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315 CH_2 - — EXAMPLE 2.2 Free Vibration Response Due to...

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Unformatted text preview: — EXAMPLE 2.2 Free Vibration Response Due to Impact A cantilever beam carries a mass M at the free end as shown in Fig. 2.1 1(a). A mass m falls from a height h on to the mass M and adheres to it without rebounding. Determine the resulting transverse vibration of the beam. Solution: When the mass m falls through a height h, it will strike the mass M with a velocity of vm = Zgh, where g is the acceleration due to gravity. Since the mass m adheres to M without rebounding, the velocity of the combined mass (M + m) immediately after the impact (X0) can be found using the principle of conservation of momentum: mvm = (M + m))&0 OI' x0=< m >vm=< m >V2‘g2 (E.1) M+m M+m The static equilibrium position of the beam with the new mass (M + m) is located at a distance of flkg below the static equilibrium position of the original mass (M) as shown in Fig. 2.11(c). Here k denotes the stiffness of the cantilever beam, given by 3E1 k=l—3 122 CHAPTER 2 FREE VIBRATION 0F SINGLE DEGREE 0F FREEDOM SYSTEMS Young‘s modulus, E Moment of inertia, I (a) (b) YY = static equilibrium position of M ZZ = static equilibrium position of M + m FIGURE 2.11 Response due to impact. Since free vibration of the beam with the new mass (M + m) occurs about its own static equilib- rium position, the initial conditions of the problem can be stated as k M+m XO 2 ‘TE, 20 = (L) V Zgh (13.2) Thus the resulting free transverse vibration of the beam can be expressed as (see Eq. (2.21)): x(t) = A cos (cont — (1)) ' 2 1/2 a)" )5 d) = tarfl< 0 ) x0wn _ k _ / 3E1 0’" M + m 13(M + m) with x0 and 1&0 given by Eq. (E2). where _ EXAMPLE 2.5 Natural Frequency of Pulley System Determine the natural frequency of the system shown in Fig. 2.13(a). Assume the pulleys to be fric- tionless and of negligible mass. Solution: To determine the natural frequency, we find the equivalent stiffness of the system and solve it as a single degree of freedom problem. Since the pulleys are frictionless and massless, the tension in the rope is constant and is equal to the weight Wof the mass m. From the static equilibrium of the pulleys and the mass (see Fig. 2.13b), it can be seen that the upward force acting on pulley l is 2W and the downward force acting on pulley 2 is 2W. The center of pulley 1 (point A) moves up by a distance 2W/k1, and the center of pulley 2 (point B) moves down by 2W/k2. Thus the total movement of the mass m (point 0) is 2 2.2V. + at kl k2 as the rope on either side of the pulley is free to move the mass downward. If keq denotes the equiv- alent spring constant of the system, Weight of the mass ‘e—T‘fl— = Net displacement of the mass Equivalent spring constant + W=4W(1+ 1)=4W(k. k2) k1k2 ._. __* .1 k” 4(k1+ k2) (E l 126 CHAPTER 2 FREE VIBRATION 0F SINGLE DEGREE 0F FREEDOM SYSTEMS k2x2 waw I l \ $31. ) j— I\_,/I v v k2x2=2W x 2w 2‘7? 2 w WW"? x=2(x1+x2> m} .——-., i :o~ . I I. _____ (a) (b) FIGURE 2.13 Pulley system. By displacing mass m from the static equilibrium position by x, the equation of motion of the mass can be written as me + keqx = o (5.2) and hence the natural frequency is given by keq 1/2 klkz 1/2 ad/ E3 to" _ m _ 4m(kl + k2) r sec ( - ) or __ w" _ I klkz ”2 f" — 27r — 4w[m(k1 + k2) cycles/sec (EA) — EXAMPLE 2.8 Effect of Mass on w" of a Spring Determine the effect of the mass of the spring on the natural frequency of the spring—mass system shown in Fig. 2.19. Solution: To find the effect of the mass of the spring on the natural frequency of the spring-mass system, we add the kinetic energy of the system to that of the attached mass and use the energy 2.5 RAYLEIGH'S ENERGYMETHOD 137 7 y z i J; dy x(t) FIGURE 2.19 Equivalent mass of a spring. method to determine the natural frequency. Let I be the total length of the spring. If x denotes the displacement of the lower end of the spring (or mass m), the displacement at distance y from the support is given by y(x/l). Similarly, if 2% denotes the velocity of the mass m, the velocity of a spring element located at distance )1 from the support is given by y(jc/l). The kinetic energy of the spring element Of length dy iS dT — 1 I d . 2 E 1 where ms is the mass of the spring. The total kinetic energy of the system can be expressed as T = kinetic energy of mass (Tm) + kinetic energy of spring (T,) 1 2-2 1 .2 1 m: yx =— + .. ._....d .__ me /y=02(1 ’X 12 > 1 .2 1m5.2 — +—~—— 13.2 2m): 2 3x ( ) The total potential energy of the system is given by U = §kx2 (12.3) By assuming a harmonic motion x(t) = X cos wnt (5-4) where X is the maximum displacement of the mass and a)" is the natural frequency, the maximum kinetic and potential energies can be expressed as Tm“ = 1(m + E>X2wi (E.5) 138 CHAPTER 2 FREE VIBRATION 0F SINGLE DEGREE 0F FREEDOM SYSTEMS 1 Umax = EkXZ (E6) By equating Tmax and Umax, we obtain the expression for the natural frequency: k 1/2 = m E.7 mu m + f ( ) Thus the effect of the mass of the spring can be accounted for by adding one-third of its mass to the main mass [2.3]. — Shock Absorber for 21 Motorcycle EXAMPLE 2.11 An underdamped shock absorber is to be designed for a motorcycle of mass 200 kg (Fig. 2.31a). When the shock absorber is subjected to an initial vertical velocity due to a road bump, the resulting displacement-time curve is to be as indicated in Fig. 2.31(b). Find the necessary stiffness and damp- ing constants of the shock absorber if the damped period of vibration is to be 2 s and the amplitude x1 is to be reduced to one-fourth in one half cycle (i.e., x15 = x1/4). Also find the minimum initial velocity that leads to a maximum displacement of 250 mm. Approach: We use the equation for the logarithmic decrement in terms of the damping ratio, equation for the damped period of vibration, time corresponding to maximum displacement for an underdamped system, and envelope passing through the maximum points of an underdamped system. Solution: Since x15 = x1/4, x2 = x1.5/4 = x1/16. Hence the logarithmic decrement becomes 5 — ln (:1) —- 1n(l6) — 2.7726 — (El) 2 W72 from which the value of g can be found as { = 0.4037. The damped period of vibration is given to be 2 5. Hence 2 277 277 = T = — = — '1 W an1 — g2 2 w” = —” = 3.4338 rad/s 2V1 — (0.4037)2 2.6 FREE VIBRATION WITH VISCOUS DAMPING (a) (b) FIGURE 2.31 Shock absorber of a motorcycle. The critical damping constant can be obtained: cc = 2mm" = 2(200)(3.4338) = 1373.54 N-s/m Thus the damping constant is given by c = {cc = (O.4037)(1373.54) = 554.4981 N-s/m and the stiffness by k = mm" = (200)(3.4338)2 = 23582652 N/m The displacement of the mass will attain its maximum value at time t], given by sin wdt1= V1 — {2 (See Problem 2.86.) This gives sin wdtl = sin m1 = V1 — (0.4037)2 = 0.9149 01' sin—1(0.9l49) t1 = —— = 0.3678 sec 7T The envelope passing through the maximum points (see Problem 2.86) is given by x = V 1 — {2Xe_§“’"’ Since x = 250 mm, Eq. (E2) gives at II 155 (E2) 156 CHAPTER 2 FREE VIBRATION 0F SINGLE DEGREE 0F FREEDOM SYSTEMS 0.25 = x /1 _ (0.4037)2 Xe—(0.4037)(3.4333)(0.3678) or X = 0.4550 in. The velocity of the mass can be obtained by differentiating the displacement x(t) = Xe‘gw sin wdt as m) = Xe‘<w~’(—;w,, sin wdt + W cos wdt) (E.3) When 1 = 0, Eq. (E3) gives x0 = de = XanI — {2 = (0.4550)(3.4338)\/1 — (0.4037)2 1.4294 m/s 180‘ = 0) ll \/ 2.92 A torsional pendulum has a natural frequency of 200 cycles/min when vibrating in a vacuum. The mass moment of inertia of the disc is 0.2 kg-mz. It is then immersed in oil and its natural frequency is found to be 180 cycles/min. Determine the damping constant. If the disc, when placed in oil, is given an initial displacement of 2°, find its displacement at the end of the first cycle. Uniform rigid bar, mass m FIGURE 2.96 2.106 The system shown in Fig. 2.97 has a natural frequency of 5 Hz for the following data: m = 10 kg, JO = 5 kg—mz, r1 — 10 cm, r2 = 25 cm. When the system is disturbed by giving it an initial displacement, the amplitude of free Vibration is reduced by 80 percent in 10 cycles. Determine the values of k and c. Pulley, mass moment of inertia JO FIGURE 2.97 — EXAMPLE 2.14 Pulley Subjected to Coulomb Damping A steel shaft of length 1 m and diameter 50 Inm is fixed at one end and carries a pulley of mass moment of inertia 25 kg-m2 at the other end. A band brake exerts a constant frictional torque of 400 N—m around the circumference of the pulley. If the pulley is displaced by 6° and released, determine (1) the number of cycles before the pulley comes to rest and (2) the final settling position of the pulley. Solution: (1) The number of half cycles that elapse before the angular motion of the pulley ceases is given by Eq. (2.121) r 2 (El) 164 CHAPTER 2 FREE VIBRATION OF SINGLE DEGREE 0F FREEDOM SYSTEMS where 00 = initial angular displacement = 6° = 0.10472 rad, k, = torsional spring constant of the shaft given by 10 l 4 (8 X 10 ){32(0.05) } k, l — 1 — 49,0875 N-m/rad and T = constant friction torque applied to the pulley = 400 N-m. Equation (E.1) gives 4 0.10472 — i— 49,0875 r 2 —— = 5.926 800 49,0875 Thus the motion ceases after six half cycles. (2) The angular displacement after six half cycles is given by Eq. (2.120). 400 0: . 42— x2— 0107 6 (49,0875 > = 0.006935 rad = 0.397340 Thus the pulley stops at 0.39734° from the equilibrium position on the same side of the initial displacement. ...
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