4.

Harmonically Excited Vibrations

1. Introduction
A dynamic system is often subjected to some type of external force or
excitation, called the forcing or exciting function. This excitation is usually timedependent. It may be harmonic, nonharmonic but periodic, nonperiodic, or
random in nature. The response of a system to a harmonic excitation is called
harmonic response. The nonperiodic excitation may have a long or short
duration. The response of a dynamic system to suddenly applied nonperiodic
excitation is called transient response.
We shall consider the dynamic response of a single degree of freedom system
under harmonic excitations of the form F(t) = F0ei(t + ) or F(t) = F0cos(t + ) or
F0sin(t + ), where F0 is the amplitude, is the frequency, and is the phase
angle of the harmonic excitation. The value of depends on the value of F(t) at t
= 0 and is usually taken to be zero. Under a harmonic excitation, the response of
the system will also be harmonic. If the frequency of excitation coincides with
the natural frequency of the system, the response of the system will be very
large. This condition, known as resonance, is to be avoided to prevent failure of
the system.
2. Equation of Motion
If a force F(t) acts on a viscously damped spring-mass system as shown in
Fig. 1, the equation of motion can be obtained using Newtons second law:

mx cx kx F t
(1)
Since this equation is nonhomogeneous, its general solution x(t) is given by the
sum of the homogeneous solution, xh(t), and the particular solution, xp(t). The
homogeneous solution, which is the solution of the homogeneous equation
mx cx kx 0
(2)
Represents the free vibration of the system and was discussed. This free
vibration dies out with time under each of the three possible conditions of
damping (underdamping, critical damping, and overdamping), and under all
1

possible initial conditions. Thus the general solution of Equation (1) eventually
reduces to the particular solution xp(t), which represents the steady-state
vibration. The steady-state motion is percent as long as the forcing function is
present. The variations of homogeneous, particular, and general solutions with
time for a typical case are shown in Fig. 2.

It can be seen that xh(t) dies out and x(t) becomes xp(t) after some time ( in Fig.
2). The part of the motion that dies out due to damping (the free vibration part)
is called transient. The rate at which the transient motion decays depends on the
values of the system parameters k, c, and m. We ignore the transient motion and
derive only the particular solution of Equation (1), which represents the steadystate response, under harmonic forcing functions.
3. Response of an Undamped System Under Harmonic Force
If a force F(t) = F0cost acts on the mass m of undamped system, the
equation of motion, Eq. (1) reduces to
mx kx F0 cost
(3)
The homogeneous solution of this equation is given by
xp t C1 cos nt C2 sin nt
(4)
where n = (k/m) is the natural frequency of the system. Because the exciting
force F(t) is harmonic, the particular solution xp(t) is also harmonic and has the
same frequency . Thus we assume a solution in the form
xp t Xcosnt
1/2

(5)
where X is a constant that denotes the maximum amplitude of xp(t). By
substituting Equation (5) into Equation (3) and solving for X, we obtain
F0
X
k m 2
(6)
Thus the total solution of Equation (3) is

x t C1 cos nt C2 sin nt

F0
cost
k m 2

Using the initial conditions x(t = 0) = x0 and

C1 x0

F0
k m 2

C2

(7)
(t = 0) =

, we find out

x 0
n

(8)

and hence

x
F0
F0

x t x0
cosnt 0 sinnt
cost
2
k m
k m 2

(9)
The maximum amplitude X in Equation (6) can also be expressed as
X
1

2
st

1
n
(10)
where st = F0/k denotes the deflection of the mass under a force F0 and is
sometimes called static deflection since F0 is a constant (static) force. The
quantity X/st, represents the ratio of the dynamic to the static amplitude of
motion and is called the magnification factor, amplification factor, or amplitude
ratio. The variation of the amplitude ratio, X/st, with the frequency ratio r = /n
(Equation 10) is shown in Figure 3). From this figure, the response of the system
can be identified to be of three types.

Case 1. When 0 < /n < 1, the denominator in Equation (10) is positive and the
response is given by Equation (5) without change. The harmonic response of the
system xp(t) is said to be in phase with the external force as shown in Figure 4.

Case 2. When /n > 1, the denominator in Equation (1) is negative, and the
steady-state solution can be expressed as
xp t Xcost
(11)
where the amplitude of motion X is redefined to be a positive quantity as
st
X
2

1
n
(12)
The variation of F(t) and xp(t) with time are shown in Figure 5.

Since xp(t) and F(t) have opposite signs, the response is said to be 180 o out of
phase with the external force. Further, as /n , X 0. Thus the response of
the system so a harmonic force of very high frequency is close to zero.
Case 3. When /n = 1, the amplitude X given by Equation (10) or (12) becomes
infinite. This condition, for which the forcing frequency is equal to the natural
frequency of the system n, is called resonance. To find the response for this
condition, we rewrite Equation (9) as

x 0
cost cosnt
x t x0 cosnt
sinnt st
2

(13)
Since the last term of this equation takes an indefinite form for = 0, we apply
LHospitals rule to evaluate the limit of this term:

d cost cosnt
cost cosnt
d

lim
2
lim
2

n
n

1 2

d
n

t sint t
n sinnt
lim

n
2
2
n2

(14)
Thus the response of the system at resonance becomes

x
t
x t x0 cosnt 0 sinnt st n sinnt
n
2
(15)
It can be seen from Equation (15) that at resonance, x(t) increases indefinitely.
The last term of Equation (15) is shown in Figure 6, from which the amplitude of
the response can be seen to increase linearly with time.

3.1Total response
The total response of the system, Equation (7) or Equation (9) can also be
expressed as
st
x t Acos nt
cost
2

1
1
n
n
; for
(16)
st
x t Acos nt
cost
2

1
1
n
n
; for
(17)
where A and can be determined. Thus the complete motion can be
expressed as the sum of two cosine curves of different frequencies. In
Equation (16), the forcing frequency is smaller than the natural frequency,
and the total response is shown in Figure 7(a).

In Equation (17), the forcing frequency, and the total response appears as
shown in Figure 7(b).
3.2Beating phenomenon
If the forcing frequency is close to, but not exactly equal to, the natural
frequency of the system, a phenomenon known as beating may occur. In this
kind of vibration, the amplitude builds up and then diminishes in a regular
pattern. The phenomenon of beating can be explained by considering the
solution given by Equation (9). If the initial conditions are taken as x0 =
= 0, Equation (9) reduces to
F m
x t 20 2 cost cosnt
n
x t

F0 m

2
n

n
n

2 sin
t sin
t

2
2

(18)
Let the forcing frequency be slightly less than the natural frequency:
n 2
(19)
where is a small positive quantity. Then n = and
n 2
(20)
8

Multiplication of Equations (19) and (20) gives

n2 2 2
(21)
Use of Equations (19) to (21) in Equation (18) gives
F m

x t 0 sint sint
2

(22)
Since is small, the function sin t varies slowly; its period, equal to 2 / is
large. Thus Equation (22) may be seen as representing vibration with period
2/ and of variable amplitude equal to
F m
x t 0 sint
2
It can also be observed that the sin wt curve will go through several cycles,
while the sin et wave goes through a single cycle, as shown in Figure 8.

Thus the amplitude builds up and dies down continuously. The time between
the points of zero amplitude or the points of maximum amplitude is called the
period of beating (b) and is given by
2
2
b

2 n
(23)
with the frequency of beating defined as
b 2 n
Example No. 1 (Plate Supporting a Pump)
A reciprocating pump, weighing 150 lb, is mounted at the middle of a steel plate
of thickness 0.5 in., width 20 in., and length 100 in., clamped along two edges as
shown in Figure 9. During operation of the pump, the plate is subjected to a
harmonic force, F(t) = 50cos 62.832t lb.

Given:
Pump weight = 150 lb
Plate dimension: thickness (t) = 0.5 in., width (w) = 20 in., and length (l) = 100
in.
Harmonic force = F(t) = 50cos 62.832 t lb.
Required:
Amplitude of vibration of the plate, X
Solution:
Plate modeled as a fixed-fixed beam having:
Youngs modulus (E) = 30 x 106 psi,
length (l) = 100 in, and
area moments of inertia () = (1/12)(20)(0.5)3 = 0.2083 in.4.
Bending stiffness of the beam:

192
E 19230 106 0.2083
k 3
1200
.0 lb in
3
l
100

The amplitude of harmonic response, by Equation (6) with:

F0 = 50 lb,
m = 150/386.4 lb-sec2/in. (neglecting the weight of the steel plate),
k = 1200.0 lb/in., and
= 62.832 rad/sec.
Thus Equation (6)
F0
50
X

0.1504in.
k m 2
150
2
1200
.0
62.832
386.4
The negative sign indicates that the response x(t) of the plate is out of phase
with the excitation F(t).
4. Response of a Damped System Under Harmonic Force
10

If the forcing function is given by F(t) = F0cost, the equation of motion becomes
mx cx kx F0 cost
(24)
The particular solution of Equation (24) is also expected to be harmonic; we
assume it in the form
xp t Xcos t
(25)
where X and are constants to be determined. X and denote the amplitude
and phase angle of the response, respectively. By substituting Equation (25) into
Equation (24), we arrive are
X k m 2 cos t c sin t F0 cost
(26)
Using the trigonometric relations
cos t cost cos sint sin

sin t sint cos cost sin

in Equation (26) and equating the coefficients of cos t and sint on both sides of
the resulting equation, we obtain
X k m 2 cos c sin F0

X k m 2 sin c cos 0

(27)

Solution of Equations (27) gives

F0
X
12
2
k m 2 c2 2

(28)

and
c
2
k m

tan1

(29)
By inserting the expressions of X and from Equations (28) and (29) into
Equation (25) we obtain the particular solution of Equation (24). Figure 10 shows
typical plots of the forcing function and (steady-state) response.

11

Dividing both the numerator and denominator of Equation (28) by k and making
the following substitutions
k
n
undamped
naturalfrequency
m

c
c
c

; 2 n
cc 2mn m

st
r

F0
deflection
underthestaticforceF0 ,and
k

frequency
ratio
n

we obtain
X

st

1
n

2 2

12

1 r

2 2

2r 2

(30)
and

tan1

tan1 2r

1 r

(31)
X/st is called the magnification factor, amplification factor, or amplitude ratio.
The variations of X/st and with the frequency ratio r and the damping ratio
are shown in Figure 11.

12

The following observations can be made from Equations (30) and (31) and from
Figure 11:
a. For an undamped system ( = 0), Equation (31) shows that the phase angle
= 0 (for r < 1) or 180o for (r > 1) and Equation (30) reduces to Equation (10).
b. The damping reduces the amplitude ratio for all values of the forcing
frequency.
c. The reduction of the amplitude ratio in the presence of damping is very
significant at or near resonance.
d. With damping, the maximum amplitude ratio occurs when
r 1 2 2

n 1 2 2

or
(32)
which is lower than the undamped natural frequency n and the damped

d n 1 2 2
natural frequency

.
r 1 2 2

e. The maximum value of X (when

X
1

st max 2 1 2
and the value of X at = n by

is given by

(33)

13

st

1
2

(34)
Equation (33) can be used for experimental determination of the measure of
damping present in the system. In a vibration test, if the maximum amplitude
of the response (X)max is measured, the damping ratio of the system can be
found using Equation (33). Conversely, if the amount of damping is known,
one can make an estimate of the maximum amplitude of vibration.
f.

For > 1/

, the graph of X has no peaks and for = 0, there is a

discontinuity at r = 1.
g. The phase angle depends on the system parameters m, c, and k and the
forcing frequency , but not on the amplitude F0 of the forcing function.
h. The phase angle by which the response x(t) or X lags the forcing function
F(t) or F0 will be very small for small values of r. For very large values of r, the
phase angle approaches 180o asymptotically. Thus the amplitude of vibration
will be in phase with the exciting force for r << and out of phase for r >> 1.
The phase angle at resonance will be 90 o for all values of damping ().
i. Below resonance ( < n), the phase angle increases with increase in
damping. Above resonance (> n), the phase angle decreases with increase
in damping.
4.1Total response
The complete solution is given by x(t) = xh(t) + xp(t). Thus
x t X0e nt cos dt 0 Xcos t
(35)
where

d 1 2 n
(36)

r
n

(37)
X and are given by Equations (30) and (31), respectively, and X0 and 0 can
be determined from the initial conditions.
4.2Quality factor and bandwidth
For small values of damping ( < 0.05), we can take
X
X
1

st max st
n

(38)
The value of the amplitude ratio at resonance is also called Q factor or quality
factor of the system, in analogy with some electrical-engineering
applications, such as the tuning circuit of a radio, where the interest lies in an
amplitude at resonance that is as large as possible. The points R1 and R2,
14

where the amplification factor falls to Q/

, are called half power points

because the power absorbed ( W) by the damper (or by the resistor in an

electrical circuit), responding harmonically at a given frequency, is
proportional to the square of the amplitude:
W cX2
(39)
The difference between the frequencies associated with the half power points
R1 and R2 is called the bandwidth of the system (see Figure 12).

To find the values of R1 and R2, we set X/st = Q/

1 r

2 2

or

2r 2

in Equation (3) so that

Q
1

2 2 2

r 4 r 2 2 4 2 1 8 2 0
(40)

The solution of Equation (40) gives

r12 1 2 2 2 1 2

r22 1 2 2 2 1 2

(41)
or small values of , Equation (41) can be approximated as
r12

R12

1
n

1 2

r22

R22

2
n

1 2

where 1 = /R1 and 2 = /R2. From Equation (42),

15

(42)

22 12 2 1 2 1 R22 R12 n2 4
(43)

using the relation

2 1 2n
(44)
Equation (43), we find that the bandwidth is given by
2 1 2 n
(45)
Combining Equation (38) and (45), we obtain
n
1
Q

2 2 1
(46)
It can be seen that the quality factor Q can be used for estimating the
equivalent viscous damping in a mechanical system.
5. Response of a Damped Systems Under F(t) = F0eit.
Let the harmonic forcing function be represented in complex form as F(t) =
F0eit so that the equation of motion becomes
mx cx kx F0eit
(47)
Since the actual excitation is given only by the real part of F(t), the response will
also be given only by the real part of x(t) where x(t) is a complex quantity
satisfying the differential equation (47). F0 in Equation (47) is, in general, a
complex number. By assuming the particular solution xp(t)
xp t Xeit
(48)
we obtain, by substituting Equation (48) into Equation (47),
F0
X
k m 2 ic
(49)
Multiplying the numerator and denominator on the right side of Equation (49) by
[(k m2) ic] and separating the real and imaginary parts, we obtain

k m 2
c
X F0
i

2
2
k m 2 c2 2
k m 2 c2 2
(50)

A x2 y2
and tan = y/x, Equation (50)

Using the relation, x + iy = Ai where

can be expressed as
F0
X
ei
12
2 2
2 2
k m c

(51)

where

16

m 2

tan1

(52)
Thus the steady-state solution, Equation (48), becomes
F0
xp t
ei t
12
2
2 2
k m c

(53)

Frequency Response
Equation (49) can be rewritten in the form
kX
1

H i
2
F0 1 r i2r
(54)
where H(i) is known as the complex frequency response of the system. The
absolute value of H(i) given by
kX
1
H i

12
2
F0
1 r2 2r 2

(55)
denotes the magnification factor defined in Equation (30). Recalling that ei =
cos + isin, we can show that Equations (54) and (55) are related:
H i H i e i
(56)
where is given by Equation (52), which can also be expressed as
2r
tan1
2
1 r
(57)
Thus Equation (53) can be expressed as
F
xp t 0 H i ei t
k
(58)
It can be seen that the complex frequency response function, H(i), contains
both the magnitude and phase of the steady state response. If F(t) = F0cost,
the corresponding steady-state solution is given by the real part of Equation
(53):
F0
xp t
cos t
12
2
k m 2 c 2

xp t Re 0 H i eit Re 0 H i ei t
k

(59)
Similarly, if F(t) = F0sint, the corresponding steady-state solution is given by the
imaginary part of Equation (53):

17

xp t

F0

k m c
2 2

12

sin t

xp t Im 0 H i eit Im 0 H i ei t
k

(60)

Complex Vector Representation of Harmonic Motion

The harmonic excitation and the response of the damped system to that
excitation can be represented graphically in the complex plane, and interesting
interpretation can be given to the resulting diagram. We first differentiate
Equation (58) with respect to time and obtain
F
p t i 0 H i ei t ixp t
velocity x
k
accelerati
on xp t i 2

F0
H i ei t 2xp t
k

(61)

Because i can be expressed as

i cos i sin ei
2
2

(62)
we can conclude that the velocity leads the displacement by the phase angle /2
and that it is multiplied by . Similarly, -1 can be written as
1 cos i sin e
(63)
Hence the acceleration leads the displacement by the phase angle , and it is
multiplied by 2.
Thus the various terms of the equation of motion (47) can be represented in
the complex plane, as shown in Fig. 13.

6. Response of a Damped System Under the Harmonic Motion of the Base

18

Sometimes the base or support of a spring-mass-damper system undergoes

harmonic motion, as shown in Fig. 14a.

Let y(t) denote the displacement of the base and x(t) the displacement of the
mass from its static equilibrium position at time t. Then the elongation of the
spring is x y and the relative velocity between the two ends of the damper is

y . From the free-body diagram shown in Fig. 14b, we obtain the

equation of motion:
mx c x y k x y 0
(64)

If y(t) = Ysint, Equation (64) becomes

mx cx kx Asint Bcost

(65)
where A = kY and B = cY. This shows that giving excitation to the base is
equivalent to applying harmonic force of magnitude (kYsint + cYcost) to the
mass. By using the solutions given in Eq. (59) and (60), the steady-state
response of the mass can be expressed as
kYsin t 1
cYcos t 1
xp t

12
12
2
2
k m 2 c 2
k m 2 c 2
(66)
The phase angle 1 will be the same for both the terms because it depends on
the values of m, c, k, and , but not on the amplitude of the excitation. Equation
(66) can be rewritten as
xp t Xcos t 1 2

xp t Y

k2 c 2

12

2
2
k m c
2

cos t 1 2

(67)
where the ratio of the amplitude of the response xp(t) to that of the base motion
y(t) is given by

19


X
k2 c 2

Y k m 2 2 c 2

12

1 2r 2

12

2
2
1 r 2r
2

(68)

and 1 and 2 by
c
2r
tan1
2
2
k m
1 r

1 tan1

1
k
1

tan
c
2r

2 tan1

(69)
The ratio X/Y is called the displacement transmissibility.
Note that if the harmonic excitation of the base is expressed in complex form
as y(t) = Re(Yeit), the response of the system can be expressed as
1 i2r it
Ye
xp t Re
2
1 r i2r

(70)
and the transmissibility as
12
X
1 2r 2 H i
Y
(71)
where |H(iw)| is given by Eq. (55).

6.1Force transmitted
In Fig. 14b, the force carried by the support F must be due to the spring
and dashpot which are connected to it. It can be determined as follows:
F k x y c x y mx
(72)
From Eq. 67, Eq. 72 can be written as
F m 2Xcos t 1 2 FT cos t 1 2
(73)
where FT is the amplitude or maximum value of the transmitted force given
by

FT
1 2r 2
r2

2
kY
1 r2 2r 2

12

(74)
The ratio (FT/kY) is known as the force transmissibility. It can be noticed that
the transmitted force is in phase with the motion of the mass x(t). The
variation of the force transmitted to the base with the frequency ratio r is
shown in Fig. 15 for different values of .

20

6.2Related motion
If z = x y denotes the motion of the mass relative to the base, the
equation of motion, Eq. (64), can be rewritten as
m 2Y sint
mz cz kz my
(75)
The steady-state solution of Eq. (75) is given by
m 2Y sin t 1
z t
Zsin t 1
2
k m 2 c 2
(76)
where Z, the amplitude of z(t), can be expressed as
m 2Y
r2
Z
Y
2
2
k m 2 c 2
1 r2 2r 2
(77)
and 1 by Eq. (69). The ratio Z/Y is shown graphically in Fig. 16.

21

Example No. 2 (Vehicle Moving on a Rough Road)

Figure 17(a) shows a simple model of a motor vehicle that can vibrate in the
vertical direction while traveling over a rough road. The vehicle has a mass of
1200 kg. The suspension system has a spring has a spring constant of 400 kN/m
and a damping ratio of = 0.5. If the vehicle speed is 100 km/hr, determine the
displacement amplitude of the vehicle. The road surface varies sinusoidally with
an amplitude of Y = 0.05 m and a wavelength of 6 m.

Given:
Vehicle model
m = 1200 kg, k = 400 kN/m, = 0.5, and speed = 100 km/hr.
Road surface: sinusoidal with Y = 0.05 m and wavelength = 6 m.
Required:
Displacement amplitude (X) of the vehicle.
Solution:
Model the vehicle as a single degree of freedom system subjected to base
motion as shown in Fig. 17b.

22

Frequency of the base excitation = the vehicle speed / length of one cycle of
road roughness (wavelength):
100kmhr 1000
m km 1

2f 2
rad sec
6 m 29.0887
3600
sechr

Natural frequency of the vehicle:

k
400 103

18.2574
rad sec
m
1200

and the frequency ratio r

29.0887
r

1.5933
n 18.2574
Amplitude ratio , Eq. 68.
X
1 2r 2

Y 1 r2 2 2r 2

12

2
1 2 0.5 1.5933

12

2
2
1 1.5933 2 0.5 1.5933
2

X
0.8493
Y
Thus, displacement amplitude of the vehicle:
X 0.8493
Y 0.8493
0.05 0.0425m
(answer)
Example No. 3
A heavy machine, weighing 3000 N, is supported on a resilient foundation. The
static deflection of the foundation due to the weight of the machine is found to
be 7.5 cm. It is observed that the machine vibrates with an amplitude of 1 cm
when the base of the foundation is subjected to harmonic oscillation at the
undamped natural frequency of the system with an amplitude of 0.25 cm. Find
(1) the damping constant of the foundation, (2) the dynamic force amplitude on
23

the base, and (3) the amplitude of the displacement of the machine relative to
the base.
Given:
Machine weight (W) = 3000 N, static deflection under W = 7.5 cm, and X = 1 cm
when v(t) = 0.25sinnt cm.
Required:
c, FT, and Z.
Solution:
(1) The stiffness of the foundation:
weightof machine 3000
k

40,000N m
st
0.075
At resonance ( = n or r = 1), Eq. (68)
1 2 2
X 0.010

4
2
Y 0.0025
2

12

0.1291
Damping constant:

2
c cc 2 km 0.1291

40,000 3000 903.0512N s m

9.81

(answer)
(2) Dynamic force amplitude on the base at r = 1 , Eq. (74).
1 4 2
FT Yk
2
4

12

kX 40,000 0.01 400N

(answer)
(3) Amplitude of the relative displacement of the machine at r = 1, Eq. (77).
Y
0.0025
Z

0.00968
m

2 2 0.1291
(answer)
It can be noticed that X = 0.01 m, Y = 0.0025 m, and Z = 0.00968 m;
therefore, Z X Y. This is due to the phase differences between x, y, and z.
7. Response of a Damped System Under Rotating Unbalance
Unbalance in rotating machinery is one of the main causes of vibration. A
simplified model of such machine is shown in Fig. 18.

24

The total mass of the machine is M, and there are two eccentric masses m/2
rotating in opposite directions with a constant angular velocity . The centrifugal
force (me2)/2 due to each mass will cause excitation of the mass M. We
consider two equal masses m/2 rotating in opposite directions in order to have
the horizontal components of excitation of the two masses cancel each other.
However, the vertical components of excitation add together and act along the
axis of symmetry A-A in Fig. 18. If the angular position of the masses is
measured from a horizontal position, the total vertical component of the
excitation is always given by F(t) = me2sint. The equation of motion can be
derived by the usual procedure:
kx me 2 sint
Mx cx
(78)
The solution of this equation will be identical to Eq. (60) if we replace m and F0
by M and me2 respectively. This solution can also be expressed as
me 2

H i ei t
xp t Xsin t Im
M n

(79)
n k M
where
and X and f denote the amplitude and the phase angle of
vibration given by
2

me

H i
X

1
2
2
M n
k M 2 c 2

me 2

(80)

and
c
2
k M

tan1

(81)

By defining z = c/cc and cc = 2Mn, Eqs. (80) and (81) can be rewritten as

25

MX
r2

r2 H i
12
2
2
me 1 r2 2r

(82)

and
2r
2
1 r

tan1

(83)
The variation of MX/me with r for different values of is shown in Fig. 16.

On the other hand, the graph of versus r remains as in Fig. 11b.

26

The following observations can be made from Eq. (82) and Fig. 16:
a. All the curves begin at zero amplitude. The amplitude near resonance ( =
n) is markedly affected by damping. Thus if the machine is to be run near
resonance, damping should be introduced purposely to avoid dangerous
amplitudes.
b. At very high speeds ( large). MX/me is almost unity, and the effect of
damping is negligible.
c. The maximum of MX/me occurs when
d MX

0
dr me
(84)
The solution of Eq. (84) gives
1
r
1
1 2 2
Accordingly, the peaks occur to the right of the resonance value of r = 1.
Example No. 4 (Francis Water Turbine)
The schematic diagram of a Francis water turbine is shown in Fig. 19 in which
water flows from A into the blades B and down into the tail race C. The rotor has
a mass of 250 kg and an unbalance (me) of 5 kgmm. The radial clearance
between the rotor and the stator is 5 mm. The turbine operates in the speed
range 600 to 6000 rpm. The steel shaft carrying the rotor can be assumed to be
clamped at the bearings. Determine the diameter of the shaft so that the rotor is
always clear of the stator at all the operating speeds of the turbine. Assume
damping to be negligible.

27

Given:
Turbine: mass (M) = 250 kg,
unbalance (me) = 5 kgmm, and
speed range = 600 - 6000 rpm.
Shaft length = 2 m and maximum radial deflection = 5 mm
Required:
Diameter of the shaft
Solution:
Equate the maximum amplitude (radial deflection) of rotor to 5 mm.
Maximum amplitude of the shaft (rotor) due to rotating unbalance, Eq. 80, set c
= 0:
me 2
X
k M 2
Limiting value of X = 5 mm
The value of ranges from
2
600rpm 600
20 rad sec
60
or
6000
rpm 6000

2
200
rad sec
60

Natural frequency of the system

k
k
n

0.0625k rad sec

M
250
If k is in N/m. For = 20 rad/sec
5.0 103 20 2
0.005
k 250 20 2

k 10.04 104 2 N m
For = 200 rad/sec
5.0 103 200
2
0.005
k 250 200
2

k 10.04 106 2 N m
From Fig. 16, the amplitude of vibration of the rotating shaft can be minimized by
making r = /n very large. This means that n must be made small compared to

28

 that is k must be made small. This can be achieved by selecting the value of
k as 10.04 x 1042 N/m.
Stiffness of a cantilever beam (shaft) supporting a load (rotor) at the end:
3EI 3E d4

k 3 3
l
l 64
Then
64kl3 64 10.04 104 2 2 3
d4

2.6005
104 m4
3E
3 2.07 1011

d 0.127m 127mm

(answer)
8. Forced Vibration with Coulomb Damping
For a single degree of freedom system with Coulomb or dry friction damping
subjected to a harmonic force F(t) = F0sinwt as in Fig. 20, the equation of motion
is given by
mx kx N F t F0 sint
(85)
where the sign of the friction force ( N) is positive (negative) when the mass
moves from left to right (right to left).

The exact solution of Eq. 85 is quite involved. However, we can except that if the
dry friction damping force is large, the motion of the mass will be discontinuous.
On the other hand, if the dry friction force is small compared to the amplitude of
the applied force F0, the steady state solution is expected to be nearly harmonic.
In this case, we can find an approximate solution of Eq. (85) by finding an
equivalent viscous damper during a full cycle of motion. If the amplitude of
motion is NX. Hence in a full cycle, the energy dissipated by dry friction
damping is given by
W 4NX
(86)
If the equivalent viscous damping constant is denoted as ceq, the energy
dissipated during a full cycle will be
W ceqX2
(87)
By equating Eqs. (86) and (87), we obtain

29

ceq

4N
X

(88)
Thus the steady-state response is given by
xp t Xsin t
(89)
where the amplitude X can be found from Eq. (60)
F0
F0 k
X

12
2
2

k m 2 ceq 2

2

1 2 2 eq
n
n

12

(90)

with

eq

ceq
cc

ceq
2mn

4N
2N

2mnX m nX
(91)

Substitution of Eq. (91) into Eq. (90) gives

F0 k
X
12
2
2

2
4N

1 2

n
kX

(92)

The solution of this equation gives the amplitude X as

2
4N
1

F0
F0

X
2 2
k
1

n2

12

(93)
As stated earlier, Eq. (93) can be used only if the friction force is small compared
to F0. In fact, the limiting value of the friction force N can be found from Eq.
(93). To avoid imaginary values of X, we need to have
4N

1
F0

F0 4

or
If this condition is not satisfied, the exact analysis is to be used.
The phase angle appearing in Eq. (89) can be found using Eq. (52):

4N

2 eq

ceq
n

1
1
1

kX
tan
tan
tan
2
2
2

k m
1 2
1 2
n
n

(94)
30

Substituting Eq. (93) into (94) for X, we obtain

4N

F0

tan1
2 12

1 4N

(95)
Equation (94) shows that tan is a constant for a given value of F0/N. F is
discontinuous at /n = 1 (resonance) since it takes a positive value for /n < 1
and a negative value for /n > 1. Thus Eq. (95) can also be expressed as

4N

F0
1

tan
2 12

1 4N

(96)
Equation (93) shows that friction serves to limit the amplitude of forced vibration
for /n 1. However, at resonance (/n = 1), the amplitude becomes infinite.
This can be explained as follows. The energy directed into the system over one
cycle when it is excited harmonically at resonance is
dx
W F dx F dt
cycle
0 dt

F0 sint Xcos t dt

(97)
Since Eq. (94) gives = 90o at resonance, Eq. (97) becomes
W F0X

sin2 tdt F0 X

(98)
The energy dissipated from the system is given by Eq. (86). SInce F0X > 4NX
for X to be real-valued, W > W at resonance (Fig. 21).

31

Thus more energy is directed into the system per cycle than is dissipated per
cycle. This extra energy is used to build up the amplitude of vibration. For the
nonresonant condition (/n 1), the energy input can be found from Eq. (97):

W F0 X

sint cos t dt F0Xsin

(99)
Due to the presence of sin in Eq. (99), the input energy curve in Fig. (21) is
made to coincide with the dissipated energy curve, so the amplitude is limited.
Thus the phase of the motion can be seen to limit the amplitude of the motion.
Example No. 5 (Spring-Mass System with Coulomb Damping
A spring-mass system, having a mass of 10 kg and a spring of stiffness of 4000
N/m, vibrates on a horizontal surface. The coefficient of friction is 0.12. When
subjected to a harmonic force of frequency 2 Hz, the mass is found to vibrate
with an amplitude of 40 mm. Find the amplitude of the harmonic force applied to
the mass.
Given:
Spring-mass system with Coulomb friction = m = 10 kg,
k = 4000 N/m, = 0.12,
harmonic force with frequency = 2 Hz, v
vibration amplitude = 40 mm
Required:
Amplitude of the applied force.
Solution:
Vertical force (weight) of the mass;
N = mg = 10 x 9.81 = 98.1 N.
Natural frequency:
k
4000
n

20rad sec
m
10
Frequency ratio:

32

 2 2

0.6283
n
20
Amplitude of vibration X is given by Eq. (93):

4N
1 F
F
0

X 0
2

1


n

12

4 0.12 98.1
1

F0
F0

0.04
2

2
4000 1 0.6283

2 12

The solution of this equation gives F0 = 97.9874 N

(answer)

9. Forced Vibration with Hysteresis Damping

Consider a single degree of freedom system with hysteresis damping and to a
harmonic force F(t) = F0sint, as indicated in Fig. 22.

The equation of motion of the mass can be derived as

k
kx F0 sint
mx
x

(100)

where (k/) x

= (h/) x

denotes the damping force. Although the solution

of Eq. (1)) is quite involved for a general forcing function F(t), our intention is to
find the response under a harmonic force.
The steady-state solution of Eq. (100) can be assumed:
xp t Xsin t
(101)
By substituting Eq. (101) into (100), we obtain

33

F0

2
1 2 2
n

12

(102)

and

2
1

n2

tan1

(103)
Equation (102) and (103) are shown plotted in Fig. 23 for several value of .

A comparison of Fig. 23 with Fig. 11 for viscous damping reveals the following:
a. The amplitude ratio
X
F0 k
attains its maximum value of F0/k at the resonant frequency ( = n) case of
hysteresis damping, while

34

b. The phase angle has a value of tan -1() at = 0 in the case of hysteresis
damping, while it has a value of zero at = 0 in the case of viscous damping.
This indicates that the response can never be in phase with the forcing
further in the case of hysteresis damping.
Note that if the harmonic excitation is assumed to be F(t) = F0eit in Fig. 22,
the equation of motion becomes
k
kx F0eit
mx
x

In this case, the response

(104)

x (t) is also a harmonic function involving the factor

eit. Hence x(t) is given by ix(t), and Eq. (104) becomes

mx k 1 i x F0eit
(105)
where the quantity k(1 + i) is called the complex stiffness or complex damping.
The steady-state solution of Eq. 105 is given by the real part of
F0eit
x t
2

k 1

n
(106)
10.Force Motion with Other Types of Damping
Viscous damping is the simplest form of damping to use in practice, since it
leads to linear equations of motion. In the cases of Coulomb and hysteretic
damping, we defined equivalent viscous damping coefficients to simplify the
analysis. Even for a more complex form of damping, we define an equivalent
viscous damping coefficient, as illustrated in the following examples.
Example No. 6 (Quadratic Damping)
Quadratic or velocity squared damping is present whenever a body moves in a
turbulent fluid flow. Find the equivalent viscous damping coefficient and
amplitude of steady state vibration of a single degree of freedom system having
quadratic damping.
Given:
Velocity squared damping
Required:
Equivalent viscous damping coefficient and
amplitude of steady state vibration of a single degree of freedom system having
quadratic damping.
Solution:
Equate energies dissipated per cycle during harmonic motion.
The damping force is assumed:

35

Fd a x 2
where a is a constant, x is the relative velocity across the damper, and the
negative (positive) sign must be used when x is positive (negative).
Energy dissipated per cycle during harmonic motion x(t) = Xsint :
X
2
8
W 2 a x 2dx 2X3
a 2 cos3 td t 2aX3
X
2
3

Equating this energy to the energy dissipated in an equivalent viscous damper.

W ceqX2
Equivalent viscous damping coefficient (ceq):
8
ceq
aX
3
(answer)
It can be noted that ceq is not a constant but varies with and X.
Amplitude of the steady-state response:
X
1

st
2 2
1 r 2 r 2

eq

where r = /n and
c
c
eq eq eq
cc 2mn
Amplitude:

3m 1 r2
X

2
8ar2

1 r

2 4

12

8ar2 st

3m

(answer)

11.Self Excitation and Stability Analysis

The force acting on a vibrating system is usually external to the system and
independent of the motion. However, there are systems for which the exciting
force is a function of the motion parameters of the system, such as
displacement, velocity, or acceleration. Such systems are called self-excited
vibrating systems since the motion itself produces the exciting force. The
instability of rotating shafts, the flutter of turbine blades, the flow induced
vibration of pipes, and the automobile wheel shimmy and aerodynamically
induced motion of bridges are typical examples of self-excited vibrations.
A system is dynamically stable if the motion (or displacement) converges or
remains steady with time. On the other hand, if the amplitude of displacement
increases continuously (diverges) with time, it is said to be dynamically unstable.
The motion diverges and the system becomes unstable if energy is fed into the
system through self-excitation. To see the circumstances that lead to instability,
we consider the equation of motion of a single degree of freedom system:
36

mx cx kx 0

(107)
If a solution of the form x(t) = Cest, where C is a constant, is assumed, Eq. 107
leads to the characteristic equation
c
k
s2 s 0
m m
(108)
The roots of this equation are
c 1 c
k
s1,2

4

2m 2 m
m
2

12

(109)
Since the solution is assumed to be x(t) = Cest, the motion will be diverging and
aperiodic if the roots s1 and s2 are real and positive. This situation can be
avoided if c/m and k/m are positive. The motion will also diverge if the roots s1
and s2 are complex conjugates with positive real parts. To analyze the situation,
let the roots s1 and s2 of Eq. 108 be expressed as
s1 p iq s2 p iq
,
(110)
where p and q are real numbers so that
s s1 s s2 s2 s1 s2 s s1s2 s2 c s k 0
m m
(111)
Equations (111) and (110) give
c
k
s1 s2 2p
s1s2 p2 q2
m
m
,
(112)
Equations (112) show that for negative p, c/m must be positive and for positive
p2 + q2, k/m must be positive. Thus the system will be dynamically stable if c and
k are positive (assuming that m is positive).
Example No. 7 (Instability of a Vibrating System)
Find the value of free stream velocity u at which the airfoil section (single degree
of freedom system) shown in Fig. 24 becomes unstable.

Given:
37

Single degree of freedom airfoil section in fluid flow

Required:
Velocity of the fluid which causes instability of the airfoil (or mass m).
Solution:
Find the vertical force acting on the air foil (or mass m) and obtain the condition
that leads to zero damping.
1
F u2DCv
2
where = density of the fluid, u = free stream velocity, D = width of the cross
section normal to the fluid flow direction, and Cv = vertical force coefficient,
which can be expressed as
u2
Cv rel
CL cos CD sin
u2
where urel is the relative velocity of the fluid,
CL is the lift coefficient,
CD is the drag coefficient, and
is the angle of attack (see Fig. 24):
x
tan1
u
For small angles of attack,
x

u
and Cv can be approximated, using Taylors series expansion about = 0, as
C
Cv Cv 0 v

0
where, for small values of a, urel = u and
Cv CL cos CD sin
Then

C
CL

cos CL sin D sin CD cos

Cv CL cos CD sin 0
Cv CL 0

Cv
x Cv
CL 0
0
u 0

x CL
Cv CL 0
CD 0
u 0

Substituting
1
1
C
F u2DCL
uD v
x
2
0
0 2
38

The equation of motion of the airfoil (or mass m) is

1
1
C
mx cx kx F u2DCL
uD v
x
2
0
0 2
The first term on the right-hand side produces a static displacement and
hence only the second term can cause instability of the system. The equation of
motion, considering only the second term on the right-hand side, is

C
1
x kx 0
mx cx kx mx c uD v
2
0

Note that m includes the mass of the entrained fluid. The displacement of the
airfoil (or mass m) will grow without bound (i.e. the system becomes unstable) if
c is negative. Hence the minimum velocity of the fluid for the onset of unstable
oscillations is given by c = 0, or,

2
c

Cv
D

(answer)
Cv
2.7
0
The value of
for a square section in a steady flow.
-

END -

39