1

Harmonic Response Analysis (Frequency Response Analysis)

The amplitude of response is sought under sinusoidal excitation.

There may be multiple external forces, possibly out of phase with one another, but the loading is considered harmonic only when all loads are of the same frequency.

• •

F1=Asin5tF2=Bsin(5t-30°)

d1(t) d2(t)

Loads of different frequency require dynamic response analysis as opposed to harmonic analysis.

When first applied, harmonic loading does not invoke harmonic response.

An initial transient exists that decays to zero in time because of damping.

What remains is the steady-state response, which is thesubject of harmonic response analysis.

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Harmonic Response of sdof System

Harmonic load r(t) on the system

Equation of motion:

The steady-state displacement response:

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Parameters Affecting Response Amplitude

β here is the frequency ratio, ratio of the load frequency Ω to the natural frequency ω of the system.

It is not just the value of the load frequency but its value relative to the system’s natural frequency that governs how the system responds.

α is the phase angle, i. e., how much response lags the loads.

For small damping ratio ξ , α is either very small (for β <1) or close to 180° (for β >1).

F0 /k is the static displacement, solution of the governing equationwhen velocity and acceleration are zero.

Steady-state response amplitude is proportional to F0 /k.u

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Amplitude Response Spectra for a sdof System

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Characteristics of Amplitude Spectrum For ξ < 0.55 and when Ω is near the natural frequency ω

(frequency ratio near unity),→ response amplitude has a peak

The smaller the damping, the larger the peak.

Resonance is when Ω =ω. Then, damping limits the amplitude which would become infinite for zero damping.

As β (i.e., ω) becomes large, the displacement approaches zero:

The force varies so rapidly that the mass has no time to respond.

Multi dof structures have similar behaviour.

There are as many amplitude peaks as the number of dof.

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Harmonic Response of Multi-dof Systems: Modal Method

Step 1: Solve the eigenvalue problem first and obtain atruncated set of m undamped modal (natural) frequencies and mode shapes

Step 2: Assuming modal or proportional damping, solve for the modal coordinates zi from

Step 3: To find the actual displacements, apply mode superposition by summing the modal coordinates weighted by mode shapes:

ii D ,ω

iTi D=φ where

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Solution of the Modal Equations

Solution of the modal equations resembles the solution of displacement in a sdof system that we’ve seen before.

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Example: Modal Method

F=Bsin5t

d(t)

Given: A cantilever beam with harmonic tip loading.

The parameters are such that

Model it with a single finite element and compute the tip displacement d(t) using the modal method with a truncated and a full set of modes.

Use a 10 % modal damping for each mode. (That is, ξ=0.10)

1 ,1420 4 == ALEIAL ρρ

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Solution Step 1: Free Vibration Analysis

We previously obtained the following modal frequencies and mode shapes from undamped free vibration analysis:

However, these mode shapes are not normalized with respect to the mass matrix M

Then, using the given data

=

=LL /62.7

1 ,

/38.11

21 φφ

( ) 0.53 , 102.9420/ 9.102 2211 === φφρφφ MM TT AL

81.34 ,533.3 Hence, 21 == ωω

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Mass-Normalized Modes and Modal Forces

To mass-normalize the modes, we divide the mode shapes by (102.9)1/2 and (53.0)1/2, respectively. Hence, the normalized modes are

Step 2: The load vector is

The modal forces:

=

=LL /05.1

137.0 ,

/136.0099.0

21 DD

=0

5sin tBR

tBp

tBp

T

T

5sin137.0

5sin099.0

22

11

==

==

RD

RD

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Modal Equations and Phase AnglesModal equations:

Solution with β1= Ω/ω1=5/3.533=1.415 and β2= Ω/ω2=0.144:

Phase angles:

tBzzz

tBzzz

pzzz

5sin137.0121296.6 and

5sin099.05.12707.0

2

222

111

11211111

=++

=++⇒

=++ ωωξ

7.1144.01

144.0*2.0Arctan

1641808.15415.11

415.1*2.0Arctan

22

21

=

−=

=+−=

−=

α

α

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Solution of Modal EquationsThen,

Similarly,

( ) ( )

)1645sin(0076.0)(

)1645sin(415.1*2.0415.11

5.12099.0)(

1

2221

−=⇒

−+−

=

tBtz

tBtz

( ) ( )

1) mode of % 2 thanless (Amplitude

)7.15sin(00012.0)(

)7.15sin(144.0*2.0144.01

1212137.0)(

2

2222

−=⇒

−+−

=

tBtz

tBtz

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Solution Step 3: Mode SuperpositionTruncated set: One mode only.

Full set: Both modes.

−=⇒

−==

−

−

Lx

xtB

LtBtz

3

4

11

1003.1

1052.7)1645sin(

136.0

099.0)1645sin(0076.0)(

D

DD

1032.1

1064.1)7.15sin(

1003.1

1052.7)1645sin(

)()(

4

5

3

4

2211

−+

−=⇒

+=

−

−

−

−

Lx

xtB

Lx

xtB

tztz

D

DDD

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Observations on the Solution Excitation frequency is a little larger than the first (fundamental)

frequency but quite smaller than the second one:⇒ second mode is excited only a little as evidenced by

the amplitudes in D due to z2(t).

For the same reason, phase angle for the mode two component of the response is very small.

Therefore the truncated mode superposition is adequate.

The two components in the second superposition reach theirpeaks at different times because the phase angles aredifferent.

⇒ To find the peak of the physical displacement, must scan numerical values in one whole cycle.

Multi dof structures have similar behaviour.

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Harmonic Response of Multi-dof Systems: Direct Method

Based on use of complex numbers which simplifies harmonic response calculations.

Equation of motion (EOM) for a general system:

A harmonic forcing R(t)=R0sin(Ωt-θ) can be expressed as

We can drop the “Imaginary” symbol and treat the forcing as complex.

The solution then has the same form:

)(tRKDDCDM =++

1 and amplitudecomplex a : , Im −=Ω ie ti RR

tie Ω= DD

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Harmonic Response of Multi-dof Systems: Direct Method

Substituting these into the EOM,

Simplifying,

Direct method solves for from the above equation.

If Ω is near a modal frequency of the system, the coefficientmatrix of above is

singular if there is no damping,ill-conditioned if there is.

If the solution is needed for a number of frequencies, it is costly.

( ) ( ) ΩΩΩΩ =+Ω+Ω iiii eeeiei RDKDCDM 2

[ ] RDMCK =Ω−Ω+ 2i

D

D

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Direct Method for Multi-dof Systems: Transformed EquationThe disadvantages of directly solving the equation are overcome

if we transform the equation using mode superposition:

Substituting into the simplified equation and premultiplying by the modal matrix φT, we get, upon observing the orthogonality conditions,

where Cφ is diagonal for proportional and modal damping.

Solving for the complex amplitude of the modal coordinate vector and this time premultiplying it with the modal matrix,

Easy to invert the above diagonal matrix even near resonance.

nmzm

iii <∑ ==

= ;matrix modal : ,

1ΦΦzD φ