Lectures HD#14 Dynamic response of continuous systems. Date: April

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1 ectures -3 Date: April 4 7 Toay: Vibrations of continuous systems HD#4 Dynamic response of continuous systems Free vibrations of elastic bars an beams. Properties of normal moe functions. Force response Reaing & other assignments: Tetbook G: , Hanout # Other: HMWK #6: ue 4/ FE analysis

2 From your tetbook Chapters 6&7: Vibration of elastic bars, beams an ros Recommene problems Chapter 6 3, 9,, 4, 5, 8,38, 54 Chapter 7 3,,43,49

3 ME67 - Hanout 4 Vibrations of Continuous Systems Aial vibrations of elastic bars The figure shows a uniform elastic bar of length an cross section A. The bar material properties are its ensity ρ an elastic moulus E. One en of the bar is attache to a fie wall while the other en is free. The force P(t) acting at the free en of the bar inuces elastic isplacements u(,t) along the bar Δ P(t) u(,t) Fig. Schematic view of elastic bar unergoing aial motions From elementary strength of materials consier a) Cross-sections A remain plane an perpenicular to the main ais () of the bar. b) Material is linearly elastic c) Material properties (ρ, E ) are constant at any given cross section. The relationship between stress σ an strain ε for uniaial tension is u σ Eε E () MEEN 67 HD#4 Vibrations of Continuous Systems.. San Anrés 8

4 Consier the free boy iagram of an infinitesimally small piece of bar with length Δ, u In the FBD, Pt (, ) A( ) σ AE is the aial force at a cross section of the bar, an f ( t, ) is a istribute aial force per unit length, Δ P(,t) f(,t) P(+Δ,t) u(,t) Fig. Free boy iagram of small piece of elastic bar Applying Newton s n law of motion on the bar ifferential element gives F ma ( ρ A ) () u Δ Δ t u AΔ P P + f Δ t ρ ( +Δ, t) (, t) (, t) P Δ P P + Δ As ( +Δ, t) (, t) (3) (4a) MEEN 67 HD#4 Vibrations of Continuous Systems.. San Anrés 8

5 u P AΔ Δ + f ( t, ) Δ t ( ρ ) u P ρ A + f (, t) t u An replacing Pt (, ) AE ρ A AE f t u u + (, t) (4) (5) PDE (5) escribes the aial motions of an elastic bar. For its solution, one nees appropriate bounary conitions (BC), which are of two types (a) essential, uu *, a specifie value, at * for all times, u (b) natural, P( *,t) AE * If P, then the natural BC is a free en, i.e. specifie u * Note: PDE (5) an its BCs can be erive from the Hamiltonian principle using the efinitions for kinetic (T) an potential (V) energies. u u T A ; V EA ρ t (6) MEEN 67 HD#4 Vibrations of Continuous Systems.. San Anrés 8 3

6 Free vibrations of elastic bars Without eternal forces (point loas or istribute loa, f), PDE (5) reuces to u u ρ A AE t (7) The solution of PDE (7) is of the form u(, t) φ( ) v( t) (8) Note that u v φ ( ) φ ( ) v( t) ; t t (9) u φ v () t φ v() t With the efinitions (. ) ; ( ' t ). For a bar with uniform material properties (ρ, E) an cross section A, substitution of the prouct solution Eq. (8) into PDE (7) gives ρ u u ρ φ ( ) v( t) φ ( ) v( t) E t E Divie this epression by u(, t) φ( ) v( t) to get () v v E φ ρ φ () t ( ) () t ( ) () MEEN 67 HD#4 Vibrations of Continuous Systems.. San Anrés 8 4

7 Above, the HS is only a function of time, while the RHS is only a function of spatial coorinate. This is possible only if both sies equal to a constant, i.e. v v E φ ω ρφ () t ( ) () t ( ) Hence, the PDE is converte into two orinary ifferential equations (ODEs), i.e. v () t + ω v ( ) ( ) φ + λ φ () where λ ω (3) ρ E The solution of the ODEs () & (3) is ( ω ) ( ω ) v() t Ctcos t + Stsin t (4) φ( ) Ccos λ + Ssin λ (5) The coefficients (C, S) are etermine from satisfying the bounary conitions for the specific bar configuration an loa conition. Equation (5) is known as the funamental equation for an elastic bar, i.e. it contains the information on natural frequencies an moe shapes. MEEN 67 HD#4 Vibrations of Continuous Systems.. San Anrés 8 5

8 Eample. u(,t) A bar with one en fie an the other en free. In this case, the bounary conitions are u φ v φ t At, (, t) () ( t) () u φ v φ t At, ( t) (6) Hence, from the characteristic equation φ () C an φ( ) Ssin( λ ) (7) cos (8) At, φ λs ( λ) Note that S for a non trivial solution. Hence, the characteristic equation for aial motions of a fie en-free en elastic bar is cos λ (9) which has an infinite number of solutions, i.e. 3 5 n π, π, π λ,..., π, n,,... An hence the roots of Eq. (9) are ( n ) π () λ n n,,... MEEN 67 HD#4 Vibrations of Continuous Systems.. San Anrés 8 6

9 ρ An since λ ω E, the natural frequencies of the fie enfree en bar are ω k ( k ) π E ρ / ; k,,... () / / / i.e. π E 3π E 5π E ω, ω, ω3... ρ ρ ρ Associate to each natural frequency, there is a natural moe shape as shown in the figure below. φk ψk sin( λ k ) k,,... () φ() Function () λ / Moe Moe Moe 3 Fig. Natural moes shapes φ() for elastic bar with fie en-free en MEEN 67 HD#4 Vibrations of Continuous Systems.. San Anrés 8 7

10 See more eamples on page 3-ff. The isplacement function response u(, t) φ( ) v( t) equals to the superposition of all the foun responses, i.e. u φ v t ( t, ) k k k k φ( ) cos sin k Ck ωkt + Sk ωkt For eample (fie en free en bar) (3a) sin( λ ) cos( ω ) sin( ω ) (3b) u C t + S t an velocity: ( t, ) k k k k k k ( λ ) ω ω ( ω ) (4) u sin C sin( t) + S cos t ( t, ) k k k k k k k The set of coefficients (C k, S k ) are etermine by satisfying the initial conitions. That is at time t, ( λ ) u U sin C (,) ( ) k k k ( λ ) u U ω sin S (,) ( ) k k k k (5) MEEN 67 HD#4 Vibrations of Continuous Systems.. San Anrés 8 8

11 Orthogonality properties of the natural moes λ, ψ satisfy the characteristic Recall that the pair { k ( ) } equation (b), i.e. k k,... ψ + λψ (6) ( ) k ( ) k,... k k An consier two ifferent eigenvalues Eq. (6), i.e. λ anλ each satisfying i j ψ + λψ & ψ + λψ i i i j j j Multiply Eq. on left by ψ an Eq. on right by ψ, an integrate over the omain {, } j to get: ( ψψ j i ) λi ( ψψ j i) + ( ψψ i j ) + λj ( ψψ i j) (7) i Integrate by parts the term on the HS to obtain j i ( j i j i ψψ ψψ ψψ (8) An recall the bounary conitions for the fie en-free en bar ( ( ] ψ & ψ (9) j i MEEN 67 HD#4 Vibrations of Continuous Systems.. San Anrés 8 9

12 An write first of Eq. (7) as λi ( ψ jψi) ( ψ jψ i ) an substituting λ i ω one obtains: i ρ E i j i j i j i j i j ( A ) ( E A ) (3a) ω ρ ψ ψ ψ ψ ( A ) ( E A ) (3b) ω ρ ψψ ψψ Subtract Eq. (3b) from (3a) to obtain ( ωj ωi ) ( ρ Aψψ i j ) An since ωi ω j, it follows that (3) ( ρ Aψψ i j ) & ( EAψψ i j ) i j,,... (3) That is, the moal functions { k} k,... ij, the i th natural frequency follows from ψ are ORTHOGONA. For Κ ω i i Μi ( EAψψ ) ( ρ Aψψ ) Where Κi, Μiare the i th moe equivalent stiffness an mass coefficients. i i i i (33) MEEN 67 HD#4 Vibrations of Continuous Systems.. San Anrés 8

13 Note that the set { k} k,... functions ψ is a COMPETE SET of orthogonal Now, consier the initial conitions, Eq. (5) ( λ ) u U sin C (,) ( ) k k k ( λ ) u U ω sin S (,) ( ) k k k k ψ m sin λ m ρa an integrate over the whole omain to obtain Multiply both sies of Eq. (5) by (5) ( ρ ψ m ( ) ) k ( ρ ψ mψk) A U C A k An since Μ when m m ρ Aψ k mψk (34) when m k Then if follow that C m ( ρ ψ m ( ) ) A U, Μ m m,,... (35) An similarly S m ( ρ ψ m ( ) ) A U, ω Μ m m m,,... (36) MEEN 67 HD#4 Vibrations of Continuous Systems.. San Anrés 8

14 with ( ) ψ Μ m ρ ψm an Κ m m A E A (37) This conclues the proceure to obtain the full solution for the vibrations of a bar, i.e. φ cos( ω ) sin( ω ) k (3) u C t + S t ( t, ) ( ) k k k k k MEEN 67 HD#4 Vibrations of Continuous Systems.. San Anrés 8

15 Eample. u(,t) A bar with both ens free. The bounary conitions are u At, () ( t ) () u v t v t At, ( t) Hence, from the characteristic equation () S an ( ) Ccos sin At, C Note that enotes rigi boy motion. Hence, the characteristic equation for aial motions of an elastic bar with free-free ens is sin which has an infinite number of solutions, i.e.,,,3,..., n, n n n n,,,...,,..., the natural frequencies of the free en- An since free en bar are E MEEN 67 HD#4 Vibrations of Continuous Systems.. San Anrés 8 3

16 ω k π E k ρ / ; k,,,... Associate to each natural frequency, there is a natural moe shape An shown in the figure below. φk cos( λ k ) k,,,... φ() Function () λ / Moe Moe Moe 3 Fig. Natural moes shapes φ() for elastic bar with both ens free. First moe is rigi boy (null natural frequency) MEEN 67 HD#4 Vibrations of Continuous Systems.. San Anrés 8 4

17 Eample 3. u(,t) A bar with both ens fie. The bounary conitions are At, u(, t) φ() v( t) φ() t At, u( t, ) φ v( t) φ Hence, from the characteristic equation φ( ) Ccos( λ) Ssin( λ) then φ () C an φ( ) Ssin( λ ) At, φ ( λ ) sin +, Note that λ enotes rigi boy motion. Hence, the characteristic equation for aial motions of a fie enfie en elastic bar is sin λ which has an infinite number of solutions, i.e. λ π, π,3 π,..., nπ, n π n λ λ n n,,... ω,,... ρ An since E, the natural frequencies of the free enfree en bar are π E ωk k ρ / ; k,,... MEEN 67 HD#4 Vibrations of Continuous Systems.. San Anrés 8 5

18 Associate to each natural frequency, there is a natural moe shape An shown in the figure below. φk sin( λ k ) k,,,... φ() Function () λ / Moe Moe Moe 3 Fig. Natural moes shapes φ() for elastic bar with both ens fie. MEEN 67 HD#4 Vibrations of Continuous Systems.. San Anrés 8 6

19 ME67 - Hanout 4 (b) Vibrations of Continuous Systems ateral vibrations of elastic beams The figure shows a uniform elastic beam of length, cross section A an area moment of inertia I. The beam material properties are its ensity ρ an elastic moulus E. One en of the beam is fie to a wall while the other en is free. The iscrete force P(t) acts at a fie aial location while f(,t) represents a loa istribution per unit length. The forces inuces elastic isplacements on the beam an esignate as v(,t). y f(,t) v(,t) P(t) Fig. Schematic view of elastic beam unergoing lateral motions From elementary strength of materials consier a) Cross-sections A remain plane an perpenicular to the neutral ais () of the beam. b) Homogeneous material beam, linearly elastic, c) Material properties (ρ,e ) are constant at any given cross section. ) Stresses σ y, σ z << σ (fleural stress), i.e. along beam. MEEN 67 HD#4 Vibrations of Continuous Systems.. San Anrés 7

20 The graph below shows the free boy iagram for motion of a ifferential beam element with length. M(,t) S(,t) f(,t) S(+,t) M(+,t) v(,t) Fig. Free boy iagram of small piece of elastic beam In The FBD, S (,t) represents the shear force an M (,t) enotes the bening moment. Apply Newton s n law to the material element: Fym ayss f(, t) A (38) S v t In the limit as : A v t f ( t, ) S (39) Apply the moment equation: neglecting rotary inertia I g M I g ~ (4) MEEN 67 HD#4 Vibrations of Continuous Systems.. San Anrés 8

21 Then M M(, t) M(, t) f S M M M f S M In the limit as : S (, t ) (4) Combining Eqs. (4) an (39) gives: A v t M f ( t, ) (4) v If the slope remains small, then the beam curvature is v. From Euler s beam theory: M EI EI v (43) where I y Ais the beam area moment of inertia. Substitute Eq. (43) into (4) to obtain the equation for lateral motions of an elastic beam: v v A f ( t, ) EI t (44) MEEN 67 HD#4 Vibrations of Continuous Systems.. San Anrés 9

22 The PDE is fourth-orer in space an n orer in time. Appropriate bounary conitions are of two types: Essential BCs: v v - specifie isplacement, * v - specifie slope, * Natural BCs: - specifie moment, - specifie shear force, M M EI v * * v * S S EI See below the most typical beam configurations: v Fie en (cantilever): v & * * Pinne en v v& M * * Free en 3 v v M & S & 3 MEEN 67 HD#4 Vibrations of Continuous Systems.. San Anrés

23 v(,t) * k Spring supporte en v M S kv* EI v * Note: PDE (44) an its BCs can be erive from the Hamiltonian principle using the efinitions for kinetic (T) an potential (V) energies of an elastic beam v v ; t (45) T A V EI Free vibrations of elastic beam Without eternal forces (point loas or istribute loa, f), PDE (44) reuces to A v v EI t (46) The solution of PDE (46) is of the form v(, t) ( ) v( t) (47) et ;. Substituting Eq. (47) into Eq (46) gives. ' t A 4 EI ( ) ( ) v( t) v 4 v 4 () t EI ( ) 4 A( ) v Above, the HS is only a function of time, while the RHS is only a function of spatial coorinate. This is possible only if both sies MEEN 67 HD#4 Vibrations of Continuous Systems.. San Anrés

24 are equal to a constant, i.e.. Hence, the separation of variables gives two orinary ifferential equations 4 v() t v & 4 (48) where The solution of the ODEs is A E I (49) v() t Ctcos t Stsin t (5) ( ) Ccos Csin C3cosh C4sinh (5) where has units of [/length]. 4 / A E I (5) The coefficients (C, S) are etermine from satisfying the bounary conitions for the specific beam configuration. Equation (5) is known as the funamental moe shape for an elastic beam, i.e., it contains the information on natural frequencies an moe shapes. is iv iv 4 The solution of ODE 4 characteristic equation k k ce with MEEN 67 HD#4 Vibrations of Continuous Systems.. San Anrés

25 Eample. Pin-pin ens beam Recall cos sin cosh sinh The BCs are: 3 4 C C C C v v t (53.a) At, (, t) () ( t) () C C () 3 () v( t) () C C v M () 3 Hence, C C 3 an sin sinh v 4 C C v t At, ( t, ) ( t) sin sinh 4 C C v M v( t) (53.b) C sin C sinh 4 from this two equations, since sin ( ) C where sinh, it follows that (54) i sin when i, i,... (55) MEEN 67 HD#4 Vibrations of Continuous Systems.. San Anrés 3

26 an hence, the natural frequencies of the pin-pin beam are A A EI i EI ii ; i,... (56) Associate to each natural frequency, there is a natural moe shape i sin i sin i ; i as shown in the graph below.,,... (57) () Function () i i.75 / Moe Moe Moe 3 Fig. Natural moe shapes () for elastic beam with both ens pinne. The isplacement function response v(, t) ( ) v( t) equals to the superposition of all the foun responses, i.e. MEEN 67 HD#4 Vibrations of Continuous Systems.. San Anrés 4

27 v cos sin v t C t S t ( t, ) k k ( ) k k k k k k k sin cos( ) sin (58) v C t S t an velocity: ( t, ) k k k k k k (59) v sin C sin( t) S cos t ( t, ) k k k k k k k The set of coefficients (C k, S k ) are etermine by satisfying the initial conitions. That is at time t, v V sin C (,) ( ) k k k v V sin S (,) ( ) k k k k (6) RECA: cos sin cosh sinh ( ) C C C3 C4 Csin Ccos C3sinh C4cosh Ccos Csin C3cosh C4sinh 3 Csin Ccos C3sinh C4cosh MEEN 67 HD#4 Vibrations of Continuous Systems.. San Anrés 5

28 Eample. Fie en-free en beam Recall cos sin cosh sinh The BCs. are 3 4 C C C C v v t (6.a) At, (, t) () ( t) () () C C3 v () v( t ) () (6.b) C C () 4 At v M v( t) (6.c) Ccos C sin C cosh( ) C sinh v S 3 v( t) (6.) Csin C cos C sinh( ) C cosh 3 4 Solution of Eqs. (a)-() gives ( ) cosh i cos i isinh i sin i : (6) MEEN 67 HD#4 Vibrations of Continuous Systems.. San Anrés 6

29 where an i cosh sinh i cosi sin 3 3 i etc i (63) (64) cosh cos sinh sin () Function () _ / Moe Moe Moe 3 Fig. Natural moe shapes () for cantilever beam (fie enfree en) MEEN 67 HD#4 Vibrations of Continuous Systems.. San Anrés 7

30 Properties of the natural moes Recall that the pair k, ( ) k k k,... satisfy the ODE iv (65) 4 A where k k k EI k k k,... As in the case of aial vibrations of a bar, it is easy to show that of a fleing beam satisfy the following the natural moes k k,... ORTHOGONA properties: i fori j EA i j (66a) fori j i fori j Aij (66b) fori j For ij, the i th natural frequency follows from i i i EA i A Where i, iare the i th moe equivalent stiffness an mass coefficients. i (67) Demonstration with integration by parts (twice). MEEN 67 HD#4 Vibrations of Continuous Systems.. San Anrés 8

31 Note that kk,... functions is a COMPETE SET of orthogonal Now, consier the initial conitions for v cos sin v t C t S t ( t, ) k k ( ) k k k k k k k v V C ; v V S (,) ( ) k k (,) ( ) k k k k k (68) Using the orthogonality properties, the coefficients (C m, S m) follow from C m AmV( ), m m,,... (69a) An similarly S m AmV ( ), m m m,,... (69b) This conclues the proceure to obtain the full solution for the lateral vibrations of a beam, i.e. cos( ) sin k (7) v C t S t ( t, ) ( ) k k k k k MEEN 67 HD#4 Vibrations of Continuous Systems.. San Anrés 9

32 Force lateral vibrations of a beam Consier a beam subjecte to an arbitrary forcing function f (,t). The PDE escribing the lateral motions of the beam is v v A f ( t, ) EI t (44) et k,... k be the set of natural moes satisfying the bounary conitions of the beam configuration (pin-pin, fie-free ens, etc). A solution to Eq. (44) is of the form v Since the set k,... f (,t) can be written as q (7) (, t) ( ) k ( t) k k k is complete, then any arbitrary function where Q m f Q (7) (, t) ( ) k ( t) k Am f(, t), m k m,,... (73) Substitution of Eqs. (7, 7) into Eq. (44) gives v v A f ( t, ) EI t MEEN 67 HD#4 Vibrations of Continuous Systems.. San Anrés 3

33 iv Ak q k kqk EIk qk (74) k but recall that each of the normal moes satisfies an hence, Eq. (74) can be written as ; iv k k k Aq k Qk EIk qk k k an, since the natural moes are linearly inepenent, then it follows that (75) Aqk Qk EIk qk k,,... A astly, recall that k ; then EI EI k write (75) as A, an Q q q ; (76) k k k k A k,,... Which can be easily solve for all type of ecitations Q () t k [ See solution of unampe SDOF EOMS ectures #] MEEN 67 HD#4 Vibrations of Continuous Systems.. San Anrés 3

34 Eample 3. Free-free ens beam Recall cos sin cosh sinh The BCs are: At 3 4 C C C C () v( t) () C C (a) v M () 3 () v( t) () C C (b) 3 v S 3 () 4 At v M v( t) (6.c) Ccos Csin C3cosh( ) C4sinh (c) 3 v S 3 v( t) Csin Ccos C3sinh( ) C4cosh () Solution of Eqs. (a)-() gives ( ) cosh i cos i i sinh i sin i where i cosh sinh i cosi sin i i (63) MEEN 67 HD#4 Vibrations of Continuous Systems.. San Anrés 3

35 an etc 3 3 (64) Note that the lowest natural frequency is actually zero, i.e. a rigi boy moe. & cosh cos sinh sin () Function () 3 _ / Moe Moe Moe 3 Fig. Elastic natural moe shapes () for beam with free-free ens MEEN 67 HD#4 Vibrations of Continuous Systems.. San Anrés 33

36 Characteristic (moe shape) equation for beams: ( ) Ccos Csin C3cosh C4sinh Csin Ccos C3sinh C4cosh Ccos Csin C3cosh C4sinh 3 Csin Ccos C3sinh C4cosh MEEN 67 HD#4 Vibrations of Continuous Systems.. San Anrés 34

37 Aial vibrations of elastic bar San Anres (c) SP 8 MEEN 67 ORIGIN : The figure shows an elastic bar of length an cross-sectional area A, an with ensity an elastic moulus equal to ρ an E, respectively. The bar is rigily attache to a wall at its left en. At its right en, a rigi block or lumpe mass M is firmly attache. Note that M/M bar ε. The fiel equation for aial motions u(,t) of the bar is u u ρ A t E A a) Determine the first three natural frequencies an characteristic moes (graph the moes) of the bar as a function of (ρ, E, ). b) Using your eperience, estimate the first natural frequency of the bar an block. Eplain your assumptions. How goo is the estimate when compare to the ones erive in (a)? Solution Proceure using separation of variables, ut (, ) φ( ) v() t () A,E,, ρ M leas to the following two ODEs: an φ + λ φ (a) u(,t) v Ω + v (b) where λ Ω ρ t E The solution to the ODEs is simple, i.e.: vt () A t cos Ω t φ( ) A cos λ + B t sin Ω t + B sin λ (3) Satisfy the bounary conitions. At, u(,t) (fie en). Thus φ( ) A cos λ + B sin λ φ( ) A + B Then: A an φ( ) sin λ (4) At the right en,, the appropriate bounary conition is: aial force M accel E A u M u t (5a)

38 or: E A v φ M φ v t. Noting that v t Ω v from (a) then, at E A φ (5b) M Ω φ recall λ Ω ρ EA E φ M λ E φ ---> ρ Replacing (4) φ( ) sin λ & φ (5b) λ cos λ into (5b) gives efine ε ρ A cos λ M M, an write the characteristic equation as: ( tan( λ ρ A ). ) ελ λ sin λ where λ λ (6) from eperience or having worke other problems, using a calculator, guess values n : 3 y : # of roots 6 f( y) : tan( y) y ε ε : λ_ : root( f ( y), y) λ_ where λ_ λ 678 & Ω λ E ρ An thus, the first three natural frequencies are Ω The shape functions are φ( z) : sin λ_ z E ρ φ( z) : sin λ_ z where z φ3( z) : sin λ_ 3 z tan( Y) Y ε. Graphical solution Y tan(y) /(ye)

39 .75 Moe Moe / Moe 3 (b) Approimate first natural frequency. Using moe shape ϕ ( ) recal lots of problems worke in class an homeworks, one can easily estimate the equivalent stiffness an mass as AE ρ A K eq Meq + M ρ A ε an the estimation for natural frequency is ω appro K eq M eq E ρ 3 + ε ε 3.95 ω appro.95 E ρ which is just % higher than the eact value y E ρ

40 (c) Free vibrations response: Consier the following intitial conitions, at t u (, ) a a :. Uniform aial stretching t u no velocity rest conition The response for aial motions of the bar is ut (, ) i φ i ( A i cos( Ω i t) + B i sin( Ω i t) ) Since the initial velocity everywhere, then it follows that B i an multiplying this equation by φj an integrating over the omain gives u (, ) a i φ i ( A i) Define z i :.. n a A i φ i φ i ( z) : sin λ_ i z z sin λ_ i z A i : sin λ_ i z φ i z z a "Use orthogonality property of shape functions recall: λ_ λ_ λ A ut (, ) : n i ( ) sinλ_ i A i cos Ω i t

41 Calculate time response at various spatial points in bar: PHYSICA Parameters: Ω : λ_ E ρ E 9 N : m ρ : 78 kg m 3 : m :. m ε f n : Ω π T n : f n f n T n T Hz natural frequencies ( ) s ut (, ) : n i natural perios ( ) sinλ_ i A i cos Ω i t Response at bar: mipoint & en T ma : 5T n. u(, t) ut (, ) t time (s)

42 Aial vibrations of elastic bar () San Anres (c) SP 8 MEEN 67 The figure shows an elastic bar of length an cross-sectional area A, an with ensity an elastic moulus equal to ρ an E, respectively. The bar is rigily attache to a wall at its left en. At its right en, a massless spring K s connects the bar to another fie wall. K s /(EA) ε. The fiel equation for aial motions u(,t) of the bar is u u ρ A E A t a) Determine the first TWO natural frequencies an characteristic moes (graph the moes) of the bar as a function of (ρ, E,, anε ). [] b) Using your eperience, estimate the first natural frequency of the bar an block. Eplain your assumptions. How goo is the estimate when compare to the ones erive in (a)? [5] ORIGIN : PHYSICA Parameters: E 9 N : m ρ : 78 kg m 3 : m :. m A : π 4 AE 7 8 N m (a) natural frequencies & moe shapes using separation of variables, ut (, ) φ( ) v() t () A,E,ρ K s leas to the following two ODEs: an φ + λ φ (a) u(,t) v Ω + v (b) where λ Ω ρ t E The solution to the ODEs is simple, i.e.: vt () A t cos Ω t φ( ) A cos λ + B t sin Ω t + B sin λ (3) Satisfy the bounary conitions. At, u(,t) (fie en). Thus φ( ) A cos λ + B sin λ φ( ) A + B Then: A an φ( ) sin λ (4)

43 At the right en,, the appropriate bounary conition is: bar aial force Ks u spring force at E A u k S u (5a) or: Replacing (4) efine E A v φ k S φ v φ( ) sin λ & E A φ EA λ cos λ k S φ k S φ (5b) from (a) λ cos λ + sin( λ ) into (5b) gives where λ λ ε k S EA guess values from graphical soln, an write the characteristic equation as: tan λ. ε : n : 4 y : ( ) T # of roots f y + ε λ : tan( y) + (6) ε y λ_ where λ_ λ & An thus, the first four natural frequencies are Ω : λ_ E ρ Ω λ E ρ ra Ω T The shape functions are φ( z) : sin λ_ z φ( z) : sin λ_ z where z φ3( z) : sin λ_ 3 z φ4( z) : sin λ_ 4 z s tan( Y) Y ε λ_ : root( f ( y), y) Graphical solution Y tan(y) -(ye)

44 .75 / Moe Moe Moe 3 Moe 4 (b) Approimate first natural frequency. Using moe shape easily estimate the equivalent stiffness an mass as ϕ ( ) : K eq : EA ϕ ( ) + k S ϕ M eq : ρ A ( ϕ( ) ) ε k S EA ε K eq K eq AE AE + k S ( + ε) Meq ρ A 3 an the estimation for the funamental natural frequency is ω appro K eq M eq which is just 3% higher than the eact value E ρ + ε 3 a : + ε 3 a.936 a.9 λ_ λ_.76 E ρ

45 (c) Free vibrations response: Consier the following intitial conitions, at t u (, ) no aial stretching ut (, ) t u inf i A i ss v o velocity an The response for aial motions of the bar is φ i ( A i cos( Ω i t) + B i sin( Ω i t) ) t u (, ) v o multiplying this equation by φj an integrating over the omain gives s v o : ss : 3 inf i φ i m s Since the initial isplacement everywhere, then it follows that ( B i Ω i ) z ss z B i Ω i v o ss φ i φ i "Use orthogonality property of shape functions Define i :.. n z z ss sin( λ_ i z) z v o B i : Ω i sin( λ_ i z) z m B T ut (, ) : φ i ( z) : sin λ_ i z n i sinλ_ i B i ( sin( Ω i t) ) recall: λ_ λ.76 λ_

46 Calculate time response at various spatial points in bar: natural frequencies: Ω f n : π T natural perios f n ( ) Hz T n : f n T T n ( ) s Bar isplacement: mipoint & en ut (, ) : n i ( ) sinλ_ i B i sin Ω i t for graph only T ma : 5T n [m]. 4 u(, t) ut (, ) t time (s) Bar velocity: mipoint & en vt (, ) : [m/s] v(, t) vt (, ) n i sinλ_ i B i Ω i cos Ω i t ( ) v o m s time : s v o ss m s v(, time).837 m s t time (s)

47 The figure shows an elastic ro of length, raius R, ensity an elastic shear moulus G. The ro is rigily attache to a wall at its left en. At the ro right en, a massless spring K S connects the ro to a fie wall. The fiel equation for angular motions (,t) of an elastic ro uner torsion is where J ½ R 4 is the polar moment of area an J GJ t M GJ is the torsional moment. et K S R /(GJ). a) Fin the first TWO natural frequencies an characteristic moes (sketch the moes) of the bar as a function of (, G,, J an ). [35] b) Using eperience, estimate the first natural frequency of the bar an spring. Eplain your assumptions. How goo is the estimate when compare to the eact (first) value erive in (a)? [5] If neee use the following G 9 Pa, 78 kg/m 3, m, Rm Top view KS (,t) KS R

48 Torsional vibrations of elastic ro San Anres (c) SP MEEN 67 The figure shows an elastic ro of length, raius R, ensity an elastic shear moulus G. The ro is rigily attache to a wall at its left en. At the ro right en, a massless spring K S connects the ro to a fie wall. The fiel equation for angular motions (,t) of an elastic ro uner torsion is J G J where J ½ R 4 is the polar moment of area an M GJ is the torsional t moment. et K S R /(GJ). a) Fin the first TWO natural frequencies an characteristic moes (sketch the moes) of the bar as a function of (, G,, J an ). [35] b) Using eperience, estimate the first natural frequency of the bar an spring. Eplain your assumptions. How goo is the estimate when compare to the eact (first) value erive in (a)? [5] If neee use the following G 9 Pa, 78 kg/m 3, m, Rm PHYSICA Parameters: G 9 N m 78 kg m 3 m.m r A 4 J r 4 A J J m 4 3 Top view ORIGIN KS (,t) KS R J GJ t ()

49 (a) natural frequencies & moe shapes ( t ) ( ) v() t () using separation of variables, Substitute into the fiel Eq. () to obtain the following two ODEs: (a) an The solution to the ODEs is simple, i.e.: v v (b) t vt () A t cos t ( ) A cos B t sin t where G B sin (3) Satisfy the bounary conitions. At, (,t) (fie en - no angular eformation). Thus ( ) A cos B sin ( ) A B Then: A an ( ) sin (4) At the right en,, the appropriate bounary conition is: ro torsional moment reaction force moment arm at GJ u k S u R (5a) spring R R or: GJv k S R v GJ k S R (5b) from (a) Replacing (4) efine k S R GJ guess values from graphical soln ( ) sin & GJ k S R cos cos sin into (5b) gives where, an write the characteristic equation as: tan. n 4 y ( ) T # of roots f y tan( y) (6) y

50 _ T The shape functions are ( z) sin _ z ( z) sin _ z 3( z) sin _ 3 z 4( z) sin _ 4 z An thus, the first four natural frequencies are where _ where z & _ E G ra s tan( Y) Y _ root( f ( y) y) Graphical solution Y tan(y) -(ye).75 / Moe Moe Moe 3 Moe 4

51 (b) Approimate first natural frequency. Using moe shape ( ) easily estimate the equivalent torsional stiffness an mass moment of inertia from K eq GJ ( ) k S R I eq J ( ) K eq GJ k S R K eq GJ Ieq J 3 with k S R GJ an the estimation for the funamental natural frequency is appro K eq I eq G a where a 3 a.936 compare to the eact value: a.9.76 G, i.e just 3 % higher than the eact value GJ k S R k S.78 7 N m

52 lateral vibrations of elastic beam San Anres (c) SP 8 MEEN 67 : 5 mm iameter ρ : 78 kg ORIGIN : m 3 E N : : m length m π A : cross-sectional area 4 l,d,e,i π 4 I P : A 64 area moment of inertia (polar) h : m y h Solution Proceure free fall velocity v o : ( g h) using separation of variables, ut (, ) φ( ) v() t () leas to the following two ODEs: an 4 φ 4 ( ρ A) λ φ (a) v + Ω v (b) t v v EI t where λ Ω ρ A EI P The solution to the ODEs is simple, i.e.: φ( ) A cos β vt () A t cos Ω t + B t sin Ω t + B sin β + C cosh β + D sinh β (3) β λ Satisfy the bounary conitions. PINNED ENDs: no lateral isplacement an null bening moment At, u u M at φ( ) A cos β + C cosh β φ A cos β + C cosh β then A C so since: cos( ) cosh( ) φ( ) B sin β + D sinh β

53 at φ B sin β + D sinh β φ then D sin( β ) B sin β + D sinh β (4) thus MODE SHAPE φ( ) sin β (5) (4) is the characteristic equation an natural frequencies f Ω : π natural frequencies ra Ω T The shape functions are φ( ) : sin β with solution: β i : Ω : φ( ) : sin β ( β) EI P ρ A φ3( z) : sin β 3 z Hz f T i π i :.. n β # of roots n : s m.75 Moe Moe / Moe 3

54 (b) Approimate first natural frequency. Using moe shape recall lots of problems worke in class an homeworks, one can easily estimate the equivalent stiffness an mass as an the estimation for natural frequency is M eq : ρ A ϕ a ( ) K eq : EI P ϕ a ( ) ϕ a ( ) : sin π ϕ a ( ) M eq ρ A K eq 48 E I P 3 4 : Ω a : K eq M eq Ω a Ω 8.3 ra s ra s eact Ω a. Ω.75 Moe approimate (c) Free vibrations response: Consier the following intitial conitions, at t u (, ) No initial eformation t u v o velocity free fall velocity The response for lateral motions of a beam is

55 p ut (, ) n i φ i ( A i cos( Ω i t) + B i sin( Ω i t) ) Since the initial isplacement everywhere, then it follows that A i Calculate time response at various spatial points in beam: ut (, ) an for velocity i :.. n : mm : n mm i B i : ( B i sin( Ω i t) ) sin β i t u multiplying this equation by φj an integrating over the omain gives select number of moes to isplay results VEOCITY v o sin( β i ) sin β i i φ i v o Ω i DISPACEMENT vel(, t) B i Ω i : mm i "Use orthogonality property of shape functions Ω i ( B i cos( Ω i t) ) sin β i B i.3 m n 7 mm 7 v o 4.49 m s T n : f Natural perios for graph T ma : 5T n

56 Displacement response at beam: mipoint (/) & /4.4. u(, t) u(, t) t time (s) T n T ( ) s Velocty response at beam: mipoint (/) & /4 5 vel(, t) vel(, t) t time (s) XX : vel( XX, s).9 v o v o 4.49 m s

57 Vibrations of a string This problem ais to unerstan the tuning process of string musical instruments. The graph shows a simple moel of a taut string fie at both ens. The string vertical isplacement u(,t) is escribe by: u u T t where 4 g/m is the string mass per unit length, m is the string length, an T is the tension applie to the string. When the string is plucke at its mile, its vibration response is ominantly represente by the first moe shape at the first natural frequency f, (see the otte lines for a sketch). Then, the soun frequency components raiating from the string are ominant with frequency (f ). The tonal frequencies of the strings in a violin are G3 96 Hz, D Hz, A4 44 Hz, an E Hz. Assume the strings are mae of the same material (steel). 78 kg/m 3 Questions: a) Fin the relation between the frequency (f ), the string length (), the mass per unit length (), an the tension (T). b) Assuming all strings have the same iameter, fin the tensions T to tune each string. Fin also the stresses an elastic eformations. c) Fin how a tonal frequency scales with the iameter of a string. Using the tension foun in (b) for G3, etermine the strings iameter an elastic eformation.

58 Vibrations of a string San Anres (c) SP 4 MEEN 67 The string vertical isplacement u(,t) is escribe by PHYSICA Parameters for a steel string: u u T t () ORIGIN where T is the tension in the string an γ is the mass per unit length. E 3 9 N m γ.45 kg m m length of string ( a) natural frequencies & moe shapes Θ( t ) ϕ( ) v() t () using separation of variables, Substitute into the fiel Eq. () to obtain the following two ODEs: ϕ λ ϕ (a) an The solution to the ODEs is simple, i.e.: v Ω v (b) t vt () A t cos( Ωt) B t sin( Ωt) ϕ( ) A cos( λ) B sin( λ) where λ Ω γ T et (3) c o T γ a characteristic spee [m/s] Satisfy the bounary conitions. At, (,t) (fie en - no isplacement). Thus ϕ( ) A cos( λ) B sin( λ) ϕ( ) A B Then: A

59 an ϕ( ) sin( λ) (4) is the equation for the shape funcion. At the right en,, the string is also fie. Thus at ϕ sin( λ) (5a) which is the characteristic equation. The roots are λ j jπ Set T 4N eample An thus, the natural frequencies are Ω j λ j c o Ω j jπ T γ ra/s Ω T The shape functions are ϕ( ) sin π ϕ( ) sin π ra s ϕ3( ) sin π3 ϕ4( ) sin π4 etc Moe Moe Moe 3 Moe 4

60 (b)fin the relation between the frequency (f), the string length (), the mass per unit length (), an the tension (T). The natural frequencies in Hz (cycles/s] are f Ω π f j j The tonal frequencies for the strings in a VIOIN are Sol G D Re T ( f) f 4 γ A4 tone Hz 44 a 4 E Mi 5 k 4 ( four strings) consiering the same ensity/length for all strings, the necessary tuning tension in a string is T k f tonek γ et a string iameter mm The stresses in the strings are T (assumption as everyone knows the strings have ifferent iameters) σ T π 4 σ T Pa ε T N an the strains σ ε E G3 D4 A4 E5 T γ Hz mm ε T ( )mm Deformation for the strings are large, in particular for one with highest tonal frequency. Recall 5 mm

61 (c) Using the tension foun in (b)g3, an assuming the strings are mae of the same material (steel), fin how a tonal frequency scales with the iameter of each string Note that a string mass/unit length γ ρa ρ π 4 where ρ is the material ensity an is the iameter of a string. Hence, the tonal frequency as a function of the string iameter equals f tone T γ 4T ρπ f tone T ρπ Set T_ T Tension for soun G3 T_ 94.9 N Hence, for a given tension, a string's iameter is inversely proportional to the tonal frequency With eformation for each string δ k k f tonek π T_ k 4 E T_ ρπ mm G3 D4 A4 E5 (c) Stresses an strains δ T ( )mm m δ T The stresses in the taut strings are T σ f A tone ρ since γ ρa an the aial strains E 3 Pa σ ε f E tone ρ NOT a function of the string E iameter with elastic eformation ρ δ ε f tone 3 proportional to ^3 an tone_freq^ E