5. VIBRATIONS OF CONTINUOUS SYSTEMS

Transcription

1 5. Te vibrating string. 5. VIBRATIONS OF CONTINUOUS SYSTEMS Consider a uniform string, of mass per unit lengt ρ, stretced under tension T, between two supports distance L apart, as sown below. u(x) x dx L Equilibrium position Wit te assumption of small vibrations about equilibrium, te tension T in te spring may be considered constant, ie. independent of te time t and te position x. Te free body diagram for te string s element of infinitesimal lengt dx, laid originally in te interval [x,x+dx], is sown below. T( θ + θ x dx) T θ + θ x dx T θ dx Tθ Tus, te mass of te element is ρdx, its acceleration is t Newton s second law, applied in te u direction, gives u, and te ρ u d x T θ t = + θ θ x dx T, (5..) wic simplifies to ρ t u θ T x =. (5..)

2 But, by definition te slop θ(x) is θ = u x. (5..3) Hence, substituting (3) in () gives ρ u t u T x =, (5..4) or u t u = c, (5..5) x were c T = ρ. (5..6) Te parameter c is known as te speed propagation of waves in te string. Te dynamic of te string is determined by te differential equation (5), te boundary conditions u( 0, t) u( L, t) (5..7) and te initial conditions u( x, 0) = f ( x) u ( x, 0) = g( x) t, (5..8) were f(x) and g(x) describe te displacement and velocity of te string at te time t=0, respectively. Equations (5)-(7) caracterise te string and its configuration. Te properties of te system, ie, its natural frequencies and mode sapes of vibrations, are tus determined by (5)-(7) alone. To obtain te 3

3 actual displacement u(x,t) of (5), for eac time t and point x, we must use te initial conditions (8) as well. Let us now determine te natural frequencies and mode sapes of te string. As done in te previous capters, we assume tat te string vibrates wit armonic motion of te form u( x, t) = v( x) sinω t. (5..9) Note tat by tis assumption we ave separated te variables x and t. Substituting (9) in (5) gives or ω v sinωt = c v sin ωt, (5..0) v + ω 0, (5..) v c = were primes denotes derivatives wit respect to x. Substituting (9) in (7) gives v( 0) sinωt v( L) sinωt. (5..) An assumption tat sinωt=0 leads troug (9) to a trivial solution u(x,t)=0. Suc a solution is not of interest since it predicts te dynamics of te string only in te case were te initial conditions (8) vanis. Trivially tere is no motion at all in tis case. We tus conclude tat () implies v( 0) v( L). (5..3) Combining () and (3) we obtain te eigenvalue problem were v + λv v( 0), v( L), (5..4) 4

4 λ = ω c. (5..5) In analogy wit te discrete eigenvalue problem ( K λm ) φ = o, te continuous eigenvalue problem (4) requires evaluation of te eigenvalues λ i and teir associated eigenfunctions v i (x). Te general solution of te differential equation in (4) is v( x) = A sin λx + A cos λ x, (5..6) were A and A are arbitrary constants. Proof. It follows from (6) tat v = A λsin λx A λcos λx. Hence te differential equation in (4) gives v + λv = A λsin λx A λcos λx + λ( A sin λx + A cos λx ), and v(x) is expressed in terms of two arbitrary constants. Te left boundary condition in (4) yields so tat (6) reduces to v( 0) = A, (5..7) v( x) = A sin λ x. (5..8) Te rigt boundary condition in (4) gives v( L) = A sin L = λ. (5..9) 0 5

5 Selecting A =0 leads to te trivial solution v(x)=u(x,t)=0. We terefore coose sin λl, (5..0) and obtain te eigenvalues λ i λ i L = i π, i =,, 3,... (5..) Te natural frequencies ( ) ω n i of te string are determined by using (5) ( ω ) n i icπ =, i =,, 3,... (5..) L Natural frequencies of a fixed-fixed string By (8) teir associated eigenfunctions v i are vi ( x ) = sin iπ L x. (5..3) Eigenfunctions, or mode-sapes, of te fixed-fixed string In te discrete case te eigenvectors can be scaled arbitrarily. Similarly, te eigenfunctions of te continuous system are determined up to a scalar constant A (see (9)). In (3) we use (9) wit A =. We know tat resonance occurs wen a multi-degree-of-freedom system is excited by a armonic force wit frequency tat is equal to a natural frequency. Similarly, te string vibrates in resonance wen a armonic force F0 sinω nt is applied to te string, were F 0 is some constant and ω n is one of te natural frequencies () of te string. We now give a pysical interpretation to te mode sapes (or eigenfunction). If te initial conditions are 6

6 u( x, 0) = vi ( x) u ( x, 0) t, (5..4) were v i is te i-t mode sape of te system, ten eac point of te string. More precisely te string ω n i vibrates in a single armonic frequency ( ) vibrates in tis case according to Proof. ic u( x, t) = vi ( x) sin L t It is easily sown by direct substitution tat iπx icπt u( x, t) = sin sin L L π. (5..5) satisfies te differential equation (5), te boundary conditions (7) and te initial conditions (8) wit and iπx f ( x) = sin = L v i g( x). 7

7 5. Te axially vibrating rod. Consider te axially vibrating rod, of density ρ(x), cross-sectional area A(x), fixed at x=0 and free to oscillate at x=l, as sown below. Denote by u(x,t) te displacement of te small element, laying originally in te interval [x, x+dx]. Te free body diagram, for an instant were 0<u(x,t)<u(x+dx,t), is also sown in te figure below. ρ(x), A(x) x dx u(x,t) P( x) x=l P P ( x ) + x dx dx Noting tat te mass of te element is ρadx, its acceleration t tat te Newton s second law gives u, we find P + P + P x dx = Adx u ρ t, (5..) or P x = ρ A u t. (5..) Recall te stress-strain relation 8

8 σ = E ε, (5..3) were σ in te applied stress, E is te modulus of elasticity and ε is te resulting strain. By definition σ = P A (5..4) and ε = u x. (5..5) Hence (3) gives P = EA u x. (5..6) Substituting (6) in () we obtain te differential equation of motion for te non-uniform rod ρ x E x A x u u ( ) ( ) = ( x) A( x) x t. (5..7) Equation of motion for a non-uniform axially vibrating rod For a uniform rod, A(x)=A, E(x)=E, ρ(x)=ρ, and (7) is simplified to u t E u ρ x =. (5..8) Equation of motion for a uniform axially vibrating rod We denote te speed propagation in te uniform rod by c E = ρ (5..9) 9

9 and write (8) in te form u t u = c, (5..0) x identical to te equation governing te motion of te string (5..5). Note owever, tat for te rod c is given by (9), wereas for te string c is given by (5..6). Te boundary conditions for te fixed-free configuration of te rod imply tat tere is no displacement at x=0, and no axial force at x=l (since tere is no pysical contact from te rigt at te free end). In view of (6) P(L)=0 wen te derivative of u wit respect to x vanises. Te boundary conditions are, terefore, u( 0, t) u( L, t) x. (5..) To determine te eigenvalue problem associated wit te differential equation (0) and te boundary conditions () we assume armonic motion of te form u( x, t) = v( x) sinω t (5..) and, upon substitution of () in (0) and (), obtain v'' + λv, λ = ω c v( 0) v'( L). (5..3) Te general solution (5..6) wit te left boundary condition yields v = A sin λ x. (5..4) Te rigt boundary condition gives 30

10 v'( L) = A λcos λ L =. (5..5) 0 Since A =0 or λ=0 lead to te unwanted trivial solution u(x,t)=0, we determine te frequency equation wit te roots cos λl, (5..6) π π π λ i L = 3 5,,,... (5..7) Using (3) we determine te natural frequencies of te fixed-free uniform rod ( ω ) n i = ( i ) cπ L, i =,, 3,... (5..8) Natural frequencies of a fixed-free uniform rod were c is given by (9). Coosing A = arbitrarily te eigenfunction (4) are v i = i x sin ( ) π, i =,, 3,... L (5..9) Mode sapes of a fixed-free uniform rod We could use oter factor for A and obtain anoter factor of te eigenfunction v i. 3

11 Example 5.. Determine te frequency equation and mode sapes of te uniform axially vibrating rod of cross sectional area A, modulus of elasticity E and density ρ, fixed at x=0 and spring supported at x=l, as sown below. x E, A, ρ L k Solution. Te rod at static equilibrium, and in motion, is sown below. At static equilibrium E, A, ρ x L k In motion ku(l,t) u(l,t) Tis illustration sows tat te rigt boundary condition is σ( L, t) ku( L, t) =, (5..0) A were te minus sign in (0) is due to te fact tat compressive stress is negative. By (3) and (5) we ave σ( L, t) (, ) E u L t x =, (5..) 3

12 so tat te rigt boundary condition is E u ( L, t ) ku( L, t) =. (5..) x A It follows tat tis rod is caracterised by te differential equation (8) u t E u ρ x wit te boundary conditions =, (5..3) u( 0, t) u( L, t) = x k EA u ( L, t ) Te armonic motion assumption (). (5..4) u( x, t) = v( x) sinω t, (5..5) troug an identical analysis to tat done for te uniform fixed-free rod, leads to ρω x v( x) = A sin, (5..6) E were v(x) is a mode sape and ω is a natural frequency. Substituting (5) and (6) in te rigt boundary condition of (4) gives or ρω E A ρωl k ρω ωt ω E EA A L cos sin = sin sin t, (5..7) E tan ρlω ρ E A = ω. (5..8) E k Frequency equation for te rod By (6) te mode sapes for te rod are ρω x, (5..9) v( x) = sin were ω is any root of te frequency equation (8). E 33

13 Denote by z te left and rigt side of (8). Ten te natural frequencies ω i of ρe A L te rod are te intersections of z = ω and z = tan ω, as sown k c below. We see from tis grap tat tere is, as expected, infinite number of natural frequencies for te continuous rod. Note also tat te fixed-free rod is a special case were k 0. Te fixed-fixed configuration is anoter extreme, were k. It also follows from te grap below tat te effect of increasing te spring constant k is to increase all te natural frequencies of te system. It was observed by Lord Rayleig tat tis penomenon applies not only for te vibrating rod but to all linear vibratory systems, discrete or continuous. z z L = tan ω c cπ L cπ 3cπ cπ 5cπ 3cπ L L L L L ω ω ω ω 3 z = ρe A k ω Natural frequencies of te fixed-spring supported rod Natural frequencies of te fixed-free rod Natural frequencies of te fixed-fixed rod Lord Rayleig, Te Teory of Sound, Dover, New York,

14 5.3 Approximating natural frequencies and mode sapes of a non-uniform rod by finite differences. Consider te axially vibrating non-uniform rod of cross-sectional area A(x), modulus of elasticity E(x) and density ρ(x), fixed at x=0 and free to vibrate at x=l, as sown below. ρ(x), A(x), E(x) x x=l We found in te section 5. tat tis rod is caracterised by te differential equation of motion (5..7) ρ x E x A x u u ( ) ( ) = ( x) A( x) x t (5.3.) and te boundary conditions u( 0, t) u( L, t) x. (5.3.) Harmonic motion assumption u( x, t) = v( x) sinω t, (5.3.3) leads to te associated eigenvalue problem ( E x A x v x ) ( ) ( ) '( ) ' + λρ( x) A( x) v( x), λ = ω v( 0) v'( L) were, as usual, primes denote derivatives wit respect to x., (5.3.4) 35

15 Depending on te functions A(x), E(x) and ρ(x), te eigenvalue problem (4) may not be solvable in closed form. We may approximate some eigenvalues and eigenvectors of (4) by using finite differences as follows. Partition te rod into n elements of equal lengt = L n and denote te displacements of te rigt and left boundaries of te i-t element by v i- and v i, respectively, as sown below. Let (EA) i and (ρa) i be te flexural rigidity and linear density of te mid-section of te i-t element. v 0 v v v i- v i vi+ v n v v v i- v i v i+ v n Ten a finite difference approximation for te i-t element gives EAv EA v i ' ' v x= ( i 0. 5) ( ) i ( ) and similarly i, (5.3.5) ( EA) v i v i i vi vi ' ( EA) ( EA) v i v i+ i x= ix + i ( EA) i ( EA) i + ( EA) i ( EA) = vi vi + + i+ v i+. (5.3.6) Te finite difference sceme for te differential equation in (4) is terefore 36

16 ( EA) i ( EA) i + ( AE) i ( EA) vi v i + or upon multiplying by + i+ v i+ + λ( ρa) v i i, (5.3.7) ( EA) i ( EA) i + ( AE) i+ ( EA) i+ vi + vi vi+ λ ( ρa) v i i. (5.3.8) Define k i EA i = ( ) (5.3.9) and m = ( A) Ten (8) can be written as i ρ. (5.3.0) ( ) i k v + k + k v k v mv = i i i i+ i i+ i+ i 0 λ. (5.3.) In terms of finite differences, te left boundary condition in (4) gives v 0 (5.3.) and te rigt boundary implies tat or v v n+ vn 0, (5.3.3) = n+ = vn. (5.3.4) Wit (), equation () gives for te first element, i= 37

17 ( ) k + k v k v mv = 0 λ. (5.3.5) Wit (4), equation () gives for te last element, i=n ( ) k v + k + k v k v λmv n n n n+ n n+ n n = k v + k v λmv = n n n n n 0. (5.3.6) Equation () for i=,3,...,n-, togeter wit (5) and (6) can be written in matrix form were ( ) K M v = o K = λ, (5.3.7) k + k k k k + k k 3 3 k3 k3 + k4 k4 O O O k n k n, (5.3.8) m M = m m 3 O m n, (5.3.9) and v = ( v, v,..., ) v n T. (5.3.0) Equation (7) is te well known generalised eigenvalue problem. It can be solved by standard routines (eg. eig(k,m) in Matlab). Te eigenvalues of (7) approximate te square of te natural frequencies of te non-uniform rod. Te eigenvectors of (7) are te approximation of te corresponding mode sapes, measured at x=i, i=,,...,n. 38

18 Tis result as an interesting interpretation. We note tat m i in (0) is te mass of te i-t element. Also, k i in (9) represents te stiffness of a rod of cross-sectional area A i, lengt and modulus of elasticity E i. Te model (7)- (0) tus describes an n degree-of-freedom mass-spring system of n masses, eac equals to te mass of its associate rod element. Tey are connected by rods of lengt, eac wit a stiffness corresponding to te stiffness of te appropriate rod s element, as sown below. m n- k n m n Te finite element model wic as developed matematically in tis section can be tus obtain simply by inspection. All we need to do is to concentrate te mass of eac element in te element s center, and to connect te masses by springs wit constants equal to te element stiffnesses, as sown above. Ten te obtained discrete model can be analysed by te metods of capter 4. Tis metod of modelling as been used for many years by engineers. Te model obtained is alternatively called a lumped parameter model. 39