CHAPTER 2
RESPONSE OF A SINGLE DEGREE OF FREEDOM
SYSTEM
Seismic Analysis and Design of Buildings
By: Prof. A. A. ElDamatty
.
Page:
2-1
2-1 Equation of motion of a SDOF
Basic components of a dynamic problem:
1.
2.
3.
4.
5.
Mass (m = W/g)
Applied load P(t)
Inertia force = FI
Damping force = FD
Elastic force = FS
Single Degree of Freedom Model:
x(t)
C
M
P(t)
K
Using D’Alembert’s principle; a dynamic system can be
considered to be in equilibrium if the inertia force is included in
the free body diagram.
FD
P(t)
FI
FS
FI + FD + FS = P(t)
Where:
d 2x
FI = mass x acceleration  m 2  mx
dt
FD = Dashpot constant x velocity = cx
FS = Spring stiffness x displacement = kx
 Equation of motion:
mx  cx  kx  P(t )
Seismic Analysis and Design of Buildings
By: Prof. A. A. ElDamatty
Elastic Stiffness for Elements:
→ Force Required for Unit displacement
k = P/
1) Springs in Series:
K1
P1
P1 = P2 = P
P1 = K11 ;
P2 = K22 ;
K2
P2
 =1 + 2
KK
P
P
P


 1 2
 1   2 P1 P2 K1  K 2

K1 K 2
KK
K  1 2
K1  K 2
K
2) Springs in Parallel:
P1, 1, K1
P, , K
P2, 2, K2
1 = 2 =
K
P P1  P2 K11  K 2 2





 K  K1  K2
P, 
.
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2-2
Seismic Analysis and Design of Buildings
By: Prof. A. A. ElDamatty
.
Page:
2-3
3) Axial Stiffness of a Bar:

P
PL
EA
K 
L
EA
L
4) Torsion:
T

TL
GJ
K 
L
GJ
L
5) Beams in Flexure:
a) Cantilever:
P

PL3
3EI

K  3
L
3EI
L
b) Simple Beam:
P

PL3
48 EI

K 
L3
48EI


L


Seismic Analysis and Design of Buildings
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c)
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Page:
2-4
P

PL3
12 EI

K 
L3
12 EI
L
P
d)


PL3
3EI

K  3
3EI
L

L
e)

K1
K1 
K2
12 EI 1
L13
K2 
12 EI 2
L32
K1 and K2 are two springs in parallel (have the same displacement)
 K  K1  K 2 
12 E
I 1  I 2 
L13
Seismic Analysis and Design of Buildings
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2-5
Type of Dynamic Loading:
X 0
0.1
a) Free Vibration:
0.08
0.06
X(t)
X0
0.04
X(t)
0.02
0
-0.02 0
2
4
6
Time
8
-0.04
-0.06
-0.08
-0.1
Time
1
b) Forced Harmonic Vibration:
0.8
0.6
P(t)
P0
0.4
P(t )  P0 sin t
P(t)
0.2
0
-0.2 0
(Machine Vibration)
2
Time6
4
-0.4
-0.6
-0.8
-1
120
c) Impulse Loading:
Time
100
P(t)
P(t)
(Driving of Piles)
P0
80
60
40
20
0
Time
Time
d) Random Vibration:
15
10
P(t)
P(t)
5
0
0
200
400
600
800
1000
-5
-10
-15
Time (s)
1200
1400
1600
1800
Time
2000
8
Seismic Analysis and Design of Buildings
By: Prof. A. A. ElDamatty
.
Page:
2-6
2-2 Undamped Free Vibration Response of a SDOF
Equation of motion: mx  kx  0
 x  02 x  0
;
02 
k
m
Solution of the above D.E.:
X(t) = Asin(0t) + Bcos(0t)
Substituting with the initial conditions: X(0) = X0 and X (0)  X 0
X 0
we obtain:
X(t) =  sin(t   X 0cos(t 
0
The above equation can be written in a compact form as:
2
 X 0 
2
X(t) =     X 0 sin( t   
 0
 X 
  tan1  0 
 X 0 / 0 
X 0
0.1
T = 2/0
X(t)0.08
0.06
0.04
X0
X(t)
0.02
0
-0.02 0
2
4
-0.04
-0.06
-0.08
-0.1
Time
6
8
Seismic Analysis and Design of Buildings
By: Prof. A. A. ElDamatty
2-3 Free Vibration of a Damped System
Equation of motion:
c


Define:
2m0
mx  cx  kx  0
c
k
x  x  x  0
m
m
c
x  x  02 x  0
m
(damping ratio)
2
Equation of motion: x  20x  0 x  0
For 0 <  < 1
The solution of the above equation becomes:
X(t) = e
 0t
A1sin(Dt   A2 cos(Dt 
2
Where: D  1   0
(Frequency of vibration of a damped system)
X 0   0 X 0
A1 
D
A2  X 0
The above equation can be written in a compact form as:
X(t) =
Xe 0t sin( D t   
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2-7
Seismic Analysis and Design of Buildings
By: Prof. A. A. ElDamatty
.
Page:
2-8
1
0.8
0.6
0.4
X(t)
0.2
0
-0.2 0
5
10
15
20
-0.4
TD = 2/D
-0.6
-0.8
-1
Time
Response to Harmonic Motion
x(t)
C
P(t) = P0cost or P0sint
M
K
P0
cost _____________(*)
m
The solution consists of a homogeneous solution and a particular
one:
2
Equation of motion: x  20x  0 x 
Homogeneous: XC(t) =
e  0t A1sin(D t   A2 cos(D t 
Try a particular solution in the form:
XP(t) = Csin(t) + Dcos(t)
Seismic Analysis and Design of Buildings
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Page:
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Substituting for the particular solution into (*) and solving for the
constants C and D we obtain:
XP(t) = XP cos(t – )
Where:
XP 
P0
m02
Amplitude of
forced response
2
    2      2
1      2  
   0      0 
 
 2   
 

  tan 1   0 2 
1     
   0  


The angle at which the
response lags the applied load
Adding the complementary and particular solutions, the total
response is:
P0
cos(t   
2
m

0
X (t )  e  0t A1sin( D t   A2 cos( D t  
2
    2      2
1      2  
   0      0 
=
transient term
Decays exponentially and
occurs at the damped
frequency of the structure
+
steady state term
Takes place at the forced
frequency. Lags behind the
force by phase angle .
Seismic Analysis and Design of Buildings
By: Prof. A. A. ElDamatty
.
Page:
2-10
Practically in design the steady state response is only considered.
Define DMF Dynamic Magnification Factor:
XP 
P0
1
m 02
    2      2
1      2  
   0      0 
2
P0
P0
X


i.e. St K m 2
0
DMF 
 X P  X St .DMF
1
2
    2      2
1      2  
   0      0 
Dynamic Magnification Factor
12
Dynamic Magnification Factor
11
10

9
8
.5
7
.1
6
5
.2
4
3
2
1
0
0
0.5
1
1.5
/0
2
2.5
3
Seismic Analysis and Design of Buildings
By: Prof. A. A. ElDamatty
.
Page:
2-11
Example 2-1
A simply supported beam of length “L” carries an electric motor of
weight W = 12 KN at it mid – span.
The flexural rigidity “EI” of the beam is such that the static
deflection under the weight of the machine is 3 mm. The
equivalent viscous damping is such that after 10 cycles, the free
vibration amplitude due to initial displacement X0 is reduced to 60
% of X0. The motor operates at 600 RPM and causes a harmonic
force due to unbalance F0= 2.5 KN. Neglecting the mass of the
beam, determine:
a) The value of the undamped frequency 0.
b) The value of the damped frequency D.
c) The magnitude of the maximum dynamic displacement “X”
for the given values of the parameters due to the operation of
the machine.
d) The maximum total deflection including gravity and dynamic
loading.
Seismic Analysis and Design of Buildings
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Seismic Analysis and Design of Buildings
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Page:
2-15
2-4 Equation of Motion of a SDOF under Earthquake
Motion
C
M
X(t)
K
Ϋg(t)
The figure above shows a simulation for a SDOF system
subjected to an earthquake ground motion described by a time
history ground acceleration Ϋg(t). X(t) represents the displacement
response – relative to ground of this SDOF system.
The equation of motion of the SDOF is given as:
m( yg (t )  x(t ))  cx  kx  0
mx(t )  cx  kx  myg (t )
As such, due to earthquake, the SDOF is subjected to a dynamic
force P(t )  myg (t )
Dividing the previous equation by the mass “m”, we get:
Seismic Analysis and Design of Buildings
x(t ) 
By: Prof. A. A. ElDamatty
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c
k
x  x   yg (t )
m
m
k


0
Defining the natural frequency of the system
m and its
c


damping ratio
2m0 , the equation of motion can be written as:
x  20x  02 x   yg (t )
The solution of the
above 2nd order D.E.
can be obtained by
obtained using:
Time History
Analysis
Response Spectrum
Analysis
Provides the entire
time response.
Provides only
maximum responses.
Can be applied to
linear and non-linear
systems.
Applicable only to
linear systems.
Involves significant
computations.
Simple
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2-5 Time History Analysis
Closed form mathematical solutions exist for the free
vibration response of a SDOF system as well as the responses to
harmonic and impulse loading. Dynamic problems which involve
forces that can not be expressed in simple mathematical forms
(example earthquake forces) can not be solved in a closed form.
For such problems, the response is obtained by applying numerical
integration techniques to the equation of motion. In general, the
numerical integration techniques can be categorised to:
1) Explicit Algorithms.
2) Implicit Algorithms.
One of the most widely applied implicit algorithms will be
discussed in details in this section.
Implicit Algorithms
Newmark method
Wilson –  method
Newmark method
We will assume:


x n1  x n  1   xn  xn1 t



_______________ (1)
x n 1  x n  x n t  1   xn  xn 1 t 2
_______________ (2)
2

 and  are implicit parameters that can be determined to obtain
integration accuracy and stability.
Seismic Analysis and Design of Buildings
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 = 1/2 and  = 1/6  correspond to linear acceleration.
 = 1/2 and  = 1/4  correspond to constant average acceleration.
Equation of motion at t = (n+1)t:
mxn1  cx n1  kxn1  F n1 (t )
_______________ (3)
For constant – average acceleration scheme ( = 1/2 and  = 1/4):


x n1  x n  xn  xn1 t / 2

_______________ (4)

x n1  x n  x n t  xn  xn1 t 2 / 4

_______________ (5)
From (5):



xn1  x n1  x n  x n t 4 / t 2  xn
_______________ (6)
Substituting (6) into (4), we get:




x n1  x n  xn t / 2  x n1  x n  x n t 4 / t 2  xn t / 2 ___ (7)
Equation (7) can be simplified to give:


x n1  x n1  x n 2 / t   x n
_______________ (8)
Substituting (6) and (8) into (3), we get:
 4 x n 4 x n n   2 x n n 
 4m 2c  n1
n1
 x   c
 x  ___ (9)
 2   k  x  F (t )  m 2 
t
 t t 
 t
  t

Seismic Analysis and Design of Buildings
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Step–By–Step Solution Using Newmark method ( = ½,  = ¼)
A- Initial Calculations:
1. Evaluate k, m and c.
0
00 00
,xx , x0
2. Initialize x , xx ,and
3. Select a time step t.
4. Calculate: a0 = 4/t2, a1 = 2/t, a2 = 4/t, a3 = t/2 .
5. Evaluate kˆ  k  a0m  a1c
B- For Each time step:
1. Calculate effective load at time t = (n+1)t; Fˆ n 1

 
Fˆ n1  F n1  m a0 x n  a2 x n  xn  c a1 x n  x n

2. Solve kˆx n 1  Fˆ n 1
3. Calculate acceleration and velocity at t = (n+1)t


xn1  a0 x n1  x n  a2 x n  xn

x n 1  x n  a3 xn 1  xn

Seismic Analysis and Design of Buildings
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Example 2-2
The frame structure shown below is subjected to a strong
ground motion pulse with a peak acceleration A0 = 1.2g. The latter
is modeled as a triangular pulse load of duration t1 equal to half the
period of vibration of the structure.
Assuming zero damping, determine by numerical integration
the maximum displacement of the structure, Xmax, and also the
time, tmax, at which it occurs. Use a time step t=T/10 sec.
X
W = 200 KN
I=∞
4.5 m
3.0 m
W 250x33
Ϋg(t)
Ϋg(t)
A0
t1= 0.5T
A0= 1.2g
t1/2
t1
t
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2-6 Concept of Elastic Response Spectrum
As a central concept of earthquake engineering, the response
spectrum provides a convenient means to summarize the peak
response of all possible linear SDOF systems to a particular
component of ground motion.
Construction of an Elastic Response Spectrum
The following steps are conducted in order to construct the
response spectrum of a given ground motion.
1. Consider a SDOF subjected to a ground motion given by the
time history analysis acceleration Ϋg(t). The SDOF system
2
m
T


2

has a certain period
0
k and a damping ratio .
M
x
C
K/2
K/2
Ϋg(t)
2. Equation of Motion:
m( yg (t )  x(t ))  cx  kx  0
mx(t )  cx  kx  myg (t )
2
Or x  20x  0 x   yg (t )
x is the relative
displacement
between the
mass and the
ground
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Knowing the time history record Ϋg(t), the response of the
structure x(t) can be evaluated using any of the numerical methods.
X(t)
Time
Sd
The maximum value Sd for the response displacement is recorded.
This maximum displacement Sd(T,) depends on the period and the
damping ratio of the SDOF.
Sd(T,) is called the spectral displacement.
3. The maximum velocity response Sv(T,) can be obtained by
the relation: Sv(T,) = 0Sd(T,)
4. The maximum acceleration response Sa(T,) can be obtained
by the relation: Sa(T,) = 0Sv(T,) = 02Sd(T,)
5. Steps (1) to (4) are repeated for different values for the
natural period of the system T as well as for different
damping ratios .
Seismic Analysis and Design of Buildings
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The following observations could be noticed:
1. Systems with very short period, say T < Ta = 0.025 s have
Sa= üg0 and Sd is very small. Why??
2. Systems with very long period, say T > 15 s have Sd
approaches Yg and Sa is very small. Why??
3. For short–period systems, Sa can be idealised as constant.
4. For medium–period systems, Sv can be idealised as constant.
5. For long–period systems, Sd can be idealised as constant.
6. Sa, Sv and Sd are plotted together on a log-log scale as
function of T and .
Figure 1 Response Spectrum (=0, 2, 5 and 10%) for El Centro
earthquake
N.B. Each earthquake has its own response spectrum
Seismic Analysis and Design of Buildings
By: Prof. A. A. ElDamatty
.
Page:
2-28
Based on the above observations one can conclude the following:
 The long period region is called the displacement –
sensitive region as the structural response is most
directly related to ground displacement.
 The medium period region is called the velocity –
sensitive region as the structural response is most
directly related to ground velocity.
 The short period region is called the acceleration –
sensitive region as the structural response is most
directly related to ground acceleration.
Elastic Design Spectrum
Based on the analysis of a large ensemble of ground motions
recorded on firm ground an elastic design spectrum was developed
by researchers as shown.
Figure 2 Elastic Design Spectrum
Seismic Analysis and Design of Buildings
By: Prof. A. A. ElDamatty
.
Page:
2-29
The amplification factors for two different non exceedance
probabilities, 50% (median) and 84.1 (median plus one standard
deviation) are given in Table 1 for several values of damping.
Table 1 Amplification Factors: Elastic Design Spectra
Median
One sigma
Damping,
(50 Percentile)
(84.1 Percentile)
 (%)
A
V
D
A
V
D
1
3.21
2.31
1.82
4.38
3.38
2.73
2
2.74
2.03
1.63
3.66
2.92
2.42
5
2.12
1.65
1.59
2.71
2.30
2.01
10
1.64
1.37
1.2
1.99
1.84
1.69
20
1.17
1.08
1.01
1.26
1.37
1.38
Seismic Analysis and Design of Buildings
By: Prof. A. A. ElDamatty
.
Page:
2-30
2-7 Seismic Response of Structures Using Response
Spectrum for a SDOF System
The elevated tank shown in the opposite figure
M
can be treated as a SDOF system having a period T
given by:
H
K
2
m
T
 2
0
k
The maximum response of the tank to an earthquake (e.g. El
Centro earthquake) can be obtained by applying the following
steps:
1. A damping ratio  is assumed for the structure.
2. Using T and  the response spectrum of El Centro can be
used to evaluate the maximum relative displacement Sd
experienced by the mass “M” during the earthquake.
3. The maximum velocity Sv(T,) = 0Sd(T,)
4. The maximum acceleration Sa(T,) = 0Sv(T,)
5. The maximum base shear Vmax = MSa = M02Sd
6. The maximum overturning moment Mmax= MHSa= MH02Sd
Seismic Analysis and Design of Buildings
By: Prof. A. A. ElDamatty
.
Page:
2-31
Example 2-3
Consider a steel spherical elevated water tank supported by a
single circular pipe. The diameter of the tank is 5 m and it is made
of steel 0.01 m thick. The circular steel pipe is 1 m in diameter
with a wall thickness of 0.02 m wall thickness. The height of the
water tank (center of tank to ground level) is 22.5 m.
Assuming the structure can be treated as a SDOF system with the
total mass concentrated at the water tank level and the foundation
condition is taken as rigid,
a) Calculate the maximum displacement, base shear force and
the overturning moment caused by the 1940 El Centro
earthquake.
b) How does the maximum displacement and base shear change
if the thickness of the supporting pipe is doubled?
Seismic Analysis and Design of Buildings
By: Prof. A. A. ElDamatty
.
Page:
2-32
Seismic Analysis and Design of Buildings
By: Prof. A. A. ElDamatty
.
Page:
2-33
Seismic Analysis and Design of Buildings
By: Prof. A. A. ElDamatty
.
Page:
2-34