CHAPTER 2

FREE VIBRATION OF SDF SYSTEMS

Expected outcome:
1. Students can formulate the governing equation of a SDF system in free vibration
2. Students can calculate positions of the mass in a SDF system in free vibration
3. Students can include the effect of damping in a SDF system in free vibration and be
able to determine damping parameters of a SDF system from decay of vibration
amplitude.

2.1 Free vibration of an undamped SDF system

Vibration of a SDF system without any external force or excitation applied, i.e.,
p  t   0 , is called “free vibration.” The vibration is due to initial displacement u  0  and

initial velocity u  0  at the beginning of vibration. The equation of motion is

mu  cu  ku  0 (2.1)

First, let us consider a simpler SDF system that has no damping, c  0 . Its vibration
is called “undamped free vibration.” And the equation of motion becomes

mu  ku  0 (2.2)

u (t )

m p (t )  0
k

Figure 2.1 Undamped single-degree-of-freedom (SDF) system

Solution of the above second order differential equation has the form of

u (t )  A cos(n t )  B sin(n t ) (2.3)

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Obtain u  t  by differentiating the above u  t  twice

u(t )   n2 A cos( n t )   n2 B sin( n t )   n2 u  t  (2.4)

Then, substitute in the equation of motion to obtain

 mn2u  t   ku  t    k  mn2  u  t   0 (2.5)

Since u  t   0 is the trivial solution, let us consider the case that u  t   0 so

k  mn2  0

k
n 
m

n is called “natural (circular) frequency of vibration” in unit of radian per second (rad/sec).
Next determine A and B by using the initial conditions: initial displacement u  0  and initial

velocity u  0  .

At time t  0 ,

u  t  0   A cos  0   B sin(0)  u  0  (2.6)

A  u 0 (2.7)

u  t  0    An sin  0   Bn cos  0   u  0  (2.8)

u  0
B (2. 9)
n

Therefore, the solution of undamped free vibration is

u  0 
u (t )  u  0  cos(nt )  sin(nt ) (2.10)
n

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Plot of the above undamped free vibration solution versus time t is shown below

Figure 2.2 Free vibration of a system without damping.

The vibrating motion is sinusoidal function with constant amplitude

2
 u (0) 
uo  u (0)   
2
 (2.11)
 n 

This motion is also known as “simple harmonic motion.”

The mass moves back and forth in cycles. The time duration to complete one cycle of
vibration is called “natural period of vibration in unit of time (seconds)”.

2
Tn  (2.12)
n

Number of cycles completed in one second is called “natural cyclic frequency of vibration
(cycles per second or Hz).”

1 n
fn   (2.13)
Tn 2

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Example 2.1

A 1000 kg lumped mass is supported by a 3m-tall steel pipe cantilever column having a
circular cross section with diameter=15cm and thickness=0.15cm. Determine the natural
frequency and period of vibration of this SDF system.

Figure 2.3 An SDF system representing a lumped mass supported by a cantilevered column

Given: mass m  1000 kg

Modulus of elasticity of steel is E  2  1011 N/m 2

Moment of inertia of pipe cross section is


I
64
 0.150 4
 0.1474   1.9292 106 m4

The lateral stiffness of cantilevered column is

3EI 3  2 10 1.9292 10   42871N m

11 6

k 
L3 33

The natural frequency of vibration is

42871
n   6.5476 rad / sec
1000

The natural period of vibration is

2 2
Tn    0.96 sec
n 6.5476

The natural cyclic frequency of vibration is

1 1
fn    1.04 Hz
Tn 0.974

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2.2 Free vibration of a viscously damped SDF system

u (t )
c
m p (t )  0

Figure 2.4 A damped SDF system

For a SDF system with viscous damping, the equation of motion is

mu  cu  ku  0 (2.14)

Dividing m on both sides of the equation, we obtain

c k
u  u  u  0 (2.15)
m m

u  2nu  n2u  0 (2.16)

c c
where damping ratio    is the ratio between damping coefficient of the SDF
ccr 2mn

system to critical damping coefficient ccr  2mn  2 km .

Solution of the governing equation of motion depends on damping ratio  .

• If   1 , or c  ccr , the system will oscillate about its equilibrium position with
decreasing amplitude. Such system is an “underdamped” system.

• If   1 , or c  ccr , the system will return from initial displacement to its equilibrium
position without oscillation. Such system is a “critically damped” system.

• If   1 , or c  ccr , there is no oscillation, but the system will return to its equilibrium
position at a slower rate. Such system is a “overdamped” system, e.g., door closer.

In structural dynamics, we will only consider the underdamped systems because all
structures of interest, i.e., buildings, bridges, typically have damping ratio less than 15%.

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 Figure 2.5 Comparison of free vibration of SDF systems with different levels of damping.

Type of Construction Damping ratio

Steel 0.02 – 0.05

Concrete 0.05 – 0.07
Masonry 0.05 – 0.10
Wood 0.10 – 0.15

Solution of the differential equation governing damped free vibration is

  u (0)  nu (0)  

u (t )  ent u (0) cos Dt    sin Dt  (2.17)
  D  

where the natural frequency of damped vibration is

 D  n 1   2 (2.18)

and the natural period of damped vibration is

2 Tn
TD   (2.19)
D 1  2

Note that the natural frequency of damped vibration is slightly lower than undamped
vibration, and the natural period of damped vibration is slightly longer than undamped
vibration. However, the effect of damping on vibration frequency is negligible (2%) in
practical range of damping (  <20%).

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The plot of damped free vibration shows that amplitude of vibration becomes smaller after
every cycle of oscillation.

Figure 2.6 Comparison of undamped and damped free vibration.

The rate of amplitude decay is exponential. The envelope curves of amplitude are   e nt
where
2
 u (0)  n u (0) 
  u (0)  
2
 (2.20)
 D 

2.3 Effect of damping on the rate at which free vibration decays

Effect of damping is directly related to the rate at which free vibration decays. A
system with larger damping ratio oscillates fewer cycles before it comes to stop.

Figure 2.7 Free vibration of SDF systems with different damping ratios.

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2.4 Estimation of damping ratio from decay of motion

• Knowing the decay of consecutive peak allow us to estimate the damping ratio

Figure 2.8 Amplitude of vibration every cycle of motion due to effect of damping.

The following peak occurs at time interval TD, so the peak is smaller by a factor of

2 2 
n 
nTD (2.21)
e e n 1 2
e 1 2

The ratio of amplitude of cycle i and i+1 is then

ui  2 
 exp   (2.22)
ui 1  1 2 
 

Logarithmic decrement of each cycle of oscillation is

 ui  2
  ln   (2.23)
 ui 1  1  2

For small damping ratio  , the square root term is nearly equal to 1.

1  2  1 (2.24)

So logarithmic decrement can be approximated as

  2 (2.25)

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and the damping ratio can be estimated from the ratio of consecutive peaks as

1  ui 
  ln   (2.26)
2  ui 1 

For lightly damped system, the amplitude decay is slow, so it is desirable to relate the ratio of
two amplitudes several cycles apart, e.g., over j cycles:

1  ui 
  ln    2 (2.27)
j 
 ui  j 

1  u 
  ln  i  (2.28)
2 j  ui  j 

2.5 Free vibration test

Because it is not possible to determine analytically the damping ratio for practical
structures, this property should be determined experimentally. The structure is excited and let
to vibrate freely.

Acceleration is measured easier than displacement. For lightly damped system, it can
also be shown that

1  u 
  ln  i 
2 j  ui  j 

Figure 2.9 Acceleration amplitude of the mass also decreases every cycle of motion

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Measured period

The natural period of damped vibration TD is the time required to complete one cycle
of vibration. Comparing the measured period to the natural period calculated from stiffness
and mass of an idealized system tells us how well the idealization represents the actual
structure.

Example 2.2

Determine the natural period and damping ratio from the following acceleration
record of its free vibration.

Figure 2.10 Measured acceleration of a mass during a free vibration test.

Peak no. Time, ti (sec) Peak value (g)

1 1.20 0.905
7 3.85 0.085

3.85  1.2
TD   0.441 sec
6

1 0.905 g
 ln  0.0628  6.28%
2 (6) 0.085 g

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Example 2.3

A free vibration test is conducted on an empty water tank tower.

Figure 2.11 A water tank idealized as a mass supported by a cantilevered column.

A cable attached to the tank applies a lateral force of 16.4 kips and pulls the tank
horizontally by 2 in. The cable is suddenly cut and the resulting free vibration is recorded. At
the end of four complete cycles, the time is 2.0 sec and the amplitude is 1 in. From these data,
compute the following: (a) damping ratio; (b) natural period of undamped vibration; (c)
stiffness; (d) weight; (e) damping coefficient; and (f) number of cycles required for the
displacement amplitude to decrease to 0.2 in.

Solution
Substituting ui =2 in., j=4 and ui+j=1 in. in

1  u  1 2
 ln  i   ln    0.0276  2.76%
2 j  ui  j  2  4   1 

2.0
TD   0.5 sec ; Tn  TD  0.5 sec
4

16.4
k  8.2 kips/in.
2

2 2
n    12.57 sec
Tn 0.5

k 8.2
m   0.0519 kip  sec2 /in.
 12.57 
2 2
n

   0.0519  386  20.03kips

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  
c   2 km  0.0276  2 8.2  0.0519    0.0360 kip  sec/in.
 

1 u1
  ln ;
2 j u1 j

1 2
j ln  13.28cycles  13cycles
2  0.0276  0.2

Example 2.4

The weight of water required to fill the tank of example 2.3 is 80 kips. Determine the natural
vibration period and damping ratio of the structure with the tank full.

Solution

  20.03  80  100.03 kips

100.03
m 0.2591 kips  sec2 /in.
386

m 0.2591
Tn  2  2  1.12 sec
k 8.2

c 0.0360
    0.0123  1.23%
2 km 2 8.2  0.2591

Observe that the damping ratio is now smaller (1.23% compared to 2.76% in example 2.3)
because the mass of the full tank is larger and hence the critical damping coefficient is larger.

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