Mechanical Vibration
Iva Petríková
Department of Applied MechanicsDepartment of Applied Mechanics
Introduction, basic definitions
• Oscillatory process – alternate increases or decreases of physical
quantities (displacements, velocities, accelerations)
• Oscillatory motion is periodic motion• Oscillatory motion is periodic motion
• Vibration of turbine blades, vibration of machine tools, electrical
oscillation, sound waves, vibration of engines, torsional vibration
of shafts, vibration of automobiles etc.
• Mechanical vibration - mechanisms and machines, buildings,
bridges, vehicles, aircrafts – cause mechanical failure
• Harmonic, periodic general motion
1.3.2018 2
Elastic elements
• Discrete elements (masses) +
linear and torsional springslinear and torsional springs
• Continuos structural elements –
beams and plates
• Number of degrees of freedom
(DOF) – minimum number of
coordinates
Single degree of freedom systems
System with infinite number DOFTwo degree of freedom systems
Oscillatory motions
• Periodic motion with
harmonic components
• Harmonic motion
harmonic components
Periodic motion repeating itself The simplest form of periodic motion isPeriodic motion repeating itself
after a certain time interval.
The simplest form of periodic motion is
harmonic motion – sin, cos
Harmonic motion
• Displacement of harmonic motion is given:
( )sinx X tω ϕ= +
• x, x(t) ... displacement [m]
• X ... amplitude of displacement [m]
• ... phase
• ω ... angular velocity [s-1]
• T ... natural period of oscillation [s]
( )tω ϕ+
2T
πω
=• T ... natural period of oscillation [s]
• f ... frequency [s-1], [Hz] - Hertz
• ϕ ... phase angle
Tω
=
2 fω π=
• velocity:
• ωX ... amplitude of velocity [ms-1]
( )cosdx
x X tdt
ω ω ϕ= = +ɺ
• ωX ... amplitude of velocity [ms ]
• acceleration:
( )2
22
sind x
x X tdt
ω ω ϕ= = − +ɺɺ
• -ω2X ... amplitude of acceleration [ms-2]
Simple degree of freedom systems
mass, spring, damper, harmonic excitation
( )mx bx kx F t+ + =ɺɺ ɺForcing function –
harmonic excitation
0mx bx kx+ + =ɺɺ ɺDamped free vibration
0mx kx+ =ɺɺUndamped free vibration
Simple degree of freedom systems
( )mx bx kx F t+ + =ɺɺ ɺ
0( ) sinF t F tω=
m ... mass
b ... (viscous) damping coefficient
k ... stiffness coefficient
F0 ... amlitudes of force
ω ... frequency of harmonic force
bcr ... critical damping coefficient
ζ ... damping factor
k... natural (circular) frequency
... frequency ratio
k
mΩ =
ωη =Ω
Undamped free vibration0+ =ɺɺmx kx
Solution of the 2nd order differential equation
Assumed solution
Characteristic equation: ( ) Ω − Ω= +i t i tx t Ae Be free vibration
( ) tx t Aeλ=
Characteristic equation: 2 0λ + =m k
1,2
1,2
λ
λ
= ±
= ± Ω
ki
mi
2 0λ + =k
m2λ = − k
m
Two arbitrary constants A a B, determinedfrom initial conditions:
( ) ( )0 00 , 0= =ɺx x x v
( ) ( )i t i tx t i Ae BeΩ − Ω= Ω −ɺ
0x A B= +
( )0v i A B= Ω −1,2
Ω = k
mNatural frequencyof the single DOF systems
0k
x xm
+ =ɺɺ2 0x x+ Ω =ɺɺ
0 01
2
ix vA
i
Ω + = Ω
0 01
2
ix vB
i
Ω − = Ω
Damped free vibration0mx bx kx+ + =ɺɺ ɺ
Solution of linear differential equation of 2nd order:
2 0m b kλ λ+ + = ( ) ( ) ( )2 21 1ζ ζ ζ ζ− + − Ω − − − Ω= +
t tx t Ae Be
21,2
14
2 2
bb mk
m mλ −= ± −
2
2
1,2 2
41
2 2 242
b mk b b k bi i
m m m mm km
m
λ
− − = ± − = ± −
2
21,2 1 1
2 22
b b bi i
m mkmλ ζ− − = ± Ω − = ± Ω −
2 CR
b b
bkmζ= = damping factor
=
1
1
1
ζζζ
><=
overdamped system
underdamped system
critically damped system
0
0.02
0.04
x t( )
x1 t( )2=CRb km critical damping coefficient
21 ζΩ − = Ω Ddamped natural frequency
( )21,2 1λ ζ ζ= − ± − Ω
0 5 10
0.04−
0.02−
0x1 t( )
t
Undamped --- and damped free vibration ---
Damped free vibrationOverdamped system: displacement becomes the sum of
two decaying exponentials with initial value of A+B, no
vibration takes place, the body tends to creep back to the
equilibrium position – APERIODIC MOTION (Fig.1)
1ζ >
Underdamped system: displacement is oscillatory with
diminishing amplitude (Fig.2). Frequency of oscillation is
less than that of the undamped case by the factor .
( ) ( )2 21 1t tx Ae Be
ζ ζ ζ ζ− + − Ω − − − Ω= +
21 ζ− Fig. 1
( )
( )
2 21 11 2
cos sin
[ ]t
i t i tt
D De A t B t
x e C e C eζ
ζ ζζ
− Ω
− Ω − − Ω− Ω
Ω + Ω =
= + =
=
1ζ <
1ζ >
Critical damping:
[ ] −Ω= + tx A Bt eObr. 2Fig. 2
( )
( )2sin 1t
D D
Ce tζ ζ γ− Ω= − Ω +
1ζ =
Damped free vibration – Logarithmic decrement
Natural logarithm of the ratio of any two amplitudes
( )( )
( )( ) ( )
1
2
cos sinln ln ln
cos sin
1 2 2ln
tD D
t TD D
x t e A t B tx
x x t T e A t B t
T
ζ
ζδ
π πζζ ζ
− Ω
− Ω +
Ω + Ω= = = =
+ Ω + Ω
= = Ω = Ω =2
ln1
2
TT
Te ζ ζ ζ
ζδ πζ
− Ω= = Ω = Ω =Ω −
≐ for CRb b<<
( )( )
ln 22
x tn damping factor
x t nT n
δδ π ζ ζπ
= =+
≐
CR
b
b
Forced vibration, harmonic excitation
0
( )
sin
mx bx kx F t
mx bx kx F tω+ + =+ + =ɺɺ ɺ
ɺɺ ɺ
1. Forced undamped vibration1. Forced undamped vibration
- homogenous solution of equation
- particular solution of equation
Solution of dif. equation with right side → steady state oscillation (response)
Assumed solution in the form of harmonic function:
dif. equation:
0+ =ɺɺmx kx
( )1 2
2 21 2
sin cos
sin cos
ω ω
ω ω ω ω
= +
= − = − +ɺɺ
p
p p
x a t a t
x x a t a t
0 sin ω+ =ɺɺmx kx F t
( ) ( )2 2sin cos sinω ω ω ω ω− + − =k m a t k m a t F tdif. equation:
Comparing of coeficients at function sin a cos on the both side of the equation →
amplitude a1
( ) ( )1 2 0sin cos sinω ω ω ω ω− + − =k m a t k m a t F t
01 22
, 0ω
= =−
Fa a
k m
Particular solution:
Forced vibration, harmonic excitation0
2sin ω
ω=
−p
Fx t
k m
0 0 1 1= = = ST
F Fa a
0
10
20
xp t( )
resonance – frequency of exciting force equels
0=ST
Fa
k
2 22 11ω ηω
= = =− −−
STa amkk mk
2
1
1 η=
−ST
a
a
0 0.5 1 1.5 220−
10−
t
Vibration of the single DOF system in resonance.
resonance – frequency of exciting force equels
to natural frquency of the system
1η ω= = Ω → ∞ST
a
aResonance curve – undamped system
Forced vibration
( )0
( )
sin 1
mx bx kx F t
mx bx kx F tω+ + =+ + =ɺɺ ɺ
ɺɺ ɺharmonic force
1. homogenous solution0mx bx kx+ + =ɺɺ ɺ1. homogenous solution
2. → steady-state solution (response)
0mx bx kx+ + =ɺɺ ɺ
( )sin ω ϕ= −x a t
( ) ( )0 0
2 2 22
1F Fa
k= =
0 sin ω+ + =ɺɺ ɺmx bx kx F t
( ) ( ) ( )20sin cos sin sinω ω ϕ ω ω ϕ ω ϕ ω− − + − + − =ma t ba t ka t F t
assumed solution of differential equation (1)
( ) ( )2 2 22221
ak m bk m b
k kω ω ω ω
= = − + − +
2tan
b
k m
ωϕω
=−
Vectors’ diagram
Single degreee of freedom system
Free damped vibration Free undamped vibration Forced damped vibration
Homogenous solution Homogenous solution
( )mx bx kx F t+ + =ɺɺ ɺ0+ + =ɺɺ ɺmx bx kx 0+ =ɺɺmx kx
Homogenous solution Homogenous solution
Initial conditions Initial conditions Amplitude of steady stateoscillation, steady state
response
( )
1 2
cos sin
sin
i t i th
h
h
x C e C e
x A t B t
x C t γ
Ω − Ω= += Ω + Ω= Ω +
( )
( )
2 21 1
2
[ ]
cos sin
sin 1
i t i tth
tD D
t
x e Ae Be
e A t B t
Ce t
ζ ζζ
ζ
ζ ζ γ
− Ω − − Ω− Ω
− Ω
− Ω
= + =
= Ω + Ω =
= − Ω +
( )0 0x t x= ( )0 0x t v=ɺ
( )sinpx a tω ϕ= −
h px x x= +0 sinmx bx kx F tω+ + =ɺɺ ɺ
( )0 0x t x= ( )0 0x t v=ɺ
response
( )( )
( ) ( )
2 21 1
0
2 22
[ ]
sin
i t i ttx t e Ae Be
F t
k m b
ζ ζζ
ω ϕ
ω ω
− Ω − − Ω− Ω= +−
+− +
Forced vibration - magnification factor and phase angle
( ) phx t x x= +
( ) ( )
( ) ( )02
2 22
sinsin 1t F t
x t Ce tk m b
ζ ω ϕζ γ
ω ω− Ω
−= − Ω + +
− +
Solution of equation (1):
a STa
a STa
a STa
a STa
Pha
se A
ngle
φ
( ) ( )0 0
2 2 22 22
1
1
F Fa
k m bk m b
k k
ωω ω ω
= = − + − +
0ST
Fa
k=
Values C a are derived from initial conditions.Amplitudes of steady-state oscillation:
γ
statical deflection of the spring masssystem under the action of steady force F0
η ω=Ω frequency ratio
Ω
ω
Ω
ωΩ
ω
Ω= ωη
Ω= ωη
Ω
ω
Ω
ωΩ
ω
Ω= ωη
Ω= ωη
Pha
se A
ngle
Transient motion, under resonance
( ) ( )2 22
1
1 2ST
a
a η ζη=
− +
Ω
magnification factor
2
2arctan
1
ξηϕη
=− phase angle
Vibration isolation and transmissibilityMachine or engines rigidly attached to a supporting
structure, vibration is transmitted directly to the
support (often undiserable vibration). Disturbing
source must be isolated. Force is trasmitted through
a) ω = const. → η = const. ω >> Ω, η >> 1,
small damping factor ζ
b) ω ≠ const. → isolation, support with
dampingsprings and damper.
damping
ω=η
Pha
se A
ngle
φ
Ω= ωη
Ω
ω=η
Rotating unbalanceRotating unbalance systems (gears, wheels,
shafts disks which are not perfectly uniform,
produce unbalance force which cause
excessive vibrations.
m0 ... unbalance mass
e … eccentricity
m
0
ma
m e
b
0me
m
20
( )
sin
mx bx kx F t
mx bx kx m e tω ω+ + =+ + =
ɺɺ ɺ
ɺɺ ɺ
( ) ( )
2
2 22
01 2
am em
η
η ζη=
− +22
2tan
1
b
k m
ω ξηϕηω
= =−−
ωΩ
Base motion, relative motionForced vibration of mechanical systems can be
caused by the support motion (vehicles,
aircrafts and ships)
harmonic motion 0ω= i ty y eharmonic motion
( ) ( )
20
0
ωω
+ − + − =+ + = −
+ + =
ɺɺ ɺ ɺ
ɺɺ ɺ ɺɺ
ɺɺ ɺ
r r r
i tr r r
mx b x y k x y
mx bx kx my
mx bx kx m y e
m
0
ma
m e
0
solution
( )ω ϕ−= i tp rx a e
20m y
aω=
0
ra
y
( ) ( )
2
2 220 1 2
ra
y
η
η ζη=
− +2
2tan
1
ξηϕη
=−
( ) ( )0
2 22r
m ya
k m b
ω
ω ω=
− +