3
Free vibration of single-degree-of-freedom
systems (under-damped) in relation to
structural dynamics during earthquakes
Abstract: In this chapter, the governing equations of motion are formulated
for free vibration of single-degree-of-freedom (SDOF) (under-damped)
systems. Motion characteristics are studied for under-damped, critically
damped and over-damped systems. Vibration characteristics of an underdamped system are illustrated. Hysteresis damping and Coulomb damping
are also discussed. Programs in MATLAB and in MATHEMATICA are
listed for the vibration of various under-damped SDOF systems.
Key words: viscous damping, logarithmic decrement, critical damping,
hysteresis damping, Coulomb damping.
3.1
Introduction
It was seen in the preceding chapter that a simple oscillator under idealized
conditions without damping will oscillate indefinitely with a constant amplitude
at its natural frequency. In practice, it is not possible to have an oscillator
that vibrates indefinitely. In any practical structure frictional or damping
forces will always be present in the mechanical energy of the system, whether
potential or kinetic energy is transformed to heat energy. In order to account
for these forces, we have to make certain assumptions about these forces
based on experience.
3.2
Damping free vibrations
The oscillatory motions considered so far have been for ideal systems, i.e.
systems that oscillate indefinitely under the action of linear restoring force.
In real systems, dissipative forces, such as friction, are present and retard the
motion. Consequently, the mechanical energy of the system diminishes in
time, and the motion is said to be damped.
One common type of retarding force is proportional to the speed and acts
in the direction opposite to the motion. The damping caused by fluid friction
is called viscous damping. The presence of this damping is always modelled
by a dashpot, which consists of a piston A moving in a cylinder B as shown
in Fig. 3.1. The frictional force is proportional to velocity and is denoted by
cẋ and the constant c is called the coefficient of viscous damping.
43
44
Structural dynamics of earthquake engineering
B
A
F
F
cẋ
F
3.1 Model of a dashpot.
mẋ˙
kx
k
m
Inertia
force
cẋ
˙˙
x , x˙ , x
C
3.2 Spring–mass damper system.
Consider the damped free vibration of a spring–mass damper system
shown in Fig. 3.2. Using D’Alembert’s principle, a dynamic problem can be
converted to a static problem by considering inertia force.
m˙˙
x + cx˙ + kx = 0
3.1
Equation 3.1 is a linear, second order, homogeneous differential equation. It
has the solution of the form
x = eλ t
3.2
Substituting it in Eq. 3.2, we get
mλ2 + cλ+ k = 0
3.3
which has two roots given by
2
c 
λ 1, λ 2 = − c ± 
− k
m
2m
 2m 
3.4
Free vibration of SDOF systems (under-damped)
45
Thus the general solution of Eq. 3.4 is the sum of two exponentials λ1 and
λ2. The critical damping coefficient is given by
and
c c = 2 km
3.5
ρ = c = damping factor
cc
3.6
cc ρ
+
2m
3.7
∴ λ1 = −
ρ2
c c2
− ( k/m )
4m2
λ1 = − ρω n + ρ 2 ω n2 − ω n2
and
3.8
λ1 = ω n ( − ρ + ρ 2 − 1 )
3.9a
λ2 = ω n (− ρ − ρ 2 − 1)
3.9b
x = Ae λ1t + Be λ 2 t
3.10
There are three special cases of damping that can be distinguished with
respect to the critical damping coefficient.
3.2.1
Over-damped system
When c > cc and ρ > 1
x = Aeλ 1t + Beλ 2t
3.11
There are two constants A and B which can be evaluated using initial conditions
xt=0 = x0; vt=0 = v0.
v = dx = λ1 Ae λ1t + λ 2 Be λ 2 t
dt
x0 = A + B
3.13a
v 0 = λ 1A + λ 2B
3.13b
3.12
Solving the above two equations:
λ1 x 0 = λ1 A + λ1 B
v 0 = λ1 A + λ 2 B
λ1 x 0 − v 0 = B( λ1 − λ 2 )
∴B =
( λ1 x 0 − v 0 )
( λ x − v0 )
; A= 2 0
( λ1 − λ 2 )
( λ1 − λ 2 )
3.14
As t increases x decreases. This motion is non-vibratory or a periodic as
shown in Fig. 3.3.
46
3.2.2
Structural dynamics of earthquake engineering
Critically damped system
When c = cc and ρ = 1
3.15
λ1 = –c/2m; λ2 = –c/2m
3.16a
λ1 = –ωn; λ2 = –ωn
3.16b
When we have two repeated roots
x = ( A + Bt )e – ω n t
3.17
This motion is also non-vibratory but it is of special interest because x
decreases at the fastest possible rate without oscillation of the mass and is
shown in Fig. 3.3.
3.2.3
Under-damped system
When c < cc and ρ < 1. The roots shown in Eq. 3.7 are complex.
λ1 = ω n ( − ρ + i 1 − ρ 2 )
3.18a
λ2 = ω n(− ρ − i 1 − ρ 2 )
3.18b
∴ x = Ae − ρω n t e i 1 − ρ 2 tω n + Be − ρω n t e − i 1 − ρ 2 tω n
3.19
= e − ρω n t ( c1 sin ω n 1 − ρ 2 t + c 2 cos ω n 1 − ρ 2 t )
3.20
Equation 3.20 may be written as
x = Ae − ρω n t [sin (ω d t + φ )]
3.21
Again there are two constants, which can be evaluated using initial conditions
x 0 = x 0; v 0 = v0.
Displacment
x
c
b
a under-damped
b critically damped
c over-damped
t
a
Time
3.3 Displacement versus time.
Free vibration of SDOF systems (under-damped)
47
The constant ωd is defined as the damped natural frequency of the system,
which is expressed as
ωd = ωn 1 − ρ2
3.22
where ω n = k/m is the natural frequency of the undamped vibration.
Equation 3.21 defines the harmonic oscillations of diminishing amplitude
as shown in Fig. 3.3. The amplitude is Ae − ρω n t
x = Ae − ρω n t [sin (ω d t + φ )]
3.23a
ẋ = − ρω n Ae − ρω n t [sin (ω d t + φ )] + Ae − ρω n t ω d [cos (ω d t + φ )]
3.23b
at t = 0 x t = 0 = x 0 = A sin θ
ẋ
or
t =0
= v 0 = − ρω n A sin φ + ω d A cos φ
( v 0 + ρω n x 0 )
= A cos φ
ωd
tan φ =
A=
ω d x0
( v 0 + ρω n x 0 )
x 02 +
( v 0 + ρω n x 0 ) 2
ω d2
3.24a
3.24b
3.24c
3.25
3.26
 ( v + ρω n x 0 )

x = e − ρω n t  0
sin 1 − ρ 2 ω n t + x 0 cos 1 − ρ 2 ω n t 
 ω n 1 − ρ 2

3.27
i.e.
x = Xe − ρω n t [sin (ω n 1 − ρ 2 t + φ )]
3.3
Logarithmic decrement
3.28
A convenient way to determine the amount of damping present in a system
is to measure the rate of decay of free oscillations. The larger the damping
the greater will be the decay. Consider the damped vibration expressed by
the general equation:
x = Xe − ρω n t [sin (ω n 1 − ρ 2 t + φ )]
which is shown graphically in Fig. 3.4.
We introduce a term called logarithmic decrement defined as
3.29
48
Structural dynamics of earthquake engineering
x
x1
x2
t
τd
3.4 Damped vibration.
δ = log
τd =
− ρω n t
x1
= log X− eρω ( t + τ ) = log e ρω nτ d = ρω n τ d
n
x2
d
Xe
ωn
∴δ =
2π
1 − ρ2
3.31
ρω n 2 π
ωn 1− ρ
3.30
2
=
2 πρ
1 − ρ2
= logarithmic decrement
3.32
The above is an exact equation,
When ρ is small
δ ≅ 2 πρ
3.33
Figure 3.5 shows a plot of the exact and approximate values of δ as function
of ρ.
From Eq. 3.31 it is seen that the period of the damped vibration τd is
constant even though the amplitude decreases
τ d = 2π
ωd
3.34
The period of damped vibration is always larger than the period of the same
system without damping.
Example 3.1
A diesel engine generator of mass 1000 kg is mounted on springs with total
stiffness 400 kN/m. If the period of oscillation is 0.32 s. determine the damping
coefficient c and damping factor ρ.
Given m = 1000 kg; k = 400 × 103 N/m; T = 0.32.
Free vibration of SDOF systems (under-damped)
49
12
10
8
δ=
6
δ
2πρ
1 – ρ2
4
δ = 2πρ
2
0.2
0.4
ρ
0.6
0.8
1.0
3.5 Logarithmic decrement as a function of ρ.
c
k
m
3.6 Model of gun barrel.
Solution
T=
2π
= 0.32; ω n =
ωn 1 − ρ2
ωn =
k
m
400 × 10 3 = 20
1000
∴ 0.32 =
2π
20 1 − ρ 2
1 − ρ 2 = 2 π ; ρ = 0.19
6.4
c c = 2 km = 2 400 × 10 3 × 1000 = 2 × 20 × 10 3
c = ρ c c = 0.19 × 40 × 10 3
c = 7608 Ns/m
Example 3.2
A gun barrel (Fig. 3.6) weighing 5395.5N has a recoil spring of stiffness
300 000 N/m. If the gun barrel recoils 1.2 m on firing, determine
(a) the initial recoil velocity of the barrel;
50
Structural dynamics of earthquake engineering
(b) the critical damping coefficient of dash pot which is engaged at the end
of recoil stage;
(c) the time required for the barrel to return to a position 50 mm from its
initial position.
Solution
Weight of gun barrel = 5395.5 N, m = 550 kg, k = 300 000 N/m
Kinetic energy = potential energy in the spring
∴ 1 mv12 = 1 k x 2
2
2
1 × 550 × v 2 = 1 × 300 000 × 1.2 2
i
2
2
vi = initial recoil velocity = 28.025 m/s
Cc = 2 km = 2 300 000 × 550 = 25690 N s/ m
Since it is critical damping:
x = e − w n t ( A + Bt )
ωn =
k =
m
300 000
= 23.35 rad/s
550
x = e −23.35t ( A + Bt )
at t = 0; x0 = –1.2 m
t = 0; ẋ0 = 0
ẋ = −23.35e −23.35 t ( A + Bt ) + e −23.35 t B
ẋ
t =0
= −23.35 A + B = 0 ; B = 23.35 A
x|t=0; A = 1.2 and B = –28.02
x = e–23.35t (–1.2 – 28.02t)
x = –0.05 = e–23.35t (–1.2 – 28.02t)
Solving by trial and error t = 0.2135 second.
Example 3.3
A vibrating system shown in Fig. 3.7 consists of weight W = 9.81kN, a
spring stiffness 20 kN/cm and a dashpot with coefficient 0.071kN/cm/s.,
Find (a) damping factor, (b) logarithmic decrement and (c) ratio of any two
consecutive amplitudes.
Free vibration of SDOF systems (under-damped)
51
C
m
3.7 Vibrating system.
Solution
Given weight W = 9.81kN = 9.81 × 103 N; m = 103 kg
3
2
k = 20 × 10 × 10 = 2 × 10 6 N/m
1
c = 0.071 × 103 × 102 N/m/s
= 0.071 × 105 N/m/s
c c = 2 km = 2 2 × 10 6 × 10 3 = 0.8944 × 10 5
Since c < cc it is an under-damped system.
Damping factor c = 0.071 = 0.079
c c 08944
ωn =
k =
m
2 × 10 6 = 44.72 rad/s
10 3
x = Xe −0.079 × 44.72 t (sin 44.72 1 − 0.79 2 t + φ )
x = Xe–3.5325t [sin (44.58 t + φ)] where tan φ =
Logarithmic decrement = δ =
ω d x0
( v 0 + ρω n x 0 )
2 πρ
1 − ρ2
When ρ is small,
δ ≅ 2 πρ = 2 π × 0.79 = 0.496
x
log 0 = 0.496
x1
x0
= e 0.496 = 1.64
x1
Example 3.4
A free vibration test is carried out on an empty elevated water tank shown in
Fig. 3.8. A cable attached to the tank applies a lateral force 144 kN and pulls
52
Structural dynamics of earthquake engineering
3.8 Elevated water tank.
the tank by 0.050 m. Suddenly the cable is cut and the resulting vibration is
recorded. At the end of five complete cycles, the time is 2 seconds and the
amplitude is 0.035 m. Compute (a) stiffness, (b) damping factor, (c) undamped
natural frequency, (d) weight of the tank, (e) damping coefficient and (f)
number of cycles when the amplitude becomes 0.005 m.
Solution
Horizontal force = F = 144 kN = 144 000 N
Displacement = u = 0.05 m
(a) Stiffness = K = F/u =144 000/0.05 = 2 880 000 N/m
Time taken for 5 cycles = 2 seconds
Period of damped oscillation = 2/5 = 0.4; Td = 2π/ωd
ω d = 2 π = 5π
Td
ωn = natural frequency = 5π rad/s.
a
Logarithmic decrement = δ = 1 ln 1 = 1 ln 0.05 = 0.07133
n
a n +1
5
0.035
2πρ = δ = 0.07133
(b) ρ = 0.0113 = 1.13%
natural frequency of damped system = 2 π = 15.7 rad/s
T
ω n2 (1 − ρ 2 ) = 15.7 2
(c)
ω n = 15.7015; T( undamped ) =
2 π = 0.40016s
15.7015
2 880 000
ω n2 = k ; m = K2 =
= 11682 kg
m
ω n 15.7015 2
Free vibration of SDOF systems (under-damped)
53
(d) Weight of the tank W = 11682 × 9.81/1000 kN = 114.6kN
(e)
c c = 2 Km = 2 2 880 000 × 11682 = 366 846.889 N s/m
c = ρ cc = 0.0113 × 366 846.889 = 4145.36 Nsc/m
1 ln 50 = 2 π ρ = 2 π × 0.0113 = 0.07099
n
5
(f)
1 2.3025 = 0.070 99; n = 32.43 cycles = 33 cycles
n
Example 3.5
For small damping, show that the logarithmic decrement is expressed in
terms of vibrational energy U and the energy dissipated per cycle.
Solution
δ = ln
x1
; or
x2
x2
= e − δ = 1 − δ + δ 2 /2 + ...
x1
The vibrational energy of the system is that stored in the spring at maximum
displacement.
U1 = 1 Kx12 ;
2
U 2 = 1 Kx 22
2
x2
U1 − U 2
U
= 1 − 2 = 1 − 22 = 1 − e −2 δ = 2 δ − 4 δ 2 ...
U1
U1
x1
∆U = 2 δ
U
3.4
Hysteresis damping
Real structures and machines do not exhibit the highly idealized form of
viscous damping considered in previous sections. When materials are cyclically
stressed, energy is dissipated within the material itself due to primarily to
internal friction caused by the slipping and sliding of particles at internal
planes during deformations. Such internal damping is generally referred to
as hysteresis damping or structural damping. This form of damping results
in a phase lag between the damping force and deformation as illustrated in
Fig. 3.9. This curve is generally referred to as a hysteresis loop. The area
enclosed within the loop represents the energy loss or dissipated energy per
loading cycle.
If ∆U represents the energy loss per cycle as illustrated in Fig. 3.9 then the
energy loss can be written as
54
Structural dynamics of earthquake engineering
Energy loss ∆U
Loading
Unloading
3.9 Hysteresis loop.
x = X sin ω t ; x˙ = ω X cos ω t
∆U =
∫F
D
∫
dx = 4 c e x˙ dx
3.35a
3.35b
ηk
3.35c
ω
Based on experiments conducted on the internal damping it can be proved
that the energy dissipated per cycle is independent of frequency and proportional
to the square of the amplitude of vibration. Thus, the energy loss per cycle
may be expressed as
∆U = π c e ω x 2 = π η k x 2 ; where c e =
∆U = πηK x 2
3.36
where
η = a dimensionless structural damping coefficient for the material
k = the equivalent stiffness of the system
x = the displacement amplitude
π = a convenient proportionality constant.
2
∆U = πηk x = 2 π η = 2 δ
U
0.5 k x 2
Hence logarithmic decrement δ = πη
δ =πη =
π ce ω n
K
3.37
3.38
3.39
Hence equivalent viscous damping coefficient is given by
ce =
kη
ωn
3.40
Therefore, for a structure considered to exhibit hysteretic or structural damping
characteristics, the coefficient η can be determined by measuring successive
Free vibration of SDOF systems (under-damped)
55
amplitudes of the oscillation and then applying Eq. 3.38. Then the structure
can be analysed as an equivalent viscously damped system by calculating the
equivalent viscous damping coefficient calculated from Eq. 3.40.
Example 3.6
The main span of a bridge structure is considered as a single-degree-offreedom (SDOF) system for calculation of its fundamental frequency. From
preliminary vibration tests, the effective mass of the structure was determined
to be 400 000 kN and the effective stiffness to be 40 000 kN/m. The ratio of
successive displacement amplitude from a free vibration trace was measured
to be 1.25. Calculate the values of the structural damping coefficient and the
equivalent viscous damping coefficient.
Solution
The ratio of successive amplitude =
Logrithmic decrement = δ = ln
x1
= 1.25
x2
x1
= 0.2231 = π η
x2
η = 0.2231 = 0.071
π
The equivalent viscous damping coefficient is determined as
cc =
η K
ωn
where
ωn =
K =
m
40 000 000
= 10 rad/s
400 000
The equivalent viscous damping coefficient is
cc =
3.5
0.071 × 40 000
= 284kNs/m
10
Coulomb damping
In most of the structures, damping occurs when relative motion takes place
at interfaces or joints between adjacent members. This form of damping is
referred to as Coulomb damping or dry-friction damping. The friction forces
developed are independent of vibration amplitude and frequency. These forces
are acting in the opposite direction of motion of the mass and the magnitude
is essentially constant.
56
Structural dynamics of earthquake engineering
Frictional damping force is given by
Fd = µN = µmg
3.41
Referring to Fig. 3.10b for the mass to move from left to right U̇ > 0 . The
equation of motion is written as
or
˙˙ + KU + µ m g = 0
mU
3.42
˙˙ + ω n2 U = − µg
U
3.43
The solution of Eq. 3.42 for motion from left to right is
U ( t ) = A1 sin ω n t + B1 cos ω n t − µg/K
3.44
Referring to Fig. 3.10c for the mass to move from right to left U̇ < 0 . The
equation of motion is written as
˙˙ + KU − µ m g = 0
mU
3.45
The solution of Eq. 3.45 for motion from right to left is
U ( t ) = A2 sin ω n t + B2 cos ω n t + µg/K
3.46
Assume initial conditions as U(0) = U0 and U̇ (0) = 0 and the motion is from
right to left.
Substituting the initial conditions in Eq. 3.46, we get
U(t) = (U0 – µmg/K) cos ωnt + µmg/K 0 ≤ t ≤ π/ωn
3.47
Equation 3.47 is valid until the motion to left ceases or when the velocity is
equal to zero.
U ( π /ω n ) = ( − U 0 + 2 µ m g/K ), U̇ ( π /ω n ) = 0
KU
K
Direction of motion
u̇ > 0
mU̇˙
µ mg
N = mg
KU
Direction of motion
U̇ < 0
mU̇˙
µ mg
N = mg
3.10 Model for Coulomb damping.
Free vibration of SDOF systems (under-damped)
57
and the motion is from left to right.
Solving for constants in Eq. 3.44 using the conditions at t = π /ω n , we get
the solution for the displacement as
U(t) = –(U0 – 3µmg/K) cos ωnt – µmg/K
3.48
Substituting when t = 2π/ωn we get
U (2 π /ω n ) = (U 0 − 4 µ m g/K ), U˙ (2 π /ω n ) = 0
3.49
Figure 3.11 shows the free vibration of a system with Coulomb damping. It
is seen that the amplitude decreases by 4Fd/K after every cycle and the
amplitudes decay linearly with time.
Example 3.7
For the system shown in Fig. 3.10W = 1kN; K = 70kN/m; coefficient of
friction µ = 0.15 and the initial conditions are initial displacement is 0.15m
with zero initial velocity. Determine the vibration displacement amplitude
after four cycles and number of cycles of motion completed before the mass
comes to rest.
Solution
Fd = µN = 0.15 × 1 = 0.15kN
ωn =
70 000 × 9.81
= 26.2rad/s
1000
K =
m
T = 2 π = 2 π = 0.2398 s
ω n 26.2
3
Displacement (m)
2
1
0
–1
–2
–3
0
1
2
3
Time (sec)
4
3.11 Free vibration with Coulomb damping.
5
58
Structural dynamics of earthquake engineering
After every cycle amplitude reduction
=
4 Fd
= 4 × 0.15 = 8.57 × 10 −3 m = 8.57mm
K
70
After four cycles displacement amplitude
= 150 – 4 × 8.57 = 115.72mm
Motion will cease when the amplitude of the nth cycle such that KUn ≤ Fd or
Un ≤ 2.14228mm.
150 – n × 8.57 ≤ 2.1428
n >17.25 cycles
This indicates that motion will terminate after 17.25 cycles.
3.6
Numerical method to find response due to
initial conditions only
The dynamic equation of equilibrium for free vibration of damped system
can be written as
m ˙˙
x + c x˙ + K x = 0
3.50
˙˙
x + 2 ρ ω n x˙ + ω n2 x = 0
3.51
or
where
c/m = 2 ρω n ; K/m = ω n2 ; ω d
= ω n 1 − ρ 2 ; ω = ω n ρ; ρ =
ρ
(1 − ρ 2 )
3.52
and
b0 = 2 ρω n ; b1 = ω d2 − ω 2 ; b2 = 2ωω d
3.53
Defining
s ( t ) = e − ρω n t sin (ω d t ); c ( t ) = e − ρω n t cos (ω d t )
3.54
we get using the approach of Wilson
s˙ ( t ) = − ω s ( t ) + ω d c ( t ); c˙ (t) = − ω c ( t ) − ω d s ( t )
3.55
˙˙
s ( t ) = − b1 s ( t ) − b2 c ( t ); c˙ ( t ) = − b1 c ( t ) + b2 s ( t )
3.56
A1 ( t ) = c ( t ) + ρ s ( t ); A2 ( t ) =
s(t )
ωd
3.57
Free vibration of SDOF systems (under-damped)
59
s˙ ( t )
A3 ( t ) = A˙1 ( t ) = c˙ ( t ) + ρ s˙( t ); A4 ( t ) = A˙ 2 ( t ) =
ωd
3.58
s (t )
˙˙1 ( t ) = c˙˙( t ) + ρ ˙˙
˙˙2 ( t ) = ˙˙
A5 ( t ) = A
s ( t ); A6 ( t ) = A
ωd
3.59
x ( t ) = A1 ( t ) x 0 + A2 ( t ) ẋ 0
3.60a
x˙ ( t ) = A˙1 ( t ) x 0 + A˙ 2 ( t ) x˙ 0 = A3 ( t ) x 0 + A4 ( t ) x˙ 0
3.60b
˙˙1( t ) x 0 + A
˙˙2 ( t ) x˙ 0 = A5 ( t ) x 0 + A6 ( t ) x˙ 0
˙˙
x (t ) = A
3.60c
Equation 3.60 gives the relationship between displacement, velocity and
acceleration with time.
3.7
Program 3.1: MATLAB program for free
vibration of under-damped (SDOF) systems
clc
close all
%****************************************************
% give mass of the system
m=2;
%give stiffness of the system
k=8;
wn=sqrt(k/m);
%give damping coefficient
c1=1;
%give initial conditions - displacement and velocity
u(1)=.3;
udot(1)=.5;
uddot(1)=(-c1*udot(1)-k*u(1))/m;
%****************************************************
cc=2*sqrt(k*m);
rho=c1/cc;
wd=wn*sqrt(1-rho^2);
wba=rho*wn;
rhoba=rho/sqrt(1-rho^2);
b0=2.0*rho*wn;
b1=wd^2-wba^2;
b2=2.0*wba*wd;
dt=0.02;
t(1)=0;
for i=2:1500
t(i)=(i-1)*dt;
60
Structural dynamics of earthquake engineering
s=exp(-rho*wn*t(i))*sin(wd*t(i));
c=exp(-rho*wn*t(i))*cos(wd*t(i));
sdot=-wba*s+wd*c;
cdot=-wba*c-wd*s;
sddot=-b1*s-b2*c;
cddot=-b1*c+b2*s;
a1=c+rhoba*s;
a2=s/wd;
a3=cdot+rhoba*sdot;
a4=sdot/wd;
a5=cddot+rhoba*sddot;
a6=sddot/wd;
u(i)=a1*u(1)+a2*udot(1);
udot(i)=a3*u(1)+a4*udot(1);
uddot(i)=a5*u(1)+a6*udot(1);
end
figure(1);
plot(t,u,‘k’);
xlabel(‘ time’);
ylabel(‘ displacement ’);
title(‘ displacement - time’);
figure(2);
plot(t,udot,‘k’);
xlabel(‘ time’);
ylabel(‘ velocity’);
title(‘ velocity - time’);
figure(3);
plot(t,uddot,‘k’);
xlabel(‘ time’);
ylabel(‘ acceleration’);
title(‘ acceleration- time’);
The displacement time, velocity time and accelertion time curves are
shown in Fig. 3.12.
3.8
1
16
Program 3.2: MATHEMATICA program for free
vibration of damped SDOF systems
Free vibration of SDOF systems (under-damped)
61
0.4
Displacement (m)
0.3
0.2
0.1
0
–0.1
–0.2
–0.3
0
10
20
30
Time (sec)
(a)
0.6
0.4
Velocity (m/s)
0.2
0
–0.2
–0.4
–0.6
–0.8
0
5
10
15
20
Time (sec)
(b)
25
30
1.5
Acceleration (m/s2)
1.0
0.5
0
–0.5
–1.0
–1.5
–2.0
0
5
10
15
20
Time (sec)
(c)
25
30
3.12 (a) Displacement
time; (b) Velocity
time; (c) Acceleration
time.
62
Structural dynamics of earthquake engineering
0.8
8
0.03(1. Cos[3.97995 x]+0.51927 Sin[3.97995 x])
{{y[x] -> ————————————————————————————————————————}}
0.4 x
E
0.03 (1. Cos[3.97995 x]+0.51927 Sin[3.97995 x])
———————————————————————————————————————————————
0.4 x
E
displacement
0.03
0.02
0.01
time in secs
5
10
15
20
–0.01
–0.02
–0.03
0.03 (1. Cos[3.97995 x]+0.51927 Sin[3.97995 x])
———————————————————————————————————————————————
0.4 x
E
{0.03, -0.0212113, 0.00553537, 0.00456859,
-0.00658279, 0.00384845, -0.000492167, -0.00128801,
0.0013758, -0.00065462,
Free vibration of SDOF systems (under-damped)
63
-0.0000313427, 0.000322237, -0.000274792,
0.000101551, 0.0000324349, -0.0000747065,
0.000052398, -0.0000134052, -0.0000115266,
-6
0.0000163566, -9.48387 10
}
4
Pi
—
2
0.00553537
0.0031892
1.73566
0.551387
94.93
0.569142
0.03 (2.06667 Cos[3.97995 x]-3.97995 Sin[3.97995 x])
——————————————————————————————————————————————————
0.4x
E
0.012 (1. Cos[3.97995 x]+0.51927 Sin[3.97995 x])
– ————————————————————————————————————————————————
0.4 x
E
64
Structural dynamics of earthquake engineering
0.03 (2.06667 Cos[3.97995 x]-3.97995 Sin[3.97995 x])
——————————————————————————————————————————————————
0.4x
E
0.012 (1. Cos[3.97995 x]+0.51927 Sin[3.97995 x])
- ————————————————————————————————————————————————
0.4
E
velocity
0.05
time in secs
5
10
15
20
–0.05
–0.01
0.03 (2.06667 Cos[3.97995 x]-3.97995 Sin[3.97995 x])
——————————————————————————————————————————————————
0.4x
E
0.012 (1. Cos[3.97995 x] + 0.51927 Sin[3.97995 x])
- —————————————————————————————————————————————————
0.4x
E
{0.05, 0.0402039, -0.0585079, 0.0343856, -0.00453622,
-0.0113839, 0.0122435, -0.00586079, -0.00024736,
0.00285517, -0.00244842,
0.000912015, 0.000282556, -0.000663096, 0.000467483,
-0.000121134, -0.000101461, 0.000145385, -0.0000847438,
0.0000106441, 0.0000285358}
Free vibration of SDOF systems (under-damped)
65
0.03 (2.06667 Cos[3.97995 x]- 3.97995 Sin[3.97995 x])
——————————————————————————————————————————————————
0.4x
E
0.012 (1. Cos[3.97995 x] + 0.51927 Sin[3.97995 x])
- —————————————————————————————————————————————————
0.4x
E
0.03 (2.06667 Cos[3.97995 x]-3.97995 Sin[3.97995 x])
——————————————————————————————————————————————————
0.4x
E
0.012 (1. Cos[3.97995 x] + 0.51927 Sin[3.97995 x])
- —————————————————————————————————————————————————
0.4x
E
acceleration
0.1
0.05
time in secs
10
15
20
–0.05
–0.01
0.03 (2.06667 Cos[3.97995 x]-3.97995 Sin[3.97995 x])
——————————————————————————————————————————————————
0.4x
E
0.012 (1. Cos[3.97995 x] + 0.51927 Sin[3.97995 x])
- —————————————————————————————————————————————————
0.4x
E
66
Structural dynamics of earthquake engineering
{0.05, 0.0402039, -0.0585079, 0.0343856, -0.00453622,
-0.0113839, 0.0122435, -0.00586079, -0.00024736,
0.00285517, -0.00244842,
0.000912015, 0.000282556, -0.000663096, 0.000467483,
-0.000121134, -0.000101461, 0.000145385, -0.0000847438,
0.0000106441, 0.0000285358}
3.9
Summary
Real structures dissipate energy while undergoing vibratory motion. The
common method is to assume that dissipated energy is proportional to damping
forces. Damping forces are proportional to velocity acting in the opposite
direction of motion. The analytical expression for the solution of the governing
differential equation depends on the magnitude of the damping ratio. Three
cases are possible: (i) under-damped system, (ii) critically damped system,
and (iii) over-damped system. A practical method of determining the damping
present in a system is to evaluate experimentally the logarithmic decrement
which is defined as the natural logarithm of the ratio of two consecutive
peaks in free vibration. The damping ratio in buildings and bridges is usually
less that 20% of critical damping. For such systems the damped frequency is
equal to the undamped natural frequency.
3.10
Exercises
1. A vibration system consists of a mass 50 kg, a spring of stiffness 30 kN/
m and a damper. The damping provided is only 20% of the critical value.
Determine (a) the damping factor, (b) the critical damping coefficient,
(c) the natural frequency of the damped vibrations, (d) the logarithmic
decrement and (e) the ratio of two consecutive amplitudes.
2. Determine the time at which the mass in a damped vibrating system
would settle down to 1/50th of its initial deflection for the following
data: m = 200 kg; ρ= 0.22; k = 40 N/mm.
Also find the number of oscillations completed to reach this value of
deflection.
3. In a single degree of damped vibrating system, a suspended mass of 8 kg
makes 30 oscillations in 18 seconds. The amplitude decreases to 0.25 of
the initial value after five oscillations. Determine (a) the stiffness of the
spring, (b) logarithmic decrement, (c) the damping factor and (d) the
damping coefficient.
4. A machine mounted on springs and filled with a dashpot has a mass of
60 kg. There are three springs each of stiffness 12 N/nm. The amplitude
of vibration reduces from 45 to 8 mm in two complete oscillations.
Free vibration of SDOF systems (under-damped)
67
Assuming that the damping force varies as the velocity, determine (a)
the damping coefficient, (b) the ratio of frequencies of damped and
undamped vibrations and (c) the periodic time of damped vibration.
5. A machine weighs 18 kg and is supported on springs and dashpots. The
total stiffness of the spring is 12 N/mm and damping is 0.2 N/mm/s. The
system is initially at rest and a velocity of 120 mm/s is imparted to the
mass. Determine (a) the displacement and velocity of mass as a function
of time and (b) the displacement and velocity after 0.4 s.
6. A gun is so designed that on firing the barrel recoils against a spring. A
dashpot at the end of the recoil allows the barrel to come back to its
initial position within the minimum time without any oscillation. A gun
barrel has a mass of 500 kg and a recoil spring of stiffness 300 N/mm.
The barrel recoils 1m on firing. Determine (a) the initial recoil velocity
of the gun barrel and (b) critical damping coefficient of the dashpot
engaged at the end of the recoil stroke.
3.11
Further reading
Biggs J M (1964) Introduction to Structural Dynamics, McGraw-Hill, New York.
Chopra A K (2002) Dynamics of Structures – Theory and applications to earthquake
engineering, Eastern Economy Edition, Prentice Hall of India, New Delhi.
Clough R W and Penzien J (1974) Dynamics of Structures, McGraw-hill, New York.
DenHartog J P (1956) Mechanical Vibrations, 4th ed., McGraw-Hill, New York.
Humar J L (1990) Dynamics of Structures, Prentice Hall, Englewood Cliffs, NJ.
Jacobsen L S and Ayre R S (1958) Engineering Vibrations, McGraw-Hill Book Co.,
New York.
Meirovitch L (1980) Computational Methods in Structural Dynamics, Sijthoff and Nordhoff,
The Netherlands.
Paz M (1980) Structural Dynamics, Theory and Computation, Van Nostrand Reinhold,
New York.
Rao S S (2003) Mechanical Vibrations, 4th ed., Prentice Hall, Inc., Englewood Cliffs, NJ.
Thompson W T(1981) Theory of Vibration with Applications, 2nd ed., Prentice Hall,
Englewood, Cliffs, NJ.
Timoshenko S (1955) Vibration Problems in Engineering, Van Nostrand Company, Inc.,
Princeton, NJ.
Wilson E L (2002) Three Dimensional Static and Dynamic Analysis of Structures, Computers
and Structures, Inc., Berkeley, CA.