Review Vibrations of SDOF Mechanical Systems

ME459/659 Notes# 1 A Review to the Dynamic Response of SDOF Second Order Mechanical System with Viscous Damping

by Luis San Andrés Mast-Childs Chair Professor

ME 459/659 Sound & Vibration Measurements

Reproduced from material in ME 363 and ME617

http://rotorlab.tamu.edu

ME459/659 Notes# 1a (pp. 1-31)

Dynamic Response of SDOF Second Order Mechanical System: Viscous Damping

d2 X d X MDKXF   d t2 d t ext() t

Free Response to F(t) = 0 + initial conditions and Underdamped, Critically Damped and Overdamped Systems

Forced Response to a Step Loading F(t) = Fo

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Second Order Mechanical Translational System:

Fundamental equation of motion about equilibrium position (X=0)

dX2 FMFFF     Xdt2 ext() t D K dX FD : Viscous Damping Force D dt

FKXk  : Elastic restoring Force

( M, D ,K ) represent the system equivalent mass, viscous damping coefficient, and stiffness coefficient, respectively.

d2 X d X  equation of motion MDKXF   ext() t d t2 d t

+ Initial Conditions in velocity and displacement; at t=0:

X(0) Xoo and X (0) V

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Second Order Mechanical Torsional System:

Fundamental equation of motion about equilibrium position, θ=0

d  Torques I  Text() t  T D  T K ;    dt

Tθ D = D  : Viscous dissipation torque

T θ K = K  : Elastic restoring torque

( I, D ,K ) are the system equivalent mass moment of inertia, rotational viscous damping coefficient, and rotational (torsional) stiffness coefficient, respectively.

dd2  Equation of motion  IDKT2    ext() t d t d t + Initial Conditions in angular velocity and displacement at t=0:

(0) ooand  (0)  

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(a) Free Response of Second Order SDOF Mechanical System

Let the external force Fext=0 and the system has an initial displacement Xo and initial velocity Vo. EOM is

d2 X d X MDKX  0 d t2 d t (1)

Divide Eq. (1) by M and define:

  K n M : natural frequency of system D   : viscous damping ratio, Dcr where DKMcr 2 is known as the critical damping magnitude.

With these definitions, Eqn. (1) 

d2 X d X 20   2 X  d t2 nn d t (2)

The solution of the Homogeneous Second Order Ordinary Differential Equation with Constant Coefficients is of the form:

st X() t Ae (3) where A is a constant found from the initial conditions.

Substitute Eq. (3) into Eq. (2) and obtain:

22 s20 nn s    A  (4)

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A is not zero for a non trivial solution. Thus, Eq. (4) leads to the CHARACTERISTIC EQUATION of the SDOF system:

22 ss20 nn     (5)

The roots of this 2nd order polynomial are:

2 1/ 2 s1,2  nn   1 (6)

The nature of the roots (eigenvalues) depends on the damping ratio  (>1 or < 1). Since there are two roots, the solution is

s12 t s t X() t A12 e A e (7)

A1, A2 determined from the initial conditions in displacement and velocity.

From Eq. (6), differentiate three cases:

Underdamped System: 0 < < 1,  D < Dcr

Critically Damped System: = 1,  D = Dcr

Overdamped System: > 1,  D > Dcr

Note that 1 n  has units of time; and for practical purposes, it is akin to an equivalent time constant for the second order system.

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Free Response of Undamped 2nd Order System

For an undamped system,  = 0, i.e., a conservative system without viscous dissipation, the roots of the characteristic equation are imaginary:

s12 inn; s  i (8) where i 1

Using the complex identity eiat = cos(at) + i sin(at), renders the undamped response as:

X( t ) C12 cosnn t C sin t (9.a)

  K where n M is the natural frequency of the system.

At time t = 0, apply the initial conditions to obtain

V0 CXC1 0and 2 (9.b) n

Eq. (9.a) can be written as:

X( t ) XMn cos t  (9.c)

2 2 V0 V0 where XXM 0 2 and tan  n n X 0

XM is the maximum amplitude response.

Notes: In a purely conservative system ( = 0), the motion never dies. Motion always oscillates about the equilibrium position X = 0

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Free Response of Underdamped 2nd Order System For an underdamped system, 0 <  < 1, the roots are complex conjugate (real and imaginary parts), i.e.

2 1/ 2 si1,2  nn 1    (10)

Using the complex identity eiat = cos(at) + i sin(at), the response is:

n t X( t ) e C12 cosdd t C sin t (11)

1/ 2 where 2 is the system damped natural frequency. dn 1  

At time t = 0, applying initial conditions gives

VX00n CXC1 0and 2 (11.b) d

Eqn. (11) can be written as:

n t X( t ) e XMd cos t  (11.c)

22 C2 where XCCM 12 and tan  C1

Note that as t , X(t)  0, i.e. the equilibrium position only if  > 0;

and XM is the largest amplitude of response only if =0 (no damping).

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Free Response of Underdamped 2nd Order System: initial displacement only damping ratio varies

Xo = 1, Vo = 0, ωn = 1.0 rad/s  = 0, 0.1, 0.25

Motion decays exponentially for > 0

Faster system response as increases, i.e. faster decay towards equilibrium position X=0

Free response Xo=1, Vo=0, wn=1 rad/s damping ratio=0.0 1.5 damping ratio=0.1 damping ratio=0.25 1

0.5

X(t) 0 10 20 30 40 -0.5

-1

-1.5 time (sec)

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Free Response of Underdamper 2nd Order System: Initial velocity only damping varies

Xo = 0, Vo = 1.0 ωn = 1.0 rad/s;  = 0, 0.1, 0.25

Motion decays exponentially for > 0

Faster system response as increases, i.e. faster decay towards equilibrium position X=0 Note the initial overshoot

Free response Xo=0, Vo=1, wn=1 rad/s damping ratio=0.0 1.5 damping ratio=0.1 damping ratio=0.25 1

0.5

X(t) 0 10 20 30 40 -0.5

-1

-1.5 time (sec)

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Free Response of Overdamped 2nd Order System For an overdamped system,  > 1, the roots of the characteristic equation are real and negative, i.e.,

1/ 2 1/ 2 ss    22 1  ;         1  (12) 12nn     

The free response of an overdamped system is:

n t X( t ) e C1 cosh * t C 2 sinh * t (13) 1/ 2 where 2 has units of 1/time. Do not confuse this term *  n   1 with a frequency since the response of motion is NOT oscillatory.

At time t = 0, apply the initial conditions to get

VX00n CXC1 0and 2 (14) *

Note that as t , X(t)  0, i.e. the equilibrium position.

Note: An overdamped system does to oscillate. The larger the damping ratio >1, the longer time it takes for the system to return to its equilibrium position.

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Free Response of Critically Damped 2nd Order System For a critically damped system,  = 1, the roots are real negative and identical, i.e.

ss12 n (15)

The solution form X(t) = A est is no longer valid. For repeated roots, the theory of ODE’s dictates that the family of solutions satisfying the differential equation is

n t X() t e C12 tC  (16)

At time t = 0, applying initial conditions gives

Then CXCVX1 0and 2  0 n 0 (17)

Note that as t , X(t)  0, i.e. the equilibrium position.

A critically damped system does to oscillate, and it is the fastest to damp the response due to initial conditions.

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Free Response of 2nd order system: Comparison between underdamped, critically damped and overdamped systems

initial displacement only

Xo = 1, Vo = 0 ωn = 1.0 rad/s  = 0.1, 1.0, 2.0

Motion decays exponentially for > 0 Fastest response for = 1; i.e. fastest decay towards equilibrium position X = 0

Free response Xo=1, Vo=0, wn=1 rad/s damping ratio=0.1 1.2 damping ratio=1 1 damping ratio=2 0.8 0.6 0.4 0.2

X(t) 0 -0.2 0 5 10 15 20 25 30 35 40 -0.4 -0.6 -0.8 -1 time (sec)

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Free Response of 2nd order System: Comparison between underdamped, critically damped and overdamped systems

Initial velocity only

Xo = 0, Vo = 1.0, ωn = 1.0 rad/s  = 0.1, 1.0, 2.0

Motion decays exponentially for > 0 Fastest response for = 1.0, i.e. fastest decay towards equilibrium position X=0. note initial overshoot

Free response Xo=0, Vo=1, wn=1 rad/s damping ratio=0.1 1 damping ratio=1 damping ratio=2 0.8

0.6

0.4

0.2

X(t) 0 0 5 10 15 20 25 30 35 40 -0.2

-0.4

-0.6

-0.8 time (sec)

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E X A M P L E: A 45 gram steel ball (m) is dropped from rest through a vertical height of h=2 m. The ball impacts on a solid steel cylinder with mass M = 0.45 kg. The impact is perfectly elastic. The cylinder is supported by a soft spring with a stiffness K = 1600 N/m. The mass-spring system, initially at rest, deflects a maximum equal to  = 12 mm, from its static equilibrium position, as a result of the impact. (a) Determine the time response motion of the mass- spring system. (b) Sketch the time response of the mass-spring system. (c) Calculate the height to which the ball will rebound.

(a) Conservation of linear momentum before impact = just after impact:

mV_ mV M xo (1) where V_2 gh = 6.26 m/s is the steel ball velocity before impact

V+ = velocity of ball after impact; and xo : initial mass-spring velocity. K Mass-spring system EOM: M x + K x  0 (2) with   59.62 rad/s n M

(from static equilibrium), the initial conditions are x(0) 0 and x (0) xo (3)

x (2) & (3) lead to the undamped free response: o x( t ) sin (nn t )  sin (  t ) n given δ = 0.012 m as the largest deflection of the spring-mass system.

Hence, xon = 0.715 m/s

M m m V V  x  6.26  7.15   0.892 mo s s (upwards) and the height of rebound is

2 V h = 41 mm 2g

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The concept of logarithmic decrement for estimation of the viscous damping ratio from a free-response vibration test

The free vibration response of an underdamped 2nd order viscous system (M,K,D) due to an initial displacement Xo is a decay oscillating wave with damped natural frequency (ωd). The period of motion is Td = 2/ ωd (sec). The free vibration response is

n t x( t ) Xo e cos d t (11.c)

1/ 2   K / M 1/ 2 ;    1 2 where   DDDKM /cr , cr 2 ; n   d n  

Free response of underdamped viscous system

1.5 damping ratio=0.1

1 X1 X2 0.5 X1+n

X(t) 0 t t1 t1+n 0 10 20 30 40 -0.5 Td -1 time (sec)

Consider two peak amplitudes, say X1 and X1+n, separated by n periods of decaying motion. These peaks occur at times, t1 and

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(t1+nTd), respectively. The system response at these two times is from Eq. (1):

n t1 X1 x( t 1 ) Xo e cos  d t 1 ,

and

- t nT  X  x t  X end1 cos  t  n T , 1n 1 nTd  o d 1 d d 

Or, since dTd  2 .

 nt11  nT d    n t  nT d  X1n X o e cos d t 1  2  n  X o e cos   d t 1  (18)

Now the ratio between these two peak amplitudes is:

- t n 1 -n t1 Xe Xod e cos  t1   nT 1 nd    e (19) - t nT - t nT  X nd 1  e nd1 1n  Xod e cos  t1  2 

Take the natural logarithm of the ratio above:

22 n  ln X / X    n T  n     n   (20) 11n n d n 221/ 2 1/ 2 n 11    

Define the logarithmic decrement as:

12X1    ln (21) nX 2 1/ 2 1n 1- 

Thus, the ratio between peak response amplitudes determines a useful relationship to identify the damping ratio of an underdamped second order system, i.e., once the log dec (δ) is determined then,

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   1/2 (22), 2 2  2 

 and for small damping ratios,  ~ . 2

The logarithmic decrement method to identify viscous damping ratios should only be used if: a) the time decay response shows an oscillatory behavior (i.e. vibration) with a clear exponential envelope, i.e. damping of viscous type, b) the system is linear, 2nd order and underdamped, c) the dynamic response is very clean, i.e. without any spurious signals such as noise or with multiple frequency components, d) the dynamic response X(t) 0 as t. Sometimes measurements are taken with some DC offset. This must be removed from your signal before processing the data. e) Strongly recommend to use more than just two peak amplitudes separated n periods. In practice, it is more accurate to graph the magnitude of several peaks in a log scale and obtain the log- decrement () as the best linear fit to the following relationship [see below Eq. (23)].

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From Eq. (18),

 nt1  nT d   n nT d  X1n X ocos  d t 1 e X 1 e ,

nt1 whereX11 Xod cos  t e

 ndnT  ln(X11n ) ln X  ln( e ) 

lnX11 nd T  A  n  , where A  ln X ;

ln(X1n ) A n (23) i.e., plot the natural log of the peak magnitudes versus the period numbers (n=1,2,…) and obtain the logarithmic decrement from a straight line curve fit. In this way you will have used more than just two peaks for your identification of damping.

Always provide the correlation number (goodness of fit = R2) for the linear regression curve (y=ax+b), with y=ln(X) and x=n as variables.

The log-dec is a most important concept widely used in the characterization of damping in a mechanical system (structure, pipe system, spinning rotor, etc) as it gives a quick estimation of the available damping in a system.

In rotating machinery, API specifications provide the minimum value of logdec a machine should have to warrant its acceptance in a shop test as well as safe (and efficient) operation in the film.

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E X A M P L E

A wind turbine is modeled as a concentrated mass (the turbine) atop a weightless elastic tower of height L. To determine the dynamic properties of the system, a large crane is brought alongside the tower and a lateral force F=200 lb is exerted along the turbine axis as shown. This causes a horizontal displacement of 1.0 in. The cable attaching the turbine to the crane is instantaneously severed, and the resulting free vibration of the turbine is recorded. At the end of two complete cycles (periods) of motion, the time is 1.25 sec and the motion amplitude is 0.64 in. From the data above determine: (a) equivalent stiffness K (lb/in) (b) damping ratio  (c) undamped natural frequency ωn (rad/s) (d) equivalent mass of system (lb-s2/in)

static force 200 lb a) K    200 lb/in static deflection 1.0 in b) cycle amplitude time Use log dec to find the viscous damping ratio 0 1.0 in 0.0 sec 2 0.64 in 1.25 sec

1xo 1 1.0   ln   ln   0.2231 nx2 2 0.64 2    ;  ~  0.035 142 2 2 2

underdamped system with 3.5% of critical damping.

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1.25 s c) Damped period of motion, T =  0.625 sec/cyc d 2 cyc 2 rad Damped natural frequency, d 10.053 Td sec Natural frequency, d rad n 10.059 1 2 sec d) Equivalent mass of system:

from n  KM K200 lb /in M  = 1.976 lb/ sec2/in 2 2 1 n 10.059 sec2

E X A M P L E A loaded railroad car weighing 35,000 lb is rolling at a constant speed of 15 mph when it couples with a spring and dashpot bumper system. If the recorded displacement-time curve of the loaded railroad car after coupling is as shown, determine (a) the logarithmic decrement  (b) the damping ratio ζ (c) the natural frequency ω n (rad/sec) (d) the spring constant K of the bumper system (lb/in) (e) the damping ratio ζ of the system when the railroad car is empty. The unloaded railroad car weighs 8,000 lbs.

xo 4.8 (a) logarithmic decrement  nn  1.1631 x1 1.5

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(b) damping ratio   0.1820 2     1/ 2 1 2 2 2 2 

The natural frequency is d rad n 16.816 1 2 sec

(d) Bumper stiffness,

221 35,000 lb lb KK M  16.816   25612.5 n car sec22 386.4 in/sec in

(e) Damping ratio when car is full: D  0.182 2K Mfull

Note that the physical damping coefficient (D) does not change whether car is loaded or not, but  does change. D Damping ratio when car is empty  e  2KM empty

The ratio 1/ 2  MMempty full 35,000      0.182 e   eMM full empty 8,000

 e  0.381

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INSERT examples of identification of system damping ratio

The figure below shows the free response (amplitude vs. time) of a simple mechanical structure. Prior tests determined the system equivalent stiffness Ke=1000 lbf/in. From the measurements determine: a) damped period of motion Td (s) b) damped natural frequency d (rad/s) c) Using the log dec () concept, estimate the system damping ratio. d) the system equivalent mass Me (lbm) e) the system equivalent damping coefficient De (lbf.s/in)

DISPLACEMENT (ft) vs time (sec) 0.25 0.2 0.15 0.1 0.05 0 X [ft) 0.05 0.1 0.15 0.2 0.25 0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1 1.1 1.2 1.3 1.4 1.5 1.6 time (s)

(a) Determine damped period of motion: 0.6 sec from 4 periods of T  d 4 damped motion Td  0.15 sec (b) Determine damped natural frequency: 2 d  rad Td   41.888 d sec (c) Determine damping ratio from log-dec: Select two amplitudes of motion (well spaced) and count number od periods in between

Xo  0.23 ft after n5 periods Xn  0.05 ft

1  Xo Log-dec is derived from ratio:   ln  n  Xn   0.305 1  

from log-dec formula 2  =  0.5   2 .5 1   2 2 4    0.049  Note that approximate formula:  0.049 is a very good estimation of damping 2 ratio

(d) damped natural frequency: d   n 0.5 2 rad 1     41.937 n sec a little higher than the damped frequency (recall damping ratio is small)

(e) system equivalent mass: 3 lbf Static tests conducted on the structure show its stiffness to be K110  in and from the equation for natural frequency, the equivalent system mass is K M  e 2 n Me  219.526 lb

(f) system damping coefficient: 0.5 De  2KM e sec D  2.314 lbf  e in

Note: sec Actual values of parameters are M 220lb C 2.4 lbf  in   0.05

DISPLACEMENT (ft) vs time (sec) 0.25 0.2 0.15 0.1 0.05 0 X [ft) 0.05 0.1 0.15 0.2 0.25 0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1 1.1 1.2 1.3 1.4 1.5 1.6 time (s)

Xo  0.23 ft after n5 periods Xn  0.0776 ft 1  Xo Log-dec is derived from ratio:   ln  n  Xn   0.217 1 from log-dec formula 2  =  0.5   2 .5 1   2 2 4    0.035  Note that approximate formula:  0.035 is a very good estimation of damping 2 ratio

(d) damped natural frequency: d   n 0.5 2 rad 1     41.913 n sec a little higher than the damped frequency (recall damping ratio is small)

(f) system damping coefficient: 0.5 De  2KM e sec D  1.649 lbf  e in

Note: sec Actual values of parameters are M 220lb C 2.4 lbf  in   0.05 Note large difference in damping - WHY?

2 Measurements of acceleration due to impact load

0 acceleration (volt) 2

8 0 16 32 48 64 80 96 112 128 144 160 time (mili-sec)

80− 7 from 4 periods of (a) Determine damped period of motion: Td := ⋅sec 4⋅ 1000 damped motion

Td = 0.02sec (b) Determine damped natural frequency: 2⋅π ωd := rad Td ωd = 344.28 (c) Determine damping ratio from log-dec: sec

Select two amplitudes of motion (well spaced) and count number od periods in between

Ao := −7.34 after n4:= periods An := −1.03

1 ⎛ Ao⎞ Log-dec is derived from ratio: δ := ⋅ln⎜ ⎟ n A ⎝ n⎠ δ = 0.49 from log-dec formula 2⋅ξπ⋅ δ = 2 0.5 δ ()1 − ξ ξ := 2 2 .5 ()4⋅δπ + ξ = 0.08 δ Note that approximate formula: = 0.08 is a very good estimation of damping 2⋅π ratio

(d) Determine damped natural frequency: ωd ωn := 2 0.5 ()1 − ξ rad ωn = 345.33 sec a little higher than the damped frequency (recall damping ratio is small)

(b) Forced Response of 2nd Order Mechanical System

Step Force or Constant Force A force with constant magnitude Fo is suddenly applied at t=0. Besides the system had initial displacement X0 and velocity V0. EOM is:

d2 X d X MDKXF   o (24) d t2 d t

Using

  K n M : natural frequency of system D   : damping ratio, where DKMcr 2 = critical damping Dcr Eqn. (1) becomes:

d2 X d X FF 2   22X oo   d t2 n d t n M K n (25)

The solution of the ODE is (homogenous + particular):

s t s t F X() t X  X  A e12  A e  o (26) HP12 K  Where A’s are constants found from the initial conditions Note: Xss=Fo / K is and XP=Fo/K is the particular solution for the step load. equivalent to the static displacement if the force is applied very slowly.

MEEN 459/659 Notes 1a © Luis San Andrés (2019) 1-23

The roots of the characteristic Eq. for the system are

2 1/ 2 s1,2  nn   1 (6)

From Eq. (6), differentiate three cases:

Underdamped System: 0 <  < 1,  D < Dcr

Critically Damped System: = 1,  D = Dcr

Overdamped System: > 1,  D > Dcr

Step Forced Response of Underdamped 2nd Order System

For an underdamped system, 0 < < 1, the roots are complex conjugate ( real and imaginary parts), i.e.

2 1/ 2 si1,2  nn 1    The response is:

n t X( t ) e C12 cosd t  C sin d t  X ss (27) 1/ 2 where 2 is the system damped natural frequency. dn 1  

and XFKss o

At time t = 0, applying the initial conditions gives

VC01n CXXC1 0 ss  and 2  (28) d

Note that as t , X(t)  Xss = Fo/K for > 0,

MEEN 459/659 Notes 1a © Luis San Andrés (2019) 1-24

i.e. the system response reaches the steady state (static) equilibrium position.

The largest the damping ratio  ( 1) , the fastest the motion will damp to reach the static position Xss. In addition, the period of damped natural 22 motion will increase, i.e., T     d  2 1/2 d n 1 

Step Forced Response of Undamped 2nd Order System: For an undamped system, i.e., a conservative system, = 0, and the response is just

X( t ) C12 cosn t  C sin n t  X ss (29) with Xss = Fo/K, C1 = (X0 – Xss) and C2 = V0 /ωn (30) if the initial displacement and velocity are null, i.e. X0 = V0 = 0, then

X( t ) Xss 1 cos n t  (31)

Note that as t , X(t) does not approach Xss for = 0.

The system oscillates forever about the static equilibrium position Xss and, the maximum displacement is 2 Xss, i.e. twice the static displacement

(F0/K).

The observation reveals the great difference in response for a force slowly applied when compared to one suddenly applied. The difference explains why things break when sudden efforts act on a system.

MEEN 459/659 Notes 1a © Luis San Andrés (2019) 1-25

Forced Step Response of Underdamped Second Order System:

damping ratio varies

Xo = 0, Vo = 0, ωn = 1.0 rad/s  = 0, 0.1, 0.25

zero initial conditions Fo/K = Xss =1;

faster response as increases; i.e. as t  , X  Xss for > 0

Step response Xss=1, Xo=0, Vo=0, wn=1 rad/s damping ratio=0.0 2.5 damping ratio=0.1 damping ratio=0.25 2

1.5

X(t) 1

0.5

0 0 10 20 30 40 time (sec)

MEEN 459/659 Notes 1a © Luis San Andrés (2019) 1-26

Forced Step Response of Overdamped 2nd Order System

For an overdamped system,  > 1, the roots of the characteristic eqn. are real and negative, i.e.

1/ 2 1/ 2 ss    22 1  ;         1  (32) 12nn     

The forced response of an overdamped system is:

n t X( t ) e C1 cosh * t  C 2 sinh * t  X ss (33) 1/ 2 where 2 . Response or motion is NOT oscillatory. *  n   1 and

VC01n CXXC1 0 ss  and 2  (34) *

Note that as t , X(t)  Xss = Fo/K for  > 1, i.e. the steady-state (static) equilibrium position.

An overdamped system does not oscillate (or vibrate).

The larger the damping ratio , the longer time it takes the system to reach its final equilibrium position Xss.

MEEN 459/659 Notes 1a © Luis San Andrés (2019) 1-27

Forced Step Response of Critically Damped System

For a critically damped system,  = 1, the roots are real negative and identical, i.e.

ss12 n (35)

The step-forced response for a critically damped system is

n t X() t e C12  tC  X ss (36)

with CXXCVC1 0 ss and 2  0  n 1 (37)

Note that as t , X(t)  Xss = Fo/K for > 1, i.e. the steady-state (static) equilibrium position.

A critically damped system does not oscillate and it is the fastest to reach the steady-state value Xss.

MEEN 459/659 Notes 1a © Luis San Andrés (2019) 1-28

Forced Step Response of Second Order System: Comparison of Underdamped, Critically Damped and Overdamped system responses

Xo = 0, Vo = 0, ωn = 1.0 rad/s  = 0.1,1.0,2.0 zero initial conditions

Fo/K = Xss =1; (magnitude of s-s response)

Fastest response for = 1. As t  , X  Xss for > 0

Step response Xss=1, Xo=0, Vo=0, wn=1 rad/s damping ratio=0.1 2 damping ratio=1 1.8 damping ratio=2 1.6 1.4 1.2 1

X(t) 0.8 0.6 0.4 0.2 0 0 5 10 15 20 25 30 35 40 time (sec)

MEEN 459/659 Notes 1a © Luis San Andrés (2019) 1-29

EXAMPLE:

The equation describing the motion and initial conditions for the system shown are: (MIRXDXKXFXX /2 )    , (0)  (0)  0

Given M=2.0 kg, I=0.01 kg-m2, D=7.2 N.s/m, K=27.0 N/m, R=0.1 m; and F =5.4 N (a step force), a) Derive the differential equation of motion for the system (as given above). b) Find the system natural frequency and damping ratio c) Sketch the dynamic response of the system X(t) d) Find the steady-state value of the response Xs- s.

(a) Using free body diagrams: Note that θ= X/R is a kinematic constraint. The EOM's are:

MXFKXDXF    G (1)

IFR G (2)

 X Then from (2) FII (3) ; G RR2 (3) into (1) gives

I MXDXKX2    F (4) R or Using the Mechanical Energy Method:

12 1 2 1 I 2 (system kinetic energy):TMXIMX     (5) 2 2 2 R2

1 2 (system potential energy): VKX 2

MEEN 459/659 Notes 1a © Luis San Andrés (2019) 1-30

2 (viscous dissipated energy) E D X dt , and External work: W F dX D   d Derive identical Eqn. of motion (4) from (6) TVEW   d   0 dt

I (b) define MM   3 Kg , K = 27 N/m, D = 7.2 N.s/m eq R2 and calculate the system natural frequency and viscous damping ratio: 1/ 2 K rad D n  3 ;   0.4, underdamped system M eq sec 2KM

2 rad 2 dn  1-  2.75 , and Td 2.28 sec is the damped period sec d of motion

(d) At steady-state, no motion occurs, X = XSS, and XX0, 0

Then F 5.4 N X  X  0.2 m ss N ss K 27 m  X( t ) X 1  en t  cos t  sin t ss d2 d  1

MEEN 459/659 Notes 1a © Luis San Andrés (2019) 1-31 ME459/659 Notes# 1b (pp. 32-46)

Dynamic Response of SDOF Second Order Mechanical System: Viscous Damping

MXDXKXF   ()t

Periodic Forced Response to

2 F(t) = Fo sin( t) and F(t) = M u  sin(t)  Frequency Response Function of Second Order Systems

MEEN 459/659 Notes 1b © Luis San Andrés (2019) 1-32

Let the external force be PERIODIC with frequency  (period T=2/) and amplitude Fo . The EOM is:

d2 X d X M D  K X  Fsin  t d t2 d t o (38) with initial conditions XVXX(0)ooand (0)

The solution of the non-homogeneous ODE (38) is

s12 t s t XtXXAeAeC( )H  P 12   c cos  tC  s sin  t (39) where XH is the solution to the homogeneous form of (38) and such that (s1, s2) satisfy the characteristic equation of the system:

22 ss20 nn    

1/ 2 nd 2 The roots of this 2 order polynomial are: s1,2  nn   1 (*)

D   K   where n M is the natural frequency, and is the Dcr viscous damping ratio. Recall DKMcr 2 is the critical damping coefficient.

The value of damping ratio determines whether the system is underdamped ( <1), critically damped ( =1), overdamped ( >1).

MEEN 459/659 Notes 1b © Luis San Andrés (2019) 1-33 Response of 2nd Order Mechanical System to a Periodic Loading:

And the particular solution is

XP C ccos  t  C s sin  t (40)

Substitution of Eq. (40) into Eq. (38), and after some algebra, gives

12 MKDK    Cc  0      (41a) 2 CFK DKMK1   so     KM   K DDM Substitute above n M ;    KKDK n and Xss = Fo/K, a “pseudo” static displacement 

22 1nn    C  0  c  (41b) 22     1   CXs ss nn       Define the frequency ratio f  (42) n that relates the (external) excitation frequency (Ω) to the natural frequency of the system (ωn); i.e. when

n f 1 , the system operates below its natural frequency

n f 1 , the system operates above its natural frequency

With this definition, write Eq. (41b) as:

MEEN 459/659 Notes 1b © Luis San Andrés (2019) 1-34 1 ff2    Cc  0      (43) 2 CX ff 1  s   ss 

Solve Eq. (50) for the coefficients Cs and Cc:

2 2 2 f 1 f  CXCXc ss22; s ss (44) 1f22  2 f22 1  f   2 f 

Response of 2nd Order Mechanical System to a Periodic Load: For an underdamped system, 0 <  < 1, the homogeneous solution (free response) is

n t XH( t ) e C 1cos d t C 2 sin d t (45)

1/ 2 where 2 is the system damped natural frequency. dn 1  

By superposition, the complete response is X() t XHP X =

n t X(t ) e C 1cos d t  C 2 sin d t  C c cos  t  C s sin  t (46)

with Cs and Cc  Eq. (44).

At time t = 0, apply the initial conditions and obtain

VCC01ns   CXCC1 0 c  and 2  (47) d

As long as  > 0, the homogeneous solution (also known as the TRANSIENT or free response) will die as time elapses, i.e., for t >>0 then XH  0.

MEEN 459/659 Notes 1b © Luis San Andrés (2019) 1-35 Thus, after all transients have passed, the dynamic response of the system is just the particular response XP(t) .

Steady State – Periodic Forced Response of Underdamped System The steady-state (or quasi-stationary) response is given by:

X()t C ccos  t  C s sin  t  C sin  t  (48)

where (Cs, Cc)  Eq. (44). Define CCCCsccos ;   sin  ; where  is a phase angle; then

C 2  f c (49) tan 2 Cs 1 f 

22 and CCCXAS  C  ss , where 1 A  ; = amplitude ratio (50) 2 1ff2   2 2 (a dimensionless quantity)

Thus, the steady-state system response is just

X()t X ss Asin  t  (51)

Note there “steady-state” implies that, for excitation with a constant frequency , the system response amplitude C and the phase angle  are constant or time invariant

MEEN 459/659 Notes 1b © Luis San Andrés (2019) 1-36 Amplitude and Phase Lag of Periodic Force Response

X()()t X ss Asin  t  for F t  F o sin  t

FRF 2nd order system damping ratio=0.05 Periodic force: Fo sin(t) damping ratio=0.1 12 damping ratio=0.2 10 A damping ratio=0.5

Amplitude ratio (A) ratio Amplitude 2

0 0 0.25 0.5 0.75 1 1.25 1.5 1.75 2 frequency ratio (f)

MEEN 459/659 Notes 1b © Luis San Andrés (2019) 1-37 Regimes of Dynamic Operation:

n f 1 , the system operates below its natural frequency

2 1f  1; 2 f  0  A  1  0

X( t ) Xss sin t i.e. similar to the “static” response

 n f 1 , the system is excited at its natural frequency

2 1  1fA  0;   ;   90  22

X ss  X( t ) sin  t  22 

if  < 0.5, the amplitude ratio A > 1 and a resonance is said to occur.

n f 1 , the system operates above its natural frequency 2 1f  1; 2 f 0  A  0     180 

X( t ) Xss A sin  t    X ss A sin  t

A <<< 1, i.e. very small,

MEEN 459/659 Notes 1b © Luis San Andrés (2019) 1-38

Steady State – Periodic Forced Response of 2nd Order system: Imbalance Load

Imbalance loads are typically found in rotating machinery. In operation, due to inevitable wear, material build ups or assembly faults, the center of mass of the rotating machine does not coincide with the center of rotation (spin). Let the center of mass be located a distance (u) from the spin center, and thus, the load due to the imbalance is a centrifugal “force” with magnitude 2 Fo = M u Ω and rotating with the same frequency as the rotor speed (Ω). This force excites the system and induces vibration1. Note that the imbalance force is proportional to the frequency2 and grows rapidly with shaft speed.

Ω M

Ne that although in practice the offset distance (u) is very small (a few mil), the system response or amplitude of vibration can be quite large affecting the performance and integrity of the rotor assembly.

For example if the rotating shaft & disk has a small imbalance mass (m) located at a radius (r) from the spin center, then it is easy to determine that the center of mass offset (u) is approximately equal to (m r/M). Note that u<

rm u M m  r m  u M

1 The current analysis only describes vibration along direction X. In actuality, the imbalance force induces vibrations along the planes (X,Y) and the rotor whirls in an orbit around the center of rotor spinning. For isotropic systems, the motion in the X plane is identical to that in the Y plane but out of phase by 90o.

MEEN 459/659 Notes 1b © Luis San Andrés (2019) 1-39 The equation of motion for the system with an imbalance force is

d2 X d X M D  K X  Fsin  t  M u 2 sin   t d t2 d t o

22 Fo Mu then Xss    u2  u f KK n

with f n . The “steady-state” system response is

2 X()t X ss Asin  t   u  A sin  t  

X()t  u Bsin  t  (52)

2  f where  is a phase angle, tan , and 1 f 2 

f 2 B  (53) 2 1ff2   2 2 is the amplitude ratio (dimensionless quantity).

Educational video – watch UNBALANCE RESPONSE demonstration https://www.youtube.com/watch?v=R2hO--TIjjA

MEEN 459/659 Notes 1b © Luis San Andrés (2019) 1-40

Amplitude B and phase lag () for response to an imbalance (u)

2 X()()tt u Bsin t   for F  M u  sin  t

FRF 2nd order system damping ratio=0.05 Imbalance force: M u ^2sin(t) damping ratio=0.1 12 damping ratio=0.2 10 B damping ratio=0.5 8

Amplitude ratio (B) ratio Amplitude 2

0 0 0.25 0.5 0.75 1 1.25 1.5 1.75 2 frequency ratio (f)

MEEN 459/659 Notes 1b © Luis San Andrés (2019) 1-41

Regimes of Dynamic Operation:

n f 1 , rotor speed is below its natural frequency

22 1f  1; 2 f  0  B  f  0 and  0

2 X( t ) u f sin  t  0 i.e. little motion or response

 n f 1 , rotor speed coincides with natural frequency

2 1  1fB  0;   ;   90  22 uu X( t ) sin  t   cos  t 2 2 2 if  < 0.5, amplitude ratio B > 1 and a resonance is said to occur. Damping is needed to survive passage through a natural frequency (critical speed).

n f 1 , the rotor speed is above its natural frequency 2 1 f  2 f 221;  0 B  1   180  ff

X( t ) u sin  t    u sin  t

B ~ 1, at high frequency operation, the maximum amplitude of vibration (Xmax) equals the unbalance displacement (u).

MEEN 459/659 Notes 1b © Luis San Andrés (2019) 1-42

Note: API demands that the operating speed of a rotor system is not too close to the natural frequency to avoid too large amplitudes that could endanger the life of the system. A rotor speed cannot operate (but for very short times) 20% above and below the natural frequency.

EXAMPLE:

A cantilevered steel pole supports a small

0 wind turbine. The pole torsional stiffness is Z Me d K (N.m/rad) with a rotational damping t coefficient C (N.m.s/rad).

m 0 e The four-blade turbine rotating assembly has mass mo , and its center of gravity is Y 2 X M e cos t displaced distance e [m] from the axis of 0 d rotation of the assembly. k c m 0 2 Z Y Iz (kg.m ) is the mass moment of inertia about the z axis of the complete turbine, including rotor assembly, housing pod, and X contents. k= Torsional Tiffness e= Torional Damping coefficient The total mass of the system is m (kg). The plane in which the blades rotate is located a distance d (m) from the z axis as shown.

For a complete analysis of the vibration characteristics of the turbine system, determine:

a) Equation of motion of torsional vibration system about z axis.

b) The steady-state torsional response  (t) (after all transients die out). 2 c) For system parameter values of k=98,670 N.m/rad, Iz=25 kg.m , C = 157 N.m.s/rad, and mo = 8 kg, e = 1 cm, d = 30 cm, present graphs showing the response amplitude (in rads) and phase angle as the turbine speed (due to wind power variations) changes from 100 rpm to 1,200 rpm. d) From the results in (c), at what turbine speed should the largest vibration occur and what is its magnitude? e) Provide a design recommendation or change so as to reduce this maximum vibration amplitude value to half the original value.

Neglect any effect of the mass and bending of the pole on the torsional response, as well as any gyroscopic effects.

(a) derive the drive torque and EOM The torque or moment induced by the mass imbalance is

MEEN 459/659 Notes 1b © Luis San Andrés (2019) 1-43 2 T(t) = d x Fu = mo e d  cos(ωt), i.e., a function of frequency T()

The equation for torsional motions of the turbine-pole system is:

2 Izo C   K   m e d  cos  t  T() cos   t (e.1)

Note that all terms in the EOM represent moments or torques.

(b) The steady state periodic forced response of the system:

 tt ss cos    (e.2) 2 1/ 2 1f2  2  f 2  

T 2 () moo ed  I z m ed 2 but ss     f (e.3) KKIIzz

1 2 f  KC with f ;;   , and   tan 2 (e.4) n 1  f nzI 2 KIz

(e.3) in (e.2) leads to

2 mo ed f (tt ) 1/ 2 cos (  -  ) (e.5) I 2 2 2 z 1ff 2    

2 mo ed f Let   (e.6) B  1/ 2 (e.7) I 2 2 2 z 1ff 2   

and rewrite (e.5) as: (t )  B cos (  t  ) (e.8)

MEEN 459/659 Notes 1b © Luis San Andrés (2019) 1-44 (c) for the given physical values of the system parameters: K rad K = 98,670 N.m/rad n 62.82 Iz sec C 2 Iz = 25 kg  m  0.05 2 KIz C = 157. N.m.s/rad

mo ed 8 0.01 0.3 0.024 -5      96 10 rad Iz 25 25

And the turbine speed varies from 100 rpm to 1,200 rpm, i.e.

 = rpm /30 = 10.47 rad/s to 125.66 rad/s, i.e.

 f  = 0.167 to 2.00, n thus indicating the system will operate through resonance.

Hence, the angular response is (t ) 96.4 105 rad  B cos (  t   )

(d) Maximum amplitude of response: since  << 1, the maximum amplitude of motion will occur when the turbine speed coincides with the natural frequency of the torsional system, 1 1  i.e., at f = 1  B  and(tt )   cos   2 22 

 -2 the magnitude is θMAX = 0.964 x 10 rad , i.e. 10 times larger max 2 than  .

MEEN 459/659 Notes 1b © Luis San Andrés (2019) 1-45 The graphs below show the amplitude (radians) and phase angle (degrees) of the polo twist vs. turbine speed (RPM)

0.01

0.005

amplitudeof torsion (rad) 0 200 400 600 800 1000 1200 rotor speed (RPM)

180

120

Phaseangle (degrees) 0 200 400 600 800 1000 1200 rotor speed (RPM)

(e) Design change: DOUBLE DAMPING but first BALANCE ROTOR!

MEEN 459/659 Notes 1b © Luis San Andrés (2019) 1-46 MEEN 363 - QUIZ 3 Names:

The rotor of an electric generator weights 750 lb and is attached to a platform weighing 7750 lb. The motor has an imbalance eccentricity (a) of 2.5 mils. The platform can be modeled as shown in the figure. The equivalent stiffness (K) of the platform is 1 million-lb/in, and the equivalent damping is C=100 lb-s/in. The operating speed of the generator is 1800 rpm. (a) determine the response of the platform (amplitude and phase) at the operating speed. (b) determine the response of the platform (amplitude and phase) if the rotor spins with a speed coinciding with the system natural frequency. (c) if the platform stiffness is increased by 25%, determine the allowable amount of imbalance (a) that will give the same amplitude of motion as determined in (a). Assume that the mass of the

platform and the damping do not change appreciably by performing the stiffening.

rotor

platform

k c

KEY: System excitation due to rotating imbalance

6 lbf Given K10:= ⋅ Mrotor := 750⋅ lb Mplatform := 7750⋅ lb in sec MM:= rotor + Mplatform C:= 100⋅ lbf⋅ in 3 M= 8.5× 10 lb calculate the system natural frequency and damping ratio: .5  K  C ωn :=   ζ :=  M  2M⋅ω⋅ n

rad ωn = 213.125 ζ = 0.011 little damping s and operating frequency ratio (r) for rotor speed: RPM:= 1800 2⋅π rad ω := RPM⋅ ⋅ rad 60 s ω = 188.496 ω s r := ωn r= 0.884 − 3 The system response (amplitude and phase) for imbalance excitation a:= 2.5⋅ 10 ⋅in are: 2 Mrotor r  −ζ2⋅ ⋅r Yop()r := a⋅ ⋅ Ψ(r ):= atan  M .5 2  2   1r−  2 2 ()1r− + ()2⋅ζ⋅r  Thus, at r= 0.884 <1 − 4 180 Y (r )= 7.894× 10 in Ψ()r ⋅ = −4.947 [degrees] op π

Let: Yoper := Yop()r

(b) If the rotor should spin with a speed coinciding with the system natural frequency, r1:=

ωω:= n 60 s RPM := ω⋅ ⋅ 3 2⋅π rad RPM= 2.035× 10

the system response is

Y (1 )= 0.01 in Ψ := −90 degrees op Yop()1 = 13.111 also determined from: Yoper Mrotor 1 a⋅ ⋅ = 0.01 in M 2⋅ζ

calculate the NEW system natural frequency and damping ratio: .5  K  C ωn :=   ζ :=  M  2M⋅ω⋅ n

rad − 3 ωn = 238.281 ζ = 9.531× 10 small change in damping ratio s and operating frequency ratio (r) for rotor speed: RPM:= 1800 2⋅π rad ω := RPM⋅ ⋅ rad 60 s ω = 188.496 ω r := s r= 0.791 ωn from relationship: 2 Mrotor r Yop()r := a⋅ ⋅ M .5  2  2 2 determine the (new) allowable imbalance: ()1r− + ()2⋅ζ⋅r 

Yoper a :=  2  M r  rotor ⋅   M .5  − 3  2  a= 5.354× 10 in  2 2   ()1r− + ()2⋅ζ⋅r   i.e. ~ twice as original imbalance (eccentricity) displacement, Example: system response due to multiple frequency inputs

d2 d M⋅ Y + C⋅ Y + KY⋅ = Kzt⋅ () where dt2 dt zt():= a1⋅cos()ω1⋅t + a2⋅sin()ω2⋅t + a3⋅cos()ω3⋅t

is an external excitation displacement function Find the forced response of the system, i.e, find Y(t)

N Given the system parameters M := 100⋅kg K := 106⋅ ζ := 0.10 m 0.5 ⎛ K ⎞ calculate natural frequency ω := ⎜ ⎟ ωn n M and physical damping ⎝ ⎠ fn := 2⋅π fn = 15.915Hz C := 2⋅ωM⋅ζ⋅ n N C = 210× 3 s ω m 2 .5 d ω := ω ⋅()1 − ζ fd := d n 2⋅π 1 Td := fd damped natural period Td = 0.063s Set frequencies and amplitudes of the excitation z(t) are:

a1 := 0.05⋅m a2 := 0.05⋅m a3 := 0.05⋅m

ωn 9 ω := ω := ω ⋅ ω := 2.2⋅ω 1 2 2 n 10 3 n

assemble: zt():= a1⋅cos()ω1⋅t + a2⋅sin()ω2⋅t + a3⋅cos()ω3⋅t

excitation displacement 0.2

0 z(t) [m]

0.2 0 0.051 0.1 0.15 0.2 0.25 time (s) 4 periods of damped natural motion SYSTEM RESPONSE is: Yt():= ai⋅H⋅cos()ωi⋅φt + i

The system frequency response function: amplitude and phase angle are

1 Hr():= ⎛ 2⋅ζ⋅r ⎞ .5 φ()r := φ ← −atan⎜ ⎟ ⎡ 2 2 2⎤ 2 ⎣()1 − r + ()2⋅ζ⋅r ⎦ ⎝ 1 − r ⎠ φφπ← − if r > 1 return φ graphs of frequency response function ω Amplitude and phase lag as a function of r = frequency ratio ωn

Amplitude of FRF 6

4 Q− factor

3 1 = 5 2 2⋅ζ Amplitude H(r) 1

0 0 0.5 1 1.5 2 2.5 3 Frequency ratio

Phase angle of FRF 0

135 Phase angle (degree) 180 0 0.5 1 1.5 2 2.5 3 Frequency ratio The response of the system is given by the superposition of individual responses, i.e

Yt():= Y1⋅cos()ω1⋅φt + 1 + Y2⋅sin()ω2⋅φt + 2 + Y3⋅cos()ω3⋅φt + 3 where ω1 r1 := r1 = 0.5 for first excitation: ω φ := φ r n Hr()1 = 1.322 1 ()1

K 180 Y1 := ⋅a1⋅Hr()1 φ ⋅ = −7.595 degrees K Y1 = 0.066m 1 π

ω for second excitation: 2 r2 := Hr = 3.821 φ2 := φ()r2 r2 = 0.9 ωn ()2 180 Y2 := a2⋅Hr()2 φ ⋅ = −43.452 degrees Y2 = 0.191m 2 π ω for third excitation: 3 r3 := Hr = 0.259 φ3 := φ()r3 r3 = 2.2 ωn ()3 180 Y3 := a3⋅Hr()3 φ ⋅ = −173.463 degrees Y3 = 0.013m 3 π

Assemble physical response:

Yt():= Y1⋅cos()ω1⋅φt + 1 + Y2⋅sin()ω2⋅φt + 2 + Y3⋅cos()ω3⋅φt + 3

Now graph the response Y(t) and the excitation z(t):

excitation & response 0.4 Note: The response Y shows little motion at the 0.2 highest excitation frequency (ω3). 0 There is an obvious amplification of motion

z( & Y(t) [m] 0.2 with second frequency (ω2 ~ ωn). 0.4 0 0.051 0.1 0.15 0.2 0.25 To understand better, time (s) let's plot the actual FRF: Z Y Td = 0.063s Amplitude of response 0.3 a1 = 0.05m

0.25 r1 = 0.5 Y1 = 0.066m

0.2 r2 = 0.9 Y2 = 0.191m

0.15 r3 = 2.2 Y3 = 0.013m Amplitude [m] 0.1

0.05 Note how response amplitude for largest frequency is largely 0 0 0.5 1 1.5 2 2.5 3 attenuated Frequency ratio Y3 < a3 all r1 while amplitudes for first two r2 frequencies are amplified, in r3 particular for ω2 which is close to the natural frequency Y Phase angle of response 2 0 = 3.821 a2

Y1 = 1.322 45 a1

90 Phase angle (degree) 135

180 0 0.5 1 1.5 2 2.5 3 Frequency ratio all r1 r2 r3 ME459/659 Notes# 1c (pp. 47-55)

Dynamic Response of SDOF Second Order Mechanical System: Viscous Damping

MXDXKXF   ()t

Transmissibility: Forces transmitted to base or foundation

Response to Periodic Motion of Base or Support

MEEN 459/659 Notes 1c © Luis San Andrés (2019) 1-47

TRANSMISSIBILITY: transmitted force to base or foundation

Needed to calculate stresses on the structural supports as well as to verify the isolation characteristics of the system from its base or support frame.

The EOM for a SDOF (K,D.M) system excited by a periodic force of constant magnitude Fo and frequency (Ω) is:

d2 X d X M D  K X  Fsin  t d t2 d t o (54)

with solution X( t ) Xss A sin  t  (55)

where

F 1 2  f XAo ;  ;   tan-1 ss 1/2 2 (56) Kf1-ff2  2  2 1    

with f n as the ratio of the excitation frequency to the system natural frequency.

The dynamic force transmitted to the base or foundation is

FFFDXKXBDK    (57)

Substitution of Eq. (55) into Eq. (57) gives,

MEEN 459/659 Notes 1c © Luis San Andrés (2019) 1-48 FBo F A  sin  t   2  f cos  t   (58)

1 2 f Let cos  ; sin   1(2 ff )22 1(2 )

And after manipulation rewrite Eq. (58) as:

FB F o A T sin  t  T  (59)

1/2 1 2  f 2   ATT= 1/2 ;    ; 1-ff2  2  2     where (60) -12  f -1 tan2 ;  tan 2  f  1 f

The transmissibility (T) is the ratio of force transmitted to base or foundation |FB| to the (input) excitation force |F0 sin(t)|, i.e.,

F 1 (2 f )2 T B  A  T 2 2 2 (61) Fo (1ff ) (2 )

Regimes of operation: at low frequencies:

nT fA 01   at high frequencies:

nT f    A  2 f

MEEN 459/659 Notes 1c © Luis San Andrés (2019) 1-49

14  2  fA 1   at resonance: nT2

NOTES: At low frequencies,  < 2 , the transmitted force (to base) is larger than external force, i.e. T>1

At  = , the system shows the same transmissibility, independent of the damping coefficient or .

Operation above  > determines the lowest transmitted forces, i.e. mechanical system is ISOLATED from base (foundation). A desirable operating condition

When operation is at large frequencies,  > 2 , viscous damping causes transmitted forces to be larger than w/o damping. Damping is NOT desirable for operation at high frequencies.

FRF 2nd order system damping ratio=0.05 Periodic force: Fo sin(t) damping ratio=0.1 100 damping ratio=0.2 damping ratio=0.5

Transmissibility ratio (T) ratio Transmissibility 0.1 0 0.25 0.5 0.75 1 1.25 1.5 1.75 2 frequency ratio (f)

MEEN 459/659 Notes 1c © Luis San Andrés (2019) 1-50

Response to Periodic Motion of Base or Support

Consider the motion of a (K, D,M) system with its base (or support) moving with a known periodic displacement Z(t) = b cos (t).

The dynamic response of this system is of particular importance for the correct design and performance of vehicle suspension systems. Response to earthquake excitations as well.

For motions about a static equilibrium,Y=Z=0.The EOM is:

MYKYZDYZ(  )  (  )  0 (62a) or

MYDYKYKZDZ    (62b)

Since Zb cos ( t) is prescribed, then Zb   sin (  t)

Substitution of Z and dZ/dt into eqn. (62b) gives

MYDYKY    Kb  cos  t  2 f sin  t (63)

1 2 f Let : cos  ; sin   1(2 ff )22 1(2 )

and write Eq.(63) as:

2 MYDYKYKb   1  2 f cos  t    Fo cos  t    (64)

MEEN 459/659 Notes 1c © Luis San Andrés (2019) 1-51

After all transients die due to damping, the system periodic steady-state response or motion is:

Y()t b A B cos t B  (65)

1/2 1 2  f 2   ABB= 1/2 ;    ; 1-ff2  2  2     where (66) -12  f -1 tan2 ;  tan 2  f  1 f NOTE that AB is identical to the amplitude of FRF for transmitted force, i.e. the transmissibility ratio

FRF 2nd order system damping ratio=0.05 Support motion: z=b cos(t) damping ratio=0.1 100 damping ratio=0.2 damping ratio=0.5

Amplitude ratio for for ratio Amplitude

support motion (AB) motion support

0.1 0 0.25 0.5 0.75 1 1.25 1.5 1.75 2 frequency ratio (f)

MEEN 459/659 Notes 1c © Luis San Andrés (2019) 1-52 EXAMPLE: A 3000 lb (empty) automobile with a 10’ wheel-base has wheels which weigh 70 lb each (with tires). Each tire has an effective stiffness (contact patch to ground) of 1000 lb/in. A static test is done in which 5 passengers of total weight 800 lb climb inside and the car is found to sag (depress toward the ground) by 2”. (a) From the standpoint of the passenger comfort, what is the worst wavelength (in feet) [sine wave road] which the car (with all 5 passengers) could encounter at 65 mph? (b) For the worst case in (a) above, what percent of critical damping is required to keep the absolute amplitude of the vertical heaving oscillations less than ½ of the amplitude of the undulated road? (c) What is the viscous damping coefficient required for the shock absorber on each wheel (assume they are all the same) to produce the damping calculated in (b) above? Give the physical units of your answer. (d) State which modes of vibration you have neglected in this analysis and give justifications for doing so.

Let

The wavelength is equal to = v T, with T as the period of motion. And the frequency () of the forced motion is:

2 2 v   T 

The system mass is:

W ( 3000  800  4x70 ) M   eq g g

800 lb lbf 3,520 lbf lbf sec2 KM 400 ;   9.12 2 in ineq 386.4 in/sec2 in

MEEN 459/659 Notes 1c © Luis San Andrés (2019) 1-53 1/ 2  K  rad      6.2 (1.05 Hz) n    M eq  sec

(a) For passenger comfort, the worst wavelength (in feet) which the car could encounter at 65 mph is when the excitation frequency coincides with the system natural

frequency, i.e.  = n. Thus from

22v    . Then T  n 5,280 ft / mile 2 65 mph  2v 3,600 sec/ hour    =  = 90.47 ft = (0.0171 miles)  rad n 6.62 sec

(b) For the worst case what percent of critical damping is required to keep the absolute amplitude of the vertical heaving oscillations less than ½ of the amplitude of the Y 1 undulated road? i.e. What value of damping ratio () makes  at  ? b 2 n

Recall that at  =n, the amplitude of the support FRF is from eqn. (77):

1/2 1 (2 )2 1 13 A 2 ? 2  B (2 ) 2 4  4

The solution indicates that the damping ratio () is imaginary! This is clearly impossible. Note that the amplification ratio AB > 1 at f = 1, i.e. the amplitude of motion |Y| for the system will always be larger than the amplitude of the base excitation (b), regardless of the amount of damping.

(c) What is the viscous damping coefficient required for the shock absorber on each wheel (assume they are all the same) to produce the damping calculated in (b) above?

No value of viscous damping ratio () is available to reduce the amplitude of motion. However, if there should be one value, then

Deq 4 D 1 lb f  ; then  DM  eq n  2 MMeq  n 2 eq n 2 in/sec

MEEN 459/659 Notes 1c © Luis San Andrés (2019) 1-54

(d) State which modes of vibration you have neglected in this analysis and give justifications for doing so.

Heaving (up & down) motion is the most important mode and the one we have studied. In this example, pitching motion is not important because the road wavelength  is large. We have also neglected yawing which is not important if the car cg is low.

One important mode to consider is the one related to “tire bouncing”, i.e. the tires have a mass and spring coefficient of their own, and therefore, its natural frequency is given by

1,000 rad   74.25 n tire 70 / 386.4 sec

However, the car bouncing natural frequency is 6.62 rad/sec is much lower than the tire natural frequency, i.e.

ncar = 6.62 rad/s < < n tire = 74.25 rad/sec

Therefore, it is reasonable to neglect the “tire” bouncing mode since its frequency is so high that it can not be excited by the road wavelength specified.

MEEN 459/659 Notes 1c © Luis San Andrés (2019) 1-55 ME459/659 Notes# 1d (pp. 56-60)

Dynamic Response of SDOF Second Order Mechanical System: Viscous Damping

MXDXKXF   ()t

DYNAMIC RESPONSE OF A SDOF SYSTEM TO ARBITRARY PERIODIC LOADS

DYNAMIC RESPONSE OF SDOF SYSTEM TO ARBITRARY PERIODIC LOADS

Fourier Series fundamental period=0.236 s Forces acting on structures 20 are frequently periodic or can T  0.236s be approximated closely by 10 F superposition of periodic 0

loads. force (N ) 10 T

As illustrated , the 20 0 0.2 0.4 function F(t) is periodic but t not harmonic. tim e (sec)

Any periodic function, however, can be represented by a convergent series of harmonic functions whose frequencies are integer multiples of a certain fundamental frequency .

The integer multiples are called harmonics. The series of harmonic functions is known as a FOURIER SERIES, written as

 a0 F( t )  ann cos( n  t )  b sin( n  t ) (67) 2 nn11

with FF()()t T t and T=2/ is the fundamental period. an, bn are the coefficients of the nth harmonic, and related to F(t) by the following formulas

2 tT a F( t )cos( n  t ) dt , n  0,1,2,......  n T t (68) 2 tT b F( t )sin( n  t ) dt , n  1,2,......  n T t each representing a measure of the participation of the harmonic content of cos(nt) and sin(nt), respectively. All the a0, bm, cm have the units of force.

Note that (½ a0) is the period averaged magnitude of F(t) .

In practice, F(t) can be approximated by a relatively small number of terms. Some useful simplifications arise when

If F(t) is an EVEN fn., i.e., F(t) = F(-t) then, bn = 0 for all n

If F(t) is an ODD fn., i.e., F(t) = -F(-t) then, an = 0 for all n

The Fourier series representation, Eq. (67), can also be written as

a  F( t )0  c cos( n  t   )  nn (69) 2 n1

1/2 2 2 1 bn where cn a n  b n and n  tan , n  1,2,.....  an are the magnitude and phase angle, respectively, of the nth harmonic frequency (n).

RESPONSE OF UNDAMPED SYSTEM

For an undamped SDOF system, the steady state response (w/o the transient solution) produced by each sine and cosine term in the harmonic load (Fourier series) is

bm am K K Xsmt() 2 sin  m t , Xcmt() 2 cos m t m=1,2, (70) 1 fm 1 fm

m where fmn,  K M . For the constant force a0, the n

1 a0 s-s response is simply X 0  2 . K

Using the principle of superposition, gives the system response as the sum of the individual components:

11a  X0  acos( m  t )  b sin( m  t ) ()t 2  m m  K21 f (71) m1  m 

Note when mn , i.e., there is a harmonic frequency equal to the natural frequency of the system, then the system response will (theoretically) be UNBOUNDED (the system will fail!).

PERIODIC FORCED RESPONSE OF A DAMPED SDOF In a damped SDOF system, the steady-state response produced by each sine and cosine term in the harmonic load series is 2 a 1fmm cos( m  t )  2 f sin( m  t ) Xc  m  mt() 2 2 (72a) K 12ff2     mm

2 b 2 fmm cos( m  t )  1  f sin( m  t ) Xs  m  mt() 2 (72b) K 12ff2   2  mm

Superposition gives the total system response as 2 a 1  am12 f m  f m b m X( t )0    cos( m t ) 2 2 2 K K m112ff2   mm   2 1  bm12 f m  f m a m (73)   sin()mt 2 2 K m112ff2   mm   o r,  a 1  c X( t )0   m cos( m  t  ) 2 2 m 2 K K m112ff2   mm   (74)

m where, fmn;  K M n 2 bm12 f m  f m a m c a22 b  tan1 ,m 1,2, and m m m ; m 2 . . . .  am12 f m  f m b m

ORIGIN := 1 d2 d M⋅ Y + C⋅ Y + KY⋅ = Kzt⋅ () where dt2 dt z (t) is an external periodic excitation function

Define periodic excitation function: Example - square wave T z := 0.1⋅m T d o zt( ):= amp← z if t < T := o 2 .33333 T amp← − z if t > o 2

amp 0.2

2⋅π Ω := fundamental frequency T zt() 0

Ω amplitude (m) = 0.333 0.2 0 0.063 0.13 0.19 ωd N := 7 t F number of Fourier coefficients time(s)

Find Fourier Series coefficients for excitation z(t)

T 1 ⌠ mean value a := ⋅⎮ zt()dt 0 T ⌡ 0 a0 = 0m j := 1 .. NF T T coefs of cos & sin 2 ⌠ 2 ⌠ aj := ⋅⎮ zt( )⋅ cos() j⋅Ω⋅t dt bj := ⋅⎮ zt( )⋅ sin() j⋅Ω⋅t dt T ⌡0 T ⌡0

T T a = ()0000000m b = ()0.127 0 0.042 0 0.025 0 0.018 m

Build z(t) as a Fourier series .5 N 2 2 F z := ⎡()aj + ()bj ⎤ aFj ⎣ ⎦ ZF()t := a0 + ()aj⋅cos() j⋅Ω⋅t + bj⋅sin() j⋅Ω⋅t ∑ Amplitude j = 1

0.2 NF = 7

zt() 0 Z ()t Fourier coefficients F 0.2 amplitude (m)

zaF 0.2 j 0.1 0 0.063 0.13 0.19

t amplitude time(s) 0 actual 012345678910 Fourier Series j

Find the forced response of the system, i.e, find Y(t)

NF SYSTEM RESPONSE is: Yt():= Yo + ∑ ⎣⎡()Ycm⋅cos( m⋅Ω⋅t + Ysm⋅sin() m⋅Ω⋅t ⎦⎤ m = 1

K Y := a ⋅ o 0 K m := 1 .. NF m⋅Ω (a) set frequency ratio fm := ωn (b) build denominator 2 2 2 denm := ⎣⎡1 − ()fm ⎦⎤ + ()2⋅ζ⋅fm

2 K ⎣⎡am⋅⎣⎡1 − ()fm ⎦⎤ − 2⋅ζ⋅fm⋅bm⎦⎤ Ycm := ⋅ K denm (d) build coefficient of sin()

2 K ⎣⎡bm⋅⎣⎡1 − ()fm ⎦⎤ + 2⋅ζ⋅fm⋅am⎦⎤ Ysm := ⋅ K denm

(e) for graph of components 2 2 .5 Y := ⎡()Ycm + ()Ysm ⎤ Fm ⎣ ⎦

NF Yt():= Yo + ∑ ()Ycm⋅cos() m⋅Ω⋅t + Ysm⋅sin() m⋅Ω⋅t m = 1

Now graph the response Y(t) and the excitation (Fourier) z(t):

excitation & response T 0.4 Ω = 3 = 0.332 Td ωn 0.2 Fourier coefficients 0.2

0 0.13 z( & Y(t) [m] 0.2

amplitude 0.067

0.4 0 0.11 0.23 0.34 0.45 0.57 0 time (s) 012345678910 Z input Z Y output Y 3 periods of fundamental excitation motion Note: obtain response for inputs with increasing frequencies (periods decrease)

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excitation & response T 0.4 5 Ω = = 0.199 T d ω n 0.2 Fourier coefficients 0.2