ME67 - Handout 4 Vbratons of Contnuous Systems Axal vbratons of elastc bars The fgure shows a unform elastc bar of length and cross secton A. The bar materal propertes are ts densty ρ and elastc modulus E. One end of the bar s attached to a fxed wall whle the other end s free. The force P(t) actng at the free end of the bar nduces elastc dsplacements u(x,t) along the bar Δ x P(t) x u(x,t) Fg. Schematc vew of elastc bar undergong axal motons From elementary strength of materals consder a) Cross-sectons A reman plane and perpendcular to the man axs (x) of the bar. b) Materal s lnearly elastc c) Materal propertes (ρ, E ) are constant at any gven cross secton. The relatonshp between stress σ and stran ε for unaxal tenson s u σ = Eε= E () x MEEN 67 HD#4 Vbratons of Contnuous Systems.. San Andrés 8
Consder the free body dagram of an nfntesmally small pece of bar wth length Δ x, u In the FBD, Pxt (, ) = A( x) σ = AE s the axal force at a x cross secton of the bar, and f ( xt, ) s a dstrbuted axal force per unt length, Δ x P(x,t) f(x,t) P(x+Δx,t) u(x,t) Fg. Free body dagram of small pece of elastc bar Applyng Newton s nd law of moton on the bar dfferental element gves Fx max ( ρ A x) () x u =Δ = Δ t u AΔ x = P P + f Δx t ρ ( x+δx, t) ( x, t) ( x, t) ( ) P Δx P P + Δx x As ( x+δx, t) ( x, t) (3) (4a) MEEN 67 HD#4 Vbratons of Contnuous Systems.. San Andrés 8
u P AΔ x = Δ x + f ( xt, ) Δx t x ( ρ ) u P ρ A = + f ( x, t) t x u And replacng Pxt (, ) = AE x ρ A AE f t x x u u = + ( x, t) (4) (5) PDE (5) descrbes the axal motons of an elastc bar. For ts soluton, one needs approprate boundary condtons (BC), whch are of two types (a) essental, u=u *, a specfed value, at x=x * for all tmes, u (b) natural, P(x *,t) = AE x = x x * If P=, then the natural BC s a free end,.e. specfed u x x= x * = Note: PDE (5) and ts BCs can be derved from the Hamltonan prncple usng the defntons for netc (T) and potental (V) energes. u u T= A dx; V EA dx ρ = t x (6) MEEN 67 HD#4 Vbratons of Contnuous Systems.. San Andrés 8 3
Free vbratons of elastc bars Wthout external forces (pont loads or dstrbuted load, f=), PDE (5) reduces to u u ρ A = AE t x x (7) The soluton of PDE (7) s of the form u( x, t) = φ( x) v( t) (8) Note that u d v = φ ( x) = φ ( x) v( t) ; t dt (9) u d φ = v () t = φ v() t x d x Wth the defntons (. d ) ; ( ' dt ) = =. For a bar wth unform d dx materal propertes (ρ, E) and cross secton A, substtuton of the product soluton Eq. (8) nto PDE (7) gves ρ u u = ρ φ ( x) v( t) = φ ( x) v( t) E t x E Dvde ths expresson by u( x, t) = φ( x) v( t) to get () v v = E φ ρ φ () t ( x) () t ( x) () MEEN 67 HD#4 Vbratons of Contnuous Systems.. San Andrés 8 4
Above, the HS s only a functon of tme, whle the RHS s only a functon of spatal coordnate x. Ths s possble only f both sdes equal to a constant,.e. v v E φ = = ω ρφ () t ( x) () t ( x) Hence, the PDE s converted nto two ordnary dfferental equatons (ODEs),.e. v () t + ω v = ( x) ( x) φ + λ φ = () where λ= ω (3) ρ E The soluton of the ODEs () & (3) s ( ω ) ( ω ) v() t = Ctcos t + Stsn t (4) ( ) ( ) φ( x) = Cxcos λx + Sxsn λx (5) The coeffcents (C, S) are determned from satsfyng the boundary condtons for the specfc bar confguraton and load condton. Equaton (5) s nown as the fundamental equaton for an elastc bar,.e. t contans the nformaton on natural frequences and mode shapes. MEEN 67 HD#4 Vbratons of Contnuous Systems.. San Andrés 8 5
Example. u(x,t) x A bar wth one end fxed and the other end free. In ths case, the boundary condtons are u = = φ v φ = t At x=, (, t) () ( t) () u x = = φ v φ = t At x=, ( ) ( t) ( ) x= (6) Hence, from the characterstc equaton φ () = C x = and φ( x) = Sxsn( λ x) (7) ( ) = = xcos = (8) At x=, φ λs ( λ) Note that Sx for a non trval soluton. Hence, the characterstc equaton for axal motons of a fxed end-free end elastc bar s ( ) cos λ = (9) whch has an nfnte number of solutons,.e. 3 5 n π, π, π λ =,..., = π, n =,,... And hence the roots of Eq. (9) are ( n ) π = () λ n n=,,... MEEN 67 HD#4 Vbratons of Contnuous Systems.. San Andrés 8 6
ρ And snce λ= ω E, the natural frequences of the fxed endfree end bar are ω ( ) π E = ρ / ; =,,... () / / /.e. π E 3π E 5π E ω =, ω, ω3... ρ = = ρ ρ Assocated to each natural frequency, there s a natural mode shape as shown n the fgure below. φ = ψ = sn( λ x) =,,... () φ(x) Functon (x).5 λ =.57 4.7 7.854.996.5.5.5.75 x/ Mode Mode Mode 3 Fg. Natural modes shapes φ(x) for elastc bar wth fxed end-free end MEEN 67 HD#4 Vbratons of Contnuous Systems.. San Andrés 8 7
See more examples on page 3-ff. The dsplacement functon response u( x, t) = φ( x) v( t) equals to the superposton of all the found responses,.e. ( ) ( ) u = φ x v t = ( xt, ) = ( ) φ( x) cos( ) sn C ωt + S ωt For example (fxed end free end bar) (3a) sn( λ ) cos( ω ) sn( ω ) (3b) u = x C t + S t and velocty: ( xt, ) = ( λ ) ω ω ( ω ) (4) u = sn x C sn( t) + S cos t ( xt, ) = The set of coeffcents (C, S ) are determned by satsfyng the ntal condtons. That s at tme t=, ( λ ) u = U = sn x C ( x,) ( x) = ( λ ) u = U = ω sn x S ( x,) ( x) = (5) MEEN 67 HD#4 Vbratons of Contnuous Systems.. San Andrés 8 8
Orthogonalty propertes of the natural modes λ, ψ satsfy the characterstc Recall that the par { ( x) } equaton (b),.e. =,... ψ + λψ = = (6) ( x) ( x),... And consder two dfferent egenvalues Eq. (6),.e. λ andλ each satsfyng j ψ + λψ = & ψ + λψ = j j j Multply Eq. on left by ψ and Eq. on rght by ψ, and ntegrate over the doman x {, } j to get: ( ψψ j dx) λ ( ψψ j dx) + = ( ψψ j dx) + λj ( ψψ jdx) = (7) Integrate by parts the term on the HS to obtan x j dx = = ( j x= j ψψ ψψ ψψ dx (8) And recall the boundary condtons for the fxed end-free end bar ( ( ] ψ = & ψ = = (9) j x= x MEEN 67 HD#4 Vbratons of Contnuous Systems.. San Andrés 8 9
And wrte frst of Eq. (7) as λ ( ψ jψdx) = ( ψ jψ dx) and substtutng λ ω = one obtans: ρ E j j j j = j ( A dx) = ( E A dx) (3a) ω ρ ψ ψ ψ ψ ( A dx) ( E A dx) (3b) ω ρ ψψ ψψ Subtract Eq. (3b) from (3a) to obtan ( ωj ω ) ( ρ Aψψ j ) And snce ω ω j, t follows that dx= (3) ( ρ Aψψ j ) dx= & ( EAψψ j ) dx= j=,,... (3) That s, the modal functons { },... =j, the th natural frequency follows from ψ = are ORTHOGONA. For Κ ω = = Μ ( EAψψ ) ( ρ Aψψ ) Where Κ, Μare the th mode equvalent stffness and mass coeffcents. dx dx (33) MEEN 67 HD#4 Vbratons of Contnuous Systems.. San Andrés 8
Note that the set { },... functons ψ = s a COMPETE SET of orthogonal Now, consder the ntal condtons, Eq. (5) ( λ ) u = U = sn x C ( x,) ( x) = ( λ ) u = U = ω sn x S ( x,) ( x) = ψ m sn λ x m = x ρa and ntegrate over the whole doman to obtan Multply both sdes of Eq. (5) by ( ) (5) ( ρ ψ m ( x) ) = ( ρ ψ mψ) A U dx C A dx = And snce Μ when ( ) m m = ρ Aψ mψ dx= (34) when m Then f follow that C m = ( ρ ψ m ( x) ) A U dx, Μ m m=,,... (35) And smlarly S m = ( ρ ψ m ( x) ) A U dx, ω Μ m m m=,,... (36) MEEN 67 HD#4 Vbratons of Contnuous Systems.. San Andrés 8
wth ( ) ( ) dψ Μ m= ρ ψm and Κ m= dx m A dx E A dx (37) Ths concludes the procedure to obtan the full soluton for the vbratons of a bar,.e. φ cos( ω ) sn( ω ) (3) u = C t + S t ( xt, ) ( x) = MEEN 67 HD#4 Vbratons of Contnuous Systems.. San Andrés 8
Example. x= u(x,t) A bar wth both ends free. The boundary condtons are u x At x=, () ( t ) () u x x= = = φ v φ = t = = φ v φ = t At x=, ( ) ( t) ( ) x= Hence, from the characterstc equaton φ () = S x = and φ( x) = Cxcos( λ x) ( ) = = xsn = At x=, φ λc ( λ) Note that λ = denotes rgd body moton. Hence, the characterstc equaton for axal motons of a fxed end-free end elastc bar s ( ) sn λ = whch has an nfnte number of solutons,.e. λ =, π, π,3 π,..., = nπ, n= π = n λ λ n n=,,,... ω,,... ρ And snce = E, the natural frequences of the free endfree end bar are MEEN 67 HD#4 Vbratons of Contnuous Systems.. San Andrés 8 3
ω π E = ρ / ; =,,,... Assocated to each natural frequency, there s a natural mode shape And shown n the fgure below. φ = cos( λ x) =,,,... φ(x) Functon (x).5 λ = 3.4 6.83.5.5.5.75 x/ Mode Mode Mode 3 Fg. Natural modes shapes φ(x) for elastc bar wth both ends free. Frst mode s rgd body (null natural frequency) MEEN 67 HD#4 Vbratons of Contnuous Systems.. San Andrés 8 4
Example 3. x= u(x,t) x A bar wth both ends fxed. The boundary condtons are At x=, u(, t) = = φ() v( t) φ() = t At x=, u( t, ) = = φ( ) v( t) φ( ) = Hence, from the characterstc equaton φ( x) Cxcos( λx) Sxsn( λx) then φ () = C x = and φ( x) = Sxsn( λ x) At x=, φ ( λ ) ( ) = = sn = = +, Note that λ denotes rgd body moton. Hence, the characterstc equaton for axal motons of a fxed endfxed end elastc bar s ( ) sn λ = whch has an nfnte number of solutons,.e. λ = π, π,3 π,..., = nπ, n= π = n λ λ n n=,,... ω,,... ρ And snce = E, the natural frequences of the free endfree end bar are π E ω = ρ / ; =,,... MEEN 67 HD#4 Vbratons of Contnuous Systems.. San Andrés 8 5
Assocated to each natural frequency, there s a natural mode shape And shown n the fgure below. φ = sn( λ x) =,,,... φ(x) Functon (x).5 λ = 3.4 6.83 9.45.5.5.5.75 x/ Mode Mode Mode 3 Fg. Natural modes shapes φ(x) for elastc bar wth both ends fxed. MEEN 67 HD#4 Vbratons of Contnuous Systems.. San Andrés 8 6
ME67 - Handout 4 (b) Vbratons of Contnuous Systems ateral vbratons of elastc beams The fgure shows a unform elastc beam of length, cross secton A and area moment of nerta I. The beam materal propertes are ts densty ρ and elastc modulus E. One end of the beam s fxed to a wall whle the other end s free. The dscrete force P(t) acts at a fxed axal locaton whle f(x,t) represents a load dstrbuton per unt length. The forces nduces elastc dsplacements on the beam and desgnated as v(x,t). y f(x,t) v(x,t) P(t) x x Fg. Schematc vew of elastc beam undergong lateral motons From elementary strength of materals consder a) Cross-sectons A reman plane and perpendcular to the neutral axs (x) of the beam. b) Homogeneous materal beam, lnearly elastc, c) Materal propertes (ρ,e ) are constant at any gven cross secton. d) Stresses σ y, σ z << σ x (flexural stress),.e. along beam. MEEN 67 HD#4 Vbratons of Contnuous Systems.. San Andrés 7
The graph below shows the free body dagram for moton of a dfferental beam element wth length x. M(x,t) S(x,t) f(x,t) x S(x+x,t) M(x+x,t) v(x,t) Fg. Free body dagram of small pece of elastc beam In The FBD, S (x,t) represents the shear force and M (x,t) denotes the bendng moment. Apply Newton s nd law to the materal element: Fym ayss x f( x, t) Ax (38) x S x v t In the lmt as x : A v t f ( xt, ) S x (39) Apply the moment equaton: neglectng rotary nerta I g M I g ~ (4) MEEN 67 HD#4 Vbratons of Contnuous Systems.. San Andrés 8
Then x M M( xx, t) M( x, t) f Sx M x M xm f Sx x M In the lmt asx : S ( x, t ) x (4) Combnng Eqs. (4) and (39) gves: A v t M x f ( xt, ) (4) v If the slope x remans small, then the beam curvature s v x. From Euler s beam theory: M EI EI v x (43) where I y das the beam area moment of nerta. Substtute Eq. (43) nto (4) to obtan the equaton for lateral motons of an elastc beam: v v A f ( xt, ) EI t x x (44) MEEN 67 HD#4 Vbratons of Contnuous Systems.. San Andrés 9
The PDE s fourth-order n space and nd order n tme. Approprate boundary condtons are of two types: Essental BCs: v v - specfed dsplacement, * v x - specfed slope, * Natural BCs: - specfed moment, - specfed shear force, M M EI v * x x * v * x x x S S EI See below the most typcal beam confguratons: v Fxed end (cantlever): v & x * x * Pnned end v v& M x x * x * Free end 3 v v M & S & 3 x x MEEN 67 HD#4 Vbratons of Contnuous Systems.. San Andrés
v(x,t) x * Sprng supported end v M x S v* EI x x x v * Note: PDE (44) and ts BCs can be derved from the Hamltonan prncple usng the defntons for netc (T) and potental (V) energes of an elastc beam v v ; t x (45) T A dx V EI dx Free vbratons of elastc beam Wthout external forces (pont loads or dstrbuted load, f=), PDE (44) reduces to A v v EI t x x (46) The soluton of PDE (46) s of the form v( x, t) ( x) v( t) (47) et ; d. Substtutng Eq. (47) nto Eq (46) gves. d ' dt dx d A dx 4 EI ( x) ( x) v( t) v 4 v 4 () t EI ( x) 4 A( x) d x v d Above, the HS s only a functon of tme, whle the RHS s only a functon of spatal coordnate x. Ths s possble only f both sdes MEEN 67 HD#4 Vbratons of Contnuous Systems.. San Andrés
are equal to a constant,.e.. Hence, the separaton of varables gves two ordnary dfferental equatons 4 d v() t v= & 4 dx (48) where The soluton of the ODEs s A E I (49) v() t Ctcos t Stsn t (5) ( x) Ccos x Csn x C3cosh x C4snh x (5) where has unts of [/length]. 4 / A E I (5) The coeffcents (C, S) are determned from satsfyng the boundary condtons for the specfc beam confguraton. Equaton (5) s nown as the fundamental mode shape for an elastc beam,.e., t contans the nformaton on natural frequences and mode shapes. s v v 4 The soluton of ODE 4 characterstc equaton x ce wth MEEN 67 HD#4 Vbratons of Contnuous Systems.. San Andrés
Example. Pn-pn ends beam x= x= Recall cos sn cosh snh The BCs are: ( ) 3 4 x C x C x C x C x v v t (53.a) At x=, (, t) () ( t) () C C () 3 () v( t) () x C C v M () 3 Hence, C =C 3 = and sn snh v ( ) 4 x C x C x v t At x=, ( t, ) ( ) ( t) ( ) sn snh ( ) 4 C C v M x ( ) v( t) ( ) (53.b) x C sn C snh ( ) 4 from ths two equatons, snce sn ( x) C x where snh, t follows that (54) sn when,,... (55) MEEN 67 HD#4 Vbratons of Contnuous Systems.. San Andrés 3
and hence, the natural frequences of the pn-pn beam are A A EI EI ;,... (56) Assocated to each natural frequency, there s a natural mode shape sn sn x ; as shown n the graph below. x,,... (57) (x) Functon (x).5 3.4 6.83 9.45.5.5.5.75 x/ Mode Mode Mode 3 Fg. Natural mode shapes (x) for elastc beam wth both ends pnned. The dsplacement functon response v( x, t) ( x) v( t) equals to the superposton of all the found responses,.e. MEEN 67 HD#4 Vbratons of Contnuous Systems.. San Andrés 4
v cos( ) sn v x t C t S t ( xt, ) ( x) sn cos( ) sn (58) v x C t S t and velocty: ( xt, ) (59) v sn x C sn( t) S cos t ( xt, ) The set of coeffcents (C, S ) are determned by satsfyng the ntal condtons. That s at tme t=, v V sn x C ( x,) ( x) v V sn x S ( x,) ( x) (6) RECA: cos sn cosh snh ( x) C x C x C3 x C4 x Csn xccos xc3snh xc4cosh x Ccos x Csn x C3cosh x C4snh x 3 Csn xccos xc3snh xc4cosh x MEEN 67 HD#4 Vbratons of Contnuous Systems.. San Andrés 5
Example. Fxed end-free end beam x= x= Recall cos sn cosh snh The BCs. are ( ) 3 4 x C x C x C x C x v v t (6.a) At x=, (, t) () ( t) () () C C3 v x () v( t ) () (6.b) C C () 4 At x= v M x ( ) v( t) ( ) (6.c) x Ccos C sn C cosh( ) C snh ( ) 3 4 3 v Sx 3 ( ) v( t) ( ) (6.d) x Csn C cos C snh( ) C cosh ( ) 3 4 Soluton of Eqs. (a)-(d) gves x x x x ( x) cosh cos snh sn : (6) MEEN 67 HD#4 Vbratons of Contnuous Systems.. San Andrés 6
where and cosh snh cos sn 3 3.8754.73496 4.6944.8466 7.854757.9995 etc (63) (64) cosh x cos x x snh x sn x (x).875 4.694 7.855 Functon (x) _.734.8.999.5.5.75 x/ Mode Mode Mode 3 Fg. Natural mode shapes (x) for cantlever beam (fxed endfree end) MEEN 67 HD#4 Vbratons of Contnuous Systems.. San Andrés 7
Propertes of the natural modes Recall that the par, ( x),... satsfy the ODE v (65) 4 A where EI,... As n the case of axal vbratons of a bar, t s easy to show that of a flexng beam satsfy the followng the natural modes,... ORTHOGONA propertes: for j EA j dx (66a) for j for j Aj dx (66b) for j For =j, the th natural frequency follows from EA A Where, are the th mode equvalent stffness and mass coeffcents. dx dx (67) Demonstraton wth ntegraton by parts (twce). MEEN 67 HD#4 Vbratons of Contnuous Systems.. San Andrés 8
Note that,... functons s a COMPETE SET of orthogonal Now, consder the ntal condtons for v cos( ) sn v x t C t S t ( xt, ) ( x) v V C ; v V S ( x,) ( x) ( x,) ( x) (68) Usng the orthogonalty propertes, the coeffcents (C m, S m) follow from C m AmV( x) dx, m m,,... (69a) And smlarly S m AmV ( x) dx, m m m,,... (69b) Ths concludes the procedure to obtan the full soluton for the lateral vbratons of a beam,.e. cos( ) sn (7) v C t S t ( xt, ) ( x) MEEN 67 HD#4 Vbratons of Contnuous Systems.. San Andrés 9
Forced lateral vbratons of a beam Consder a beam subjected to an arbtrary forcng functon f (x,t). The PDE descrbng the lateral motons of the beam s v v A f ( xt, ) EI t x x (44) et,... be the set of natural modes satsfyng the boundary condtons of the beam confguraton (pn-pn, fxed-free ends, etc). A soluton to Eq. (44) s of the form v Snce the set,... f (x,t) can be wrtten as q (7) ( x, t) ( x) ( t) s complete, then any arbtrary functon where Q m f Q (7) ( x, t) ( x) ( t) Am f( x, t) dx, m m,,... (73) Substtuton of Eqs. (7, 7) nto Eq. (44) gves v v A f ( xt, ) EI t x x MEEN 67 HD#4 Vbratons of Contnuous Systems.. San Andrés 3
v A q Q EI q (74) but recall that each of the normal modes satsfes and hence, Eq. (74) can be wrtten as ; v Aq Q EI q and, snce the natural modes are lnearly ndependent, then t follows that (75) Aq Q EI q,,... A astly, recall that ; then EI EI wrte (75) as A, and Q q q ; (76) A,,... Whch can be easly solved for all type of exctatons Q () t [ See soluton of undamped SDOF EOMS ectures #] MEEN 67 HD#4 Vbratons of Contnuous Systems.. San Andrés 3
Example 3. Free-free ends beam x= x= Recall cos sn cosh snh The BCs are: At x= ( ) 3 4 x C x C x C x C x () v( t) () x C C (a) v M () 3 () v( t) () x C C (b) 3 v S 3 x () 4 At x= v M x ( ) v( t) ( ) (6.c) x ( ) Ccos Csn C3cosh( ) C4snh (c) 3 v Sx 3 ( ) v( t) ( ) x ( ) Csn Ccos C3snh( ) C4cosh (d) Soluton of Eqs. (a)-(d) gves ( x) cosh x cos x snh x sn x where cosh snh cos sn (63) MEEN 67 HD#4 Vbratons of Contnuous Systems.. San Andrés 3
and 4.734.985 7.8535.777.9956.999966 etc 3 3 (64) Note that the lowest natural frequency s actually zero,.e. a rgd body mode. = & cosh x cos x x snh x sn x (x) Functon (x) 3 _ 4.73 7.853.996.983. 3.5.5.75 x/ Mode Mode Mode 3 Fg. Elastc natural mode shapes (x) for beam wth free-free ends MEEN 67 HD#4 Vbratons of Contnuous Systems.. San Andrés 33
Characterstc (mode shape) equaton for beams: ( x) Ccos x Csn x C3cosh x C4snh x Csn xccos xc3snh xc4cosh x Ccos x Csn x C3cosh x C4snh x 3 Csn xccos xc3snh xc4cosh x MEEN 67 HD#4 Vbratons of Contnuous Systems.. San Andrés 34