2 ω 2 +4 c 0 ωi +2k 2 0 5) Question 2: The bem remins t rest, if x 1 t) 0. As seen from 4) this is the cse t the circulr eigenfrequency ω = 2k0 6) 6) is the eigenfrequency of the mss if point C is fixed. With ω given by 6) the determinnt D becomes, cf. 5) D = 2k 2 0 7) Then, the mplitude of the mss becomes X 2 = 2k0 2 f 0 = f 0 2 8) With X 1 = 0, the interction force becomes X 2 cosωt). This force must blnce the externl force f 0 cosωt) cting t point B t ll times, which is obtined with the mplitude X 2 given by 8).

7 7 PROBLEM 2 Question 1:.. -mx 1 m A B C.. -mx 2 m x mx 3 x 1 x 2 D Fig. 1: Definition of degrees of freedom. The bems of the structure re mssless nd inextensible. Hence, the structure hs 3 degrees of freedom, which re selected s the verticl displcements x 1 t) ndx 2 t) of the points A nd C, nd the common horizontl displcement x 3 t) ofthebemabc with signs s defined in fig. 1. The ltter degree of freedom is relted with the mss 2m. Next, inertil forces mẍ 1, mẍ 2 nd 2mẍ 3 re pplied s externl forces in the direction of the defined degrees of freedom. Then, the equtions of motion red, cf. 3.1) x 1 t) =δ 11 mẍ 1 )+δ 12 mẍ 2 )+δ 13 2mẍ 3 ) x 2 t) =δ 21 mẍ 1 )+δ 22 mẍ 2 )+δ 23 2mẍ 3 ) 1) x 3 t) =δ 31 mẍ 1 )+δ 32 mẍ 2 )+δ 33 2mẍ 3 ) The flexibility coefficients re given s D = δ 11 δ 12 δ 13 δ 21 δ 22 δ 23 = ) 6 δ 31 δ 32 δ The equtions of motion my then be written in the following mtrix form Mẍ + Kx = 0 xt) = x 1t) x 2 t), M = m x 3 t) 0 0 2, K = D 1 = ) )

8 8 The undmped circulr eigenfrequencies nd eigenmodes Φ j)t = [Φ j) 1 Φj) 2 Φj) 3 ]re obtined s non-trivil solutions of the homogeneous liner equtions, cf. 3-42) 7 λ j 3 6 Φ j) λ j 6 Φ j) 2 = 0 0 5) λ j Φ j) 0 3 where λ j = 10 3 m 3 ω2 j 6) The chrcteristic eqution becomes det 7 λ λ 6 = 2λ 3 j +56λ 2 j 400λ j = λ j 9 61, j =1 λ j = 10, j = , j = ω j = m 3, j =1 m 3, j =2 m 3, j =3 7) 8) The eigenmodes re normlized s follows 1 Φ j) = Φ j) 2 Φ j) 3 9) Next, the components Φ j) 2 nd Φ j) 3 re determined from the 2nd nd 3rd equtions of 5) [ 7 λj λ j [ j) ] Φ 2 Φ j) = 3 ][ j) ] Φ 2 Φ j) = 3 [ ] 3 6 [ ] λj 2λ 2 j 42λ j λ j 10)

9 9 Insertion of 7) then provides the following undmped eigenmodes Φ 1) = ) Φ 2) = Φ 3) = ) 11) PROBLEM 3 Question 1: A,ì C Öx) x 2 ux,t) D u2,t) kx-u) 0 2 x t) 2 x -u) 2 x 2 Fig. 1: Definitions. The bem hs constnt bending stiffness, constnt mss per unit length µ, nd the norml force in the sttic equilibrium stte is N = 0. Hence, the eigenmode is given by 4.18) nd 4.19) Φx) =A sin λ x ) + B cos λ x ) + C sinh λ x ) + D cosh λ x ) 1) λ 4 = µω2 2) 2 2) The boundry conditions t point Ax = 0) become, cf. 4-23) Φ0) = d2 Φ0) = 0 3) dx2 Insertion of 1) into 3) provides B = D = 0, cf. 4-20). Then 1) reduces to Φx) =A sin λ x ) + C sinh λ x ) 2 2 4)

10 10 Q2,t) C x -u) 2 M2,t) k x t)-u2,t)) 0 2 x 2 x 2 Fig. 2: Forces on free end point C, nd free mss. The mechnicl boundry conditions t point Cx = 2) specify tht the bending moment is 0, nd the sher force must blnce the interction for x2 t) u2, t) ) of the supporting spring system, where x 2 t) denotes the verticl displcement of the mss m, ndu2, t) is the displcement of the bem t point C. The eqution of motion of the mss reds, see fig. 2 ẍ 2 = x2 u2, t) ) x 2 5) Under hrmonic vibrtions, where ux, t) =Φx)cosωt) ndx 2 t) =X 2 cosωt), 5) provides the following reltions between the mplitudes X 2 nd Φ2) ω 2 X 2 = X2 Φ2) ) X 2 X 2 = 2 ω 2 Φ2) 6) Hence the interction force will be hrmonic vrying with the mplitude where X2 Φ2) ) = k 1 ω)φ2) 7) k 1 ω) = ω 2 2 ω 2 8) k 1 ω) my be interpreted s n equivlent liner spring of the spring system under hrmonic motion. For =0orω = 0), 8) ttins the vlue k 1 0) = 1 2, corresponding to the sttic replcement spring of the two springs in series. For = or ω = ), 8) provides the solution k 1 ) =. In this cse the mss is t rest, nd only the upper spring is effective. Notice tht for frequencies in the rnge m <ω2 < 2, k 1 ω) becomes negtive. In this frequency bnd the spring system will push t the bem, rther thn support it. The boundry conditions t x =2 now become, cf. 4-13) d 2 Φ2) dx 2 =0 d3 Φ2) dx 3 = k 1 ω)φ2)

11 11 d 2 Φ1) dξ 2 =0 d 3 Φ1) dξ 3 = κ κ αλ4 2κ αλ 4 Φ1) 9) where ξ = x 2, κ = 2) 3, α = µ2 10) κ nd α re non-dimensionl prmeters for the concentrted spring stiffness nd mss. By inserting 4) into 9) the following homogeneous equtions re obtined for the determintion of A nd C λ 2 A sin λ + C sinh λ) =0 λ 3 A cos λ + C cosh λ) =κ 1 λ)a sin λ + C sinh λ) =0 [ sin λ sinh λ λ 3 cos λ κ 1 λ) sinh λ 3 cosh λ κ 1 λ) sinh λ } ][ ] A = C [ ] ) where κ 1 λ) =κ κ αλ4 2κ αλ 4 12) The frequency condition then becomes sin λ λ 3 cosh λ κ 1 λ) sinh λ ) sinh λ λ 3 cos λ κ 1 λ) sin λ ) =0 λ 3 cot λ λ 3 coth λ +2κ 1 λ) = 0 13) For κ 0 =20ndα = 1 the 6 lowest solutions of 13) become , j = , j = , j =3 λ j = , j = , j = , j =6

12 12 PROBLEM 4 Question 1: yt) mg m k V m kuvt,t)-y) Vt x Rt) ux,t) l,ì Fig. 1: Definition of structurl system nd forces on free vehicle mss. The sttic equilibrium stte of the vehicle mss is defined s the position of the mss, when the vehicle is t rest on horizontl plne outside the bridge. The dynmic displcement yt) of the mss is mesured from this position with sign s shown in fig. 1. The sttic force mg is directly trnsmitted through the spring nd mkes up prt of the rection force Rt). The displcement of the bridge t position x t the time t is denoted ux, t) with sign s defined in fig. 1. Then, t the time t [ ] 0, l V the displcement of the bridge t the position x = Vt of the vehicle becomes uvt,t). Hence, the elongtion of the vehicle spring is uvt,t) yt). The vehicle mss is cut free from the spring, nd the spring force k uvt,t) yt) ) is pplied with sign s shown in fig. 1. Then, Newton s 2nd lw of motion for the free vehicle mss provides mÿ = k uvt,t) y ),t [0, l V ] } mÿ = ky, t ] l V, [ with the initil conditions y0) = ẏ0) = 0 2) 1) The initil conditions 2) nd the second eqution in 1) follow from the ssumption of the surfce being smooth nd horizontl on both sides of the bridge. Then, the rection force between the bridge nd the vehicle my be written { ) mg + k yt) uvt,t), t [0, l Rt) = V ] } 0, t ] l V, [ The bridge is modelled s single-degree-of-freedom system. Hence, the following pproximtion pplies, cf. 4-52) ux, t) Φ 1) x)q 1 t) 4) 3)

13 The fundmentl eigenmode for the simply supported bem with homogeneous cross sections becomes, cf. 4-31) Φ 1) x) = sin π x ) 5) l 13 Since the bridge is free of dmping, nd t rest t the time t = 0, the equtions of motion nd the initil vlues of the modl coordinte q 1 t) become, cf. 4-53) q 1 + ω 2 1q 1 = 1 M 1 F 1 t) 6) q 1 0) = q 1 0) = 0 7) where, cf. 4-33), 4-65), 4-69) ω1 2 = π4 l 4 µ M 1 = 1 2 µl 8) 9) where F 1 = { Rt) sinωt), t [0, l V ] 0, t ] l V, [ 10) ω = π V l 11) Then, insertion of uvt,t)=φ 1) Vt)q 1 t) = sinωt)q 1 t) into 1) provides the following eqution of motion for the vehicle mss ÿ + ω0y 2 ω0 2 sinωt)q 1 =0, t [0, l } V ÿ + ω0y 2 =0, t ] l V, [ 12) where ω 2 0 = k m 13) ω 0 is the circulr eigenfrequency of the vehicle t stndstill position. However, in 12) ω 0 should merely be considered s prmeter. Similrly, 3), 6), 9), 10) provide the following eqution of motion for the bridge ) q 1 + ω1q 2 1 = 2m µl g + ω0 ) } 2 y sinωt)q1 sinωt), t [0, l V ] q 1 + ω1q 2 1 =0, t ] l V, [

14 14 q 1 + ω αω 2 0 sin 2 ωt) ) q 1 αω 2 0 sinωt)y = αg sinωt), t [0, l V ] q 1 + ω 2 1q 1 =0, t ] l V, [ } 14) where α denotes the following mss rtio α = 2m µl 15) 12) nd 14) with the initil vlues 2) nd 7) cn only be solved numericlly. This is consequence of time-dependent coefficients of the differentil equtions, for which no generl solution strtegy is vilble for obtining fundmentl system of solutions to the corresponding homogeneous equtions.