BENDING FREQUENCIES OF BEAMS, RODS, AND PIPES Revision S

Transcription

1 BENDING FREQUENCIES OF BEAMS, RODS, AND PIPES Revisio S By Tom Irvie tom@vibratioata.com November, Itrouctio The fuametal frequecies for typical beam cofiguratios are give i Table. Higher frequecies are give for selecte cofiguratios. Table. Fuametal Beig Frequecies Cofiguratio Frequecy (Hz) f =.556 Catilever f = 6.68 f f = 7.56 f Catilever with E Mass m f =.5 m Simply-Supporte at both Es (Pie-Pie) f = f =, =,,,. (rigi-boy moe) Free-Free f =.7 f =.757 f f = 5. f

2 Table. Fuametal Beig Frequecies (cotiue) Cofiguratio Fixe-Fixe Frequecy (Hz) Same as free-free beam except there is o rigi-boy moe for the fixe-fixe beam. Fixe - Pie f = 5.8 where E I P is the moulus of elasticity is the area momet of iertia is the legth is the mass esity (mass/legth) is the applie force Note that the free-free a fixe-fixe have the same formula. The erivatios a examples are give i the appeices per Table. Table. Table of Cotets Appeix Title Mass Solutio A B Catilever Beam I Catilever Beam II E mass. Beam mass is egligible Beam mass oly Approximate Approximate C Catilever Beam III Both beam mass a the e mass are sigificat Approximate D Catilever Beam IV Beam mass oly Eigevalue E F Beam Simply-Supporte at Both Es I Beam Simply-Supporte at Both Es II Ceter mass. Beam mass is egligible. Beam mass oly Approximate Eigevalue

3 Table. Table of Cotets (cotiue) Appeix Title Mass Solutio G Free-Free Beam Beam mass oly Eigevalue H I Steel Pipe example, Simply Supporte a Fixe-Fixe Cases Rocket Vehicle Example, Free-free Beam Beam mass oly Beam mass oly Approximate Approximate J Fixe-Fixe Beam Beam mass oly Eigevalue K Fixe-Pie Beam Beam mass oly Eigevalue Referece. T. Irvie, Applicatio of the Newto-Raphso Metho to Vibratio Problems, Revisio E, Vibratioata,.

4 APPENDIX A Catilever Beam I Cosier a mass moute o the e of a catilever beam. Assume that the e-mass is much greater tha the mass of the beam. m g Figure A-. E is the moulus of elasticity. I is the area momet of iertia. is the legth. g is gravity. m is the mass. The free-boy iagram of the system is M R R mg Figure A-. R is the reactio force. M R is the reactio beig momet. Apply Newto s law for static equilibrium. forces (A-) R - mg = (A-)

5 R = mg (A-) At the left bouary, momets (A-) M R - mg = M R = mg (A-5) (A-6) Now cosier a segmet of the beam, startig from the left bouary. V y x M R R M Figure A-. V is the shear force. M is the beig momet. y is the eflectio at positio x. Sum the momets at the right sie of the segmet. momets (A-7) M R - R x - M = M = M R - R x (A-8) (A-9) The momet M a the eflectio y are relate by the equatio M y (A-) 5

6 Itegratig, y M R R x (A-) y mg mg x (A-) y mg x (A-) mg y x (A-) y mg x x a (A-5) Note that a is a itegratio costat. Itegratig agai, mg y( x) x x ax b 6 (A-6) A bouary coitio at the left e is y() = (zero isplacemet) (A-7) Thus b = (A-8) Aother bouary coitio is y' (zero slope) (A-9) 6

7 Applyig the bouary coitio to equatio (A-6) yiels, a = (A-) The resultig eflectio equatio is mg y( x) x x 6 (A-) The eflectio at the right e is mg y( ) 6 mg y( ) (A-) (A-) Recall Hooke s law for a liear sprig, F = k y (A-) F is the force. k is the stiffess. The stiffess is thus k = F / y (A-5) The force at the e of the beam is mg. The stiffess at the e of the beam is k mg mg k (A-6) (A-7) 7

8 The formula for the atural frequecy f of a sigle-egree-of-freeom system is k f (A-8) m The mass term m is simply the mass at the e of the beam. The atural frequecy of the catilever beam with the e-mass is fou by substitutig equatio (A-7) ito (A-8). f (A-9) m 8

9 APPENDIX B Catilever Beam II Cosier a catilever beam with mass per legth. Assume that the beam has a uiform cross sectio. Determie the atural frequecy. Also fi the effective mass, where the istribute mass is represete by a iscrete, e-mass., Figure B-. The goverig ifferetial equatio is y x y t (B-) The bouary coitios at the fixe e x = are y() = (zero isplacemet) (B-) y x (zero slope) (B-) The bouary coitios at the free e x = are y x (zero beig momet) (B-) y x (zero shear force) (B-5) Propose a quarter cosie wave solutio. 9

10 y x yo x ( ) cos (B-6) y y o x si (B-7) y yo x cos (B-8) y yo x x si (B-9) The propose solutio meets all of the bouary coitios expect for the zero shear force at the right e. The propose solutio is accepte as a approximate solutio for the eflectio shape, espite oe eficiecy. The Rayleigh metho is use to fi the atural frequecy. The total potetial eergy a the total kietic eergy must be etermie. The total potetial eergy P i the beam is P y (B-) By substitutio, P y o x cos (B-) P yo x cos (B-) P yo x cos (B-)

11 P y o x x si (B-) P yo (B-5) P y o 6 (B-6) The total kietic eergy T is T y (B-7) T yo x cos (B-8) T yo x x cos cos (B-9) T yo x x cos cos (B-) T yo x x cos cos (B-) T yo x x cos cos (B-) T y o x x x si si (B-) T yo (B-)

12 T yo 8 (B-5) Now equate the potetial a the kietic eergy terms. 8 6 y o y o (B-6) 8 6 (B-7) 6 8 (B-8) 6 8 / (B-9) f 6 8 / (B-) f 6 8 / (B-)

13 / f 8 (B-) f 66. (B-) Recall that the stiffess at the free of the catilever beam is k (B-) The effective mass m eff at the e of the beam is thus meff k f (B-5) m eff 66. (B-6) m eff 5. (B-7) m eff 5. (B-8)

14 APPENDIX C Catilever Beam III Cosier a catilever beam where both the beam mass a the e-mass are sigificat., m g Figure C-. The total mass m t ca be calculate usig equatio (B-8). m t 5. m (C-) Agai, the stiffess at the free of the catilever beam is k (C-) The atural frequecy is thus f. 5 m (C-)

15 APPENDIX D Catilever Beam IV This is a repeat of part II except that a exact solutio is fou for the ifferetial equatio. The ifferetial equatio itself is oly a approximatio of reality, however., Figure D-. The goverig ifferetial equatio is y x y t (D-) Note that this equatio eglects shear eformatio a rotary iertia. Separate the epeet variable. y( x, t) Y( x) T(t) (D-) Y( x) T( t) Y ( x) T( t) x t (D-) T(t) Y( x) Y( x) T(t) (D-) t 5

16 Y( x) Y( x) T(t) t T(t) (D-5) et c be a costat Y( x) Y( x) T(t) t c T(t) (D-6) Separate the time variable. T(t) t c (D-7) T(t) T(t) c T(t) (D-8) t Separate the spatial variable. Y( x) c Y( x) (D-9) Y x c ( ) Y ( x ) (D-) A solutio for equatio (D-) is Y( x) a sih x a cosh x a si x a cos x (D-) 6

17 Y( x) a cosh x asih x a cos x asi x (D-) Y( x) a x a x a x a x sih cosh si cos (D-) Y( x) a x a x a x a x cosh sih cos si (D-) Y( x) a x a x a x a x sih cosh si cos (D-5) Substitute (D-5) a (D-) ito (D-). a sih x a cosh x a si x a cos x c a x a x a x a x sih cosh si cos (D-6) a sih x a cosh x a si x a cos x c a sih x a cosh x a si x a cos x The equatio is satisfie if c (D-7) (D-8) c / (D-9) 7

18 The bouary coitios at the fixe e x = are Y() = (zero isplacemet) (D-) Y x (zero slope) (D-) The bouary coitios at the free e x = are Y x Y x (zero beig momet) (D-) (zero shear force) (D-) Apply equatio (D-) to (D-). a a (D-) a a (D-5) Apply equatio (D-) to (D-). a a (D-6) a a (D-7) Apply equatio (D-) to (D-). a sih a cosh a si a cos (D-8) Apply equatio (D-) to (D-). a cosh a sih a cos a si (D-9) Apply (D-5) a (D-7) to (D-8). a sih a cosh a si a cos (D-) 8

19 si sih cos cosh a a (D-) Apply (D-5) a (D-7) to (D-9). a cosh a sih a cos a si (D-) cos cosh si sih a a (D-) Form (D-) a (D-) ito a matrix format. si sih cos cosh cos cosh si sih a a (D-) By ispectio, equatio (D-) ca oly be satisfie if a = a a =. Set the etermiat to zero i orer to obtai a otrivial solutio. si sih cos cosh (D-5) si sih cos cos cosh cosh (D-6) si sih cos cos cosh cosh (D-7) cosh cos (D-8) cosh cos (D-9) cos cosh (D-) There are multiple roots which satisfy equatio (D-). Thus, a subscript shoul be ae as show i equatio (D-). cos cosh (D-) 9

20 The subscript is a iteger iex. The roots ca be etermie through a combiatio of graphig a umerical methos. The Newto-Rhapso metho is a example of a appropriate umerical metho. The roots of equatio (D-) are summarize i Table D-, as take from Referece. Table D-. Roots Iex =.875 =.699 > (-)/ Note: the root value formula for > is approximate. Rearrage equatio (D-9) as follows c (D-) Substitute (D-) ito (D-8). T(t ) T(t) t (D-) Equatio (D-) is satisfie by T(t) b si t b cos t (D-)

21 The atural frequecy term is thus (D-5) Substitute the value for the fuametal frequecy from Table D (D-6).556 f (D-7) Substitute the value for the seco root from Table D (D-8). f (D-9) f 6.68f (D-5) Compare equatio (D-7) with the approximate equatio (B-). SDOF Moel Approximatio The effective mass m eff at the e of the beam for the fuametal moe is thus meff k f (D-5)

22 m eff 556. (D-5) m eff. 596 (D-5) m eff 7. (SDOF Approximatio) (D-5) Eigevalues (-)/ Note that the root value formula for > 5 is approximate. Normalize Eigevectors Mass ormalize the eigevectors as follows Y (x) (D-55)

23 The calculatio steps are omitte for brevity. The resultig ormalize eigevectors are Y (x) cosh x cos x.7 sih x si x (D-56) Y (x) cosh x cos x.87 sih x si x (D-57) Y (x) cosh x cos x.999 sih x si x (D-58) Y (x) cosh x cos x. sih x si x (D-59) The ormalize moe shapes ca be represete as Yi (x) cosh i x cos i x Di sih i x si i x (D-6) where Di i cosh i sih cos (D-6) si i i Participatio Factors The participatio factors for costat mass esity are Y (x) (D-6)

24 The participatio factors from a umerical calculatio are.78 (D-6).9 (D-6).5 (D-65).88 (D-66) The participatio factors are o-imesioal. Effective Moal Mass The effective moal mass is Y (x) m eff, (D-67) Y (x) The eigevectors are alreay ormalize such that Y (x) (D-68) Thus, m eff, Y (x) (D-69) The effective moal mass values are obtaie umerically. m eff, meff, meff,.6 (D-7).88 (D-7).67 (D-7) meff,.6 (D-7)

25 APPENDIX E Beam Simply-Supporte at Both Es I Cosier a simply-supporte beam with a iscrete mass locate at the mile. Assume that the mass of the beam itself is egligible. m g Figure E-. The free-boy iagram of the system is Ra mg Rb Figure E-. Apply Newto s law for static equilibrium. forces (E-) Ra + Rb - mg = Ra = mg - Rb (E-) (E-) At the left bouary, 5

26 momets (E-) Rb - mg = Rb = mg ( / ) Rb = (/) mg (E-5) (E-6a) (E-6b) Substitute equatio (E-6) ito (E-). Ra = mg (/)mg Ra = (/)mg (E-7) (E-8) x V y Ra mg M Sum the momets at the right sie of the segmet. momets (E-9) - R a x + mg <x- > - M = (E-) 6

27 Note that < x- > eotes a step fuctio as follows, for x x x, for x M = - R a x + mg <x- > (E-) (E-) M = - (/)mg x + mg <x- > M = [ - (/) x + <x- > ][ mg ] (E-) (E-) y [ - (/) x x - ][ mg ] (E-5) y(x) mg y [ - (/) x x - ] (E-6) mg y [ - x x - ] a (E-7) mg - x x - ax b 6 (E-8) The bouary coitio at the left sie is y() = (E-9a) This requires b = (E-9b) Thus y(x) mg - x x - ax 6 (E-) The bouary coitio o the right sie is y() = (E-) 7

28 8 a mg (E-) a mg 8 - (E-) a mg (E-) a mg 8 - (E-5) a mg 6 - (E-6) mg 6 a (E-7) mg 6 a (E-8) Now substitute the costat ito the isplacemet fuctio x mg 6 mg x - 6 x - y(x) (E-9) mg x - 6 x 6 x - y(x) (E-) The isplacemet at the ceter is mg y (E-)

29 9 mg 96 - y (E-) mg y (E-) mg 96 y (E-) mg 8 y (E-5) Recall Hooke s law for a liear sprig, F = k y (E-6) F is the force. k is the stiffess. The stiffess is thus k = F / y (E-7) The force at the ceter of the beam is mg. The stiffess at the ceter of the beam is 8 mg mg k (E-8) 8 k (E-9) The formula for the atural frequecy f of a sigle-egree-of-freeom system is

30 f k m (E-) The mass term m is simply the mass at the ceter of the beam. 8 f (E-) m f 6.98 (E-) m

31 APPENDIX F Beam Simply-Supporte at Both Es II Cosier a simply-supporte beam as show i Figure F-., Figure F-. Recall that the goverig ifferetial equatio is y x y t (F-) The spatial solutio from sectio D is Y( x) a sih x a cosh x a si x a cos x (F-) Y( x) a x a x a x a x sih cosh si cos (F-) The bouary coitios at the left e x = are Y() = (zero isplacemet) (F-) Y x (zero beig momet) (F-5)

32 The bouary coitios at the right e x = are Y() = (zero isplacemet) (F-6) Y x (zero beig momet) (F-7) Apply bouary coitio (F-) to (F-). a a (F-8) a a (F-9) Apply bouary coitio (F-5) to (F-). a a (F-) a a (F-) Equatios (F-8) a (F-) ca oly be satisfie if a a (F-) a (F-) The spatial equatios thus simplify to Y( x) a sih x a si x (F-) Y( x) a x a x sih si (F-5) Apply bouary coitio (F-6) to (F-). si a sih a (F-6)

33 Apply bouary coitio (F-7) to (F-5). si a sih a (F-7) si a sih a (F-8) sih sih si si a a (F-9) By ispectio, equatio (F-9) ca oly be satisfie if a = a a =. Set the etermiat to zero i orer to obtai a otrivial solutio. si sih si sih (F-) sih si (F-) si sih Equatio (F-) is satisfie if (F-),,,,... (F-),,,,... (F-) The atural frequecy term is (F-5),,,,... (F-6) f,,,,... (F-7)

34 f,,,,... (F-8) SDOF Approximatio Now calculate effective mass at the ceter of the beam for the fuametal frequecy. (F-9) Recall the atural frequecy equatio for a sigle-egree-of-freeom system. k m (F-) Recall the beam stiffess at the ceter from equatio (E-9). k 8 (F-) Substitute equatio (F-) ito (F-). 8 m (F-) Substitute (F-) ito (F-9). 8 m (F-) 8 m (F-) 8 m m 8 (F-5) (F-6)

35 The effective mass at the ceter of the beam for the first moe is m 8 (SDOF Approximatio) (F-7) Normalize Eigevectors The eigevector a its seco erivative at this poit are Y( x) a sih x a si x (F-8) Y(x) a sih x a six (F-9) The eigevector erivatio requires some creativity. Recall Y() = (zero isplacemet) (F-) Y x (zero beig momet) (F-) Thus, Y Y for x= a, =,,, (F-) a sih a si, =,,, (F-) The si() term is always zero. Thus a =. The eigevector for all moes is Y (x) a si x / (F-) 5

36 Mass ormalize the eigevectors as follows Y (x) (F-5) a si x / (F-6) a cos( x / ) (F-7) a x si(x / ) (F-8) a (F-9) a (F-5) a (F-5) Y (x) si x / (F-5) 6

37 Participatio Factors The participatio factors for costat mass esity are Y (x) (F-5) si x / (F-5) si x / (F-5) cos x / (F-55) cos, =,,,. (F-56) Effective Moal Mass The effective moal mass is Y (x) m eff, (F-57) Y (x) The eigevectors are alreay ormalize such that Y (x) (F-58) 7

38 Thus, m eff, Y (x) (F-59) m eff, (F-6) cos m eff, cos, =,,,. (F-6) 8

39 APPENDIX G Free-Free Beam Cosier a uiform beam with free-free bouary coitios., Figure G-. The goverig ifferetial equatio is y x y t (G-) Note that this equatio eglects shear eformatio a rotary iertia. The followig equatio is obtai usig the metho i Appeix D Y x c ( ) Y ( x ) (G-) The propose solutio is Y( x) a sih x a cosh x a si x a cos x (G-) Y( x) a cosh x asih x a cos x asi x (G-) Y( x) a x a x a x a x sih cosh si cos (G-5) Y(x) a cosh x a sih x a cos x a six (G-6) 9

40 Apply the bouary coitios. Y x (zero beig momet) (G-7) a a (G-8) a a (G-9) Y x (zero shear force) (G-) a a (G-) a a (G-) Y(x) a sih x si x a cosh x cos x (G-) Y(x) a cosh x cos x a sih x six (G-) Y (zero beig momet) (G-5) x

41 si cosh cos a sih a (G-6) Y x (zero shear force) (G-7) cos sih si a cosh a (G-8) Equatio (G-6) a (G-8) ca be arrage i matrix form. sih cosh si cosh cos a cos sih si a (G-9) Set the etermiat equal to zero. sih si sih si cosh cos (G-) sih (G-) si cosh cosh cos cos cos cosh (G-) cos cosh (G-) The roots ca be fou via the Newto-Raphso metho, Referece. The free-free beam has a rigi-boy moe with a frequecy of zero, correspoig to.

42 The seco root is.7 (G-) (G-5).7 (G-6).7 (G-7) The thir root is 7.85 (G-8) (G-9) 7.85 (G-) 6.67 (G-).757 (G-) The fourth root is.9956 (G-)

43 (G-).9956 (G-5).9 (G-6) 5. (G-7) Eigevalues for 5 (G-8) The followig moe shape a coefficiet erivatio applies oly to the elastic moes where. Equatio (G-8) ca be expresse as cos si cosh a a (G-9) sih Recall a a (G-)

44 a a (G-) The isplacemet moe shape is thus Y(x) x six cosh x cos x a sih a (G-) cos si cosh Y(x) a sih x six cosh x cos x (G-) sih Moify the moe shapes as follows. Y(x) â sih x six cosh sih cos si cosh x cos x (G-) Normalize the eigevectors with respect to mass. Y (x) (G-5) The eigevectors are mass-ormalize for â =. Thus Y (x) sih x six cosh sih cos si cosh x cos x (G-6)

45 The first erivative is y cosh x cos x cosh sih cos si sih x six (G-7) The seco erivative is y sih x six cosh sih cos si cosh x cos x (G-8) The participatio factors for costat mass esity are Y (x) (G-9) The participatio factors are calculate umerically. As a result of the rigi-boy moe, for > (G-5) 5

46 APPENDIX H Pipe Example Cosier a steel pipe with a outer iameter of. iches a a wall thickess of.6 iches. The legth is feet. Fi the atural frequecy for two bouary coitio cases: simply-supporte a fixe-fixe. The area momet of iertia is I D o D i (H-) 6 D o. i (H-) D i D i. (.6) i (H-).. i (H-) D i. i (H-5).. i I (H-6) 6 I. i (H-7) The elastic moulus is 6 lbf E (H-8) i The mass esity is mass per uit legth. (H-9).. i lbm.8 (H-) i 6

47 lbm.85 (H-) i 6 lbf i. slugft /sec i i lbf ft.85 lbm slug i. lbm.5 5 i sec (H-) (H-) The atural frequecy for the simply-supporte case is f,,,,... (H-) f 5 i.5 (H-5) i sec ft ft f. Hz (simply-supporte) (H-6) The atural frequecy for the fixe-fixe case is.7 f (H-7) 7

48 8 sec i.5 ft i ft.7 f 5 (H-8) 7.58 Hz f (fixe-fixe) (H-9)

49 APPENDIX I Suborbital Rocket Vehicle Cosier a rocket vehicle with the followig properties. mass = 78.9 lbm (at time = sec) = 7. iches lbm 7. iches 7.87 lbm i The average stiffess is = 6 ( 6 ) lbf i^ The vehicle behaves as a free-free beam i flight. Thus.7 f (I-) f.7 7 i 6e 6 lbf i slug ft / sec i. lbm lbf ft slugs lbm 7.87 i (I-) f =.6 Hz (at time = sec) (I-) Note that the fuametal frequecy ecreases i flight as the vehicle expels propellat mass. 9

50 APPENDIX J Fixe-Fixe Beam Cosier a fixe-fixe beam with a uiform mass esity a a uiform cross-sectio. The goverig ifferetial equatio is y x The spatial equatio is y t (J-) Y(x) c Y(x) x (J-) The bouary coitios for the fixe-fixe beam are: Y() = (J-) Y(x) x (J-) Y()= (J-5) Y(x) x (J-6) The eigevector has the form Y( x) a sih x a cosh x a si x a cos x (J-7) Y( x) a cosh x asih x a cos x asi x (J-8) 5

51 Y( x) a x a x a x a x sih cosh si cos (J-9) Y() = a + a = (J-) (J-) - a = a (J-) Y(x) x (J-) a + a = a + a = (J-) (J-5) - a = a (J-6) sih x six a cosh x cos x Y(x) a (J-7) Y(x) cosh x cos x a sih x si x a (J-8) Y() = (J-9) sih si a cosh cos a (J-) Y(x) x (J-) cosh cos a sih si a (J-) 5

52 cosh cos a sih si a (J-) sih cosh si cosh cos a cos sih si a (J-) sih et cosh si cosh cos cos sih si (J-5) si ][sih si ] [cosh cos ] [sih (J-6) si cosh cos cosh cos sih (J-7) cosh cos (J-8) cosh cos (J-9) The roots ca be fou via the Newto-Raphso metho, Referece. The first root is.7 (J-) (J-).7 (J-).7 (J-) 5

53 .7 f (J-) cosh cos a sih si a (J-5) et a = (J-6) a cosh cos sih si (J-7) si cos sih a (J-8) cosh Y(x) si cos sih cosh x cosx sihx six (J-9) cosh Y(x) si cos sih cosh x cosx sihx six (J-) cosh The u-ormalize moe shape for a fixe-fixe beam is cosh x cos x sih x si x Ŷ (x) (J-) where si cos sih (J-) cosh 5

54 The eigevalues are For > 5 (J-) Normalize Eigevectors Mass ormalize the eigevectors as follows Y (x) (J-) The mass ormalizatio is satisfie by Y (x) cosh x cos x sih x si x (J-5) where si cos sih (J-6) cosh 5

55 The first erivative is Y (x) sih x si x cosh x cos x (J-7) The seco erivative is Y (x) cosh x cos x sih x si x (J-8) Participatio Factors The participatio factors for costat mass esity are Y (x) (J-9) cosh x cos x sih x si x (J-5) sih x si x cosh x cos x (J-5) sih si cosh cos (J-5) sih si cosh cos (J-5) 55

56 The participatio factors from a umerical calculatio are.89 (J-5) (J-55).68 (J-56) (J-57).5 (J-58) 5 The participatio factors are o-imesioal. 56

57 APPENDIX K Beam Fixe Pie Cosier a fixe pie beam as show i Figure K-., Figure K-. Recall that the goverig ifferetial equatio is y x The spatial solutio is y t (K-) Y( x) a sih x a cosh x a si x a cos x (K-) Y( x) a cosh x asih x a cos x asi x (K-) Y( x) a x a x a x a x sih cosh si cos (K-) The bouary coitios at the left e x = are Y() = (zero isplacemet) (K-5) 57

58 Y(x) x (zero slope) (K-6) The bouary coitios at the right e x = are Y() = (zero isplacemet) (K-7) Y x (zero beig momet) (K-8) Apply bouary coitio (K-5). a a (K-9) a a (K-) Apply bouary coitio (K-6). a a (K-) a a (K-) Apply the left bouary results to the isplacemet fuctio. sih x six a cosh x cos x Y(x) a (K-) Apply bouary coitio (K-7). sih si a cosh cos a (K-) 58

59 Apply the left bouary results to the seco erivative of the isplacemet fuctio. Y(x) a sih x si x a cosh x cosx (K-5) Apply bouary coitio (K-7). sih si a cosh cos a (K-6) sih si a cosh cos a (K-7) sih sih si cosh cos a si cosh cos a (K-8) sih et sih si cosh cos si cosh cos (K-9) sih si cosh cos sih si cosh cos (K-) sih sih cosh sih cos si cosh si cos cosh sih cos si cosh si cos (K-) cos si cosh sih (K-) 59

60 cos si cosh sih (K-) ta tah (K-) The eigevalues are For >5, (K-5) si cos sih a a (K-6) cosh The uscale eigevector is si cos sih Y(x) a sih x six coshx cosx (K-7) cosh The mass-ormalize eigevector is si cos sih Y (x) sih x six coshx cosx (K-8) cosh 6

61 Participatio Factors The participatio factors for costat mass esity are Y (x) (K-9) sih x si x cosh x cos x (K-) si cos sih (K-) cosh cosh x cos x sih x si x (K-) cosh cos sih si (K-) cosh cos sih si (K-) The participatio factors from a umerical calculatio are (K-5) -.86 (K-6) -. (K-7) -.7 (K-8) (K-9) The participatio factors are o-imesioal. 6