MECH 364MECHANICAL MECHANICAL VIBRATIONSVIBRATIONS

Clarence W. de Silva, Ph.D., D.Eng. (hc), P.Eng. Professor of Mechanical EngineeringThe University of British Columbia

e-mail: desilva@mech.ubc.cahttp:// www.mech.ubc.ca/~ial

C.W. de Silva

Presentation Part 3

PlanPlan

Time Response AnalysisTime Response AnalysisTo Study:• Element Behavior• Naturally Oscillating Systems• Undamped Oscillator• Damped Oscillator• Free Response• Forced Response

MECH 364 Road Map

Time Response AnalysisTime Response Analysis• Time Domain:

– Independent variable is time t– System is represented by differential equations

• Free Response:– No external forcing input– Response is due to an initial-condition excitation– Exhibits “natural” response– E.g.: system shut-down conditions– Useful in measuring natural frequency and damping ratio

• Forced Response:– Response is due to a forcing excitation– E.g.: system start-up and operating conditions– Initial-condition effects are not present (or, decayed due to

damping)– Useful in determining resonant conditions

Element BehaviorElement Behavior

Mechanical ElementsMechanical ElementsMass/Inertia Element

Spring/Flexibility Element

k

b

Damper/Dissipation Element

Mass (Inertia) Element

Constitutive Equation (Newton’s 2nd Law): m dv

dtf=

; m = mass (inertia) Power = fv = rate of change of energy è

E fv dt m dvdt

v dt mv dv= = = ∫∫∫ è Energy E mv=12

2 (Kinetic Energy)

Note: Energy storage element

Integrate constitutive equation è v t vm

f dtt

( ) ( )= +−

−∫0 1

0

Set t = +0 è v v( ) ( )0 0+ −= unless force is infinite. Note: 0− denotes instant just before t = 0 and 0+ denotes instant just after t = 0. Observations:

1. Velocity (across variable) represents the state of an inertia element è “A=Type Element” Notes: 1. Velocity at any t is completely determined from initial velocity and the

applied force; 2. Energy of inertia element is represented by v alone. 2. Velocity across an inertia element cannot change instantaneously unless infinite

force/torque applied. 3. A finite force cannot cause an infinite acceleration in inertia element. A finite

instantaneous change (step) in velocity needs an infinite force è v is a natural output (or state) variable and f is a natural input variable for inertia element.

Spring (Flexibility) Element

Constitutive Equation (Hook’s Law): dfdt

kv= ; k = stiffness Note: Differentiated version of familiar force-deflection Hooke’s law in order to use velocity (as for inertia element)

Energy E fv dt fk

dfdt

dtk

f df= = = ∫∫∫1 1

è Energy E fk

=12

2

(Elastic potential energy)

f t f k v dtt

( ) ( )= +−

−∫00

è f f( ) ( )0 0+ −= unless an infinite velocity is applied

Note: Energy storage element Observations: 1. Force (through variable) represents state of spring element è “T-Type Element” Justified because: 1. Spring force of a spring at time t is completely

determined from initial force and applied velocity; 2. Spring energy is represented by f alone.

2. Force through stiffness element cannot change instantaneously unless an infinite velocity is applied to it.

3. Force f is a natural output (state) variable and v is a natural input variable for spring.

Note: Displacement x may be used in place of force f in the above discussion.

Damping (Dissipation) Element

Constitutive Equation: f bv= b = damping constant (damping coefficient); for viscous damping

Observations:

1. This is an energy dissipating element (D-Type Element) 2. Either f or v may represent the state 3. No new state variable is defined by this element.

Gravitation Potential EnergyGravitation Potential EnergyWork done by external force in raising an object against gravity

Lumped mass m is raised to height y by a constant external force f(Note: f = mg in order to avoid acceleration). Work done by the external force è Energy:

Gravitational Potential Energy: PE = mgy

E fdy mgdy mgy= = =∫ ∫

m

y

f = mg

mg

GroundReference

Constitutive (Physical) Equations

Constitutive Relations for:

System Energy Storage Elements Energy Dissipating Elements

Type A-Type (Across) Element

T-Type (Through) Element

D-Type (Dissipative) Element

Translatory- Mechanical v = velocity f = force

Mass m dv

dtf=

(Newton’s 2nd Law) m = mass

Spring dfdt

kv=

(Hooke’s Law) k = stiffness

Viscous Damper f bv=

b = damping constant

Electrical v = voltage i = current

Capacitor C dv

dti=

C = capacitance

Inductor L di

dtv=

L = inductance

Resistor Ri v= R = resistance

Thermal T = temperature difference Q = heat transfer rate

Thermal Capacitor C dT

dtQt =

Ct = thermal capacitance

None Thermal Resistor R Q Tt =

Rt = thermal resistance

Fluid P = pressure

difference Q = volume flow rate

Fluid Capacitor C dP

dtQf =

Cf = fluid capacitance

Fluid Inertor I dQ

dtPf =

If = inertance

Fluid Resistor R Q Pf =

Rf = fluid resistance

Naturally Oscillating SystemsNaturally Oscillating Systems

km

x

(a)

(d)

l

mg

θ

(b)

KJ θ

(c)x

m

k1 k2k = k1+ k2

y

yh

l

Mass density = ρ

A(e)

(f)i

vC

VL

L

C+ -

Six Examples of Single-D.O.F Oscillatory Systems: (a) Translatory, (b) Rotatory, (c) Flexural, (d) Pendulous, (e) Liquid slosh, and (f) Electrical.

UndampedUndamped OscillatorOscillator

Mass-Spring System(Simplified model of a rail car impacting against a snubber)

Conservation of Energy:Differentiate: Generally at all tè

è

Natural frequency:

UndampedUndamped OscillatorOscillator

2 21 12 2mx kx constant+ =&

mxx kxx&&& &+ = 0&x ≠ 0 &&x k

mx+ = 0

&&x xn+ =ω 2 0ω n

km=

k

m

x

General Solution of Equation of Motion:Proof: Satisfies the equation; Has two unknowns (A and )Initial Conditions:A = amplitude; = phase angle

Free Free UndampedUndamped ResponseResponsesin( )nx A tω φ= +

φ(0) sin ; (0) coso o nx x A x v Aφ ω φ= = = =&

φA x

vo

o

n

= +22

2ω

φω

= −tan 1 n o

o

xv

Natural Frequencies of Natural Frequencies of Six Types of Systems Six Types of Systems

Approaches for Equations of MotionApproaches for Equations of Motion

Example: Conveyor System OscillationsExample: Conveyor System Oscillations

Tracked Conveyor System

Free-Body Diagram

( )1 22 0Jm x k k xr

+ + + =

&&

( )1 2 2/nJk k mr

ω = + +

k k keq = +1 2m m Jreq = + 2

( )K r k keq = +21 2J r m Jeq = +2

Heavy SpringHeavy Spring

Distributed Parameter (Continuous) System: A Heavy Spring

Note: Mass and flexibility are distributed throughout the spring

(not at a few discrete points)

Lumped Model of a Heavy Spring

ms = mass of spring; k = stiffness of spring; l = length of spring One end fixed and the other end moving at velocity v

Kinetic Energy Equivalence

Local speed of element δx = vlx . Element mass = x

lms δ è

Element kinetic energy KE = 2)(21 v

lxx

lms δ

As δx → dx, Total KE = 321

21)(

21 2

0

23

2

0

2 vmdxxlvmv

lxdx

lm s

ls

ls ∫∫ ==

è Equivalent lumped mass concentrated at free end = ×31 spring mass

Assumption: Conditions are uniform along the spring.

Lumped-Parameter Approximation of a Heavy Spring

Note: Natural frequency equivalence may give a different approximate model

Free (Unforced) Response of Free (Unforced) Response of Damped OscillatorDamped Oscillator

Natural Frequency:

Damped Oscillator (Free)Damped Oscillator (Free)

kx

b

k

m

m

mx bx kx&& &+ + = 0

&& &x x xn n+ + =2 02ζω ω

ω nkm=

2ζω nbm

=

ζ = 12

bkmDamping Ratio:

Damping constantdamping ratioDamping constant for critically damped conditions

ζ = =Note:

Steps:1. Substitute into equation of motion (i.e., seek) solution of the

form:2. Solve the resulting characteristic equation:

to give roots:

3. General response:

4. Determine the unknowns using initial conditions (ICs)

Free Response DeterminationFree Response Determination

Three Cases of Roots:1. Real and unequal è Overdamped (Nonoscillatory)2. Real and equal è Critically damped (Nonoscillatory)3. Complex conjugate è Underdamped (Oscillatory)

x Ce t= λ

λ ζω λ ω2 22 0+ + =n n

λ ζω ζ ω= − ± −n n2 1 1 2 and λ λ=

1 21 2

t tx C e C eλ λ= + λ λ1 2≠

x C e C tet t= +1 2λ λ λ λ λ1 2= =

Free Response ResultsFree Response Results

Typical Time ResponsesTypical Time Responses

0 0.2 0.4 0.6 0.8 1

-0.4

-0.2

0

0.2

0.4

0.6Responsex (m)

ζ = 0.5

Time t (s)

(a)

0 0.2 0.4 0.6 0.8 1

-0.4

-0.2

0

0.2

0.4

0.6Responsex (m) ζ = 1.0

Time t (s)

ζ = 2.0ζ = 4.0

(b)

Period:

Logarithmic Decrement Method ofLogarithmic Decrement Method ofDamping MeasurementDamping Measurement

Decay Ratio During r Periods:

Td

=2πω

( )

sin( )( )( ) sin[ ( ) ]

n

n

td

t rTd

Ae tx tx t rT Ae t rT

ζω

ζω

ω φω φ

−

− +

+=

+ + +

Note:2

2

1( )

nn

n

rt

rTt rT

e e ee

π ζζω

ζζωζω

−−

− += = = 2 and d rT rω π=2

2 21

n nd

T π πω ω

ω ζ= =

−

2

2exp1

i

i r

A rA

π ζ

ζ+

=

− Using peaks è

è

2

Logarithmic decrement per cycle:

1 2ln1

i

i r

Ar A

πζδ

ζ+

= =

− ( )2

1 = Logarithmic decrement per radian21 2 /

δζ

ππ δ= ≈

+è

Dependence of Free Response Dependence of Free Response (Stability) on Pole Location(Stability) on Pole Location

A and B are stable; C is marginally stable; D and E are unstable

Im

Re

s-Plane

(Eigenvalue Plane)

E

E

A

A

DB

C

C

Forced Response of Forced Response of Damped OscillatorDamped Oscillator

Equation of Motion:

Forced OscillatorForced Oscillator

è

Massm

Viscous Damperb

Springk

y

f(t)

( )my by ky f t+ + =&& &

&& & ( )y y y u tn n n+ + =2 2 2ζω ω ω

Note: In control systems nomenclature, u = input; y = output

Normalized force u(t) = f(t)/k (has units of displacement)

Forced Response ComponentsForced Response Components

Particular Solution Method1. Add a suitable particular solution (a solution that satisfies the forced

equation) to the homogeneous solution (general solution of free response)2. Determine the unknown constants (in the homogeneous solution) by

using ICs

Convolution Integral Method1. Determine the impulse response h(t)—the response to a unit impulse

(assumes zero ICs)2. Use convolution integral to determine the zero-IC response:

3. Add the zero-input response (i.e., IC response)

Laplace Transform Method1. In the equation of motion represent time-derivative by Laplace variable

s and corresponding ICs2. Express the forcing function (input) by its Laplace transform 3. Express the response in terms of s (algebraic manipulations)4. Using Laplace tables determine the inverse Laplace

Methods of Forced Response DeterminationMethods of Forced Response Determination

0 0( ) ( ) ( ) ( ) ( )y t h t u d h u t dτ τ τ τ τ τ

∞ ∞= − = −∫ ∫

Particular Solutions for Useful Input Functions

Useful Results on Useful Results on Forced Response Forced Response