INDIAN INSTITUTE OF TECHNOLOGY ROORKEE
Dynamic Behaviour of Bridges
by
Prof. P.C. Ashwin Kumar
Department of Earthquake Engineering, IIT Roorkee
India, 247667
• Any load is dynamic if
Force
Structural Dynamics
Time
• As designers we are interested in maximum displacement umax
Kumax = Pmax
• When a body is acted upon by a force, the time rate of change of its momentum
equals the force
d  du 
d 2u
p (t )   m   m 2
dt  dt 
dt
• D’ Alembert’s Principle: At any instant of time the unbalanced force equals to mass
times the acceleration
ku(t)
m
p(t)
p(t )  ku (t )  mu(t )

mu(t )  ku (t )  p (t )
2
Structural Dynamics
Equation of Motion for a SDOF system
fc = Reactive force due to damper (cu)
fk = Reactive force due to spring (ku)
mü = Pseudo force
Hence, by force equilibrium,
mü  fc  fk  p(t )
mü  cu  ku  p(t )
In case of an incoming earthquake,
the exciting force is in the form of
ground acceleration. So as to
consider its effect in the inertial
frame, we denote the equation of
motion as:
mü  fc  fk  müg
3
Structural Dynamics
Degree of freedom
The number of independent coordinates required to
define the motion/deformation of a structure is
called the degree of freedom of the structure. Skillful
determination of degree of freedom is the key to
reducing problem complexity and duration of
analysis.
Single degree of freedom (SDOF) system
A SDOF system is one whose
complete trajectory of motion can be
defined by defining only one degree
of freedom.
4
Free Vibration of SDOF
Free Vibration-A structure is said to be undergoing free vibration when it is disturbed
from its static equilibrium position and then allowed to vibrate without any external
dynamic excitation. The free vibration can either be an output of an initial velocity or
displacement. Where wn is determined as:
wn 
k
m
The time period of oscillation is related to the circular natural frequency by the following
relation:
Tn 
2
wn
The cyclic natural frequency is related to the circular natural frequency and natural period of
oscillation by the following relation:
fn 
wn 1

2 Tn
5
6
Free Vibration of SDOF
• Equation of motion for free vibration with some initial condition u0 and u0
mu  ku  0
• Solution of second order linear differential equation
• Trivial solution u = 0, physically if the structure is not subjected to load it will not
vibrate
• Consider a function to describe the displacement
u (t )  u (0) cos w nt 
u (0)
wn
sin w nt
• The amplitude of this motion given by:
 u (0) 
A  {u (0)}2  

w
n


2
7
Forced Vibration of SDOF
Forced Vibration- The vibration induced in a structure due to the virtue of an external
exciting agent is called forced vibration. The response of such vibration varies with
frequency of the exciting force i.e. steady state solution in addition to its
natural/transient counterparts:
u (t )  ut (t )  us (t )
In case of a damped vibration case, the transient response is varies with the
damped natural frequency of the structure, given by:
w D  wn 1   2
Here, ζ is the damping ratio of the SDOF system, given by:
 
c
c

2mw n 2 km
8
Forced Vibration of SDOF
Forced Vibration- The vibration induced in a structure due to the virtue of an external
exciting agent is called forced vibration. The response of such vibration varies with
frequency of the exciting force i.e. steady state solution in addition to its
natural/transient counterparts:
mu  ku  p (t )
p (t )  p0 sin w t
u (t )  ut (t )  us (t )
us (t )  G1 sin w t  G2 cos wt
ut (t )  C1 sin wt  C2 cos wt




w
u (0) po
1

w n  sin w nt  po
u (t )  u (0) cos w nt  

sin wt
2 
2
wn
k 1 w
k 1 w


wn 
wn
transient




steady  state
9
Forced Un-damped Vibration of SDOF
p0
wn p0
w

 0.2 u (0)  0.5
u0 
wn
k
k
10
Forced Damped Vibration of SDOF
p0
wn p0
w

 0.2 u (0)  0.5
u0 
  0.05
wn
k
k
11
El-Centro Ground Motion
•
Horizontal ground acceleration
recorded during 1940 Imperial
Valley earthquake
•
Ground velocity and ground
displacement computed by
integrating ground acceleration
12
SDOF Response
•
Deformation response of three linear SDF systems to El-Centro earthquake
13
System Force Demand Determination
•
Once deformation response history, u(t), has been evaluated, internal
forces/stresses can be determined by static analysis of structure at each time instant
𝒔
𝟐
𝒏
•
A(t) pseudo-acceleration response of the system
14
Response Spectrum
•
Introduced by M. A. Biot in 1932, popularized by G. W. Housner as a practical
means of characterizing ground motions and their effects on structures
•
Definition: A plot of peak response (displacement, pseudo-velocity, pseudo
acceleration) of a series of linear SDF oscillators of varying time period but same
damping ratio for a particular ground motion.
•
Displacement spectra used to compute peak values of deformation and internal
forces
•
Pseudo-velocity and pseudo-acceleration useful in studying characteristics of
response spectra, constructing design spectra and relating structural dynamics
results to building codes
15
Displacement Response Spectrum
16
𝒏
•
V has units of velocity and is related to the peak value of strain energy stored in the
system during earthquake
𝟐
𝟐
𝒐
•
𝟐
𝒏
𝟐
V, Peak relative pseudo velocity or peak pseudo velocity. Pseudo- prefix is used
because V is not equal to peak velocity ůo
17
Pseudo-acceleration Response Spectrum
𝟐
𝒏
•
A has units of acceleration and is related to the peak value of base shear
𝒃
𝒔
𝒃
•
A/g interpreted as base shear coefficient or lateral force coefficient
18
Response Spectrum
19
D-V-A Spectrum
𝒏
𝒏
•
Plotted by using pseudo-velocity
spectrum plot shown on slide 9
•
For Tn = 2 secs, D and A ?
•
Spectrum can be developed for
large range of fundamental time
period and damping ratios
20
Steps to Develop Spectrum
•
Response spectrum for a given ground motion can be developed by:
•
Ground motion ordinates are defined every, say, 0.02 secs
•
Select a Tn and damping ratio for the SDF system
•
Compute the deformation response u(t) of this SDF system due to the ground
motion by any of the numerical methods
•
Determine u0, the peak value of u(t)
•
The spectral ordinates are D = u0, V = (2π/Tn)D, A = (2π/Tn)2 D
•
Repeat the above four steps for a range of Tn and damping ratios
•
Prepare either a single spectrum or a combined spectra
21
Spectrum Characteristics
•
Response spectrum curve for 5% damping taken from above
- Tn< 0.035secs: A=
- Tn> 15secs: D= Ugo
- 0.035 < Tn> 0.5 secs: A exceeds
Amplification depends on Tn and ξ
- 0.125 < Tn> 0.5 Amplification depends
mainly on ξ
- 3 < Tn> 15 secs: D exceeds Ugo
Amplification depends on Tn and ξ
- 3 < Tn> 10 Amplification depends
mainly on ξ
22
Spectrum Characteristics
•
Response spectrum curve for 5% damping taken from above
- Ta, Tb, Te and Tf on the idealized
spectrum is independent of damping
- Idealized process not precise
- Values of Ta, Tb, etc., and the
amplification factors for b-c, c-d and d-e
are not unique
- Reflects the inherent differences in
ground motion even if recorded under
similar condition
- However, response trends matches
with other ground motions
23
Elastic Design Spectrum
•
Design spectrum is based on statistical analysis of response spectra for ensemble of
ground motions
•
Mean and (mean+ std. deviation) curves are
smoother than individual response spectrum
•
Straight line idealization lends more
credibility
24
Procedure for Elastic Design Spectrum
•
•
Plot 3 dashed lines corresponding to the peak values of ground acceleration,
velocity and displacement for design ground motion
Obtain αA, αV, and αD for the selected ξ
25
Procedure for Elastic Design Spectrum
•
Multiply peak ground acceleration by αA to obtain straight line b-c representing
constant value of pseudo-acceleration A
•
Multiply peak ground velocity by αV to obtain straight line c-d representing
constant value of pseudo-velocity V
•
Multiply peak ground displacement by αD to obtain straight line d-e representing
constant value of deformation D
•
Draw a line A equal to peak ground accl. for periods shorter than Ta and D equal
peak ground displacement for periods longer than Tf
•
Transition lines a-b and e-f complete the spectrum
•
Periods for a,b,e and f are fixed here, the values here are for firm ground
26
Procedure for Elastic Design Spectrum
27
Elastic Design Spectrum vs Response Spectrum
28
MDOF System
29
Free Vibration of MDOF
• Modal analysis or free vibration analysis is performed to obtain natural frequencies or mode
shapes of a structure.
• Modal analysis is a subset of the general equation of motion
mü  ku  0
m
u2
k
2m
u1
2k
30
Free Vibration
m
u2
k
2m
u1
2k
31
Free Vibration
• Two characteristic deflected shapes exist for this two-DOF system such that if it is displaced
in one of these shapes and released, it will vibrate in simple harmonic motion, maintaining
the initial deflected shape
• Both floors vibrate in the same phase, i.e., they pass through their zero, maximum, or
minimum displacement positions at the same instant of time
• Each characteristic deflected shape is called a natural mode of vibration of an MDF system
• A natural period of vibration Tn of an MDF system is the time required for one cycle of the
simple harmonic motion in one of these natural modes
32
Eigen Value Problem
• Solution for the problem gives natural frequency and mode shapes of a system
• Free vibration for the undamped system as shown above can be written as
u (t )  qn (t )n (t )
does not vary with time
Can be described by simple harmonic function
• An and Bn can be obtained through initial condition that initiated the motion
• Substituting above two relations in the equation of motion
 wn 2 mn  kn  qn (t )  0
• To satisfy this equation
kn  wn 2 mn
• For non-trivial solution
𝒏
𝟐
Matrix Eigenvalue Problem
Characteristic Equation
33
Eigen Value Problem
34
Eigen Value Problem
FIN. RL
PIER CAP
TOP
PILE CAP
TOP
35
Eigen Value Problem
• Solving the characteristic equation provides eigenvalues which are also the natural
frequencies of the system
• With known natural frequencies, the corresponding eigenvector or mode shape can be
obtained
• Eigenvalue problem does not fix the absolute amplitude of the vectors , only the shape of
the vector is obtained
“Eigenvalues and Eigenvectors are natural properties of the system in free vibration and depends
only on mass and stiffness”
0
0 
64.65
 0
 ton
Mass =
64.65
0


 0
0
64.65
0 
 71286 35643


Stiffness =  35643 71286 35643 kN/m
 0
35643 35643 
Characteristic Equation
270212.433 w   744871245.6 w   4.9284  1011 w 2   4.5285  1013  0
2 3
2 2
36
Eigen Value Problem
• Natural Frequency or Eigenvalue
1
= 10.45 rad/s
1
= 0.60 s
2
= 29.28 rad/s
2
= 0.21 s
3
= 42.31 rad/s
3
= 0.15 s
• Eigenvectors
m
k
m
k
m
u3
0.44


1  0.80
1.00 


1.26 


2  0.56 
 1.00 


 1.81 


3  2.25
 1.00 


u2
u1
k
37
Eigen Value Problem
• Participation Factor
n
W 
pk  in1
i i
2
W

 ii
i 1
1
= 1.22
2
= -0.28
3
= 0.059
1
= 177.19
2
= 14.98
3
= 2.12
• Modal Mass
2


W

 i i 

M k   in1

2
g   Wi i  
 i 1

n
Mode 1 = 91.36%
Mode 2 = 7.7%
Mode 3 = 1.1%
38
Modal Analysis
• The design lateral force (Qik) at floor i in mode k
Qik  Akik PW
k i
Ak is design horizontal acceleration spectrum using natural period of vibration Tk of mode k
• Design lateral force in each mode
Qi1   A1 P1i1Wi 
• Similarly Qi2, Qi3, Qi4,….. Qin
 A1 P111W1 
 A P W 
Qi1   1 1 21 2 
 ......... 


A
P

W
 1 1 n1 n 1 
39
Modal Analysis
• So displacement vector can be expanded in terms of modal contribution
N
u (t )   qr (t )r
r 1
• Coupled equation can then be transformed into a set of uncoupled equation with modal
coordinates qn(t) as the unknown
• Substituting in the basic equation of motion
N
N
 m q (t )   k q (t )  p(t )
r
r 1
r
r 1
r
r
• Pre-multiplying each term by n
T
N
N
  m q (t )    k q (t )   p(t )
r 1
T
n
r
r
r 1
T
n
r
r
T
n
• Considering the orthogonality condition
nT mn qn (t )  nT kn qn (t )  nT p(t )
M n qn (t )  K n qn (t )  Pn (t )
40
Modal Analysis
M n qn (t )  K n qn (t )  Pn (t )
Equation governing the response qn(t) of a single degree of freedom system
With mass Mn being the generalized mass for the nth natural mode
stiffness Kn being the stiffness for the nth natural mode
Pn being the generalized force for the nth natural mode
These parameters depend only on the n-th mode n
So, from N coupled differential equation in nodal displacement has been transformed into set of
N uncoupled equation in modal coordinates
Only thing left to do is combination of results!!!
41
Modal Analysis
Modal Combination
Since the maximum values for each mode do not occur at the same time, individual
mode responses can be superimposed by ‘Square Root of the Sum of the Squares’ (SRSS)
or ‘Compete Quadratic Combination’ (CQC) method to calculate the maximum values
of member forces and displacement
42
Modal Analysis
Modal Combination
SRSS
2 1/2
n
Rmax   R  R  .....  R 
2
1
2
2
Provides good results for structures with well-distributed natural frequency. Can over- or underestimate results for structural system with close natural frequencies eg: multi-span bridge with
close short span
CQC (Complete quadratic combination)
1/2


Rmax    Ri ij R j 
 i 1 j 1

N
N
Considers probabilistic correlation between modes and applies correlation factor for close natural
frequencies
ABS (Absolute sum)
Rmax  R1  R2  ....  Rn
Overestimates the response results
43
Time History Analysis
• Analytical solution of the equation of motion is usually not possible if applied external forces
p(t) vary arbitrarily with time or if system is non-linear
• Numerical time stepping methods can be used for integration of differential equations
mü  cu  ku  p(t )
Subjected to initial condition (0) = 0 and u(0) =0
• The applied force p(t) can be taken as set of discrete values pi = p(ti), i= 0 to N
• Response is determined numerically at these discreet time instants
44
Time History Analysis
• One of the popular methods is Newmark’s method
ui 1  ui  1    t  ui   t  ui 1
2
ui 1  ui   t  ui   0.5    t   ui 1


• Parameters β and γ define the variation of acceleration over a time step and determine stability
and accuracy characteristics of the method
• Typically γ =0.5 and 1    1
6
We have,
4
mui 1  cui 1  kui 1  pi 1
So, an iterative procedure has to be used to find the response of the system at ti+1 using the data
known at ti
45
Time History Analysis
Constant Average
Acceleration Method
• Special case for γ = 0.5 and   0.25
ui 1  ui  1    t  ui   t  ui 1
2
ui 1  ui   t  ui   0.5    t   ui 1


• Special case for γ = 0.5 and  
1
6
Linear Acceleration Method
46
Time History Analysis
47
Time History Analysis
48
49