Lecture 11_Dynamic of structures_2019_MP.pdf

Design of offshore structures and foundations – Prof. C. Mazzotti, Eng. M. Palermo
1
Michele Palermo
DICAM
University of Bologna
DYNAMICS OF STRUCTURES

2
 Dynamic response of SDOF systems
• Undamped free vibrations
• Damped free vibration
• Harmonic undamped vibrations
• Harmonic damped vibrations
 Dynamic response of MDOF systems
• Methods of solutions
• Modal analysis
 Response spectrum
• Elastic spectrum
• Inelastic spectrum
 Seismic hazard: basic concepts
OUTLINE

3
Dynamic of SDOF
systems

Single-Degree of Freedom Systems: Mathematical model
1

Structures well approximted by Single-Degree of Freedom Systems
The concept of SDOF system will
be also used to reduce the
behaviour of a complex structure
to simpler ones.
1

6
RC frame structures: superposition of SDOF systems
piano 4°
piano 3°
piano 2°
piano 1°
30 x 30
35 x 35
40 x 40
45 x 45
30 x 30
35 x 35
40 x 40
40 x 40
30 x 40
35 x 40
40 x 40
40 x 40
m4
m3
m2
m1
k4
k3
k2
k1
(a) (b)
4-storey frame 4DOF model Superposition of 4
SDOF

Forces in a SDOF system
MASS
SPRING (STIFFNESS)
a) Undamped SDOF;
b) SDOF with damping;
DAMPER
3

APPLIED FORCE p(t)
Different excitations of a SDOF system
GROUND MOTION
ACCELERATION
DISPLACEMENT
DISPLACEMENTS
total
relative
)
(t
ug


4

Parameters to describe a Single Degree of Freedom system:
•STIFFNESS : Linear – elastic relationship
K depends on beam
stiffness EIb
Real case Rigid beam No stiffness beam
STIFFNESS
)
(t
u
k
fs 

Elastic restoring force

Parameters to describe a Single Degree of Freedom:
•DAMPER : Damping force
c : viscous damping coefficient [F t / L]
It cannot be calculated from geometrical parameters of the structure because
it is not possible to identify all the mechanisms that dissipate vibrational
energy of actual structure.
VISCOUS DAMPING COEFFICIENT
Damping force )
(t
u
c
fD




Free-body diagram at time t.
The mass is replaced by its inertia force
Newton’s second law of Motion
Dynamic equilibrium, the excitation is p(t)
)
(t
p
f
f
u
m s
D 




)
(t
p
ku
u
c
u
m 

 


u
m
fI 


)
(t
u
c
fD



)
(t
u
k
fs 

APPLIED FORCE p(t)

12
Dynamic equation of motion for different external actions
Undamped free vibrations:     0
mv t kv t
 
Damped free vibrations:       0
mv t cv t kv t
  
Undamped Vibrations under harmonic input::     0 sin
mv t kv t p t

 
      0 sin
mv t cv t kv t p t

  
Damped Vibrations under harmonic input::
       
mv t cv t kv t p t
  

FREE VIBRATION
Differential equation of motion of undamped system with external force
equal to zero:
Initial conditions (at t =0):
u=u(0) and
General solution:
Enforcing the initial conditions:
where
Natural circular
frequency of vibration
Natural period
of vibration
Natural frequency of
vibration
t
B
t
A
t
u n
n 
 sin
cos
)
( 

0

ku
u
m 

t
u
t
u
t
u n
n
n 

 sin
)
0
(
cos
)
0
(
)
(



m
k
n 

n
n
T


2



2
n
n
f 
FREE VIBRATION - Undamped System
)
0
(
u
u 
 

FREE VIBRATION
Solution: displacement as function of time
t
u
t
u
t
u n
n
n 



 sin
)
0
(
cos
)
0
(
)
(

m
k
n 

n
n
Π
T

2

 
2
2
0
)
0
(
)
0
( 







n
u
u
u


FREE VIBRATION - Undamped System
Initial conditions
period

18
Fundamental periods: examples
2
2 [sec]
m
T
k



 
Example: 4 storey RC frame
T = 0.59 sec
For an n-storey RC frame
T = n ·0.1 sec
piano 4°
piano 3°
piano 2°
piano 1°
30 x 30
35 x 35
40 x 40
45 x 45
30 x 30
35 x 35
40 x 40
40 x 40
30 x 40
35 x 40
40 x 40
40 x 40
m4
m3
m2
m1
k4
k3
k2
k1
(a) (b)

19
2
2 [sec]
m
T
k



 
Hospital Maggiore in Bologna
18 storey, steel frame + RC
cores
T (dir N-S) = 1.40 sec
T (dir E-W) = 0.95 sec
Fundamental period: examples

20
2
2 [sec]
m
T
k



 
Alcoa building, San Francisco,
California:
26 storey steel braced frame
T (torsionale) = 1.12 sec
N-S
E-W

21
2
2 [sec]
m
T
k



 
Transamerica building, San
Francisco, California:
60 storey (steel)

22
2
2 [sec]
m
T
k



 
Golden Gate, San Francisco,
California
Suspended bridge
span= 1300 m
T (vibr. trasv.) = 18.2 sec
T (vibr. vert.) = 10.9 sec
T (vibr. long.) = 3.81 sec
T (vibr. tors.) = 4.43 sec

23
2
2 [sec]
m
T
k



 
Fundamental period: Perla Steel jacket
67m
approximately 12 Km off Gela

FREE VIBRATION
Differential equation of motion of damped system in free
vibration:
Dividing by m:
where
Critical damping
coefficient
Damping ratio
0
)
(
)
(
)
( 

 t
u
k
t
u
c
t
u
m 


0
)
(
)
(
2
)
(
2


 t
u
t
u
t
u n
n 
 


cr
n c
c
m
c




2
n
n
cr
k
m
k
m
c


2
2
2 


FREE VIBRATION - SDOF with damping
m
k
n 

Natural circular
frequency of vibration
(of the undamped system)

Form of the solution:
Substituting in the differential equation gives:
It is satisfied for all values of time t, if :
In general:
Three cases are possible:
cr
n c
c
m
c




2
0
)
(
)
(
2
)
(
2


 t
u
t
u
t
u n
n 
 


st
e
u
0
)
2
(
2
2


 st
n
n e
s
s 

0
2
2
2


 n
n s
s 

 
2
1,2 1
n
s     
Free vibrations - SDOF with damping
 
 
1 2
2
1,2
2
1,2
1
1 1
1 1
n
n
n
s s
s
i
s
     
        
       
Critically damped
vibrations
Over-damped
vibrations
Under-damped
vibrations

FREE VIBRATION
Overdamped
Critically damped
Underdamped (buildings, bridges, civil structures)
cr
n c
c
m
c




2
1


1


1


  st
u t e

 
2 2
2 0
st
s s e
 
   2
1,2 1
s   
   
Characteristic equation:

FREE VIBRATION
Underdamped:
Solution:
The second part can rewritten in terms of trigonometric functions (Euler Formula):
t
i
t
e D
D
t
i D



sin
cos 


 
t
B
t
A
e
t
u D
D
t
n 


 

sin
cos
)
( Solution
Euler Formula
1


2
1 



 n
D
 
t
i
t
i
t D
D
n e
A
e
A
e
t
u 





 2
1
)
(





 





 2
2
,
1 1
i
s n
Circular frequency
of damped system
where A and B are constants to be determined using the initial conditions:

FREE VIBRATION
Initial conditions – for t = 0
The unknown coefficients
are:
 
t
B
t
A
e
t
u D
D
t
n 


 

sin
cos
)
(
2
1 



 n
D
)
0
(
u
A
D
n u
u
B


 )
0
(
)
0
( 














 
t
u
u
t
u
e
t
u D
D
n
D
t
n






sin
)
0
(
)
0
(
cos
)
0
(
)
(

Solution:
Underdamped:
1


)
0
(
),
0
( u
u 
Circular frequency of damped
system (related to the natural
frequency without damping)
Note: for we obtain the same solution of undamped system
0



FREE VIBRATION
Underdamped: for or 5%
Natural period of
damped system
(related to the
natural period
without damping)
05
.
0


cr
n c
c
m
c




2













 

t
u
u
t
u
e
t
u D
n
n
D
t
n
sin
)
0
(
)
0
(
cos
)
0
(
)
(

2
1 

 n
D
T
T

FREE VIBRATION
Effects of damping on the natural vibration frequency
For most structures, the
damped properties are
approximately equal to those
of undamped structure.
2
1 



 n
D 2
1 

 n
D
T
T
n
D T
T  n
D 



FREE VIBRATION
Effects of damping on the rate at which free vibrations decay
Systems subjected to the same initial displacement, different damping ratios:

FREE VIBRATION
Effect of damping on the rate of free vibration decay
Logarithmic decrement
















 
2
1 1
2
exp
)
exp(
)
(
)
(
D
n
i
i
D
T
u
u
T
t
u
t
u
2
1 1
2
ln







i
i
u
u
If is small,
 1
1 2






 2
 
t
B
t
A
e
t
u D
D
t
n



 

sin
cos
)
(
D
T

35
1
 
Steel structures with welded joints 1 – 3 %
Steel structures with bolted joints 3 – 5 %
Uncracked RC buildings 1 – 1.5 %
Cracked RC buildings 3 – 5 %
Sources of damping:
Inherent damping: dissipating phenomena in the elastic field (for design typically a
5% inhrent damping ratio is assumed).
Hysteretic damping: due to excursion into the inelastic field.
Added viscous damping: due to energy dissipation devices
Damping ratios for civil structures (material damping)

We will analyze the following cases:
 UNDAMPED SYSTEM SUBJECTED TO HARMONIC FORCE
 DAMPED SYSTEM SUBJECTED TO HARMONIC FORCE
 ENERGY DISSIPATED IN VISCOUS DAMPING
 RESPONSE TO IMPULSES (for UNDAMPED & DAMPED SYSTEM)
 RESPONSE TO A Ground Motion
 NUMERICAL SOLUTION METHODS
Forced Vibration of SDOF
p(t)

41
Why harmonic forces?
    0 sin
mv t kv t p t

 
  0 sin
p t p t

 Harmonic force
Fourier:
A dynamic force (like earthquake ground
motion) can be decomposed in a series of
harmonic components
   
1
cos
n n n
n
p t A t
 


 


42
Fourier
Why harmonic forces?

 
t
k
p
t
B
t
A
t
u
n
n
n 







 sin
/
1
1
sin
cos
)
( 2
0
FORCED VIBRATION
Equation of motion: Applied force p(t)
A, B are determined by imposing the initial condition for t=0
Complete solution:
Complementary solution Particular solution
(see the following)
t
p
ku
u
m 

 sin
0


t
k
p
u
u n
n 



 sin
0
2
2


Forced Vibration of SDOF - UNDAMPED
p(t)
n



Frequency of
harmonic force
Frequency
of SDOF

By imposing the initial condition we find
the two unknown constants A and B
From equation of motion ignoring
the dynamic effect, we obtain the
static deformation:
max
Transient
component
continues forever for
undamped system,
but in real system
due to the damping
the free vibration
decay with t.
 
 
t
k
p
u
t
u
t
u n
n
n
n
n 





 sin
1
)
0
(
cos
)
0
( 2
0











 
t
k
p
n
o



sin
1
1
2


TRANSIENT
STEADY STATE
t
k
p
ust 
sin
0
  
k
p
ust
0
0

Forced Vibration of SDOF - undamped
)
0
(
u
u  )
0
(
u
u 
 

where
Deformation Response Factor:
phase
In phase
Out of phase
u and p have the
same sign
u and p have the
opposite sign
The steady-state can be also written as:
   
 
t
u
t
u
n
st 


sin
1
1
2
0 







  )
(
sin
)
(
)
(
sin 0
0 


 


 t
R
u
t
u
t
u d
st
1
/ 
n


  2
0
0
)
/
(
1
1
n
st
d
u
u
R









 

n
n







180
0
Or :
where  
k
p
ust
0
0

p(t) u(t)

Deformation or Displacement
Response Factor:
is the ratio between dynamic and static
deformation.
dynamic deformation < static deformation
Then, dynamic deformation due to a rapid
force is very small.
similar to 1
dynamic deformation >> static deformation
1) For small value of frequency ratio,
Rd is similar to 1;
2)
3)
2)
1) 3)
  2
0
0
)
/
(
1
1
n
st
d
u
u
R





1
/ 
n


2
/ 
n


If 1

d
R
n

/
If 1

d
R
n

/

47
Dynamic amplification
  0
2
1
sin
1
p
p
v t t
k


 
  

 




0
0
static displacement under
p
p
k

  0
INPUT:
sin
p t p t


1
 

48
Some examples:
  0
INPUT:
sin
2
con 12.57 0.5sec
sec
p t p t
rad
T





   
EXTERNAL FORCE

49
 
  0
2
CONDIZIONI INIZIALI:
0 0
0
1
v
p
v
k



 

 

  
 
 

0.1sec 0.2
T


  
0.2sec 0.4
T


  
0.3sec 0.6
T


  
p0/k
-p0/k

50
 
  0
2
0 0
0
1
v
p
v
k



 

 

  
 
 

0.5sec 1
T


  
0.6sec 1.2
T


  
0.4sec 0.8
T


  
p0/k
-p0/k

51
 
  0
2
0 0
0
1
v
p
v
k



 

 

  
 
 

0.7sec 1.4
T


  
0.8sec 1.6
T


  
0.9sec 1.8
T


  
p0/k
-p0/k

What happens at the resonant
frequency?
The displacement grows infinitely but it becomes infinite after an infinite time.
If the particular solution defined previously is no longer valid because it is
contained in the complementary solution. The new particular solution is:
The new complete solution, considering at rest conditions at t =0, is:
n

 

Damped Systems subjected to Harmonic Force
Equation of motion:
Applied force:
Initial condition:
t
p
ku
u
c
u
m 
sin
0


 


)
0
(
u
u  )
0
(
u
u 
 
t
p
t
p 
 sin
)
( 0
n



t
p 
sin
0

Complete solution:
By imposing initial condition (at
t=0) we find the two unknown
constants A and B.
Applied force
n



Equation of motion:
t
p
ku
u
c
u
m 
sin
0


 


Forced Vibration of SDOF - damped
  t
D
t
C
t
B
t
A
e
t
u D
D
t
n





cos
sin
sin
cos
)
( 


 
2
1 



 n
D  
 
   
 2
2
2
2
0
2
1
/
1
n
n
n
k
p
C











 
   
 2
2
2
0
2
1
2
n
n
n
k
p
D












TRANSIENT STEADY STATE

Transiet component
decay with t for
damped system, at a
rate depending on
frequency ratio and
damping ratio.
Solution:
  t
D
t
C
t
B
t
A
e
t
u D
D
t
n







 

cos
sin
sin
cos
)
(
TRANSIENT STEADY STATE

56
The same example:
  0
INPUT:
sin
2
con 12.57 0.5sec
sec
p t p t
rad
T





   
FORZANTE ARMONICA ESTERNA

57
0.2sec 0.4
5%
T



  

0.2sec 0.4
10%
T



  

0.2sec 0.4
20%
T



  

p0/k
-p0/k

58
0.4sec 0.8
5%
T



  

0.4sec 0.8
10%
T



  

0.4sec 0.8
20%
T



  

p0/k
-p0/k

59
0.5sec 1
5%
T



  

0.5sec 1
10%
T



  

0.5sec 1
20%
T



  

p0/k
-p0/k
D=10
D=5
D=2.5

60
0.6sec 1.2
5%
T



  

0.6sec 1.2
10%
T



  

0.6sec 1.2
20%
T



  

p0/k
-p0/k

61
0.8sec 1.6
5%
T



  

0.8sec 1.6
10%
T



  

0.8sec 1.6
20%
T



  

p0/k
-p0/k

In comparison with
undamped system
the damping
lowers each peak
and limits the
response.
Bounded value
Response for:
n



   

























 

t
t
t
e
u
t
u n
D
D
t
o
st
n
cos
sin
1
cos
2
1
2
 
 


2
o
st
u
t
u
for or 5%
05
.
0



Deformation or Displacement
Response Factor:
dynamic deformation < static deformation,
indipendent from damping.
Then, dynamic deformation due to a
rapidly force is very small.
dynamic deformation >> static deformation
1) For small value of frequency ratio,Rd is
similar to 1. Then, dynamic and static
deformations are essentially the same
indipendent from damping.
2)
3)
1) 2)
3)
In this area the
response is controlled
by damping.
1
/ 
n


n

/
2
/ 
n


If 1

d
R
similar to 1
n

/
If 1

d
R

Phase Lag: With damping,the phase is not necessarily only 0° or 180° as
in the undamped case.
u and p have the
opposite sign.
p(t) u(t) p(t) u(t)
In phase
In phase
Out of phase
u and p have the
same sign.
Out of phase
ϕ = 𝑡𝑎𝑛−1
2ζ(𝜔 𝜔𝑛)
1 − 𝜔 𝜔𝑛
2

BASE GROUND-MOTION
The total displacement, velocity and acceleration of the mass are:
𝑢𝑡 𝑡 = 𝑢 𝑡 + 𝑢𝑔(𝑡)
Hence, the inertial force becomes:
𝑓𝐼 = 𝑚𝑢𝑡 𝑡 = 𝑚𝑢 𝑡 + 𝑚𝑢𝑔(𝑡)
The equation of motion becomes:
𝑚𝑢𝑡 𝑡 + 𝑐𝑢 𝑡 + 𝑘𝑢 𝑡 = 0
𝑚[𝑢 𝑡 + 𝑢𝑔 𝑡 ] + 𝑐𝑢 𝑡 + 𝑘𝑢 𝑡 = 0
𝑚𝑢 𝑡 + 𝑐𝑢 𝑡 + 𝑘𝑢 𝑡 = −𝑚𝑢𝑔 𝑡
𝑢𝑔(𝑡)
𝑢 𝑡
𝑢𝑡 𝑡
-1.5
-1
-0.5
0
0.5
1
1.5
2
0 10 20 30 40
Ground
acceleration
[m/s
2
]
Time [s]

.
Equation of motion :
Main Dynamic Analysis Methods :
1) Duhamel’s Integral;
2) Numerical Methods;
)
(t
p
ku
u
c
u
m 

 


Methods of solution for systems under arbitrary forces
Initial conditions
Initial conditions must be specified in order to define the problem completely.
Typically, the structure is «at rest» before a dynamic excitation, hence the initial
velocity and displacement are zero.

The response of a SDOF subjected to an arbitrary force p(t) can be
obtained decomposing the force in a sequence of infinitesimally short
impulses.
Hence, the response of the system at time t can be obtained by adding the
responses of all impulses up to that time.
This method can be used ONLY for LINEAR SYSTEMS
The impulse can be modeled as a Dirac delta function δ(t) for continuous-
time systems.
δ = 0 for t ≠
δ = 1 for t =


Systems under arbitrary forces: Duhamel’s Integral

According to Newton’s second law of motion:
That, integrated with respect to t, gives:
Hence, a unit impulse (magnitude equal 1) imposes to mass m the following
velocity:
This formulation is also applicable to spring or dampers: in fact, if the
force acts for an infinitesimally short duration, the spring and the damper
has no time to respond.
Impulse Theorem:
the change in
momentum equals
the impulse
Magnitude of
the impulse I:
m
u
1
)
( 


𝑝 𝑡 = 𝑚𝑢
𝑝 𝑡 𝑑𝑡 = 𝑚(𝑢2 − 𝑢1
𝑡2
𝑡1
) = 𝑚∆𝑢

Once the response of the dynamic system to a unit impulse is known, the solution can be
computed for a force varying arbitrarily with time.
The force p(t) can be represented as a sequence of infinitesimally short impulses.
The response of a linear dynamic system to the impulse at time  of magnitude p()d, is
this magnitude times the unit impulse-response function.
𝑑𝑢 𝑡 = 𝑝 𝜏 𝑑𝜏 ℎ(𝑡 − 𝜏) 𝑡 > 𝜏
The response of the system at time t is the sum of the responses to all impulses up to
that time:
𝑢 𝑡 = 𝑝 𝜏 ℎ 𝑡 − 𝜏 𝑑𝜏
𝑡
0
Convolution
Integral

Substituting the unit impulse response for the
damped system, we obtain Duhamel’s
Integral:
𝑢 𝑡 =
1
𝑚𝜔𝐷
𝑝 𝜏 𝑒−𝜁𝜔𝑛(𝑡−𝜏)
𝑡
0
𝑠𝑖𝑛 𝜔𝐷(𝑡 − 𝜏) 𝑑𝜏
assuming «at rest» initial conditions:
𝑢 0 = 0 𝑢 0 = 0
Duhamel’s Integral is based on the
principle of superposition of effects,
valid only for linear elastic systems.

For more general solution of the equation motion of SDOF systems
subjected to an arbitrary force it is possible to use TIME-STEPPING
METHODS.
In the present section we will consider only two numerical methods
for solving II order differential equation:
1. CENTRAL DIFFERENCE
2. NEWMARK’s METHOD
They can be used for linear and non linear systems.
Systems under arbitrary forces: Numerical Methods

The considered methods are based on the discretization of the applied force p, the
acceleration and velocity with respect to time.
In fact, the continuous function p(t) is substituted with discrete values pi, computed at
the discrete time instant ti.
A constant time interval is commonly adopted:
54
i
i
i t
t
t 

 1
As a result, the response of
the system (displacements,
accelerations or velocities) are
also obtained for the discrete
time instant ti.

At each time i the following equation has to be satisfied:
𝑚𝑢𝑖 + 𝑐𝑢𝑖 + (𝑓𝑠)𝑖= 𝑝𝑖
Where (𝑓𝑠)𝑖 represents the resisting force at time i (𝑘𝑢𝑖) if the system is linear elastic,
but depends on the previous history of displacement if the system is nonlinear.
The time-stepping methods allow to solve the previous equation for each instant i,
giving the desired response at each ti.
Approximate numerical procedures are adopted to step from time i to i+1.
The three important requirements for a numerical procedure are:
1. CONVERGENCE - as the time step decreases, the numerical solution should approach
the exact solution;
2. STABILITY - the numerical solution should be stable in the presence of numerical
round-off errors;
3. ACCURACY - the numerical procedure should provide results that are close enough
to the exact solution.

Dynamic of MDOF
systems
78

79
MULTI DEGREES OF FREEDOM
(MDOF)

80
Definition of EQUATIONS OF MOTION for
Planar systems (2-D)
Three cases will be introduced:
1. Free Vibration
MDOF without damping
MDOF with damping
2. Systems subjected to External Force
3. Systems subjected to Ground Motion
Dynamics of Structures

81
Systems of equations of motion for
MDOF in matrix form
)
(
)
(
)
(
)
( t
t
t
t p
Ku
u
C
u
M 

 


       
t
p
t
ku
t
u
c
t
u
m 

 


SDOF
Equation of motion for
Single Degree of Freedom
MDOF
m
Equations of Motion - MDOF
N
N
N
N
N

82
FREE VIBRATION
of MDOF SYSTEMS
without DAMPING
By free vibration we mean the motion of a structure without any
dynamic excitation – external forces or support motion. Free vibration
is initiated by disturbing the structure from its equilibrium position by
some initial displacements and/or by imparting some initial velocities.

Initial conditions:
Equation of motion in matrix form for undamped system
0
Ku
u
M 
 )
(
)
( t
t


)
0
(
)
0
(
u
u
u
u

 

The solution has the following form:
MDOF – Free Vibration
   
n n
u t q t


n-th circular frequency

84
Natural modes of vibration
A vibrating system with N degrees of Freedom has N natural vibration frequencies arranged
from smallest to largest
Natural periods:
  0
M
K 

 2
det n
Matrix eigenvalue problem
n

Natural frequencies
N





 ...
2
1
N
T
T
T 

 ...
2
1
n
n
T



2
Vector grouping the N
natural frequencies
TN
TN-2
TN-3
T1

85
Natural modes of vibration
A vibrating system with N degrees of Freedom has N natural vibration frequencies arranged
from smallest to largest
Natural periods:
  0
M
K 

 2
det n
Matrix eigenvalue problem
n

Natural frequencies Natural modes of vibration
N
vectors
N





 ...
2
1
N
T
T
T 

 ...
2
1
n
n
T



2
  0
φ
M
K 

 n
n
2
1
φ
2
φ
3
φ
N
φ

86
Modal matrix – each column is a natural mode
The modal vectors represent the vibration modes of the structure (natural modes
or principal modes). For each natural frequency there is a vibration mode.
The modal vectors are linearly independent. They are defined up to within a
multiplicative constant then the mode shapes are unique but indeterminate in
amplitude. The choice of the constant is called normalization. There are different
types of normalization.
1
φ
N
φ
1
φ 2
φ 3
φ N
φ
ij i DOF j MODE
  

87
MODAL SHAPES
= matrix of modal shape vector
= vector of modal coordinates
Vector of displacement as function of modal coordinates
(modal expansion);
 
t
t Φq
u 
)
(
 
t
q
  )
(
)
(
)
(
)
( 3
3
2
2
1
1 t
q
t
q
t
q
t
t 





 Φq
u
 
t
qn

88
FREE VIBRATION
of MDOF SYSTEMS
with DAMPING

89
Initial conditions
Equation of motion in matrix form for damped system
0
Ku
u
C
u
M 

 )
(
)
(
)
( t
t
t 


)
0
(
)
0
(
u
u
u
u

 

Φ
C
Φ
C
KΦ
Φ
K
MΦ
Φ
M
K
M
0
q
K
q
C
q
M
Φ
0
q
Φ
K
q
Φ
C
q
Φ
M
Φq
u
T
T
T
T
and
matrix
diagonal
matrix
diagonal
,
and
matrices
diagonal
gives
stiffness
the
to
and
mass
the
to
respect
ith
property w
ity
orthogonal
the
t
t
t
property)
lity
(orthogona
by
ying
premultipl
t
t
t
t
t
nt
displaceme
of
expansion
modal
the
uting
By substit










ˆ
ˆ
ˆ
ˆ
ˆ
)
(
ˆ
)
(
ˆ
)
(
ˆ
)
(
)
(
)
(
)
(
)
(






IN MODAL COORDINATES

90
Equation of motion in matrix form for damped system
0
Ku
u
C
u
M 

 )
(
)
(
)
( t
t
t 


CΦ
Φ
C
0
q
K
q
C
q
M
T
t
t
t




ˆ
)
(
ˆ
)
(
ˆ
)
(
ˆ 


Matrix may be diagonal or not.
1) If is diagonal, the system will be constituted by uncoupled differential
equations and the system will be said to have classical damping.
The classical modal analysis is applicable to classically damped systems.
The system has the same natural modes of undamped system.
2) If matrix is non diagonal, the system has non-classical damping.
The classical modal analysis is not applicable to this system (complex damping
theory is needed or approximate methods).
The system has not the same natural modes of undamped system.
Ĉ
Ĉ
Ĉ

91
RESPONSE SPECTRA
- elastic response -

ELASTIC RESPONSE of SDOF system
Chopra, Dynamics of Structures: Theory and Applications to Earthquake Engineering, 2nd Ed.
Accelerogram –
ground motion recording
Response of the SDOF
structure
u(t)
u
[mm]

93
Spectrum of the Seismic input
The earthquake ground motion time-history can be decomposed into
a series of harmonic contributions, each one having an amplitude, a
frequency and a phase
   
cos i t
n n n n
n n
p t A t Pe
 
 
 
  
 

94
Fourier Transform
Spectrum of the Seismic input

95
Spectrum of a recorded earthquake

96

97

98

99
Frequency content
• Typically, most of the input energy is concentrated in a range of
frequencies between 1-10 Hz corresponding to periods in the
range of 0.1-1 sec

From the equation of motion:
where:
is the total displacement of the top of the
structure with respect to the bedrock.
Recalling that 𝜔𝑛 =
𝑘
𝑚
, the elastic forces can
be rewritten as:
𝑓𝑠 t = ku t = 𝜔𝑛
2
𝑢 𝑡 𝑚 = 𝐴 𝑡 𝑚
𝜔𝑛
2𝑢 𝑡 is called pseudo-acceleration: 𝐴 𝑡
0
)
( 


 ku
u
c
u
u
m g





g
t u
u
u 

u(t)
Elastic Response of SDOF system
𝐴 𝑡
-4
-3
-2
-1
0
1
2
3
4
5
0 5 10 15 20
Ground
acceleration
[m/s
2
]
time [s]
-0.06
-0.04
-0.02
0
0.02
0.04
0.06
0 5 10 15 20
Displacement
[m]
time [s]
-12
-10
-8
-6
-4
-2
0
2
4
6
8
10
12
0 5 10 15 20
Pseudo-acceleration
[m/s
2
]
time [s]
SDOF: T=0.5s; 𝜉=5%
Pseudo-acceleration

For undamped system:
𝑚 𝑢(𝑡) + 𝑢𝑔(𝑡) + 𝑘𝑢(𝑡) = 0
𝑚 𝑢(𝑡) + 𝑢𝑔(𝑡) = −𝑘𝑢(𝑡)
𝑚 𝑢(𝑡) + 𝑢𝑔(𝑡) = −𝜔𝑛
2𝑢 𝑡 𝑚
𝑚 𝑢(𝑡) + 𝑢𝑔(𝑡) = −𝑚𝐴(𝑡)
If damping = 0, pseudo-acceleration = total acceleration of the system.
If damping ≠ 0, pseudo-acceleration ≠ total acceleration:
𝑚 𝑢(𝑡) + 𝑢𝑔(𝑡) + 𝑐𝑢 𝑡 + 𝑘𝑢(𝑡) = 0
𝑚 𝑢(𝑡) + 𝑢𝑔(𝑡) + 𝑐𝑢 𝑡 = −𝑘𝑢(𝑡)
𝑚 𝑢(𝑡) + 𝑢𝑔(𝑡) + 𝑐𝑢 𝑡 = −𝑚𝐴(𝑡)
u(t)
Elastic Response of SDOF system
Pseudo-acceleration

The maximum force is achieved when we have the maximum displacement during
time. Hence:
𝑓𝑠,𝑚𝑎𝑥 = max 𝑓𝑠 t = 𝜔𝑛
2 max |𝑢 𝑡 | 𝑚 = max |𝐴 𝑡 |𝑚
𝑓𝑠,𝑚𝑎𝑥 = 𝜔𝑛
2
𝑆𝐷𝑚 = 𝑆𝐴𝑚
Multiplying the maximum pseudo- acceleration (also called Peak pseudo-
acceleration) with the mass, the Equivalent static force is obtained.
Equivalent static force is the force that if applied to the structure induces on it the
same maximum displacement produced by the seismic action.
Elastic Response of SDOF system:
maximum equivalent static force
𝑆𝐷 𝑆𝐴
Pseudo-acceleration

Peak displacement
Peak Pseudo - velocity
Peak Pseudo – acceleration
Peak total acceleration
Peak relative velocity
Definitons:
𝑆𝐷 = max |𝑢 𝑡 |
𝑆𝑉 = max |𝑉 𝑡 | = 𝜔𝑛 𝑆𝐷
𝑆𝐴 = max |𝐴 𝑡 | = 𝜔𝑛
2𝑆𝐷
𝑆𝐴,𝑡𝑜𝑡 = max |𝑢𝑡 t |
𝑆𝑉,𝑟𝑒𝑙 = max |𝑢 𝑡 |
Where 𝜔𝑛 =
2𝜋
𝑇𝑛
Elastic response spectra
The Elastic Response Spectrum is the plot of a peak response quantity (displacement, pseudo-
velocity, pseudo-acceleration, …) of a SDOF system subjected a ground-motion, as a function of its
natural period T (or natural frequency), for a fixed damping ratio  (usually 5%).
The Response Spectrum is a very useful tool as it provides the maximum value of the structural
dynamic response without performing a time-history analysis.
Once the maximum structural response is known, it‘s then possible to evaluate the maximum elastic
forces in the structure due to the earthquake ground motion.
Peak response quantities

2 1
M
T s
K
  
Accelerogram selected
Structure (1 SDOF)
response
Evaluation of displacement response spectra
𝑆𝐷

2 1
M
T s
K
   2s
Structure (1 SDOF)
response
𝑆𝐷

By repeating the same procedure for
SDOF systems characterized by
different natural periods, the elastic
response spectrum is obtained.
The spectrum depends on the
selected damping ratio and on the
selected accelerogram.
𝑆𝐷

11
1. For a given accelerogram;
2. Select the damping ratio for the
structure ;
3. Select a value of the natural period T of
the SDOF;
4. Compute the structure (SDOF) response
to the accelerogram
5. Take the maximum response in terms of
displacement Sd = max [ u(t) ]
6. Determine the maximum pseudo-velocity
and the maximum pseudo-acceleration:
SV =  Sd = (2/T) Sd
SA = 2 Sd = (2/T)2 Sd
7. Repeat this process from step 3 to 6 for
every period T to be considered.
S
A
,
g
S
V
,
in/sec
:SUMMARY
S
D
,
in
Evaluation of displacement response spectra: summary

Accelerogram
mi
Spettro di
Risposta
Pseudo -Acceleration
Response Spectrum
Accelerograms
(acceleration time-histories)
Different Accelerograms Different response spectra
Elastic response spectra - different accelerograms

ζ = 0.1
ζ= 0.05
ζ= 0.02
Elastic response spectra - different damping ratios

Response Spectra
Displacement Pseudo - velocity Pseudo - acceleration
Response Spectra
can be defined in terms of
Related to the
peak deformation
of a system
Related to the peak strain
energy stored in the
system during the
earthquake:
𝐸𝑚𝑎𝑥 =
1
2
𝑘|𝑢𝑚𝑎𝑥|2
𝐸𝑚𝑎𝑥 =
1
2
𝜔𝑛
2𝑚𝑆𝑑
2
𝐸𝑚𝑎𝑥 =
1
2
𝑆𝑉
2
𝑚
Related to the peak
value of the equivalent
static force

It is convenient to display the spectra on a log-log graph, with the abscissa representing
the natural period of the system (or some dimensionless measure of it) and the ordinate
representing the pseudo velocity ,SV (in a dimensional or dimensionless form).
Response Spectra: the tripartite plot
Period T
S
V
El Centro Earthquake,
N-S component (1940)

On such a plot, 2 further axes can be identified: SD and SA.
Displ. = K
accell. = K
Period T
Velocity
Tripartite
Plots

Advantages:
• The response spectrum can be approximated more readily and accurately in terms of
all three quantities rather than in terms of a single quantity and an arithmetic plot.
• In certain regions of the spectrum the spectral deformations can more conveniently
be expressed indirectly in terms of SV or SA rather than directly in terms of Sd. All
these values can be read off directly from the logarithmic plot.

Period T
Velocity
Spectral Region:
Constant pseudo - velocity
Constant displacement
Constant pseudo - acceleration
Tripartite
Plots

Velocity
Acceleration approximately equal to
ground acceleration - Relative
displacement very small
Acceleration very small - Relative
displacement approximately equal to
ground displacement
Period T
Low values of T
High values of T
ug
ug

m = 43 000 kg
k = 40 000 kN/m
ζ= 0.02
s
T
s
rad
m
k
n
n
n
206
.
0
5
.
30
1415
.
3
2
2
/
5
.
30
43000
1000
40000











Example of use of the response spectrum

1.08
0.206

m = 43 000 kg
k = 40 000 kN/m
ζ = 0.02
from response spectrum we obtain:
Max pseudo acceleration = SA (T=0.206s) = 1.08 g
Maximum Displacement =SD (T=0.206s) = SA (T=0.206s)/𝜔n
2=
11.4 mm
Maximum force acting on the structure:
Fs,max = m*SA =43 000 x 1.08 x 9.81/1000 = 456 kN
Fs,max = k*SD= 40 000 x 11.4/1000 = 456 kN
s
T
s
rad
m
k
n
n
n
206
.
0
5
.
30
1415
.
3
2
2
/
5
.
30
43000
1000
40000











)
,
(
)
,
(
)
,
(
2
max
, 





 n
A
n
D
n
n
D
S S
m
S
m
kS
F 




RESPONSE SPECTRA
- inelastic spectra -

1
SDOF with non-linear behaviour
Steel structure
Masonry wall
RC frame
DUCTILITY OF STRUCTURES – see the
Advanced Design of Structures course

2
• Ductile structures can withstand significant deformations beyond the elastic limit,
before reaching the failures.
This ability - DUCTILITY - is a property of structures as important as strength.
• The Equal displacement rule ad Equal energy rule are commonly adopted to relate
force reduction factor (or behavior factor R) and ductility demand:
 The maximum displacement of the inelastic system under an earthquake is
assumed equal to that of the corresponding linear-elastic system
 The maximum displacement of the inelastic system under an earthquake is
obtained according to energy criterion
Seismic Response of Structures
Elastic behaviour
y
u
d
d


dy du
H
Inelastic behaviour

3
Elastic SDOF Inelastic SDOF
dy du
H
de
H

4
Elastic Behaviour
Inelastic Behaviour
Comparison in
terms of
displacements

5
SDOF: non-linear behaviour
Yield strength reduction
factor or structural
reduction factor
Ductility demand
y
m
u
u


1
0 

y
y
f
f
R
uy um
H
u0
H

6
Yield reduction factor:
Ductility demand:
Ratio of peak
deformations
1
0
0



y
y
y
u
u
f
f
R
y
m
u
u


y
m
R
u
u 

0
y
m
y
m
y
m
R
u
u
u
u
u
u
u
u
u
u
0
0
0
0
0




1) Equal disp.
2) Equal energy
1
2 

 d
y q
R 
2)
1
2 

 d
y q
R 
d
1)

Ductility demand:
um uu
Seismic Demand
= Demand from
seismic action
Capacity
DUCTILITY CAPACITY:
Depends only on
the structural
properties
y
u
C
u
u


y
m
D
u
u


Comparison with the
available ductility limit
D
C 


m
u u
u 
Capacity > Demand

Chopra, Dynamics of Structures: Theory and Applications
to Earthquake Engineering, 2nd Ed. 11
0
1 f
  
1 y
f
  
Spectra with different
levels of DUCTILITY
DEMAND
Elastic response
spectrum

29
0
0.2
0.4
0.6
0.8
1
1.2
0 0.5 1 1.5 2 2.5 3 3.5 4
T [sec]
Sd
[g]
Serie
Serie5 q
S
S el
d 
EUROCODE 8 – Code Inelastic Response Spectra

31
EUROCODE 8 – Behaviour Factor

Example : Design Response Spectra for RC structures
SDOF Multi storey structures MDOF
0
0.2
0.4
0.6
0.8
1
1.2
0 0.5 1 1.5 2 2.5 3 3.5 4
T [sec]
Sd
[g]
Frame structures q=5.85
Elastic Spectrum
Wall systems q=4.4
Uncoupled wall systems q=4.8
Torsionally flexible systems q=3
36
Elastic Spectrum
Frame structure
High Ductility Class

37
Italian code (NTC2008) adopts a linear relationship instead of the
criterion of equal energy.
q
T
  = q
Linear relationship
between  and q
Tc
elastic





Ee
d
E d
d 

 
q
d 

1
)
1
(
1 T
T
q C
d 





Seismic hazard
133

SA (0.5 s) Return period 475 yrs SA (1.0 s) Return period 475 yrs

Seismic risk
Different earthquake effects lead to different
kinds of risk.
In general we can define risk as the
combination of:
RISK = HAZARD (*) VULNERABILITY (*) EXPOSED VALUE

Seismic risk
RISK = HAZARD (*) VULNERABILITY (*) EXPOSED VALUE
• Seismic Hazard is the probability of occurrence of a specified level
of ground shaking in a specified period of time. But a more general
definition includes anything associated with an earthquake that may
affect the normal activities of people, i.e. surface faulting, ground
shaking, landslides, liquefaction, tectonic deformation, and
tsunamis.
• Vulnerability is the degree of damage caused by various levels of
loading. The vulnerability may be calculated in a probabilistic or
deterministic way for a single structure or groups of structures.
• Seismic Risk is expressed in terms of economic costs, loss of lives
or environmental damage per unit of time.

Hazard and Vulnerability
475years in terms of a macroseismic intensity on the MCS
ucantoni et al. 2001)
Seismic hazard maps for a return period of 475
years in terms of peak ground acceleration
Fig. 2 Seismic hazard maps for a return period of 475years in terms o
scale, b peak ground acceleration (adapted from Lucantoni et al. 2001
Fig. 3 Vulnerability maps produced in a the SAVE Project (Zuccaro
Vulnerability map for residential buildings. A =
highest vulnerability
Lucantoni A, Bosi V, Bramerini F, De Marco R, Lo Presti T, Naso G, Sabetta F (2001) Seismic risk in Italy.
Ing Sismica XVII(1):5–36
0
0.2
0.4
0.6
0.8
1
0 0.1 0.2 0.3 0.4 0.5
P(limit
state)
Sa(T1) [g]
Life safety
Collapse
(c)

Hazard x Vulnerability
Number of collapsed dwellings per municipality due to ground motions
with a 10% probability of exceedance in 50years
of 475years in terms of a macroseismic intensity on the MCS
m Lucantoni et al. 2001)
AVE Project (Zuccaro 2004), and b Lucantoni et al. (2001)
Fig. 2 Seismic hazard maps for a return period of 475years in terms of a macroseismic intensity on the MCS
scale, b peak ground acceleration (adapted from Lucantoni et al. 2001)
Fig. 3 Vulnerability maps produced in a the SAVE Project (Zuccaro 2004), and b Lucantoni et al. (2001)
percentage distribution of buildings belonging to vulnerability class A within each munici-

LIMIT STATE Exceedance
probability in the
reference period VR
Serviceability Fully Operational 81%
Damage 63%
Ultimate Life Safe 10%
Near Collapse 5%
VR
=VN
×CU
Nominal life Occupancy factor
Reference life

Structure Nominal life VN
Temporary, Provisional, etc. ≤ 10
Normal buildings, bridges, normal
infrastructures, normal dams, etc.
≥50
Special structures, strategic bridges,
strategic infrastructures, strategic
dams
≥100
CLASS I II III IV
OCCUPANCY
FACTOR
0.7 1.0 1.5 2.0
Class I: Agricultural facilities, etc.
Class II: Normal buildings
Class III: Buildings with large crowds, etc.
Class IV: Public facilities (e.g. hospitals), strategic bridges, etc.

PROBABILITY OF
EXCEEDANCE DURING
SERVICE LIFE
SERVICE LIFE: VN*CU
RETURN PERIOD (years)
HAZARD ANALYSIS AND ITALIAN CODE
 
1
1
ln 



n
R
R
P
V
T
 
R
n
V
P 1
1
ln 




LIMIT STATES
SLS
ULS

ELASTIC RESPONSE SPECTRA FOR REFERENCE RETURN PERIODS
Spectral
acceleration
[g]
T [s]

PGA
Site response
*
C
C
C T
C
T 

3
/
C
B T
T 
6
.
1
4 

g
a
T
g
D
T
S S
S
S 

Topographic effect
Stratigraphic
effect
0
g
a S F
 
0
F
*
0
, ,
g C
a F T
DEPEND ON SITE LOCATION
(MAPS)
(%)
DR

HAZARD ANALYSIS AND ITALIAN CODE
SOIL

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