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3
Response to Harmonic and
Periodic Excitations

PREVIEW

The response of SDF systems to harmonic excitation is a classical topic in structural dy-
namics, not only because such excitations are encountered in engineering systems (e.g.,
force due to unbalanced rotating machinery), but also because understanding the response
of structures to harmonic excitation provides insight into how the system will respond to
other types of forces. Furthermore, the theory of forced harmonic vibration has several
useful applications in earthquake engineering.
In Part A of this chapter the basic results for response of SDF systems to harmonic
force are presented, including the concepts of steady-state response, frequency-response
curve, and resonance. Applications of these results to experimental evaluation of the
natural vibration frequency and damping ratio of a structure, to isolation of vibration,
and to the design of vibration-measuring instruments is the subject of Part B; also in-
cluded is the concept of equivalent viscous damping. This concept is used in Part C to
obtain approximate solutions for the response of systems with rate-independent damp-
ing or Coulomb friction; these results are then shown to be good approximations to the
“exact” solutions. A procedure to determine the response of SDF systems to periodic
excitation is presented in Part D. A Fourier series representation of the excitation, com-
bined with the results for response to harmonic excitations, provides the desired proce-
dure.

65
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66 Response to Harmonic and Periodic Excitations Chap. 3

PART A: VISCOUSLY DAMPED SYSTEMS: BASIC RESULTS

3.1 HARMONIC VIBRATION OF UNDAMPED SYSTEMS

A harmonic force is p(t) = po sin ωt or po cos ωt, where po is the amplitude or maximum
value of the force and its frequency ω is called the exciting frequency or forcing frequency;
T = 2π/ω is the exciting period or forcing period (Fig. 3.2.1a). The response of SDF
systems to a sinusoidal force will be presented in some detail, along with only brief com-
ments on the response to a cosine force because the concepts involved are similar in the
two cases.
Setting p(t) = po sin ωt in Eq. (1.5.2) gives the differential equation governing the
forced harmonic vibration of the system, which for systems without damping specializes to
m ü + ku = po sin ωt (3.1.1)
This equation is to be solved for the displacement or deformation u(t) subject to the initial
conditions
u = u(0) u̇ = u̇(0) (3.1.2)
where u(0) and u̇(0) are the displacement and velocity at the time instant the force is
applied. The particular solution to this differential equation is (see Derivation 3.1)
po 1
u p (t) = sin ωt ω = ωn (3.1.3)
k 1 − (ω/ωn )2
The complementary solution of Eq. (3.1.1) is the free vibration response determined in
Eq. (d) of Derivation 2.1:
u c (t) = A cos ωn t + B sin ωn t (3.1.4)
and the complete solution is the sum of the complementary and particular solutions
po 1
u(t) = A cos ωn t + B sin ωn t + sin ωt (3.1.5)
k 1 − (ω/ωn )2
The constants A and B are determined by imposing the initial conditions, Eq. (3.1.2), to
obtain the final result (see Derivation 3.1)

 
u̇(0) po ω/ωn
u(t) = u(0) cos ωn t + − sin ωn t
ωn k 1 − (ω/ωn )2
  
transient
po 1
+ sin ωt (3.1.6a)
k 1 − (ω/ωn )2
  
steady state
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Sec. 3.1 Harmonic Vibration of Undamped Systems 67

Amplitude, po

(a)

Period, T = 2π/ω

2 Total response
Steady-state response
1
u(t) / (ust)o

(b) 0

-1

-2

0 0.5 1 1.5 2
t/T
Figure 3.1.1 (a) Harmonic force; (b) response of undamped system to harmonic force; ω/ωn = 0.2,
u(0) = 0, and u̇(0) = ωn po /k.

Equation (3.1.6a) has been plotted for ω/ωn = 0.2, u(0) = 0, and u̇(0) = ωn po /k as
the solid line in Fig. 3.1.1. The sin ωt term in this equation is the particular solution of
Eq. (3.1.3) and is shown by the dashed line.
Equation (3.1.6a) and Fig. 3.1.1 show that u(t) contains two distinct vibration com-
ponents: (1) the sin ωt term, giving an oscillation at the forcing or exciting frequency;
and (2) the sin ωn t and cos ωn t terms, giving an oscillation at the natural frequency of the
system. The first of these is the forced vibration or steady-state vibration, for it is present
because of the applied force no matter what the initial conditions. The latter is the tran-
sient vibration, which depends on the initial displacement and velocity. It exists even if
u(0) = u̇(0) = 0, in which case Eq. (3.1.6a) specializes to

po 1 ω
u(t) = sin ωt − sin ωn t (3.1.6b)
k 1 − (ω/ωn )2 ωn
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68 Response to Harmonic and Periodic Excitations Chap. 3

The transient component is shown as the difference between the solid and dashed lines in
Fig. 3.1.1, where it is seen to continue forever. This is only an academic point because
the damping inevitably present in real systems makes the free vibration decay with time
(Section 3.2). It is for this reason that this component is called transient vibration.
The steady-state dynamic response, a sinusoidal oscillation at the forcing frequency,
may be expressed as
 
1
u(t) = (u st )o sin ωt (3.1.7)
1 − (ω/ωn )2
Ignoring the dynamic effects signified by the acceleration term in Eq. (3.1.1) gives the
static deformation (indicated by the subscript “st”) at each instant:
po
u st (t) = sin ωt (3.1.8)
k
The maximum value of the static deformation is
po
(u st )o = (3.1.9)
k
which may be interpreted as the static deformation due to the amplitude po of the force; for
brevity we will refer to (u st )o as the static deformation. The factor in brackets in Eq. (3.1.7)
has been plotted in Fig. 3.1.2 against ω/ωn , the ratio of the forcing frequency to the natural
frequency. For ω/ωn < 1 or ω < ωn this factor is positive, indicating that u(t) and p(t)
have the same algebraic sign (i.e., when the force in Fig. 1.2.1a acts to the right, the system
would also be displaced to the right). The displacement is said to be in phase with the
applied force. For ω/ωn > 1 or ω > ωn this factor is negative, indicating that u(t) and

2
[ 1 - (ω/ωn)2 ] -1

-1

-2

-3

-4

-5
0 1 2 3
Frequency ratio ω / ωn Figure 3.1.2
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Sec. 3.1 Harmonic Vibration of Undamped Systems 69

p(t) have opposing algebraic signs (i.e., when the force acts to the right, the system would
be displaced to the left). The displacement is said to be out of phase relative to the applied
force.
To describe this notion of phase mathematically, Eq. (3.1.7) is rewritten in terms of
the amplitude u o of the vibratory displacement u(t) and phase angle φ:

u(t) = u o sin(ωt − φ) = (u st )o Rd sin(ωt − φ) (3.1.10)

where

uo 1 0◦ ω < ωn
Rd = = and φ= (3.1.11)
(u st )o | 1 − (ω/ωn )2 | 180◦ ω > ωn

For ω < ωn , φ = 0◦ , implying that the displacement varies as sin ωt, in phase with the
applied force. For ω > ωn , φ = 180◦ , indicating that the displacement varies as −sin ωt,
out of phase relative to the force. This phase angle is shown in Fig. 3.1.3 as a function of
the frequency ratio ω/ωn .

5
Deformation response factor Rd = uo / (ust)o

0
180o
Phase angle φ

90o

0o Figure 3.1.3 Deformation response factor

0 1 2 3 and phase angle for an undamped system
Frequency ratio ω / ωn excited by harmonic force.
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70 Response to Harmonic and Periodic Excitations Chap. 3

The deformation (or displacement) response factor Rd is the ratio of the amplitude
u o of the dynamic (or vibratory) deformation to the static deformation (u st )o . Figure 3.1.3,
which shows Eq. (3.1.11a) for Rd plotted as a function of the frequency ratio ω/ωn , permits
several observations: If ω/ωn is small (i.e., the force is “slowly varying”), Rd is only
slightly larger than 1 and the amplitude√of the dynamic deformation is √essentially the same
as the static deformation. If ω/ωn > 2 (i.e., ω is higher than ωn 2), Rd < 1 and the
dynamic√deformation amplitude is less than the static deformation. As ω/ωn increases
beyond 2, Rd becomes smaller and approaches zero as ω/ωn → ∞, implying that the
vibratory deformation due to a “rapidly varying” force is very small. If ω/ωn is close to
1 (i.e., ω is close to ωn ), Rd is many times larger than 1, implying that the deformation
amplitude is much larger than the static deformation.
The resonant frequency is defined as the forcing frequency at which Rd is maxi-
mum. For an undamped system the resonant frequency is ωn and Rd is unbounded at this
frequency. The vibratory deformation does not become unbounded immediately, however,
but gradually, as we demonstrate next.
If ω = ωn , the solution given by Eq. (3.1.6b) is no longer valid. In this case the
choice of the function C sin ωt for a particular solution fails because it is also a part of the
complementary solution. The particular solution now is
po
u p (t) = − ωn t cos ωn t ω = ωn (3.1.12)
2k
and the complete solution for at-rest initial conditions, u(0) = u̇(0) = 0, is (see Deriva-
tion 3.2)
1 po
u(t) = − (ωn t cos ωn t − sin ωn t) (3.1.13a)
2 k
or

u(t) 1 2π t 2π t 2π t
=− cos − sin (3.1.13b)
(u st )o 2 Tn Tn Tn
This result is plotted in Fig. 3.1.4, which shows that the time taken to complete one
cycle of vibration is Tn . The local maxima of u(t), which occur at t = j (Tn /2),
are π( j − 1/2)(u st )o — j = 1, 2, 3, . . .—and the local minima, which occur at t = j Tn ,
are −π j (u st )o — j = 1, 2, 3, . . . . In each cycle the deformation amplitude increases by
πpo
| u j+1 | − | u j |= (u st )o [π ( j + 1) − π j] =
k
The deformation amplitude grows indefinitely, but it becomes infinite only after an in-
finitely long time.
This is an academic result and should be interpreted appropriately for real structures.
As the deformation continues to increase, at some point in time the system would fail if
it is brittle. On the other hand, the system would yield if it is ductile, its stiffness would
decrease, and its “natural frequency” would no longer be equal to the forcing frequency,
and Eq. (3.1.13) or Fig. 3.1.4 would no longer be valid.
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Sec. 3.1 Harmonic Vibration of Undamped Systems 71

30

20 Envelope curve π

• •
10
u(t) / (ust)o

-10
uj
π

• •
-20
uj+1
-30
0 2 4 6 8 10
t / Tn

Figure 3.1.4 Response of undamped system to sinusoidal force of frequency ω = ωn ; u(0) =

u̇(0) = 0.

Derivation 3.1
The particular solution of Eq. (3.1.1), a linear second-order differential equation, is of the form

u p (t) = C sin ωt (a)

Differentiating this twice gives

ü p (t) = −ω2 C sin ωt (b)

Substituting Eqs. (a) and (b) in the differential equation (3.1.1) leads to a solution for C:

po 1
C= (c)
k 1 − (ω/ωn )2
which is combined with Eq. (a) to obtain the particular solution presented in Eq. (3.1.3).
To determine the constants A and B in Eq. (3.1.5), it is differentiated

po ω
u̇(t) = −ωn A sin ωn t + ωn B cos ωn t + cos ωt (d)
k 1 − (ω/ωn )2
Evaluating Eqs. (3.1.5) and (d) at t = 0 gives

po ω
u(0) = A u̇(0) = ωn B + (e)
k 1 − (ω/ωn )2
These two equations give
u̇(0) po ω/ωn
A = u(0) B= − (f)
ωn k 1 − (ω/ωn )2
which are substituted in Eq. (3.1.5) to obtain Eq. (3.1.6a).
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72 Response to Harmonic and Periodic Excitations Chap. 3

Derivation 3.2
If ω = ωn , the particular solution of Eq. (3.1.1) is of the form
u p (t) = Ct cos ωn t (a)
Substituting Eq. (a) in Eq. (3.1.1) and solving for C yields
po
C = − ωn (b)
2k
which is combined with Eq. (a) to obtain the particular solution, Eq. (3.1.12).
Thus the complete solution is
po
u(t) = A cos ωn t + B sin ωn t − ωn t cos ωn t (c)
2k
and the corresponding velocity is
po po 2
u̇(t) = −ωn A sin ωn t + ωn B cos ωn t − ωn cos ωn t + ω t sin ωn t (d)
2k 2k n
Evaluating Eqs. (c) and (d) at t = 0 and solving the resulting algebraic equations gives
u̇(0) po
A = u(0) B= +
ωn 2k
Specializing for at-rest initial conditions gives
po
A=0 B=
2k
which are substituted in Eq. (c) to obtain Eq. (3.1.13a) .

3.2 HARMONIC VIBRATION WITH VISCOUS DAMPING

3.2.1 Steady-State and Transient Responses

Including viscous damping the differential equation governing the response of SDF
systems to harmonic force is
m ü + cu̇ + ku = po sin ωt (3.2.1)
This equation is to be solved subject to the initial conditions
u = u(0) u̇ = u̇(0) (3.2.2)
The particular solution of this differential equation is (from Derivation 3.3)
u p (t) = C sin ωt + D cos ωt (3.2.3)
where
po 1 − (ω/ωn )2
C=
k [1 − (ω/ωn )2 ]2 + [2ζ (ω/ωn )]2
(3.2.4)
po −2ζ ω/ωn
D=
k [1 − (ω/ωn )2 ]2 + [2ζ (ω/ωn )]2
The complementary solution of Eq. (3.2.1) is the free vibration response given by Eq. (f)
of Derivation 2.2:
u c (t) = e−ζ ωn t (A cos ω D t + B sin ω D t)
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Sec. 3.2 Harmonic Vibration with Viscous Damping 73


where ω D = ωn 1 − ζ 2 . The complete solution of Eq. (3.2.1) is
u(t) = e−ζ ωn t (A cos ω D t + B sin ω D t) + C sin ωt +
 D cos ωt (3.2.5)
   
transient steady state

where the constants A and B can be determined by standard procedures (e.g., see Deriva-
tion 3.1) in terms of the initial displacement u(0) and initial velocity u̇(0). As noted in
Section 3.1, u(t) contains two distinct vibration components: forced vibration or steady-
state vibration, and transient vibration.
Equation (3.2.5) is plotted in Fig. 3.2.1 for ω/ωn = 0.2, ζ = 0.05, u(0) = 0, and
u̇(0) = ωn po /k; the total response is shown by the solid line and the steady-state response
by the dashed line. The difference between the two is the transient response, which decays
exponentially with time at a rate depending on ω/ωn and ζ ; compare this with no decay
for undamped systems in Fig. 3.1.1. After awhile, essentially the forced response remains,
and we therefore call it steady-state response and focus on it for the rest of this chapter
(after Section 3.2.2). It should be recognized, however, that the largest deformation peak
may occur before the system has reached steady state; see Fig. 3.2.1.

2 Total response
Steady-state response
1
u(t) / (ust)o

-1

-2

0 0.5 1 1.5 2
t/T

Figure 3.2.1 Response of damped system to harmonic force; ω/ωn = 0.2, ζ = 0.05,
u(0) = 0, and u̇(0) = ωn po /k.

Derivation 3.3
Dividing Eq. (3.2.1) by m gives
po
ü + 2ζ ωn u̇ + ωn2 u = sin ωt (a)
m
The particular solution of Eq. (a) is of the form
u p (t) = C sin ωt + D cos ωt (b)
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74 Response to Harmonic and Periodic Excitations Chap. 3

Substituting Eq. (b) and its first and second derivatives in Eq. (a) gives
po
[(ωn2 − ω2 )C − 2ζ ωn ωD] sin ωt + [2ζ ωn ωC + (ωn2 − ω2 )D] cos ωt = sin ωt (c)
m
For Eq. (c) to be valid for all t, the coefficients of the sine and cosine terms on the two sides
of the equation must be equal. This requirement gives two equations in C and D which, after
dividing by ωn2 and using the relation k = ωn2 m, become

  2 
ω ω po
1− C − 2ζ D= (d)
ωn ωn k

   2
ω ω
2ζ C + 1− D=0 (e)
ωn ωn

Solving the two algebraic equations (d) and (e) leads to Eq. (3.2.4).

3.2.2 Response for ω = ωn

In this section we examine the role of damping in the rate at which steady-state response
is attained and in limiting the magnitude of this response when the forcing frequency is
the same as the natural frequency. For ω = ωn , Eq. (3.2.4) gives C = 0 and D =
−(u st )o /2ζ ; for ω = ωn and zero initial conditions, the constants A and B in Eq. (3.2.5)
can be determined: A = (u st )o /2ζ and B = (u st )o /2 1 − ζ 2 . With these solutions for A,
B, C, and D, Eq. (3.2.5) becomes
  
1 ζ
u(t) = (u st )o e−ζ ωn t cos ω D t +  sin ω D t − cos ωn t (3.2.6)
2ζ 1 − ζ2
This result is plotted in Fig. 3.2.2 for a system with ζ = 0.05. A comparison of Fig. 3.2.2
for damped systems and Fig. 3.1.4 for undamped systems shows that damping lowers each
peak and limits the response to the bounded value:
(u st )o
uo = (3.2.7)
2ζ
For lightly damped systems the sinusoidal term in Eq. (3.2.6) is small and ω D
ωn ; thus
1
u(t) (u st )o (e−ζ ωn t − 1) cos ωn t (3.2.8)
2ζ
  
envelope function

The deformation varies with time as a cosine function, with its amplitude increasing with
time according to the envelope function shown by dashed lines in Fig. 3.2.2.
The amplitude of the steady-state deformation of a system to a harmonic force with
ω = ωn and the rate at which steady state is attained is strongly influenced by damping.
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Sec. 3.2 Harmonic Vibration with Viscous Damping 75

20
Envelope curves Steady-state amplitude

10 1/2ζ
u(t) / (ust)o

-10 1/2ζ

-20

0 2 4 6 8 10
t / Tn

Figure 3.2.2 Response of damped system with ζ = 0.05 to sinusoidal force of frequency ω = ωn ;
u(0) = u̇(0) = 0.
30
ζ = 0.01
20

10 ζ = 0.05
u(t) / (ust)o

ζ = 0.1
0

-10

-20

-30

0 2 4 6 8 10
t / Tn

Figure 3.2.3 Response of three systems—ζ = 0.01, 0.05, and 0.1—to sinusoidal force
of frequency ω = ωn ; u(0) = u̇(0) = 0.

The important influence of the damping ratio on the amplitude is seen in Fig. 3.2.3, where
Eq. (3.2.6) is plotted for three damping ratios: ζ = 0.01, 0.05, and 0.1. To study how the
response builds up to steady state, we examine the peak u j after j cycles of vibration. A
relation between u j and j can be written by substituting t = j Tn in Eq. (3.2.8), setting
cos ωn t = 1, and using Eq. (3.2.7) to obtain
| uj |
= 1 − e−2π ζ j (3.2.9)
uo
This relation is plotted in Fig. 3.2.4 for ζ = 0.01, 0.02, 0.05, 0.10, and 0.20. The discrete
points are joined by curves to identify trends, but only integer values of j are meaningful.
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76 Response to Harmonic and Periodic Excitations Chap. 3

0.2
1
••••••••• • • • • • •• • • • • ••
• ••0.1•••0.05 •• • • •
• • • • •
0.8 • • 0.02 •
• • • •• ζ = 0.01
• • ••
| uj | / uo 0.6 • • •
• •
• • • ••
0.4 • ••
•
• • •••
0.2
•
••
•
0 •
0 10 20 30 40 50
j = number of cycles

Figure 3.2.4 Variation of response amplitude with number of cycles of harmonic force
with frequency ω = ωn .

The lighter the damping, the larger is the number of cycles required to reach a certain
percentage of u o , the steady-state amplitude. For example, the number of cycles required
to reach 95% of u o is 48 for ζ = 0.01, 24 for ζ = 0.02, 10 for ζ = 0.05, 5 for ζ = 0.10,
and 2 for ζ = 0.20.

3.2.3 Maximum Deformation and Phase Lag

The steady-state deformation of the system due to harmonic force, described by Eqs. (3.2.3)
and (3.2.4), can be rewritten as
u(t) = u o sin(ωt − φ) = (u st )o Rd sin(ωt − φ) (3.2.10)
√
where u o = C 2 + D 2 and φ = tan−1 (−D/C). Substituting for C and D gives

uo 1
Rd = = (3.2.11)
(u st )o [1 − (ω/ωn ) ]2 + [2ζ (ω/ωn )]2
2

2ζ (ω/ωn )
φ = tan−1 (3.2.12)
1 − (ω/ωn )2
Equation (3.2.10) is plotted in Fig. 3.2.5 for three values of ω/ωn and a fixed value of
ζ = 0.20. The values of Rd and φ computed from Eqs. (3.2.11) and (3.2.12) are identified.
Also shown by dashed lines is the static deformation [Eq. (3.1.8)] due to p(t), which varies
with time just as does the applied force, except for the constant k. The steady-state motion
is seen to occur at the forcing period T = 2π/ω, but with a time lag = φ/2π; φ is called
the phase angle or phase lag.
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Sec. 3.2 Harmonic Vibration with Viscous Damping 77

3 (a) ω / ωn = 0.5

2 Dynamic: u(t) / (ust)o

Rd = 1.29
Static: ust(t) / (ust)o
1

-1 T
φ/2π = 0.041 T
-2

-3
3 (b) ω / ωn = 1 Rd = 2.5

-1
φ/2π = 0.25
-2

-3
3 (c) ω / ωn = 2

1
Rd = 0.32
0

-1
φ/2π = 0.46
-2

-3
0 1 2 3
t/T

Figure 3.2.5 Steady-state response of damped systems (ζ = 0.2) to sinusoidal force for three values
of the frequency ratio: (a) ω/ωn = 0.5, (b) ω/ωn = 1, (c) ω/ωn = 2.
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78 Response to Harmonic and Periodic Excitations Chap. 3

5
ζ = 0.01
0.1

4
Deformation response factor Rd = uo / (ust)o

0.2

1
0.7
ζ=1

0
180°
ζ = 0.01 0.2
Phase angle φ

0.1 0.7
ζ=1
90°

0°
0 1 2 3
Frequency ratio ω / ωn

Figure 3.2.6 Deformation response factor and phase angle for a damped system excited
by harmonic force.

A plot of the amplitude of a response quantity against the excitation frequency is

called a frequency-response curve. Such a plot for deformation u is given by Fig. 3.2.6,
wherein the deformation response factor Rd [from Eq. (3.2.11)] is plotted as a function
of ω/ωn for a few values of ζ ; all the curves are below the ζ = 0 curve in Fig. 3.1.3.
Damping reduces Rd and hence the deformation amplitude at all excitation frequencies.
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Sec. 3.2 Harmonic Vibration with Viscous Damping 79

The magnitude of this reduction is strongly dependent on the excitation frequency and is
examined next for three regions of the excitation-frequency scale:
1. If the frequency ratio ω/ωn  1 (i.e., T  Tn , implying that the force is “slowly
varying”), Rd is only slightly larger than 1 and is essentially independent of damping. Thus
po
u o (u st )o = (3.2.13)
k
This result implies that the amplitude of dynamic response is essentially the same as the
static deformation and is controlled by the stiffness of the system.
2. If ω/ωn  1 (i.e., T  T n, implying that the force is “rapidly varying”), Rd
tends to zero as ω/ωn increases and is essentially unaffected by damping. For large values
of ω/ωn , the (ω/ωn )4 term is dominant in Eq. (3.2.11), which can be approximated by
ωn2 po
u o (u st )o = (3.2.14)
ω2 mω2
This result implies that the response is controlled by the mass of the system.
3. If ω/ωn 1 (i.e., the forcing frequency is close to the natural frequency of the
system), Rd is very sensitive to damping and, for the smaller damping values, Rd can be
several times larger than 1, implying that the amplitude of dynamic response can be much
larger than the static deformation. If ω = ωn , Eq. (3.2.11) gives
(u st )o po
uo = = (3.2.15)
2ζ cωn
This result implies that the response is controlled by the damping of the system.
The phase angle φ, which defines the time by which the response lags behind the
force, varies with ω/ωn as shown in Fig. 3.2.6. It is examined next for the same three
regions of the excitation-frequency scale:
1. If ω/ωn  1 (i.e., the force is “slowly varying”), φ is close to 0◦ and the dis-
placement is essentially in phase with the applied force, as in Fig. 3.2.5a. When the force
in Fig. 1.2.1a acts to the right, the system would also be displaced to the right.
2. If ω/ωn  1 (i.e., the force is “rapidly varying”), φ is close to 180◦ and the
displacement is essentially out of phase relative to the applied force, as in Fig. 3.2.5c.
When the force acts to the right, the system would be displaced to the left.
3. If ω/ωn = 1 (i.e., the forcing frequency is equal to the natural frequency), φ = 90◦
for all values of ζ , and the displacement attains its peaks when the force passes through
zeros, as in Fig. 3.2.5b.
Example 3.1
The displacement amplitude u o of an SDF system due to harmonic force is known for two
excitation frequencies. At ω = ωn , u o = 5 in.; at ω = 5ωn , u o = 0.02 in. Estimate the
damping ratio of the system.
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80 Response to Harmonic and Periodic Excitations Chap. 3

Solution At ω = ωn , from Eq. (3.2.15),

1
u o = (u st )o =5 (a)
2ζ
At ω = 5ωn , from Eq. (3.2.14),
1 (u st )o
u o (u st )o = = 0.02 (b)
(ω/ωn )2 25
From Eq. (b), (u st )o = 0.5 in. Substituting in Eq. (a) gives ζ = 0.05.

3.2.4 Dynamic Response Factors

In this section we introduce deformation (or displacement), velocity, and acceleration

response factors that are dimensionless and define the amplitude of these three response
quantities. The steady-state displacement of Eq. (3.2.10) is repeated for convenience:
u(t)
= Rd sin(ωt − φ) (3.2.16)
po /k
where the deformation response factor Rd is the ratio of the amplitude u o of the dynamic
(or vibratory) deformation to the static deformation (u st )o ; see Eq. (3.2.11).
Differentiating Eq. (3.2.16) gives an equation for the velocity response:
u̇(t)
√ = Rv cos(ωt − φ) (3.2.17)
po / km
where the velocity response factor Rv is related to Rd by
ω
Rv = Rd (3.2.18)
ωn
Differentiating Eq. (3.2.17) gives an equation for the acceleration response:
ü(t)
= −Ra sin(ωt − φ) (3.2.19)
po /m
where the acceleration response factor Ra is related to Rd by
 2
ω
Ra = Rd (3.2.20)
ωn
Observe from Eq. (3.2.19) that Ra is the ratio of the amplitude of the vibratory acceleration
to the acceleration due to force po acting on the mass.
The dynamic response factors Rd , Rv , and Ra are plotted as functions of ω/ωn in
Fig. 3.2.7. The plots of Rv and Ra are new, but the one for Rd is the same as that in
Fig. 3.2.6. The deformation response factor Rd is unity at ω/ωn = 0, peaks at ω/ωn < 1,
and approaches zero as ω/ωn → ∞. The velocity response factor Rv is zero at ω/ωn = 0,
peaks at ω/ωn = 1, and approaches zero as ω/ωn → ∞. The acceleration response factor
√ at ω/ωn = 0, peaks at ω/ωn > 1, and approaches unity as ω/ωn → ∞. For
Ra is zero
ζ > 1/ 2 no peak occurs for Rd and Ra .
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Sec. 3.2 Harmonic Vibration with Viscous Damping 81

5
ζ = 0.01
4 0.1

3
Rd (a)
0.2
2
0.7
1
ζ=1
0

5
ζ = 0.01
4 0.1

3
Rv 0.2 (b)
2
0.7
1
ζ=1
0

5
ζ = 0.01
4 0.1

3
Ra 0.2 (c)
2 0.7

1
ζ=1
0
0 1 2 3
Frequency ratio ω / ωn

Figure 3.2.7 Deformation, velocity, and acceleration response factors for a damped
system excited by harmonic force.

The simple relations among the dynamic response factors

Ra ω
= Rv = Rd (3.2.21)
ω/ωn ωn
make it possible to present all three factors in a single graph. The Rv –ω/ωn data in the
linear plot of Fig. 3.2.7b are replotted as shown in Fig. 3.2.8 on four-way logarithmic
graph paper. The Rd and Ra values can be read from the diagonally oriented logarithmic
scales that are different from the vertical scale for Rv . This compact presentation makes it
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82 Response to Harmonic and Periodic Excitations Chap. 3

10
ζ = 0.01

50

50

a
R
or
ζ = 0.1

D
5

ct
ef

fa
or

se
m
ζ = 0.2

on
at

10

10
Velocity response factor Rv

io

sp
n

re
re

n
sp

io
5

5
on

at
er
se

el
fa

cc
ct

A
or
Rd
1

0.

5
0.
5
ζ = 0.7
0.5
ζ=1
0.

1
0.
1
0.

05
05

0.
0.1
0.1 0.05 1 5 10
Frequency ratio ω / ωn

Figure 3.2.8 Four-way logarithmic plot of deformation, velocity, and acceleration

response factors for a damped system excited by harmonic force.

possible to replace the three linear plots of Fig. 3.2.7 by a single plot. The concepts under-
lying construction of this four-way logarithmic graph paper are presented in Appendix 3.

3.2.5 Resonant Frequencies and Resonant Responses

A resonant frequency is defined as the forcing frequency at which the largest response
amplitude occurs. Figure 3.2.7 shows that the peaks in the frequency-response curves
for displacement, velocity, and acceleration occur at slightly different frequencies. These
resonant frequencies can be determined
√ by setting to zero the first derivative of Rd , Rv , and
Ra with respect to ω/ωn ; for ζ < 1/ 2 they are:

Displacement resonant frequency: ωn 1 − 2ζ 2
Velocity resonant frequency: ωn

Acceleration resonant frequency: ωn ÷ 1 − 2ζ 2

For an undamped system the three resonant frequencies are identical and equal to the
natural frequency ωn of the system. Intuition might suggest that the resonant frequencies
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Sec. 3.2 Harmonic Vibration with Viscous Damping 83


for a damped system should be at its natural frequency ω D = ωn 1 − ζ 2 , but this does not
happen. The difference is small, however; for the degree of damping usually embodied in
structures, typically well below 20%, the differences among the three resonant frequencies
and the natural frequency are negligible.
The three dynamic response factors at their respective resonant frequencies are
1 1 1
Rd =  Rv = Ra =  (3.2.22)
2ζ 1 − ζ2 2ζ 2ζ 1 − ζ 2

3.2.6 Half-Power Bandwidth

An important property of the frequency response curve for Rd is shown in Fig. 3.2.9, where
the half-power bandwidth is defined. If ωa and ωb are the forcing frequencies on either side

5
Deformation response factor R

Resonant amplitude
2) Resonant amplitude
3

1 2ζ = Half-power bandwidth
(1/√

0
0 1 2 3 4

Frequency ratio ω / ωn

Figure 3.2.9 Definition of half-power bandwidth.

√
of the resonant frequency at which the amplitude u o is 1/ 2 times the resonant amplitude,
then for small ζ
ωb − ωa
= 2ζ (3.2.23)
ωn
This result, derived in Derivation 3.4, can be rewritten as
ωb − ωa fb − fa
ζ = or ζ = (3.2.24)
2ωn 2 fn
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84 Response to Harmonic and Periodic Excitations Chap. 3

where f = ω/2π is the cyclic frequency. This important result enables evaluation of dam-
ping from forced vibration tests without knowing the applied force (Section 3.4).
Derivation 3.4
√
Equating Rd from Eq. (3.2.11) and 1/ 2 times the resonant amplitude of Rd given by
Eq. (3.2.22), by definition, the forcing frequencies ωa and ωb satisfy the condition
1 1 1
 2 = √  (a)
2 2ζ 1 − ζ 2
1 − (ω/ωn )2 + [2ζ (ω/ωn )]2
Inverting both sides, squaring them, and rearranging terms gives
 4  2
ω 2 ω
− 2(1 − 2ζ ) + 1 − 8ζ 2 (1 − ζ 2 ) = 0 (b)
ωn ωn

Equation (b) is a quadratic equation in (ω/ωn )2 , the roots of which are

 2 
ω
= (1 − 2ζ 2 ) ± 2ζ 1 − ζ2 (c)
ωn
where the positive sign gives the larger root ωb and the negative sign corresponds to the smaller
root ωa .
For the small damping ratios representative of practical structures, the two terms con-
taining ζ 2 can be dropped and
ω
(1 ± 2ζ )1/2 (d)
ωn
Taking only the first term in the Taylor series expansion of the right side gives
ω
1±ζ (e)
ωn
Subtracting the smaller root from the larger one gives
ωb − ωa
2ζ (f)
ωn

3.2.7 Steady-State Response to Cosine Force

The differential equation to be solved is

m ü + cu̇ + ku = po cos ωt (3.2.25)
The particular solution given by Eq. (3.2.3) still applies, but in this case the constants C
and D are
po 2ζ (ω/ωn )
C=  
k 1 − (ω/ωn )2 2 + [2ζ (ω/ωn )]2
(3.2.26)
po 1 − (ω/ωn )2
D=  
k 1 − (ω/ωn )2 2 + [2ζ (ω/ωn )]2
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Sec. 3.3 Response to Vibration Generator 85

These are determined by the procedure of Derivation 3.3. The steady-state response given
by Eqs. (3.2.3) and (3.2.26) can be expressed as
u(t) = u o cos(ωt − φ) = (u st )o Rd cos(ωt − φ) (3.2.27)
where the amplitude u o , the deformation response factor Rd , and the phase angle φ are the
same as those derived in Section 3.2.3 for a sinusoidal force. This similarity in the steady-
state responses to the two harmonic forces is not surprising since the two excitations are
the same except for a time shift.

PART B: VISCOUSLY DAMPED SYSTEMS: APPLICATIONS

3.3 RESPONSE TO VIBRATION GENERATOR

Vibration generators (or shaking machines) were developed to provide a source of har-
monic excitation appropriate for testing full-scale structures. In this section theoreti-
cal results for the steady-state response of an SDF system to a harmonic force caused
by a vibration generator are presented. These results provide a basis for evalua
ting the natural frequency and damping of a structure from experimental data
(Section 3.4).

3.3.1 Vibration Generator

Figure 3.3.1 shows a vibration generator having the form of two flat baskets rotating in
opposite directions about a vertical axis. By placing various numbers of lead weights in
the baskets, the magnitudes of the rotating weights can be altered. The two counterrotat-
ing masses, m e /2, are shown schematically in Fig. 3.3.2 as lumped masses with eccentri-
city = e; their locations at t = 0 are shown in (a) and at some time t in (b). The x-
components of the inertia forces of the rotating masses cancel out, and the y-components
combine to produce a force
p(t) = (m e eω2 ) sin ωt (3.3.1)

By bolting the vibration generator to the structure to be excited, this force can be transmit-
ted to the structure. The amplitude of this harmonic force is proportional to the square of
the excitation frequency ω. Therefore, it is difficult to generate force at low frequencies
and impractical to obtain the static response of a structure.

3.3.2 Structural Response

Assuming that the eccentric mass m e is small compared to the mass m of the structure, the
equation governing the motion of an SDF system excited by a vibration generator is
 
m ü + cu̇ + ku = m e eω2 sin ωt (3.3.2)
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86 Response to Harmonic and Periodic Excitations Chap. 3

Figure 3.3.1 Counterrotating eccentric weight vibration generator. (Courtesy of D. E.

Hudson.)

The amplitudes of steady-state displacement and of steady-state acceleration of the SDF

system are given by the maximum values of Eqs. (3.2.16) and (3.2.19) with po = m e eω2 .
Thus

mee 2 mee ω 2
uo = ω Rd = Rd (3.3.3)
k m ωn

mee 2 m e eωn2 ω 2
ü o = ω Ra = Ra (3.3.4)
m m ωn

ω ω meeω2/2
me /2 me /2 meeω2/2
ωt ωt

e
• • p(t) = (meeω2) sin ωt

(a) (b)

Figure 3.3.2 Vibration generator: (a) initial position; (b) position and forces at time t.
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Sec. 3.4 Natural Frequency and Damping from Harmonic Tests 87

10

8 ζ = 0.01

üo 6
me eω2 n /m 0.1
4
0.2
2

0
0 1 2 3
Frequency ratio ω / ωn

Figure 3.3.3

The acceleration amplitude of Eq. (3.3.4) is plotted as a function of the frequency ratio
ω/ωn in Fig. 3.3.3. For forcing frequencies ω greater than the natural frequency ωn of the
system, the acceleration increases rapidly with increasing ω because the amplitude of the
exciting force, Eq. (3.3.1), is proportional to ω2 .

3.4 NATURAL FREQUENCY AND DAMPING FROM

HARMONIC TESTS

The theory of forced harmonic vibration, presented in the preceding sections of this chap-
ter, provides a basis to determine the natural frequency and damping of a structure from
its measured response to a vibration generator. The measured damping provides data for
an important structural property that cannot be computed from the design of the structure.
The measured value of the natural frequency is the “actual” property of a structure against
which values computed from the stiffness and mass properties of structural idealizations
can be compared. Such research investigations have led to better procedures for developing
structural idealizations that are representative of actual structures.

3.4.1 Resonance Testing

The concept of resonance testing is based on the result of Eq. (3.2.15), rewritten as
1 (u st )o
ζ = (3.4.1)
2 (u o )ω=ωn
The damping ratio ζ is calculated from experimentally determined values of (u st )o and of
u o at forcing frequency equal to the natural frequency of the system.† Usually, the accel-
eration amplitude is measured and u o = ü o /ω2 . This seems straightforward except that
the true value ωn of the natural frequency is unknown. The natural frequency is detected

† Strictly speaking, this is not the resonant frequency; see Section 3.2.5.
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88 Response to Harmonic and Periodic Excitations Chap. 3

experimentally by utilizing the earlier result that the phase angle is 90◦ if ω = ωn . Thus the
structure is excited at forcing frequency ω, the phase angle is measured, and the exciting
frequency is adjusted progressively until the phase angle is 90◦ .
If the displacement due to static force po —the amplitude of the harmonic force—
can be obtained, Eq. (3.4.1) provides the damping ratio. As mentioned earlier, it is difficult
for a vibration generator to produce a force at low frequencies and impractical to obtain
a significant static force. An alternative is to measure the static response by some other
means, such as by pulling on the structure. In this case, Eq. (3.4.1) should be modified to
recognize any differences in the force applied in the static test relative to the amplitude of
the harmonic force.

3.4.2 Frequency-Response Curve

Because of the difficulty in obtaining the static structural response using a vibration gener-
ator, the natural frequency and damping ratio of a structure are usually determined by ob-
taining the frequency-response curve experimentally. The vibration generator is operated
at a selected frequency, the structural response is observed until the transient part damps
out, and the amplitude of the steady-state acceleration is measured. The frequency of the
vibration generator is adjusted to a new value and the measurements are repeated. The
forcing frequency is varied over a range that includes the natural frequency of the system.
A frequency-response curve in the form of acceleration amplitude versus frequency may
be plotted directly from the measured data. This curve is for a force with amplitude propor-
tional to ω2 and would resemble the frequency-response curve of Fig. 3.3.3. If each mea-
sured acceleration amplitude is divided by ω2 , we obtain the frequency–acceleration curve

rres
Response amplitude r

rres / √2

fn

fa fb
Forcing frequency f

Figure 3.4.1 Evaluating damping from frequency-response curve.

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Sec. 3.4 Natural Frequency and Damping from Harmonic Tests 89

for a constant-amplitude force. This curve from measured data wouldresemble a curve
in Fig. 3.2.7c. If the measured accelerations are divided by ω4 , the resulting frequency–
displacement curve for a constant-amplitude force would be an experimental version of the
curve in Fig. 3.2.7a.
The natural frequency and damping ratio can be determined from any one of the
experimentally obtained versions of the frequency-response curves of Figs. 3.3.3, 3.2.7c,
and 3.2.7a. For the practical range of damping the natural frequency f n is essentially equal
to the forcing frequency at resonance. The damping ratio is calculated by Eq. (3.2.24) us-
ing the frequencies f a and f b , determined, as illustrated in Fig. 3.4.1, from the experimen-
tal curve shown schematically. Although this equation was derived from the frequency–
displacement curve for a constant-amplitude harmonic force, it is approximately valid for
the other response curves mentioned earlier as long as the structure is lightly damped.
Example 3.2
The plexiglass frame model of Fig. 1.1.4 is mounted on a shaking table which can apply
harmonic base motions of specified frequencies and amplitudes. At each excitation frequency
ω, acceleration amplitudes ü go and ü to of the table and the top of the frame, respectively, are
recorded. The transmissibility TR = ü to /ü go is compiled and the data are plotted in Fig. E3.2.
Determine the natural frequency and damping ratio of the plexiglass frame from these data.

Solution The peak of the frequency-response curve occurs at 3.59 Hz. Assuming that the
damping is small, the natural frequency f n = 3.59 Hz.
√The peak value of the transmissibility curve is 12.8. Now draw a horizontal line at
12.8/ 2 = 9.05 as shown. This line intersects the frequency-response curve at f b = 3.74 Hz
and f a = 3.44 Hz. Therefore, from Eq. (3.2.24),
3.74 − 3.44
ζ = = 0.042 = 4.2%
2 (3.59)

14
12.8
12

10
Transmissibility

9.05
8

4
fn = 3.59
2
fa = 3.44 fb = 3.74
0
3.0 3.2 3.4 3.6 3.8 4.0
f, Hz

Figure E3.2
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90 Response to Harmonic and Periodic Excitations Chap. 3

This damping value is slightly higher than the 3.96% determined from a free vibration test on
the model (Example 2.4).
Note that we have used Eq. (3.2.24) to determine the damping ratio of the system from
its transmissibility (TR) curve, whereas this equation had been derived from the frequency–
displacement curve. This approximation is appropriate because at excitation frequencies in
the range f a to f b , the numerical values of TR and Rd are close; this is left for the reader to
verify after an equation for TR is presented in Section 3.6.

3.5 FORCE TRANSMISSION AND VIBRATION ISOLATION

Consider the mass–spring–damper system shown in the left inset in Fig. 3.5.1 subjected to
a harmonic force. The force transmitted to the base is
f T = f S + f D = ku(t) + cu̇(t) (3.5.1)

100

50 p(t) = po sin ωt ut
Transmissibility (TR) = (fT)o / po = üot / ügo

m m
ζ = 0.01 ug
k c k c
fT
0.05
10

0.1
5

0.2

0.7
1
ζ=1
0.5


√2
0.1
0.1 0.5 1 5 10
Frequency ratio ω / ωn

Figure 3.5.1 Transmissibility for harmonic excitation. Force transmissibility and ground
motion transmissibility are identical.
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Sec. 3.6 Response to Ground Motion and Vibration Isolation 91

Substituting Eq. (3.2.10) for u(t) and Eq. (3.2.17) for u̇(t) and using Eq. (3.2.18) gives
f T (t) = (u st )o Rd [k sin(ωt − φ) + cω cos(ωt − φ)] (3.5.2)
The maximum value of f T (t) over t is

( f T )o = (u st )o Rd k 2 + c2 ω2
which, after using (u st )o = po /k and ζ = c/2mωn , can be expressed as


( f T )o ω 2
= Rd 1 + 2ζ
po ωn
Substituting Eq. (3.2.11) for Rd gives an equation for the ratio of the maximum transmitted
force to the amplitude po of the applied force, known as the transmissibility (TR) of the
system:

1/2
1 + [2ζ (ω/ωn )]2
TR = (3.5.3)
[1 − (ω/ωn )2 ]2 + [2ζ (ω/ωn )]2
The transmissibility is plotted in Fig. 3.5.1 as a function of the frequency ratio ω/ωn
for several values of the damping ratio ζ . Logarithmic scales have been chosen to highlight
the curves for large ω/ωn , the region of interest. While damping decreases the amplitude
of motion at all excitation
√ frequencies (Fig. 3.2.6), damping decreases the transmitted force
only if ω/ωn < 2. For the transmitted force to be less than the applied force, the stiffness
of the support
√ system and hence the natural frequency should be small enough so that
ω/ωn > 2. No damping is desired in the support system because, in this frequency
range, damping increases the transmitted force. This implies a trade-off between a soft
spring to reduce the transmitted force and an acceptable static displacement.
If the applied force arises from a rotating machine, its frequency will vary as it starts
to rotate and increases its speed to reach the operating frequency. In this case the choice of
a flexible support system to minimize the transmitted force must be a compromise. It must
have sufficient damping to limit the force transmitted while passing through resonance,
but not enough to add significantly to the force transmitted at operating speeds. Luckily,
natural rubber is a very satisfactory material and is often used for the isolation of vibration.

3.6 RESPONSE TO GROUND MOTION AND VIBRATION ISOLATION

In this section we determine the response of an SDF system (see the right inset in Fig. 3.5.1)
to harmonic ground motion:
ü g (t) = ü go sin ωt (3.6.1)
For this excitation the governing equation is Eq. (1.7.4), where the forcing function is
peff (t) = −m ü g (t) = −m ü go sin ωt, the same as Eq. (3.2.1) for an applied harmonic force
with po replaced by −m ü go . Making this substitution in Eqs. (3.1.9) and (3.2.10) gives
−m ü go
u(t) = Rd sin(ωt − φ) (3.6.2)
k
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92 Response to Harmonic and Periodic Excitations Chap. 3

The acceleration of the mass is

ü t (t) = ü g (t) + ü(t) (3.6.3)
Substituting Eq. (3.6.1) and the second derivative of Eq. (3.6.2) gives an equation for ü t (t)
from which the amplitude or maximum value ü to can be determined (see Derivation 3.5):
 1/2
ü to 1 + [2ζ (ω/ωn )]2
TR = =  2 (3.6.4)
ü go 1 − (ω/ωn )2 + [2ζ (ω/ωn )]2
The ratio of acceleration ü to transmitted to the mass and amplitude ü go of ground accelera-
tion is also known as the transmissibility (TR) of the system. From Eqs. (3.6.4) and (3.5.3)
it is clear that the transmissibility for the ground excitation problem is the same as for the
applied force problem.
Therefore, Fig. 3.5.1 also gives the ratio ü to /ü go as a function of the frequency ratio
ω/ωn . If the excitation frequency ω is much smaller than the natural frequency ωn of the
system, ü to ü go (i.e., the mass moves rigidly with the ground, both undergoing the same
acceleration). If the excitation frequency ω is much higher than the natural frequency ωn of
the system, ü to 0 (i.e., the mass stays still while the ground beneath it moves). This is the
basic concept underlying isolation of a mass from a moving base by using a very flexible
support system. For example, buildings have been mounted on natural rubber bearings to
isolate them from ground-borne vertical vibration—typically with frequencies that range
from 25 to 50 Hz—due to rail traffic.
Before closing this section, we mention without derivation the results of a related
problem. If the ground motion is defined as u g (t) = u go sin ωt, it can be shown that the
amplitude u to of the total displacement u t (t) of the mass is given by
 1/2
u to 1 + [2ζ (ω/ωn )]2
TR = =  2 (3.6.5)
u go 1 − (ω/ωn )2 + [2ζ (ω/ωn )]2
Comparing this with Eq. (3.6.4) indicates that the transmissibility for displacements and
accelerations is identical.
Example 3.3
A sensitive instrument with weight 100 lb is to be installed at a location where the vertical
acceleration is 0.1g at a frequency of 10 Hz. This instrument is mounted on a rubber pad of
stiffness 80 lb/in. and damping such that the damping ratio for the system is 10%. (a) What
acceleration is transmitted to the instrument? (b) If the instrument can tolerate only an accel-
eration of 0.005g, suggest a solution assuming that the same rubber pad is to be used. Provide
numerical results.
Solution (a) Determine TR.

80
ωn = = 17.58 rad/sec
100/386
ω 2π(10)
= = 3.575
ωn 17.58
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Sec. 3.6 Response to Ground Motion and Vibration Isolation 93

Substituting these in Eq. (3.6.4) gives


ü t 1 + [2(0.1)(3.575)]2
TR = o = = 0.104
ü go [1 − (3.575)2 ]2 + [2(0.1)(3.575)]2
Therefore, ü to = (0.104)ü go = (0.104)0.1g = 0.01g.
(b) Determine the added mass to reduce acceleration. The transmitted acceleration can
be reduced by increasing ω/ωn , which requires reducing ωn by mounting the instrument on
mass m b . Suppose that we add a mass m b = 150 lb/g; the total mass = 250 lb/g, and

80 ω
ωn = = 11.11 rad/sec = 5.655
250/386 ωn
To determine the damping ratio for the system with added mass, we need the damping coeffi-
cient for the rubber pad:

100
c = ζ (2mωn ) = 0.1(2) 17.58 = 0.911 lb-sec/in.
386
Then
c 0.911
ζ = = = 0.063
2(m + m b )ωn 2(250/386)11.11
Substituting for ω/ωn and ζ  in Eq. (3.6.4) gives ü to /ü go = 0.04; ü to = 0.004g, which is
satisfactory because it is less than 0.005g.
Instead of selecting an added mass by judgment, it is possible to set up a quadratic
equation for the unknown mass which will give ü to /ü go = 0.005g.

Example 3.4
An automobile is traveling along a multispan elevated roadway supported every 100 ft. Long-
term creep has resulted in a 6-in. deflection at the middle of each span (Fig. E3.4a). The
roadway profile can be approximated as sinusoidal with an amplitude of 3 in. and period of
100 ft. The SDF system shown is a simple idealization of an automobile, appropriate for a
“first approximation” study of the ride quality of the vehicle. When fully loaded, the weight of
the automobile is 4 kips. The stiffness of the automobile suspension system is 800 lb/in., and
its viscous damping coefficient is such that the damping ratio of the system is 40%. Determine

(a) (b)
ut

3″ w
••

k c
ug

100′ 100′
• • •

Figure E3.4
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94 Response to Harmonic and Periodic Excitations Chap. 3

(a) the amplitude u to of vertical motion u t (t) when the automobile is traveling at 40 mph, and
(b) the speed of the vehicle that would produce a resonant condition for u to .
Solution Assuming that the tires are infinitely stiff and they remain in contact with the road,
the problem can be idealized as shown in Fig. E3.4b. The vertical displacement of the tires is
u g (t) = u go sin ωt, where u go = 3 in. The forcing frequency ω = 2π/T , where the forcing
period T = L/v, the time taken by the automobile to cross the span; therefore, ω = 2πv/L.
(a) Determine u to .
2π(58.67)
v = 40 mph = 58.67 ft/sec ω= = 3.686 rad/sec
100
 
k 800 ω
ωn = = = 8.786 rad/sec = 0.420
m 4000/386 ωn
Substituting these data in Eq. (3.6.5) gives

1/2
u to 1 + [2(0.4)(0.420)]2
= = 1.186
u go [1 − (0.420)2 ]2 + [2(0.4)(0.420)]2

u to = 1.186u go = 1.186(3) = 3.56 in.

(b) Determine the speed at resonance. If ζ were small, resonance would occur ap-
proximately at ω/ωn = 1. However, automobile suspensions have heavy damping, to reduce
vibration. In this case, ζ = 0.4, and for such large damping the resonant frequency is sig-
nificantly different from ωn . By definition, resonance occurs for u to when TR (or TR2 ) is
maximum over all ω. Substituting ζ = 0.4 in Eq. (3.6.5) and introducing β = ω/ωn gives
1 + 0.64β 2 1 + 0.64β 2
TR2 = = 4
(1 − 2β + β ) + 0.64β
2 4 2 β − 1.36β 2 + 1
d(TR)2
= 0 ⇒ β = 0.893 ⇒ ω = 0.893ωn = 0.893(8.786) = 7.846 rad/sec
dβ
Resonance occurs at this forcing frequency, which implies a speed of
ωL (7.846)100
v= = = 124.9 ft/sec = 85 mph
2π 2π

Example 3.5
Repeat part (a) of Example 3.4 if the vehicle is empty (driver only) with a total weight of 3
kips.
Solution Since the damping coefficient c does not change but the mass m does, we need to
recompute the damping ratio for an empty vehicle from
 
c = 2ζ f km f = 2ζe km e
where the subscripts f and e denote full and empty conditions, respectively. Thus
 1/2  1/2
mf 4
ζe = ζ f = 0.4 = 0.462
me 3
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Sec. 3.7 Vibration-Measuring Instruments 95

For an empty vehicle

 
k 800
ωn = = = 10.15 rad/sec
m 3000/386
ω 3.686
= = 0.363
ωn 10.15
Substituting for ω/ωn and ζ in Eq. (3.6.5) gives

1/2
u to 1 + [2(0.462)(0.363)]2
= = 1.133
u go [1 − (0.363)2 ]2 + [2(0.462)(0.363)]2

u to = 1.133u go = 1.133(3) = 3.40 in.

Derivation 3.5
Equation (3.6.2) is first rewritten as a linear combination of sine and cosine functions. This
can be accomplished by substituting Eqs. (3.2.11) and (3.2.12) for the Rd and φ, respectively,
or by replacing po in Eq. (3.2.4) with −m ü go and substituting in Eq. (3.2.3). Either way the
relative displacement is


−m ü go [1 − (ω/ωn )2 ] sin ωt − [2ζ (ω/ωn )] cos ωt
u(t) = (a)
k [1 − (ω/ωn )2 ]2 + [2ζ (ω/ωn )]2
Differentiating this twice and substituting it in Eq. (3.6.3) together with Eq. (3.6.1) gives
ü t (t) = ü go (C1 sin ωt + D1 cos ωt) (b)
where
1 − (ω/ωn )2 + 4ζ 2 (ω/ωn )2 −2ζ (ω/ωn )3
C1 = D1 = (c)
[1 − (ω/ωn )2 ]2 + [2ζ (ω/ωn )]2 [1 − (ω/ωn )2 ]2 + [2ζ (ω/ωn )]2
The acceleration amplitude is

ü to = ü go C12 + D12 (d)

This result, after substituting for C1 and D1 from Eq. (c) and some simplification, leads to
Eq. (3.6.4).

3.7 VIBRATION-MEASURING INSTRUMENTS

Measurement of vibration is of great interest in many aspects of structural engineering.

For example, measurement of ground shaking during an earthquake provides basic data
for earthquake engineering, and records of the resulting motions of a structure provide
insight into how structures respond during earthquakes. Although measuring instruments
are highly developed and intricate, the basic element of these instruments is some form of
a transducer. In its simplest form a transducer is a mass–spring–damper system mounted
inside a rigid frame that is attached to the surface whose motion is to be measured.
Figure 3.7.1 shows a schematic drawing of such an instrument to record the horizontal
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96 Response to Harmonic and Periodic Excitations Chap. 3

u (magnified)
u
c k
m t

ug(t)

Figure 3.7.1 Schematic drawing of a vibration-measuring instrument and recorded motion.

motion of a support point; three separate transducers are required to measure the three com-
ponents of motion. When subjected to motion of the support point, the transducer mass
moves relative to the frame, and this relative displacement is recorded after suitable mag-
nification. It is the objective of this brief presentation to discuss the principle underlying
the design of vibration-measuring instruments so that the measured relative displacement
provides the desired support motion—acceleration or displacement.

3.7.1 Measurement of Acceleration

The motion to be measured generally varies arbitrarily with time and may include many
harmonic components covering a wide range of frequencies. It is instructive, however, to
consider first the measurement of simple harmonic motion described by Eq. (3.6.1). The
displacement of the instrument mass relative to the moving frame is given by Eq. (3.6.2),
which can be rewritten as
 
1 φ
u(t) = − Rd ü g t − (3.7.1)
ωn2 ω
The recorded u(t) is the base acceleration modified by a factor −Rd /ωn2 and recorded with
a time lag φ/ω. As shown in Fig. 3.2.6, Rd and φ vary with the forcing frequency ω, but
ωn2 is an instrument constant independent of the support motion.
The object of the instrument design is to make Rd and φ/ω as independent of excita-
tion frequency as possible because then each harmonic component of acceleration will be
recorded with the same modifying factor and the same time lag. Then, even if the motion to
be recorded consists of many harmonic components, the recorded u(t) will have the same
shape as the support motion with a constant shift of time. This constant time shift simply
moves the time scale a little, which is usually not important. According to Fig. 3.7.2 (which
is a magnified plot of Fig. 3.2.6 with additional damping values), if ζ = 0.7, then over the
frequency range 0 ≤ ω/ωn ≤ 0.50, Rd is close to 1 (less than 2.5% error) and the variation
of φ with ω is close to linear, implying that φ/ω is essentially constant. Thus an instru-
ment with a natural frequency of 50 Hz and damping ratio of 0.7 has a useful frequency
range from 0 to 25 Hz with negligible error. These are the properties of modern, com-
mercially available instruments designed to measure earthquake-induced ground accelera-
tion. Because the measured amplitude of u(t) is proportional to Rd /ωn2 , a high-frequency
instrument will result in a very small displacement that is substantially magnified in these
instruments for proper measurement.
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Sec. 3.7 Vibration-Measuring Instruments 97

1.05
ζ = 0.6

0.65
Rd
1

0.7
0.75
0.95
0.0 0.2 0.4 0.6 0.8 1.0

90°

0.75
0.65
45°
φ

0.7
Linear
ζ = 0.6

0°
0.0 0.2 0.4 0.6 0.8 1.0
Frequency ratio ω/ωn

Figure 3.7.2 Variation of Rd and φ with frequency ratio ω/ωn for ζ = 0.6, 0.65, 0.7,
and 0.75.

Figure 3.7.3 shows a comparison of the actual ground acceleration ü g (t) =

0.1g sin(2π f t) and the measured relative displacement of Rd ü g (t − φ/ω), except for the
instrument constant −1/ωn2 in Eq. (3.7.1). For excitation frequencies f = 20 and 10 Hz,
the measured motion has accurate amplitude, but the error at f = 40 Hz is noticeable; and
the time shift, although not identical for the three frequencies, is similar. If the ground ac-
celeration is the sum of the three harmonic components, this figure shows that the recorded
motion matches the ground acceleration in amplitude and shape reasonably well but not
perfectly.
The accuracy of the recorded motion u(t) can be improved by separating u(t) into its
harmonic components and correcting one component at a time, by calculating ü g (t − φ/ω)
from the measured u(t) using Eq. (3.7.1) with Rd determined from Eq. (3.2.11) and known
instrument properties ωn and ζ . Such corrections are repeated for each harmonic compo-
nent in u(t), and the corrected components are then synthesized to obtain ü g (t). These
computations can be carried out by discrete Fourier transform procedures, which are intro-
duced in Appendix A of this book.
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98 Response to Harmonic and Periodic Excitations Chap. 3

Actual: üg(t) Measured: Rd üg(t - φ/ω)
üg f = 40 Hz
0.1g

-0.1g 0.005 sec

f = 20 Hz
0.1g Measured
Actual

-0.1g 0.0047 sec

f = 10 Hz Actual
0.1g
Measured

0.0045 sec
-0.1g

Measured
0.4g Actual: üg(t) = 0.1 g Σ sin 2πft

-0.4g
0 0.05 0.1 0.15
Time, sec

Figure 3.7.3 Comparison of actual ground acceleration and measured motion by an

instrument with f n = 50 Hz and ζ = 0.7.

3.7.2 Measurement of Displacement

It is desired to design the transducer so that the relative displacement u(t) measures the
support displacement u g (t). This is achieved by making the transducer spring so flexible
or the transducer mass so large, or both, that the mass stays still while the support beneath
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Sec. 3.8 Energy Dissipated in Viscous Damping 99

it moves. Such an instrument is unwieldy because of the heavy mass and soft spring, and
because it must accommodate the anticipated support displacement, which may be as large
as 36 in. during earthquakes.
To examine the basic concept further, consider harmonic support displacement
u g (t) = u go sin ωt (3.7.2)
With the forcing function peff (t) = −m ü g (t) = mω2 u go sin ωt, Eq. (1.7.4) governs the
relative displacement of the mass; this governing equation is the same as Eq. (3.2.1) for
applied harmonic force with po replaced by mω2 u go . Making this substitution in
Eq. (3.2.10) and using Eqs. (3.1.9) and (3.2.20) gives
u(t) = Ra u go sin(ωt − φ) (3.7.3)
For excitation frequencies ω much higher than the natural frequency ωn , Ra is close to
unity (Fig. 3.2.7c) and φ is close to 180◦ , and Eq. (3.7.3) becomes
u(t) = −u go sin ωt (3.7.4)
This recorded displacement is the same as the support displacement [Eq. (3.7.2)] except
for the negative sign, which is usually inconsequential. Damping of the instrument is not
a critical parameter because it has little effect on the recorded motion if ω/ωn is very
large.

3.8 ENERGY DISSIPATED IN VISCOUS DAMPING

Consider the steady-state motion of an SDF system due to p(t) = po sin ωt. The energy
dissipated by viscous damping in one cycle of harmonic vibration is

  2π/ω  2π/ω
ED = f D du = (cu̇)u̇ dt = cu̇ 2 dt
0 0
 2π/ω
ω 2
=c [ωu o cos(ωt − φ)]2 dt = π cωu 2o = 2π ζ ku (3.8.1)
0 ωn o

The energy dissipated is proportional to the square of the amplitude of motion. It

is not a constant value for any given amount of damping and amplitude since the energy
dissipated increases linearly with excitation frequency.
In steady-state vibration, the energy input to the system due to the applied force is
dissipated in viscous damping. The external force p(t) inputs energy to the system, which
for each cycle of vibration is
  2π/ω
EI = p(t) du = p(t)u̇ dt
0
 2π/ω
= [ po sin ωt][ωu o cos(ωt − φ)] dt = π po u o sin φ (3.8.2)
0
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Utilizing Eq. (3.2.12) for phase angle, this equation can be rewritten as (see Derivation 3.6)
ω
E I = 2π ζ ku 2o (3.8.3)
ωn
Equations (3.8.1) and (3.8.3) indicate that E I = E D .
What about the potential energy and kinetic energy? Over each cycle of harmonic
vibration the changes in potential energy (equal to the strain energy of the spring) and
kinetic energy are zero. This can be confirmed as follows:
  2π/ω
ES = f S du = (ku)u̇ dt
0
 2π/ω
= k[u o sin(ωt − φ)][ωu o cos(ωt − φ)] dt = 0
0
  2π/ω
EK = f I du = (m ü)u̇ dt
0
 2π/ω
= m[−ω2 u o sin(ωt − φ)][ωu o cos(ωt − φ)] dt = 0
0

The preceding energy concepts help explain the growth of the displacement ampli-
tude caused by harmonic force with ω = ωn until steady state is attained (Fig. 3.2.2). For
ω = ωn , φ = 90◦ and Eq. (3.8.2) gives
E I = π po u o (3.8.4)
The input energy varies linearly with the displacement amplitude (Fig. 3.8.1). In contrast,
the dissipated energy varies quadratically with the displacement amplitude (Eq. 3.8.1). As
shown in Fig. 3.8.1, before steady state is reached, the input energy per cycle exceeds
the energy dissipated during the cycle by damping, leading to a larger amplitude of dis-
placement in the next cycle. With growing displacement amplitude, the dissipated energy
increases more rapidly than does the input energy. Eventually, the input and dissipated
energies will match at the steady-state displacement amplitude u o , which will be bounded
no matter how small the damping. This energy balance provides an alternative means of

ED EI

EI = ED
Energy

uo
Figure 3.8.1 Input energy E I and energy
Amplitude dissipated E D in viscous damping.
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Sec. 3.8 Energy Dissipated in Viscous Damping 101

finding u o due to harmonic force with ω = ωn ; equating Eqs. (3.8.1) and (3.8.4) gives
π po u o = π cωn u 2o (3.8.5)
Solving for u o leads to
po
uo = (3.8.6)
cωn
This result agrees with Eq. (3.2.5), obtained by solving the equation of motion.
We will now present a graphical interpretation for the energy dissipated in viscous
damping. For this purpose we first derive an equation relating the damping force f D to the
displacement u:
f D = cu̇(t) = cωu o cos(ωt − φ)

= cω u 2o − u 2o sin2 (ωt − φ)

= cω u 2o − [u(t)]2

This can be rewritten as

 2  2
u fD
+ =1 (3.8.7)
uo cωu o
which is the equation of the ellipse shown in Fig. 3.8.2a. Observe that the f D –u curve
is not a single-valued function but a loop known as a hysteresis loop. The area enclosed
by the ellipse is π(u o )(cωu o ) = π cωu 2o , which is the same as Eq. (3.8.1). Thus the area
within the hysteresis loop gives the dissipated energy.

fD + fS
fS = ku
kuo
fD
u˙ > 0 cωuo cωuo
Loading, u̇ > 0
u u
uo uo

u˙ < 0 Unloading, u̇ < 0

(a) (b)

Figure 3.8.2 Hysteresis loops for (a) viscous damper; (b) spring and viscous damper in parallel.
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102 Response to Harmonic and Periodic Excitations Chap. 3

It is of interest to examine the total (elastic plus damping) resisting force because
this is the force that is measured in an experiment:
f S + f D = ku(t) + cu̇(t)

= ku + cω u 2o − u 2 (3.8.8)

A plot of f S + f D against u is the ellipse of Fig. 3.8.2a rotated as shown in Fig. 3.8.2b
because of the ku term in Eq. (3.8.8). The energy dissipated by damping is still the area en-
closed by the ellipse because the area enclosed by the single-valued elastic force, f S = ku,
is zero.
The hysteresis loop associated with viscous damping is the result of dynamic hystere-
sis since it is related to the dynamic nature of the loading. The loop area is proportional
to excitation frequency; this implies that the force–deformation curve becomes a single-
valued curve (no hysteresis loop) if the cyclic load is applied slowly enough (ω = 0). A
distinguishing characteristic of dynamic hysteresis is that the hysteresis loops tend to be
elliptical in shape rather than pointed, as in Fig. 1.3.1c, if they are associated with plastic
deformations. In the latter case, the hysteresis loops develop even under static cyclic loads;
this phenomenon is therefore known as static hysteresis because the force–deformation
curve is insensitive to deformation rate.
In passing, we mention two measures of damping: specific damping capacity and
the specific damping factor. The specific damping capacity, E D /E So , is that fractional part
of the strain energy, E So = ku 2o /2, which is dissipated during each cycle of motion; both
E D and E So are shown in Fig. 3.8.3. The specific damping factor, also known as the loss
factor, is defined as
1 ED
ξ= (3.8.9)
2π E So
If the energy could be removed at a uniform rate during a cycle of simple harmonic motion
(such a mechanism is not realistic), E D /2π could be interpreted as the energy loss per
Resisting force

ESo

Deformation
ED

Figure 3.8.3 Definition of energy loss

E D in a cycle of harmonic vibration and
maximum strain energy E So .
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Sec. 3.9 Equivalent Viscous Damping 103

radian divided by the total energy in the system. These two measures of damping are not
often used in structural vibration since they are most useful for very light damping (e.g.,
they are useful in comparing the damping capacity of materials).
Derivation 3.6
Equation (3.8.2) gives the input energy per cycle where the phase angle, defined by Eq. (3.2.12),
can be expressed as
 
ω ω uo
sin φ = 2ζ Rd = 2ζ
ωn ωn po /k
Substituting this in Eq. (3.8.2) gives Eq. (3.8.3).

3.9 EQUIVALENT VISCOUS DAMPING

As introduced in Section 1.4, damping in actual structures is usually represented by equiv-

alent viscous damping. It is the simplest form of damping to use since the governing
differential equation of motion is linear and hence amenable to analytical solution, as seen
in earlier sections of this chapter and in Chapter 2. The advantage of using a linear equation
of motion usually outweighs whatever compromises are necessary in the viscous damping
approximation. In this section we determine the damping coefficient for viscous damping
so that it is equivalent in some sense to the combined effect of all damping mechanisms
present in the actual structure; these were mentioned in Section 1.4.
The simplest definition of equivalent viscous damping is based on the measured re-
sponse of a system to harmonic force at exciting frequency ω equal to the natural frequency
ωn of the system. The damping ratio ζeq is calculated from Eq. (3.4.1) using measured val-
ues of u o and (u st )o . This is the equivalent viscous damping since it accounts for all the
energy-dissipating mechanisms that existed in the experiments.
Another definition of equivalent viscous damping is that it is the amount of damping
that provides the same bandwidth in the frequency-response curve as obtained experimen-
tally for an actual system. The damping ratio ζeq is calculated from Eq. (3.2.24) using the
excitation frequencies f a , f b , and f n (Fig. 3.4.1) obtained from an experimentally deter-
mined frequency-response curve.
The most common method for defining equivalent viscous damping is to equate the
energy dissipated in a vibration cycle of the actual structure and an equivalent viscous
system. For an actual structure the force-displacement relation obtained from an experi-
ment under cyclic loading with displacement amplitude u o is determined; such a relation
of arbitrary shape is shown schematically in Fig. 3.9.1. The energy dissipated in the actual
structure is given by the area E D enclosed by the hysteresis loop. Equating this to the
energy dissipated in viscous damping given by Eq. (3.8.1) leads to
ω 1 1 ED
4πζeq E So = E D or ζeq = (3.9.1)
ωn 4π ω/ωn E So
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104 Response to Harmonic and Periodic Excitations Chap. 3

Resisting force
ESo

uo
ED Deformation

Figure 3.9.1 Energy dissipated E D in a

cycle of harmonic vibration determined from
experiment.

where the strain energy, E So = ku 2o /2, is calculated from the stiffness k determined by
experiment.
The experiment leading to the force–deformation curve of Fig. 3.9.1 and hence E D
should be conducted at ω = ωn , where the response of the system is most sensitive to
damping. Thus Eq. (3.9.1) specializes to

1 ED
ζeq = (3.9.2)
4π E So
The damping ratio ζeq determined from a test at ω = ωn would not be correct at any other
exciting frequency, but it would be a satisfactory approximation (Section 3.10.2).
It is widely accepted that this procedure can be extended to model the damping in
systems with many degrees of freedom. An equivalent viscous damping ratio is assigned
to each natural vibration mode of the system (defined in Chapter 10) in such a way that the
energy dissipated in viscous damping matches the actual energy dissipated in the system
when the system vibrates in that mode at its natural frequency.
In this book the concept of equivalent viscous damping is restricted to systems vi-
brating at amplitudes within the linearly elastic limit of the overall structure. The energy
dissipated in inelastic deformations of the structure has also been modeled as equivalent
viscous damping in some research studies. This idealization is generally not satisfactory,
however, for the large inelastic deformations of structures expected during strong earth-
quakes. We shall account for these inelastic deformations and the associated energy dis-
sipation by nonlinear force–deformation relations, such as those shown in Fig. 1.3.4 (see
Chapters 5 and 7).
Example 3.6
A body moving through a fluid experiences a resisting force that is proportional to the square
of the speed, f D = ±a u̇ 2 , where the positive sign applies to positive u̇ and the negative sign
to negative u̇. Determine the equivalent viscous damping coefficient ceq for such forces acting
on an oscillatory system undergoing harmonic motion of amplitude u o and frequency ω. Also
find its displacement amplitude at ω = ωn .
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Sec. 3.10 Harmonic Vibration with Rate-Independent Damping 105

Solution If time is measured from the position of largest negative displacement, the har-
monic motion is
u(t) = −u o cos ωt
The energy dissipated in one cycle of motion is
  2π/ω  π/ω
ED = f D du = f D u̇ dt = 2 f D u̇ dt
0 0
 π/ω  π/ω
=2 (a u̇ 2 )u̇ dt = 2aω3 u 3o sin3 ωt dt = 83 aω2 u 3o
0 0
Equating this to the energy dissipated in viscous damping [Eq. (3.8.1)] gives
8 2 3 8
πceq ωu 2o =
aω u o or ceq = aωu o (a)
3 3π
Substituting ω = ωn in Eq. (a) and the ceq for c in Eq. (3.2.15) gives
 1/2
3π po
uo = (b)
8a ωn2

PART C: SYSTEMS WITH NONVISCOUS DAMPING

3.10 HARMONIC VIBRATION WITH RATE-INDEPENDENT DAMPING

3.10.1 Rate-Independent Damping

Experiments on structural metals indicate that the energy dissipated internally in cyclic
straining of the material is essentially independent of the cyclic frequency. Similarly,
forced vibration tests on structures indicate that the equivalent viscous damping ratio is
roughly the same for all natural modes and frequencies. Thus we refer to this type of
damping as rate-independent linear damping. Other terms used for this mechanism of in-
ternal damping are structural damping, solid damping, and hysteretic damping. We prefer
not to use these terms because the first two are not especially meaningful and the third is
ambiguous because hysteresis is a characteristic of all materials or structural systems that
dissipate energy.
Rate-independent damping is associated with static hysteresis due to plastic strain,
localized plastic deformation, crystal plasticity, and plastic flow in a range of stresses
within the apparent elastic limit. On the microscopic scale the inhomogeneity of stress dis-
tribution within crystals and stress concentration at crystal boundary intersections
produce local stress high enough to cause local plastic strain even though the average
(macroscopic) stress may be well below the elastic limit. This damping mechanism does
not include the energy dissipation in macroscopic plastic deformations, which as men-
tioned earlier, is handled by a nonlinear relationship between force f S and deformation u.
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106 Response to Harmonic and Periodic Excitations Chap. 3

The simplest device that can be used to represent rate-independent linear damping
during harmonic motion at frequency ω is to assume that the damping force is proportional
to velocity and inversely proportional to frequency:
ηk
fD = u̇ (3.10.1)
ω
where k is the stiffness of the structure and η is a damping coefficient. The energy dissi-
pated by this type of damping in a cycle of vibration at frequency ω is independent of ω
(Fig. 3.10.1). It is given by Eq. (3.8.1) with c replaced by ηk/ω:

E D = πηku 2o = 2πηESo (3.10.2)

In contrast, the energy dissipated in viscous damping [Eq. (3.8.1)] increases linearly with
the forcing frequency as shown in Fig. 3.10.1.

Viscous damping
Energy dissipated ED

Rate-independent damping

ωn
Figure 3.10.1 Energy dissipated in viscous
Forcing frequency ω damping and rate-independent damping.

Rate-independent damping is easily described if the excitation is harmonic and we

are interested only in the steady-state response of this system. Difficulties arise in trans-
lating this damping mechanism back to the time domain. Thus it is most useful in the
frequency-domain method of analysis (Appendix A).

3.10.2 Steady-State Response to Harmonic Force

The equation governing harmonic motion of an SDF system with rate-independent linear
damping, denoted by a crossed box in Fig. 3.10.2, is Eq. (3.2.1) with the damping term
replaced by Eq. (3.10.1):
ηk
m ü + u̇ + ku = p(t) (3.10.3)
ω
The mathematical solution of this equation is quite complex for arbitrary p(t).
Here we consider only the steady-state motion due to a sinusoidal forcing function,
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Sec. 3.10 Harmonic Vibration with Rate-Independent Damping 107

u
η
m p(t)
k Friction-free surface Figure 3.10.2 SDF system with
rate-independent linear damping.

p(t) = po sin ωt, which is described by

u(t) = u o sin(ωt − φ) (3.10.4)
The amplitude u o and phase angle φ are
1
u o = (u st )o  2 (3.10.5)
1 − (ω/ωn )2 + η2
η
φ = tan−1 (3.10.6)
1 − (ω/ωn )2
These results are obtained by modifying the viscous damping ratio in Eqs. (3.2.11) and
(3.2.12) to reflect the damping force associated with rate-independent damping, Eq. (3.10.1).
In particular, ζ was replaced by
c ηk/ω η
ζ = = = (3.10.7)
cc 2mωn 2(ω/ωn )
Shown in Fig. 3.10.3 by solid lines are plots of u o /(u st )o and φ as a function of the
frequency ratio ω/ωn for damping coefficient η = 0, 0.2, and 0.4; the dashed lines are
described in the next section. Comparing these results with those in Fig. 3.2.6 for viscous
damping, two differences are apparent: First, resonance (maximum amplitude) occurs at
ω = ωn , not at ω < ωn . Second, the phase angle for ω = 0 is φ = tan−1 η instead of zero
for viscous damping; this implies that motion with rate-independent damping can never be
in phase with the forcing function.
These differences between forced vibration with rate-independent damping and forced
vibration with viscous damping are not significant, but they are the source of some diffi-
culty in reconciling physical data. In most damped vibration, damping is not viscous, and
to assume that it is without knowing its real physical characteristics is an assumption of
some error. In the next section this error is shown to be small when the real damping is
rate independent.

3.10.3 Solution Using Equivalent Viscous Damping

In this section an approximate solution for the steady-state harmonic response of a system
with rate-independent damping is obtained by modeling this damping mechanism as equiv-
alent viscous damping.
Matching dissipated energies at ω = ωn led to Eq. (3.9.2), where E D is given by
Eq. (3.10.2), leading to the equivalent viscous damping ratio:
η
ζeq = (3.10.8)
2
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108 Response to Harmonic and Periodic Excitations Chap. 3

5
η=ζ=0
ζ = 0.1 η = 0.2

4
Deformation response factor Rd = uo / (ust)o

Rate-independent damping
Equivalent viscous damping
3

ζ = 0.2 η = 0.4

0
180°
ζ = 0.1
ζ = 0.2
η = 0.2
Phase angle φ

η = 0.4
90°

η=ζ=0
0°
0 1 2 3
Frequency ratio ω / ωn

Figure 3.10.3 Response of system with rate-independent damping: exact solution and
approximate solution using equivalent viscous damping.

Substituting this ζeq for ζ in Eqs. (3.2.10) to (3.2.12) gives the system response. The
resulting amplitude u o and phase angle φ are shown by the dashed lines in Fig. 3.10.3.
This approximate solution matches the exact result at ω = ωn because that was the cri-
terion used in selecting ζeq (Fig. 3.10.1). Over a wide range of excitation frequencies the
approximate solution is seen to be accurate enough for many engineering applications.
Thus Eq. (3.10.3)—which is difficult to solve for arbitrary force p(t) that contains many
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Sec. 3.11 Harmonic Vibration with Coulomb Friction 109

harmonic components of different frequencies ω—can be replaced by the simpler Eq. (3.2.1)
for a system with equivalent viscous damping defined by Eq. (3.10.8). This is the basic ad-
vantage of equivalent viscous damping.

3.11 HARMONIC VIBRATION WITH COULOMB FRICTION

3.11.1 Equation of Motion

Shown in Fig. 3.11.1 is a mass–spring system with Coulomb friction force F = µN that
opposes sliding of the mass. As defined in Section 2.4, the coefficients of static and kinetic
friction are assumed to be equal to µ, and N is the normal force across the sliding surfaces.
The equation of motion is obtained by including the exciting force in Eqs. (2.4.1) and
(2.4.2) governing the free vibration of the system:
m ü + ku ± F = p(t) (3.11.1)
The sign of the friction force changes with the direction of motion; the positive sign applies
if the motion is from left to right (u̇ > 0) and the negative sign is for motion from right
to left (u̇ < 0). Each of the two differential equations is linear, but the overall problem is
nonlinear because the governing equation changes every half-cycle of motion. Therefore,
exact analytical solutions would not be possible except in special cases.

k
m p(t)
Figure 3.11.1 SDF system with Coulomb
Friction force ± F
friction.

3.11.2 Steady-State Response to Harmonic Force

An exact analytical solution for the steady-state response of the system of Fig. 3.11.1
subjected to harmonic force was developed by J. P. Den Hartog in 1933. The analysis
is not included here, but his results are shown by solid lines in Fig. 3.11.2; the dashed lines
are described in the next section. The displacement amplitude u o , normalized relative to
(u st )o = po /k, and the phase angle φ are plotted as a function of the frequency ratio ω/ωn
for three values of F/ po . If there is no friction, F = 0 and u o /(u st )o = (Rd )ζ =0 , the same
as in Eq. (3.1.11) for an undamped system. The friction force reduces the displacement
amplitude u o with the reduction depending on the frequency ratio ω/ωn .
At ω = ωn the amplitude of motion is not limited by Coulomb friction if
F π
< (3.11.2)
po 4
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110 Response to Harmonic and Periodic Excitations Chap. 3

3.5

0 Coulomb friction
Equivalent viscous damping
3.0 0.5
F / po = 0
Deformation response factor Rd = uo / (ust)o

0.7
2.5 0.5
π/4

2.0
0.7

1.5

0.8
1.0

0.5
π/4

0.0
180°
0.5
Phase angle φ

0.7
90°
0.8
0.7

0.5 F / po = 0
0°
0.0 0.5 1.0 1.5 2.0
Frequency ratio ω / ωn

Figure 3.11.2 Deformation response factor and phase angle of a system with Coulomb
friction excited by harmonic force. Exact solution from J. P. Den Hartog; approximate
solution is based on equivalent viscous damping.

which is surprising since F = (π/4) po represents a large friction force, but can be ex-
plained by comparing the energy E F dissipated in friction against the input energy E I .
The energy dissipated by Coulomb friction in one cycle of vibration with displacement
amplitude u o is the area of the hysteresis loop enclosed by the friction force–displacement
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Sec. 3.11 Harmonic Vibration with Coulomb Friction 111

diagram (Fig. 3.11.3):

E F = 4Fu o (3.11.3)

Friction
force

−uo uo Displacement

−F
Figure 3.11.3 Hysteresis loop for
Coulomb friction.

Observe that the dissipated energy in a vibration cycle is proportional to the ampli-
tude of the cycle. The energy E I input by the harmonic force applied at ω = ωn is also
proportional to the displacement amplitude. If Eq. (3.11.2) is satisfied, it can be shown that

EF < EI

that is, the energy dissipated in friction per cycle is less than the input energy (Fig. 3.11.4).
Therefore, the displacement amplitude would increase cycle after cycle and grow with-
out bound. This behavior is quite different from that of systems with viscous damping or
rate-independent damping. For these forms of damping, as shown in Section 3.8, the dis-
sipated energy increases quadratically with displacement amplitude, and the displacement
amplitude is bounded no matter how small the damping. In connection with the fact that
infinite amplitudes occur at ω = ωn if Eq. (3.11.2) is satisfied, the phase angle shows a
discontinuous jump at ω = ωn (Fig. 3.11.2).

EI
Energy

EF

Figure 3.11.4 Input energy E I and energy

Amplitude dissipated E F by Coulomb friction.
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112 Response to Harmonic and Periodic Excitations Chap. 3

3.11.3 Solution Using Equivalent Viscous Damping

In this section an approximate solution for the steady-state harmonic response of a system
with Coulomb friction is obtained by modeling this damping mechanism by equivalent
viscous damping. Substituting E F , the energy dissipated by Coulomb friction given by
Eq. (3.11.3), for E D in Eq. (3.9.1) provides the equivalent viscous damping ratio
2 1 uF
ζeq = (3.11.4)
π ω/ωn u o
where u F = F/k. The approximate solution for the displacement amplitude u o is obtained
by substituting ζeq for ζ in Eq. (3.2.11):
uo 1
=  2 1/2
(u st )o
1 − (ω/ωn )2 + [(4/π )(u F /u o )]2

This contains u o on the right side also. Squaring and solving algebraically, the normalized
displacement amplitude is
 1/2
uo 1 − [(4/π )(F/ po )]2
= (3.11.5)
(u st )o 1 − (ω/ωn )2
This approximate result is valid provided that F/ po < π/4. The approximate solution
cannot be used if F/ po > π/4 because then the quantity under the radical is negative and
the numerator is imaginary.
These approximate and exact solutions are compared in Fig. 3.11.2. If the friction
force is small enough to permit continuous motion, this motion is practically sinusoidal
and the approximate solution is close to the exact solution. If the friction force is large,
discontinuous motion with stops and starts results, which is much distorted relative to a
sinusoid, and the approximate solution is poor.
The approximate solution for the phase angle is obtained by substituting ζeq for ζ in
Eq. (3.2.12):
(4/π )(u F /u o )
tan φ =
1 − (ω/ωn )2
Substituting for u o from Eq. (3.11.5) gives
(4/π )(F/ po )
tan φ = ±  1/2 (3.11.6)
1 − [(4/π )(F/ po )]2
For a given value of F/ po the tan φ is constant but with a positive value if ω/ωn < 1 and
a negative value if ω/ωn > 1. This is shown in Fig. 3.11.2, where it is seen that the phase
angle is discontinuous at ω = ωn for Coulomb friction.
Example 3.7
The structure of Example 2.7 with friction devices deflects 2 in. under a lateral force of
p = 500 kips. What would be the approximate amplitude of motion if the lateral force is
replaced by the harmonic force p(t) = 500 sin ωt, where the forcing period T = 1 sec?
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Sec. 3.11 Harmonic Vibration with Coulomb Friction 113

Solution The data (given and from Example 2.7) are

po
(u st )o = = 2 in. u F = 0.15 in.
k
ω Tn 0.5
= = = 0.5
ωn T 1
Calculate u o from Eq. (3.11.5).
F F/k uF 0.15
= = = = 0.075
po po /k (u st )o 2

Substituting for F/ po in Eq. (3.11.5) gives

 1/2
uo 1 − [(4/π)0.075]2
= = 1.327
(u st )o 1 − (0.5)2
u o = 1.327(2) = 2.654 in.

PART D: RESPONSE TO PERIODIC EXCITATION

A periodic function is one in which the portion defined over T0 repeats itself indefinitely
(Fig. 3.12.1). Many forces are periodic or nearly periodic. Under certain conditions,
propeller forces on a ship, wave loading on an offshore platform, and wind forces induced
by vortex shedding on tall, slender structures are nearly periodic. Earthquake ground mo-
tion usually has no resemblance to a periodic function. However, the base excitation arising
from an automobile traveling on an elevated freeway that has settled because of long-term
creep may be nearly periodic.

T0 T0 T0
• • • •

Figure 3.12.1 Periodic excitation.

We are interested in analyzing the response to periodic excitation for yet another
reason. The analysis can be extended to arbitrary excitations utilizing discrete Fourier
transform techniques. These are introduced in Appendix A of this book.
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114 Response to Harmonic and Periodic Excitations Chap. 3

3.12 FOURIER SERIES REPRESENTATION

A function p(t) is said to be periodic with period T0 if it satisfies the following relation-
ship:
p(t + j T0 ) = p(t) j = −∞, . . . , −3, −2, −1, 0, 1, 2, 3, . . . , ∞
A periodic function can be separated into its harmonic components using the Fourier
series:
∞ 
∞
p(t) = a0 + a j cos( jω0 t) + b j sin( jω0 t) (3.12.1)
j=1 j=1

where the fundamental harmonic in the excitation has the frequency

2π
ω0 = (3.12.2)
T0
The coefficients in the Fourier series can be expressed in terms of p(t) because the sine
and cosine functions are orthogonal:

 T0
1
a0 = p(t) dt (3.12.3)
T0 0
 T0
2
aj = p(t) cos( jω0 t) dt j = 1, 2, 3, . . . (3.12.4)
T0 0
 T0
2
bj = p(t) sin( jω0 t) dt j = 1, 2, 3, . . . (3.12.5)
T0 0

The coefficient a0 is the average value of p(t); coefficients a j and b j are the amplitudes of
the jth harmonics of frequency jω0 .
Theoretically, an infinite number of terms is required for the Fourier series to con-
verge to p(t). In practice, however, a few terms are sufficient for good convergence. At
a discontinuity, the Fourier series converges to a value that is the average of the values
immediately to the left and to the right of the discontinuity.

3.13 RESPONSE TO PERIODIC FORCE

A periodic excitation implies that the excitation has been in existence for a long time, by
which time the transient response associated with the initial displacement and velocity has
decayed. Thus, just as for harmonic excitation, we are interested in finding the steady-state
response. The response of a linear system to a periodic force can be determined by com-
bining the responses to individual excitation terms in the Fourier series.
The response of an undamped system to constant force p(t) = a0 is given by Eq. (f)
of Example 1.5, in which the cos ωt term will decay because of damping (see Section 4.3),
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Sec. 3.13 Response to Periodic Force 115

leaving the steady-state solution.†

a0
u 0 (t) = (3.13.1)
k
The steady-state response of a viscously damped SDF system to harmonic
cosine force p(t) = a j cos( jω0 t) is given by Eqs. (3.2.3) and (3.2.26) with ω replaced
by jω0 :
a j 2ζβ j sin( jω0 t) + (1 − β j ) cos( jω0 t)
2
u cj (t) = (3.13.2)
k (1 − β j2 )2 + (2ζβ j )2
where
jω0
βj = (3.13.3)
ωn
Similarly, the steady-state response of the system to sinusoidal force p(t) = b j sin( jω0 t)
is given by Eqs. (3.2.3) and (3.2.4) with ω replaced by jω0 :
b j (1 − β j ) sin( jω0 t) − 2ζβ j cos( jω0 t)
2
u sj (t) = (3.13.4)
k (1 − β j2 )2 + (2ζβ j )2
If ζ = 0 and one of β j = 1, the steady-state response is unbounded and not meaningful
because the transient response never decays (see Section 3.1); in the following it is assumed
that ζ = 0 and β j = 1.
The steady-state response of a system with damping to periodic excitation p(t) is the
combination of responses to individual terms in the Fourier series:

∞ 
∞
u(t) = u 0 (t) + u cj (t) + u sj (t) (3.13.5)
j=1 j=1

Substituting Eqs. (3.13.1), (3.13.2), and (3.13.4) into (3.13.5) gives

a0  ∞
1 1  
u(t) = + a j (2ζβ j ) + b j (1 − β 2
j ) sin( jω0 t)
k j=1
k (1 − β j )2 + (2ζβ j )2
2

  
+ a j (1 − β j2 ) − b j (2ζβ j ) cos( jω0 t) (3.13.6)

The response u(t) is a periodic function with period T0 .

The relative contributions of the various harmonic terms in Eq. (3.13.6) depend on
two factors: (1) the amplitudes a j and b j of the harmonic components of the forcing func-
tion p(t), and (2) the frequency ratio β j . The response will be dominated by those har-
monic components for which β j is close to unity [i.e., the forcing frequency jω0 is close
to the natural frequency (see Fig. 3.2.6)].

† The notation u used here includes the subscript zero consistent with a ; this should not be confused with
0 0
u o with the subscript “oh” used earlier to denote the maximum value of u(t).
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116 Response to Harmonic and Periodic Excitations Chap. 3

Example 3.8
The periodic force shown in Fig. E3.8a is defined by

po 0 ≤ t ≤ T0 /2
p(t) = (a)
− po T0 /2 ≤ t ≤ T0

t / T0
0 1 2

(a)

j=1 j=3
1.5 2 j=1
uj(t) / (ust)o

j=5
pj(t) / po

j=3 j=7 j=7

0 0

j=5
-1.5 (b) -2 (d)

Three terms
Four terms
1.5 Three terms Four terms 2
u(t) / (ust)o
p(t) / po

0 0

-1.5 -2

0 0.25 0.5 0 0.25 0.5

t / T0 t / T0
(c) (e)

Figure E3.8
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Sec. 3.13 Response to Periodic Force 117

Substituting this in Eqs. (3.12.3) to (3.12.5) gives the Fourier series coefficients:

 T0
1
a0 = p(t) dt = 0 (b)
T0 0
 T0
2
aj = p(t) cos( jω0 t) dt
T0 0
  T0 /2  T0 
2
= po cos( jω0 t) dt + (− po ) cos( jω0 t) dt = 0 (c)
T0 0 T0 /2

 T0
2
bj = p(t) sin( jω0 t) dt
T0 0
  T0 /2  T0 
2
= po sin( jω0 t) dt + (− po ) sin( jω0 t) dt
T0 0 T0 /2

0 j even
= 4 po / jπ j odd (d)

Thus the Fourier series representation of p(t) is

 4 po  1
∞
p(t) = p j (t) = sin( jω0 t) (e)
π j
j=1,3,5

The first four terms of this series are shown in Fig. E3.8b, where the frequencies and
relative amplitudes—1, 13 , 15 , and 17 —of the four harmonics are apparent. The cumulative sum
of the Fourier terms is shown in Fig. E3.8c, where four terms provide a reasonable represen-
tation of the forcing function. At t = T0 /2, where p(t) is discontinuous, the Fourier series
converges to zero, the average value of p(T0 /2).
The response of an SDF system to the forcing function of Eq. (e) is obtained by substi-
tuting Eqs. (b), (c), and (d) in Eq. (3.13.6) to obtain

4  1 (1 − β j ) sin( jω0 t) − 2ζβ j cos( jω0 t)

∞ 2
u(t) = (u st )o (f)
π j (1 − β j2 )2 + (2ζβ j )2
j=1,3,5

Shown in Fig. E3.8d are the responses of an SDF system with natural period Tn = T0 /4 and
damping ratio ζ = 5% to the first four loading terms in the Fourier series of Eq. (e). These
are plots of individual terms in Eq. (f) with β j = jω0 /ωn = j Tn /T0 = j/4. The relative
amplitudes of these terms are apparent. None of them is especially large because none of the
β j values is especially close to unity; note that β j = 14 , 34 , 54 , 74 , and so on. The cumulative
sum of the individual response terms of Eq. (f) is shown in Fig. E3.8e, where the contribution
of the fourth term is seen to be small. The higher terms would be even smaller because the
amplitudes of the harmonic components of p(t) decrease with j and β j would be even farther
from unity.
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118 Response to Harmonic and Periodic Excitations Chap. 3

FURTHER READING

Blake, R. E., “Basic Vibration Theory,” Chapter 2 in Shock and Vibration Handbook, 3rd ed. (ed.
C. M. Harris), McGraw-Hill, New York, 1988.
Hudson, D. E., Reading and Interpreting Strong Motion Accelerograms, Earthquake Engineering
Research Institute, Berkeley, Calif., 1979.
Jacobsen, L. S., and Ayre, R. S., Engineering Vibrations, McGraw-Hill, New York, 1958, Sec-
tion 5.8.

APPENDIX 3: FOUR-WAY LOGARITHMIC GRAPH PAPER

Rv is plotted as a function of ω/ωn on log-log graph paper [i.e., log Rv is the ordinate and
log(ω/ωn ) the abscissa]. Equation (3.2.21) gives
ω
log Rv = log + log Rd (A3.1)
ωn
If Rd is a constant, Eq. (A3.1) represents a straight line with slope of +1. Grid lines
showing constant Rd would therefore be straight lines of slope +1, and the Rd -axis would
be perpendicular to them (Fig. A3.1). Equation (3.2.21) also gives
ω
log Rv = − log + log Ra (A3.2)
ωn

If Ra is a constant, Eq. (A3.2) represents a straight line with slope of −1. Grid lines
showing constant Ra would be straight lines of slope −1, and the Ra -axis would be per-
pendicular to them (Fig. A3.1).
With reference to Fig. A3.1, the scales are established as follows:

1. With the point (Rv = 1, ω/ωn = 1) as the origin, draw a vertical Rv -axis and a
horizontal ω/ωn -axis with equal logarithmic scales.
2. The mark A on the Ra -axis would be located at the point (Rv = A1/2 , ω/ωn = A1/2 )
in order to satisfy
ω
Ra = Rv (A3.3)
ωn
Rv and ω/ωn are taken to be equal because the Ra -axis has a slope of +1. This
procedure is shown for A = 9, leading to the scale marks 3 on the Rv and ω/ωn
axes.
3. The mark D on the Rd -axis would be located at the point (Rv = D 1/2 , ω/ωn =
D −1/2 ) in order to satisfy
ω
Rd = Rv ÷ (A3.4)
ωn
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Chap. 3 Problems 119

10
Grid lines of Grid lines of

50

50
constant Rd constant Ra

a
R
5

ef

r
or

to
m

ac
at

ef
io

10

10

ns
n
A1/2 A

o
3

re

sp
sp
Velocity response factor Rv

re
o
ns

n
5

io
ef
D1/2

at
2 D

ac

er
to

el
cc
rRd

A
1

0.

5
0.
5
0.5
0.

1
0.
1
0.

05
05

0.
D−1/2 A1/2
0.1
0.1 0.5 1 3 5 10
Frequency ratio ω / ωn

Figure A3.1 Construction of four-way logarithmic graph paper.

and the condition that the Rd -axis has a slope of −1. This procedure is shown for
D = 4, leading to the scale mark 2 on the Rv -axis and to the scale mark 12 on the
ω/ωn -axis.

The logarithmic scales along the Rd and Ra axes are equal but not the same as the
Rv and ω/ωn scales.

PROBLEMS

Part A

3.1 The mass m, stiffness k, and natural frequency ωn of an undamped SDF system are unknown.
These properties are to be determined by harmonic excitation tests. At an excitation frequency
of 4 Hz, the response tends to increase without bound (i.e., a resonant condition). Next, a
weight w = 5 lb is attached to the mass m and the resonance test is repeated. This time
resonance occurs at f = 3 Hz. Determine the mass and the stiffness of the system.
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120 Response to Harmonic and Periodic Excitations Chap. 3

3.2 An SDF system is excited by a sinusoidal force. At resonance the amplitude of displacement
was measured to be 2 in. At an exciting frequency of one-tenth the natural frequency of the
system, the displacement amplitude was measured to be 0.2 in. Estimate the damping ratio of
the system.
3.3 In a forced vibration test under harmonic excitation it was noted that the amplitude of motion
at resonance was exactly four times the amplitude at an excitation frequency 20% higher than
the resonant frequency. Determine the damping ratio of the system.
3.4 A machine is supported on four steel springs for which damping can be neglected. The natural
frequency of vertical vibration of the machine–spring system is 200 cycles per minute. The
machine generates a vertical force p(t) = p0 sin ωt. The amplitude of the resulting steady-
state vertical displacement of the machine is u o = 0.2 in. when the machine is running at 20
revolutions per minute (rpm), 1.042 in. at 180 rpm, and 0.0248 in. at 600 rpm. Calculate the
amplitude of vertical motion of the machine if the steel springs are replaced by four rubber
isolators which provide the same stiffness, but introduce damping equivalent to ζ = 25% for
the system. Comment on the effectiveness of the isolators at various machine speeds.
3.5 An air-conditioning unit weighing 1200 lb is bolted at the middle of two parallel simply
supported steel beams (Fig. P3.5). The clear span of the beams is 8 ft. The second mo-
ment of cross-sectional area of each beam is 10 in4 . The motor in the unit runs at 300 rpm
and produces an unbalanced force of 60 lb at this speed. Neglect the weight of the beams and
assume 1% viscous damping in the system; for steel E = 30, 000 ksi. Determine the am-
plitudes of steady-state deflection and steady-state acceleration (in g’s) of the beams at their
midpoints which result from the unbalanced force.

Air-conditioning unit
Steel beams

4′ 4′
• • • Figure P3.5

3.6 (a) Show that the steady-state response of an SDF system to a cosine force, p(t) = po cos ωt,
is given by
 
po 1 − (ω/ωn ) cos ωt + [2ζ (ω/ωn )] sin ωt
2
u(t) =  2
k 1 − (ω/ω )2 + [2ζ (ω/ω )]2
n n

(b) Show that the maximum deformation due to cosine force is the same as that due to sinu-
soidal force.
3.7 (a) Show that ωr = ωn (1 − 2ζ 2 )1/2 is the resonant frequency for displacement amplitude of
an SDF system.
(b) Determine the displacement amplitude at resonance.
3.8 (a) Show that ωr = ωn (1 − 2ζ 2 )−1/2 is the resonant frequency for acceleration amplitude of
an SDF system.
(b) Determine the acceleration amplitude at resonance.
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Chap. 3 Problems 121

3.9 (a) Show that ωr = ωn is the resonant frequency for velocity amplitude of an SDF system.
(b) Determine the velocity amplitude at resonance.

Part B

3.10 A one-story reinforced concrete building has a roof mass of 500 kips/g, and its natural fre-
quency is 4 Hz. This building is excited by a vibration generator with two weights, each 50 lb,
rotating about a vertical axis at an eccentricity of 12 in. When the vibration generator runs
at the natural frequency of the building, the amplitude of roof acceleration is measured to be
0.02g. Determine the damping of the structure.
3.11 The steady-state acceleration amplitude of a structure caused by an eccentric-mass vibration
generator was measured for several excitation frequencies. These data are as follows:

Frequency (Hz) Acceleration (10−3 g) Frequency (Hz) Acceleration (10−3 g)

1.337 0.68 1.500 7.10
1.378 0.90 1.513 5.40
1.400 1.15 1.520 4.70
1.417 1.50 1.530 3.80
1.438 2.20 1.540 3.40
1.453 3.05 1.550 3.10
1.462 4.00 1.567 2.60
1.477 7.00 1.605 1.95
1.487 8.60 1.628 1.70
1.493 8.15 1.658 1.50
1.497 7.60

Determine the natural frequency and damping ratio of the structure.

3.12 Consider an industrial machine of mass m supported on spring-type isolators of total stiffness
k. The machine operates at a frequency of f hertz with a force unbalance po .
(a) Determine an expression giving the fraction of force transmitted to the foundation as a
function of the forcing frequency f and the static deflection δst = mg/k. Consider only the
steady-state response.
(b) Determine the static deflection δst for the force transmitted to be 10% of po if f = 20 Hz.
3.13 For the automobile in Example 3.4, determine the amplitude of the force developed in the
spring of the suspension system when the automobile is traveling at 40 mph.
3.14 Determine the speed of the automobile in Example 3.4 that would produce a resonant condi-
tion for the spring force in the suspension system.
3.15 A vibration isolation block is to be installed in a laboratory so that the vibration from adjacent
factory operations will not disturb certain experiments (Fig. P3.15). If the isolation block
weighs 2000 lb and the surrounding floor and foundation vibrate at 1500 cycles per minute,
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122 Response to Harmonic and Periodic Excitations Chap. 3

determine the stiffness of the isolation system such that the motion of the isolation block is
limited to 10% of the floor vibration; neglect damping.

Isolation block

Figure P3.15

3.16 An SDF system is subjected to support displacement u g (t) = u go sin ωt. Show that the
amplitude u to of the total displacement of the mass is given by Eq. (3.6.5).
3.17 The natural frequency of an accelerometer is 50 Hz, and its damping ratio is 70%. Com-
pute the recorded acceleration as a function of time if the input acceleration is ü g (t) =
0.1g sin(2πft) for f = 10, 20, and 40 Hz. A comparison of the input and recorded
accelerations was presented in Fig. 3.7.3. The accelerometer is calibrated to read the input
acceleration correctly at very low values of the excitation frequency. What would be the error
in the measured amplitude at each of the given excitation frequencies?
3.18 An accelerometer has the natural frequency f n = 25 Hz and damping ratio ζ = 60%. Write an
equation for the response u(t) of the instrument as a function of time if the input acceleration
is ü g (t) = ü go sin(2π f t). Sketch the ratio ωn2 u o /ü go as a function of f / f n . The accelerom-
eter is calibrated to read the input acceleration correctly at very low values of the excitation
frequency. Determine the range of frequencies for which the acceleration amplitude can be
measured with an accuracy of ±1%. Identify this frequency range on the above-mentioned
plot.
3.19 The natural frequency of an accelerometer is f n = 50 Hz, and its damping ratio is ζ = 70%.
Solve Problem 3.18 for this accelerometer.
3.20 If a displacement-measuring instrument is used to determine amplitudes of vibration at fre-
quencies very much higher than its own natural frequency, what would be the optimum instru-
ment damping for maximum accuracy?
3.21 A displacement meter has a natural frequency f n = 0.5 Hz and a damping ratio ζ = 0.6.
Determine the range of frequencies for which the displacement amplitude can be measured
with an accuracy of ±1%.
3.22 Repeat Problem 3.21 for ζ = 0.7.
3.23 Show that the energy dissipated per cycle for viscous damping can be expressed by

πpo2 2ζ (ω/ωn )
ED =
k 1 − (ω/ω )2 2 + [2ζ (ω/ω )]2

n n

3.24 Show that for viscous damping the loss factor ξ is independent of the amplitude and propor-
tional to the frequency.
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Chap. 3 Problems 123

Part C

3.25 The properties of the SDF system of Fig. P2.20 are as follows: w = 500 kips, F = 50 kips,
and Tn = 0.25 sec. Determine an approximate value for the displacement amplitude due to
harmonic force with amplitude 100 kips and period 0.30 sec.

Part D

3.26 An SDF system with natural period Tn and damping ratio ζ is subjected to the periodic force
shown in Fig. P3.26 with an amplitude po and period T0 .
(a) Expand the forcing function in its Fourier series.
(b) Determine the steady-state response of an undamped system. For what values of T0 is the
solution indeterminate?
(c) For T0 /Tn = 2, determine and plot the response to individual terms in the Fourier series.
How many terms are necessary to obtain reasonable convergence of the series solution?

po

t
−To −To/2 0 To/2 To 2To Figure P3.26
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