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recurrence relations under the condition that a displacement or a force must be zero. EXAMPLE 3.14 Using the Holzer method, determine the natural frequencies for the mechan- ical model in Fig. 3.18. Solution Write successive Eqs. (3.41): F 1 q 1 m 1 o 2 À k 0 q 1 q 0 À F 0 k 0 F 1 F 0 m 1 o 2 q 1 X V ` X But q 0 0, and F 0 Àk 0 q 1 results in q 2 q 1 À F 1 k 1 F 2 F 1 m 2 o 2 q 2 X V ` X Replacing F 1 as a function of q 1 and q 2 in an F 2 expression, we obtain F 2 q 1 m 1 o 2 À k 0 m 2 o 2 1 À 1 k 1 m 1 o 2 À k 0 ! q 3 q 2 À F 2 k 2 F 3 F 2 m 3 o 2 q 3 X V ` X Replacing q 2 as a function of q 1 and F 2 as a function of q 1 in the expression for q 3 , and then in the F 3 expression, one can obtain F 3 : q 4 q 3 À F 2 k 2 F 4 F 3 m 4 o 2 q 4 X V ` X Replacing q 3 as a function of q 1 and F 3 as a function of q 1 in the expression for q 4 , and q 4 0, in the end one can calculate F 4 . m Figure 3.18 Mechanical model for Example 3.14. 408 Theory of Vibration Vibration EXAMPLE 3.15 Using the Holzer method, determine the natural frequencies of the mechan- ical model in Fig. 3.19. Solution Write successive Eqs. (3.41) to obtain q 2 q 1 À F 1 k 1 F 1 F 0 m 1 o 2 q 1 X V ` X But F 0 0, so q 2 q 1 1 À mo 2 ak 1 ÀÁ and q 3 q 2 À F 2 k 2 F 2 F 1 m 2 o 2 q 2 X V ` X Replacing q 2 as a function of q 1 and F 2 as a function of q 1 in the expression for q 3 , one obtains F 3 F 2 m 3 o 2 q 3 X Replacing F 2 as a function of q 1 and q 3 as a function of q 1 in the expression for F 3 , and given that F 3 0, we obtain the expression for the natural frequencies of the mechanical system. m 3.5.2 APPLICATION OF THE HOLZER METHOD TO A MECHANICAL MODEL WITH ROTORS AND HOOKE MODELS In this case, vectors of state attached to a shaft section are used. One vector of state is zjY M T , where j is the angle of rotation of the section and M is the torsion moment (torque) in some section. Because in the zone of the rotor the diagram of moments is discontin- uous, M l i T M r i , where M l i is the torsion moment at left and M r i at right (Fig. 3.20). Because the rotor is rigid, j l i j r i j i (the rotation at left is equal to the rotation at right). Vectors of state at left and right of section i are z l i j l i M l i ! Y and z r i j r i M r i ! X The equation of motion of rotor i is J i j i M r i À M l i Y 3X42 Figure 3.19 Mechanical model for Example 3.15. 3. Linear Systems with Finite Numbers of Degrees of Freedom 409 Vibration and with j i tj i sinot j, M l i M l i sinot j, M r i M r i sinot j, we obtain Ào 2 J i j i M r i À M l i X 3X43 Therefore, j l i j r i Y M r i M l i À o 2 J i j i Y 3X44 and the result is the recurrence relation between vectors of state, z r i A i z l i Y 3X45 where A i 10 Ào 2 J i 1 ! X For the zone between two successive rotors, M l i M r iÀ1 Y j l i À j r iÀ1 M l i k iÀ1 M r iÀ1 k iÀ1 X 3X46 There results the following recurrence relation between vectors of state at the extremity of a shaft section between two successive rotors: z l i B i z r iÀ1 Y 3X47 where B i 1 1 k iÀ1 01 P R Q S X In conclusion, the recurrence relations between vectors of state are z r i A i z l i in a second of rigid rotors z l i B i z r iÀ1 in a section between two successive rotorsX EXAMPLE 3.16 With the Holzer method, determine the natural frequencies for the mechan- ical model from Fig. 3.21. Figure 3.20 Torsional mechanical model. 410 Theory of Vibration Vibration Solution We can write, successively, z l 3 B 3 z r 2 Y z r 2 A 2 z l 2 z l 2 B 2 z r 1 Y z r 1 A 1 z l 1 Y z l 1 B 1 z r 0 Y where B 3 1 1 k 3 01 P R Q S Y B 2 1 1 k 2 01 P R Q S Y B 1 1 1 k 1 01 P R Q S A 1 À 10 Ào 2 J 1 1 ! Y A 2 10 Ào 2 J 2 1 ! X Reuniting the preceding relations, one can write z l 3 B 3 A 2 B 2 A 1 B 1 z r 0 Y but z l 3 0 M l 3 ! Y z r 0 0 M r 0 ! Y and 0 M l 3 ! B 3 A 2 B 2 A 1 B 1 0 M r 0 ! X With M r 0 0 one can obtain the equation for natural frequencies. m 3.5.3 RAYLEIGH METHOD The dynamic response of free undamped vibrations is an overlapping of harmonic vibrations with natural frequencies o 1 Y FFFY o n , resulting in q 1 tA 11 sino 1 t j 1 ÁÁÁA 1n sino n t j n q 2 tA 21 sinw 1 t j 1 ÁÁÁA 2n sino n t j n F F F q n tA n1 sino 1 t j 1 ÁÁÁA nn sino n t j n X V b b b ` b b b X 3X48 One can denote A 21 m 21 A 11 Y A 31 m 31 A 11 Y FFFY A n1 m n1 A 11 A 22 m 22 A 12 Y A 32 m 32 A 12 Y FFFY A n2 m n2 A 12 F F F A 2n m 2n A 1n Y A 3n m 3n A 1n Y FFFY A nn m nn A 1n Y V b b b ` b b b X 3X49 Figure 3.21 Mechanical model for Example 3.16. 3. Linear Systems with Finite Numbers of Degrees of Freedom 411 Vibration where m ik are called the coef®cients of distribution. The vectors of distribution can be introduced: m 1 1 m 21 F F F m n1 P T T T R Q U U U S Y m 2 1 m 22 F F F m n2 P T T T R Q U U U S Y FFFY m n 1 m 2n F F F m nn P T T T R Q U U U S X It easy to demonstrate that the vectors of distribution m k verify the system RÀo 2 k M m k 0X 3X50 The natural modes of vibration associated with the natural frequency are the column vectors q k A 1k sino k t j k A 2k sino k t j k F F F A nk sino k t j k P T T T R Q U U U S m k A 1k sino k t j k X 3X51 The kinetic energy of a system with more than one degree of freedom can be calculated using the relation T 1 2 q T M qY 3X52 and the potential of elastic forces attached to Hooke models can be determined using the relation V 1 2 q T RqX 3X53 Corresponding to the k mode of vibration: Kinetic energy T k 1 2 q T k M q k Y Potential of elastic forces V k 1 2 q T k Rq k X Replacing q k with Eq. (3.51), the following results are obtained: T k 1 2 m T k M m k A 2 1k o 2 k cos 2 o k t j k 3X54 and V k 1 2 m T k Rm k A 2 1k sin 2 o k t j k X 3X55 The system is undamped; therefore, T k max V k max , and there results the relation of calculus for the natural frequency, o k m T k Rm k m T k M m k s X 3X56 The following is the methodology for working with the Rayleigh method: j Adopt an expression for m k and determine, using Eq. (3.56), a ®rst value for o k . 412 Theory of Vibration Vibration j With the natural frequency calculated in this way, introduce the torsors of inertia (forces or torques) and determine the new displace- ment, namely a new expression for m k , which is reintroduced in Eq. (3.56). The Rayleigh method gives the minimum natural frequency superior to the real value, and maximum natural frequency inferior to the real value. EXAMPLE 3.17 Using the Rayleigh method, determine one natural frequency for the mechanical model from Fig. 3.22, where k 1 k, k 2 2k, k 3 k, m 1 2m, m 2 3m, m 3 m. Solution The mathematical model is m 1 q 1 k 1 q 1 k 2 q 2 0 m 2 q 2 À k 1 q 1 k 1 k 2 q 2 À k 2 q 3 0Y m 3 q 3 À k 2 q 2 k 2 k 3 q 3 0Y V b ` b X where the matrix of inertia is M m 1 00 0 m 2 0 00m 3 P R Q S Y and the matrix of rigidity (stiffness) is R k 1 Àk 1 0 Àk 1 k 1 k 2 Àk 2 0 Àk 2 k 2 k 3 P R Q S X The vectors of distribution is m k 1Y m 2k Y m 3k T . Figure 3.22 Mechanical model for Example 3.17. 3. Linear Systems with Finite Numbers of Degrees of Freedom 413 Vibration Replacing this in Eq. (3.49), one ®nds o k 1Y m 2k Y m 3k k 1 Àk 1 0 Àk 1 k 1 k 2 Àk 2 0 Àk 2 k 2 k 3 P R Q S 1 m 2k m 3k P R Q S 1Y m 2k Y m 3k m 1 00 0 m 2 0 00m 3 P R Q S 1 m 2k m 3k ! V b b b b b b ` b b b b b b X W b b b b b b a b b b b b b Y 1a2 X If the masses vibrate in phase, m k 1Y 1Y 1 T , which in particular cases results in o o k 0X166 k m r X If two neighboring masses vibrate in phase, and the third in opposition of phase, one can have the cases m k 1Y À1Y À1 T when o o k 1X5 k m r Y m k 1Y 1Y À1 T when o o k 0X833 k m r X If the neighboring masses vibrate in opposition of phase, m k 1Y À1Y 1 T when o o k 2X15 k m r X m 3.5.4 ANALYSIS OF STABILITY OF VIBRANT SYSTEM A vibrant system with more than one degree of freedom is a multivariable open linear system with [F ]orf as inputs and [q] as output. The matrix of transfer for the open system is H C T sI ÀA À1 BY 3X57 where matrices [A], [B], [C ] are given as functions of the inertia matrix [M], damping matrix [D], and stiffness matrix [R]. The transfer matrix becomes H I Y 0 Á sI ÀA À1 00 I 0 ! Y 3X58 where [I] is the unit matrix and [0] is the null matrix, both with the dimensions n  n. The characteristic polynomial for the open system is P sdetsI ÀA 3X59 or P sdet sI ÀI M À1 R sI M À1 D ! Y 3X60 414 Theory of Vibration Vibration which becomes P sdets 2 M sDR 0X 3X61 The stability of vibrant systems with n degrees of freedom depends on the position of the roots of the polynomial Eq. (3.61). The stability criteria are algebraic (Routh, Hurwitz), or grapho-analytical criteria (Cramer, Leonhard). The use of polar diagrams (Nyquist) or Bode diagram requests a procedure to reduce the multivariable system to a monovariable system. EXAMPLE 3.18 Determine the conditions of stability of motion for the vibrant system presented in Example 3.8. Solution With M J 1 0 0 J 2 0 00J 3 P T R Q U S Y D c 1 c 2 Àc 2 0 Àc 2 c 2 c 3 Àc 3 0 Àc 3 c 3 c 4 P T R Q U S Y R k 1 k 2 Àk 2 0 Àk 2 k 2 k 3 Àk 3 0 Àk 3 k 3 k 4 P T R Q U S Y the characteristic polynomial becomes a 1 Àc 2 s À k 2 0 Àc 2 s À k 2 a 2 Àc 3 s À k 3 0 Àc 3 s À k 3 a 3 0Y where a 1 s 2 J 1 sc 1 c 2 k 1 k 2 , a 2 s 2 J 2 sc 2 c 3 k 2 k 3 , a 3 s 2 J 3 sc 3 c 4 k 3 k 4 . Developing this, we obtain a sixth-order polynomial, and the Hurwitz criterion can be applied to study stability. Stability criterion Nyquist starts with the construction of the polar diagram. The polar diagram (Fig. 3.23) has the following characteristics: j Point C, which corresponds to o 0, gives the speci®c displacement of the system in the static regime [maN]. j The number of loops of polar diagram is the number of vibration modes (implicitly the number of degrees of freedom). For the diagram in Fig. 3.23, these are the three modes of vibration corresponding to loops 1, 2, 3. j The value of o that corresponds to the point on the loop placed at maximum distance from origin is the natural frequency associated with the mode of vibration that corresponds to the respective loop. For the polar diagram in Fig. 3.23, point B corresponds to one angular 3. Linear Systems with Finite Numbers of Degrees of Freedom 415 Vibration frequency o B , giving the natural frequency o B that is attached to the vibration mode corresponding to loop 2. j The points of polar diagram in proximity to the origin give information about the behavior of the vibrant system at high frequencies of excitation. j The maximum de¯ection of the vibrant system, for each mode of vibration, is equal to the diameter of a circle that better approximates the associated loop of points of maximum distance from origin (i.e., diameter A 1 A 2 of the circle associated with loop 1 is the maximum de¯ection for the ®rst mode of vibration). j With the preceding circle diameters one can appreciate the damping level of the respective mode of vibration. The polar diagram can be traced for a vibrant physical system using experimental data. In addition, this type of diagram indicates the linearity or nonlinearity of the vibrant system. m 4. Machine-Tool Vibrations 4.1 The Machine Tool as a System The machine tool is a system (Fig. 4.1) with the following characteristics: j Elastic subsystem (ES): workpiece, tool, device, and elastic structure of the machine tool j Actuator subsystem (AS); electric motors and hydraulic motors j Subsystem due to the friction process and dissipation effects (FS) j Subsystem due to the cutting process (CPS) that includes processes from the contact between workpiece and tool Figure 3.23 Polar diagram. 416 Theory of Vibration Vibration The interconnection of the different subsystems is made with the help of the elastic subsystem (ES) (Fig. 4.1). The deformations of ES produce the variation of chip thickness, namely the variation of the cutting force (ESÀ3CPS. The variation of the cutting force produces the modi®cation of stresses and deformations of the elastic structure CPSÀ3ES. In engineering practice, equivalent systems are used: j Equivalent subsystem SDE1±AS (Fig. 4.2a). This subsystem is used for modeling the actuator's processes. As a mechanical model it has lumped masses, rotors, Hooke models, and linear dampers. j Equivalent subsystem SDE2±FS (Fig. 4.2b). This subsystem is used for modeling the friction processes, with the same mechanical model as SDE1±AS. j Equivalent subsystem SDE3±CPS (Fig. 4.2c). This subsystem is used for modeling the cutting process. The preceding systems can be analyzed using the system presented in Fig. 4.3 where z is eliminated. Figure 4.1 The machine- tool system. Figure 4.2 Equivalent subsystem. (a) Subsystem SDE1±AS; (b) subsystem SDE2±FS; (c) subsystem SDE3±CPS. Figure 4.3 Open system. 4. Machine-Tool Vibrations 417 Vibration [...]... approximated with a full cylinder b b ` 8 4X7 JR mD 2 b b b 4 for rotors that can be approximated with a ring X with mean diameter D where m is the mass of the rotor in (kg) Calculate the moments of inertia for each shaft with respect to the geometric axis of the shaft: V 2 b mD b for full section shafts ` 8 4X8 X JA b m D 2 À d 2 b X for tubular shafts 8 For rotors placed at the extremity of... reduction of the damping coef®cients to the reference shaft is given by the relation cr ci 2 X 4X21 Step 8 Reduce the outside torques M (that act on a rotor placed on other shafts than the reference shaft) to the reference shaft: Mr MiX 4X22 Using step 1 to step 8 one may obtain an equivalent mechanical model with rotors placed on the reference shaft The reduction of the rotors (and therefore the... For the mechanical model obtained after step 7 or step 8 one can write the mathematical model A complex kinematic chain can be transformed into a Figure 4.6 Reduction of number of rotors (and therefore of degrees of freedom) 4 Machine-Tool Vibrations 423 simpler model as shown in Fig 4.7 A point of rami®cation P as a rotor with inertia moment JP 0 and rotation qP can be considered Figure 4.7 Mechanical. .. motor H JEM GD 2 kg m2 Y 8g where g is the gravitational acceleration, g 9X81 mas2 , and D is the diameter of the rotor of the motor The inertia moment of the clutch is neglected The inertia moments for the spur gears are m1 r1 o2 1 Y 2 H JR2 m 2 r2 o 2 2 Y 2 H JR3 m3 r2 o2 3 Y 2 H JR4 m4 r4 o2 4 kg m2 X 2 H 2 The inertia moment for the rotor is JR MdR a8 kg m 2 ) Compute the inertia... 2 2 Y 2 H JR3 m3 r2 o2 3 Y 2 H JR4 m4 r4 o2 4 kg m2 X 2 H 2 The inertia moment for the rotor is JR MdR a8 kg m 2 ) Compute the inertia moments for shafts JI MI dI2 Y 8 JII 2 MII dII Y 8 JIII 2 MIII dIII kg m2 X 8 The corrected inertia moments for extremity rotors are computed using the relations H JME JME 1 JI Y 6 H JR3 JR3 1 JII Y 6 Step 3 H JR1 JR1 1 JI Y 6 H JRA JR4... lK 5 X The elastic constants can be recalculated as kI GIPI Y LI kII GIPII Y LII kIII GIPIII N maradX LIII The mechanical model is shown in Fig 4.9 The damping constants cI , cII , cIII can be determined with Eq (3.7) Vibration lEM 4 28 Theory of Vibration Figure 4.9 Mechanical model Step 6 Shaft III is chosen as the reference shaft The reduction of the inertia moments of rotors placed on... reference shaft: 2 k1 kI i1À3 Y 2 k2 kII i2À3 Y k3 kIII X Vibration Step 8 The damping coef®cients cI and cII are reduced to reference shaft: 2 c1 cI Á i1À3 Y Step 9 2 c2 cII Á i2À3 Y c3 cIII X The external torque M acts on the reference shaft The mechanical model reduced to the reference shaft is shown in Fig 4.10 The mechanical model has four degrees of freedom To simplify the calculations... J3 A new mechanical model is obtained (Fig 4.11) The mathematical model corresponding to the new mechanical model, shown in Fig 4.11, is V b J1 Q1 Àk1 Q1 À Q2 À c1 Q1 À Q2 ` J H Q Àk1 Q2 À Q1 À c1 Q2 À Q1 À k23 Q2 À Q3 À c23 Q2 À Q3 b 2 2 X H J4 Q3 Àk23 Q3 À Q2 À c23 Q3 À Q2 M0 sin ot X 429 4 Machine-Tool Vibrations Figure 4.10 Reduced mechanical. .. Mmax (Nam), rotation for the unload function (null couple) n0 (revamin), rotation corresponding to Mmax from mechanical characteristic ncr , frequency of the alternative current fr (AC), moment of ¯y wheel GD 2 (Nam2 ) Shaft characteristics are given in Table 4.1 424 Theory of Vibration Figure 4 .8 Gear train with two stages Table 4.1 Shaft Characteristics Shaft number I II III Diameter (m) dI dII dIII... mechanical model Figure 4.11 Mechanical model with three degrees of freedom The kinematic chain just analyzed is speci®c to a gear box and a feed box of a machine tool To obtain a linear motion, the following kinematic chains can be used: screw nuts (Fig 4.12a), screw nuts with rolling element (Fig 4.12b), pinion rack (Fig 4.12c), and linear hydraulic motor (Fig 4.12d) For the mechanical model shown in . moment for the rotor is J H R Md 2 R a8kg m 2 ). Compute the inertia moments for shafts J I M I d 2 I 8 Y J II M II d 2 II 8 Y J III M III d 2 III 8 kg m 2 X The corrected inertia moments. respect to the geometric axis of the shaft: J A mD 2 8 for full section shafts mD 2 À d 2 8 for tubular shafts X V b b ` b b X 4X8 For rotors placed at the extremity of the shaft, the. expression for q 4 , and q 4 0, in the end one can calculate F 4 . m Figure 3. 18 Mechanical model for Example 3.14. 4 08 Theory of Vibration Vibration EXAMPLE 3.15 Using the Holzer method, determine