Base excitation of dynamic systems

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Published on March 5, 2014

Author: SondiponAdhikari

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Base excitation of SDOF systems, dynamics due to rotating unbalance

2.4 Base Excitation • Important class of vibration analysis – Preventing excitations from passing from a vibrating base through its mount into a structure • Vibration isolation – Vibrations in your car – Satellite operation – Building under earthquake – Disk drives, etc. © Eng. Vib, 3rd Ed. 1/27 @ProfAdhikari, #EG260

FBD of SDOF Base Excitation System Sketch x(t) m k y(t) System FBD m c   k ( x − y ) c(x − y) base   x ∑ F =-k(x-y)-c(x-y)=m  m x + kx = cy + ky x+c  © Eng. Vib, 3rd Ed. 2/27 College of Engineering (2.61)

SDOF Base Excitation (cont) Assume: y(t) = Y sin(ω t) and plug into Equation(2.61) m x + kx = cω + kY sin(ω t) (2.63) x+c  Y cos(ω t)     harmonic forcing functions For a car, ω= 2π 2πV = τ λ The steady-state solution is just the superposition of the two individual particular solutions (system is linear). f0 s     2 2  ζω n x + ω n x = 2ζω nω Y cos(ω t) + ω n Y sin(ω t)  x+2      f0 c © Eng. Vib, 3rd Ed. 3/27 College of Engineering (2.64)

Particular Solution (sine term) With a sine for the forcing function, 2  ζω n x + ω n x =f0 s sin ω t  x+2 x ps = As cos ω t + Bs sin ω t = X s sin(ω t − φ s ) where As = Bs = © Eng. Vib, 3rd Ed. 4/27 −2ζω nω f0 s (ω − ω ) + ( 2ζω nω ) 2 n 2 2 2 (ω − ω ) f0 s 2 n 2 2 2 (ω − ω ) + ( 2ζω nω ) 2 n 2 College of Engineering Use rectangular form to make it easier to add the cos term

Particular Solution (cos term) With a cosine for the forcing function, we showed 2  ζω n x + ω n x =f0c cos ω t  x+2 x pc = Ac cosω t + Bc sin ω t = Xc cos(ω t − φ c ) where Ac = Bc = © Eng. Vib, 3rd Ed. 5/27 (ω − ω ) f0c 2 n 2 2 2 (ω − ω ) + ( 2ζω nω ) 2 n 2 2ζω nω f0c (ω − ω ) + ( 2ζω nω ) 2 n 2 2 2 College of Engineering

Magnitude X/Y Now add the sin and cos terms to get the magnitude of the full particular solution X = f 02c + f 02s (ω − ω ) + ( 2ζω nω ) 2 n 2 2 2 = ω nY (2ζω ) 2 + ω n2 (ω − ω ) + ( 2ζω nω ) 2 n 2 2 2 where f 0 c = 2 ζω nωY and f 0 s = ω n2Y if we define r = ω ωn this becomes X =Y (1 − r ) + ( 2ζ r ) X 1 + (2ζ r ) 2 = 2 2 2 Y (1 − r ) + ( 2ζ r ) © Eng. Vib, 3rd Ed. 6/27 1 + (2ζ r ) 2 College of Engineering 2 2 (2.71) 2 (2.70)

The relative magnitude plot of X/Y versus frequency ratio: Called the Displacement Transmissibility 40 ζ =0.01 ζ =0.1 ζ =0.3 ζ =0.7 30 X/Y (dB) 20 10 0 -10 -20 0 © Eng. Vib, 3rd Ed. 7/27 0.5 Figure 2.13 1 1.5 2 Frequency ratio r College of Engineering 2.5 3

From the plot of relative Displacement Transmissibility observe that: • X/Y is called Displacement Transmissibility Ratio • Potentially severe amplification at resonance • Attenuation for r > sqrt(2) Isolation Zone • If r < sqrt(2) transmissibility decreases with damping ratio Amplification Zone • If r >> 1 then transmissibility increases with damping ratio Xp~2Yξ /r © Eng. Vib, 3rd Ed. 8/27 College of Engineering

Next examine the Force Transmitted to the mass as a function of the frequency ratio   FT = −k(x − y) − c( x − y ) = m x From FBD At steady state, x(t) = X cos(ω t − φ ),  ω 2 X cos(ω t − φ ) so x=x(t) FT = mω X = kr X 2 m 2 FT k c y(t) © Eng. Vib, 3rd Ed. 9/27 College of Engineering base

Plot of Force Transmissibility (in dB) versus frequency ratio 40 ζ =0.01 ζ =0.1 F/kY (dB) 30 ζ =0.3 ζ =0.7 20 10 0 -10 -20 0 0.5 Figure 2.14 © Eng. Vib, 3rd Ed. 10/27 1 1.5 2 Frequency ratio r College of Engineering 2.5 3

Figure 2.15 Comparison between force and displacement transmissibility Force Transmissibility Displacement Transmissibility © Eng. Vib, 3rd Ed. 11/27 College of Engineering

Example 2.4.1: Effect of speed on the amplitude of car vibration © Eng. Vib, 3rd Ed. 12/27 College of Engineering

Model the road as a sinusoidal input to base motion of the car model Approximation of road surface: y(t) = (0.01 m)sin ω bt 1    hour   2π rad  ω b = v(km/hr)     = 0.2909v rad/s  0.006 km   3600 s   cycle  ω b (20km/hr) = 5.818 rad/s From the data given, determine the frequency and damping ratio of the car suspension: ωn = k = m c ζ= = 2 km 2 © Eng. Vib, 3rd Ed. 13/27 4 × 10 4 N/m = 6.303 rad/s ( ≈ 1 Hz) 1007 kg 2000 Ns/m (4 × 10 4 ) N/m (1007 kg ) College of Engineering = 0.158

From the input frequency, input amplitude, natural frequency and damping ratio use equation (2.70) to compute the amplitude of the response: r= ω b 5.818 = ω 6.303 1 + (2ζ r)2 X =Y (1 − r 2 )2 + (2ζ r)2 = (0.01 m ) 1 + [2(0.158)(0.923)] 2 (1 − (0.923) ) + (2 (0.158)(0.923)) 2 2 2 = 0.0319 m What happens as the car goes faster? See Table 2.1. © Eng. Vib, 3rd Ed. 14/27 College of Engineering

Example 2.4.2: Compute the force transmitted to a machine through base motion at resonance From (2.77) at r =1: 1/2 FT  1 + (2ζ )  = kY  (2ζ )2   2 ⇒ FT = kY 1 + 4ζ 2 2ζ From given m, c, and k: ζ = c 900 = ≅ 0.04 2 km 2 40, 000g 3000 From measured excitation Y = 0.001 m: kY (40, 000 N/m)(0.001 m) 2 FT = 1 + 4ζ = 1 + 4(0.04)2 = 501.6 N 2ζ 2(0.04) © Eng. Vib, 3rd Ed. 15/27 College of Engineering

2.5 Rotating Unbalance • • • • Gyros Cryo-coolers Tires Washing machines m0 e Machine of total mass m i.e. m0 ω rt included in m e = eccentricity mo = mass unbalance ω r t =rotation frequency © Eng. Vib, 3rd Ed. 16/27 College of Engineering k c

Rotating Unbalance (cont) Rx ω rt What force is imparted on the structure? Note it rotates with x component: m0 e θ Ry xr = e sin ω r t ⇒ a x = r = −eω r2 sin ω r t x From the dynamics, R 0 = es = es r = x m i −ω t a ω m iω n θ n x m− 2 o r 2 o r 2 o r 2 o r R 0 = ec = ec r = y m o −ωω a ωθm ot s s y m− © Eng. Vib, 3rd Ed. 17/27 College of Engineering

Rotating Unbalance (cont) The problem is now just like any other SDOF system with a harmonic excitation m0 eω r2 sin(ω r t)  + cx + kx = mo eω r2 sin ω r t mx  x(t) m k c © Eng. Vib, 3rd Ed. 18/27 (2.82) mo 2  or  + 2ζω n x + ω x = x eω r sin ω r t m 2 n Note the influences on the forcing function (we are assuming that the mass m is held in place in the y direction as indicated in Figure 2.18) College of Engineering

Rotating Unbalance (cont) • Just another SDOF oscillator with a harmonic forcing function • Expressed in terms of frequency ratio r x p (t ) = X sin(ω r t − φ ) (2.83) moe r2 X= m (1 − r 2 )2 + (2ζ r )2 (2.84)  2ζ r  φ = tan  2 1− r  (2.85) −1 © Eng. Vib, 3rd Ed. 19/27 College of Engineering

Figure 2.20: Displacement magnitude vs frequency caused by rotating unbalance © Eng. Vib, 3rd Ed. 20/27 College of Engineering

Example 2.5.1:Given the deflection at resonance (0.1m), ζ = 0.05 and a 10% out of balance, compute e and the amount of added mass needed to reduce the maximum amplitude to 0.01 m. At resonance r = 1 and mX 1 1 0.1 m 1 = = ⇒ 10 = = 10 ⇒ e = 0.1 m m0 e 2ζ 2(0.05) e 2ζ Now to compute the added mass, again at resonance; m X    = 10 m0  0.1 m  Use this to find Δm so that X is 0.01: m + ∆m  0.01 m  m + ∆m = 100 ⇒ ∆m = 9m   = 10 ⇒ m0  0.1 m  (0.1)m Here m0 is 10%m or 0.1m © Eng. Vib, 3rd Ed. 21/27 College of Engineering

Example 2.5.2 Helicopter rotor unbalance Given Fig 2.21 k = 1 × 10 5 N/m mtail = 60 kg mrot = 20 kg m0 = 0.5 kg ζ = 0.01 Fig 2.22 Compute the deflection at 1500 rpm and find the rotor speed at which the deflection is maximum © Eng. Vib, 3rd Ed. 22/27 College of Engineering

Example 2.5.2 Solution The rotating mass is 20 + 0.5 or 20.5. The stiffness is provided by the Tail section and the corresponding mass is that determined in the example of a heavy beam. So the system natural frequency is k ωn = = 33 m+ mtail 140 The frequency of rotation is 10 5 N/m = 53.72 rad/s 33 20.5 + 60 kg 140 rev min 2π rad ω r = 1500 rpm = 1500 = 157 rad/s min 60 s rev 157 rad/s ⇒ r= = 2.92 53.96 rad/s © Eng. Vib, 3rd Ed. 23/27 College of Engineering

Now compute the deflection at r = 2.91 and ζ =0.01 using eq (2.84) m0 e r2 X= m (1 − r 2 )2 + (2ζ r)2 (0.5 kg )(0.15 m ) = 34.64 kg (2.92 )2 2 2 (1 − (2.92) ) − (2(0.01)(2.92))2 = 0.002 m At around r = 1, the max deflection occurs: rad rev 60 s r = 1 ⇒ ω r = 53.72 rad/s = 53.72 = 515.1 rpm s 2π rad min At r = 1: m0 e 1 (0.5 kg )(0.15 m ) 1 X= = = 0.108 m or 10.8 cm meq 2ζ 34.34 kg 2(0.01) © Eng. Vib, 3rd Ed. 24/27 College of Engineering

2.6 Measurement Devices • A basic transducer used in vibration measurement is the accelerometer.   x • This device can be ∑ F = - k(x-y) - c(x-y) = m modeled using the ⇒ m = -c( x − y) - k(x − y)   x base equations (2.86) and (2.61) developed in the Here, y(t) is the measured previous section response of the structure © Eng. Vib, 3rd Ed. 25/27 College of Engineering

Base motion applied to measurement devices Let z(t) = x(t) − y(t) (2.87) : ⇒ 2 m + cz(t) + kz(t) = mω b Y cosω bt (2.88) z  Z r2 ⇒ = Y (1− r 2 )2 + (2ζ r)2 (2.90) Accelerometer and  2ζ r  θ = tan −1  2÷  1− r  (2.91) These equations should be familiar from base motion. Here they describe measurement! © Eng. Vib, 3rd Ed. 26/27 Strain Gauge College of Engineering

Magnitude and sensitivity plots for accelerometers. Effect of damping on proportionality constant Fig 2.27 Fig 2.26 Magnitude plot showing Regions of measurement In the accel region, output voltage is nearly proportional to displacement © Eng. Vib, 3rd Ed. 27/27 College of Engineering

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