Problem 2 (Exam 2 Fall 02) Identification of parameters MEEN 363/LSA/ FALL 07

The figure shows the dynamic free response (amplitude [ft] versus time [sec]) of a simple mechanical structure. Static load measurements determined the structure stiffness K=1000 lbf/in. From the measurements, determine a) damped period of motion Td (sec) b) damped natural frequency ωd [rad/s], c) Using the concept of log-dec (δ), if applicable, determine the system damping ratio ξ. Explain your method d) Undamped natural frequency ωn [rad/s], e) Estimate the system equivalent mass, Me [lb] f) Estimate the system damping coefficient, Ce [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⋅π ωd := rad Td ω = 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: δ := ⋅ln⎜ ⎟ n X ⎝ n⎠ δ = 0.305

from log-dec formula 2⋅ξπ⋅ δ = δ 0.5 ξ := 2 .5 ()1 − ξ 2 2 ()4⋅δπ + ξ = 0.049 1 δ Note that approximate formula: = 0.049 is a very good estimation of damping 2⋅π ratio

(d) Determine damped natural frequency: ωd ω := n 0.5 2 rad ()1 − ξ ω = 41.937 n sec a little higher than the damped frequency (recall damping ratio is small)

(e) Determine system 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 ωn Me = 219.526 lb

(f) Determine system damping coefficient: 0.5 Ce := ξ⋅2⋅()KM⋅ e sec C = 2.314 lbf ⋅ e in

Note: sec Actual values of parameters are M= 220lb C= 2.4 lbf ⋅ in ζ = 0.05

2 MEEN 363/502 FALL 06: PROBLEM 1

The car of mass M=500 kg is traveling at constant speed Vo= 50 kilometer/hour when it hits a rigid wall. A spring (K) and a viscous dashpot (C) represent the car front bumper system. The system natural frequency fn=3 Hz, and the dashpot provides critical damping. Disregard friction on the car wheels and ground. a) Derive the EOM for the car after the collision using the coordinate Y(t) that has its origin at the car location when the bumper first touches the wall. Provide initial conditions in speed and displacement. [10] b) State the solution to the EOM, i.e. give the system response Y(t) as a function of the system parameters [natural frequency, damping ratio, etc] and initial conditions. [10] c) Find the maximum bumper deflection and the time, after collision, when this event occurs. Compare the calculated deflection with the maximum deflection for a system without damping. Comment on your findings. [5] d) Sketch the motion Y(t) vs. time (to 0.5 sec) labeling the axes with physical units and noting important parameters. Will the car bounce from the wall? Explain your answer. [5] 1000⋅ m kph := a) EQN of motion for car after collision, use Y coordinate: (Y=0, no bumper deflection) 3600⋅ s 2 d d F = C⋅ Y Y(t) M⋅ Y = −F − F damper 2 damper spring dt dt V F = KY⋅()− 0 spring K bumper Rigid Thus the EOM is: 2 M wall d d M⋅ Y + C⋅ Y + KY⋅ = 0 (1) car 2 dt dt C

with I.C's at t=0 Y0()= Y0 = 0m⋅ no bumper deflection, and d m and Y = V0 V0 := 50⋅ kph= V0 = 13.889 car speed at instant of collision with wall. dt s Known: natural frequency and mass of car fn := 3Hz⋅ M:= 500⋅ kg ωn := fn⋅π2⋅ 2 natural period of motion calculate stiffness: KM:= ⋅ω 5 N n K= 1.777× 10 1 m Tn := fn Transient response of critically damped system, step force Fo=0 ζ=1

2 at t=0 d d m M⋅ Y + C⋅ Y + KY⋅ = Fo + initial conditions Y0 := 0m⋅ V0 = 13.889 and Fo := 0N⋅ F 2 dt s o dt Yss := K From cheat sheet, ss⋅ t Yt()= e ⋅ A + tA⋅ + Y (2) ss := −ω root of characteristic eqn. soln is: ()1 2 ss where n 1 A1 := Y0 − Yss ()A1⋅s + A2 = V0 ss= − 18.85 s A2 := V0 − A1⋅ss m A1 = 0m A2 = 13.889 s =Vo The dynamic response of the car Y(t) is given by: ss⋅ t Yt():= e ⋅t⋅A2 (3)

The graph below displays the car motion for a time equal to 3 periods of natural motion (undamped). Note the overshoot and largely damped response w/o oscillations. The car does not bounce from the wall. Tmax := 1.5⋅ Tn without damping, max bumper deflection is

1 2 1 2 V0 Based on PCME ⋅M⋅V = KX X := 0 max max X = 0.737 m 2 2 ωn max system response after collision 0.74 fn = 3Hz

0.55 Tn = 0.333 s

0.37 Calculation of maximum bumper deflection 0.18 It occurs when velocity is 0 m/s

Displacement Y(t) [m] Displacement (Take time derivative of soln, Eq. (3), to obtain) 0 0 0.13 0.25 0.38 0.5 ss⋅ t time (sec) Vt():= A2⋅()1sst+ ⋅ ⋅e

Graph not for exam Bumper velocity is null at time ta

()1sst+ ⋅ a = 0 system response after collision −1 13.89 Hence ta := ss ta = 0.053 s

Maximum deflection: Yt()a = 0.271 m 0 compare to undamped system: Xmax = 0.737 m Velocity V(t) [m/s] 10 Xmax 0 0.13 0.25 0.38 0.5 = 2.718 A significant reduction in deflection. time (sec) Yt()a Note how quickly the dashpot dissipates the initial kinetic energy. Comparison not for exam for completeness, compare the response with that of a lightly UNDERDAMPED system

ζ := 0.05 2 V0 −ωζ⋅ n⋅t Note that model with little ω := ω ⋅ 1 − ζ Xt():= ⋅e ⋅sin ω ⋅t d n ()d damping predicts the car will ωd bounce from wall. system response after collision 0.74 Let's find the forces transmitted to the wall - certainly same as force "felt" by car 0 Fw()t := KYt⋅ ()+ CVt⋅ ()

0.42 Felas_max := KYt⋅ ()a Displacement Y(t) [m] Displacement

1 Felas_no_damping := KX⋅ max 0 0.13 0.25 0.38 0.5 4 time (sec) Felas_max = 4.816× 10 N critically damped lightly damped 5 kN Felas_no_damping = 1.309× 10 N Bumper forces after collision 48.16 5 N K= 1.777× 10 m

24.08 Note how large is the force in the bumper. No wonder Force to wall in kN why a car crash is always a 0 0 0.13 0.25 0.38 0.5 disastrous event! time (sec) CAR BUMPER SYSTEM RESPONSE TO AN IMPULSE MEEN 363 - FA03 LSA(c)

The sketch shows a test stand for automobile bumpers. Each of the two shock absorbers consists of a spring having a stiffness K=3000 N/m that acts concentrically with a dashpot (C). The bumper mass (M) is 40 kg. The system is at rest in the static equilibrium position when a force F having an impulse of 2000 N-s acting over a short interval is applied at the centerline of the bumper.

(a) determine the value C that will bring the bumper to rest in the shortest possible time without rebound (b) If C= 1500 Ns/m determine the displacement x(t) of the bumper. (c) Find the bumper’s maximum displacement from the equilibrium position and its time of occurrence. Is the result reasonable? Explain.

Assume: F(t) bumper is rigid - undeformable X=0 denotes SEP. M (bumper) X(t)

C K C K For motions from the static equilibrium position, Shock The equation of motion is: absorber

2 d d M⋅ X + 2C⋅ ⋅ X + 2K⋅ ⋅X = F 2 dt dt

where M40kg:= ⋅ bumper mass, regarded as rigid - non deformable N K:= 3000⋅ absorber stiffness and damping coefficients m

C (damping coefficient to be determined) .5  2K⋅  rad The natural frequency equals ω :=   ω = 12.247 n  M  n s ω := n = fn fn 1.949 Hz 2⋅π (a) The system must be critically damped to bring the damper to rest in the shortest possible time. Thus, the amount of physical damping C must equal (for each shock absorber)

.5 C := 22K⋅()⋅ ⋅M s crit C = 979.796 N⋅ crit m

1 Ns⋅ C := ⋅C C= 489.898 Amount of damping on each absorber 2 crit m

Ns⋅ 2C⋅ (b) If C:= 1500⋅ , the damping ratio is ξ := ξ = 3.062 i.e. an overdamped system m Ccrit The solution of the EOM, with RHS = 0, is:

s ⋅t s ⋅t s ⋅t s ⋅t ⋅ 1 + ⋅ 2 d ⋅ ⋅ 1 + ⋅ ⋅ 2 xt()= Ae Be x = As1 e Bs2 e dt The roots of the characteristic equation are .5 := −ωξ⋅ω+ ⋅()ξ2 − s1 n n 1 1 s = −2.056 1 s .5 2 1 s := −ωξ⋅ω− ⋅()ξ − 1 s = −72.944 2 n n 2 s

satisfying the initial conditions. At t=0 s x0()= 0= AB+ Thus, BA= − no initial displacement

The impulse produces an initial velocity equal to Imp v0()= v = = As⋅()− s as explained in class Imp:= 2000⋅ N⋅s 0 M 1 2

:= Imp vo = m M vo 50 initial velocity is quite large! s (180 km/hour!)

vo A := − ()s1 s2 A= 0.705 m for graphing thus, the dynamic response of the bumper after the impuse acts is: := 1 Tn fn DISPLACEMENT ⋅ ⋅  s1 t s2 t T := 3T⋅ xt():= A⋅ e − e  max n

1 Note large overshoot (max deflection) and return xt() 0.5 to equilibrium with no oscillations. Disp X [m]

0 0 0.51 1.03 1.54 2.05 2.56 3.08 t time (s)

VELOCITY ⋅ ⋅  s1 t s2 t  vt():= A⋅ e ⋅s − e ⋅s   1 2 m v0s()50⋅ = s s

50

vt() 0 Vel V [m/s]

50 0 0.51 1.03 1.54 t time (s)

Max bumper displacement (deflection) occurs when velocity equals 0, i.e

 s ⋅t s ⋅t  ⋅ 1 M⋅ − 2 M⋅  v0= = Ae s1 e s2

⋅ s1 tM e s2 = ⋅ s2 tM s1 e

take natural log of both sides to find  s2   s2  ln  ()s − s ⋅t = ln  1 2 M  s1   s1  t := M − s1 s2

= tM 0.05 s

and the maximum bumper deflection is = xt()M 0.618 m

This is a large deflection! probably bumper will deform plastically before this occurs. Example Problem: Seismic instrument MEEN 363 FALL 02 kinetics of M-K-C system The figure shows a seismic instrument with its case rigidly attached L San Andres to a vibrating surface. The instrument displays the motion Y(t) Case of mass Mc RELATIVE to the case motion Z(t). a) Derive the equation of motion for the instrument using the Vibrating K D relative motion Y(t) as the output variable. Show all surface assumptions and modeling steps for full credit. M

b) Determine the instrument natural frequency ωn [rad/s]and critical damping Dc [lb.s/in] for M=0.02 lb and K=13 lb/in. c) When will the instrument show Y=0, i.e. no motion? Y(t) Z(t) d) If the vibrating surface moves with a constant acceleration of 3 g, what will the instrument display at s-s? Give value in inches. Note: the sensor works under all attachment configurations, i.e. vertical, horizontal, top, bottom, etc. a) EQN of motion using Y (relative) coordinate:

The relationship between the absolute motion X(t) of the sensor mass and the case motion is

Xt()= Zt()+ Yt() (1)

PCLM (Newton's Law) refers to inertial (absolute) references:

2 No effect of gravity since d M⋅ X = −F − F sensor is positioned horizontal. 2 damper spring (2) dt Actually, not important for any dynamic motion.

with d Fdamper = D⋅ Y Fspring = KY⋅ (3) dt

Substitution of (3) and (1) into (2) gives the EOM as:

2 2 d d d M⋅ Y + D⋅ Y + KY⋅ = −M⋅ Z (4) 2 2 dt dt dt

b) Determine system parameters: in W:= 0.02⋅ lbf g= 386.089 2 W s M := g lbf lbs K13:= DD= ⋅ Damping coeffic.not in in specified

.5 natural frequency  K  rad ωn :=   ωn = 500.957  M  s

.5 lbf⋅ s Critical damping Dc := 2KM()⋅ Dc = 0.052 in Damping ratio. D ζ := .5 2KM()⋅

c) Instrument will record NO motion Y=0, if Z(t)= cte or dZ/dt=cte, i.e. the "vibrating surface" remains stationary or moves at constant speed.

d) if the vibrating surface moves with a constant acceleration equal to 3 g's. Then, the instrument will display, at steady state (after transients have disappeared due to damping): M Ys := −3⋅g⋅ K − 3 Ys = −4.615 × 10 in

Let's calculate the instrument dynamic response:

For example, let ζ := 0.10

Then the sensor response for a sudden (3g) acceleration becomes with zero initial conditions: .5 m ()2 Y0 := 0m⋅ V0 := 0⋅ ωd := ωn⋅ 1 − ζ s

V0 + ζω⋅ n⋅()Y0 − Ys  C1 := ()Y0 − Ys C2 := ωd  −ωζ⋅ ⋅t n π Yt():=  e ⋅()C1⋅cos()ωd⋅t + C2⋅sin()ωd⋅t + Ys  2  Td :=    ωd 

0 Ys = −0.117 mm

0.1

Y(t) [mm] 0.2

0.3 0 0.05 0.1 time [s] Response of seismic instrument to sudden acceleration

SEE Video_seismic_instrument The spring-mass-damper system represents a MEEN 363 SP08 EXAM 2: PROBLEM 1 package cushioning an electronic component. The package rests on a hard floor. A pulse force F(t) of very, very short duration is exerted on the F(t) package as shown. The component vibrates without rebounding. Let K=60 lb/in, M= 5.9 lb, and C=0.04 lb-sec/in, Fo=2000 lbf a) Calculate the system natural frequency (Hz) and damping ratio [8]. M b) Provide an engineering estimation (value) for K the maximum system velocity (ft/sec). C t c) The maximum deflection (ft) of the spring (displacement of system) and the time when it occurs. To =0.001sec d) Draw a graph of the system displacement x vs. time. Label all physical coordinates.

KEY: KNOWLEDGE base: DOES APPLIED FORCE ACT OVER A LONG TIME OR NOT? IS time To very long?

Pulse: force magnitude F:= 2000⋅ lbf applied during To := 0.001⋅ sec (a) Find the natural frequency, natural period of motion and viscous damping ratio:

lbf sec K60:= ⋅ M:= 5.9⋅ lb C:= 0.04⋅ lbf ⋅ in in

NATURAL FREQUENCY: K rad ω := ωn ω = 62.66 n M n sec fn := fn = 9.973 Hz 2⋅π 1 Tn := Tn = 0.1 sec natural period of motion DAMPING RATIO: fn C ζ := 2M⋅ω⋅ ζ = 0.021 a small value n .5 2 rad ω := ω ⋅()1 − ζ ω = 62.647 DAMPED NATURAL FREQUENCY: d n d sec 2⋅π Td := T = 0.1 sec ωd d Tn The damping ratio is very small, thus the system is very lightly damped. = 100.274 To (b) engineering judgment: The (damped) natural period of motion is very large compared to the time the load acts. Hence, the pulse can be treated as an impulse, which changes "instantaneously" the initial velocity of the system. This initial velocity equals Impulse:= F⋅ T Impulse (1) o Vo := in M V = 130.877 o sec This initial velocity is (of course) the maximum ever, since damping will remove the initial kinetic energy due to the impact until the system returns to rest.

The motion starts from rest, X(0)=0.m. Thus, the response of the system becomes (cheat sheet):

V o −ωζ⋅ n⋅t Xt():= ⋅e ⋅sin()ωd⋅t (2) ωd

However, since damping is so small ζ~0, the engineering approximation leads to: V o with velocity V ()t := V ⋅cos ω ⋅t (3) Xa()t := ⋅sin()ωn⋅t a o ()n ωn

The maximum displacement is: Vo Xmax := ωn Xmax = 2.089 in

which occurs at a time ~ 1/4 natural period of motion, i.e Tn = 0.025 sec 4 Let's graph the responses (displacement and velocity):

Displacement vs time 4 Note how close the engineering approximation is with respect to 2 the exact solution, in particular during the first period of motion. 0 Xmax = 2.089 in 2 displacement X (in) Note that Xmax can also be easily 4 0 0.05 0.1 0.15 0.2 0.25 0.3 determined from PCME: T=V time (s) Xa 1 2 1 2 ⋅M⋅V = ⋅K⋅X X exact 2 o 2 max

ft V = 10.906 o sec Velocity vs time 20

10

0

velocity V (m/s) 10

20 0 0.05 0.1 0.15 0.2 0.25 0.3 time (s) Va Vexact V d ⎛ o −ωζ⋅ n⋅t ⎞ Maxiimum displacement happens when speed =0, i.e. ⎜ ⋅e ⋅sin()ωd⋅t ⎟ = 0 dt⎝ ωd ⎠ Take time derivative of X(t) and obtain time t_

−ωζ⋅ n⋅ωsin()ωd⋅t_ + d⋅cos()ωd⋅t_ = 0 0.5 2 ωd ()1 − ζ tan()ωd⋅t_ = = ωn⋅ζ ζ

⎡ 0.5⎤ ⎢ 2 ⎥ ()1 − ζ 180 Let: θ_:= atan⎢ ⎥ θ_⋅ = 88.803 ⎣ ζ ⎦ π θ_ and time for maximum displacement is t_ := t_ ω d = 0.247 t_= 0.025 sec Tn

and the maximum deflection is Xt_( )= 2.022 in Xt_() Xmax = 2.089 in compare it to engineering approx: = 0.968 Xmax

excellent agreement! Q3: Description of motion in a moving reference frame

An instrument package installed in the nose of a rocket is cushioned Luis San Andres, MEEN 363 (c) FALL 2010 against vibration with a soft spring-damper. The rocket, fired vertically from rest, has a constant acceleration ao. The instrument mass is M, and the support stiffness is K with damping C. The instrument-support system is underdamped. The relative motion of rocket head the instrument with respect to the rocket is of importance. a) Derive the equation of relative motion for the instrument b) Give or find an analytical expression for the relative displacement sensor of the instrument vs. time. Express your answer with well defined parameters and variables. M ACME ACME c) Given M=1 kg, K=1 N/mm, and damping ratio ζ=0.10. Find the K C natural frequency & damping coefficient of the system d) For ao=3g, find the steady-state displacement of the instrument, relative to rocket and absolute.

ao

Definitions: coordinate systems Xt() Absolute displacement of instrument recorded from ground M Zt() Absolute displacement of rocket from ground. K C Y=X-Z YXZ= − displacement of instrument relative to rocket az=ao X d2 aZ = Z = ao acceleration of rocket fired from REST Z dt2

N ground sensor parametes: M1kg:= ⋅ K := 1000⋅ m

ao := 3g⋅ ζ := 0.10

Equation of motion for instrument Newton's Laws are applicable to inertial CS from free body diagram, let Y=(X-Z)>0 Free body diagram d2 M⋅ X = −W − Fs − Fd (1) where dt2 M

Fs = −W + KX⋅()− Z= −W + KY⋅ (2a) W is the spring force supporting instrument. The dashpot force is Fs=-W + K (X-Z) d Fd=-C d(X-Y)/dt Fd = C⋅ ()XZ− ()2b dt Substitute Eqs. (2) into Eq.(1): d2 d 2 M⋅ X = −W + W − KY⋅ − C⋅ Y d d 2 dt M⋅ X = −K⋅Y − C⋅ Y dt dt2 dt But interest is in the relative motion Y; d2 d hence, substitute X=Y+Z M⋅ ()YZ+ + KY⋅ + C⋅ Y = 0 dt2 dt

⎛ d2 d ⎞ M⋅⎜ Y + C⋅ Y⎟ + KY⋅ = −M⋅a = −M⋅a (3) is the desired EOM. ⎜ 2 ⎟ Z o ⎝ dt dt ⎠ Find natural frequency and damping coefficient natural frequency of sensor is: .5 ⎛ K ⎞ ω := ⎜ ⎟ rad n ⎝ M⎠ ωn = 31.623 2⋅π s Natural period: Tn := ωn damping coefficient. 0.5 C := ζ⋅2⋅()KM⋅ s C= 6.325 N⋅ T = 0.199 s m n

damped natural 2 0.5 frequency ωd := ωn⋅()1 − ζ rad ω = 31.464 d s

Motion starts from rest Solution of ODE - prediction of relative motion m Yo := 0m⋅ Vo := 0⋅ The solution of ODE Eq. (3) with null initial conditions since s motion starts from rest is (Use cheat sheet) −M⋅ao Y := − ζ⋅ωn⋅t s K YY= s + e ⋅()C1⋅cos()ωd⋅t + C2⋅sin()ωd⋅t V + ζω⋅ ⋅C C := Y − Y ()o n 1 is the formula describing 1 o s C2 := the motion of instrument ωd relative to rocket − 3 C1 = 0.029 m C2 = 2.957× 10 m and, after long time Y approaches: Ys = −0.029 m To find the instrument absolute displacement, first determine the absolute motion of the rocket, i.e. t2 velocity V ()t := a ⋅t and displacement Zt():= a ⋅ z o o 2

The absolute displacement of the sensor is XYZ= + after very-long tines, times removes the homogenous (transient) response; and the sensor reaches its steady state motion XSS()t := Z() t + Ys not for Quiz Lets graph the relative and absolute displacements of the sensor

without damping − ζ⋅ωn⋅t Yt():= Ys + e ⋅()C1⋅cos()ωd⋅t + C2⋅sin()ωd⋅t Y_() t := Ys⋅()1− cos()ωn⋅t

for plots, set Tmax := 7T⋅ n

Relative displacement of sensor w/r to rocket

Y5T()⋅ n = −0.028 m 0 0 note the effect of damping − 0.02 Y Ys = −0.029 m Ys [m] − 0.04

− 0.06 0 0.5 1 time (s) Underdamped UNDAMPED

The absolute displacement of the sensor isXYZ= + where

Displacements - Sensor (X) and Rocket (Z)

30 1 m kmh := ⋅ [m] 3.6 s X, 20

Z 10

0 0 0.5 1 time (s) X Z Application: a torsional pendulum LSA for MEEN 363 FA10 A device designed to determine the moment of inertia of a wheel-tire assembly consists of a 2 mm diameter steel suspension wire, 2 m long, and a mounting plate, to which it is attached the wheel-tire assembly. The suspension wire is fixed at its upper end and hangs vertically. When the system oscillates as a torsional pendulum, the period of oscillation without the wheel tire assembly is 4 seconds. With the wheel tire assembly mounted to the support plate, the period of oscillation is 25 seconds. Determine the mass moment of inertia of the wheel tire assembly. Recall that the shear modulus for steel is G= 82.7 x 109 N/m2 and the torsional 4 stiffness of the cable is Kθ=(πd /32)G/L.

Consider the system does not have any damping - and there is no external moment acting' θ The general EOM for rotations θ(t) about a fixed axis (o-o) is

d2 Io⋅ θ + kθ⋅θ = 0 (1) dt2

where Io is a mass moment of inertia and Kθ is a torsional stiffness

The natural frequency of the system is 1 K 2 ⎛ θ ⎞ ⎛ 2⋅π ⎞ (2) ωn = ⎜ ⎟ = ⎜ ⎟ where Tn is the natural period of motion ⎝ Io ⎠ ⎝ Tn ⎠

Let: N G:= 82.7⋅ 109⋅ Shear modulus for steel m2 L2m:= ⋅ cable length d2mm:= ⋅ cable diameter ⎛ 4 ⎞ G π⋅d m Kθ := ⋅⎜ ⎟ K = 0.065 N⋅ L ⎝ 32 ⎠ θ rad

2 From (2) ⎛ Tn ⎞ Io = ⎜ ⎟ ⋅Kθ ⎝ 2⋅π ⎠ first case: period of oscillation of assembly alone Tn1 := 4s⋅ (w/o tire) Find moment of inertia of support plate 2 ⎛ Tn1 ⎞ Iplate := ⎜ ⎟ ⋅Kθ ⎝ 2⋅π ⎠

2 Iplate = 0.026 kg m second case: period of oscillation of assembly with wheel included Tn2 := 25⋅ s

Find moment of inertia of system (plate+wheel) 2 ⎛ Tn2 ⎞ Isystem := ⎜ ⎟ ⋅Kθ ⎝ 2⋅π ⎠

2 Isystem = 1.028 kg m

Solve: find mass moment of inertia of wheel-tire

Iwheel_tire := Isystem − Iplate 2 Iwheel_tire = 1.002 kg m Application example A two-blade composite-aluminum of a wind turbine is supported on a shaft & bearing system so that it is free to rotate about its centroidal axis. A weight W=100 kg is taped to one of the blades at a distance R=20 m from the axis of rotation as shown. When a blade is pushed a small angle from the vertical position and released, the wind turbine is found to oscillate 2 periods of natural motion in 1 minute. Assume there is no friction at the support bearings. * Determine the wind turbine centroidal mass moment of inertia (IT) in (kg.m2). * The turbine weighs 2500 kg. Determine the radius of gyration.

g

R

W

MEEN 363 – FA10 – Mass Moment of Inertia 1 Make a free body diagram of turbine swinging (oscillating) about pivot O with angle θ(t)

g Assume: WT • no drag (frictionless bearings & no Wind turbine weight air drag) o • center of mass of turbine = center Rx, Ry of rotation O Reaction forces R

Let IT: mass moment of inertia of

W turbine; hence EOM (moments) about Added weight O is θ FREE BODY DIAGRAM 2   ()IMRITO+==−θ θθ WR sin (1)

For small θ angles, Eq. (1) reduces to the linear form  IMgRO θθ+ =0 (2) i.e. the typical equation of a pendulum with natural period given as 1/2 2π ⎛⎞Io Tn ==2π ⎜⎟ (3) ωn ⎝⎠MgR Hence, the wind turbine mass moment of inertia can be easily determined from the measured period of oscillation 2 2 ⎛⎞Tn IMRMgRT =− +⎜⎟ ⎝⎠2π (4) Given

MEEN 363 – FA10 – Mass Moment of Inertia 2 W=100 kgf taped to one of the blades at a distance R=20 m from the axis of rotation. Natural period Tn= 60 seconds / 2 periods of natural motion = 30 sec. Note M=W/g = 100 kg 2 T 5 2 2 ⎛⎞n I = 4.071× 10 kg m * IMRMgRT =− +⎜⎟ T ⎝⎠2π :

2 Since the turbine weighs WT=2500 kgf, and from IMrTTk= , the radius of gyration is

1/2 ⎡⎤I r = T rk = 12.761m * k ⎢⎥ = ⎣⎦MT

MEEN 363 – FA10 – Mass Moment of Inertia 3 Example - Cushioning of package MEEN 363/617 01/31/08 (c) Luis San Andres TAMU The EOM after the package first contacts hard ground is:

d2 d M := 2⋅kg M⋅ z + c⋅ z + kz⋅ = mg⋅ v dt2 dt [1] N m k := 50⋅ z mm k represent the stiffness of the packaging k, c material. M is the component mass.

Assume no damping, c=0 h d with I.C's: z()0 = 0 z = vo dt Schematic view of package falling and Equation [1] is valid for z>0, i.e. as long as packaging impacting ground (hard surface). material is being compressed.

Note that if package rebounds, then Fspring=0 h := 250⋅mm height of drop

0.5 ⎛ k ⎞ rad ωn ωn := ⎜ ⎟ M ωn = 158.114 fn := ⎝ ⎠ s 2⋅π fn = 25.165Hz Natural frequency of sytem π Natural period: Tn := 2⋅ T = 0.04s ωn n ()Mg⋅ z := [2] z = 0.392mm s-s response - after transients die out ss k ss 0.5 from free fall, assume no air drag. the package velocity when touching ground is vo := ()2⋅g⋅h

The EOM of motion w/o damping is: zo := 0⋅m d2 M⋅ z + kz⋅ = W [3] with 2 IC. m dt vo = 2.214 s

Dynamic response for undamped system, ξ := 0

The solution of this ODE is very simple, i.e. the superposition of the particular solution (zss) and the homogenous solution (periodic with natural frequency).

zt()= zss + Ac⋅cos()ωn⋅t + As⋅sin()ωn⋅t [4a] displacement of package

d [4b] velocity of package vt()= z = ωn⋅()−Ac⋅sin()ωn⋅t + As⋅cos()ωn⋅t dt d 2 at()= v = −ωn ⋅()−Ac⋅cos()ωn⋅t + As⋅sin()ωn⋅t [4c] acceleration of package dt The constants Ac and As are determined from the initial conditions. At time t=0 s, the pakage materi is not (yet) deflected z(0)=0 and the pakage initial velocity is that of the free fall, v(0)=vo From Eq [4a]: z()0 = z = 0 = z + A o ss c Ac := −zss [5a] From Eq [4b] vo()= v = A ⋅ω o s n vo As := [5b] ωn zt():= zss + Ac⋅cos()ωn⋅t + As⋅sin()ωn⋅t [4a]

NOTE that package REBOUNDS when it crosses z=0 on its upward motion (remember when you jump on a trampolin.)

What is the maximum dynamic displacement (deflection)? Prior to obtaining this deflection, let's recall some trigonometry:

.5 Let: 0.5 2 amplitude of dyanmic 2 2 ⎡ vo ⎤ AA= ()c + As ⎢ 2 ⎛ ⎞ ⎥ motion Az:= ⎢ ss + ⎜ ⎟ ⎥ ⎣ ⎝ ωn⎠ ⎦

and in the formula Ac⋅cos()ωn⋅t + As⋅sin()ωn⋅t , multiply by A and divide by A

⎛ Ac As ⎞ A⋅⎜ ⋅cos()ωn⋅t + ⋅sin()ωn⋅t ⎟ [*] A ⎝ A A ⎠ As

Now, define a phase angle Φ such that Φ

Ac As Ac cos()Φ = sin()Φ = A A As vo with tan()Φ = = and write expression [*] above as Ac ωn⋅()−zss

A⋅() cos()Φ ⋅cos()ωn⋅t + sin()Φ ⋅sin()ωn⋅t or A⋅( cos()ωn⋅Φt − ⎡ vo ⎤ where Φ := atan⎢ ⎥ + π ⎣ ωn⋅()−zss ⎦ Φ = 1.599 radians

Hence, with the aid of the simple trigonometry "trick" we write the displacement z(t), Eqn [4a] as

zt():= zss + Ac⋅cos()ωn⋅t + As⋅sin()ωn⋅t [4a] for graphs zt():= zss + A⋅() cos()ωn⋅Φt − [6a] displacement of package Tnn := 0.04⋅s Let's return to the question, what is the maximum displacement?

Clearly, the trig function cos(x) can be at most +1 or -1, hence the z()0⋅s = 0m

Max dynamic z := z + A displacement is max ss [7a] .5 ⎡ v 2⎤ ⎢ 2 ⎛ o ⎞ ⎥ zmax := zss + ⎢zss + ⎜ ⎟ ⎥ ⎣ ⎝ ωn⎠ ⎦ [m] package displacement z 0.0144 zmax = 14.403mm

0.0072 zss = 0.392mm

0

z [m] z>0 means travel downwards, i.e. compression of pakaking 0.0072 material. Motion does not die since 0.0144 there is no damping. 0 0.01 0.02 0.03 0.04 time (s)

The response found is strictly valid for z>0, i.e. when packaging material "spring" is compressed. "Spring" of package can not be stretched.

Velocity [m/s], v(t)=dz/dt

vt():= −ω A⋅ n⋅()sin()ωn⋅Φt − [6b]

2 .5 ⎡ ⎛ ωn⎞ ⎤ m what is maximum speed? vmax := A⋅ωn ⎢ ⎥ = ⎢1 + ⎜zss⋅ ⎟ ⎥ ⋅vo = 2.215 [7b] ⎣ ⎝ vo ⎠ ⎦ s velocity of package [m/s] 2.21 m 1.11 vmax = 2.215 s 0 vmax = 1 velocity (m/s) 1.11 v m o v()0⋅s = 2.214 s 2.21 0 0.01 0.02 0.03 0.04 m vo = 2.214 time (s) s acceleration [m/s2], a(t)=dv/dt,

2 at():= −ω A⋅ n ⋅()cos()ωn⋅Φt − [6c]

2 m what is maximum accel? a := A⋅ω a = 350.256 max n max 2 s 2 .5 since ⎡ v ⎤ ⎢ 2 ⎛ o ⎞ ⎥ Az:= ⎢ ss + ⎜ ⎟ ⎥ ⎣ ⎝ ωn⎠ ⎦

then .5 2 ⎡ 2 4 2⎤ amax = A⋅ωn = ⎣zss ⋅ωn + ()vo⋅ωn ⎦

but 2 ⎛ Mg⋅ ⎞ k zss⋅ωn = ⎜ ⎟⋅ = g ⎝ k ⎠ M 2 ⎛ vo⋅ωn⎞ k 1 2⋅h ⎜ ⎟ = 2⋅g⋅h⋅ ⋅ = ⎝ g ⎠ Mg⋅ g zss Hence: ⎛ 2⋅h⎞ W amax := g⋅ ⎜1 + ⎟ [7c] where zss = ⎝ zss ⎠ k Note that if packaging stiffness (k) is very large then zss is very small, and hence, the maximum (peak) acceleration can be quite large!!! Max acceleration m a a = 350.256 max if no damping: max 2 = 35.716 s g

[m/s2] Acceleration of package 350.26 peak decelerations can 175.13 be much larger than 1g! m a()0⋅s = 9.807 0 2 s

Acceleration (m/s2) 175.13 negative (peak) acceleration denotes upward force

350.26 0 0.01 0.02 0.03 0.04 time (s) The force from the cushioning into the package is F = -k z - c v = - W + M acc

Ft():= ()M− g+ at() ⋅ Fmax := Ma⋅()max + g Cushioning force 720.13 Fmax = 720.125N Mg⋅ = 19.613N 360.06 F<0 means upwards force - spring being compressed

0 Fmax = 36.716 Mg⋅ 360.06 Cushioning Force (N) Qute large!! Much larger than component weight (W) 720.13 0 0.01 0.02 0.03 0.04 time (s)

The package will REBOUND when z=0 (z<0) and dz/dt=V < 0.

Approximately at Tn trebound := 2 trebound = 0.02s

Note: The material above is copyrighted by Dr. Luis San Andres. This means you cannot distribute or copy the material w/o the consent of its author, i.e. Dr. San Andres. The contents above are for use by MEEN students in course MEEN 363 or 617