VOM

1. 1
Mechanical Vibrations - Introduction
V. Rouillard © 2003 - 2013
Mechanical Vibrations
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Some Figures Courtesy Addison Wesley

2. 2
Mechanical Vibrations - Introduction
V. Rouillard © 2003 - 2013
CONTENT
• Fundamentals of vibrations
• Single degree-of-freedom systems
• Free vibrations
• Harmonic forcing functions
• General forcing functions
• Two degree-of-freedom systems
• Free vibrations
• Forced vibrations
• Multi degree-of-freedom systems
• Free vibrations
• Forced vibrations
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3. 3
Mechanical Vibrations - Introduction
Mechanical vibrations
•
•
•
Defined as oscillatory motion of bodies in response to disturbance.
Oscillations occur due to the presence of a restoring force
Vibrations are everywhere:
•
•
•
•
•
Vehicles: residual imbalance of engines, locomotive wheels
Rotating machinery: Turbines, pumps, fans, reciprocating machines
Musical instruments
Excessive vibrations can have detrimental effects:
•
•
•
•
•
•
Human body: eardrums, vocal cords, walking and running
Noise
Loosening of fasteners
Tool chatter
Fatigue failure
Discomfort
When vibration frequency coincides with natural frequency, resonance occurs.
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V. Rouillard © 2003 - 2013

4. 4
Mechanical Vibrations - Introduction
V. Rouillard © 2003 - 2013
Mechanical vibrations
•
Aeolian, wind-induced or vortex-induced vibration of the Tacoma Narrows bridge on 7 November 1940 caused it
to resonate resulting in catastrophic failure.
Tacoma Narrows Bridge Collapse Video
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5. 5
Mechanical Vibrations - Introduction
V. Rouillard © 2003 - 2013
Mechanical vibrations
•
Millennium Bridge, London: Pedestrians, in reaction to lateral motion of the bridge, altered their gait and started
behaving in concert to induce the structure to resonate further (forced periodic excitation):
Video link
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6. 6
Mechanical Vibrations - Introduction
V. Rouillard © 2003 - 2013
Fundamentals
•
In simple terms, a vibratory system involves the transfer of potential energy to kinetic energy and vice-versa in
alternating fashion.
•
•
When there is a mechanism for dissipating energy (damping) the oscillation gradually diminishes.
In general, a vibratory system consists of three basic components:
•
•
•
A means of storing potential energy (spring, gravity)
A means of storing kinetic energy (mass, inertial component)
A means to dissipate vibrational energy (damper)
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7. 7
Mechanical Vibrations - Introduction
V. Rouillard © 2003 - 2013
Fundamentals
•
•
This can be observed with a pendulum:
At position 1: the kinetic energy is zero and the potential energy is
mgl(1 − cos θ )
•
•
•
At position 2: the kinetic energy is at its
maximum
At position 3: the kinetic energy is again
zero and the potential energy at its
maximum.
In this case the oscillation will eventually stop
due to aerodynamic drag and pivot friction →
HEAT
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8. 8
Mechanical Vibrations - Introduction
V. Rouillard © 2003 - 2013
Degrees of Freedom
•
The number of degrees of freedom : number of independent coordinates required to completely determine the
motion of all parts of the system at any time.
•
Examples of single degree of freedom systems:
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9. 9
Mechanical Vibrations - Introduction
Degrees of Freedom
•
Examples of two degree of freedom systems:
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10. 10
Mechanical Vibrations - Introduction
Degrees of Freedom
•
Examples of three degree of freedom systems:
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11. 11
Mechanical Vibrations - Introduction
V. Rouillard © 2003 - 2013
Discrete and continuous systems
•
Many practical systems small and large or structures can be describe with a finite number of DoF. These are
referred to as discrete or lumped parameter systems
•
Some large structures (especially with continuous elastic elements) have an infinite number of DoF These are
referred to as continuous or distributed systems.
•
In most cases, for practical reasons, continuous systems are approximated as discrete systems with sufficiently
large numbers lumped masses, springs and dampers. This equates to a large number of degrees of freedom
which affords better accuracy.
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12. 12
Mechanical Vibrations - Introduction
V. Rouillard © 2003 - 2013
Classification of Vibration
•
Free and Forced vibrations
•
•
•
Free vibration: Initial disturbance, system left to vibrate without influence of external forces.
Forced vibration: Vibrating system is stimulated by external forces. If excitation frequency coincides with
natural frequency, resonance occurs.
Undamped and damped vibration
•
•
•
Undamped vibration: No dissipation of energy. In many cases, damping is (negligibly) small (steel 1 –
1.5%). However small, damping has critical importance when analysing systems at or near resonance.
Damped vibration: Dissipation of energy occurs - vibration amplitude decays.
Linear and nonlinear vibration
•
Linear vibration: Elements (mass, spring, damper) behave linearly. Superposition holds - double
excitation level = double response level, mathematical solutions well defined.
•
Nonlinear vibration: One or more element behave in nonlinear fashion (examples). Superposition does
not hold, and analysis technique not clearly defined.
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13. 13
Mechanical Vibrations - Introduction
V. Rouillard © 2003 - 2013
Classification of Vibration
•
Deterministic and Random vibrations
•
•
Deterministic vibration: Can be described by implicit mathematical function as a function of time.
Random vibration: Cannot be predicted. Process can be described by statistical means.
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14. 14
Mechanical Vibrations - Introduction
Vibration Analysis
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•
•
•
Input (excitation) and output (response) are wrt time
Response depend on initial conditions and external forces
Most practical systems very complex – (mathematical) modelling requires simplification
Procedure:
→
→
→
→
Mathematical modelling
Derivation / statement of governing equations
Solving of equations for specific boundary conditions and external forces
Interpretation of solution(s)
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15. 15
Mechanical Vibrations - Introduction
Vibration Analysis
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Example (1.3 Ed.3)
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16. 16
Mechanical Vibrations - Introduction
V. Rouillard © 2003 - 2013
Spring Elements
•
•
Pure spring element considered to have negligible mass and damping
Force proportional to spring deflection (relative motion between ends):
F = k ∆x
•
For linear springs, the potential energy stored is:
U = 1 k ( ∆x )
2
•
Actual springs sometimes behave in
nonlinear fashion
•
Important to recognize the presence and
significance (magnitude) of nonlinearity
•
Desirable to generate linear estimate
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17. 17
Mechanical Vibrations - Introduction
Spring Elements
•
Equivalent spring constant.
•
•
Eg: cantilever beam: Mass of beam assumed negligible cf lumped mass
Deflection at free end:
mgl 3
δ=
3EI
•
Stiffness (Force/defln):
k=
•
mg 3EI
= 3
δ
l
This procedure can be applied for various geometries and
boundary conditions. (see appendix)
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V. Rouillard © 2003 - 2013

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Spring Elements
•
Equivalent spring constant.
•
Springs in parallel:
w =mg=kδ +k 2
δ
1
w=mg=keqδ
•
where
keq =k1 + k2
•
In general, for n springs in parallel:
i=n
keq = ∑ ki
i=1
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V. Rouillard © 2003 - 2013

19. Mechanical Vibrations - Introduction
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Spring Elements
•
Equivalent spring constant.
•
Springs in series:
δt =δ1 + δ2
•
Both springs are subjected to the same
force:
mg = k1δ 1 = k2δ 2
mg=keqδ t
•
Combining the above equations:
k1δ 1 = k2δ 2 = keqδ t
δ 1=
keqδ t
k
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and δ 2 =
keqδ t
k2
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20. 20
Mechanical Vibrations - Introduction
V. Rouillard © 2003 - 2013
Spring Elements
•
Springs in series (cont’d):
•
Substituting into first eqn:
δt =
•
keqδ t
k1
+
keqδ t
k2
Dividing by keqδt throughout:
1
1
1
=
+
keq k1
k2
•
For n springs in series:
1 i=n  1 
=∑  
keq i=1  ki 
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21. 21
Mechanical Vibrations - Introduction
V. Rouillard © 2003 - 2013
Spring Elements
•
Equivalent spring constant.
•
•
When springs are connected to rigid components such as pulleys and gears, the energy equivalence
principle must be used.
Example:
Example (1.10 Ed.3)
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Mechanical Vibrations - Introduction
Mass / Inertia Elements
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•
•
•
•
Mass or inertia element assumed rigid (lumped mass)
•
Modelling with lumped mass elements. Example: assume
frame mass is negligible cf mass of floors.
Its energy (kinetic) is proportional to velocity.
Force ∝ mass * acceleration
Work = force * displacement
Work done on mass is stored as Kinetic Energy
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V. Rouillard © 2003 - 2013

23. 23
Mechanical Vibrations - Introduction
V. Rouillard © 2003 - 2013
Mass / Inertia Elements
•
Equivalent mass - example:
•
The velocities of the mass elements can be written as:
l2
x
x
&2 = &1
l1
•
and
l3
x
x
&3 = &1
l1
To determine the equivalent mass at position l1:
x
x
& eq = & 1
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24. Mechanical Vibrations - Introduction
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V. Rouillard © 2003 - 2013
Mass / Inertia Elements
•
Equivalent mass – example (cont’d)
•
Equating the kinetic energies:
1m & 2
x
2 1 1
•
1
+ 2 m2& 2 + 2 m3& 3 = 2 meq& eq
x2 1 x2 1
x2
Substituting for the velocity terms:
meq
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2
2
l 
l 
= m1 +  2 ÷ m2 +  3 ÷ m3
 l1 
 l1 

25. 25
Mechanical Vibrations - Introduction
V. Rouillard © 2003 - 2013
Damping Elements
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•
•
Absorbs energy from vibratory system → vibration amplitude decays.
Damping element considered to have no mass or elasticity
Real damping systems very complex, damping modelled as:
•
Viscous damping:
•
•
•
•
Based on viscous fluid flowing through gap or orifice.
Damping force ∝ relative velocity between ends
Eg: film between sliding surfaces, flow b/w piston & cylinder, flow thru orifice, film around journal
bearing.
Coulomb (dry Friction) damping:
•
•
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Based on friction between unlubricated surfaces
Damping force is constant and opposite the direction of motion

26. 26
Mechanical Vibrations - Introduction
V. Rouillard © 2003 - 2013
Damping Elements
•
Hysteretic (material or solid) damping:
•
•
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Based on plastic deformation of materials (energy loss due to slippage b/w grains)
Energy lost due to hysteresis loop in force-deflection (stress-strain) curve of element when load is
applied:

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Mechanical Vibrations - Introduction
V. Rouillard © 2003 - 2013
Damping Elements
•
Equivalent damping element:
•
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Combinations of damping elements can be replace by equivalent damper using same procedures as
for spring and mass/inertia elements.

28. Mechanical Vibrations - Introduction
V. Rouillard © 2003 - 2013
Damping Elements
FORD AU IRS
2
1
0
-2
Force [KN]
28
-1.5
-1
-0.5
0
-1
-2
-3
-4
-5
Velocity [m/s]
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0.5
1
1.5
2

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Mechanical Vibrations - Introduction
Harmonic Motion
•
Harmonic motion: simplest form of periodic motion
(deterministic).
•
•
Pure sinusoidal (co-sinusoidal) motion
•
The motion of mass m is described by:
Eg: Scotch-yoke mechanism rotating with angular
velocity ω - simple harmonic motion:
x = Asin( θ ) = A sin( ωt )
•
Its velocity and acceleration are:
dx
= ω A cos( ωt )
dt
and
d 2x
dt 2
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= − ω 2 A sin( ωt ) = − ω 2 x
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30. 30
Mechanical Vibrations - Introduction
V. Rouillard © 2003 - 2013
Harmonic Motion
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•
•
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Sinusoidal motion emanates from cyclic motion
The rotating vector generates a sinusoidal and a co-sinusoidal components along
mutually perpendicular axes.
Can be represented by a vector (OP) with a magnitude, angular velocity
(frequency) and phase.

31. Mechanical Vibrations - Introduction
31
Harmonic Motion
•
Often convenient to represent sinusoidal and co-sinusoidal components (mutually
perpendicular) in complex number format
•
•
Where a and b denote the sinusoidal (x) and co-sinusoidal (y) components
a and b = real and imaginary parted of vector X
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32. Mechanical Vibrations - Introduction
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V. Rouillard © 2003 - 2013
Harmonic Motion
Definition of terms:
•
Cycle: motion of body from equilibrium position → extreme position → equilibrium position → extreme position
in other direction → equilibrium position .
•
•
Amplitude: Maximum value of motion from equilibrium. (Peak – Peak = 2 x amplitude)
Period: Time taken to complete one cycle
τ=
ω = circular frequency
•
Frequency: number of cycles per unit time.
f =
ω : radians/s
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f Hertz (cycles /s)
1 ω
=
τ 2π
2π
ω

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V. Rouillard © 2003 - 2013
Harmonic Motion
•
Phase angle: the difference in angle (lead or lag) by which two harmonic motions of the same frequency
reach their corresponding value (maxima, minima, zero up-cross, zero down-cross)
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34. Mechanical Vibrations - Introduction
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V. Rouillard © 2003 - 2013
Harmonic Motion
•
Phase angle: the difference in angle (lead or lag) by which two harmonic motions of the same frequency
reach their corresponding value (maxima, minima, zero up-cross, zero down-cross)
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35. Mechanical Vibrations - Introduction
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V. Rouillard © 2003 - 2013
Harmonic Motion
•
Natural frequency: the frequency at which a system vibrates without external forces after an initial
disturbance. The number of natural frequencies always matches the number of DoF.
•
Beats: the effect produced by adding two harmonic motions with similar (close) frequencies.
x1 = A sin( ωt )
x2 = A sin( ωt + δωt )
xt = x1 + x2 = A [sin( ωt ) + sin( ωt + δωt )]
M +N
M −N
cos
2
2
δωt 
 δωt 
xt = 2 A sin  ωt +
cos 

÷
÷
2 

 2 
Since
sin M + sin N = 2 sin
Eg: ω=40 Hz and δ= -0.075
•
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In mechanical vibratory systems, beats occur when the (harmonic) excitation (forcing) frequency is close to
the natural frequency.

36. Mechanical Vibrations - Introduction
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V. Rouillard © 2003 - 2013
Harmonic Motion
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•
•
Octave: doubling of any quantity. Used mainly for frequency.
Octave band (frequency): maximum is double of minimum. Eg: 64 – 128 Hz, 1000 – 2000 Hz.
Decibel: defined as 10 x log(power ratio)
P
dB = 10Log  ÷
 P0 
In electrical systems (as in mechanical vibratory systems) power is proportional to the value squared hence:
 X 
dB = 20Log 
÷
 X0 
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37. Mechanical Vibrations - Introduction
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V. Rouillard © 2003 - 2013
Harmonic (Fourier) Analysis
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•
Many vibratory systems not harmonic but often periodic
Any periodic function can be represented by the Fourier series – infinite sum of sinusoids and co-sinusoids.
ao
+ a1 cos( ωt ) + a2 cos( 2ωt ) + ........
2
+ b1 sin( ωt ) + b2 sin( 2ωt ) + .......
x( t ) =
ao ∞
= + ∑ [an cos( nωt ) + bn sin( nωt )]
2 n =1
•
To obtain an and bn the series is multiplied by cos(nωt) and sin(nωt) respectively and integrated over one
period.
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38. Mechanical Vibrations - Introduction
38
Harmonic (Fourier) Analysis
•
Example:
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39. Mechanical Vibrations - Introduction
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Harmonic (Fourier) Analysis
•
Example:
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40. Mechanical Vibrations - Introduction
40
Harmonic (Fourier) Analysis
•
As for simple harmonic motion, Fourier series can be expressed with complex numbers:
eiωt = cos( ωt ) + i sin( ωt )
e −iωt = cos( ωt ) − i sin( ωt )
eiωt + e −iωt
cos( ωt ) =
2
eiωt − e −iωt
sin( ωt ) =
2i
•
The Fourier series:
ao ∞
x( t ) = + ∑ [an cos( nω t ) + bn sin( nω t )]
2 n =1
Can be written as:
 eiω t − e−iω t  
ao ∞   eiω t + e −iω t 


x( t ) = + ∑ an 
÷+ bn 
÷
÷

÷
2 n =1  
2
2i



 

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41. Mechanical Vibrations - Introduction
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V. Rouillard © 2003 - 2013
Harmonic (Fourier) Analysis
•
Defining the complex Fourier coefficients
cn =
•
an − ibn
2
and
cn −1 =
The (complex) Fourier series is simplified to:
x( t ) =
∞
∑
n =−∞
04:36:37
cn einωt
an + ibn
2

42. Mechanical Vibrations - Introduction
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V. Rouillard © 2003 - 2013
Harmonic (Fourier) Analysis
ao ∞
x( t ) = + ∑ [an cos( nω t ) + bn sin( nω t )]
2 n =1
•
•
The Fourier series is made-up of harmonics.
Their amplitudes and phases are defined as:
2
2
An = ( an + bn )
b 
φn = a tan  n ÷
 an 
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harmonics

43. Mechanical Vibrations - Introduction
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V. Rouillard © 2003 - 2013
Harmonic (Fourier) Analysis
•
The amplitudes (magnitudes) and phases of the harmonics can be plotted as a function of frequency to form
the frequency spectrum of spectral diagram:
An
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44. 44
Mechanical Vibrations – Single Degree-of-Freedom systems
V. Rouillard © 2003 - 2013
Free undamped vibration single DoF
•
•
•
Recall: Free vibrations → system given initial disturbance and oscillates free of external forces.
Undamped: no decay of vibration amplitude
Single DoF:
•
•
•
•
mass treated as rigid, limped (particle)
Elasticity idealised by single spring
only one natural frequency.
The equation of motion can be derived using
•
•
•
•
Newton’s second law of motion
D’Alembert’s Principle,
The principle of virtual displacements and,
The principle of conservation of energy.
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45. 45
Mechanical Vibrations – Single Degree-of-Freedom systems
Free undamped vibration single DoF
•
Using Newton’s second law of motion to develop the equation of motion.
1.
Select suitable coordinates
2.
Establish (static) equilibrium position
3.
Draw free-body-diagram of mass
4.
Use FBD to apply Newton’s second law of motion:
“Rate of change of momentum = applied force”
F( t ) =
As m is constant
F( t ) = m
For rotational motion
d  dx( t ) 
m
÷
dt 
dt 
d 2 x( t )
dt
2
= mx
&&
M ( t ) = J &&
θ
For the free, undamped single DoF system
04:36:37
F( t ) = −kx = mx
&&
or
mx + kx = 0
&&
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46. 46
Mechanical Vibrations – Single Degree-of-Freedom systems
V. Rouillard © 2003 - 2013
Free undamped vibration single DoF
Principle of virtual displacements:
•
“When a system in equilibrium under the influence of forces is given a virtual displacement. The total
work done by the virtual forces = 0”
•
Displacement is imaginary, infinitesimal, instantaneous and compatible with the system
•
When a virtual displacement dx is applied, the sum of work done by the spring force and the inertia force
are set to zero:
−( kx )δ x − ( mx )δ x = 0
&&
•
Since dx ≠ 0 the equation of motion is written as:
04:36:37
kx + mx = 0
&&

47. 47
Mechanical Vibrations – Single Degree-of-Freedom systems
Free undamped vibration single DoF
Principle of conservation of energy:
•
•
•
No energy is lost due to friction or other energy-dissipating mechanisms.
If no work is done by external forces, the system total energy = constant
For mechanical vibratory systems:
KE + PE = cons tan t
or
d
( KE + PE ) = 0
dt
•
Since
1
KE = mx 2
&
2
then
and
d 1 2 1 2
&
 mx + kx ÷ = 0
dt  2
2

or
mx + kx = 0
&&
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1
PE = kx 2
2
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48. 48
Mechanical Vibrations – Single Degree-of-Freedom systems
Free undamped vibration single DoF
Vertical mass-spring system:
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49. Mechanical Vibrations – Single Degree-of-Freedom systems
49
Free undamped vibration single DoF
Vertical mass-spring system:
mg
•
From the free body diagram:, using Newton’s second law of motion:
mx = − k( x + δ st ) + mg
&&
sin ce
kδ st = mg
mx + kx = 0
&&
•
∀
•
Note that this is the same as the eqn. of motion for the horizontal mass-spring system
∴ if x is measured from the static equilibrium position, gravity (weight) can be ignored
04:36:37 be also derived by the other three alternative methods.
This can
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Mechanical Vibrations – Single Degree-of-Freedom systems
V. Rouillard © 2003 - 2013
Free undamped vibration single DoF
•
•
The solution to the differential eqn. of motion.
As we anticipate oscillatory motion, we may propose a solution in the form:
x( t ) = Acos( ωn t ) + B sin( ωn t )
or
x( t ) = Aeiωnt + Be −iωnt
alternatively,if we let s = ±iωn
x( t ) = C e ± st
•
By substituting for x(t) in the eqn. of motion:
C( ms 2 + k ) = 0
sin ce c ≠ 0,
ms 2 + k = 0
and
s = ±iωn = ±
or
04:36:37
ωn =
k
m
¬ Characteristic equation
k
m
¬ roots = eigenvalues

51. Mechanical Vibrations – Single Degree-of-Freedom systems
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Free undamped vibration single DoF
•
•
The solution to the differential eqn. of motion.
Applying the initial conditions to the general solution:
x( t =0 ) = A = x0
& ( t =0 ) = Bωn = & 0
x
x
•
The solution becomes:
x( t ) = Acos( ωn t ) + B sin( ωn t )
initial displacement
initial velocity
x( t ) = x0 cos( ωn t ) +
&o
x
sin( ωn t )
ωn

x
2 & 
if we let A0 =  x0 +  0 ÷ 

 ωn  


x( t ) = A0 sin( ωn t + φ )
1
2 2
•
x ω 
and φ = a tan  0 n ÷ then
x
 &o 
This describes motion of harmonic oscillator:
•
•
•
Symmetric about equilibrium position
Thru equilibrium: velocity is maximum & acceleration is zero
At peaks and valleys, velocity is zero and acceleration is maximum
∀ ωn = √(k/m) is the natural frequency
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52. 52
Mechanical Vibrations – Single Degree-of-Freedom systems
Free undamped vibration single DoF
•
Note: for vertical systems, the natural frequency can be written as:
ωn =
k
m
sin ce k =
ωn =
04:36:37
g
δ st
mg
δ st
or
fn =
1
g
2π δ st
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53. 53
Mechanical Vibrations – Single Degree-of-Freedom systems
Free undamped vibration single DoF
•
•
Torsional vibration.
Approach same as for translational system. Laboratory exercise.
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Mechanical Vibrations – Single Degree-of-Freedom systems
Free undamped vibration single DoF
•
•
Compound pendulum.
•
Assume rigid body → single DoF
Given an initial angular displacement or velocity, system will
oscillate due to gravitational acceleration.
Restoring torque:
mgd sin θ
∴ Equation of motion :
J o&& + mgd sin θ = 0
θ
¬ nonlinear2nd order ODE
Linearity is approximated if sin θ ≈ θ Therefore :
J o&& + mgdθ = 0
θ
Natural frequency :
ωn =
04:36:37
mgd
Jo
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55. 55
Mechanical Vibrations – Single Degree-of-Freedom systems
Free undamped vibration single DoF
Natural frequency :
mgd
ωn =
Jo
sin ce for a simple pendulum
g
ωn =
l
J
2
Then, l = o and since J o = mko then
md
2
ko
gd
ωn =
and l =
2
d
ko
2
2
Applying the parallel axis theorem ko = kG + d 2
2
kG
l=
+d
d
Let l = GA + d = OA
ωn =
g
2
ko / d
=
g
g
=
l
OA
2

kG 
04:36:37
The location A  GA =
÷ is the " centre of percussion ′′

d ÷


V. Rouillard © 2003 - 2013

56. 56
Mechanical Vibrations – Single Degree-of-Freedom systems
Free undamped vibration single DoF
•
•
Stability.
Some systems may have inherent instability
04:36:37
V. Rouillard © 2003 - 2013

57. 57
Mechanical Vibrations – Single Degree-of-Freedom systems
Free undamped vibration single DoF
•
•
•
Stability.
Some systems may have inherent instability
When the bar is deflected by θ,
The spring force is :
2kl sin θ
The gravitational force thru G is :
mg
The inertial moment about O due to the angular acceleration && is :
θ
2
ml &&
J o&& =
θ
θ
3
The eqn. of motion is written as :
ml 2 &&
l
θ + ( 2kl sin θ ) l cos θ − mg sin θ = 0
3
2
04:36:37
V. Rouillard © 2003 - 2013

58. 58
Mechanical Vibrations – Single Degree-of-Freedom systems
Free undamped vibration single DoF
For small oscillations, sinθ = θ and cos θ = 1 .Therefore
ml 2
mgl
θ + 2kl 2θ −
θ =0
3
2
or
 12kl 2 − 3mgl 
&& + 
θ
÷θ = 0
2

÷
2ml


The solution to the eqn. of motion depends of the sign of ( )
(1) If ( ) >0, the resulting motion is oscillatory (simple harmonic)
with a natural frequency
 12kl 2 − 3mgl 
ωn = 
÷

÷
2ml 2


04:36:37
V. Rouillard © 2003 - 2013

59. 59
Mechanical Vibrations – Single Degree-of-Freedom systems
V. Rouillard © 2003 - 2013
Free undamped vibration single DoF
2


&& +  12kl − 3mgl ÷θ = 0
θ

÷
2ml 2


(2) If ( ) =0, the eqn. of motion reduces to:
&&
θ =0
The solution is obtained by int egrating twice yielding :
θ ( t ) = C1t + C2
&
&
Applying initial conditions θ ( t = 0 ) = θ0 and θ ( t = 0 ) = θ0
&
θ ( t ) = θ0 t + θ0
Which shows a linear increase of angular displ. at cons tan t velocity.
&
And if θ = 0 the bar remains in static equilibrium at θ ( t ) = θ
0
04:36:37
0

60. 60
Mechanical Vibrations – Single Degree-of-Freedom systems
Free undamped vibration single DoF
 12kl 2 − 3mgl 
&&
θ + 
÷θ = 0
2

÷
2ml


(3) If ( ) < 0, we define:
 12kl 2 − 3mgl   3mgl − 12kl 2 
α = −
÷= 
÷
2
2
2ml
2ml

 

The solution of the eq.of motion is :
θ ( t ) = B1eα t + B2 e −α t
&
&
Applying initial conditions θ ( t = 0 ) = θ0 and θ ( t = 0 ) = θ0
1 
&
&
( αθ0 + θ0 ) eα t + ( αθ0 − θ0 ) e−αt 

2α 
which shows that θ ( t ) increases exp onentially with time
and is therefore unstable because the restoring moment ( springs )
θ( t ) =
is less than the non − restoring moment due to gravity.
04:36:37
V. Rouillard © 2003 - 2013

61. 61
Mechanical Vibrations – Single Degree-of-Freedom systems
V. Rouillard © 2003 - 2013
Free undamped vibration single DoF
•
•
Rayleigh’s Energy method to determine natural frequency
Recall: Principle of conservation of energy:
T1 + U1 = T2 + U 2
•
Where T1 and U1 represent the energy components at the time when the kinetic energy is at its maximum
(∴ U1=0) and T2 and U2 the energy components at the time when the potential energy is at its maximum
(∴ T2=0)
T1 + 0 = 0 + U 2
•
For harmonic motion
Tmax = U max
04:36:37

62. Mechanical Vibrations – Single Degree-of-Freedom systems
62
Free undamped vibration single DoF
•
Rayleigh’s Energy method to determine natural frequency: Application example:
•
Find minimum length of mercury u-tube manometer tube so that f n of
fluid column < 2 Hz.
•
•
Determine Umax and Tmax:
Umax = potential energy of raised fluid column + potential energy of
depressed fluid column.
U = mg
x
x
+ mg
2 raised
2 depressed
= ( Axγ )
x
x
+ ( Axγ )
2 raised
2 depressed
= Aγ x 2
A : cross sec tional area and γ : specific weight of mercury
•
Kinetic energy:
04:36:37
1
T = ( mass of mercury col ) vel 2
2
1  Alγ  2
= 
&
x
2 g ÷

V. Rouillard © 2003 - 2013

63. 63
Mechanical Vibrations – Single Degree-of-Freedom systems
Free undamped vibration single DoF
•
Rayleigh’s Energy method to determine natural frequency: Application example:
•
If we assume harmonic motion:
x( t ) = X cos( 2π f n t )
where X is the max . displacement
& t ) = 2π f n X sin( 2π f n t ) where 2π f n X is the max . velocity
x(
•
Substituting for the maximum displacement and velocity:
U max = Aγ X 2
U max = Tmax
fn =
•
and
∴
1  2g 
 ÷
2π  l 
Minimum length of column:
04:36:37
1  Alγ 
Tmax = 
( 2π f n ) 2 X 2
2 g ÷

1  Alγ 
Aγ X 2 = 
( 2π f n ) 2 X 2
2 g ÷

1  2g 
 ÷ ≤ 1.5 Hz
2π  l 
l ≥ 0.221 m
fn =
V. Rouillard © 2003 - 2013

64. 64
Mechanical Vibrations – Single Degree-of-Freedom systems
Free single DoF vibration + viscous damping
•
Recall: viscous damping force ∝ velocity:
F = −cx
&
c = damping cons tan t or coefficient [ Ns / m ]
Applying Newton' s sec ond law of motion to obtain the eqn.of motion :
mx = − cx − kx
&&
&
or
mx + cx + kx = 0
&& &
If the solution is assumed to take the form :
x( t ) = Ce st
where s = ±iωn
then : & t ) = sCe st and &x( t ) = s 2Ce st
x(
&
Substituting for x, & and && in the eqn.of motion
x
x
ms 2 + cs + k = 0
The root of the characteristic eqn. are :
2
−c ± c 2 − 4mk
c
c  k
s1,2 =
=−
± 

÷ − ÷
2m
2m
2m   m 

The two solutions are :
x1 ( t ) = C1e s1t
04:36:37
and
x 2 ( t ) = C 2 e s 2t
V. Rouillard © 2003 - 2013

65. 65
Mechanical Vibrations – Single Degree-of-Freedom systems
Free single DoF vibration + viscous damping
•
The general solution to the Eqn. Of motion is:
x( t ) = C1e s1t + C2 e s2t
or
x( t ) = C1


2
 c
 c   k 
− + 
÷ − ÷t
2m  2m   m  


e
+ C2
where C1 and C2 are arbitrary cons tan ts
det er min ed from the initial conditions .
04:36:37


2
 c
 c   k 
− 
−
÷ − ÷t
2m
 2m   m  


e
V. Rouillard © 2003 - 2013

66. Mechanical Vibrations – Single Degree-of-Freedom systems
66
V. Rouillard © 2003 - 2013
Free single DoF vibration + viscous damping
•
Critical damping (cc): value of c for which the radical in the general solution is zero:
2
 cc   k 

÷ −  ÷= 0
 2m   m 
•
k
= 2mωn = 2 km
m
cc = 2m
or
Damping ratio (ζ): damping coefficient : critical damping coefficient.
ζ =
c
cc
c
c cc
=
= ζω n
2m cc 2m
or
The roots can be re − written :
2
(
)
c
 c  k
2
s1,2 = −
± 
÷ −  ÷ = −ζ ± ζ − 1 ω n
2m
 2m   m 
And the solution becomes :
 −ζ + ζ 2 −1 ω t

÷ n


x( t ) = C1e
•
+ C2
 −ζ − ζ 2 −1 ω t

÷ n


e
The response x(t) depends on the roots s 1 and s2 → the behaviour of the system is dependent on the
damping ratio ζ.
04:36:37

67. Mechanical Vibrations – Single Degree-of-Freedom systems
67
V. Rouillard © 2003 - 2013
Free single DoF vibration + viscous damping
 −ζ + ζ 2 −1 ω t

÷ n

x( t ) = C1e
•
+ C2
2 

 −ζ − ζ −1 ÷ωn t

e
When ζ <1, the system is underdamped. (ζ2-1) is negative and the roots can be written as:
(
)
s1 = −ζ + i 1 − ζ 2 ω n
and
(
)
s2 = −ζ − i 1 − ζ 2 ω n
And the solution becomes :
x( t ) = C1
x( t ) = e
 −ζ +i 1−ζ 2 ω t

÷ n

e
−ζω nt
x( t ) = e −ζω nt
x( t ) = e −ζω nt
 i
 
C1e


+ C2
1−ζ 2 ω nt
÷

 −ζ −i 1−ζ 2 ω t

÷ n

e
+ C2
 −i 1−ζ 2 ω t 

÷ n 

e



(
)
(
{
{ C cos ( 1 − ζ ω t ) + C sin ( 1 − ζ ω t ) }
sin ( 1 − ζ ω t + φ )
or x( t ) = X e
( C1 + C2 ) cos
x( t ) = Xe −ζω nt
'
1
1 − ζ 2 ω nt + i ( C1 − C2 ) sin
2
2
n
n
'
2
2
1 − ζ 2ωnt
)}
n
0
−ζωn t
cos
(
1 − ζ 2 ω n t − φo
04:36:37
Where C’1, C’2; X, φ and Xo, φo are arbitrary constant determined from initial conditions.
)

68. Mechanical Vibrations – Single Degree-of-Freedom systems
68
V. Rouillard © 2003 - 2013
Free single DoF vibration + viscous damping
{
'
x( t ) = e −ζωnt C1 cos
•
(
)
1 − ζ 2 ωn t + C'2 sin
(
1 − ζ 2 ωn t
)}
For the initial conditions:
x( t = 0 ) = x0 and & t = 0 ) = & 0
x(
x
Then
'
C1 = x0
and
C'2 =
& 0 + ζωn x0
x
1 − ζ 2ωn
Therefore the solution becomes
x( t ) = e
•
−ζωn t


 x0 cos


(
2
)
1 − ζ ωnt +
x
& 0 + ζωn x0
2
1 − ζ ωn
sin
(
)


1 − ζ ωnt 


2
This represents a decaying (damped) harmonic motion with angular frequency √(1-ζ2)ωn also known as
the damped natural frequency. The factor e -( ) causes the exponential decay.
04:36:37

69. 69
Mechanical Vibrations – Single Degree-of-Freedom systems
V. Rouillard © 2003 - 2013
Free single DoF vibration + viscous damping
τd =
2π
ωd
Xe −ζωnt
Exponentially decaying harmonic – free SDoF vibration with viscous damping .
04:36:37
Underdamped oscillatory motion and has important engineering applications.

70. 70
Mechanical Vibrations – Single Degree-of-Freedom systems
V. Rouillard © 2003 - 2013
Free single DoF vibration + viscous damping
x( t ) = Xe −ζωnt sin
(
1 − ζ 2 ωn t + φ
)
or
x( t ) = X 0 e −ζωnt cos
(
1 − ζ 2 ωnt − φo
The cons tan ts ( X ,φ ) and ( X 0 ,φ0 ) representing the magnitude and phase become :
X = X0 =
( ) ( )
'
C1
'
 C1 
φ = a tan  ' ÷
 C2 
04:36:37
2
+ C'2
and
2
 C'2 
φ0 = a tan  − ' ÷
 C1 
)

71. Mechanical Vibrations – Single Degree-of-Freedom systems
71
V. Rouillard © 2003 - 2013
Free single DoF vibration + viscous damping
When ζ = 1, c=cc , system is critically damped and the two roots to the eqn. of motion become:
•
s1 = s2 = −
cc
= −ωn
2m
and solution is
x( t ) = ( C1 + C2t )e −ωnt
Applying the initial conditions x( t = 0 ) = x0 and & t = 0 ) = & 0 yields
x(
x
C1 = x0
C2 = & 0 + ωn x0
x
The solution becomes :
x( t ) = [ x0 + ( & 0 + ωn x0 ) t ] e −ωnt
x
•
As t→∞ , the exponential term diminished toward zero and depicts aperiodic motion
04:36:37

72. 72
Mechanical Vibrations – Single Degree-of-Freedom systems
V. Rouillard © 2003 - 2013
Free single DoF vibration + viscous damping
•
When ζ > 1, c>cc , system is overdamped and the two roots to the eqn. of motion are real and negative:
(
= ( −ζ −
)
−1 ) ω < 0
s1 = −ζ + ζ 2 − 1 ωn < 0
s2
ζ2
n
with s2 = s1 and the initial conditions x( t = 0 ) = x0 and & t = 0 ) = & 0
x(
x
the solution becomes :
 −ζ + ζ 2 −1

x( t ) = C1e
where
C1 =
C2 =
ω t
÷ n

(
+ C2
 −ζ − ζ 2 −1

e
)
x0ωn −ζ + ζ 2 − 1 + & 0
x
2ωn ζ 2 − 1
(
)
− x0ωn −ζ − ζ 2 − 1 − & 0
x
2ωn ζ 2 − 1
04:36:37
Which shows aperiodic motion which diminishes exponentially with time.
ω t
÷ n


73. 73
Mechanical Vibrations – Single Degree-of-Freedom systems
V. Rouillard © 2003 - 2013
Free single DoF vibration + viscous damping
Underdamped ( ζ = 0 )
Overdamped ( ζ > 1 )
Critically
damped ( ζ = 1 )
Underdamped ( ζ < 1 )
2π
ωn
2π
ωd
Critically damped systems have lowest required damping for aperiodic motion and mass returns to equilibrium
position in shortest possible time.
04:36:37

74. Mechanical Vibrations – Single Degree-of-Freedom systems
V. Rouillard © 2003 - 2013
Free single DoF vibration + viscous damping
Example
1.2
1
0.8
-]
0.6
0.4
0.2
0
[
t
n
m
e
c
a
l
p
s
i
D
74
-0.2
0
0.5
1
1.5
2
-0.4
-0.6
-0.8
-1
04:36:37
Elapsed Time [s]
2.5
3
3.5

75. 75
Mechanical Vibrations – Single Degree-of-Freedom systems
V. Rouillard © 2003 - 2013
Free single DoF vibration + viscous damping
•
Logarithmic decrement: Natural logarithm of ratio of two successive peaks (or troughs) in an
exponentially decaying harmonic response.
•
•
•
Represents the rate of decay
Used to determine damping constant from experimental data.
Using the solution for underdamped systems:
x1 X 0 e −ζωnt1 cos( ωd t1 − φ0 )
=
x2 X 0 e −ζωnt2 cos( ωd t2 − φ0 )
Let
t2 = t1 + τ d = t1 +
x1
x2
2π
then
ωd
cos( ωd t2 − φ0 ) = cos( 2π + ωd t1 − φ0 ) = cos( ωd t1 − φ0 )
and
x1
e −ζωnt1
=
= eζωnτ d
x2 e −ζωn ( t1 +τ d )
Applying the natural ln on both sides,
the log arithmic decrement δ is obtained :
x 
2π
2πζ
2πζ
δ = ln  1 ÷ = ζωnτ d = ζωn
=
=
04:36:37  x2 
1 − ζ 2 ωn
1 − ζ 2 ωd
τd
t1
t2

76. 76
Mechanical Vibrations – Single Degree-of-Freedom systems
V. Rouillard © 2003 - 2013
Free single DoF vibration + viscous damping
•
Logarithmic decrement:
For low damping ( ζ << 1 )
14
x 
δ = ln  1 ÷ = 2πζ
 x2 
Valid for ζ < .3
12
10
δ
8
6
4
2
0
0
0.2
0.4
0.6
ζ
04:36:37
0.8
1

77. 77
Mechanical Vibrations – Single Degree-of-Freedom systems
V. Rouillard © 2003 - 2013
Free single DoF vibration + viscous damping
•
Logarithmic decrement after n cycles:
x1
•
Since the period of oscillation is
constant:
x
x x x
= 1 2 3 .... m
xm +1 x2 x3 x4 xm +1
xj
Since
= eζω nτ d then
x j +1
x1
x1
xm +1
(
)
ζω nτ d m
= e
= emζω nτ d
The log arithmic decrement can therefore
be obtained from a number m of
successive decaying oscillations
δ=
1  x1 
ln 
m  xm +1 ÷

04:36:37
x2
Xm+1

78. 78
Mechanical Vibrations – Single Degree-of-Freedom systems
V. Rouillard © 2003 - 2013
Free single DoF vibration + Coulomb damping
•
•
•
Coulomb or dry friction dampers are simple and convenient
Occurs when components slide / rub
Force proportional to normal force:
F = µN
F = µ mg
for free − s tan ding systems
where µ is the coefficient of friction.
•
•
Force acts in opposite direction to velocity and is independent of displacement and velocity.
Consider SDOF system with dry friction:
04:36:37
Case 1.
Case 2.

79. 79
Mechanical Vibrations – Single Degree-of-Freedom systems
V. Rouillard © 2003 - 2013
Free single DoF vibration + Coulomb damping
•
•
Case 1: Mass moves from left to right. x = positive and x’ is positive or x = negative and x’ is positive.
The eqn. of motion is:
¬ 2nd order hom ogeneous DE
mx = −kx − µ N or
&&
mx + kx = − µ N
&&
For which the general solution is :
x( t ) = A1 cos( ωn t ) + A2 sin( ωn t ) −
where the fre quency of vibration ωn is
conditions of this portion of the cycle.
•
•
µN
k
(1)
k
and A1 and A2 are constants dependent on the initial
m
Case 2: Mass moves from right to left. x = positive and x’ is negative or x = negative and x’ is negative.
The eqn. of motion is:
mx = − kx + µ N or
&&
mx + kx = µ N
&&
For which the general solution is :
µN
x( t ) = A3 cos( ωn t ) + A4 sin( ωn t ) +
k
(2)
k
and A3 and A4 are constants dependent
m
04:36:37
on the initial conditions of this portion of the cycle.
where the fre quency of vibration ωn is again

80. 80
Mechanical Vibrations – Single Degree-of-Freedom systems
V. Rouillard © 2003 - 2013
Free single DoF vibration + Coulomb damping
•
The term µN/k [m] is a constant representing the virtual displacement of the spring k under force µN. The
equilibrium position oscillates between +µN/k and -µN/k 1 for each harmonic half cycle of motion.
04:36:37

81. 81
Mechanical Vibrations – Single Degree-of-Freedom systems
Free single DoF vibration + Coulomb damping
•
To find a more specific solution to the eqn. of motion we apply the simple initial conditions:
x( t = 0 ) = x0
& t =0)=&0
x(
x
The motion starts from the extreme right ( ie. velocity is zero )
Substituting int o
µN
x( t ) = A3 cos( ωn t ) + A4 sin( ωn t ) +
(2)
k
and
& t ) = − A3ωn sin( ωn t ) + A4ωn cos( ωn t ) + 0
x(
gives
µN
A3 = x0 −
and
A4 = 0
k
Eqn.( 2 ) becomes
µN 
µN
x( t ) =  x0 −
cos( ωn t ) +
( 2a ) valid for 0 ≤ t ≤ π / ωn

÷
k 
k

04:36:37
and
V. Rouillard © 2003 - 2013

82. 82
Mechanical Vibrations – Single Degree-of-Freedom systems
V. Rouillard © 2003 - 2013
Free single DoF vibration + Coulomb damping
µN 
µN
x( t ) =  x0 −

÷cos( ωn t ) +
k 
k

valid for 0 ≤ t ≤ π / ωn
= Initial displacement for next half cycle
04:36:37

83. 83
Mechanical Vibrations – Single Degree-of-Freedom systems
V. Rouillard © 2003 - 2013
Free single DoF vibration + Coulomb damping
•
The displacement at π/ωn becomes the initial displacement for the next half cycle, x 1.

π  
µN 
µN
2µ N 
− x1 = x  t =
=  x0 −
cos( π ) +
= −  x0 −
÷

÷
÷
k 
k
k 

 ωn  

π 
and the initial velocity & ( t = 0 ) is = &  t =
x
x
÷ in eqn ( 2a )
 ωn 
Substituting these initial conditions int o eqn.( 1)
µN
x( t ) = A1 cos( ωn t ) + A2 sin( ωn t ) −
( 1)
k
and its derivative
& t ) = −ωn A1 sin( ωn t ) + ωn A2 cos( ωn t )
x(
gives
3µ N
and
k
such that eqn.( 1 ) becomes :
A1 = x0 −
A2 = 0
3µ N 
µN
x( t ) =  x0 −
cos( ωn t ) −

÷
k 
k

04:36:37
( 1a )
valid for π / ωn ≤ t ≤ π 2 / ωn

84. 84
Mechanical Vibrations – Single Degree-of-Freedom systems
V. Rouillard © 2003 - 2013
Free single DoF vibration + Coulomb damping
µN 
µN
x( t ) =  x0 −
valid for 0 ≤ t ≤ π / ωn

÷cos( ωn t ) +
k 
k

3µ N 
µN
x( t ) =  x0 −
valid for π / ωn ≤ t ≤ π 2 / ωn

÷cos( ωn t ) −
k 
k

= Initial displacement for next half cycle
04:36:37
This method can be applied to successive half cycles until the motion stops.

85. Mechanical Vibrations – Single Degree-of-Freedom systems
85
Free single DoF vibration + Coulomb damping
During each half period π/ωn the reduction in magnitude (peak height) is 2µN/k
•
•
Any two succesive peaks are related by:
4µ N 
xm = xm −1 − 

÷
 k 
•
•
The motion will stop when xn < µN/k
The total number of half vibration cycles, r, is obtained from:
2µ N   µ N 
x0 − r 

÷≤ 
÷
 k   k 
or

µN 
x0 −



k 
r≥

 2µ N  

÷
 k 


04:36:37
V. Rouillard © 2003 - 2013

86. 86
Mechanical Vibrations – Single Degree-of-Freedom systems
V. Rouillard © 2003 - 2013
Free single DoF vibration + Coulomb damping
µN 
µN
x( t ) =  x0 −
valid for 0 ≤ t ≤ π / ωn

÷cos( ωn t ) +
k 
k

3µ N 
µN
x( t ) =  x0 −
valid for π / ωn ≤ t ≤ π 2 / ωn

÷cos( ωn t ) −
k 
k

Final position
= Initial displacement for next half cycle
04:36:37

87. 87
Mechanical Vibrations – Single Degree-of-Freedom systems
V. Rouillard © 2003 - 2013
Free single DoF vibration + Coulomb damping
•
Important features of Coulomb damping:
1.
The equation of motion is nonlinear (cf. linear for viscous damping)
2.
Coulomb damping does not alter the system’s natural frequency (cf. damped natural frequency for viscous
damping).
3.
The motion is always periodic (cf. overdamped for viscous systems)
4.
Amplitude reduces linearly (cf. exponential decay for viscous systems)
5.
System eventually comes to rest – number of vibration cycles finite (cf. sustained vibration with viscous
damping)
6.
The final position is the permanent displacement (not equilibrium) equivalent to the friction force (cf.
approaches zero for viscous systems)
04:36:37

88. Mechanical Vibrations – Single Degree-of-Freedom systems
88
V. Rouillard © 2003 - 2013
Forced (harmonically excited) single DoF vibration
•
•
External energy supplied to system as applied force or imposed motion (displacement, velocity or acceleration)
•
Harmonic forcing function takes the form:
This section deals only with harmonic excitation which results in harmonic response (cf. steady-state or
transient response from non-harmonic excitation).
F( t ) = F0 ei ( ω t +φ )
•
•
•
•
or
F( t ) = F0 cos( ωt + φ )
or
F( t ) = F0 sin( ωt + φ )
Where F0 is the amplitude, ω the frequency and φ the phase angle.
The response of a linear system subjected to harmonic excitation is also harmonic.
The response amplitude depends on the ratio of the excitation frequency to the natural frequency.
Some “common” harmonic forcing functions are:
•
•
Rotating machine / element with (large) residual imbalance
•
•
Vehicle travelling on pavement corrugations or sinusoidal surfaces
Regular shedding of vortices caused by laminar flow across slender structures (VIV) – ie: chimneys,
bridges, overhead cables, mooring cables, tethers, pylons…
Structures excited by regular (very narrow banded) ocean / water waves
04:36:37

89. Mechanical Vibrations – Single Degree-of-Freedom systems
89
V. Rouillard © 2003 - 2013
Forced (harmonically excited) single DoF vibration
•
Equation of motion when a force is applied to a viscously damped SDOF system is:
mx + cx + kx = F ( t )
&& &
•
¬ non hom ogeneous differential eqn.
The general solution to a nonhomogeneous DE is the sum if the homogeneous solution x h(t) and the particular
solution xp(t).
•
The homogeneous solution represents the solution to the free SDOF which is known to decay over time for all
conditions (underdamped, critically damped and overdamped).
•
The general solution therefore reduces to the particular solution x p(t) which represents the steady-state vibration
which exists as long as the forcing function is applied.
04:36:37

90. Mechanical Vibrations – Single Degree-of-Freedom systems
90
V. Rouillard © 2003 - 2013
Forced (harmonically excited) damped single DoF vibration
•
Example of solution to harmonically excited damped SDOF system:
Homogenous solution: decaying vibration @ natural frequency
Particular solution: steady-state vibration @ excitation
frequency
Complete solution
04:36:37

91. Mechanical Vibrations – Single Degree-of-Freedom systems
91
V. Rouillard © 2003 - 2013
Forced (harmonically excited) single DoF vibration – undamped.
•
Let the forcing function acting on the mass of an undamped SDOF system be:
F( t ) = F0 cos( ωt )
•
The eqn. of motion reduces to:
mx + kx = F0 cos( ωt )
&&
•
Where the homogeneous solution is:
xh ( t ) = C1 cos( ωn t ) + C2 sin( ωn t )
where ωn = k / m
•
As the excitation is harmonic, the particular solution is also harmonic with the same frequency:
x p ( t ) = X cos( ωt )
•
Substituting xp(t) in the eqn. of motion and solving for X gives:
X=
•
F0
k − mω 2
The complete solution becomes
04:36:37
x( t ) = xh ( t ) + x p ( t ) = C1 cos( ωnt ) + C 2 sin( ωnt ) +
F0
k − mω
2
cos( ωt )

92. Mechanical Vibrations – Single Degree-of-Freedom systems
92
V. Rouillard © 2003 - 2013
Forced (harmonically excited) single DoF vibration – undamped.
•
Applying the initial conditions
x( t = 0 ) = x0
C1 = x0 −
•
and
F0
& t =0)=&0
x(
x
and
k − mω 2
C2 =
gives:
&0
x
ωn
The complete solution becomes:
& 
F0 
x
F0

x( t ) =  x0 −
cos( ωnt ) +  0 ÷sin( ωn t ) +
cos( ωt )
÷
2
2
ωn 

k − mω 
k − mω

•
The maximum amplitude of the steady-state solution can be written as:
X
=
δ st
•
1
2
ω 
1−  ÷
 ωn 
F
where δ st = 0
k
X/δst is the ratio of the dynamic to the static amplitude and is known as the amplification factor or amplification
ratio and is dependent on the frequency ratio r = ω/ωn.
04:36:37

93. Mechanical Vibrations – Single Degree-of-Freedom systems
93
V. Rouillard © 2003 - 2013
Forced (harmonically excited) single DoF vibration – undamped.
•
When ω/ωn < 1 the denominator of the steady-
X / δ st
state amplitude is positive and the amplification
factor increases as ω approaches the natural
frequency ωn. The response is in-phase with
the excitation.
•
When ω/ωn > 1 the denominator of the steadystate amplitude is negative an the amplification
factor is redefined as:
X
1
=
δ st  ω  2
ω ÷ −1
 n
and the steady − state response becomes :
x p ( t ) = − X cos( ωt )
which shows that the response is out-of-phase with
the excitation and decreases (→ zero ) as ω increases
(→ ∞)
04:36:37
r=
ω
ωn

94. Mechanical Vibrations – Single Degree-of-Freedom systems
94
Forced (harmonically excited) single DoF vibration – undamped.
•
When ω/ωn < 1 the denominator of the steadystate amplitude is positive and the amplification
factor increases as ω approaches the natural
frequency ωn. The response is in-phase with
the excitation.
•
When ω/ωn > 1 the denominator of the steadystate amplitude is negative an the amplification
factor is redefined as:
X
1
=
δ st  ω  2
ω ÷ −1
 n
and the steady − state response becomes :
x p ( t ) = − X cos( ωt )
which shows that the response is out-of-phase with
the excitation and decreases (→ zero ) as ω increases
(→ ∞)
04:36:37
X / δ st
V. Rouillard © 2003 - 2013

95. Mechanical Vibrations – Single Degree-of-Freedom systems
95
Forced (harmonically excited) single DoF vibration – undamped.
•
When ω/ωn = 1 the denominator of the steadystate amplitude is zero an the response
becomes infinitely large. This condition when
ω=ωn is known as resonance.
X / δ st
04:36:37
V. Rouillard © 2003 - 2013

96. Mechanical Vibrations – Single Degree-of-Freedom systems
96
V. Rouillard © 2003 - 2013
Forced (harmonically excited) single DoF vibration – undamped.
•
The complete solution
& 
F0 
x
F0

x( t ) =  x0 −
cos( ωn t ) +  0 ÷sin( ωn t ) +
cos( ωt )
2÷
2

k − mω 
k − mω
 ωn 
can be written as:
x( t ) = Acos( ωn t + φ ) +
x( t ) = Acos( ωn t + φ ) −
δ st
2
cos( ωt )
for ω / ωn < 1
2
cos( ωt )
for ω / ωn > 1
ω 
1−  ÷
 ωn 
δ st
ω 
1−  ÷
 ωn 
where A and φ are functions of x0 and & 0 as before.
x
•
The complete solution is a sum of two cosines with frequencies corresponding to the natural and forcing
(excitation) frequencies.
04:36:37

97. 97
Mechanical Vibrations – Single Degree-of-Freedom systems
Forced (harmonically excited) single DoF vibration – undamped.
ω /ωn < 1
ω /ωn > 1
04:36:37
V. Rouillard © 2003 - 2013

98. Mechanical Vibrations – Single Degree-of-Freedom systems
98
V. Rouillard © 2003 - 2013
Forced (harmonically excited) single DoF vibration – undamped.
•
When the excitation frequency ω is close but not exactly equal to the natural frequency ωn beating may
occur.
•
Letting the initial conditions x0= x’0 =0 , the complete solution:
& 
F0 
x
F0

x( t ) =  x0 −
cos( ωn t ) +  0 ÷sin( ωnt ) +
cos( ωt )
2÷
2
ωn 

k − mω 
k − mω

reduces to :
x( t ) =
( F0 / m )
( ωn2 − ω 2 )
[ c os( ωnt ) − cos( ωt )] =
( F0 / m ) 
  ω + ωn  
  ω − ωn   
2 sin 
÷t  ×sin  
÷t  
2
2 
( ωn − ω ) 

2
 
If we let the excitation frequency be slightly less than the natural frequency:
ωn − ω = 2ε
where ε is a small positive number. Then
ωn ≈ ω and
ωn + ω = 2ω
therefore :
2
( ω n − ω ) ( ωn + ω ) = ω n − ω 2 = 4 ε ω
2
Substituting for ωn − ω , ωn + ω and ωn − ω 2 in the complete solution yields :
04:36:37
x( t ) =
( F0 / m )
sin ( ε t ) ×sin ( ωt )
( 2εω )

2
 

99. Mechanical Vibrations – Single Degree-of-Freedom systems
99
V. Rouillard © 2003 - 2013
Forced (harmonically excited) single DoF vibration – undamped.
x( t ) =
•
( F0 / m ) sin ε t ×sin ωt
( )
( )
( 2εω )
Since ε is small, sin(ε t) has a long period. The solution can then be considered as harmonic motion with a
principal frequency ω an a variable amplitude equal to
X(t ) =
04:36:37
( F0 / m ) sin ε t
( )
( 2εω )

100. 100
Mechanical Vibrations – Single Degree-of-Freedom systems
V. Rouillard © 2003 - 2013
Forced (harmonically excited) single DoF vibration – Damped.
•
•
Steady-state Solution
•
The equation of motion of a SDOF system with viscous damping is:
If the forcing function is harmonic:
F( t ) = F0 cos( ωt )
mx + cx + kx = F0 cos( ωt )
&& &
•
The steady-state response is given by the particular solution which is also expected to be harmonic:
x p ( t ) = X cos( ωt − φ )
where the amplitude X and the phase angle φ are to be det er min ed
04:36:37

101. 101
Mechanical Vibrations – Single Degree-of-Freedom systems
Forced (harmonically excited) single DoF vibration – Damped.
•
Substituting xp into the steady-state eqn. of motion yields:
(
)
X  k − mω 2 cos( ωt − φ ) − cω sin( ωt − φ ) = F0 cos( ωt )


applying the trigonometric relationships :
cos( ωt − φ ) = cos( ωt )cos( φ ) + sin( ωt ) sin( φ )
sin( ωt − φ ) = sin( ωt )cos( φ ) − cos( ωt ) sin( φ )
we obtain :
(
)
X ( k − mω 2 ) sin( φ ) − cω cos( φ ) = 0


X  k − mω 2 cos( φ ) + cω sin( φ ) = F0


which gives :
X=
F0
(
)

2
k − mω
− ( cω ) 



for the particular solution
2 2
x p ( t ) = X cos( ωt − φ )
04:36:37
1
and
2
 cω 
φ = a tan 
÷
 k − mω 2 
V. Rouillard © 2003 - 2013

102. 102
Mechanical Vibrations – Single Degree-of-Freedom systems
V. Rouillard © 2003 - 2013
Forced (harmonically excited) single DoF vibration – Damped.
•
Alternatively, the amplitude and phase can be written in terms of the frequency ratio r = ω/ωn and the
damping coefficient ζ:
X
=
δ st
1

2
ω 

ω 

 1 −  ÷  +  2ζ
 
ωn    ωn  
 



2 2

ω 
 2ζ
÷
ωn ÷
φ = a tan 
=
2÷

1−  ω  ÷
 ω ÷ ÷
  n 
04:36:37
1
=
2
 2ζ r 
a tan 
÷
 1 − r2 
1

2
 1 − r  + [ 2ζ r ] 


2 2
1
2

103. 103
Mechanical Vibrations – Single Degree-of-Freedom systems
Forced (harmonically excited) single DoF vibration – Damped.
X
=
δ st
04:36:37
1
1

2 2
2 2
 1 − r  + [ 2ζ r ] 



 2ζ r 
φ = a tan 
÷
 1 − r2 
V. Rouillard © 2003 - 2013

104. 104
Mechanical Vibrations – Single Degree-of-Freedom systems
V. Rouillard © 2003 - 2013
Forced (harmonically excited) single DoF vibration – Damped.
•
The magnification ratio at all frequencies is reduced
with increased damping.
•
The effect of damping on the magnification ratio is
greatest at or near resonance.
•
The magnification ratio approaches 1 as the frequency
ratio approaches 0 (DC)
•
The magnification ratio approaches 0 as the frequency
ratio approaches ∞
•
For 0 < ζ < 1/ √2 the magnification ratio maximum
occurs at r = √(1 - 2ζ2) or ω = ωn √(1 - 2ζ2) which is
lower than both the undamped natural frequency ωn and
the damped natural frequency ωd = ω n √(1 - ζ2)
•
When r = √(1 - 2ζ2) Mmax= 1/[2ζ √(1 - ζ2)] → if Mmax can
be measured, the damping ratio can be determined.
•
•
04:36:37
When ζ = 1/√2 dM/dr = 0 at r = 0.
When ζ > 1/√2 M decreases monotonically with
increasing frequency.

105. 105
Mechanical Vibrations – Single Degree-of-Freedom systems
V. Rouillard © 2003 - 2013
Forced (harmonically excited) single DoF vibration – Damped.
•
•
For damped systems (ζ > 0) when r < 1 the phase
angle is less than 90o and response lags the excitation
and when r >1 the phase angle is greater than 90 o and
the response leads the excitation (approaches 180 o for
large frequency ratios..
•
04:36:37
For undamped systems (ζ = 0) the phase angle is 0o
(response in phase with excitation) for r<1 and 180 o
(response out of phase with excitation) for r>1.
For damped systems (ζ > 0) when r =1 the phase lag is
always 90o.

106. 106
Mechanical Vibrations – Single Degree-of-Freedom systems
V. Rouillard © 2003 - 2013
Forced (harmonically excited) single DoF vibration – Damped.
•
Complete Solution
•
The complete solution is the sum of the homogeneous solution xh(t) and the particular solution xp(t):
x( t ) = X 0 e −ζωnt cos( ωd t − φ0 ) + X cos( ωt − φ )
where ωd = ωn 1 − ζ 2 , X and φ are given as before, and X 0 and φ0 are det er min ed from
the initial conditions
04:36:37

107. Mechanical Vibrations – Single Degree-of-Freedom systems
107
V. Rouillard © 2003 - 2013
Forced (harmonically excited) single DoF vibration – Damped.
•
Quality Factor & Bandwidth
•
•
When damping is small (ζ < 0.05) the peak magnification ratio corresponds with resonance ( ω =ωn).
The value of the magnification ratio (Quality factor or Q factor) becomes:
 X 
1
1
Q=
=
=
÷
1
2ζ
 δ st ω =ωn 
2
 2
2
2

ω 
  ω 
 1 −  ÷  +  2ζ
 
  ωn    ωn  




•
The points where the magnification ratio falls
below Q/√2, are called the half power points R 1
and R2. (Power is proportional to amplitude
squared: Power = Fv = cv2 = c(dx/dt)2
•
The difference between the half power
frequencies is called the bandwidth.
1
2ζ
Q
2
The Quality factor Q can be used to estimate
the equivalent viscous damping of systems.
•
Q=
04:36:37
Bandwidth
R1 1 R2

108. 108
Mechanical Vibrations – Single Degree-of-Freedom systems
V. Rouillard © 2003 - 2013
Forced (harmonically excited) single DoF vibration – Damped.
•
The values of the half power frequencies are determined as follows:
 X 
1
Q
1
=
=
=
δ ÷
2
2 2 2ζ
 st 
2
  ω 2  
ω
1 −  ÷  +  2ζ

  ωn    ωn 


In terms of the frequency ratio r :
r 4 − r 2 ( 2 − 4ζ 2 ) + ( 1 − 8ζ 2 ) = 0
Which, when solved gives :
r12 = 1 − 2ζ 2 − 2ζ 1 + ζ 2
and
2
r2 = 1 − 2ζ 2 + 2ζ 1 + ζ 2
When ζ is small , ζ 2 is negligible and the solutions can be reduced to :
2
2 ω 
r12 = R1 =  1 ÷ ; 1 − 2ζ
 ωn 
∴
(
)
2
and
2
2
2
2
2
2
ω2 − ω1 = R2 − R1 ωn ; 4ζωn
04:36:37
2
2 ω 
r2 = R2 =  2 ÷ ; 1 + 2ζ
 ωn 

109. 109
Mechanical Vibrations – Single Degree-of-Freedom systems
V. Rouillard © 2003 - 2013
Forced (harmonically excited) single DoF vibration – Damped.
ω2 + ω1
2
2
= ωn and ω2 − ω1 = ( ω2 + ω1 ) ( ω2 − ω1 ) ,
2
the bandwidth ∆ω = ω2 − ω1 can be written as :
Since
2
2
2
ω2 − ω1 4ζωn
∆ω =
;
; 2ζωn
ω2 + ω1
2ωn
The qualily factor Q can then be exp ressed in terms of the natural frequency and bandwidth :
Q;
ω
1
; n
2ζ ∆ω
04:36:37

110. 110
Mechanical Vibrations – Single Degree-of-Freedom systems
V. Rouillard © 2003 - 2013
Forced (harmonically excited) single DoF vibration – Damped.
•
•
Complex notation.
Recall that a harmonic function may expressed as follows:
F( t ) = F0 cos( ωt + φ )
•
•
•
=
F0 sin( ωt + φ )
F0 ei( ωt +φ )
If the harmonic forcing function is expressed in complex form:
F = F0 eiωt
The equation of motion for a damped SDOF system becomes:
mx + cx + kx = F0 eiωt
&& &
The actual excitation function is real and is represented by the real part of the complex function.
Consequently, the steady-state response is also real and is represented by the real part of the complex
particular solution which takes the form:
x p ( t ) = Xeiωt
Therefore :
& p ( t ) = iω Xeiωt
x
•
=
and && p ( t ) = −ω Xeiωt
x
Substituting in the eqn. of motion gives:
− mω 2 Xeiωt + icω Xeiωt + kXeiωt = F0 eiωt
04:36:37

111. 111
Mechanical Vibrations – Single Degree-of-Freedom systems
V. Rouillard © 2003 - 2013
Forced (harmonically excited) single DoF vibration – Damped.
•
The response amplitude becomes:
X=
F0
¬ X / F0 is called the RECEPTANCE ( Dynamic compliance )
 k − mω 2 + icω 


multiplying the numerator & deno min ator on the RHS by k − mω 2 − icω
and separating real and imaginary components :
(
)
(
)


k − mω 2
cω

X = F0 
−i
2
2


k − mω 2 + c 2ω 2
k − mω 2 + c 2ω 2 



(
)
(
)
e −iφ
where
y
applying the complex relationships : x + iy = Aeiφ where A = x 2 + y 2 and φ = a tan  
 ÷
x
The magnitude of the response can be written as :
X=
F0
(
1
)

 2
k − mω
+ c 2ω 2 



And the steady − state solution becomes :
xp( t ) =
04:36:37
2 2
F0
(

 k − mω

)
2 2

+ c 2ω 2 

1
ei( ωt −φ )
2
 cω 
φ = a tan 
÷
 k − mω 2 

112. 112
Mechanical Vibrations – Single Degree-of-Freedom systems
V. Rouillard © 2003 - 2013
Forced (harmonically excited) single DoF vibration – Damped.
•
As before the response amplitude:
X=
(
F0
)
 k − mω 2 + icω 


can be written in terms of the frequency ratio r and the damping ratio ζ :
kX
1
=
≡ H( iω ) ¬ Complex Frequency Re sponse Function ( FRF )
F0 1 − r 2 + i2ζ r
The magnitude of H( iω ) is given by :
H( iω ) =
kX
=
F0
1
( 1− r )
2 2
which is the same as the magnification ratio M :
+ ( 2ζ r )
2
It can be shown that the complex FRF and its magnitude are related by :
H( iω ) = H( iω ) e −iφ
where e −iφ = cos φ + i sin φ and
 2ζ r 
φ = a tan 
÷
 1 − r2 
The steady − state response can therefore be exp ressed as :
F
x p ( t ) = 0 H( iω ) ei( ωt −φ )
k
•
Measurements of the magnitude FRF can be used to experimentally determine the values of m, c and k.
04:36:37

113. 113
Mechanical Vibrations – Single Degree-of-Freedom systems
V. Rouillard © 2003 - 2013
Forced (harmonically excited) single DoF vibration – Damped.
•
•
When the excitation function is described by: F( t ) = F0 cos( ωt )
The steady-state response is given by the real part of the solution:
xp( t ) =
F0
(
)
1
ei( ωt −φ ) =
2

 2
k − mω 2 + c 2ω 2 



F
= Re  0 H( iω )eiωt 
k



F
= Re  0 H( iω ) ei( ωt −φ ) 
k



•
•
F0
(

k − mω 2


Conversely, when the excitation function is described by:
)
2

+ c 2ω 2 

1
cos( ωt − φ )
2
F( t ) = F0 sin( ωt )
The steady-state response is given by the imaginary part of the solution:
xp( t ) =
F0
(
)
1
2

 2
k − mω 2 + c 2ω 2 



F
= Im  0 H( iω )eiωt 
k



 F0 H( iω ) ei( ωt −φ ) 
=
04:36:37 Im
k



ei( ωt −φ ) =
F0
(

k − mω 2


)
2

+ c 2ω 2 

1
sin( ωt − φ )
2

114. 114
Mechanical Vibrations – Single Degree-of-Freedom systems
Forced (harmonically excited) single DoF vibration – Damped.
•
•
Complex Vector Notation of Harmonic Motion:
Harmonic excitation and response can be represented in the complex plane
Steady − state displacement :
F
x p ( t ) = 0 H( iω ) ei( ω t −φ )
k
Steady − state velocity :
F
& p ( t ) = iω 0 H( iω ) ei( ω t −φ ) = iω x p ( t )
x
k
Steady − state acceleration :
F0
H( iω ) ei( ω t −φ ) = −ω 2 x p ( t )
k
Since i and − 1 respectively can be written as :
2
x
&& p ( t ) = ( iω )
π
π 
π  i2
i = cos  ÷+ i sin  ÷ = e
2
2
•
and
− 1 = cos ( π ) + i sin ( π ) = eiπ
It can be seen that:
•
•
The velocity leads the displacement by 90 o and is multiplied by ω.
The acceleration leads the displacement by 180 o and is multiplied by ω2.
04:36:37
V. Rouillard © 2003 - 2013

115. 115
Mechanical Vibrations – Single Degree-of-Freedom systems
V. Rouillard © 2003 - 2013
Forced (harmonically excited) single DoF vibration – Damped.
•
Complex Vector Notation of Harmonic Motion:
xp( t ) =
04:36:37
F0
H( iω ) ei( ω t −φ )
k
x
& p ( t ) = iω x p ( t )
x
&& p ( t ) = −ω 2 x p ( t )

116. Mechanical Vibrations – Single Degree-of-Freedom systems
116
V. Rouillard © 2003 - 2013
Forced (harmonically excited) single DoF vibration – Damped.
•
•
•
Response due to base motion (harmonic)
•
At any time, the length of the spring is x – y and
the relative velocity between the two ends of the
damper is x’ – y’.
•
The equation of motion is:
In this case, the excitation is provided by the imposed harmonic motion of the supporting base.
The displacement of the base about a neutral position is denoted by y(t) and the response of the mass from
its static equilibrium position by x(t).
mx + c( & − & ) + k( x − y ) = 0
&&
x y
If y( t ) = Y sin( ωt ) the eqn.of motion becomes :
mx + cx + kx = cy + ky
&& &
&
k( x − y )
= cωY cos( ωt ) + kY sin( ωt )
= A sin( ωt − α )
cω
where A = Y k 2 + ( cω )2 and α = a tan  − 

÷
 k 
•
The applied displacement has the same effect of
applying a harmonic force of magnitude A to the
mass.
04:36:37
c( & − &y )
x
y( t ) = Y sin( ω t )

117. 117
Mechanical Vibrations – Single Degree-of-Freedom systems
V. Rouillard © 2003 - 2013
Forced (harmonically excited) single DoF vibration – Damped.
•
The steady-state response of the mass is given by the particular solution x p(t):
xp( t ) =
Y k 2 + ( cω )2
(
1
)
sin( ωt − φ1 − α )
2

 2
k − mω 2 + c 2ω 2 



cω
 cω 
where α = a tan  −  and φ1 = a tan 

÷
÷
 k 
 k − mω 2 
The solution can be simplified to :
x p ( t ) = X sin( ωt − φ )
where


2
2
X 
k + ( cω )

=
Y
2 2
2 2
+c ω 
 k − mω


(
)
1
2


2
1 + ( 2ζ r )

=

2 2
2
+ ( 2ζ r ) 
 1− r


(
)
1
2
¬ Displacement Transmissibility
and




mcω 3
2ζ r 3

÷ = a tan 
φ = a tan
÷
 k k − mω 2 + ( cω )2 ÷
1 + ( 4ζ 2 − 1)r 2 



04:36:37
(
)

118. 118
Mechanical Vibrations – Single Degree-of-Freedom systems
Forced (harmonically excited) single DoF vibration – Damped.


2
X 
1 + ( 2ζ r )

=
Y
2 2
2
+ ( 2ζ r ) 
 1− r


(
04:36:37
)
1
2
and


2ζ r 3
φ = a tan 
÷
1 + ( 4ζ 2 − 1 )r 2 

V. Rouillard © 2003 - 2013

119. 119
Mechanical Vibrations – Single Degree-of-Freedom systems
Forced (harmonically excited) single DoF vibration – Damped.
•
Characteristics of the displacement
transmissibility:
•
The transmissibility is 1 when r = 0 (DC) and
close to 1 when r is small.
•
For undamped systems (ζ = 0), Td → ∞ at
resonance (r = 1)
•
For all damping values Td<1 for r >√2 and
Td = 1 for r = √2
•
For r <√2 Td is inversely proportional to ζ
•
For r >√2 Td is proportional to ζ
04:36:37
V. Rouillard © 2003 - 2013

120. Mechanical Vibrations – Single Degree-of-Freedom systems
120
V. Rouillard © 2003 - 2013
Forced (harmonically excited) single DoF vibration – Damped.
•
•
Transmitted Force
The force transmitted to the base/support is caused by the reaction of the spring and damper:
F = k( x − y ) + c( & − & ) = − mx
x y
&&
Since the steady − state ( particular ) solution is x p ( t ) = X sin( ωt − φ ) ,F can be written as :
F = mω 2 X sin( ω t − φ ) = FT sin( ωt − φ )
•
Where FT is the amplitude of the transmitted force and is given by:
1
2
 2
FT
1 + ( 2ζ r )
2
=r 
¬ Force Transmissibility
2 2
2
kY
 ( 1 − r ) + ( 2ζ r ) 
•
Note that the transmitted force is always in–phase with
the motion of the mass x(t):
k( x − y )
c( & − &y )
x
y( t ) = Y sin( ω t )
04:36:37

121. 121
Mechanical Vibrations – Single Degree-of-Freedom systems
V. Rouillard © 2003 - 2013
Forced (harmonically excited) single DoF vibration – Damped.
 1 + ( 2ζ r )

FT
= r2 

kY
( 1 − r 2 ) + ( 2ζ r )2 

2
04:36:37
1
2

122. 122
Mechanical Vibrations – Single Degree-of-Freedom systems
V. Rouillard © 2003 - 2013
Forced (harmonically excited) single DoF vibration – Damped.
•
•
Relative Motion
If z = x – y represents the motion of the mass relative to the base, the eqn. of motion:
k( x − y )
mx + c( & − & ) + k( x − y ) = 0
&&
x y
can be written as :
mz + cz + kz = − my = mω 2Y sin( ωt )
&&
&& &
The ( steady − state ) solution of which is :
z( t ) =
mω 2Y sin( ωt − φ1 )
(
)

2
2 2
k − mω
+ ( cω ) 



where the amplitude Z is given by :
Z=
mω 2Y
(
)

2
k − mω
+ ( cω ) 



and the phase φ1 is given by :
2 2
1
1
2
= Z sin( ωt − φ1 )
=Y
2
c( & − &y )
x
r2
(

 1− r

 cω 
 2ζ r 
φ1 = a tan 
= a tan 
÷
÷
 1 − r2 
 k − mω 2 
04:36:37
)
2 2
2
+ ( 2ζ r ) 

1
2

123. 123
Mechanical Vibrations – Single Degree-of-Freedom systems
Forced (harmonically excited) single DoF vibration – Damped.
•
Relative Motion
Z
=
Y
r2
(

 1− r

)
2
+ ( 2ζ r ) 

 2ζ r 
φ1 = a tan 
÷
 1 − r2 
04:36:37
2 2
1
2
V. Rouillard © 2003 - 2013

124. 124
Mechanical Vibrations – Single Degree-of-Freedom systems
V. Rouillard © 2003 - 2013
Forced (harmonically excited) single DoF vibration – Damped.
•
•
Rotating Imbalance Excitation
With the horizontal components cancelled the vertical component of the excitation is:
2
F( t ) = meω sin( ω t )
The eqn. of motion is :
Mx + cx + kx = meω 2 sin( ω t )
&& &
and the steady − state solution becomes :
 me  ω 2

i( ω t −φ ) 
x p ( t ) = X sin( ω t − φ ) = Im  
÷ H( iω ) e
 M  ωn 



The response amplitude and phase are given by :
X =
2
meω 2
(
)
1
me  ω 
MX
=
 ω ÷ H( iω ) or me =
M  n

2
k − Mω
+ ( cω ) 



 cω 
 2ζ r 
φ = a tan 
= a tan 
÷
÷
 1 − r2 
 k − Mω 2 
04:36:37
2 2
2
r2
(

 1− r

)
2 2
2
+ ( 2ζ r ) 

1
= r 2 H( iω )
2

125. 125
Mechanical Vibrations – Single Degree-of-Freedom systems
Forced (harmonically excited) single DoF vibration – Damped.
•
Rotating Imbalance Excitation
MX
=
me
r2
(

 1− r

)
2 2
= r 2 H( iω )
 2ζ r 
φ = a tan 
÷
 1 − r2 
04:36:37
1
 2
+ ( 2ζ r ) 2 

V. Rouillard © 2003 - 2013

126. Mechanical Vibrations – Single Degree-of-Freedom systems
126
V. Rouillard © 2003 - 2013
Forced (harmonically excited) single DoF vibration – Damped.
•
•
Forced Vibration with Coulomb Damping
The equation of motion for a SDOF with Coulomb damping subjected to a harmonic force is:
Mx + kx ± µ N = F0 sin( ω t )
&&
•
•
Solution complicated.
•
If µN << F0 motion of mass m will approximate harmonic motion
•
When µN << F0 an approximate solution to eqn. of motion may be used to determine equivalent viscous
If µN is large cf F0, motion of mass m is discontinuous
damping ratio.
•
•
This is achieved by equating dissipated energy for both cases.
For Coulomb damping, the energy dissipated during a cycle of amplitude X is:
∆W = 4 ( µ NX ) − 4 quarter cycles
•
For viscous damping, the energy dissipated during a cycle of amplitude X is:
∆W =
2π / ω
∫
Fv dt =
t =0
= π ceqω X 2
04:36:37
2π / ω
∫
t =0
2
2π
dx
ceq   dt = ∫ ceq X 2ω cos 2 ( ωt ) d( ωt )
 ÷
 dt 
t =0

127. Mechanical Vibrations – Single Degree-of-Freedom systems
127
Forced (harmonically excited) single DoF vibration – Damped.
•
Equating the dissipated energies:
ceq =
•
πω X 2
And the equivalent damping ratio is defined as:
ζ eq =
•
4µ N
ceq
cc
=
ceq
2mωn
=
4µ N
2µ N
=
2mωnπω X π mωnω X
The amplitude X and the phase φ of the response becomes:
12



2

 4µ N  
1− 

÷ 
F0 
 π F0  
X=
k 
2 2 
 1 −  ω   
   ωn ÷  


    
•
These approximations are only valid for
04:36:37



4µ N
±1 −


π F0

φ = a tan 
1 2

2
  4µ N   

 1−
 


π F0 ÷  
  

 
µ N << F0
V. Rouillard © 2003 - 2013

128. 128
Mechanical Vibrations – Single Degree-of-Freedom systems
V. Rouillard © 2003 - 2013
SDoF systems – General forcing functions
•
•
•
•
Methods to solve response due to general (nonharmonic) forcing functions.
•
When forcing function is periodic (not harmonic), it can be described with a series (sum) of harmonic or
Fourier components.
General forcing function may be periodic (nonharmonic) or aperiodic.
Aperiodic forcing functions may be finite or infinite
When the duration of a transient forcing function << natural period of system, forcing function called
SHOCK.
04:36:37

129. 129
Mechanical Vibrations – Single Degree-of-Freedom systems
V. Rouillard © 2003 - 2013
Types of deterministic forcing functions.
Deterministic
Deterministic
Periodic
Periodic
Sinusoidal
Sinusoidal
04:36:37
Complex Periodic
Complex Periodic
Non-periodic
Non-periodic
Almost Periodic
Almost Periodic
Transient
Transient

130. 130
Mechanical Vibrations – Single Degree-of-Freedom systems
V. Rouillard © 2003 - 2013
Types of deterministic forcing functions.
Deterministic
Deterministic
Periodic
Periodic
Sinusoidal
Sinusoidal
Non-periodic
Non-periodic
Complex
Complex
Periodic
Periodic
Almost Periodic
Almost Periodic
Transient
Transient
Can be defined mathematically. Waveform contains harmonics which are multiples if the
04:36:37
fundamental frequency (show spectrum) Signal factory.vee

131. 131
Mechanical Vibrations – Single Degree-of-Freedom systems
V. Rouillard © 2003 - 2013
Types of deterministic forcing functions.
Deterministic
Deterministic
Periodic
Periodic
Sinusoidal
Sinusoidal
Complex Periodic
Complex Periodic
Non-periodic
Non-periodic
Almost Periodic
Almost Periodic
Transient
Transient
Contains sine wave of arbitrary frequencies which frequency ratios are not rational numbers (show
04:36:37
spectrum) Signal factory.vee

132. 132
Mechanical Vibrations – Single Degree-of-Freedom systems
V. Rouillard © 2003 - 2013
Types of deterministic forcing functions.
Deterministic
Deterministic
Periodic
Periodic
Sinusoidal
Sinusoidal
Complex Periodic
Complex Periodic
Half-sine pulse
04:36:37
Non-periodic
Non-periodic
Transient
Transient
Almost Periodic
Almost Periodic
Sin(x)/x
Exp(T-x)
All other deterministic data that can be described by a suitable function

133. 133
Mechanical Vibrations – Single Degree-of-Freedom systems
V. Rouillard © 2003 - 2013
SDoF systems – General forcing functions - Periodic
•
•
For periodic forcing functions, the response of system is obtained by using the principle of superposition:
•
The periodic forcing function (period τ = 2π/ω) can be expressed as a Fourier series:
The total response consists of sum of response functions due to individual harmonic functions in forcing
function.
a
F( t ) = o +
2
∞
∑ a j cos( jω t )
j =1
+
∞
∑ b j sin( jωt )
j =1
where
τ
2
a j = ∫ F( t )cos( jω t ) dt
τ0
for
τ
2
b j = ∫ F( t ) sin( jω t ) dt,
τ0
•
j = 1, 2, 3.....
for
The eqn. of motion can be written as:
a
mx + cx + kx = o +
&& &
2
•
j = 0, 1, 2.....
∞
∑ a j cos( jω t )
j =1
+
∞
∑ b j sin( jω t )
j =1
The RHS
04:36:37 is a constant + a sum of harmonic functions.

134. 134
Mechanical Vibrations – Single Degree-of-Freedom systems
V. Rouillard © 2003 - 2013
SDoF systems – General forcing functions - Periodic
•
Using the principle of superposition, the steady-state solution is the sum of the steady-state solution for the
following equations:
mx + cx + kx =
&& &
mx + cx + kx =
&& &
ao
2
(1)
∞
∑ a j cos( jω t )
(2)
j =1
mx + cx + kx =
&& &
∞
∑ b j sin( jω t )
(3)
j =1
•
The steady-state solutions of (1), (2) and (3) are
xp( t ) =
xp( t ) =
xp( t ) =
04:36:37
ao
2k
aj k
( 1− j r )
2 2 2
+ ( 2ζ jr )2
bj k
( 1− j r )
2 2 2
+ ( 2ζ jr )2
cos( jω t − φ j )
sin( jω t − φ j )

135. 135
Mechanical Vibrations – Single Degree-of-Freedom systems
V. Rouillard © 2003 - 2013
SDoF systems – General forcing functions - Periodic
•
The entire steady-state solution is given by:
a
xp( t ) = o +
2k
∞
∑
j =1
aj k
( 1− j r )
2 2 2
+ ( 2ζ jr )2
cos( jω t − φ j ) +
∞
∑
j =1
bj k
( 1− j r )
2 2 2
+ ( 2ζ jr )2
sin( jω t − φ j )
where
 2ζ jr 
ω
φ j = a tan 
and r =
÷
ωn
1 − j2r2 

•
•
•
The response amplitude and phase for each harmonic (j th term) depend on j.
•
•
Complete Solution
•
This requires setting the complete solution and its derivative to the specified initial displacement and velocity
which produces a complicated expression for the transient part of the solution.
When r = 1 the response amplitude is relatively high for any value j (more so when both j and ζ are small)
As j becomes larger (higher harmonics) the amplitude response becomes smaller → the first few terms are
usually needed to generate a reasonably accurate response.
The complete solution is obtained by including the transient part of the solution which is dependent on the
initial conditions.
04:36:37
Example: Triangular forcing function. Vee & Excel

136. 136
Mechanical Vibrations – Single Degree-of-Freedom systems
V. Rouillard © 2003 - 2013
SDoF systems – General forcing functions - Periodic
•
Situation sometimes arises when the periodic forcing function is given (obtained) experimentally (eg: wave,
wind , seismic, topography..) and represented by discrete measurement data.
•
•
When the (measured) data cannot be readily described by a mathematical function
The discrete measurement data can be integrated numerically to obtain the Fourier coefficients.
2 N
a0 = ∑ Fi
N i =1
•
2 N
 2 jπ ti 
a j = ∑ Fi cos 
÷ and
N i =1
 τ 
2 N
 2 jπ ti 
b j = ∑ Fi sin 
÷
N i =1
 τ 
for
j = 1, 2.....
The Fourier coefficients can then be used to find the solution with the excitation frequency taken as the
lowest frequency component of the data.
04:36:37
2π
ω=
τ

137. 137
Mechanical Vibrations – Single Degree-of-Freedom systems
V. Rouillard © 2003 - 2013
SDoF systems – General forcing functions – Nonperiodic
•
When the forcing function is arbitrary and nonperiodic (aperiodic) it cannot be represented with a Fourier
series
•
Alternative methods for determining the response must be used:
•
•
•
•
Representation of the excitation function with a Convolution integral
Using Laplace Transformations
Approximating F(t) with a suitable interpolation method then using a numerical procedure
Numerical integration of the equations of motion.
04:36:37

138. 138
Mechanical Vibrations – Single Degree-of-Freedom systems
V. Rouillard © 2003 - 2013
SDoF systems – General forcing functions – Nonperiodic
•
When the forcing function is arbitrary and nonperiodic (aperiodic) it cannot be represented with a Fourier
series
•
Alternative methods for determining the response must be used:
•
•
•
•
Representation of the excitation function with a Convolution integral
Using Laplace Transformations
Approximating F(t) with a suitable interpolation method then using a numerical procedure
Numerical integration of the equations of motion.
04:36:37

139. 139
Mechanical Vibrations – Single Degree-of-Freedom systems
V. Rouillard © 2003 - 2013
SDoF systems – General forcing functions – Nonperiodic
•
•
Convolution integral
•
An impulse can be measured by the resulting change in momentum:
Consider one of the simplest nonperiodic exciting force: Impulsive force: which has a large magnitude F
which acts for a very short time ∆t.
Im pulse = F ∆t = m& 2 − mx1
x
&
where & 1 and & 2 represent the velocity of the lumped mass before and after the impulse .
x
x
•
The magnitude of the impulse F∆t is represented by
F=
%
t + ∆t
∫
F dt
t
and a unit impulse is defined as
f = lim
% ∆t →0
•
t + ∆t
∫
F dt = Fdt = 1
t
For Fdt to have a finite value, F approaches infinity as ∆t nears zero.
04:36:37

140. 140
Mechanical Vibrations – Single Degree-of-Freedom systems
V. Rouillard © 2003 - 2013
SDoF systems – General forcing functions – Nonperiodic
•
•
•
Convolution integral – Impulse response
Consider a (viscously) damped SDoF (mass-spring-damper system) subjected to an impulse at t=0.
For an underdamped system, the eqn. of motion is:
mx + cx + kx = 0
&& &
•
And its solution:
−ζωn t
x( t ) = e


& 0 + ζωn x0
x


sin ( ωd t ) 
 x0 cos ( ωd t ) +
1 − ζ 2 ωn




where
c
ζ =
2mωn
•
ωd = ωn 1 − ζ
•
2
k  c 
=
−
÷
m  2m 
ωn =
If, prior to the impulse load being applied, the mass is at rest, then:
−
x( t < 0 ) = 0 and & t < 0 ) = 0
x(
•
2
or
x( t = 0 ) = 0 and & t = 0 − ) = 0
x(
The impulse-momentum equation gives:
−
f = 1 = mx( t = 0 ) − mx( t = 0 ) = mx 0
&
&
&
%
And the initial conditions are given by:
x( t
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= 0 ) = x0 = 0
and
k
m
& t =0)=&0 =
x(
x
1
m

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Mechanical Vibrations – Single Degree-of-Freedom systems
V. Rouillard © 2003 - 2013
SDoF systems – General forcing functions – Nonperiodic
•
•
Convolution integral – Impulse response
The solution reduces to:
e −ζωnt
x( t ) = g( t ) =
sin ( ωd t )
mωd
•
g(t) is the impulse response function an represents the response of a viscously damped single degree of
freedom system subjected to a unit impulse.
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Mechanical Vibrations – Single Degree-of-Freedom systems
SDoF systems – General forcing functions – Nonperiodic
•
•
Convolution integral – Impulse response
If the magnitude of the impulse is F instead of unity, the initial
velocity x’0 = F/m and the response becomes:
Fe −ζωnt
x( t ) = %
sin ( ωd t ) = F g( t )
%
mωd
•
If the impulse is applied to a stationary system at an
arbitrary time t = τ the response is
x( t ) = F g( t −τ )
%
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143. 143
Mechanical Vibrations – Single Degree-of-Freedom systems
V. Rouillard © 2003 - 2013
SDoF systems – General forcing functions – Nonperiodic
•
•
Convolution integral – Arbitrary exciting force
•
•
•
The impulse acting at t = τ is given by F(τ )∆τ.
If we consider the arbitrary force to comprise of a series of impulses of varying magnitudes such that at time
τ, the force F(τ) acts on the system for a short period ∆ τ.
At any time t the elapsed time is t - τ
The system response at t due to the impulse is
x( t ) = F g( t −τ ) = F( τ )∆τ g( t −τ )
%
•
The total response at time t is determined by summing
the responses caused by the impulses acting al all
times τ :
x( t ) = ∑ F( τ ) g( t − τ ) ∆τ
Making ∆τ → 0 the response can be exp ressed as :
t
x( t ) = ∫ F( τ ) g( t − τ ) dτ
0
Substituting the impulse response function g( t − τ ) :
t
•
•
1
−ζω ( t −τ )
x( t ) =
sin [ ωd ( t − τ ) ] dτ ¬ Convolution or Duhamel int egral
∫ F( τ )e n
mωd 0
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This solution does not account for initial conditions.
Can be integrated explicitly or numerically depending on F(t)

144. 144
Mechanical Vibrations – Single Degree-of-Freedom systems
V. Rouillard © 2003 - 2013
SDoF systems – General forcing functions – Nonperiodic
•
•
Convolution integral – Arbitrary exciting force
In the case where the excitation is provided by an arbitrary imposed motion of the base, y(t), the relative
displacement is given by:
t
1
z( t ) =
y( −ζω ( t −τ ) sin [ ωd ( t − τ ) ] dτ
∫&& τ )e n
ωd 0
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Example: Step load

145. 145
Mechanical Vibrations – Single Degree-of-Freedom systems
V. Rouillard © 2003 - 2013
SDoF systems – General forcing functions – Nonperiodic
•
When the forcing function is arbitrary and nonperiodic (aperiodic) it cannot be represented with a Fourier
series
•
Alternative methods for determining the response must be used:
•
•
•
•
Representation of the excitation function with a Convolution integral
Using Laplace Transformations
Approximating F(t) with a suitable interpolation method then using a numerical procedure
Numerical integration of the equations of motion.
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146. 146
Mechanical Vibrations – Single Degree-of-Freedom systems
SDoF systems – General forcing functions – Nonperiodic
•
•
•
•
•
•
•
Laplace Transformation
Efficient method to generate solution of linear differential equations
Converts differential equations into algebraic equations to facilitate solving
Can be applied to discontinuous functions
Can be used for any type of excitation including periodic & harmonic
Automatically accounts for initial conditions
The Laplace transform of x(t) is given by:
∞
x( s ) = L x( t ) = ∫ e − st x( t ) dt
0
•
Where s the subsidiary variable and is usually complex.
•
To use Laplace Transform:
1.
Write the equation of motion
2.
Compute or look-up the Laplace transform of each term using known initial conditions
3.
Solve the transformed (algebraic ) equation of motion
4.
Use the inverse Laplace transform to obtain the response (solution)
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147. 147
Mechanical Vibrations – Single Degree-of-Freedom systems
V. Rouillard © 2003 - 2013
SDoF systems – General forcing functions – Nonperiodic
•
When the forcing function is arbitrary and nonperiodic (aperiodic) it cannot be represented with a Fourier
series
•
Alternative methods for determining the response must be used:
•
•
•
•
Representation of the excitation function with a Convolution integral
Using Laplace Transformations
Approximating F(t) with a suitable interpolation method then using a numerical procedure
Numerical integration of the equations of motion.
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Mechanical Vibrations – Single Degree-of-Freedom systems
V. Rouillard © 2003 - 2013
SDoF systems – General forcing functions – Nonperiodic
•
•
•
Numerical Methods (interpolation)
•
Often more practical to represent the digitised data with a series of incremental functions:
•
•
Step functions
Used when the nonperiodic forcing function cannot be described mathematically
It may be possible to “fit” a mathematical approximation (say polynomial) to data then use the convolution
integral
The arbitrary function is
represented by a series of step
functions of varying magnitudes
∆F1, ∆F2, ∆F3… and start times t1,
t2, t3….
•
Note that the polarity of ∆F
changes with the slope of the
function
•
Smaller intervals yield better
accuracy.
•
The approximation is also
improved by choosing the
subsequent start times so that
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F(t) intersects the step at midheight of the step.

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Mechanical Vibrations – Single Degree-of-Freedom systems
V. Rouillard © 2003 - 2013
SDoF systems – General forcing functions – Nonperiodic
•
Numerical Methods (interpolation) - Step functions
•
The system response due to a step excitation ∆Fi for any time interval ti - 1 < t < ti (i = 1, 2, 3 …..j-1) can
be determined from the previous example:


1 j −1
x( t ) =
∆ Fi 1 − e −ζωn ( t −ti )  cos
k i =1


∑
•
( ωd ( t − ti ) ) +

ζωn
sin ( ωd ( t − ti ) )  
ωd

When t = tj the response is:


−ζω ( t − t ) 
ζω
1 j −1
x( t ) = ∑ ∆ Fi 1 − e n j i  cos ωd ( t j − ti ) + n sin ωd ( t j − ti )  
k i =1
ωd



(
04:36:37
)
(
)

150. 150
Mechanical Vibrations – Single Degree-of-Freedom systems
V. Rouillard © 2003 - 2013
SDoF systems – General forcing functions – Nonperiodic
•
•
Numerical Methods (interpolation) - Rectangular impulses
•
The response of the system in any time interval ti - 1 < t < ti is obtained by adding the response caused by Fj
The arbitrary function is represented by a series of rectangular impulses Fi the polarity of which depends on the
polarity of F(t) at that instant.
(applied over ∆tj to the response at t = tj which represent the initial condition:
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151. 151
Mechanical Vibrations – Single Degree-of-Freedom systems
V. Rouillard © 2003 - 2013
SDoF systems – General forcing functions – Nonperiodic
•
•
Numerical Methods (interpolation) – Ramps (linear) approximation
The arbitrary function is represented by a series of linear functions and the response of the system in any time
interval ti - 1 < t < ti is obtained by adding the response caused by the linear (ramp) during a specified interval to
the response due to the previous ramp (initial condition)
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152. 152
Mechanical Vibrations – Single Degree-of-Freedom systems
V. Rouillard © 2003 - 2013
SDoF systems – General forcing functions – Nonperiodic
•
•
(Shock) Response Spectrum
•
The Shock Response Spectrum (SRS) is plotted for a range of natural frequencies usually at fractional
octave intervals.
•
The SRS is used to determine the effect of a particular (shock) excitation function on damped SDoF
systems.
•
Given the nature of real shocks, the SRS is usually computed using numerical means.
Shows the variation in maximum response of a damped SDOF due to a particular transient (shock)
excitation.
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153. 153
•
Mechanical Vibrations – Two Degree-of-Freedom systems
Two degree of freedom systems:
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V. Rouillard © 2003 - 2013

154. 154
•
Mechanical Vibrations – Two Degree-of-Freedom systems
Two degree of freedom systems:
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V. Rouillard © 2003 - 2013

155. 155
Mechanical Vibrations – Two Degree-of-Freedom systems
V. Rouillard © 2003 - 2013
•
•
•
•
No. of DoF of system = No. of mass elements x number of motion types for each mass
•
Under an arbitrary initial disturbance, the system will vibrate freely such that the two normal modes are
superimposed.
•
Under sustained harmonic excitation, the system will vibrate at the excitation frequency. Resonance occurs
if the excitation frequency corresponds to one of the natural frequencies of the system
For each degree of freedom there exists an equation of motion – usually coupled differential equations.
Coupled means that the motion in one coordinate system depends on the other
If harmonic solution is assumed, the equations produce two natural frequencies and the amplitudes of the
two degrees of freedom are related by the natural, principal or normal mode of vibration.
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156. 156
Mechanical Vibrations – Two Degree-of-Freedom systems
•
•
Equations of motion
•
•
V. Rouillard © 2003 - 2013
Motion of system described by position x1(t) and x2(t) of masses m1 and m2
Consider a viscously damped system:
The free-body diagram is used to develop the equations of motion using Newton’s second law
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157. 157
•
Mechanical Vibrations – Two Degree-of-Freedom systems
Equations of motion
m1&& 1 + c1& 1 + k1 x1 − c2 ( & 2 − & 1 ) − k 2 ( x2 − x1 ) = F1
x
x
x x
m2&& 2 + c2 ( & 2 − & 1 ) + k2 ( x2 − x1 ) + c3& 2 + k3 x2 = F2
x
x x
x
or
m1&& 1 + ( c1 + c2 )x1 − c2& 2 + ( k1 + k 2 )x1 − k 2 x2 = F1
x
&
x
m2&& 2 − c2& 1 + ( c2 + c3 )x2 − k2 x1 + ( k2 + k3 )x2 = F2
x
x
&
•
•
The differential equations of motion for mass m 1 and mass m2 are coupled.
The motion of each mass is influenced by the motion of the other.
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158. 158
•
Mechanical Vibrations – Two Degree-of-Freedom systems
V. Rouillard © 2003 - 2013
Equations of motion
m1&& 1 + ( c1 + c2 )x1 − c2& 2 + ( k1 + k 2 )x1 − k 2 x2 = F1
x
&
x
m2&& 2 − c2& 1 + ( c2 + c3 )x2 − k 2 x1 + ( k 2 + k3 )x2 = F2
x
x
&
•
The coupled differential eqns. of motion can be written in matrix form:
r
r
r
&& t ) + [ c ] &r t ) + [ k ] x( t ) = F( t )
x(
[ m] x(
where [ m] , [ c ] and [ k ] are the mass, damping and stiffness matrices respectively and are given by:
 m1 0 
[ m] = 

 0 m2 
− c2 
 c1 + c2
[ c] = 
c2 + c3 
 − c2

 k1 + k2
[ k] = 
 − k2
r
r r r
x(t), & && and F(t) are the displacement, velocity, acceleration and force vectors
x(t), x(t)
− k2 
k 2 + k3 

respectively and are given by :
x
x
 x1( t )  r
 & 1( t )  r
&& 1( t ) 
r
x( t ) = 
x(
x(
 & t)= 
 && t ) = 
 and
x2 ( t ) 
& 2 ( t )
x
&& 2 ( t ) 
x



•
r
 F1( t ) 
F( t ) = 

F2 ( t ) 

Note: the mass, damping and stiffness matrices are all square and symmetric [m] = [m] T and consist of the
mass, damping and stiffness constants.
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159. 159
•
•
Mechanical Vibrations – Two Degree-of-Freedom systems
V. Rouillard © 2003 - 2013
Free vibrations of undamped systems
The eqns. of motion for a free and undamped TDoF system become:
m1&& 1 + ( k1 + k2 )x1 − k2 x2 = 0
x
m2&& 2 − k 2 x1 + ( k 2 + k3 )x2 = 0
x
•
Let us assume that the resulting motion of each mass is harmonic: For simplicity, we will also assume that
the response frequencies and phase will be the same:
x1( t ) = X 1 cos( ωt + φ )
•
and
x2 ( t ) = X 2 cos( ωt + φ )
Substituting the assumed solutions into the eqns. of motion:
{
}
 − m1ω 2 + ( k1 + k 2 ) X 1 − k 2 X 2  cos( ω t + φ ) = 0


{
}
 − k2 X 1 + − m2ω 2 + ( k2 + k3 ) X 2  cos( ωt + φ ) = 0


As these equations must be zero for all values of t, the cosine terms cannot be zero. Therefore:
{ −m1ω 2 + ( k1 + k2 ) } X 1 − k2 X 2 = 0
− k2 X 1 + { − m2ω 2 + ( k 2 + k3 ) } X 2 = 0
•
Represent two simultaneous algebraic equations with a trivial solution when X1 and X2 are both zero – no vibration.
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160. 160
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Mechanical Vibrations – Two Degree-of-Freedom systems
V. Rouillard © 2003 - 2013
Free vibrations of undamped systems
Written in matrix form it can be seen that the solution exists when the determinant of the mass / stiffness
matrix is zero:
 − m1ω 2 + ( k1 + k 2 )

− k2

  X1  = 0
 

− k2
− m2ω 2 + ( k2 + k2 )   X 2 




{
}
{
}
or
2
m1m2ω 4 − { ( k1 + k 2 ) m2 + ( k 2 + k3 ) m1 } ω 2 + ( k1 + k2 ) ( k 2 + k2 ) − k 2 = 0
•
•
The solution to the characteristic equation yields the natural frequencies of the system.
The roots of the characteristic equation are:
2
2 1  ( k + k ) m + ( k 2 + k3 ) m1 
ω1 , ω 2 =  1 2 2

2
m1m2

1
2  2
 ( k1 + k 2 ) ( k 2 + k 3 ) − k 2 

1   ( k1 + k2 ) m2 + ( k 2 + k3 ) m1 
± 
 −4
2 
m1m2




2
•
m1m2
This shows that the homogenous solution is harmonic with natural frequencies
04:36:37



ω1 and ω2