Dynamic response to harmonic excitation

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Published on February 24, 2014

Author: SondiponAdhikari

Source: slideshare.net

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Response of dynamic systems to harmonic excitation is discussed. Single degree of freedom systems are considered. For general damped multi degree of freedom systems, see my book Structural Dynamic Analysis with Generalized Damping Models: Analysis (e.g., in Amazon http://buff.ly/NqwHEE)

Chapter 2 Response to Harmonic Excitation Introduces the important concept of resonance © Eng. Vib, 3rd Ed. 1/49 @ProfAdhikari, #EG260

2.1 Harmonic Excitation of Undamped Systems • Consider the usual spring mass damper system with applied force F(t)=F0cosωt • • ω is the driving frequency F0 is the magnitude of the applied force • We take c = 0 to start with Displacement x F=F0cosωt © Eng. Vib, 3rd Ed. 2/49 College of Engineering M k

Equations of motion Figure 2.1 • Solution is the sum of homogenous and particular solution • The particular solution assumes form of forcing m = −kx(t) + F0 cos(ω t) x(t) function (physically 2 the input wins):  + ω n x(t) = f0 cos(ω t) x(t) x p (t ) = X cos(ωt ) F0 where f0 = © Eng. Vib, 3rd Ed. 3/49 College of Engineering m , ωn = k m

Substitute particular solution into the equation of motion: x (t ) = X cos ( ωt ) p p x 2 ωn x p      2 −ω 2 X cosω t + ω n X cosω t = f0 cos ω t f0 solving yields: X = 2 ωn − ω 2 Thus the particular solution has the form: f0 x p (t ) = 2 cos(ωt ) 2 ωn − ω © Eng. Vib, 3rd Ed. 4/49 College of Engineering

Add particular and homogeneous solutions to get general solution: x(t) = particular     f0 A1 sin ω n t + A2 cos ω n t + 2 cosω t 2     ω −ω n homogeneous A1 and A2 are constants of integration. © Eng. Vib, 3rd Ed. 5/49 College of Engineering

Apply the initial conditions to evaluate the constants f0 f cos 0 = A2 + 2 0 2 = x0 2 ωn − ω 2 ωn − ω f ⇒ A2 = x0 − 2 0 2 ωn − ω x(0) = A1 sin 0 + A2 cos0 +  x(0) = ω n (A1 cos0 − A2 sin 0) − f0 sin 0 = ω n A1 = v0 2 2 ωn − ω ⇒ A1 = v0 ωn ⇒  v0 f0  f x(t) = sin ω n t +  x0 − 2 cos ω n t + 2 0 2 cos ω t 2÷ ωn ωn − ω  ωn − ω  © Eng. Vib, 3rd Ed. 6/49 College of Engineering (2.11)

Comparison of free and forced response • Sum of two harmonic terms of different frequency • Free response has amplitude and phase effected by forcing function • Our solution is not defined for ωn = ω because it produces division by 0. • If forcing frequency is close to natural frequency the amplitude of particular solution is very large © Eng. Vib, 3rd Ed. 7/49 College of Engineering

Response for m=100 kg, k=1000 N/m, F=100 N, ω = ωn +5 v0=0.1m/s and x0= -0.02 m. Displacement (x) 0.05 0 -0.05 0 2 4 6 Time (sec) 8 10 Note the obvious presence of two harmonic signals © Eng. Vib, 3rd Ed. 8/49 College of Engineering Go to code demo

What happens when ω is near ωn? x (t ) = 2 f0  ω − ω   ωn + ω  sin  n t ÷sin  t ÷ (2.13) 2 2 ωn − ω  2   2  When the drive frequency and natural frequency are close a beating phenomena occurs Displacement (x) 1 2 f0  ω −ω  sin  n t÷ 2 2 ωn − ω  2  0.5 0 Larger amplitude -0.5 -1 0 © Eng. Vib, 3rd Ed. 9/49 5 10 15 20 Time (sec) College of Engineering 25 30 Called beating

What happens when ω is ωn? x p (t ) = tX sin(ωt ) grows without bound substitute into eq. and solve for X X=  f x(t) = A1 sin ω t + A2 cos ω t + 0 t sin(ω t) 2ω f0 2ω When the drive frequency and natural frequency are the same the amplitude of the vibration grows without bounds. This is known as a resonance condition. The most important concept in Chapter 2! © Eng. Vib, 3rd Ed. 10/49 Displacement (x) 5 0 -5 0 5 10 15 Time (sec) College of Engineering 20 25 30

Example 2.1.1: Compute and plot the response for m=10 kg, k=1000 N/m, x0=0,v0=0.2 m/s, F=23 N, ω =2 ω n. k 1000 N/m ωn = = = 10 rad/s, ω = 2ω n = 20 rad/s m 10 kg F 23 N v0 0.2 m/s f0 = = = 2.3 N/kg, = = 0.02 m m 10 kg ω n 10 rad/s f0 2.3 N/kg = = −7.9667 ×10 −3 m 2 ω n − ω 2 (10 2 − 20 2 ) rad 2 / s2 Equation (2.11) then yields: x(t) = 0.02sin10t + 7.9667 ×10−3 (cos10t − cos20t) © Eng. Vib, 3rd Ed. 11/49 College of Engineering

Example 2.1.2 Given zero initial conditions a harmonic input of 10 Hz with 20 N magnitude and k= 2000 N/m, and measured response amplitude of 0.1m, compute the mass of the system. f0 cos20π t − cosω n t ) for zero initial conditions 2 2 ( ωn − ω  ωn − ω   ωn + ω  2 f0 trig identity ⇒ x(t) = 2 sin  t ÷sin  t÷ 2  2   2  ωn − ω    x(t) = 0.1 m ⇒ 2 f0 2(20 / m) = 0.1 ⇒ = 0.1 2 2 2 2000 − (20π ) ωn − ω m m = 0.45 kg © Eng. Vib, 3rd Ed. 12/49 ( ) College of Engineering

Example 2.1.3 Design a rectangular mount for a security camera. 0.02 x 0.02 m in cross section Compute l > 0.5 m so that the mount keeps the camera from vibrating more then 0.01 m of maximum amplitude under a wind load of 15 N at 10 Hz. The mass of the camera is 3 kg. © Eng. Vib, 3rd Ed. 13/49 College of Engineering

Solution:Modeling the mount and camera as a beam with a tip mass, and the wind as harmonic, the equation of motion becomes: 3EI m + x From strength of materials:  3 x(t) = F0 cosω t bh 3 I= 12 Thus the frequency expression is: 3Ebh 3 Ebh 3 2 ωn = = 3 12m 4m3 Here we are interested computing  that will make the amplitude less then 0.01m:  (a) 2 f0  < 0.01 ⇒  2 ωn − ω 2  (b)   © Eng. Vib, 3rd Ed. 14/49 − 0.01 < 2 f0 2 , for ωn − ω 2 < 0 2 ωn − ω 2 2 f0 2 < 0.01, for ωn − ω 2 > 0 2 ωn − ω 2 College of Engineering

Case (a) (assume aluminum for the material): 2 f0 Ebh3 2 −0.01 < 2 ⇒ 2 f 0 < 0.01ω 2 − 0.01ωn ⇒ 0.01ω 2 − 2 f 0 > 0.01 ωn − ω 2 4m3 Ebh3 ⇒  > 0.01 = 0.321 ⇒  > 0.6848 m 2 4m(0.01ω − 2 f 0 ) 3 Case (b): 2 f0 Ebh3 2 2 2 < 0.01 ⇒ 2 f 0 < 0.01ωn − 0.01ω ⇒ 2 f 0 + 0.01ω < 0.01 2 2 ωn − ω 4m3 Ebh3 ⇒  < 0.01 = 0.191 ⇒  < 0.576 m 2 4m(2 f 0 + 0.01ω ) 3 © Eng. Vib, 3rd Ed. 15/49 College of Engineering

Remembering the constraint that the length must be at least 0.5 m, (a) and (b) yield 0.5 <  < 0.576, or  > 0.6848 m Less material is usually desired, so chose case b, say  = 0.55 m. To check, note that 3Ebh3 2 2 2 ωn − ω = − ( 20π ) = 1742 > 0 3 12m Thus the case a condition is met. Next check the mass of the designed beam to insure it does not change the frequency. Note it is less then m. © Eng. Vib, 3rd Ed. 16/49 m = ρ bh = (2.7 × 103 )(0.55)(0.02)(0.02) = 0.594 kg College of Engineering

A harmonic force may also be represented by sine or a complex exponential. How does this change the solution? m + kx(t) = F0 sin ω t or  + ω n x(t) = f 0 sin ω t x(t) x(t) 2 (2.18) The particular solution then becomes a sine: x p (t ) = X sin ωt (2.19) Substitution of (2.19) into (2.18) yields: x p (t ) = f0 sin ωt 2 2 ωn − ω Solving for the homogenous solution and evaluating the constants yields v f0  f ω x(t ) = x0 cos ωnt +  0 − sin ωnt + 2 0 2 sin ωt (2.25) ÷ ωn ωn ωn2 − ω 2  ωn − ω  © Eng. Vib, 3rd Ed. 17/49 College of Engineering

Section 2.2 Harmonic Excitation of Damped Systems Extending resonance and response calculation to damped systems © Eng. Vib, 3rd Ed. 18/49 College of Engineering

2.2 Harmonic excitation of damped systems m + cx (t) + kx(t) = F0 cos ω t x(t)   + 2ζω n x (t) + ω x(t) = f0 cosω t  x(t) x p (t) = cos(ω) X   t − θ  2 n now includes a phase shift Displacement x F=F0cosωt k M c © Eng. Vib, 3rd Ed. 19/49 College of Engineering

Let xp have the form: x p (t) = As cos ω t + Bs sin ω t  Bs  2 2 −1 X = As + Bs , θ = tan  ÷  As   x p = −ω As sin ω t + ω Bs cosω t p = −ω As cosω t − ω Bs sin ω t x 2 2 Note that we are using the rectangular form, but we could use one of the other forms of the solution. © Eng. Vib, 3rd Ed. 20/49 College of Engineering

Substitute into the equations of motion (−ω As + 2ζωnω Bs + ω As − f 0 ) cos ωt 2 2 n + ( −ω Bs + 2ζωnω As + ω Bs ) sin ωt = 0 2 2 n for all time. Specifically for t = 0, 2π / ω ⇒ (ω − ω ) As + (2ζωnω ) Bs = f 0 2 n 2 (−2ζωnω ) As + (ω − ω ) Bs = 0 2 n © Eng. Vib, 3rd Ed. 21/49 2 College of Engineering

Write as a matrix equation: 2 (ωn − ω 2 ) 2ζωnω   As   f0   =  2 2   −2ζωnω (ωn − ω )   Bs   0  Solving for As and Bs: (ω − ω ) f0 As = 2 2 (ωn − ω ) + (2ζωnω ) 2 n 2 2 2 2ζωnω f0 Bs = 2 (ωn − ω 2 )2 + (2ζωnω )2 © Eng. Vib, 3rd Ed. 22/49 College of Engineering

Substitute the values of As and Bs into xp:  2ζω nω  x p (t) = cos(ω t − tan  2 ) 2÷ 2 2 2 2  ωn − ω  (ω n − ω ) + (2ζω nω )         f0 −1 θ X Add homogeneous and particular to get total solution: x(t) = Ae−ζω nt sin(ω d t + φ ) + cos(ω) X   t − θ      homogeneous or transient solution particular or steady state solution Note: that A and φ will not have the same values as in Ch 1, for the free response. Also as t gets large, transient dies out. © Eng. Vib, 3rd Ed. 23/49 College of Engineering

Things to notice about damped forced response • If ζ = 0, undamped equations result • Steady state solution prevails for large t • Often we ignore the transient term (how large is ζ, how long is t?) • Coefficients of transient terms (constants of integration) are effected by the initial conditions AND the forcing function • For underdamped systems at resonance, the amplitude is finite. © Eng. Vib, 3rd Ed. 24/49 College of Engineering

Example 2.2.1: ωn = 10 rad/s, ω = 5 rad/s, ζ = 0.01, F0= 1000 N, m = 100 kg, and the initial conditions x0 = 0.05 m and v0 = 0. Compare A and φ for forced and unforced case: Using the equations on slide 23: X = 0.133, θ = −0.013 x (t ) = Ae −0.1t sin(9.99t + φ ) + 0.133cos(5t − 0.013) Differentiating yields: v(t ) = − 0.01Ae −0.1t sin(9.999t + φ ) + 9.999 Ae −0.1t cos(9.999t + φ ) − 0.665sin(5t − 0.013) applying the intial conditions : A = − 0.083 (0.05), φ = 1.55 (1.561) © Eng. Vib, 3rd Ed. 25/49 College of Engineering x0 0 5 ( )= . 0 = sn + . 3c s−. 1) Ai φ 0 3o( 0 3 1 0 v0 0 −. 1 sn ( )= = 0 Ai φ 0 + . 9Aoφ 0 6 sn . 1 9 9 c s + . 5i 0 3 9 6 0 The numbers in ( ) are those obtained by incorrectly using the free response values

Proceeding with ignoring the transient • Always check to make sure the transient is not significant • For example, transients are very important in earthquakes • However, in many machine applications transients may be ignored © Eng. Vib, 3rd Ed. 26/49 College of Engineering

Proceeding with ignoring the transient Magnitude: X= f0 2 (ωn − ω 2 )2 + (2ζωnω )2 Frequency ratio: ω r= ωn Non dimensional Form: (2.39) 2 Xk Xωn 1 = = F0 f0 (1 − r 2 )2 + (2ζ r )2 Phase: © Eng. Vib, 3rd Ed. 27/49 ( θ = tan −1 2ζ r College of Engineering 1 − r2 ) (2.40)

Magnitude plot • • • Resonance is close to r=1 For ζ = 0, r =1 defines resonance As ζ grows resonance moves r <1, and X decreases The exact value of r, can be found from differentiating the magnitude X= 30 20 10 0 -10 -20 0 © Eng. Vib, 3rd Ed. 28/49 (1 − r 2 ) 2 + (2ζ r )2 40 X (dB) • 1 0.5 College of Engineering 1 r 1.5 Fig 2.7 2

Phase plot • • Resonance occurs at Φ = π/2 The phase changes more rapidly when the damping is small From low to high values of r the phase always changes by 1800 or π radians ( 1− r2 ) 3.5 ζ =0.01 ζ =0.1 ζ =0.3 ζ =0.5 ζ =1 3 2.5 Phase (rad) • θ = tan −1 2ζ r 2 1.5 1 0.5 0 0.5 College of Engineering 1 1.5 r © Eng. Vib, 3rd Ed. 29/49 0 Fig 2.7 2

Example 2.2.3 Compute max peak by differentiating: d  Xk  d  1  ÷=  dr  F0  dr  (1 − r 2 )2 + (2ζ r ) 2   ÷= 0 ⇒ ÷  rpeak = 1 − 2ζ < 1 ⇒ ζ < 1 / 2 (2.41)  Xk  1  ÷ = F0  max 2ζ 1 − ζ 2  (2.42) 2 © Eng. Vib, 3rd Ed. 30/49 College of Engineering

Effect of Damping on Peak Value 30 • The top plot shows how the peak value becomes very large when the damping level is small • The lower plot shows how the frequency at which the peak value occurs reduces with increased damping • Note that the peak value is only defined for values ζ<0.707 25 Xk (dB) F0 20 15 10 5 0 0 0.2 0.4 0 0.2 0.4 ζ 0.6 0.8 0.6 0.8 1 0.8 rpeak 0.6 0.4 0.2 0 © Eng. Vib, 3rd Ed. 31/49 College of Engineering ζ Fig 2.9

Section 2.3 Alternative Representations • A variety methods for solving differential equations • So far, we used the method of undetermined coefficients • Now we look at 2 alternatives: a frequency response approach a transform approach • These also give us some insight and additional useful tools. © Eng. Vib, 3rd Ed. 32/49 College of Engineering

2.3.2 Complex response method Ae jω t = t + (Asin ω t) j A cosω       real part (2.47) imaginary part jω t m + cx (t) + kx(t) = F0 e x(t)   (2.48) harmonic input Real part of this complex solution corresponds to the physical solution © Eng. Vib, 3rd Ed. 35/49 College of Engineering

Choose complex exponential as a solution x p (t) = Xe jω t (2.49) (−ω m + cjω + k)Xe 2 jω t = F0 e jω t F0 X= = H ( jω )F0 2 (k − mω ) + (cω ) j 1 H ( jω ) = 2 (k − mωj ) + (cω )   the frequency response function Note: These are all complex functions © Eng. Vib, 3rd Ed. 36/49 College of Engineering (2.50) (2.51) (2.52)

Using complex arithmetic: X= F0 (k − mω 2 )2 + (cω ) 2 e − jθ  cω  θ = tan  2 ÷  k − mω  F0 j (ω t −θ ) x p (t ) = e 2 2 2 (k − mω ) + (cω ) −1 Has real part = to previous solution © Eng. Vib, 3rd Ed. 37/49 College of Engineering (2.53) (2.54) (2.55)

Comments: • It is the real part of this complex solution that is physical • The approach is useful in more complicated problems © Eng. Vib, 3rd Ed. 38/49 College of Engineering

Example 2.3.1: Use the frequency response approach to compute the particular solution of an undamped system The equation of motion is written as 2 m + kx(t) = F0 e jωt ⇒  + ω n x(t) = f0 e jωt x(t) x(t) Let x p (t) = Xe jω t 2 ⇒ ( −ω 2 + ω n ) Xe jωt = f0 e jωt f0 ⇒X= 2 ωn − ω 2 ) ( © Eng. Vib, 3rd Ed. 39/49 College of Engineering

2.3.3 Transfer Function Method The Laplace Transform • Changes ODE into algebraic equation • Solve algebraic equation then compute the inverse transform • Rule and table based in many cases • Is used extensively in control analysis to examine the response • Related to the frequency response function © Eng. Vib, 3rd Ed. 40/49 College of Engineering

The Laplace Transform approach: • See appendix B and section 3.4 for details • Transforms the time variable into an algebraic, complex variable • Transforms differential equations into an algebraic equation • Related to the frequency response method ∞ X(s) = L(x(t)) = ∫ x(t)e dt 0 © Eng. Vib, 3rd Ed. 41/49 College of Engineering − st

Take the transform of the equation of motion: m + cx + kx = F0 cosω t ⇒ x  F0 s 2 (ms + cs + k)X(s) = 2 s +ω2 Now solve algebraic equation in s for X(s) F0 s X (s) = (ms 2 + cs + k )(s 2 + ω 2 ) To get the time response this must be “inverse transformed” © Eng. Vib, 3rd Ed. 42/49 College of Engineering

Transfer Function Method With zero initial conditions: 2 (ms + cs + k ) X (s) = F (s) ⇒ The transfer X ( s) 1 function = H ( s) = 2 F ( s) ms + cs + k (2.59) 1 H ( jω ) = 2 (2.60) k − mω + cω j = frequency response function © Eng. Vib, 3rd Ed. 43/49 College of Engineering

Example 2.3.2 Compute forced response of the suspension system shown using the Laplace transform Summing moments about the shaft:  Jθ (t) + kθ (t) = aF sin ω t 0 Taking the Laplace transform: ω s2 + ω 2 1 ⇒ X ( s) = aω F0 2 ( s + ω 2 ) ( Js 2 + k ) Js 2 X (s) + kX (s) = aF0 Taking the inverse Laplace transform:   1 ÷ θ (t ) = aω F0 L  2 2 2  ( s + ω ) ( Js + k ) ÷    aω F 1 1 1 k ⇒ θ (t ) = sin ωt − sin ωnt ÷, ωn = 2  J ω 2 − ωn  ω ωn J  −1 © Eng. Vib, 3rd Ed. 44/49 College of Engineering

Notes on Phase for Homogeneous and Particular Solutions • Equation (2.37) gives the full solution for a harmonically driven underdamped SDOF oscillator to be x(t) = Ae sin ( ω d t + θ ) + X cos ( ω t − φ )         −ζω nt homogeneous solution particular solution How do we interpret these phase angles? Why is one added and the other subtracted? © Eng. Vib, 3rd Ed. 45/49 College of Engineering

Non-Zero initial conditions x0 ≠ 0 v0 ≠ 0 sin (ω d t + θ ) xoω d θ = tan vo + ζω n xo −1 © Eng. Vib, 3rd Ed. 46/49 College of Engineering

Zero initial displacement © Eng. Vib, 3rd Ed. 47/49 College of Engineering

Zero initial velocity © Eng. Vib, 3rd Ed. 48/49 College of Engineering

Phase on Particular Solution • Simple “atan” gives -π/2 < Φ < π/2 • Four-quadrant “atan2” gives 0 < Φ < π © Eng. Vib, 3rd Ed. 49/49 College of Engineering

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