Lobontiu: System Dynamics for Engineering Students Website Chapter 3 1. z b z

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1 Chapter W3 Mechanica Systems II Introduction This companion website chapter anayzes the foowing topics in connection to the printed-book Chapter 3: Lumped-parameter inertia fractions of basic compiant members Lumped-parameter stiffness fractions of basic compiant members Lagrange s method for deriving mathematica modes of conservative and nonconservative mechanica systems Coordinate couping and decouping methods in mutipe-dof mechanica systems W3.1 Lumped-Parameter Inertia Fractions from Distributed Inertia ( 3.1.1) The vibration response of structura members such as bars (that undergo either aia or torsiona deformations), beams (which are deformed in bending), membranes or pates is generay defined by using distributed parameters to formuate the corresponding mathematica mode. For these eastic members it is possibe, however, to equivaenty transform the continuous structure into a umped-parameter one. Studied here are the cases of beams in bending, bars in torsion and bars under aia oading. W3.1.1 Bending of Beams Figure W3.1 suggests how a rea cantiever beam with distributed inertia is equivaenty transformed into a point-ike mass that is ocated at the cantiever s free end. As indicated in Fig. W3.1(a), the distributed-parameter cantiever has a defection, which depends on both the time t and the position on the cantiever. The foowing reationship eists between a generic defection (measured at an abscissa from the free end), denoted by u z (, t), and the defection at the free end, which is denoted by u z (t), of the form: u( t,) = f( u ) () t (W3.1) z b z Website Chapter 3 1

2 where f b () is the bending-reated distribution function. This distribution function can be found by appying a force at the free end of the cantiever and by cacuating the defection at that point, as we as at another point of abscissa. bending cantiever z massess cantiever umped mass u z m e u z (a) (b) Figure W3.1 Vibrating cantiever with: (a) distributed inertia; (b) equivaent umped inertia It can be shown by using basic mechanics of materias cacuation that for a constant cross-section cantiever the bending-reated distribution function is: 3 1 fb( ) = 1 + It is aso known from mechanics of materias that the kinetic energy of the distributedparameter cantiever is epressed as: z (W3.) 1 uz( t,) 1 u() t T = = [ ( )] ρ A d A fb d t t ρ (W3.3) where Eq. (W3.1) has been used. The umped-parameter equivaent mode consisting of the tip mass of Fig. W3.1(b) has the kinetic energy: 1 Te = me uz () t t (W3.4) The two kinetic energies of Eqs. (W3.3) and (W3.4) need to be identica, which resuts in the foowing equivaent mass: m = [ ( )] e ρ A fb d (W3.5) 0 For a constant cross-section cantiever for which the bending distribution function is given in Eq. (W3.), the bending-equivaent mass of Fig. W3.1(b) is: Website Chapter 3

3 33 me = m (W3.6) 140 where m is the tota mass of the cantiever. The bridge (aso known as camped-camped beam) is another common design where distributed-parameter inertia can be rendered into equivaent umped-parameter inertia. Figure W3. iustrates the main parameters that define the two systems. bending bridge z massess bridge umped mass u z m e u z / / (a) (b) Figure W3. Vibrating bridge with: (a) distributed inertia; (b) equivaent umped inertia The derivation eaborated for the cantiever is aso vaid for the bridge with the mention that the bending distribution function is determined by epressing the defection u z () at an arbitrary ocation in terms of the maimum defection u z at the midspan (with the reference frame positioned as shown in Fig. W3.(a)). It can be shown by appying mechanics of materias notions that for a constant cross-section bridge the distribution function is: f ( ) b 4 3 4,0 = , (W3.7) By substituting Eq. (W3.7) into Eq. (W3.5), the umped-parameter mass, which is equivaent to the distributed-parameter inertia bridge, is: 13 me = m (W3.8) 35 where m is the bridge s mass. Website Chapter 3 3

4 W3.1. Torsion of Bars Figure W3.3 shows a distributed-inertia camped-free bar in torsion as we as its equivaent system, which coects a umped-parameter body of moment of inertia J e at its free end. bar in torsion massess bar umped mass θ J e θ (a) (b) Figure W3.3 Bar undergoing torsiona vibrations with: (a) distributed inertia; (b) equivaent umped inertia The rotation ange θ () at an arbitrary distance measured from the free end can be epressed as a function of the maimum rotation ange θ at the free end, namey: θ ( t,) = f( ) θ () t (W3.9) t where f t () is the torsion-reated distribution function. This distribution function can be found by appying a torque at the free end of the bar and cacuating the rotation at that point as we as at a point of arbitrary abscissa. It can be shown using basic mechanics of materias reationships that for a constant cross-section fied-free bar the torsion distribution function is: ft ( ) = 1 (W3.10) It is aso known from mechanics of materias that the kinetic energy of the distributedparameter bar undergoing torsiona vibrations is epressed as: θ( t,) θ() t t t t 0 t t T = ρi d = ρ I [ f ( ) ] d (W3.11) where I t is the cross-section torsion moment of inertia more detais can be found in Appendi D of the printed book. The umped-parameter equivaent mode consisting of the tip body of Fig. W3.3(b) has the kinetic energy: Website Chapter 3 4

5 1 θ () t Te = Je t (W3.1) The two kinetic energies of Eqs. (W3.11) and (W3.1) are identica, which resuts in the foowing equivaent mass moment of inertia: J = I [ f ( ) ] d ρ (W3.13) e t t 0 For a constant cross-section cantiever for which the torsion distribution function is given in Eq. (W3.10), the equivaent mass moment of inertia is: 1 1 Je = ρ It = Jt (W3.14) 3 3 where J t is the mass moment of inertia of the bar about its ongitudina ais. The bridge (a camped-camped bar for torsion) is another common design where distributed-parameter inertia can be transformed into equivaent umped-parameter inertia. Figure W3.4 iustrates the main parameters that define the two systems. bridge in torsion z massess bridge umped inertia z θ J e / / θ (a) (b) Figure W3.4 Torsionay vibrating bridge with: (a) distributed inertia; (b) equivaent umped inertia The derivation eaborated for the fied-free bar is aso vaid for the bridge with the mention that the bending distribution function is determined by epressing the rotation ange θ () at an arbitrary ocation in terms of the maimum rotation ange θ at the midspan (with the reference frame positioned as shown in Fig. W3.4(a)). It can be shown by appying mechanics of materias notions that the distribution function is:,0 ft ( ) = 1, (W3.15) Website Chapter 3 5

6 Substitution of Eq. (W3.15) into Eq. (W3.13) gives the umped-parameter mass moment of inertia, which is equivaent to the distributed-parameter inertia bridge: 1 Je = J t (W3.16) 3 and this is identica to the equivaent mass moment of inertia of a camped-free bar undergoing torsiona motion. W3.1.3 Aia Loading of Bars Heavy Springs in Transatory Motion Heavy springs undergoing transatory motion possess inertia properties in addition to their stiffness so it shoud be possibe to determine an equivaent umped-parameter mass to be paced at the free end of the spring, which is camped at the opposite end and undergoes aia vibrations. Figure W3.5 shows schematicay the equivaence process mentioned above whereby the heavy spring is transformed into a massess spring of stiffness k and a body of mass m e. k m e Figure W3.5 Heavy spring in transatory motion and its equivaent spring-mass system Consider that the actua heica spring is straightened into an aia bar (of tota ength L) that is fied at one end and acted upon by a force in the aia direction at the other end. The motion of an arbitrary point at distance from the free end of the rea spring, u (, t), is connected to the free end point dispacement u (t) as: u( t,) = f( u ) () t (W3.17) a where f a () is the distribution function reated to the aia deformations. It can be demonstrated that f a () is identica to f t () the distribution function corresponding to the torsion of a camped-free bar in Eq. (W3.10). The kinetic energy of the distributedparameter spring which vibrates aiay can be epressed as: Website Chapter 3 6

7 L L 1 u( t,) 1 u() t T = A d = A f d t t ρ ρ [ a ( )] (W3.18) 0 0 Again, the assumption has been appied that the aia vibrations of the actua spring are identica to those of a straight bar having its ength equa to that of the spring wire. The umped-parameter equivaent mass m e has a kinetic energy of: 1 u () t Te = me t The two kinetic energies of Eqs. (W3.18) and (W3.19) are identica, which gives an equivaent mass of: 0 (W3.19) L m = [ ( )] e ρ A fa d (W3.0) For a constant cross-section bar under aia oad, the equivaent mass becomes: where m is the tota mass of the spring. 1 me = m (W3.1) 3 W3. Lumped-Parameter Stiffness from Distributed Stiffness ( 3.1.) Simiar to the process of deriving equivaent umped-parameter inertia properties of vibrating beams and bars, which have been obtained from the origina distributed inertia properties, equivaent umped-parameter stiffness properties can aso be derived from distributed ones, as shown net. W3..1 Beams in Bending Figure W3.6 indicates how a rea cantiever beam with distributed stiffness is equivaenty transformed into a fied-position, point-ike spring that is ocated at the cantiever s free end. Website Chapter 3 7

8 bending cantiever z u z u z spring k e (a) (b) Figure W3.6 Vibrating cantiever with: (a) distributed stiffness; (b) equivaent umped stiffness It is known from mechanics of materias that the strain energy stored by a deformed beam is epressed as: u(,) z t d fb y [ z() ] y d ( ) U = EI d = u t EI d (W3.) where I y is the cross-section moment of inertia about the y ais. Equation (W3.1) showing the reationship between the defection u z () at an arbitrary abscissa and the maimum defection at the free end u z has been used to derive Eq. (W3.). The distributed stiffness of the rea beam is equivaent to the umped-parameter stiffness of a transation spring which is paced at the cantiever s free end, as iustrated in Fig. W3.6(b). The mobie end of the spring is assumed to have the motion of the rea cantiever s free end u z. The strain energy of the spring is: U 1 [ ()] e = ke uz t (W3.3) Comparing Eqs. (W3.) and (W3.3) and assuming that the strain energies of the actua and equivaent systems are identica, the stiffness of the equivaent spring is obtained as: d f ( ) b ke = EI y d (W3.4) 0 d For a constant cross-section cantiever whose distribution function is given in Eq. (W3.), Eq. (W3.4) simpifies to: 3EI y ke = (W3.5) 3 Website Chapter 3 8

9 which is an equation we-known from mechanics of materias. There are situations, particuary in MEMS, where a beam is camped at both ends, one of which is connected to a moving body, as sketched in Fig. W3.7. deformed beam transating mass Figure W3.7 Camped-guided beam as a spring eement As known from mechanics of materias, in designs where the beam is actuay camped at the eft (immobie) end and guided at the right (moving) end, the equivaent stiffness of a spring that needs to be paced at the guided end is: k e 1EI = 3 y (W3.6) which indicates that by changing the boundary condition at one end from free to guided, the equivaent stiffness increased four times (compare Eqs. (W3.5) and (W3.6)). For bridges, a simiar formuation can be given to determine the equivaent stiffness of a spring ocated at the midspan, as iustrated in Fig. W3.8. bending bridge z u z / u z / spring k e (a) (b) Figure W3.8 Vibrating bridge with: (a) distributed stiffness; (b) equivaent umped stiffness Website Chapter 3 9

10 The derivation eaborated for the cantiever is aso vaid for the bridge, and therefore Eq. (W3.4) does appy. For a constant cross-section bridge, with the distribution function of Eq. (W3.7), the equivaent stiffness of the umped-parameter spring of Fig. W3.8(b) becomes: k e 19EI = 3 which is the equation provided by mechanics of materias tets. y (W3.7) Eampe W3.1 Cacuate the equivaent stiffness of the foded-beam suspension arrangement sketched in Fig. W3.9 by considering the transating motion of the centra body is perpendicuar to the four beams ongitudina dimensions. rigid connector short beam ong beam transating body fied anchor 1 Figure W3.9 Foded-beam suspension for transatory motion Soution: The eastic suspension of the centra body is formed of four identica onger beams, which attach the body to two side rigid connectors and four identica shorter beams which attach the side connectors to two fied anchors (supports). On each side of the moving body, two identica arrangements are formed comprising each two onger beams and two shorter ones. It shoud aso be noticed that a beams are fied-guided. As such, the equivaent spring system is the one sketched in Fig. W3.10, whereby the eft eastic beam group of Fig. W3.9 is paced above the centra body and the right group underneath the body but the order is not important since the two eastic groups are identica. Website Chapter 3 10

11 k 1 k 1 k k k k k 1 k 1 Figure W3.10 Equivaent spring system of the foded-beam suspension The stiffness denoted by k 1 is generated by the either of two shorter beams on each side of the centra body. Simiary, the stiffness k is produced by either of the two onger beams on each side of the body. The group of the k 1 springs is in series with the k group. It shoud be noted that the spring group to the eft of the centra body and the spring group to the right of it (which are dispaced to the top and bottom of the centra body in Fig. W3.10) are connected in parae. As a consequence: with k e k1 k 4kk 1 = = k + k k + k 1 1 1EI k = 1EI k = y y 3 (W3.8) (W3.9) where I y1 and I y are the moments of inertia of the two beams cross-sections with respect to aes that are perpendicuar to the motion pane. Substituting the two stiffnesses of Eqs. (W3.9) into Eq. (W3.8) produces the foowing equivaent stiffness is: k e 48EI y1i y 3 3 y1+ Iy1 = I (W3.30) Website Chapter 3 11

12 W3.. Bars in Torsion A feibe bar camped at one end and free at the other (as the one of Fig. W3.3(a)), which is under the action of twisting moment (torque) acting about the bar s ongitudina ais, is a distributed-stiffness member. The aim is to transform this actua member into a rigid bar that has an equivaent spring at the origina free end, as sketched in Fig. W3.11. The strain energy stored by a bar that is torsionay deformed can be epressed as: θ (,) t dft() t [ () ] t 0 0 d 1 1 U = GI d = θ t GI d (W3.31) where, again, the anguar rotation at an abscissa measured from the free end, θ (), is epressed in terms of the (maimum) rotation at the free end, θ, by means of the torsion distribution function f t () that is provided in Eq. (W3.10). end bearing torsiona spring rigid bar θ k e Figure W3.11 Bar undergoing torsiona vibrations with torsiona spring representing equivaent umped stiffness The distributed stiffness of the rea bar is made equivaent to the umped-parameter stiffness of a rotation spring which is paced at the bar s free end, as iustrated in Fig. W3.11. The mobie end of the spring is assumed to have the motion of the rea bar s free end θ. The strain energy of the spring is: U 1 [ ()] e = ke θ t (W3.3) Website Chapter 3 1

13 Comparing Eqs. (W3.31) and (W3.3) and aso assuming that the strain energies of the actua and equivaent systems are identica, the stiffness of the equivaent spring is obtained as: dft ( ) ke = GIt d d 0 (W3.33) For a constant cross-section cantiever whose distribution function is given in Eq. (W3.10), Eq. (W3.33) reduces to: GIt ke = (W3.34) which is we-known from mechanics of materias. For a camped-camped bar undergoing torsiona vibrations, as the one sketched in Fig.W3.4(a), a simiar equivaence procedure is appied which transforms the actua, distributed-stiffness member into an equivaent rigid rod and two end torsiona springs, as indicated in Fig. W3.1. torsion spring rigid bar torsion spring θ k e / k e Figure W3.1 Equivaent of douby camped bar undergoing torsiona vibrations with two torsiona springs representing equivaent umped stiffness The vaue of the equivaent stiffness of the torsiona springs (that are connected in parae in Fig. W3.1, as epained previousy in this section) is determined by means of Eqs. (W3.15) and (W3.33), and for a constant cross-section bar it is found to be equa to: k e GIt = (W3.35) which is aso known from mechanics of materias. Pease note that there are two springs in Fig. W3.1, which indicates that the tota equivaent stiffness is twice the one of Eq. (W3.35) as the two end torsiona springs act in parae. Website Chapter 3 13

14 W3..3 Bars in Aia Vibration A feibe bar of ength, which is camped at one end and free at the other (as the one of Fig. W3.13) is subjected to aia vibrations such that the motion of its free end is u (t) and the motion of a point situated at a distance from the free end is u (, t). They are reated by means of Eq. (W3.17). bar u spring k e Figure W3.13 Bar undergoing aia vibrations with equivaent umped stiffness at its free end. The strain energy of the distributed-stiffness bar is u(,) t dfa() [ ()] 0 0 d 1 1 U = EA d = u t EA d The eastic energy stored in the equivaent spring of stiffness k e is: (W3.36) U = 1 [ (,)] e ke u t (W3.37) Equating the energies of Eqs. (W3.36) and (W3.37), yieds: dfa (,) t ke = EA d d 0 (W3.38) With the distribution function of Eq. (W3.10), which corresponds to a constant crosssection bar, the equivaent stiffness of Eq. ($3.38) becomes: EA ke = with is an equation we-known in mechanics of materias. (W3.39) Website Chapter 3 14

15 W3.3 Lagrange s Equations Method ( 3.3.) Another energy-based method that enabes formuation of mathematica modes for dynamic systems is provided by the Lagrange s equations. In the case of nonconservative systems, Lagrange s equations are: d L L = Qi (W3.40) dt q i qi For mechanica systems, q i (i = 1,,, n), the generaized coordinates, usuay represent the minimum number of geometric coordinates (inear and/or anguar) which fuy define the state of the systems, and can be identica to the degrees of freedom, defined in Chapter 3. Q i are known as generaized forces, and physicay they can be forces or moments of nonconservative character. The amount L is named Lagrangian and is defined as the difference between the kinetic and the potentia energy: L= T U (W3.41) For conservative mechanica systems there are no generaized forces acting on the system and therefore: d L L = 0 dt q i q i (W3.38) Eampe W3. Derive the mathematica mode of the mechanica system sketched in Fig. W3.14 by using Lagrange s equations method. Assume a friction force acts on the body defined by 1 whereas the motion of the other body is frictioness. f f m 1 k m f f f m 1 f d f d m f c Figure W3.14 Two-DOF mechanica system and free-body diagrams with nonconservative forces Soution: The two-dof mechanica system has the foowing Lagrangian: Website Chapter 3 15

16 1 1 1 L = T U = m ( ) 1 + m k 1 (W3.43) This is a non-conservative system, as forces other than eastic and gravitationa act on the two bodies. The two DOF being 1 and, there are two non-conservative forces acting on each of the two bodies, as shown in the corresponding free-body diagrams. The damping and friction force, according to the way they were drawn in Fig. W3.14 are: ( ) fd = c f f = µ mg 1 and the generaized forces Q 1 (corresponding to the generaized coordinate q 1 = 1 ) and Q (corresponding to the generaized coordinate q = ) are: ( ) ( ) Q1 = fd f f = c 1 µ mg Q = f fd = f c 1 (W3.44) (W3.45) By taking the necessary derivatives, according to Lagrange s equation definition eft hand side of Eq. (W3.40), the foowing equations are obtained: d L L = m 1+ k 1 dt 1 1 d L L = m + k dt ( ) ( ) 1 (W3.46) Using the fu form of Eqs. (W3.40), which means combining Eqs. (W3.45) and (W3.46), yieds the foowing equations which represent the mathematica mode of the two-dof system of Fig. W3.14: ( ) ( ) ( ) ( ) m 1+ c 1 + k 1 = µ mg m + c 1 + k 1 = f Equations (W3.47) can aso be written into vector-matri form as: m 0 1 c c 1 k k 1 µ mg 0 m + + = c c k k f which can succincty be epressed as: [ ]{ } + [ ]{ } + [ ]{ } = { } (W3.47) (W3.48) M C K F (W3.49) where [M] is the mass matri, [C] is the damping matri, [K] is the stiffness matri, {F} is the forcing vector, and {} = { 1 } t is the dispacement vector (the one defined by the generaized coordinates or DOF). Website Chapter 3 16

17 Eampe W3.3 By using Lagrange s equations, determine the mathematica mode of the doube-penduum shown in Fig. W3.15 assuming the system motion takes pace in a vertica pane. Aso assume the rods are massess. datum ine y y 1 θ 1 1 m 1 θ 1 1 v 1 v 1 v v 1 θ θ 1 θ m θ v 1 v 1 y (a) (b) (c) Figure W3.15 Doube penduum in vertica pane: (a) schematic with position parameters; (b) schematic with bob veocities; (c) veocity poygon Soution: This is a two-dof mechanica system and the anguar coordinates θ 1 and θ fuy define its state, and therefore, the generaized coordinates are seected as: q 1 = θ 1 and q = θ. Since the rods are massess, the kinetic energy ony resuts from the two bobs which have absoute veocities v 1 and v, and is of the form: 1 1 T = mv 1 1+ mv (W3.50) The veocity of bob # 1 comes from its pure rotation about the fied pivot point and is therefore equa to: v = θ (W3.51) 1 11 The veocity of bob #, as known from panar motion, is found by vector addition of the veocity of bob # 1 and the reative veocity of bob # rotating around bob # 1, namey: where the reative veocity can be found as: v = v1 + v 1 (W3.5) v = θ (W3.53) 1 Website Chapter 3 17

18 As Fig. W3.15(c) shows, and according to the generaized Pythagoras theorem, the magnitude of v is cacuated as: ( ) v = v + v vv cos θ θ (W3.54) Substituting Eqs. (W3.51), (W3.53) and (W3.54) into Eq. (W3.50), the kinetic energy becomes: = 1 T m + m θ m θ θ θθ + m θ (W3.55) ( ) cos( ) The potentia energy of the system resuts from the two bobs changing their vertica position with respect to the datum ine, as sketched in Fig. W3.15(a), and is epressed as: ( ) ( ) U = m gy m gy = m g cosθ m g cosθ + cosθ = m + m g cosθ m g cosθ (W3.56) where the minus sign indicates the positions of the two bobs are aways underneath the datum ine. The system Lagrangian can now be assembed as: 1 L= T U= ( m1+ m) 1θ1 m 1 cos( θ θ1) θθ 1 + m θ m + m g cosθ + m g cosθ ( ) (W3.57) The system is conservative one and therefore Eq. (W3.4) yieds the Lagrange s equations to appy here. For the first generaized coordinate, θ 1, the foowing are the terms obtained after performing the necessary partia derivatives and time derivatives: d L = + L = m1 θθ 1 sin ( θ θ1) ( m1+ m) g1sinθ1 θ1 ( m m ) θ m θ cos( θ θ ) θ ( θ θ ) sin ( θ θ ) dt θ 1 (W3.58) As can easiy be checked out, the differentia Eqs. (W3.58) are non-inear in their variabes, as in addition to terms simpy containing θ 1 and θ and their first-order and second-order derivatives, others factors invoving sine and cosine of the anges, as we first-order derivatives to the power of do enter the above equations. Equations (W3.54) are inearized by assuming sma anges during vibration, namey: sinθ θ (W3.59) cosθ 1 and by negecting the second powers of the time derivatives in the two Eqs. (W3.58). As a consequence, the Lagrange equation Eq. (W3.4) corresponding to the coordinate θ 1 is ( ) ( ) m m m m m g (W3.60) 1+ 1θ1 θ + 1+ θ1 = 0 Website Chapter 3 18

19 Simiar cacuations are performed corresponding to the second coordinate θ ; after inearization, Eq. (W3.38) yieds: + + g = (W3.61) 1θ1 θ θ 0 Equations (W3.60) and (W3.61) form the mathematica mode of the doube penduum of Fig. W3.15. In matri form, the mathematica mode can be written as: ( m + m ) m θ ( m m ) g 0 θ = θ 0 g θ 0 1 (W3.6) W3.4 Static and Dynamic Coordinate Couping The notion of couping is now addressed with respect to the inertia and stiffness matrices of a muti-dof system whose free response is anayzed. Let us consider a two- DOF system whose dynamic equations describing its free response are: mq kq 11 1 = 0 (W3.63) mq + kq = 0 Each of the two Eqs. (W3.63) can be soved independenty as each coordinate appears soey in one differentia equation. This specific format describes a system whose coordinates are uncouped (conditions aso referred to as decouped). When the vectormatri notation is used, Eq. (W3.63) is reformuated as: [ ]{ } + [ ]{ } = { 0} where the mass and stiffness matrices are of diagona form: M q K q (W3.64) m11 0 k11 0 = = 0 m 0 k [ M] ; [ K] (W3.65) In situations where ony the inertia matri is diagona, the condition of static (or eastic) couping occurs, whereas for modes where ony the stiffness matri is diagona, the condition is known as dynamic (or inertia) couping. In severa instances, the mass and stiffness matrices are of the genera form: m m k k = 11 1 = 11 1 m m k k [ M] ; [ K] 1 1 (W3.66) Website Chapter 3 19

20 which indicates that neither of the two matrices are diagona and therefore, the corresponding system is both staticay and dynamicay (or fuy) couped. The definitions of static and dynamic couping that have been introduced for two-dof systems are obviousy vaid for systems possessing more than two DOF as we. For couped systems there are a few uncouping procedures incuding the coordinate transformation (mainy through matri rotation) and the proper choice of generaized coordinates (DOF) these methods wi briefy be anayzed net. The method of natura coordinates is frequenty appied in mechanica vibrations to decoupe systems by empoying more advanced notions and therefore, is not discussed here. W3.4.1 Coordinate Transformation Let us consider, for the sake of simpicity, a two-dof umped-parameter conservative mechanica system that is defined by the foowing mass and stiffness matrices: m m k k = = [ M] ; [ K] 1 1 m m3 k k3 (W3.67) and whose variabes that render it fuy couped are coected in a two-dimensiona vector t { } = { } q q q. The mass and stiffness matrices of Eq. (W3.66) are symmetric, a 1 condition which is normay obtained for couped conservative mechanica systems. We want to determine the conditions that are necessary to find another vector t { } = { } as: y y y that fuy decoupe the system. Let us assume the two vectors are reated 1 where [A] is a square matri of the form: { q} = [ A]{ y } (W3.68) [ A] a 1 = a 3 a4 and whose coefficients need to be determined. Substitution of Eq. (W3.68) into Eq. (W3.64) resuts in: [ ][ ]{ } + [ ][ ]{ } = { 0} a (W3.69) M A y K A y (W3.70) Website Chapter 3 0

21 with: ma 1 1+ ma 3 ma 1 + ma 4 [ M][ A] = ma 1 ma 3 3 ma ma (W3.71) ka 1 1+ ka 3 ka 1 + ka 4 [ K][ A] = ka 1 ka 3 3 ka ka In order for the Eq. (W3.70) to define an uncouped system, it is necessary that the two matrices of Eq. (W3.71) be diagona, namey: ma 1+ ma 3 3= 0 ma 1 + ma 4 = 0 (W3.7) ka 1 + ka 4 = 0 ka 1+ ka 3 3= 0 Equations (W3.7) yied the foowing conditions: m k a = = m k a m k a = = m k a (W3.73) In other words, in order to decoupe a two-dof system through coordinate rotation, it is necessary that the generaized mass and stiffness coefficients in a row be proportiona. Two coefficients of the rotation matri [A] can be seected arbitrariy but the other two coefficients have to be determined with satisfaction of Eq. (W3.73). W3.4. Choice of coordinates The choice of coordinates determines the nature of couping for a given mechanica system. It is possibe to render various couping conditions through specific coordinate seection, as shown in the net eampe. Eampe W3.4 The two-dof mechanica system shown in Fig. W3.16 consists of two point masses m 1 and m that are connected by a rigid, massess rod of ength, and two inear springs of stiffnesses k 1 and k. Anayze the system couping corresponding to the foowing coordinate seection: (a) The coordinates are the point masses vertica dispacements, 1 and. (b) The coordinates are the dispacement of the center of gravity (CG) and the tit ange θ of the rod. Website Chapter 3 1

22 1 1 θ m m 1 CG k 1 k Figure W3.16 Two-DOF mechanica system for coordinate couping anaysis Soution: The position of the center of gravity CG can be found by using the conditions: which resuts in: (a) m = m 1 + = = = 11 m 1 m1+ m m 1 m1+ m (W3.74) (W3.75) Let us use Lagrange s equations method to derive the mathematica mode of this mechanica system when the generaized coordinates are 1 and. The kinetic energy is: 1 1 T= m 1 1+ m (W3.76) and the potentia energy when ony considering the eastic spring contributions (gravity effects are negected) is: 1 1 U= k 1 1+ k (W3.77) The terms of Lagrange s eft-hand side equation corresponding to coordinate 1 are: Website Chapter 3

23 d T = m 1 1 dt U = k and, simiary, those connected to coordinate are: (W3.78) d T = m dt U = k (W3.79) such that the two Lagrange s equations (which form the mathematica mode of the mechanica system) are: m 1 1+ k 1 1= 0 (W3.80) m + k = 0 The two differentia Eqs. (W3.80) are independent and therefore, the system is fuy uncouped. (b) Let us now use the energy method to derive the mathematica mode of the same mechanica system. When the coordinates of motion are (the CG vertica dispacement) and θ (the rod s tit ange), it is necessary to epress the coordinates 1 and in terms of and θ. For sma motions, tanθ is approimated to θ, and therefore, 1 and are found from Fig. W3.16 as: 1 = + 1θ (W3.81) = θ Equations (W3.81) are substituted into Eqs. (W3.76) and (W3.77), which give the foowing epressions for the kinetic and potentia energies: 1 1 T = m1( + 1θ) + m( θ) (W3.8) and 1 1 U = k1( + 1θ) + k( θ ) (W3.83) The system being conservative, the time derivative of the tota energy (the sum of kinetic and potentia energies) is zero, which eads to the equation: Website Chapter 3 3

24 ( + ) + ( ) θ + ( + ) + ( ) m1 m m 11 m k1 k k 11 k θ + θ ( m m ) + ( m + m ) θ + ( k k ) + ( k + k ) θ = (W3.84) Because the veocities pertaining to the coordinates and θ cannot be zero at a times, the ony aternative for the eft-hand side of Eq. (W3.84) to be zero is when: ( m m ) ( m m ) θ ( k k ) ( k k ) θ = 0 ( m 11 m ) + ( m 11 + m ) θ + ( k 11 k) + ( k 11 + k) θ = 0 (W3.85) Equations (W3.85) are the mathematica mode of the mechanica system of Fig. W3.16 when the coordinates of motion are and θ. When epressing Eqs. (W3.85) in vector-matri form, the corresponding mass and stiffness matrices are: m + m m m k + k k k = = m 11 m m 11 + m k 11 k k 11 + k [ M] ; [ K] (W3.86) It can be seen that neither of the two matrices in Eqs. (W3.86) are diagona and therefore, the mechanica system is both staticay and dynamicay (so fuy) couped for the particuar choice of coordinates. The system is dynamicay decouped when m 1 1 = m as the mass matri of Eqs. (W3.86) becomes diagona. Simiary, the system is staticay decouped for k 1 1 = k because the stiffness matri of Eq. (W3.86) becomes diagona. Website Chapter 3 4