EG260-Harmonic excitation 2

EG260-Harmonic excitation 2 - A.K. Slone EG-260 Dynamics...

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Unformatted text preview: A.K. Slone EG-260 Dynamics (1) EG-260 DYNAMICS I – Harmonic Excitation 2 EG-260 DYNAMICS I – Harmonic Excitation 2 .......................................................... 1 1. Damped SDOF harmonically excited systems ................................................ 2 1. The particular solution .................................................................................... 3 2. The complete solution ..................................................................................... 7 2.1. Example 2.1 ..................................................................................................... 8 3. The transient and steady state responses. ...................................................... 11 3.1. Example 3.1 ................................................................................................... 16 4. Resonance in damped systems .................................................................. 18 5. Alternative solution methods ........................................................................ 21 5.1. The geometrical method. ............................................................................... 21 5.2. The frequency response method. ................................................................... 22 5.3. The transform method. .................................................................................. 26 6. Harmonic excitation of the base .................................................................... 28 6.1. Force transmitted to the mass ........................................................................ 35 6.2. Example 6.2 ................................................................................................... 36 7. Rotating imbalance ........................................................................................ 37 7.1. Force transmitted to the base ......................................................................... 41 7.2. Example 7.2 ................................................................................................... 42 7.3. Example 7.3 ................................................................................................... 45 8. Measurement. ................................................................................................ 49 8.1. Measurement devices .................................................................................... 49 ©a.k.slone 2010 1 of 1 A.K. Slone EG-260 Dynamics (1) 1. Damped SDOF harmonically excited systems x(t) k F(t) m fk fc F N c Friction-free surface mg Free body diagram Figure 1 Viscously damped SDOF spring-mass system In this document the case of a viscously damped SDOF system subject to harmonic excitation is considered. Using Newton’s Second Law of motion the equation of motion for the system shown in From Figure 1 is: m&& + cx + kx = F (t ) x & (1) For harmonic excitation the forcing term is assumed to be F (t ) = F0 cos (ωt ) thus the equation of motion is: m&& + cx + kx = F0 cos(ωt ) x & (2) Equation (2) may be divided by m to give: && + x ©a.k.slone 2010 c k & x + x = f 0 cos(ωt ) m m (3) 2 of 2 A.K. Slone where EG-260 Dynamics (1) f0 = F0 m . From the formulae for the natural frequency and the m c 2 damping ratio k = ωn and m = 2ζω n . Hence equation (3) may be expressed as : 2 && + 2ζω n x + ωn x = f 0 cos (ωt ) & x (4) As with any second order differential equation the solution to the non-homogenous equation (4) is formed by obtaining a solution to the homogenous equation and the particular solution and adding the two solutions together. 1. The particular solution The particular solution of equation (4) will be of a similar form to the right hand side, i.e. the forcing term. But here the frequency of the forced response will be of the same frequency as the forcing term but with different amplitude and phase. The phase shift is due to the effect of the damping force. Thus the particular solution is assumed to be of the form: x p (t ) = X cos(ωt − θ ) (5) Successive differentiation of equation (5) gives: ©a.k.slone 2010 3 of 3 A.K. Slone EG-260 Dynamics (1) & x p (t ) = − Xω sin(ωt − θ ) &&p (t ) = − Xω 2 cos(ωt − θ ) x and (6) (7) Substituting equations (5), (6) and (7) into equation (4) gives: 2 − Xω 2 cos(ωt − θ ) − X 2ζω nω sin(ωt − θ ) + Xω n cos(ωt − θ ) = f 0 cos ωt (8) Collecting terms: {( ) } 2 X ωn − ω 2 cos(ωt − θ ) − 2ζω nω sin(ωt − θ ) (9) = f 0 cos ωt Using the trigonometrical relationships: sin(ωt − θ ) = sin ωt cos θ − cos ωt sinθ cos(ωt − θ ) = cos ωt cos θ + sin ωt sinθ Equation (9) may be expressed as: {( ) } 2 X ω n − ω 2 (cos ωt cos θ + sin ωt sin θ ) − 2ζω n ω (sin ωt cos θ − cos ωt sin θ ) = f 0 cos ωt (10) Collecting cosine and sine terms equation (10) gives: [( [( ) ) 2 ⎧ ω n − ω 2 cos θ + 2ζω n ω sin θ cos ωt + ⎫ ⎪ ⎪ X⎨ 2 ⎬ 2 ⎪ ω n − ω sin θ − 2ζω n ω cos θ sin ωt ⎪ ⎩ ⎭ = f 0 cos ωt ©a.k.slone 2010 (11) 4 of 4 A.K. Slone EG-260 Dynamics (1) Comparing right-hand side and left-hand side terms gives: [( X [(ω ) − ω )sin θ − 2ζω ω cos θ ] = 0 2 X ω n − ω 2 cos θ + 2ζω n ω sin θ = f 0 2 n 2 n (12) (13) Equations (12) and (13) may be expressed in matrix form as: ( ) 2ζω n ω ⎤ ⎡ X cos θ ⎤ ⎡ f 0 ⎤ = ⎥ 2 ωn − ω 2 ⎦ ⎢ X sin θ ⎥ ⎢ 0 ⎥ ⎣ ⎦ ⎣ ⎦ 2 ⎡ ωn − ω 2 ⎢ ⎣ − 2ζω n ω ( ) (14) The matrix equation (14) may be solved using standard matrix methods, i.e., the solution of the matrix equation -1 Ax = b is x = A −1b where A is the inverse of matrix A. (note: This is only a 2x2 matrix so the inverse is easy to find and the equation may also be solved using simultaneous equation methods, but for more complex matrix equations packages such as MATLAB may be used.) Solving equation (14) for θ gives: ⎛ 2ζω n ω 2 2 ⎝ ωn − ω θ = tan −1 ⎜ ⎜ From equation (14) (ω ©a.k.slone 2010 2 n −ω2 ) 2 ( ) ⎞ ⎟ ⎟ ⎠ sin 2 = 2ζω n ω cos 2 θ (15) 2 (16) 5 of 5 A.K. Slone EG-260 Dynamics (1) 2 2 2 Adding (ωn − ω ) cos θ to both sides of equation (16) gives: 2 (ω − ω ) sin + (ω − ω ) cos θ = (2ζω ω ) cos θ + (ω − ω ) cos 2 2 2 n 2 2 2 2 2 n 2 2 2 2 n n 2 2 (17) θ Collecting terms gives: {(2ζω ω ) + (ω 2 2 n 2 n −ω2 ) }cos θ = (ω 2 2 2 n −ω2 ) (sin + cos θ ) 2 2 2 (18) Hence cos θ = Similarly adding (2ζω n ω ) (16) gives: 2 ((ω 2 n −ω2 ) + (2ζω ω ) )sin 2 2 2 n (ω (ω 2 n sin 2 θ −ω2 2 n −ω2 ) ) + (2ζω ω ) 2 2 (19) n to both sides of equation θ = (2ζω n ω ) (cos 2 θ + sin 2 θ ) 2 (20) Hence: sin θ = (ω 2ζω n ω 2 n − ω2 ) + (2ζω ω ) 2 (21) 2 n Substituting equations (19) and (21) into equation (12) gives: ⎡ X⎢ ⎢ ⎣ (ω 2 n −ω2 (2ζω ω ) + (ω ©a.k.slone 2010 2 n ) 2 2 n (2ζω ω ) (2ζω ω ) + (ω 2 −ω2 ) 2 + n 2 n 2 n −ω2 ) 2 ⎤ ⎥= f 0 ⎥ ⎦ (22) 6 of 6 A.K. Slone EG-260 Dynamics (1) Thus (ω X ) ( − ω 2 + 2ζωnω 2 2 n ) 2 = f0 (23) which on rearrangement gives the forcing term amplitude as: X= (ω 2 n ) f0 − ω 2 + (2ζω nω ) 2 2 (24) Substitution into equation (5) gives the particular solution as: x p (t ) = (ω f0 2 n −ω ) + (2ζω ω ) 2 2 2 cos(ωt − θ ) (25) n From equations (19) and (21) the forcing term phase angle is: ⎛ 2ζω nω ⎞ ⎟ 2 ωn − ω 2 ⎟ ⎝ ⎠ θ = tan −1 ⎜ ⎜ (26) 2. The complete solution For the under-damped case (0< ζ <1), the solution to the complementary function, i.e. the case of free vibration, is: x(t ) = Ae−ζω t sin(ωd t + φ ) n ©a.k.slone 2010 (27) 7 of 7 A.K. Slone EG-260 Dynamics (1) Thus the complete solution is: x(t ) = Ae −ζωnt sin(ω d t + φ ) + X cos (ωt − θ ) (28) where X and θ are given by equations (25) and (26) The solution of equation (28) requires the evaluation of the constants A and φ, which are determined from initial conditions. Successive differentiation of equation (28) gives: x(0 ) = A sin(φ ) + X cos(θ ) & x(0 ) = −ζω n A sin (φ ) + ωd A cos (φ ) + ωX sin (θ ) (29) (30) From which A and φ, may be determined from initial conditions. 2.1. Example 2.1 Compute the constant of integration for an undamped system subject to a harmonic excitation 1000 cos(5t ) and compare these values for the unforced case, given the parameters ω n = 10 rad/s, ζ = 0.01, m = 100 kg and the initial conditions xo = 0.05 m and v0 = 0 m/s. Solution f0 = 1000 = 10 ζ = 0.01 and ω = 5 rad/s 100 ©a.k.slone 2010 8 of 8 A.K. Slone EG-260 Dynamics (1) From equation (25) X = = (ω f0 2 n ) − ω 2 + (2ζω nω ) 10 2 2 + (100 − 25) + (2 × 0.01 × 10 × 5) X = 0.133 m 2 2 (31) From equation (26) the forcing term phase angle is: ⎛ 2ζω nω ⎞ ⎟ 2 ωn − ω 2 ⎟ ⎝ ⎠ θ = tan −1 ⎜ ⎜ ⎛ 2 × 0.01 × 10 × 5 ⎞ = tan −1 ⎜ ⎟ ⎝ 100 − 25 ⎠ (32) θ = tan −1 (0.01333) Thus: θ = 0.013 rad (33) The damped natural frequency is: ωd = ω n 1 − ζ 2 = 10 1 − 0.012 = 9.999 (34) Substituting into equation (28) gives: x(t ) = Ae −0.1t sin(9.999t + φ ) + 0.133 cos(5t − 0.013) (35) Differentiating gives: & x(t ) = −0.1Ae −0.1t sin(9.999t + φ ) + 9.999 Ae −0.1t cos(9.999t + φ ) − 0.665 sin(5t − 0.013) ©a.k.slone 2010 (36) 9 of 9 A.K. Slone EG-260 Dynamics (1) Setting t = 0 in equation (35) gives: x(0 ) = 0.05 = A sin(φ ) + 0.133 cos(0.013) Hence: A= (37) 0.05 − 0.133 cos (0.013) − 0.083 = sin(φ ) sin(φ ) (38) Setting t = 0 in equation (36) gives: & x(0) = −0.1A sin(φ ) + 9.999 A cos(φ ) (39) − 0.665 sin(− 0.013) Substituting for A from equation (38) gives: & x(0) = 0 = −0.1(− 0.083) + 9.999(− 0.083) cot (φ ) (40) + 0.665 sin(0.013) From equation (40) φ = 1.55 rad and so A = −0.083 m For the free vibration case F0 =0, the response for the under-damped case is: x(t ) = Ae −0.1t sin(9.999t + φ ) (41) Setting t =0 in equation (41) gives: x(0 ) = 0.05 = A sin(φ ) (42) Differentiating equation (41) gives; ©a.k.slone 2010 10 of 10 A.K. Slone EG-260 Dynamics (1) & x(t ) = −0.1Ae −0.1t sin(9.999t + φ ) + 9.99 Ae −0.1t cos(9.999t + φ ) (43) Setting t = 0 gives: & x(0 ) = 0 = −0.1A sin(φ ) + 9.99 A cos(φ ) (44) 9.99 = 0.999 0.1 φ = 1.56 rad (45) So tan(φ ) = From equation (42) 0.05 sin(φ ) A = 0.05 m A= (46) Comparing the values of A and φ in equations (45) and (46) for the case of free vibration with those when the system is subject to a driving force where φ = 1.55 rad and so A = −0.083 m, the effect of the driving force has been that the amplitude is greatly reduced and the phase angle is increased. 3. The transient and steady state responses. Equation (28) can be considered as a combination of a transient response and a steady state response, as shown below: ©a.k.slone 2010 11 of 11 A.K. Slone EG-260 Dynamics (1) x(t ) = Ae −ζωnt sin(ω d t + φ ) + X cos (ωt − θ ) 14 244 4 3 144 2444 4 3 transient response steady state response (47) If the system has relatively large damping, the term e −ζωnt causes the transient response to quickly die out, in some cases within a fraction of a second and so need not be considered. However, if the system is lightly damped, i.e. ζ very small, the transient term in the solution may last a sufficiently long time so that it could be significant. In a number of applications, the damping is sufficiently large that the transient term may be ignored. But there are a number of applications, such as earthquake analysis and satellite analysis where the transient term is significant. The transient response may also be significant if the amplitude is large. The usual practice in engineering is to design and analyse the system based on the steady state response alone. But in all cases the transient response must be checked to see if it is reasonable to ignore it. Assuming that the transient term may be ignored, the frequency ratio r is defined as r = ω , which is a ωn dimensionless quantity and may be used in equations (26) and (to give: ⎛ ω ⎞ ⎜ 2ζ n ω ⎟ 2 ⎛ 2ζω ω ⎞ θ = tan −1 ⎜ 2 n 2 ⎟ = tan −1 ⎜ 2 ω 2 ⎟ (48) ⎜ω −ω ⎟ ⎜ ωn ω ⎟ ⎝ n ⎠ ⎜ 2− 2⎟ ⎝ω ω ⎠ ©a.k.slone 2010 12 of 12 A.K. Slone EG-260 Dynamics (1) ⎛ 2ζr ⎞ 2 ⎟ ⎝1− r ⎠ θ = tan −1 ⎜ (49) and the forcing term amplitude is: X = (ω f0 2 n − ω2 ) + (2ζω ω ) 2 2 (50) n 2 Dividing top and bottom by ω n X = 2 f 0 ωn (51) (1 − r ) + (2ζr ) 2 2 2 Thus 2 Xω n Xk = = fo F0 1 (1 − r ) + (2ζr ) 2 2 2 (52) 2 Xω n Xk The term or in equation (52) is called the F0 f0 normalised amplitude and is plotted against the frequency ratio in Figure 2 for several values of the damping ratio ζ; similarly the phase of the steady state response is plotted against the frequency ratio in Figure 3 As the driving frequency ω approaches the undamped natural frequency ωn, i.e. r → 1 , the magnitude of the amplitude reaches a maximum for light values of damping, ( ζ ≤ 0.1) ©a.k.slone 2010 13 of 13 A.K. Slone EG-260 Dynamics (1) Figure 2 Normalised Amplitude - steady state damped system In Figure 3 as the driving frequency approaches the undamped natural frequency, i.e. the definition of resonance for the damped case, the phase shift crosses through π 2 • For ζ > 0 and 0 < r < 1, π the phase angle is given by 0 < θ < , implying the 2 response lags the excitation. • For ζ > 0 and r > 1, π the phase angle is given by < θ < π , implying the 2 response leads the excitation. • For ζ > 0 and r = 1, ©a.k.slone 2010 14 of 14 A.K. Slone EG-260 Dynamics (1) π the phase angle is θ = , implying the phase 2 difference between the excitation and the response is π 2 • For ζ > 0 and large value of r, the phase angle approaches π, implying that the excitation and the response are out of phase. Figure 3 Phase of steady state response for damped system It is important in a design phase to examine how the amplitude of the steady state vibration is affected by changing the damping ratio. Figure 4 shows the same information as in Figure 2 except that it is plotted on a log scale so that curves for ©a.k.slone 2010 15 of 15 A.K. Slone EG-260 Dynamics (1) both small and large values of damping may be shown on the same graph. Figure 4 shows that as the damping ratio ζ increases the peak in amplitude decreases until it disappears. Conversely as the damping ratio decreases the peak value increase and become much sharper. In the limit as ζ → 0 , the peak reaches an infinite value, which is consistent with the undamped response at resonance. Figure 4 Log scale magnitude of steady state response. 3.1. Example 3.1 A small spring 30 mm long is welded to a stationary table so that it is fixed at the point of contact, with a 12 ©a.k.slone 2010 16 of 16 A.K. Slone EG-260 Dynamics (1) mm bolt welded to the other end, which s free to move. The mass of the system is 49.2x10-3 kg, the spring constant is measured to be 857.8 N/m and the damping rate of the spring is measured to be 0.11 kg/s. The spring is initially deflected 10mm. Calculate the steady state response given ω = 132 rad/s, f0 = 10 N/kg. Calculate the change in amplitude if ω = 125 rad/s Solution Using the data given the undamped natural frequency and damping ratio are: k = 857.8 m = 49.2 × 10 −3 ∴ ωn = 857.8 = 132.04 rad/s 49.2 × 10 −3 c 0.11 = 2 mk 2 857.8 × 49.2 × 10 −3 ζ = 0.0085 c = 0.11 ζ = The magnitude of the particular solution is x p (t ) = X , the amplitude of the steady state response, which from equation ( 25) is: ©a.k.slone 2010 17 of 17 A.K. Slone X= EG-260 Dynamics (1) f0 (ω −ω ) + (2ζω ω) 2 n = 2 2 2 n 10 2 −132 ⎞2 + ⎛ 2× 0.0085 13204×132⎞2 2⎟ ⎜ ⎟ 13204 . × . ⎟ ⎝ ⎠ ⎟ ⎛ ⎜ ⎜ ⎜ ⎝ (53) ⎠ X = 0.0337m If the driving frequency is changed to 125 rad/s, the amplitude is: X = (132.04 2 ) 10 − 1252 + (2 × 0.085 × 132.04 × 125) 2 2 (54) X = 0.005 m Comparing the two values of the amplitude of the steady state response, given by equations (53) and (54), shows that a slight change in the driving frequency, from one near resonance at 132 rad/s to driving frequency of 125 rad/s, causes an order of magnitude change in the amplitude of the steady state response. 4. Resonance in damped systems Resonance is defined to occur if ω = ωn , i.e. when the driving frequency is equal to the undamped natural frequency, which corresponds to a phase shift of π rads. 2 This definition, which is equivalent to r = 1 or ω = ωn is frequently applied to damped systems, but Figure 4 ©a.k.slone 2010 18 of 18 A.K. Slone EG-260 Dynamics (1) shows that the maximum value of Xk occurs at different F0 values of r depending on the value of the damping ratio ζ. • If ζ > 1 2 Maximum value of Xk at r = 0 F0 1 • If 0 ≤ ζ ≤ 2 = 0.707 Maximum value of Xk at r = 1 − 2ζ 2 F0 The value of driving frequency corresponding to the maximum value of Xk is called the driving frequency, F0 ω p where: ω p = ωn 1 − 2ζ 2 Hence as damping decreases ω p → ωn , this gives the undamped resonance condition. As ζ increases, i.e. as the damping increases, the curves in Figure 4 have peaks that are increasingly further away from the line r = 1. Eventually the damping increases so that ζ > 1 and the maximum value of 2 Xk occurs at F0 r = 0. ©a.k.slone 2010 19 of 19 A.K. Slone EG-260 Dynamics (1) In many practical situations there is only light damping so that ζ is small such that 1 − 2ζ 2 → 1 and so ω p = ωn 1 − 2ζ 2 → ωn , i.e. r = 1. So that, in lightly damped cases, the undamped resonance condition ω = ωn is often used. Figure 5 shows the change in peak frequency with the damping ratio. As the damping increases in the range 0≥ζ ≤ 1 = 0.707 , the size of the peak frequency ratio 2 decreases more sharply as ζ approaches 0.707. 1 Figure 5 Peak Amplitude for ζ ≤ 2 Figure 6 shows on a log scale shows how the peak response given by equation (52) decreases as damping increases, where it may be seen that the peak value changes in orders of magnitude as the damping increases ©a.k.slone 2010 20 of 20 A.K. Slone EG-260 Dynamics (1) Figure 6 Normalise Amplitude versus ζ 5. Alternative solution methods So far we have only considered one method for solving the equations of motion for a harmonically excited SDOF spring-mass-damper system, the method of undetermined coefficients. Alternative methods, such as a geometric approach, a frequency approach and a transform approach are described below. 5.1. The geometrical method. The geometrical approach uses the fact that position, velocity and acceleration will be out of phase with each other by π/2 radians. Therefore express displacement, velocity and acceleration as a vector. Compute X in terms of F0 via vector addition This method is of mathematical rather than practical interest to the practicing engineer. If you are interested in further explanation see the section on complex vector representation of harmonic motion in Rao1 1 Mechanical Vibrations, S. S. Rao, ISBN 0-13-196751-7 ©a.k.slone 2010 21 of 21 A.K. Slone EG-260 Dynamics (1) 5.2. The frequency response method. If the forcing function F (t )for a harmonically excited system is expressed in complex form as: F (t ) = F0 e jωt (55) Then the equation of motion for the spring-mass-damper system in Figure 7 may be expressed as: m&& + cx + kx = F0 e jωt x & x(t) k (56) fk mg Mass, m c Friction-free surface N Figure 7 Spring-mass-damper SDOF system But the excitation is real, so the response will also be the real part of x(t), where x(t) is a complex function that by definition must satisfy the equation of motion as given by equation (56) The particular solution must also be a complex number as it is of the same form as the right hand side of equation (56), thus: ©a.k.slone 2010 22 of 22 A.K. Slone EG-260 Dynamics (1) x p (t ) = Xe jωt (57) Differentiating gives: & x p (t ) = Xjωe jωt (58) && p (t ) = − Xω 2 e jωt x (59) Substituting equations (57) (58) and (59) into equation (56) gives: − mω 2 Xe jωt + jcωXe jωt + kXe jωt = F0 e jωt (60) Collecting terms and simplifying gives: {( ) } X k − mω 2 + jcω = F0 (61) Hence X= F0 (k − mω ) + jcω 2 (62) X = H ( jω )F0 (63) and where the frequency response function, H(jω) is: 1 H ( jω ) = (k − mω 2 ) + jcω ©a.k.slone 2010 (64) 23 of 23 A.K. Slone EG-260 Dynamics (1) Using standard methods for complex number manipulation, multiply the numerator and denominator of the right hand side (RHS) of equation (62) by the complex conjugate of the RHS denominator gives: ( ( ) ( ) ) F0 ((k − mω 2 ) − cjω ) X= 2 2 (k − mω ) + (cω )2 ⎛ ⎞⎛ k − mω 2 − cjω ⎞ F0 ⎟⎜ ⎟ X =⎜ 2 2 ⎜ k − mω + cjω ⎟⎜ k − mω − cjω ⎟ ⎝ ⎠⎝ ⎠ (65) (66) Separating the real and imaginary parts gives: ⎡ ⎤ k − mω 2 cω −i X = F0 ⎢ ⎥ 2 2 2 2 k − mω 2 + (cω ) ⎥ ⎢ k − mω 2 + (cω ) ⎣ ⎦ ( ) ( ) (67) Using the relationship: α − βj = Ae− jθ (68) where A= α2 + β2 tan(θ ) = β α (69) Let α= k − mω 2 (k − mω ) 2 2 + (cω ) 2 β= cω (k − mω ) 2 2 + (cω ) 2 (70) Then ©a.k.slone 2010 24 of 24 A.K. Slone EG-260 Dynamics (1) A= (k − mω ) + (cω ) {(k − mω ) + (cω ) } 2 2 2 2 2 2 (71) 2 Thus A= 1 (k − mω ) 2 2 + (cω ) 2 (72) Hence the complex number: (k − mω 2 ) − j (cω ) 1 = e − jθ (k − mω 2 )2 + (cω )2 (k − mω 2 )2 + (cω )2 (k − mω 2 )2 + (cω )2 (73) Substituting equation (73) into equation (67) gives: F0 X= (k − mω 2 )2 + (cω )2 e − jθ (74) Substituting into equation (57) gives the particular solution: ⎡ ⎤ F0 ⎢ ⎥ x p (t ) = ⎢ e − jθ ⎥ e jωt 2 ⎢ k − mω 2 + (cω )2 ⎥ ⎣ ⎦ F0 x p (t ) = e j (ωt −θ ) ( ) (75) (k − mω 2 )2 + (cω )2 ©a.k.slone 2010 25 of 25 A.K. Slone EG-260 Dynamics (1) Using the Euler relationship: eαj = cos(α ) + j sin(α ) (76) Equation (75) may be expressed as x p (t ) = F0 (k − mω ) 2 2 + (cω )2 (cos(ωt − θ ) + j sin(ωt − θ )) (77) Comparing equation (77) with equation (25) it can be seen that considering only the real part of the solution as given by equation (77), both methods yield the same result. The frequency response approach is very similar to the eigenvalue approach that is commonly used in aerodynamics. 5.3. The transform method. Consider the equation for a spring-mass damper system subject to harmonic excitation. & m&& + cx + kx = F0 cos ωt x So far we have only considered solutions for SDOF systems. An alternative method is the Laplace transform method, which is also applicable to multi-degree-pf freedom, (MDOF) systems. ©a.k.slone 2010 26 of 26 A.K. Slone EG-260 Dynamics (1) The equation of motion may be transformed using the Laplace transform: ∞ X ( s ) = L( x( t )) = ∫ x( t )e − st dt (78) 0 To give: ( ms 2 + cs + k ) X ( s ) = F0 s s2 + ω 2 (79) Equation (79) is an algebraic equation in s, which may be solved for X(s) to give: X( s ) = F0 s ( ms 2 + cs + k )( s 2 + ω 2 ) (80) Then an inverse transformation must be performed to obtain the solution in the time domain. For zero initial conditions the transformation gives: ( ms 2 + cs + k ) X ( s ) = F ( s ) X( s ) 1 = 2 = H (s ) F ( s ) ms + cs + k (81) (82) 1 where H ( s ) = ms 2 + cs + k ©a.k.slone 2010 27 of 27 A.K. Slone EG-260 Dynamics (1) But this is directly comparable to the frequency response function, as given in equation (64) with (jω) = s 6. Harmonic excitation of the base Figure 8 shows a spring- mass-damper SDOF system undergoing harmonic excitation of the base. The displacement of the base is denoted by y(t) and the displacement of the mass from its static equilibrium position at time t is denoted by x(t). +x m k c +y y(t)= Ysin(ωt) Figure 8 Base excitation spring-mass-damper SDOF system The net elongation from the equilibrium position is (x –y) and the relative velocity between the two ends of the & & damper is ( x − y ) . ©a.k.slone 2010 28 of 28 A.K. Slone EG-260 Dynamics (1) From Figure 9 the equation of motion is: & & m&& + c( x − y ) + k (x − y ) = 0 x (83) + && x +x M k (x − y ) & & c( x − y ) Figure 9 Free body diagram If the base is moving under harmonic motion such that y(t ) = Y sin (ω t ) then equation (83) becomes: & & m&& + cx + kx = ky + cy x ⇒ & m&& + cx + kx = kY sin (ωb t ) + cωbY cos (ωb t ) x ⇒ 2 2 && + 2ζωn x + ωn x = ωn Y sin (ωb t ) + 2ζωn ωbY cos (ωb t ) & x (84) One interpretation of equation (84) is that system is acted upon by two harmonic forces kY sin (ωt ) and cω cos(ωt ) ©a.k.slone 2010 29 of 29 A.K. Slone EG-260 Dynamics (1) From Equation (24) the magnitude of the particular solution for a cosine excitation is: X= (ω 2 n −ω2 ) f0 2 + (2ζω nω )2 (24) Substituting f 0 = 2ζω n ω b in equation (24) gives the particular solution for the cosine forcing function: 2ζω nωbY x1 (t ) = cos(ωbt − θ1 ) p (85) 2 2 2 2 ωn − ωb + (2ζω nωb ) ( ) ⎛ 2ζω n ω b ⎜ ω 2 −ω 2 b ⎝ n θ 1 = tan −1 ⎜ ⎞ ⎟ ⎟ ⎠ To calculate the particular response to the sine forcing function, the same method of undetermined coefficients is used. Applying the procedures used to calculate 2 equation (24) for the harmonic input ωn Y sin (ωbt ) gives: x2 p (t ) = ( 2 ωn Y ) + (2ζω ω ) 2 2 2 ωn − ωb n b 2 sin (ωbt − θ1 ) (86) Here the particular solution is assumed to be of the form: x 2 (t ) = X sin (ωbt − θ1 ) note that the phase angle is the p same as in equation (85). ©a.k.slone 2010 30 of 30 A.K. Slone EG-260 Dynamics (1) By linear superposition the two particular solutions may 1 2 be added i.e. x p (t ) = x p (t ) + x p (t ) solution: x p (t ) = ω n Y ( 2 ω n + (2ζω n ω b )2 2 ωn ) 2 2 − ωb + (2ζω n ω b ) to give the particular cos(ω b t − θ 1 − θ 2 ) 2 ⎛ ω n θ 2 = tan ⎜ 2ζω b ⎝ −1 ⎜ Where (87) ⎞ ⎟ ⎟ ⎠ f = F0 was the amplitude of the driving force of the syste m, but here the amplitude of the harmonic excitation is A, by equation (Error! Reference source not found., thus the magnitude of the harmonic excitation is: Y k 2 + (cω ) 2 X1 = (k − mω ) + (cω ) 2 2 2 (88) Equation (15) gives the corresponding phase angle φ1. ⎛ cω ⎞ ⎟ ⎝ k − mω 2 ⎠ φ1 =tan −1 ⎜ (89) Thus the particular solution for base excitation is ©a.k.slone 2010 31 of 31 A.K. Slone EG-260 Dynamics (1) x p (t ) = Y k 2 + (cω ) 2 sin(ωt − φ1 − α ) (k − mω ) + (cω ) 2 2 2 (90) where φ1 and α are given by equations (89) and (Error! Reference source not found. respectively. Using trigonometrical identities Equation (90) may be written in a more simple form as: x p (t ) = X sin(ωt − φ ) (91) where k 2 + (cω ) 2 X = Y (k − mω ) + (cω ) 2 2 (92) 2 and −1 ⎛ ⎜ ⎞ mcω 3 ⎟ φ = tan ⎜ 2 ⎟ 2 ⎝ k k − mω + (cω ) ⎠ Using ζ = c 2 mω n ( and ) r= (93) ω ω n equations (92) and (93) may be written as: X =Y 1 + (2ζr ) 2 (1 − r ) + (2ζr ) 2 2 2 (94) and ©a.k.slone 2010 32 of 32 A.K. Slone EG-260 Dynamics (1) ⎛ 2ζr 3 φ = tan ⎜ ⎜ 1 + 4ζ 2 − 1 r 2 ⎝ −1 ( ) ⎞ ⎟ ⎟ ⎠ (95) Equation (94) may be expressed as the ratio of the maximum response amplitude and the input displacement amplitude: X = Y 1 + (2ζr ) 2 (1 − r ) + (2ζr ) 2 2 2 (96) X Y = Td is called the displacement transmissibility and describes how motion is transmitted from the base to the mass of the system as function of the frequency ratio , r = ω ω n where ω is the frequency of the base. A plot of Td against the frequency ratio, r, is shown in Comparing Figure 10 with the plot of normalised magnitude for harmonic excitation given in Figure 4 it may be seen that the frequency ratio r in both is independent of the damping ratio. However, the shape of the two plots varies slightly due to the effect of the numerator having an extra term of 2ζr for base excitation, which is due to the load transmitted to the damper in base excitation. ©a.k.slone 2010 33 of 33 A.K. Slone EG-260 Dynamics (1) Figure 10 Displacement transmissibility as a function of frequency ratio From Figure 10 the following further observations may be made: For r = 0 Td ≡ 1 For small values of r, Td is close to 1 For r > 2 Td < 1 for all values of ζ For r = 2 Td = 1 for all values of ζ For r < 2 smaller values of ζ lead to larger values of Td ©a.k.slone 2010 34 of 34 A.K. Slone EG-260 Dynamics (1) For r > 2 smaller values of ζ lead to smaller values of Td For an undamped system ζ = 0 and r = 1, i.e. resonance, Td → ∞ then 6.1. Force transmitted to the mass For the system shown in Figure 8 a force is transmitted to the mass due to the reaction from the spring and dashpot. This force, F(t) is given by: & & F (t ) = k ( x − y ) + c( x − y ) (97) By Newton’s second law, this force must balance the inertial force, hence: & & F (t ) = k ( x − y ) + c(x − y ) = −m&& x (98) The steady state solution is: Fp (t ) = FT sin(ωt − φ ) (99) Where FT is the amplitude or maximum value of the force transmitted and is given by: FT = r2 kY ©a.k.slone 2010 1 + (2ζr ) 2 (1 − r ) + (2ζr ) 2 2 2 (100) 35 of 35 A.K. Slone EG-260 Dynamics (1) The ratio FT kY is the force transmissibility. From equation (99) is can be seen that the force transmitted is in phase with the motion of the mass x(t). Figure 11 shows the normalised force transmitted to the mass F versus the frequency ratio for differing damping T kY ratios showing how the normalised force varies with frequency ratio. Figure 11 Log Plot of force transmitted to the base 6.2. Example 6.2 A floor deflects 5 cm under the weight of a 2000-N machine resting on rubber mounts with ζ=0.01. If the ©a.k.slone 2010 36 of 36 A.K. Slone EG-260 Dynamics (1) floor is vibrating with amplitude of 0.2 cm very near this machine’s resonance, calculate the machine’s displacement amplitude and transmitted force. Solution Y = 2cm = 0.002m and the stiffness of the floor is: k= 2000 N = 40kN/m 0.05m Assume r ≈ 1 FT = kYr 2 2 1 + 4ζ 2 1 + (2ζ r ) ≈ kY 2 2ζ ( 1 − r 2 )2 + (2ζ r ) FT ≈ (40000 N/m )(0.002 m ) 1 + 4(0.01) = 4001 N 2(0.01) 2 7. Rotating imbalance Rotating imbalance is present in a variety of engineering applications having rotating parts such as gyros, cryocoolers, tyres and washing machines. The imbalance is due to irregularities in the distribution of the rotating component, which may cause substantial vibrations. A schematic of rotating imbalance of mass m0 a distance e , eccentricity, from the centre of rotation is shown in Figure 12 ©a.k.slone 2010 37 of 37 A.K. Slone EG-260 Dynamics (1) Guide Guide x(t) Machine of total mass m m0 e ωt r c k Rubber floor mounting modelled as a spring and damper Friction free surface Figure 12 Schematic of a machine with rotating imbalance A free body diagram is shown in Figure 13, where ωr is the frequency of the rotating machinery and m is the total mass including both machine and imbalance mass. ©a.k.slone 2010 38 of 38 A.K. Slone EG-260 Dynamics (1) a b Figure 13 Free body diagram of the imbalance (a), and machine (b) In this example the vertical direction is denoted by x, and summing the forces in the x direction for the rotating imbalance in Figure 13a gives: m0 (&& + &&r ) = − Fr x x (101) Summing the forces for the machine in Figure 13b gives: & (m − m0 )&& = Fr − cx − kx x (102) Adding equations (101) and (102) and rearranging gives: & m&& + m0 &&r + cx + kx = 0 x x (103) It is assumed that there is no motion in the horizontal direction due to the presence of the guides. ©a.k.slone 2010 39 of 39 A.K. Slone EG-260 Dynamics (1) x m0 ωrt e sin(ωrt) e θ y e cos(ωrt) Figure 14 Rotating of mass, m0 From Figure 14, the vertical, i.e. x, component of the motion of the mass is xr = e sin(ω r t ) (104) So that the acceleration is &&r = −eω r 2 sin(ω r t ) x (105) Substituting in equation (103) and rearranging gives: 2 m&& + cx + kx = m0 eω r sin(ω r t ) x & (106) Dividing through by total mass, m, gives: m eω && + 2ζω n x + ω x = 0 r sin(ω r t ) & x m 2 2 n ©a.k.slone 2010 (107) 40 of 40 A.K. Slone EG-260 Dynamics (1) From which it is apparent that both the mass of the imbalance and the total mass influence the forcing function. Equation (107) represents a SDOF oscillator with a harmonic forcing function. Thus we can use the methods and solutions found in the previous section, where: x p ( t ) = X sin( ω r t − φ ) (108) ω r Setting r = ω gives the magnitude and the phase of the n steady state motion of the mass m due to the rotating imbalance of mass m0 moe r2 X= m (1 − r 2 ) 2 + (2ζr )2 and r⎞ ⎛ 2ζ ⎟ 2 ⎝ 1− r ⎠ φ = tan −1⎜ (109) Thus the steady state solution is: mo e r2 r⎞ ⎛ 2ζ x p (t ) = sin(ωr t − φ ) where φ = tan −1 ⎜ ⎟ m (1 − r 2 ) 2 + (2ζr )2 ⎝ 1− r 2 ⎠ (110) 7.1. Force transmitted to the base The force transmitted to the base of the machine is: & F (t ) = kx + cx ©a.k.slone 2010 41 of 41 A.K. Slone EG-260 Dynamics (1) where the steady state displacement is given by equation (110). Differentiating equation (110) for velocity gives the force transmitted to base of machine as: me F (t ) = 0 m kr 2 (1 − r ) 2 2 + (2ζr ) 2 me sin (ωt ) + 0 m cω 2 r 2 (1 − r ) 2 2 + (2ζr ) cos(ωt ) 2 (111) Thus the magnitude of the force transmitted to the base is: r 2 k 2 + c 2ω 2 me F0 = 0 m (1 − r ) 2 2 + (2ζr )2 (112) which may be simplified as: F0 = m0eω 7.2. 2 1 + (2ζr )2 (1 − r ) + (2ζr ) 2 2 2 (113) Example 7.2 A machine with rotating balance, as shown in Figure 12 has a maximum measured deflection at resonance of 0.1 m. The damping ratio is estimated to be ζ = 0.05. From the manufacturing process the out of balance mass is estimated to be 10%. Estimate the eccentricity e and hence the approximate location of the imbalance mass. Also find what additional mass should be added ©a.k.slone 2010 42 of 42 A.K. Slone EG-260 Dynamics (1) uniformly to the system to reduce the deflection at resonance to 0.01 m ©a.k.slone 2010 43 of 43 A.K. Slone EG-260 Dynamics (1) Solution Let total mass be M kg then rotating imbalance mass is 0.1M kg. ζ = 0.05, at resonance X = 0.1 m At resonance r = 1 and using equation (109) me 1 mo e r2 = o X = m 2ζ m ( 1 − r 2 )2 + (2ζ r )2 0.1M 1 e M 2 × 0.05 e = 0. 1 m 0. 1 = (114) Now it is required to reduce the maximum amplitude to 0.01 m by means of an added mass say ∆m, again at resonance: mo e 1 mo e r2 = X = m 2ζ m ( 1 − r 2 )2 + (2ζ r )2 ⇒ 0.01 = ©a.k.slone 2010 0.1M 1 × 0.1 × M + ΔM 2 × 0.05 (115) 44 of 44 A.K. Slone EG-260 Dynamics (1) Thus rearranging gives: 0.1M M + ΔM M + ΔM = 10 M ΔM = 9 M 0.01 = (116) Hence, the total mass must be increased by a factor of 9 to reduce the maximum deflection to 0.1 m or 1 cm. Rotating imbalance is also important in helicopters and prop aeroplanes. The tail rotor of a helicopter is used to provide yaw control and torque balance. The main rotor provides lift and horizontal motion and in doing so introduces a clockwise thrust, which is overcome by the counter clockwise thrust of the tail motor. An out of balance rotor may cause damaging vibrations and limit the performance of the helicopter. 7.3. Example 7.3 Figure 15 shows a schematic of a helicopter tail section, where the stiffness is provided by tail section in the vertical direction, where k = 1 x 105 N/m and the corresponding mass is 60 kg. The tail rotor has mass of 20 kg. A 500 g mass is attached to one of the blades at a distance of 15 cm from the axis of rotation. Calculate the magnitude of the deflection of the tail section of the helicopter as the tail rotates at 1500 rpm, assuming a damping ratio of 0.01. At what rotor speed is ©a.k.slone 2010 45 of 45 A.K. Slone EG-260 Dynamics (1) deflection a deflection. maximum? Calculate the maximum Figure 15 Schematic of helicopter tail section Solution This problem only models the vertical vibration of the tail section. The rotor system may be modelled as a machine of total mass 20.5 kg attached to a spring of mass 60 kg, stiffness 1 x 105 N/m, as shown in Figure 16. m0 m m0 ωrt m X ms , k ms , k Figure 16 Vertical vibration of helicopter tail section ©a.k.slone 2010 46 of 46 A.K. Slone EG-260 Dynamics (1) From the problem statement the mass of the spring is 60 kg and the imbalance mass, m0 is 0.5 kg. The total mass of the tail rotor including the added mass is: m = 20 + 0.5 = 20.5 kg The damping ratio ζ is 0.01 and the rotator rotates at 157 rpm. In the example of significant spring mass, (see modelling and energy methods on blackboard), the natural frequency of the spring-mass system was found to be: ωn = m+ k ms 105 = 20.5 + 60 3 3 ωn = 46.69 rad/s (117) The frequency of the rotation is found from the rpm. 1500 × 2π rad/s 60 ωr = 157.6 rad/s ωr = (118) Thus the frequency ratio is: r= ωr 157.6 = ωn 46.69 r = 3.16 ©a.k.slone 2010 (119) 47 of 47 A.K. Slone EG-260 Dynamics (1) The magnitude of the oscillation is given by equation (109) as: mo e r2 X= m ( 1 − r 2 )2 + (2ζ r )2 0.5 × 0.15 3.16 2 = 2 20.5 ( 1 − 3.16 2 )2 + (2 × 0.01× 3.16 ) (120) = 0.004 m The maximum deflection will occur at resonance, i.e. r~1, so that: ωr = ωn = 49.69 rad/s 60 rpm 2π = 474.5 rpm ωr = 49.69 × (121) From equation (109) the magnitude of the oscillation for r=1 moe 1 m 2ζ 0.5 × 0.15 1 = 20.5 2 × 0.01 = 0.183m = 18.3 cm X= (122) 18.3 cm represents an unacceptably large deflection of the rotor; hence the tail rotor should not be allowed to rotate at 474.5 rpm ©a.k.slone 2010 48 of 48 A.K. Slone EG-260 Dynamics (1) 8. Measurement. Dynamics testing is a valuable complement to dynamics analysis, this is achieved through either: • Modal testing, in terms of - Natural frequencies, useful for selecting operational speeds of nearby machinery to avoid resonant conditions. - Damping - Mode shapes • Operational testing - In-service loads (often random) 8.1. Measurement devices Transducers are used to measure vibration. A transducer is device which converts mechanical vibration into an electric signal such as voltage proportional to acceleration. The size of the transducer is important and should not influence the structure’s dynamics through added mass or stiffness. In analytic models the mass of the transducer is included in the analysis. There are a number of different types of transducers: • Response Transducers. Two broad types based on either velocity or on displacement. Response transducers based on displacement may be further categorised as linear variable differential transformers (LVDT), rotary variable differential transformers (RVDT) and optical sensors. ©a.k.slone 2010 49 of 49 A.K. Slone EG-260 Dynamics (1) LVDT are based on a system of coils and magnets. They are often noisy and are best for lowfrequency, high-displacement application. Optical sensors are applicable where nocontact is required. Accelerometers. As the name implies they convert the acceleration into an electric signal. Accelerometers are based on piezoelectric elements and are spring-mass-damper systems; therefore internal resonance is a potential problem and has to be considered when constructing such a measurement device. A simple accelerometer is shown in Figure 17a • a) b) casing k + + m&& x Voltage m m y(t c k(x-y) & & c(x − y ) Figure 17 Schematic of accelerometer mounted on structure Figure 17b shows the forces acting on the mass m. Both the damper and the spring, by means of the casing, are connected to the ground hence the force ©a.k.slone 2010 50 of 50 A.K. Slone EG-260 Dynamics (1) k(x –y) is parallel to the damping force. Balance of forces gives: & & m&& = −k ( x − y ) − c(x − y ) x (123) Assuming the base is undergoing SHM such that: y (t ) = Y cos ωt (124) The relative motion of the accelerometer mass is defined as: z (t ) = x(t ) − y (t ) (125) Equation (123) may be rewritten in terms of z to give: & m&& + cz + kz = mω 2Y cos (ωt ) z (126) which is of the same form as equation (2), the only difference is that equation (126) is in terms of the relative displacement whereas equation (2) is in terms of the absolute displacement. Bearing this in mind, the solution to equation (126) will follow that for the preceding paragraphs. ©a.k.slone 2010 51 of 51 ...
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This note was uploaded on 05/18/2010 for the course ENGINEERIN EG 260 taught by Professor Stone during the Spring '10 term at Swansea UK.

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