315 CH_5 - — EXAMPLE 5.2 Initial Conditions to Excite...

Info iconThis preview shows pages 1–10. Sign up to view the full content.

View Full Document Right Arrow Icon
Background image of page 1

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
Background image of page 2
Background image of page 3

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
Background image of page 4
Background image of page 5

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
Background image of page 6
Background image of page 7

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
Background image of page 8
Background image of page 9

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
Background image of page 10
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: — EXAMPLE 5.2 Initial Conditions to Excite Specific Mode Find the initial conditions that need to be applied to the system shown in Fig. 5.4 so as to make it Vibrate in (a) the first mode, and (b) the second mode. Solution: Approach: Specify the solution to be obtained for the first or second mode from the general solution for arbitrary initial conditions and solve the resulting equations. For arbitrary initial conditions, the motion of the masses is described by Eq. (5.15). In the pres- ent case, r1 = 1 and r2 = — 1, so Eq. (5.15) reduces to Eq. (E.10) of Example 5.1: 261(1) = Xfil)c0s (.13! + (151) + XSDCOS (.lfit + (1)2) m m 2620:) = XSDCOS (,lfit + (bl) — XSDCOS (11%1‘ + (112) m m Assuming the initial conditions as in Eq. (5.16), the constants X S”, X 32), (1)1. and (152 can be obtained from Eq. (5.18), using r1 = l and r2 = —1: (El) 1 . ' 1/2 ' Xi” = —5{[x1(0) + x2<0>12 + %[x1(0> + mow} (E 2) 2) 1 2 m - ~ 2 U2 Xi = —5 [—x1(0) + x2(0)1 + §[x1(0)- x2(0)] (13.3) (b _ W1 {—Wwo) + MON} (E 4) 1 VI? MD) + x2(0)l ' 392 CHAPTER 5 TWO DEGREE OF FREEDOM SYSTEMS #1 W7 mo) — 19(0)] (1)2 — tan (E.5) V 3k [—X1(0) + X2(0)l a. The first normal mode of the system is given by Eq. (E.8) of Example 5.1: X31) cos <1 [gt + (1)1) x<”(t) = (E.6) m Xfil)cos< it + dbl) Comparison of Eqs. (El) and (E6) shows that the motion of the system is identical with the first normal mode only if X 32) = 0. This requires that (from Eq. E.3) X1(0) = 162(0) and i1(0) = i2(0) (E-7) b. The second normal mode of the system is given by Eq. (E9) of Example 5.1: 38%) = (13.8) Comparison of Eqs. (El) and (E8) shows that the motion of the system coincides with the second normal mode only if X3” = 0. This implies that (from Eq. E.2) 161(0) = —xz(0) and IMO) = 'i2(0) (E-9) 5.4 TORSIONAL SYSTEM 397 — Natural Frequencies of a Marine Engine Propeller EXAMPLE 5.5 The schematic diagram of a marine engine connected to a propeller through gears is shown in Fig. 5.8(a). The mass moments of inertia of the flywheel, engine, gear 1, gear 2, and the propeller (in kg-mz) are 9000, 1000, 250, 150, and 2000, respectively. Find the natural frequencies and mode shapes of the system in torsional Vibration. Solution Approach: Find the equivalent mass moments of inertia of all rotors with respect to one rotor and use a two degree of freedom model. Assumptions 1. The flywheel can be considered to be stationary (fixed) since its mass moment of inertia is very large compared to that of other rotors. 2. The engine and gears can be replaced by a single equivalent rotor. Gear 1,40 teeth Propeller Flywheel FIGURE 5.8 Marine engine propeller system. 398 CHAPTER 5 TWO DEGREE 0F FREEDOM SYSTEMS Since gears l and 2 have 40 and 20 teeth, shaft 2 rotates at twice the speed of shaft 1. Thus the mass moments of inertia of gear 2 and the propeller, referred to the engine, are given by (162)..1 = (2)2050) = 600 1(ng (7%, = (2)2(2000) = 8000 kg-m2 Since the distance between the engine and the gear unit is small, the engine and the two gears can be replaced by a single rotor with a mass moment of inertia of 11 = JE + JG1 + (102)eq = 1000 + 250 + 600 = 1850 kg-mz Assuming a shear modulus of 80 X 109 N/m2 for steel, the torsional stiffnesses of shafts 1 and 2 can be determined as GI d4 80 x 109 0.104 0‘ G<W 1) - ( )(Wx ) — 981,750.0 N-m/rad k” 11 — 11 32 (0.8)(32) GI d4 80 x 109 0.15 4 — 02 G<7T 2) = ( WX ) — 3,976,087.5 N-m/rad ’2 I2 712 32 (1.0)(32) Since the length of shaft 2 is not negligible, the propeller is assumed to be a rotor connected at the end of shaft 2. Thus the system can be represented as a two degree of freedom torsional system, as indicated in Fig. 5.8(b). By setting k3 = 0, k1 = k,1,k2 = k,2, m1 = 11, and m2 = 12 in Eq. (5.10), the natural frequencies of the system can be found as 1112 i (kn + [(0)12 + kt2 Jr 2 _ 4 (kn + kt2)kt2 — kt22 “2 1112 1112 = {(191 + ktZ) +£} l{(k:1+ kt2)‘,2 + ktZ Jr} 2 J] 2 J2 (kn + ktZ) ktZ 2 knkrz 1’2 i — + — — E.1 [{ 2 11 2 12 11 12 ( ) Since (ktl + kg) + ktz (98.1750 + 397.6087) X 104 + 397.6087 X 104 211 212 2 X 1850 2 X 8000 = 1588.46 and k,l k,2 (98.1750 x 104) (397.6087 x 104) — 26.37 0 x 104 J1 12 (1850) (8000) 5 5.5 COORDINATE COUPLING AND PRINCIPAL COORDINATES 399 Eq. (E. 1) gives (6%, wg 1588.46 :1: [(1588.46)2 — 26.3750 x 104]“2 1588.46 j: 1503.1483 Thus a)? = 85.3117 or w, = 9.2364 rad/sec (0% = 3091.6083 or (02 = 55.6022 rad/sec For the mode shapes, we set k1 = k,1,k2 = kg, k3 = 0, m1 = 11, and m2 = J2 in Eq. (5.11) to obtain r _ ‘11“)1 + (kn + ktz)_— kt2 —(1850) (85.3117) ‘1‘ (495.7837 X 104) _ 4 — 1.2072 397.6087 X 10 and _ _]1w2 + (kn + kt2) r2 — — km —(1850) (3091.6083) + (495.7837 X 104) — 0.1916 397.6087 X 104 Thus the mode shapes can be determined from an equation similar to Eq. (5.12) as g (1)_ i _ 1 @z _ r1 ‘12072 and — Frequencies and Modes of an Automobile E XAM P L E 5 . 7 ~ Determine the pitch (angular motion) and bounce (up and down linear motion) frequencies and the location of oscillation centers (nodes) of an automobile with the following data (see Fig. 5.11): Mass (m) = 1000 kg Radius of gyration (r) = 0.9 m Distance between front axle and CG. ([1) = 1.0 m Distance between rear axle and CG. ([2) = 1.5 In Front spring stiffness (kf) = 18 kN/m Rear spring stiffness (k,) = 22 kN/m Solution: If x and 0 are used as independent coordinates, the equations of motion are given by Eq. (5.23) with k1 = kf, k2 = k,, and J0 = mrz. For free vibration, we assume a harmonic solution: x(t) = X cos(wt + ¢), 0 (t) = (9 cos(wt + (1)) (El) FIGURE 5.11 Pitch and bounce motions of an automobile. 5.5 COORDINATE COUPLING AND PRINCIPAL COORDINATES Using Eqs. (El) and (5.23), we obtain (— mwz + k1 + k2) (— k1]1 + kzlz) X 0 2 2 2 = (E2) (_ [(111 + [(212) (_ 10(1) + [(111 + [(212) G) 0 For the known data, Eq. (E.2) becomes (—1000w2 + 40,000) 15,000 X _ 0 (E 3) 15,000 (~810w2 + 67,500) 0 0 ' from which the frequency equation can be derived: 8.10:4 — 99909 + 24,750 = 0 (13.4) The natural frequencies can be found from Eq. (E4): (01 = 5.8593 rad/s, (02 = 9.4341 rad/s (E5) With these values, the ratio of amplitudes can be found from Eq. (E.3): X0) X0) @(1) . , W = 0.3061 (E6) The node locations can be obtained by noting that the tangent of a small angle is approximately equal to the angle itself. Thus, from Fig. 5.12, we find that the distance between the CG. and the node is —2.6461 In for (01 and 0.3061 In for (02. The mode shapes are shown by dashed lines in Fig. 5.12. FIGURE 5.12 Mode shapes of an automobile. — Steady-State Response of Spring-Mass System E XA M P L E 5 . 8 Find the steady—state response of the system shown in Fig. 5.13 when the mass m1 is excited by the force RC) = F10 cos wt. Also, plot its frequency response curve. Solution: The equations of motion of the system can be expressed as m 0 361 2k —k x1 _ Flocoswt [0 m]{x2}+[—k did-i o } Comparison of Eq. (E.1) with Eq. (5.27) shows that mm = "'22 = 7", m12 = 0, 011: C12 2 C22 2 0, k11:k22=2k, k12=_k, F1=F10C03wt, F220 408 CHAPTER 5 TWO DEGREE OF FREEDOM SYSTEMS IFIU) = F10 cos wt FIGURE 5.13 A two—mass system subjected to harmonic force. We assume the solution to be as follows:2 xj(t) = X]- cos wt; j = 1, 2 Equation (5.31) gives 211(0)) = 222(0)) = ‘mwz + 2k’ 212(0)) 2 ‘k Hence X1 and X2 are given by Eq. (5.35): (—wzm + 2k)F10 (—wzm + 2k)F10 X10") _ (— wzm + 2k)2 — k2 _ (— mm2 + 3k)(—mw2 + k) X _ = 2(a)) (—mw2 + 2k)2 — k2 (—mw2 + 3k)(—mw2 + k) By defining a)? = k/m and (0% = 3k/m, Eqs. (E4) and (E5) can be expressed as X1(w) _ (E2) (E3) (E4) (E5) (E6) 2Since Flocos out = Real(F10ei""), we shall assume the solution also to be xj=Real()(jei‘"’)= X j cos wt, 1' = l, 2. It can be verified that are real for an undamped system. 5.7 SEMIDEFINITE SYSTEMS 409 (a) (b) FIGURE 5.14 Frequency response curves of Example 5.8. F10 2 2 2 w a) w (1)] (01 (01 The responses X1 and X2 are shown in Fig. 5.14 in terms of the dimensionless parameter w/wl. In the dimensionless parameter, w/wl, ml was selected arbitrarily; (1)2 could have been selected just as easily. It can be seen that the amplitudes X1 and X2 become infinite when w2 = a)? or (02 = (0%. Thus there are two resonance conditions for the system: one at w] and another at (02. At all other val- ues of w, the amplitudes of vibration are finite. It can be noted from Fig. 5.14 that there is a partic- ular value of the frequency a) at which the Vibration of the first mass ml, to which the force F1(t) is applied, is reduced to zero. This characteristic forms the basis of the dynamic vibration absorber dis- cussed in Chapter 9. X2(w) — (E-7) ...
View Full Document

This note was uploaded on 09/08/2010 for the course MAE 315 taught by Professor Wu during the Summer '08 term at N.C. State.

Page1 / 10

315 CH_5 - — EXAMPLE 5.2 Initial Conditions to Excite...

This preview shows document pages 1 - 10. Sign up to view the full document.

View Full Document Right Arrow Icon
Ask a homework question - tutors are online