Figuren viser et linert system af 1 frihedsgrad med fjederkonstanten k, dmpningskonstanten

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1 SKRIFTLIG OMPRVE I SVINGNINGSTEORI Bygge- og Anlgskonstruktion,. semester Torsdg den 6. september kl..-6. Alle hjlpemidler er tilldt OPGAVE Figuren viser et linert system f frihedsgrd med fjederkonstnten k, dmpningskonstnten c og mssen m, der pvirkes f en periodisk dynmisk belstning, f(t), med perioden T. Belstningen er linert voksende over et tidsintervl T med mksimlvrdi f, efterfulgt f et ubelstet intervl, ligeledes f lngden T. Belstningens tidsvrition fremgr ivrigt f guren. Sprgsml (%, =4:4%) ) Bestem den sttionre periodiske bevgelse f mssen fr dennes sttiske ligevgtsposition. Sprgsml (%, =:5%) Bestem bevgelsen f mssen, nr denne strter med begyndelsesbetingelserne x() = _x() =.

2 OPGAVE Figuren viser et system f punktformige msser f strrelsen m, m og m, der 4 bevger sig i vndret retning i smme pln. Mssen m pvirkes f en hrmonisk vrierende krft f(t) = f cos(!t) med mplitude f og cirkulr frekvens!, virkende i mssernes bevgelsesretning. Msserne er indbyrdes forbundet med linert elstiske fjedre med fjederkonstnten k, og mssen m er endvidere forbundet til endnu en linert elstisk fjeder med fjederkonstnten k, der er fstnet til et fst vederlg i den nden ende. Der betrgtes kun sm svingninger f msserne ud fr den sttiske ligevgtstilstnd. Endvidere ses der bort fr dmpning. Sprgsml (5%, =4:%) Opstil ligningerne til bestemmelse f mssernes bevgelse ud fr den sttiske ligevgtstilstnd. Sprgsml (%, =6:6%) Bestem de cirkulre frekvenser for hvilke mssen m er i ro i den sttionre tilstnd, nr bevgelsen fr eventuelle begyndelsesbetingelser er klinget bort.

3 OPGAVE Figuren viser en horizontl, pln smmenst bjlke bestende f delbjlkerne AB og BC. Begge delbjlker er Bernoulli-Euler bjlker f lngden, med konstnt bjningsstivhed og konstnt msse pr. lngdeenhed. Bjlkerne er fst indspndt i punkterne A og C, og bjningsstift forbundet i punkt B. Bjlkerne er uendeligt stive over for xildeformtioner. I punkt B er nordnet en udbredt msse f strrelsen m og med msseinertimonentet J om en kse vinkelret p bjlkernes bjningscenter. Der betrgtes kun sm lodrette svingninger omkring bjlkens sttiske ligevgtstilstnd. Sprgsml (5%, =:%) Vis, t symmetriske og nti-symmetriske udmpede egensvingninger omkring punktet B kn bestemmes vh. de p guren viste kvivlente systemer. Sprgsml (5%, =5:%) Formuler betingelser til bestemmelse f cirkulre egenfrekvenser f henholdsvis symmetriske og nti-symmetriske udmpede egensvingninger. Frekvensbetingelserne forlnges ikke lst. OPGAVE 4 Sprgsml (5%, =5:4%) Problemet i opgve betrgtes igen. Nu nskes problemet i stedet lst ved hjlp f en elementmetode, idet der benyttes et enkelt element for hver f delbjlkerne AB og BC. Bestem de to lveste cirkulre egenfrekvenser for m = of J =.

4 4 SOLUTIONS PROBLEM Question : Fig. : Hlf-rnge expnsion of lod. Expnsion in sine series. The lod f(t) is dened for t>. However, we my extend the rnge of denition to the negtive t-xis to obtin n odd function s shown in g.. Then the lod my be represented by Fourier sine series f(t) = b n = T T Z T X n= Z T,T f t T b n sin n t T f(t) sin n t dt = T T sin n t T dt = f n Z T f(t) sin n t dt = T n sin n, cos Alterntively, the extension of the intervl of denition to the negtive t-xis my be performed in wy so tht n even function is obtined, or in wy so tht the time vrition of f(t) =f(t + T ) for t ], T;[ s pplied in Exmple -5. In these cses cosine series or full Fouries series is obtined, respectively. () my be written in the following form f(t) = X b n cos n t T, = X n Re b n e i(n tt, ) = n= n= X, Re F n e i! nt ; F n = b n e,i = b n e,i n ; n = n= ;! n = n T () is on the form (-9) with =. Then, the sttionry periodic motion follows from (-95), (-96) nd (-9) D.G. Zill nd M.R. Cullen: Dierentil Eutions with Boundry-Vlue Problems, 4th Ed. Brooks/Cole Publishing Compny, 997. () () ()

5 5 x () (t) = X n= n = (! n )+ jx n j cos(! n t, n) (4) ; (! n ) = rctn!! n! Tn! o, = rctn! n! o T, n (5) jx n j = b n m p (!,! n ) +4!! n = b n T m p (! T, n ) +4! T n (6) where! nd signify the undmped circulr eigenfreuency nd the dmping rtio s given by (-7) nd (-9). Insertion of (), (5) nd (6) into (4) provides the following solution for the sttionry periodic motion x () (t) = f T m X n=,, cos n sin, n n n p n (! T, n ) +4! T sin t n T, (! n ) (7) Question : The complete solution my be written x(t) =x () (t) +x () (t) () where x () (t) is dmped eigenvibrtion s given by (-4). The combined solution fulls the initil conditions x() = _x() =, which mens tht the eigenvibrtion must full the initil conditions x () () =,x () () ; _x () () =, _x () () (9) Insertion of (9) nd (-4) then provides the following solution x(t) =x () (t), e,! t x () () cos, p,! t + _x() () +! x () ()! p, sin, p,! t () The initil vlues x () () nd _x () () cn be clculted from (7).

6 6 PROBLEM Question : Fig. : Denition of degrees of freedom. Forces on free msses. The system hs degrees of freedom, which re dened s shown in g.. The msses re cut free from the springs nd the internl spring forces re pplied s externl forces. Then, Newton's nd lw of motion for ech of the free msses provides the following eutions of motion mx = f cos(!t), kx + k(x, x ) mx =,k(x, x )+k(x, x ) 4 mx =,k(x, x ) 9 >= >; ) Mx + Kx = f cos(!t) () x(t) = 4 x (t) x (t) x (t) 5 ; f = f 4 5 ; M = m ; K = k 4,,,5 (), Question : Since the system is undmped, the msses will be in phse or in counter-phse to the excittion in the sttionry stte, where the response from possible initil vlues hs dissipted wy. Then, the hrmonic response is on the form x(t) =X cos(!t) () X = 4 X X X 5 =, K,! M, f =,, 4,,,,, 5, 4 5 f k =

7 7 f kd(!) 4, 4 +, 5 (4) where = m! 4k (5) D(!) =(, ), (, )(, ), +,, (, ) =,6 +6, 5 + (6) The mss m will be t rest if X =, which mens tht, 4 + = ) j = (, p ; j = p + ; j = (7) From (5) nd (7) it follows tht the mss m will be t rest for the circulr vibrtion freuencies! j = < :, p + p k m ; j = k m ; j = () PROBLEM Question : Fig.: ) Euivlent system for symmetric eigenfreuencies. b) Euivlent system for nti-symmetric eigenfreuencies.

8 For symmetric eigenvibrtions the ngulr rottion t the point B becomes B. The inertil moment from the mss moment of inerti J then becomes,j B. Conseuently, J cn be ignored for symmetric eigenvibrtions. The inertil force,mu B is shred eully between the bems AB nd BC. All the indicted boundry conditions t the point B re represented by the euivlent system shown in g.. For nti-symmetric eigenvibrtions the displcement t the point B becomes u B. The inertil force from the mss m then becomes,mu B. Conseuently, m cn be ignored for nti-symmetric eigenvibrtions. The inertil moment,j B is shred eully between the bems AB nd BC. All the indicted boundry conditions t the point B re represented by the euivlent system shown in g. b. Question : Eigenvibrtions of the systems shown in gs. nd b, representing symmetric nd nti-symmetric eigenvibrtions of the underlying system, respectively, re determined by the following eigenvlue problems, cf. (4-) d4 dx,! (x) = ; x ];[ () = ; d dx () = d d dx() = ; dx () =,! m() d4 dx,! (x) = ; x ];[ () = ; d dx () = () = ; d dx () =! J d dx () 9 >= >; 9 >= >; () () The solution to the dierentil eutions of () nd () becomes, cf. (4-), (4-9) (x) =A sin x + B cos x + C sinh x + D cosh x 4 =! 4 The boundry conditions t x = provide B + D = D =,B (A + C) = ) C =,A Then, () reduces to (x) =A sin x, sinh x + B cos x, cosh x () (4) (5) (6)

9 9 In cse of symmetric eigenvibrtions, (6) is inserted into the boundry conditions t x = in (), which provides the following liner eutions for the determintion of A nd B (cos, cosh ),(sin + sinh ) (sin, sinh ), (cos + cosh ) + 4 m (sin, sinh ) + 4 m (cos, cosh ) A B (7) is homogeneous system of liner eutions. Non-trivil solutions A 6= _ B 6= re obtined if the determinnt of the coecient mtrix is zero. = is not n eigenvlue. After some reduction the following freuency condition my then be derived, sin cosh, cos sinh + m (7),, cos cosh = () For m = the lowest three solutions to () nd the circulr eigenfreuencies of the corresponding symmetric eigenvibrtions become j =! j = < : >< >: :7 ; j = 4:977 ; j =4 7:96446 ; j =6 :95455 :999 6:46 4 ; j = 4 ; j =4 4 ; j =6 In cse of nti-symmetric eigenvibrtions, (6) is inserted into the boundry conditions t x = in (), which provides the following liner eutions for the determintion of A nd B sin, sinh cos, cosh (sin + sinh )+ 5 J (cos, cosh ) (cos + cosh ), 5 J (sin + sinh ) From the determinnt condition for obtining non-trivil solutions the following freuency condition my be derived, sin cosh + cos sinh + J (9) () = A = B (),, cos cosh = () For J = the lowest three solutions to () nd the circulr eigenfreuencies of the corresponding nti-symmetric eigenvibrtions become j = < : :444 ; j = 4:795 ; j = 7:557 ; j =5 ()

10 ! j = >< >: :994 4 ; j = : ; j = 6:759 4 ; j =5 (4) PROBLEM 4 Fig. : Finite element model of bem. Globl node numbering nd denition of globl degrees of freedom. The nite element model is shown in g.. The globl degrees of freedom hve been indicted. Locl nd globl coordinte systems re co-directionl. The locl stiness nd consistent mss mtrices for the two elements become, cf. (5-4) nd (5-4) k = k = 6 4 m = m = 4 6, 6 6 4,6,,6,6 6, , 4, 54 56,,,, 4 5 () 7 5 () The globl stiness nd mss mtrices with no correction for geometricl boundry conditions or for the concentrted mss elements t point B re ssembled s indicted in (5-7) nd (5-7). Due to the geometricl constrints u A (t) = u C (t) = A (t) = C (t) = the system hs but globl degrees of freedom x(t) =[u B (t); B (t)] T. The corresponding globl stiness mtrix nd globl consistent mss mtrix become K = 4 ()

11 M = 56 + m 4 + J (4) As seen, the globl mss mtrix hs been djusted by dding the discrete mss of mgnitude m = in the min digonl t u B (t) nd the mss moment of inerti J in the min digonl t B (t), cf. (5-6). For m = nd J = the following freuency condition is obtined, cf. (-4) det 4, 576j, 44 = ; j =; (5) j j = 4! j ; j =; (6) (5) nd (6) give j =! j = < : >< >: 5 ; j = 4 ; j = : ; j = :954 4 ; j = () (9) The corresponding nlyticl solutions re indicted in es. () nd (4) of the solution to Problem! j = >< >: :994 4 ; j = : ; j = ()