CHAPTER 8. TRANSVERSE VIBRATIONS-IV: MULTI-DOFs ROTOR SYSTEMS

Transcription

1 CHAPTER 8 TRANSVERSE VIBRATIONS-IV: MULTI-DOFs ROTOR SYSTEMS Transverse vibrations have been considered previousl in great detail for mainl single mass rotor sstems. The thin disc and long rigid rotors were considered with various complexities at supports; for example, the rigid disc mounted on flexible mass less shaft with rigid bearings (e.g., the simpl supported, overhung, etc.), the flexible bearings (anisotropic and cross-coupled stiffness and damping properties), and flexible foundations. Higher order effects of the groscopic moment on the rotor, for most general case of motion, was also described in detail. However, in the actual case, as we have seen in previous two chapters for torsional vibrations, the rotor sstem can have several masses (e.g., turbine blades, propellers, flwheels, gears, etc.) or distributed mass and stiffness properties, and multiple supports, and other such components like coupling, seals, etc. While dealing with torsional vibrations, we did consider multi-dof rotor sstems. Mainl three methods were dealt for multi-dof sstems, that is, the Newton s second law of motion (or the D Alembert principle), the transfer matrix method, and the finite element method. We will be extending the idea of these methods from torsional vibrations to transverse vibrations along with some additional methods, which are suitable for the analsis of multi-dof rotor sstem transverse vibrations. In the present chapter, we will consider the analsis of multi-dof rotor sstems b the influence coefficient method, transfer matrix method, and Dunkerle s method. The main focus of these methods would be to estimate the rotor sstem natural frequencies, mode shapes, and forced responses. The relative merit and demerit of these methods are discussed. The continuous sstem and finite element transverse vibration analsis of multi-dof rotor sstems will be treated in subsequent chapters. Conversional methods of vibrations like the modal analsis, Raleigh-Ritz method, weighted sum approach, collocation method, mechanical impedance (or receptance) method, dnamic stiffness method, etc. are not covered exclusivel in the present text book, since it is readil available elsewhere (Meirovitch, 986; Thomson and Dahleh, 998). However, the basic concepts of these we will be using it directl whenever it will be required with proper references. 8. Influence Coefficient Method In transverse vibrations due to coupling of the linear and angular (tilting) displacements the analsis becomes more complex as compared to torsional vibrations. A force in a shaft can produce the linear as well as angular displacements; similarl a moment can produce the angular as well as linear displacements. Influence coefficients could be used to relate these parameters (the force, the moment, and the linear and angular displacements) relativel easil. In the present section, the influence

2 4 coefficient method is used to calculate natural frequencies and forced responses of rotating machines. Up to three-dof rotor sstems the hand calculation is feasible, however, for more than three-dof the help of computer routines (e.g., MATLAB, etc.) are necessar. The method is described for multi- DOF i.e., n number of discs mounted on a flexible shaft (Figure 8.) and supported b rigid bearings; which can be extended for the multi-dof rotor sstem with flexible supports. Figure 8. A multi-dof rotor sstem mounted on rigid bearings 8.. The static case: Let f, f,, and f n are static forces on discs,,, and n respectivel, and x, x,, x n are the corresponding shaft deflections at discs. The reference for the measurement of the shaft displacement is from the static equilibrium and the sstem under consideration is linear, so that gravit effect will not appear in equilibrium equations. If a force f is applied to the disc of mass m, then deflection of m will be proportional to f, i.e. x f or x = α f (8.) where α is a constant, which depends upon the elastic properties of the shaft and support conditions. It should be noted that we will have deflections at other disc locations (i.e.,, 3,, n) as well due to force at disc due to the coupling. Now the same force f is applied to the disc of mass m instead of mass m, then the deflection of m will still be proportional to the force, i.e. x f or = f (8.) x α where, α is another constant (the first subscript represents displacement position and the second subscript represent the force location). In general we have α α. Similarl, if force f is applied to the disc of mass m n, then the deflection at m will be x = α f (8.3) n

3 4 where, α n is a constant. If forces f, f,..., and f n are applied at the locations of all the masses simultaneousl, then the total deflection at m, will be summation of all the three displacements obtained above b the use of superposition theorem, as x = α f + α f + + α f (8.4) n n In the equation, it has been assumed that displacements are small so that a linear relation exists between the force applied and corresponding displacement produced. Similarl, we can write displacement at other disc locations as and x = α f + α f + + α f (8.5) n n x = α f + α f + + α f (8.6) n n n nn n Here α j,, α nj, where j =,,, n are another sets of constants and can be defined as described for α i above. Hence, in general α ij is defined as a displacement at i th station due to a unit external force at station j th and keeping all other external forces to zero. Equations (8.4) to (8.6) can be combined in a matrix form as x α α αn f x α α α n f = x n α n α n α nn f n (8.7) It should be noted that due to the transverse force actuall both the linear and angular displacements take place, i.e., a coupling exists between the linear and angular displacements. We have alread seen such coupling in Chapter due to bending of the shaft and due to groscopic couples, respectivel. Moreover, we have seen in Chapters 4 and 5 couplings between the horizontal and vertical plane linear motions (x and ) due to dnamic properties of fluid-film bearings and between the horizontal and vertical plane angular motions ( x and ), respectivel. Similarl, a moment gives the angular displacement as well as the linear displacement. The method can be extended to account for the angular displacement (tilting),, of the disc, and for the application of the point moment, M, at various disc locations along the shaft a part of loading on discs. Then, the equation will take the following form { d} = [ α ]{ f } (8.8)

4 43 with { d} x x n = n ; { f } f f n = M M n and α α α α α α α α α ( n) ( n) ( n) ( n) ( n)( n) (8.9) which gives { f } = [ α] { d} (8.) where α ij, with i, j =,,,n are influence coefficients. The first subscript defines the linear (or angular) displacement location and the second subscript defines the force (or moment) location. The analsis so far has referred onl to static loads applied to the shaft. When the displacement of the disc is changing rapidl with time, the applied force has to overcome the disc inertia as well as to deform the shaft. 8.. The dnamic case: In Figure 8. free bod diagrams of a disc and the shaft is shown. Let and M (not shown in free bod diagrams for brevit) be the external force and moment on the disc m f whereas and M are the reaction force and moment transmitted to the shaft (which is equal and opposite to the reaction force and moment of the shaft on the disc). From the force and moment balance of disc, we have f f f = m x and M M = I (8.) d where I d is the diametral mass moment of inertia the disc. Figure 8. Free bod diagrams of (a) a disc and (b) the shaft

5 44 Similarl at other disc locations, we can write f f = mx and M M = I (8.) d and f n fn = m nxn and M M = I (8.3) n n dn n Substituting for f, f,, fn, M, M, and M n from equations (8.)-(8.3) and remembering that for the simple harmonic motion of discs x = ω x and = ω (when no external excitation is present then ω is the natural frequenc of the sstem ω nf, and when there is an external excitation then it is equal to the excitation frequenc, ω), equation (8.8) gives f + ω m x f n + ω mn x n = M ω Id + M n + ω Id n { d} [ α ] (8.4) which can be expanded as m x mn x n = + nf Id Id n n { d} [ α ]{ f } ω [ α ] with { f } f f n = M M n (8.5) In view of equation (8.9), equation (8.5) can be rearranged as α m α m α I α I α m α m α I α I n n ( n+ ) d ( n) dn n nn n n( n+ ) d n( n) dn { d} = [ ]{ f } + { d} α ωnf α ( n+ )m α ( n+ ) nmn α ( n+ )( n+ ) Id α ( n+ )( n) Idn α m α m α I α I ( n) ( n) n n ( n)( n+ ) d ( n)( n) dn (8.6)

6 45 which can be written in more compact form as with ω ω [ A] [ I ] { d} = [ α ]{ f } αm α nmn α( n+ ) Id α ( n) Id n α m α m α I α I = α( n+ )m α( n+ ) nmn α( n+ )( n+ ) Id α ( n+ )( n) Idn α( n)m α( n) nmn α( n)( n+ ) Id α ( n)( n) I dn n nn n n( n+ ) d n( n) dn [ A] { d} (8.7) (8.8) Disc displacements x and can be calculated for known applied loads (e.g., the unbalance forces and moments) as with { d} = [ R]{ f } ω ω [ R] = [ A] [ I ] [ α ] (8.9) (8.) where R represents the receptance matrix and for the present case it contains onl real elements. For free vibrations the right hand side of equation (8.7) will be zero, i.e. ω [ A] [ I ] { d} = { } (8.) which onl satisf when = (8.) ω [ A] [ I ] { } and it will give the frequenc equation and sstem natural frequencies could be calculated from this. Alternativel, through eigen value analsis of matrix [A] sstem natural frequencies and mode shapes could be obtained directl. In general, the receptance matrix, [R], ma contain complex elements when damping forces also act upon the shaft, in which case applied forces and disc displacements will not all be in phase with one another. Hence, a more general form of equation (8.9) would be that

7 46 which indicates both the real and imaginar parts of x, f and R. In such case the real and imaginar parts of each equations need to be separated and then these can be assembled again to into a matrix form, which will have double the size that with complex quantities. Some of the steps are described below { dr} + j{ di} = ([ Rr ] + j[ Ri ])({ f r} + j{ f i} ) where r and i refer to the real and imaginar parts, respectivel. Above equation can be expanded as { dr} + j{ di} = ([ Rr ]{ f r} [ Ri ]{ f i} ) + j( [ Ri ]{ f r} + [ Rr ]{ f i} ) Now on equating the real and imaginar parts on both sides of equations, we get { d } = [ R ]{ f } [ R ]{ f } and { d } = [ R ]{ f } + [ R ]{ f } r r r i i i i r r i Above equations can be combined again as [ ] [ ] [ ] [ ] dr Rr Ri f r = di Ri Rr f i (8.3) It can be observed that now the size of the matrix and vectors are double as that of equation (8.9). Now through simple numerical examples (for the two or more DOFs) some of the basic concepts of the present method will be illustrated.

8 47 Example 8. Obtain transverse natural frequencies of a rotor sstem as shown in Figure 8.3. Take the mass of the disc, m = kg and the diametral mass moment of inertia, I d =. kg-m. The disc is placed at.5 m from the right support. The shaft has a diameter of mm and a span length of m. The shaft is assumed to be massless. Take the Young s modulus E =. N/m of the shaft. Consider a single plane motion onl. Figure 8.3 A rotor sstem Solution: Influence coefficients for a simpl supported shaft are defined as with α f α = α M α x z and 3 a b 3 ; a l a α = α al = 3EIl 3EIl ab ( b a ) 3 3 ; al a α = α l = 3EIl 3EIl It should be noted that subscript represents corresponding to a force or a linear displacement, and subscript represents a moment or an angular displacement. To obtain natural frequencies of the rotor sstem having a single disc, from equation (8.), we have αm αid ω nf [ A] [ I ] = ω nf αm αid ω nf (a) Hence, the determinant of the above matrix would give the frequenc equation as ( ) ( ) 4 midωnf αα α ωnf αm + α Id + = (b)

9 48 For the present problem, we have α = m/n; α = α = 3.3 m/n; α =.4 m/n Equation (a) becomes, ω 8.55 ω = nf nf which gives two natural frequencies of the sstem, as ω = 9.4 nf rad/sec and ω = 9 rad/sec nf It should be noted that the linear and angular motions are coupled for the present transverse vibrations since the disc is offset from its mid span; however, when the disc is at the mid span then the linear and angular motions will be decoupled (i.e., α = α = ). Natural frequencies for such case would be ω = = =.44 nf 4 mα. rad/s for the pure translation motion of the disc, and ω = = = rad/s α..884 nf 3 I d for the pure rotational motion of the disc. These expressions also of course can be obtained from frequenc equation (a) for α = α =. Example 8. Find the transverse natural frequenc of a stepped shaft rotor sstem as shown in Figure 8.4. Consider the shaft as massless and is made of steel with the Young s modulus, E =. () N/m. The disc could be considered as a point mass of kg. The circular shaft is simpl supported at ends (In Figure 8.4 all dimensions are in cm). Figure 8.4 A simpl supported stepped shaft

10 49 Solution: To simpl the analsis let us consider that the angular displacement of the disc is negligibl small. From equation (8.3), we have [ A] [ I ] = αm ω nf ωnf from which the natural frequenc is given as ω nf = mα Hence, now the next step would be to obtain the influence coefficient, this influence coefficient is obtained as follows α. Using the energ method Figure 8.5 A free bod diagram of the rotor sstem For a load F when it acts at the disc, reaction forces at bearings can be obtained as (Figure 8.5) and + M = F F.6 = F =.6 F A B B + F = F + F = F F =.4F A B A Figure 8.6 Free bod diagram of the shaft section z.6 From Figure 8.6, the bending moment in the shaft can be obtained as M C = M x.4 Fz = M.4 Fz x = (a)

11 43 Figure 8.7 Free bod diagram of the shaft section.6 z. From Figure 8.7, the bending moment in the shaft can be obtained as M D = M x + F ( z.6).4 Fz = M.6 F ( z ) x = (b) The strain energ stored in the shaft from bending moments can be obtained as U.6 M x dz. M x dz = + EI EI.6 From the Castigliano s theorem the linear displacement can be obtained as δ M U = F dz + F EI M x x M.6 x M. x = F EI.6 dz (c) On substituting equations (a) and (b) into equation (c), we get.6 Fz z. {.6 F( z) }{.6( z) } (.4 )(.4 ) δ = dz + dz EI EI.6.6.6Fz..36 F( z z + ).5F.768F = dz + dz = EI + EI EI EI.6 The stiffness of the beam given as F k = = = + α δ EI EI From above in fact it can be observed that two shaft segment can be thought as connected parallel to each other at disc location.

12 43 With π π E = N/m ; I =. = 4.97 m ; I =.3 = m , we have k = = α m/n Hence the natural frequenc is given as ωn = = = mα rad/sec The influence coefficient can be also obtained b the singular function approach and for more details readers are referred to Timoshenko and Young (968). Example 8.3 Find transverse natural frequencies and mode shapes of a rotor sstem shown in Figure 8.8. B is a fixed end, and D and D are rigid discs. The shaft is made of the steel with the Young s modulus E =. N/m and a uniform diameter d = mm. Shaft lengths are: BD = 5 mm, and D D = 75 mm. The mass of discs are: m = kg and m = 5 kg. Consider the shaft as massless and neglect the diametral mass moment of inertia of discs. Figure 8.8 Solution: For simplicit of the analsis, the shaft is considered as massless and disc masses are considered as point masses (i.e., the diametral mass moment of inertia of discs are neglected). The first step would be to obtain the influence coefficients corresponding to two disc locations acted b concentrated forces. Basicall, we need to derive linear displacements at two disc locations due to forces F and F acting at these locations as shown in Fig These deflection relations are often available in a tabular form in standard handbooks (e.g., Young and Budnas, ). However, for the present problem the calculation of influence coefficients is explained b the energ method and b the singularit function to clarif the basic concept for the completeness.

13 43 Figure 8.9 A shaft with two concentrated forces Fig. 8. The free-bod diagram of a shaft segment for z a Energ method: In this method, we need to obtain the strain energ due to bending moments in the shaft. Bending moments at different segments of the shaft can be obtained as (i) Shaft segment for z a From Fig. 8., the bending moment can be written as M = F z (a) z (ii) Shaft segment for a z L From Fig. 8., the bending moment can be written as M z = F ( ) z F z a (b) Fig. 8. The free-bod diagram of a shaft segment for a z L The strain energ is given b

14 433 a M z dz L M z dz U = + EI a EI (c) Using the Castigliano theorem, the linear deflection at station can be written as M M M dz M dz = = + z z z z U a F L F F a EI EI (d) On substituting equations (a) and (b) into equation (d), we get ( )( ) L { ( )}( ) a U a F z z dz F z F z a z dz = = + F EI EI 3 3 ( L a ) a( L a ) a L L F z F z F z az F F = + + = ( a + L a ) + EI 3 EI 3 EI 3 3EI EI 3 a a which finall takes the form 3 3 ( L a ) a( L a ) = F + F 3EI 3EI EI 3 L (e) Equation (e) has the following form = α F + α F (f) f f Hence, for a =.75 m, L =.5 m, EI =3. N-m, we have α f = m/n and α f =.3-6 m/n (g) The deflection at station can be obtained as M M M dz M dz = = + z z z z U a F L F F a EI EI (h) On substituting equations (a) and (b) into equation (h), we get

15 434 ( )( ) L { ( )}( ) + a U a F z dz F z F z a z a dz = = + F EI EI L 3 3 z az F z az F = a z EI 3 EI 3 a L a which finall takes the form ( L a ) a( L a ) ( L a ) a( L a ) a ( L a) = F + + F 3EI EI 3EI EI EI (i) Equation (i) has the following form = α F + α F (j) f f Hence, for a =.75 m, L =.5 m, EI =3. N-m, we have α f = m/n and α f =.9-7 m/n (k) Method of the singularit function: Now the influence coefficients are obtain b the method of singularit function for illustration. The singularit function (< >) is defined as < f > = f(z) if f(z) > The bending moment can be written as = if f (z) < (l) EI = F < z > + F.75 < z > (m) On integrating twice equation (m), we get following expressions and EI = F < z > + F < z.75 > + c (n) 3 < z > F = (o) EI F < z.75 > c z c

16 435 where the integration constants c and c are obtained b the boundar conditions, and are give as (i) = from equation (n), we have.5 z = which gives <.5 > <.5 > F + F + c = c =.78F.3F (p) (ii) z = = from equation (o), we have.5 which gives 3 3 <.5 > <.5 > F + F F F + c = 6 6 c = 6.5 F F (q) 4 4 Finall equation (o) becomes 3 < z > F EI = F.75 (.78.3 ) < z > + F F z F + + F 6 6 (r) On evaluating the deflection at station for z = (i.e., at the free end), we have = 6.5 z F F = EI 4 4 { } which has the following form = α F + α F f f with α f = m/n and α f =.34-6 m/n The deflection at station for z =.75 m, we get 3.75 = F.75 ( z ) F = EI 6 which has the following form 4 4 (s)

17 436 = α F + α F f f with α f =.34-6 m/n and α f =.44-6 m/n which is same as obtained b the energ method. For free vibrations from equation (8.7), we have α f m α f m ω nf = α f m α f m ω nf (t) For a non-trial solution, it gives α m m f f ωnf α m α = α m (u) f f ωnf which give the frequenc equation as ( ) ( ) mm ω α α α ω α m + α m + = (v) 4 nf f f f nf f f For the present problem, we have α f = m/n, m = kg and m = 5 kg α = α =.34-6 m/n and f f α f =.44-6 m/n Equation (u) becomes, ω.77 ω +.9 = 4 6 nf nf which gives two natural frequencies of the sstem for the pure translator motion, as ω = rad/sec and nf ω = 34. rad/sec nf

18 437 B D D () Relative linear displacements () Shaft length (mm) Figure 8. Mode shapes; () - first mode shape, () - second mode shape The mode shapes corresponding to these natural frequencies are plotted in Fig. 8.. While plotting mode shapes the maximum displacement corresponding to a particular mode has been chosen as unit and other displacements have been normalized accordingl. It should be noted that for the case when discs have appreciable diametral mass moment of inertia, then along with the forces at stations and, the moments also need to be considered while deriving influence coefficients. In that case, we will have sixteen influence coefficients, i.e., the size of the influence coefficient matrix would be 4 4, however, the smmetr conditions of influence coefficients will still prevail. Correspondingl, we would have four natural frequencies and corresponding mode shapes. Example 8.4 Find transverse natural frequencies and mode shapes of the rotor sstem shown in Figure 8.3. B is a fixed bearing, which provide fixed support end condition; and D, D, D 3 and D 4 are rigid discs. The shaft is made of the steel with the Young s modulus E =. () N/m and the uniform diameter d = mm. Various shaft lengths are as follows: D D = 5 mm, D D 3 = 5 mm, D 3 D 4 = 5 mm and D 4 B = 5 mm. The mass of discs are: m = 4 kg, m = 5 kg, m 3 = 6 kg and m 4 = 7 kg. Consider the shaft as massless. Consider the disc as point masses, i.e., neglect the diametral and polar mass moment of inertia of all discs.

19 438 Figure 8.3 A multi-disc overhung rotor Solution: The first step would be to obtain influence coefficients. We have the linear deflection,, and force relations from the strength of material for a cantilever shaft with a concentrated force at an point (Fig. 8.4) as f z = (3 a x) for < x < a and 6EI f a = (3 x a) for a < x < L 6EI Figure 8.4 A cantilever shaft with a concentrated force at an point Let us take stations,, 3, and 4 at each of the disc locations. Figure 8.5 shows a cantilever shaft with forces at stations and. Station is at free end and it has an intermediate station between station and the fixed support. Lengths L, a, and b are shown in Fig. 8.5 and for this case we have L = a + b. Between two stations, it will have the following influence coefficients (which relates the force to linear transverse displacement) f 3 α = = ; L 3EI f 3 α = = ; ( ) α α 3L a a 3EI a = = = = (a) f f 6EI similar relations would also be valid between stations (, 3) and (, 4). Figure 8.5 A cantilever shaft with a force at the free end and another at the intermediate point

20 439 Figure 8.5 shows a cantilever shaft with a force at station, and has a station 3 between the force and the fixed end. Between two stations, it will have the following influence coefficients ( a + b) 3 α = ; 3EI 33 3 a 3EI a 3 = 3 = + 3 (b) 6EI α = ; α α ( a b) These relations could also be used between stations (, 4) and (3, 4). Figure 8.6 A cantilever shaft with two forces intermediate stations S.N. Table 8. A calculation procedure of influence coefficients in a multi-discs cantilever shaft Disc location* L i (m) a ij (m) b ij (m) α ii (m/n) α ij = α ji (m/n) α jj (m/n) (, ) (, ) (3, 3) (4, 4) (, ) (, 3) (, 4) e-7 8 (, 3) (, 4) (3, 4) * Numbers in bracket represent the force station numbers It should be noted that in Fig. 8.6 the shaft segment from the force f to the free end would act as rigid shaft. It will not contribute in the deformation of the shaft at station, it will act as if it is not present at all (this is true onl for mass-less shaft assumption).

21 44 When disc and is present, we have L i = L =.3 m, a ij = a =.5 m. Hence the influence coefficients takes the following form: α = m/n, α = α = m/n, α = m/n. On the similar lines other influence coefficients can be calculated and it is summarized in Table 8. Now for free vibrations, the governing equation has the following form α f m α f m α f m 3 3 α f m 4 4 α f m α f m 3 3 α f m 4 4 α 3 f m 3 3 α 3 f m = 4 4 ω nf 3 sm α 4 4 f m 4 4 (c) which can be written as for the data of the present problem, as ω [ A] [ I ] { } = { } nf (d) with α f m α f m α f m 3 3 α f m α f m α f m 3 3 α f m [ A] = = α 3 f m 3 3 α 3 f m sm α f m 4 4 The eigen value and eigen vector of the above matrix is obtained as and / ω nf.576 / ω 4. nf [/ ωnf ] = = ; / ω. nf3 / ω. nf 4 [ Y ] Y Y Y Y Y Y Y Y = = Y3 Y3 Y3 Y 3 Y Y Y Y 4 ω nf ω nf ω nf 3 ωnf Hence, the transverse natural frequencies are ω nf = 4.36 rad/s; ω nf = 99.9 rad/s; ω nf = 36.8 rad/s, ω nf 4 = 77.7 rad/s. The corresponding eigen vectors (mode shapes) can be plotted.

22 44 8. Transfer Matrix Method In the influence coefficient method, governing equations are derived b considering the equilibrium of the whole sstem. Difficulties with such an approach are that for a large sstem it becomes ver complex and we need to evaluate influence coefficients for each case. While in the transfer matrix method (TMM), which is also called the Mklestand & Prohl Method, the shaft is divided into a number of imaginar smaller beam elements and governing equations are derived for each of these elements in order to determine the overall sstem behavior. The TMM has an advantage over the method of influence coefficient in that the size of matrices being handled does not increase with the number of DOFs. It demands less computer memor and associated ill-conditioning problems (i.e., nearl singular matrices) are less. Also this method is relativel simple and straightforward in application. For the present case discs are considered as point masses. In the case when the mass of the shaft is appreciable then it is divided up into a number of smaller masses concentrated (or lumped) at junctions (or stations) of beam elements so that concentrated masses and the shaft ma be modeled as shown in Figure 8.7. The station number can be assigned wherever there is change in the state vector (i.e., the linear and angular displacements, shear forces, and bending moments) for example at step change in shaft diameter, at discs, coupling, and support positions, etc. Station numbers are given from to (n+). Figure 8.7 Modeling a real rotor with discrete elements Figure 8.8 A free bod diagram of an i th shaft segment Figure 8.9 A cantilever beam 8.. A field matrix: Figure 8.8 shows the free bod diagram of the i th shaft segment, which is between stations (i - ) th and i th. Let S be the shear force in -direction, M z is the bending moment in

23 44 -z plane, is the linear displacement in -direction, and is the angular displacement in -z plane. The back-subscript R refers to the right of a particular mass and L refers to the left of a particular mass. Directions for the force, the moment, and displacements are chosen according to the positive sign conversion of the strength of material (Timoshenko and Young, 968). x The displacement (linear or angular) at i th station will be equal to the sum of the relative displacement between i th and (i - ) th stations and the absolute displacement of (i - ) th station. As a first step for obtaining the relative displacement, (i - ) th station can be considered to be fixed end as shown in Figure 8.9. Hence, the displacement and the slope at the free end are related to the applied moment and the shear force at free end b considering the beam as though it were a cantilever (Timoshenko and Young, 968). Then as a second step, the displacement and the slope of the fixed end is considered b considering the beam as a rigid. On assumption of small displacements, hence, finall above two steps could be superimposed to get the total displacement and slope of the i th shaft segment at the left of i th station (i.e. i th mass), as and 3 L M z l i L S l i = l + (8.4) EI 3EI L i R i Rxi L M z l i L S l i = + (8.5) EI EI Lxi Rxi where l is the length of shaft segment, and EI is the flexural rigidit of the shaft segment. It should be noted that the first two terms in right hand side of equation (8.4) and first term in the right hand side of equation (8.5) are related to the second step in which shaft is considered as rigid. Last two terms on the right hand side of equations (8.4) and (8.5) are from the first step in which the shaft is considered to be cantilevered. From the free bod diagram (Figure 8.8) the shear force and the bending moment at either ends of the i th shaft segment are related as and L S i R S i = (8.6) M = M + S l (8.7) L z i R z i L i On substituting for L S and M from equations (8.6) and (8.7) into equations (8.4) and (8.5), i i L z these equations could be rewritten as

24 443 L i R i R xi ( + ) M S l l S l = l + EI 3EI L xi R xi 3 R z i R i R i ( + ) M S l l S l = + EI EI R z i R i R i (8.8) (8.9) and L S = i R S i M = M + S l L z i R z i R i (8.3) (8.3) These equations can be rearranged and expressed in a matrix form as with { S} = [ F] { S} L i i R i 3 l l l EI 6EI l l = = = l S x { } ; [ ] L S i F i EI EI ; R{ S} i M z x M z S L i R i i (8.3) (8.33) where [ F ] is the field matrix for the i th shaft segment, and is the state vector at i th i { S} i mass. The filed matrix, [ F] i, transforms the state vector from the left of a shaft segment to the right of the shaft segment. Equation (8.3) is for the motion in the vertical plane. Similar set of equations ma be written for motion in the horizontal plane. In general, state vectors will have both real and imaginar components due to two plane motion (and/or due to the damping and groscopic moments, however, the damping is not considered in the present analsis and because of this field matrix remains real. The effect of groscopic moments will be considered subsequentl). Equation (8.3) is expanded to give a more general form as with { S } = [ F ] { S } L i i R i (8.34)

25 444 { S} h [ ] { S} r F h r { S} h [ ] { } j F S hj { S } = { S} ; [ F ] = [ F] ; { S } = { S} v r { S} [ F] v { S} j v j L i vr i R i i L i R i (8.35) where { S } is the modified state vector of the size 7 and [ ] is the modified field matrix of the size 7 7. Subscripts h and v refer to the horizontal and vertical directions respectivel; and r and j refer to real and imaginar parts, respectivel. The last equation of an identit has been added to facilitate inclusion of the unbalance in the analsis, as will be made clear later (it was used for the torsional vibrations also in Chapter 6). It should be noted that in the simplest case for the single plane motion and without damping (i.e., imaginar components will be zero) the size of the modified field F matrix * F i will be 5 5 and is given as 3 l l l EI 6EI l l x = = M z l = S S (8.36) x * EI EI * L{ S } i M z ; [ F] i ; R{ S } i L i R i i 8.. A point matrix: Figure 8. shows the free bod diagram of the point mass m i, where u i is the unbalance force at i th location. The relationship between forces and displacements at the concentrated mass is given b its equations of motion S S + u = m = m ω (8.37) R i L i i i i i i where ω is the spin speed of the shaft. It should be noted that a snchronous whirl is considered here.

26 445 Figure 8. Forces, moments and linear & angular displacements on a concentrated mass Similarl, for the moment and the angular displacement, we have the following relationship M M = I = ω I (8.38) R zi L zi di xi di xi where I d is the diametral mass moment of inertia of the disc. Moreover, since the linear and angular (slope) displacements on each side of mass are equal. Hence, we can rewrite all governing equations for a concentrated mass as = R i L i (8.39) = R xi L xi (8.4) and M = ω I + M R zi di L xi L zi S = m ω + S u R i i R i L i i (8.4) (8.4) Hence, we can combine these conditions including in a matrix form as { S} = [ P] { S} + { u} R i i L i i (8.43) with x x R{ S} i = ; [ P] i = ; { } ; { } L S i = u i = M z ω I M z d S mω S R i i L u i i (8.44)

27 446 where [ P] is the point matrix and is the unbalance vector of the i th i {u} i mass. The point matrix, [P] i, transforms the state vector from the left of a disc to the right of the disc. Equation (8.43) can be written for the horizontal plane also. In general since unbalance will also have both the real and imaginar components, equation (8.43) can be expanded to a more general form as with { S } = [ P ] { S } R i i L i { S} h [ P] { u} { } r h S r h r { S} h [ ] { } { } j P u hj S hj { S } = { } ; [ P ] S ν = [ P] { u} ; { S } { S} r ν r = ν r { S} ν [ P] { u} { S} j ν j ν j R i i L i R i i L i (8.45) (8.46) where [ P ] i is the modified point matrix and the size of the matrix is 7 7. B putting in the last row of the state vector, unbalance force components have been accommodated in the last column of the modified point matrix itself. It should be noted that in the simplest case for a single plane motion the size of the modified point matrix [ P ] i will be 5 5 and is given as (8.47) * [ P ] i = ω Id mω u i 8..3 An overall transfer matrix: Once we have the point and field matrices, we can use these to form the overall transfer matrix to relate the state vector at one extreme end station (sa left) to the other extreme end (right), provided we do not have intermediate supports that will be dealt subsequentl. Equations (8.34) and (8.45) can be combined as with { S } = [ P ] { S } = [ P ] [ F ] { S } = [ U ] { S } R i i L i i i R i i R i [ U ] = [ P ] [ F ] i i i (8.48)

28 447 where [ U ] i is the modified transfer matrix for i th segment, which transforms the state vector from the right of (i-) th station to the right of i th station. The transfer matrix for all (n+) stations (i.e., from th to n th as shown in Fig. 8.) in the sstem ma be obtained in the similar manner as equation (8.48) as follows R * * * { S } = { } U S R * * * * * * { S } = { } { } U S = U U S * * * * * * * { S } = U { S } = U U U { S } R R R R 3 3 R 3 R { } { } { } S = [ U ] S = [ U ] [ U ] [ U ] S = [ T ] { S } * * * * * * * * * R n n R n n n R R (8.49) where * [ T ] matrix is 7 7. is the overall modified transfer matrix for the complete rotor sstem and the size of the Now the transfer matrix method is illustrated for a ver simple case when there is no coupling between the vertical and horizontal planes (e.g., from groscopic effects or from bearings, then onl single plane motion can be considered) and no damping in the sstem (i.e., in a single plane displacements and forces are in phase with each other), then the overall transfer matrix [T] will take a size of 5 5. Equation (8.49) can be written in the expanded form as t, t, t,3 t,4 t,5 x t, t, t,3 t,4 t,5 x M z =t3, t3, t3,3 t3,4 t 3,5 M z S t4, t4, t4,3 t4,4 t 4,5 S n (8.5) To determine the sstem characteristics it is first necessar to define the sstem boundar conditions, which describe the sstem support conditions. Figure 8. A simpl supported multi-dof rotor sstem

29 448 For illustration of the TMM, let us consider a simpl supported multi-dof rotor sstem as shown in Figure 8., for which linear displacements and moments are zero at supports (i.e., M = M = ). These states are corresponding to st and 3 rd rows in the state vector {S} and above z z n equation take the following form = n = and t, t, t,3 t,4 t,5 x t, t, t,3 t,4 t,5 x =t3, t3, t3,3 t3,4 t 3,5 S t4, t4, t4,3 t4,4 t 4,5 S n (8.5) On considering onl st and 3 rd set of equations from the matrix equation (8.5), since the have zero on the left hand side vector, hence, we get t, t,4 x t,5 t t = S t 3, 3,4 3,5 (8.5) The remaining two equations, i.e., nd and 4 th set of equations from the matrix equation (8.5) can be written as x S n t, t,4 x t,5 = t t + S t 4, 4,4 4,5 (8.53) 8..4 Free vibrations: For free vibrations, the frequenc of vibration, ω, in equation (8.44) will be the natural frequenc, ω nf, of the sstem. For the present illustration of the simpl supported shaft, equation (8.5) becomes a homogeneous equation since right hand side terms (i.e., terms of t s with second subscript as 5) are related to unbalance forces, which are zero in free vibrations. Hence, we have t, t,4 x t t = S 3, 3,4 (8.54) For non-trivial solution the determinant of the above matrix will be zero, hence ( nf ),,,, f ω = t t t t = (8.55)

30 449 Equation (8.55) is a frequenc equation in the form of a polnomial and it contains natural frequencies as unknown. In case of simple sstems, the frequenc can be obtained in an explicit form. As b hand calculations it is feasible to obtain roots of a polnomial of onl third degree with the help of closed form solutions. However, for complex sstems an convenient root-searching techniques of the numerical analsis (e.g., the incremental method, the bisection method, the Newton-Raphson method, etc.) could be used. In such cases there is no need to multiple the field and point matrices in the smbolic form, whilst these matrices should be multiplied in the numerical form b choosing suitable guess value of the natural frequenc and condition of equation (8.55) ma be checked with certain acceptable limits after the final multiplications. Based on the residue of equation (8.55) or its derivatives, the next guess value of the natural frequenc ma be decided and the process ma be repeated till the residue is reduced to the desired level of accurac. Since equation (8.55) has, in general, several roots the procedure ma be repeated to obtain the remaining roots. Care should be exercised in obtaining all the roots (natural frequencies) without stepping over an of them. Torsional vibrations b using the TMM ma be referred for more details of the root searching algorithm. To obtain mode shapes, from equation (8.54) we have S t t = =, 3, x x t, 4 t3, 4 (8.56) Let us take a reference value of the angular displacement at station as =. On substituting the x second term (the third term can also be used) of equation (8.56) into equation (8.53) for free vibrations, we get x t, t,4 = S t4, t 4,4 t, / t,4 n (8.57) Now the state vector at n th station is known completel. B back substitution of the state vector at n th station into equations (8.49), we can get state vectors at other stations. It should be noted that t s are function of the natural frequenc, hence from the procedure described above we will get a mode shape corresponding to a particular natural frequenc. The procedure can be repeated for each of the sstem natural frequencies to get the corresponding mode shapes. In general, for a sstem of n number of degrees of freedom, it will have n number of natural frequencies; and that man number of mode shapes. Moreover, the mode shape for a particular natural frequenc has the unique relative

31 45 amplitudes. For other kinds of boundar conditions Table 8. provides frequenc equations and equations to obtain mode shapes. Table 8. Equations for the calculation of natural frequencies and mode shapes. S.N. Boundar conditions Station numbers Equations to get natural frequencies Equations to get mode shapes Simpl supported (Pinned-roller support) : Pinned end, n: roller support t, t,4 x t t = S 3, 3,4 x S R n t = t t,,4 t x S 4, 4,4 Cantilever (Fixed-free) : Fixed end, n: free end t t M z = t t S t M 3 t4 z = t t S R x n Fixed-fixed, n: Fixed ends 4 Free-free, n: Free ends 5 Fixed-pined : Fixed end, n: pinned end t,3 t,4 M z t t = S,3,4 t3, t3, t4, t = 4, z t,3 t,4 M z t t = S 3,3 3,4 M z S R n z n x S n t = t t = t t = t t 3,3 3,4 t M z S 4,3 4,4 t,, t,, z t,3,4 t M z S 4,3 4, Forced response: It should be noted that the extreme right hand side vector of equations (8.5) and (8.53) are unbalance forcing terms (i.e., terms of t s with second subscript as 5). Other terms of t s which contain ω is the spin speed of the shaft. Equation (8.5) can be used to obtain the state vector at station. Subsequentl, state vectors at the n th and other locations can be obtained b equations (8.53) and (8.49), respectivel. These state vectors contain both the angular and linear displacements, and the reaction force and the moment. Hence, unbalance responses and reaction forces are known, which can be obtained for various speeds. Equations (8.5) and (8.53) are tabulated in Table 8.3 along with similar equations for other standard boundar conditions for a read reference.

32 45 Table 8.3 Governing equations for forced vibration with different boundar conditions S.N. Boundar conditions Simpl supported (pinnedroller supports) Cantilever (fixed-free) Station numbers : pinned end, n: roller support : fixed end, n: free end 3 Fixed-fixed, n: fixed ends 4 Free-free, n: free ends 5 Fixedpined : fixed end, n: pinned end t, t,4 x t,5 t t = S t 3, 3,4 3,5 t t M z t3,5 = t t S t 4, t,3 t,4 M z t,5 t t = S t,3,4,5 t3, t3, t3,5 t t = t 4, 4, z 4,5 t,3 t,4 M z t,5 t t = S t 3,3 3,4 3,5 Equations to get state vectors x S R n t, t,4 x t,5 = t t + S t t t 4, 4,4 4,5 M t 3 4 z,5 = R x t S n 3 t + 4 t,5 M z S R n z n x S n t3,3 t3,4 M z t3,5 = t t + S t 4,3 4,4 4,5 t, t, t,5 = t t t,, z,5 t,3 t,4 M z t,5 = t t S t 4,3 4,4 4, Groscopic effects: If groscopic effects are allowed for, in equation (8.45) the modified point matrix, [P * ], will get some extra terms, however, the modified field matrix, [F * ], will not be affected. For free vibrations, the formulation would have more general form since equations will contain both the natural whirl frequenc, ν (while considering groscopic effects we tried to distinguish the whirl natural frequenc smbol b ν in place of ω nf ; while the former is speed-dependent ν ( ω ), the latter is speed independent so that ν ( ω = ) = ωnf ), and the spin speed, ω, of the shaft. This leads to speed dependenc of the natural whirl frequenc. Equilibrium equations for the moment in the -z vertical and z-x horizontal planes, considering groscopic moments also, are given as (noting equation (8.38) and for groscopic terms refer Chapter 5) and M M = I + I ω (8.58) R zi L zi di xi pi i M M = I I ω (8.59) R zxi L zxi di i pi xi where is the polar mass moment of inertia of the disc. In the modified matrix [ P ], following rows I p and columns element will be affected: (i) Equations (8.58) and (8.59) are equations of moments, and

33 45 in the state vector {S}, 3 rd, 7 th, th and 5 th rows are for moment equations. Hence, onl these rows in the modified matrix [ P ] will be affected. Now in equations (8.58) and (8.59), additional terms for slopes, and (see Figure 8.), are appearing because of groscopic effects. In the modified x state vector {S * } the angular displacements are at nd, 6 th, th and 4 th rows. Hence in the modified point matrix [P * ] columns nd, 6 th, th and 4 th will be affected. For more the clarit of the above explanation let us express angular displacements and moments as = ( Φ j ) j r + Φ j e νt ; ( j ) j r j t M = M + M e ν (8.6) where ν be the natural whirl frequenc with groscopic effects, and subscripts r and j represent the real and imaginar parts of a complex quantit, respectivel (for brevit the subscript i corresponding to i th mass has been dropped). On substituting equation (8.6) into equation (8.58), we get ( j ) ( j ) ( j ) j ( j R M z ) r R M z j L M zr L M z I j dν i xr x I j p ων i r j + + = Φ + Φ + Φ + Φ (8.6) Separating the real and imaginar parts from equation (8.6), we get and M M = I ν Φ I ωνφ (8.6) R zr L zr di xr pi j M M = I ν Φ + I ωνφ (8.63) R z j L z j di x j pi r It should be noted that equation (8.6) corresponds to effects term (i.e., I ων, which is the coefficient of p i j row in equation (8.45) and groscopic that is in the 6 th row of the modified state vector {S * }) in equation (8.6) will corresponds to 6 th column in equation (8.45). Similarl, from equation (8.63), it can be seen that groscopic effects will introduce another term (i.e., row and nd column corresponding to M z j th and r, respectivel. Hence, we have I p i ων ) at 5 th P = I ων and P5, = I p ων (8.64) i,6 pi where P i,j represent the additional element at the i th row and the j th column of the modified point matrix. Similarl, using equation (8.59), we get the additional element of the modified point matrix as

34 453 P3,4 = P7, = I pων (8.65) Figure 8. Coordinate axes and positive conventions for angular displacements (a) Rectangular 3- dimensional axis sstem (b) tilting of the shaft axis in z-x plane (c) tilting of shaft axis in -z plane Example 8.4 Obtain the unbalance response and transverse critical speeds of an overhang rotor sstem as shown in Figure 8.3. End B of the shaft is fixed. The length of the shaft is. m and the diameter is. m. The disc is thin and has kg of mass with the radius of 3. cm. Neglect the mass of the shaft and the groscopic effect. Take an unbalance of 3 gm at a radius of cm. Choose the shaft speed suitabl so as to cover two critical speeds in the unbalance response. E =. N/m. Figure 8.3 Solution: Let station be the fixed end and station be the free end. Consider onl a single plane motion. The overall transformation can be written as with { * } * { * } S = T S R (a) * * * T = P F (b)

35 454 [ P ] = Idω ; mω u x * = M z ; { S } S 3 l l l EI 6EI l l EI EI [ F ] = l (c) (d) 3 l l l EI 6EI l l EI EI = l [ P][ F] Idω mω u 3 l l l EI 6EI l l EI EI ω I dl ω Idl = ω Id l EI EI 3 mω l mω l mω mω l + u EI 6EI (e) Boundar conditions for the present case are that all the linear and angular displacements at station are zero and all the moments and shear forces are zero at right of station. Hence, the state vector at the station and right of will have the following form and * { S } { M z S } = T (f) * { } { x } = T (g) S R From free vibrations, unbalance, u, is zero. Hence, 3 rd and 4 th rows will give the eigen value problem of the following form

36 455 ωnf Idl ωnf Idl l EI EI M z = 3 mω S nf l mωnf l EI + 6EI (h) which gives a polnomial of the following form ml + I E I ω + = (i) 4 3 d nf EI ω 3 nf 4 ml Id ml Id For the present problem, we have following data m = kg, r =.3 m, Id 4 = mr = kg-m, l =. m, 4.5 E =. N/m, d =. m, and I π 4 = d / 64 = 4.9 m 4 Hence from equation (i), we get ( ) ωnf ( ) ω = 4 6 nf which gives natural frequencies as ω = 95.4 rad/s nf ω = 346. rad/s nf Now for the unbalance response from equations (a), (e), (g) and (f); we have 3 l l l EI 6EI l l EI EI x ω I dl ω Idl ω I M z d l = EI EI S 3 mω mω l mω l + mω l ( 6 ) u EI EI (j) The first two rows and the last three rows will give

37 456 and 3 l l EI 6EI M z = l l S x EI EI ω I dl ω Idl l EI EI M z 3 mω l mω l ( ) u + 6 S EI EI = (k) (l) The form of equations (k) and (l) is similar to that given in Table 8.3. Equation (k) can be written as or ω I dl ω Idl l EI EI M z = 3 S u mω l mω l + EI ( 6EI ) ω I dl ω Idl M z l EI EI = S 3 mω l mω l + EI ( 6EI ) u (m) (n) Equation (n) can be used to get the bending moment and the shear force due to unbalance force. Hence, on substituting equation (n) into equation (k), we get 3 l l ω I dl ω Idl l EI 6EI EI EI = x l l 3 EI EI mω l + mω l EI ( 6EI ) u (o) Equation (o) can be solved for a particular spin speed to get the unbalance response ( and φ x ). Then in the similar wa the spin speed can be varied to get the variation of the unbalance response with the spin speed. A plot of the and φ x with respect to the spin speed of the rotor is given in Figures 8.4(a and b). The resonant condition can be seen as large amplitudes of vibration and it indicate critical speeds. A similar plot can be obtained from equation (n) for M and S and are shown in Figs. 8.4(c and d). It can be observed from all four plots that critical speeds are same as natural frequencies z

38 457 obtained b free vibration analsis ( ω = 95.4 rad/s and nf ω = 346. rad/s). In the plots of the nf shear force and the bending moment at support, anti-resonances can be seen in between the two critical speeds. This indicates that two modes of vibrations have cancelling effects on the shear force and the bending moment. The frequenc at which the anti-resonance occurs for the shear force is not same as that of the bending moment. It indicates that it is not a sstem characteristics and its location ma change for different unbalance force, however, critical speed have fixed frequencies since it is sstem characteristics. Linear displacement,, (m) Spin speed, ω, (rad/s) Figure 8.4(a) Variation of the unbalance linear response with the spin speed 5 Angular displacement, x rad/s Spin speed, ω, (rad/s) Figure 8.4(b) Variation of the unbalance angular response with the spin speed

39 458 5 Bending moment, M, (Nm) Spin speed, ω, (rad/s) Figure 8.4(c) Variation of the support bending moment with the spin speed Shearing force, S, (N) Spin speed (rad/s) Figure 8.4(d) Variation of the support shearing force with the spin speed Example 8.5 Obtain transverse natural frequencies and corresponding mode shapes of a rotor sstem as shown in Figure 8.5. Take the mass of the disc, m = kg, the diametral mass moment of inertia, I d =. kg-m and the disc is placed at.5 m from the right support. The shaft has the diameter of mm and the span length of m. The shaft material has the Young s modulus E =. N/m and consider the shaft as mass-less. Neglect groscopic effects and take one plane motion onl. Compare the results b the influence coefficient method.

40 459 Figure 8.5 A simpl supported rotor sstem Solution: Station numbers, and can be assigned at the left support, at the disc and at the right support; respectivel. Then, the overall transformation of states is given as { } = [ ] [ ] [ ] { } R S F P F S (a) which can be expanded as R with l.5βl βl l.5βl βl 6 6 β.5βl β.5βl = ω I { S} { S} l nf d l mω nf β = EI and β = l / EI. l / (b) Subscripts and outside the matrices belong to the following parameters: β, l, and I d. On multiplication of matrices, we get R + m βlω 6 nf l.5id β lωnf.5βl β 6 l l.5βl βl 6.5m β l ω I β ω β.5β l β.5βl = m l ω I ω l l nf d nf { S} { S} nf d nf m ωnf which finall takes the following form x M z S t t t t t t t t x = t M 3 t3 t33 t 34 z t t t t S R (c)

41 46 with = + β ω 6 nf ; = ( + β ω 6 nf ) + (.5 d β ω ) nf ; =.5β ( + β ω 6 nf ) + β (.5 d β ω ).5 ; nf + β ( nf ) ( d ) ( ) nf =.5m βlωnf ; t = l (.5mβ lωnf ) + ( Id β ) ωnf ; =.5β (.5 β ωnf ) + β ( d β ω ) ; nf + β = β (.5 β ωnf ) +.5β ( d β ωnf ) + ( β ) +.5 β ; t m l t l m l l I l t l m l l I l l 3 t = β l + m β l ω +.5β l l.5i β l ω + l.5 β l + β l ; t t l m l I 3 t l m l l I l l 4 6 = 6 ( nf ) + ( d ) nf ( ) + = ωnf ; = ( ωnf ) + ( d ω ); nf =.5β ( ωnf ) + β ( d ω ) ; nf + = β 6 ( ωnf ) + β ( d ω ) nf + + = ωnf = ( ω ) ( ) ( nf t43 βl mω nf t44 βl mω nf ) t β l.5m β l ω.5β l I β ω l β.5 β l ; 4 t m l t l m l I 3 3 t l m l I 33 t l m l.5 l I l l ; t m ; 34 4 t l m 4 ; =.5 ; = + ; 6 (d) The following boundar conditions are applied for the present case (Figure 8.5) M M = = R = R = ; (e) On application of boundar conditions in equation (c), the following set of equations is obtained t t4 x = t t S 3 34 and x S t = t 4 R 4 44 t t x S (f) From the first set of equations (e), the frequenc equation takes the following form t t 34 t4t3 = (g) On substituting equation (e) into equation (g), we get { ( β ) ( )} ω ( ) ( 6 nf.5 d β ω ) nf β ω 6 nf.5β d ω nf 6 ( 6 ) ( ) ( ) { } { β }{ ( ) ( ) } l mβ lωnf βl l Id βlωnf l βl βl l ml ωnf Id ωnf l m l l I l l m l l I l l = which simplifies to