13th World Conference on Earthquake Engineering
Vancouver, B.C., Canada
August 1-6, 2004
Paper No. 884
NUMEREICAL SIMULATION OF POUNDING OF BRIDGE DECKS
Gakuho WATANABE1, Kazuhiko KAWASHIMA2
SUMMARY
Pounding of bridge decks often resulted in unseating of the decks under near-field ground motions. It is
required to take account of the impact effect in a seismic analysis of bridges. The impact spring is often
used to represent the impact effect. This paper clarifies the application of the impact spring to a numerical
analysis of a pounding of two elastic rods in their longitudinal direction in comparison with the exact
solutions based on the wave propagation theory. It is found that the idealization of a pounding in terms of
the impact spring provides a good estimate to obtain an overall behavior of a pounding except the
acceleration response.
INTRODUCTION
Different responses among several bridges which were induced by the difference of column stiffness or
incoherent ground motions resulted in unseating of the decks due to collisions between adjacent decks in
past earthquakes. Although the direct damage of decks due to collisions was limited, a collision
transferred a large inertia force from one deck to other decks, which resulted in damage of bearings or
columns. It is known that a pounding force between adjacent decks is significant when an initial gap
between the two decks is in the range of 0.4-0.6 times the maximum relative displacement between the
decks [1, 2].
In a dynamic response analysis of bridges, the impact spring is often used to idealize the pounding effects.
It is a simple idealization, but provides a realistic response of bridge decks subjected to pounding effects.
The idealization of poundings using the impact springs was first used by Tseng and Penzien for plastic
collisions and Kawashima and Penzien for elastic collisions [3, 4].
In a dynamic response analysis of poundings of decks, the magnitude of impact spring stiffness and the
time interval of numerical integration have to be appropriately chosen. Since an entire process of a
pounding from a contact to a separation of two rods which collide in their longitudinal direction can be
predicted based on the wave propagation theory, a numerical simulation is conducted to clarify the
appropriate stiffness of impact spring and the time interval of numerical integration using the impact
spring.
1
2
Research Associate, Tokyo Institute of Technology, Tokyo, Japan. Email: [email protected]
Professor, Tokyo Institute of Technology, Tokyo, Japan. Email:[email protected]
BASIC THEORY OF COLLISION OF TWO RODS
Wave Propagation Theory
Collisions of two adjacent decks in their longitudinal direction during an earthquake may be simplified as
a collision of two elastic rods as shown in Fig. 1. An entire process of a collision from a contact to a
separation of two rods which are initially propagating with velocity of V1 and V2 in their longitudinal
direction can be analyzed using the classical wave propagation theory [5].
x
V1
Rod 1
u1
Collision
u2
V2
Rod 2
: Propagation of Compression Wave
Fig. 1 Pounding and Stress Wave Propagation
As well known, the equation of motion of the rods is written as
∂ 2u
∂ 2u
C
=
∂t 2
∂x 2
(1)
where, u : the displacement of the rods, and C : velocity of wave propagation defined as C = E ρ , in
which E and ρ represent the elastic modulus and mass density of the rods. Using the D’Alembert
solution in the form of
u = f (x + Ct ) + g (x − Ct )
(2)
the displacement of rod 1 and rod 2 can be written as
u1 = f ( x + C1t ) ; u 2 = g (x − C2t )
(3)
When the rod i (i=1 and 2) has a section Ai , an elastic modulus Ei and a velocity Vi , the stress of rods 1
and 2 is σ 1 = E1 ∂u1 ∂x = E1 f ′( x + C1t ) and σ 2 = E2 ∂u 2 ∂x = E2 g ′( x − C2t ) , respectively, and the
particle velocity of rods 1 and 2 is v1 = V1 + ∂u1 ∂t =V1 + C1 f ′(c + Ct ) and v2 = V2 − C 2 g ′( x − C2t ) .
Introducing a continuity condition at the contact surface, one obtains
σ1 = −
A2 E1E2 (V1 − V2 )
A E E (V − V )
; σ2 = − 1 1 2 1 2
A2 E2C1 + A1E1C2
A2 E2C1 + A1E1C 2
A E C V + A2 E2C1V2
ν1 = ν 2 = 1 1 2 1
A2 E2C1 + A1E1C2
On the other hand, the acceleration of the rod can be given by
(4)
(5)
∂ 2u
∂t
2
=
Cσ 0 
x
δ t − 
E  C
(6)
where, δ (t − x C ) is a Dirac delta function defined as infinitive at x − Ct = 0 . This means that the
acceleration at x = l s becomes infinitive when a compression wave propagates at t = l s C .
The inertia force FI between a short distance ∆x at x = l s is
FI = − ρA
2
∂ u
∂t
2
ls +
∆x =
∆x
2
σ 
x
ρAC 0 δ t −  dx = − Aσ 0
E  C
∆x
∫
ls −
(7)
2
ANALYSIS OF POUNDING BY IMPACT SPRING
A process of pounding from a contact to a separation of the two rods is idealized as shown in Fig. 2 using
an impact spring. The impact spring and impact force PI are given by
k L∆u ≤ −uG
k = I
 0 L∆u > −uG
(8)
k (∆u − uG )L∆u ≤ −uG
PI =  I
0
L∆u > −uG

(9)
where, ∆u : relative displacement between two rods defined as ∆u = u 2 − u1 , uG : initial gap between two
rods, and k I : impact spring stiffness.
P
u1
uG
u2
0
k
∆u
kI
uG
1
Fig. 2 Idealization of a Pounding by Impact Spring
Kawashima and Penzien analyzed elastic collisions of a model curved bridge, and found that the collisions
of the model can be well represented by choosing the impact spring stiffness k I so that a parameter
defined as
k
k L
γ= I = I
(10)
k B nEA
is nearly 1.0, in which E , A and L are the elastic modulus, section and length of a deck, and n
represents number of elements of the rod [4].
The direct integration analysis using the constant acceleration method was used. Numerical iterations
were used to maintain the equilibrium of the equation of motions when an unbalance force is significant.
Idealization of the poundings by the impact spring stiffness is studied here.
COLLISIONS OF TWO RODS WITH THE SAME LENGTH
Stress and Particle Velocity
When two rods with the same length, section, elastic modulus and mass density propagate with the same
velocity in the opposite direction and collide, representing A1 = A2 = A , E1 = E2 = E and C1 = C 2 = C , the
stress σ 0 and particle velocity v 0 become from Eqs. (4) and (5) as
σ0 =
V +V
E (V1-V2 )
; ν0 = 1 2
2C0
2
(11)
2L
C0
(12)
The duration of contact TI is given as
TI =
Assuming E =100, A =1, ρ =0.1, L =10, V1 = V0 =0.1, and V2 = −V0 =-0.2, the duration of contact TI and
the stress σ 0 become from Eqs. (11) and (12) as
TI = 10 5 ; σ 0 = 10 10
(13)
It should be noted that any unit can be used for E , A , ρ , L and V0 . In the following analysis, TI by
Eq. (13) is expressed as a 0.2 unit time.
The impact spring stiffness k I in Eqs. (8) and (9) was set so that γ by Eq. (10) is equal to 1.0. A very
week linear spring with a stiffness of 0.01 was set at two the other ends of the contact surface so that
numerical instability does not occur when two rods are not in contact. Because the natural period of the
rods due to this weak spring is about 20 times the unit time (about 100 times the contact duration TI ), the
effect of the weak springs on the total response of rods is limited. Two rods are idealized by 10 linear
beam elements each. No damping was provided in the analysis. The numerical time increment ∆t was set
as 10 2000 (=0.00158) times TI .
Fig. 3 compares the stress and particle velocity along the two rods at 6 specific unit times between the
classical wave propagation theory and the numerical analysis. Based on the wave propagation theory, a
compression wave generated at the instance of contact propagates from the contact surface to the other
end in both rods. The compression waves reach at the other ends at 0.1 unit time, and then reflect as a
tension wave to reach the contact surface at 0.2 unit time. Associated with the propagation of the
compression wave from the contact surface toward the other end, the initial particle velocity of V0 and V0 in rods 1 and 2, respectively, becomes zero. After the tension wave starts to propagate from the other
ends to the contact surface, the particle velocity changes to - V0 and V0 in rods 1 and 2, respectively,
associated with the propagation of tension wave. At the instance when the tension wave reaches the
contact surface, the stress becomes zero along the entire length of rods 1 and 2, and the particle velocity
becomes - V0 and V0 in rods 1 and 2, respectively. As a consequence, the contact of two rods complete
and a separation starts to occur.
Unit Time
1.0
0.00
-1.0
0.05
-1.0
0.10
-1.0
0.15
-1.0
0.20
1.0
0.25
-1.0
(a) Stress σ 0
(b) Particle Velocity V0
Fig. 3 Propagation of Stress and Particle Velocity during a Collision of Two Rods with Equal Length
The analysis using the impact spring predicts the overall stress and particle velocity responses of the two
rods from a contact to a separation with satisfactory accuracy. Only a difference is that the change of the
computed stress wave and particle velocity is not sharp enough at the wave fronts. Fig. 4 compares the
variation of stress and particle velocity at the contact end and the other side surface. Although small
numerical oscillations take place in the analysis in both stress and particle velocity, the overall behavior is
well represented.
Because the mass of a rod m is ρAL , the kinematic energy E and momentum P are mV0 2 and 0,
respectively, before a collision. On the other hand, after the collision they become
E = 1 2 ⋅ m(− V0 )2 + 1 2 ⋅ mV0 2 = mV0 2 ; P = mV0 − mV0 = 0
(14)
Consequently, the kinematic energy and momentum are preserved during the collision.
Stiffness of Impact Spring
Figs. 5, 6 and 7 shows the effect of magnitude of impact spring stiffness k I on the relative displacement
between two rods ∆u , particle velocity, and stress at the contact surface. Exact solution by the wave
propagation theory is presented here for comparison. If k I is not large enough, an overlap between two
rods is excessive, which is not appropriate to simulate a pounding. On the other hand, if k I is too stiff,
contacts and separations repeat between two rods during the contact duration, which also results in an
excessive oscillation in stress and particle velocity responses. Fig. 8 shows the impact force induced at the
contact surface. If we set γ larger than 10, the predicted impact force becomes excessive compared to the
exact solution by Eq. (7). Based on these clarifications, it is obvious that the impact spring with γ nearly
1.0 provides the most appropriate estimate to the exact solution.
Stress / σ0
Particle Velocity / V0
Numerical Solution
Exact Solution
1
0
1
0
-1
-1
0
0.2
0.4
Unit Time
0.6
0
(a) Stress σ 0
0.2
0.4
Unit Time
0.6
(b) Particle Velocity V0
Particle Velocity / V0
(1) Contact Surface
Stress / σ0
1
0
1
0
-1
-1
0
0.2
0.4
Unit Time
0.6
0
0.2
0.4
Unit Time
0.6
(a) Stress σ 0
γ=1
γ=0.5
γ=0.2
γ=0.1
0.05
Relative Displacement
Relative Displacement
(b) Particle Velocity V0
(2) Free Surface
Fig. 4 Variation of Stress and Particle Velocity
0
-0.05
0
0.2
Unit Time
γ=1
γ=10
γ=100
0.01
0
-0.01
0
0.2
Unit Time
(a) γ =0.1-1
(b) γ =10-100
Fig. 5 Effect of Magnitude of Impact Spring Stiffness on Relative Displacement between Two Rods
Number of Elements
Two rods were idealized by 10 elements each in the above analysis. The stiffness of impact spring k I was
set so that γ was equal to 1.0. Fig. 9 shows how the stress at the contact surface varies if the number of
element n changes between 5 and 20. The magnitude of stress and a manner of oscillation at the contact
surface are not sensitive to n. However, the peak acceleration at the contact surface increases as 2.59, 5.18
and 10.36 as n increases as 5, 10 and 20, respectively. It is noted that the peak acceleration increases
almost linearly with the increase of n. This is because if the mass at a nodal point becomes a half, an
acceleration to produce the same magnitude of inertia force needs to be doubled. It should be noted
therefore that accuracy of the acceleration during a collision is not reliably evaluated by the idealization
using the impact spring.
1
0
-1
0
Patricle Velocity / V 0
Patricle Velocity / V 0
γ = 0.1
γ = 0.5
γ=1
Exact Solution
γ=10
γ=100
Exact Solution
1
0
-1
0.2
Unit Time
0
(a) γ =0.1-1
0.2
Unit Time
(b) γ =10-100
Fig. 6 Effect of Magnitude of Impact Spring Stiffness on Particle Velocity at Contact Surface
γ = 0.1
γ = 0.5
γ=1
Exact Solution
γ = 10
1
Stress / σ0
Stress / σ0
1
Exact Solution
0
-1
γ = 100
0
-1
0
0.2
Unit Time
0
(a) γ =0.1-1
0.2
Unit Time
(b) γ =10-100
-1
γ = 0.1
γ = 0.5
γ=1
-2
0
0.2
0
-1
-10
-20
Unit Time
γ = 10
γ = 100
0
0.2
Unit Time
(a) γ =0.1-1
(b) γ =10-100
Fig. 8 Effect of Magnitude of Impact Spring Stiffness on Impact Force at Contact Surface
n=5
n = 10
n = 20
Exact Solution
1
Stress / σ0
Impact Force
0
Relative Displacement
Fig. 7 Effect of Magnitude of Impact Spring Stiffness on Stress at Contact Surface
0
-1
0
0.2
Unit Time
Fig. 9 Effect of Number of Element on Stress at Contact Surface
Effect of Time Interval in Numerical Integration
Although time interval of numerical integration ∆t was set 10 2000 (=0.00158) times the duration of
contact TI in the above analysis, Fig. 10 shows the effect of ∆t in terms of the responses at the contact
surface. The stiffness of impact spring k I was set so that γ was equal to 1.0, and the number of element
0.1
0
-0.1-1
0
Unit Time
1
Particle Velocity
/ V0
Relative
Displacement
n was assumed to 10. By increasing ∆t TI from 10 2000 to 0.1, the accuracy of predicting the
variation of the relative displacement between two rods, particle velocity, stress and acceleration
decreases, but the overall response of a pounding between two rods is evaluated satisfactorily even at
∆t TI =0.1. This is also the case at ∆t TI =1 and 5, however variation of responses during a collision and
the duration of contact TI are poorly predicted.
2
1
0
-1
-2-1
1
(b) Particle Velocity
5
Acceleration
Stress / σ0
(a) Relative Displacement
2
1
0
-1
-2
-1
0
Unit Time
0
Unit Time
0
-5
-1
1
(c) Stress
0
Unit Time
1
(d) Acceleration
0.1
Particle Velocity
/ V0
Relative
Displacement
(1) ∆t TI = 10 2000
0
-0.1-1
0
Unit Time
1
2
1
0
-1
-2-1
0
Unit Time
1
(b) Particle Velocity
Acceleration
Stress / σ0
(a) Relative Displacement
2
1
0
-1
-2
-1
0
Unit Time
1
5
0
-5
-1
(c) Stress
0
Unit Time
(d) Acceleration
(2) ∆t TI =0.1
Fig. 10(a) Effect of Time Interval on the Responses at Contact Surface
1
0
-0.1-1
0
Unit Time
1
Particle Velocity
/ V0
Relative
Displacement
0.1
2
1
0
-1
-2-1
0
Unit Time
1
(b) Particle Velocity
Acceleration
Stress / σ0
(a) Relative Displacement
2
1
0
-1
-2
-1
0
Unit Time
1
5
0
-5
-1
(c) Stress
0
Unit Time
1
(d) Acceleration
0.1
0
-0.1-1
0
Unit Time
1
Particle Velocity
/ V0
Relative
Displacement
(3) ∆t TI =1
2
1
0
-1
-2-1
0
Unit Time
1
(b) Particle Velocity
Acceleration
Stress / σ0
(a) Relative Displacement
2
1
0
-1
-2
-1
0
Unit Time
1
5
0
-5
-1
(c) Stress
0
Unit Time
1
(d) Acceleration
(4) ∆t TI =5
Fig. 10(b) Effect of Time Interval on the Responses at Contact Surface
COLLISIONS OF TWO RODS WITH DIFFERENT LENGTH
Stress and Particle Velocity
If the rods 1 and 2 have 10 and 20 unit length, and the initial velocity of V1 = 2V0 =0.2 unit velocity and
V2 = V0 =0.1 unit velocity, respectively, the duration of contact TI becomes
TI = 4 L C0 = 2 10 5
(15)
This duration of contact is two times the duration of contact by Eq. (13). Consequently, the period of
contact by Eq. (15) is expressed as a 0.4 unit time in the following analysis.
Fig. 11 shows the stress and particle velocity of two rods during a pounding. Exact solution by the wave
propagation theory and a numerical analysis are compared. In the exact analysis, a compression wave
generated at the instance of contact reaches the other end, and starts to reflect in rod 1 at 0.1 unit time,
while a compression wave is still at a half length in rod 2. At 0.2 unit time, the tension wave which
Unit Time
0.00
-1.0
0.05
-1.0
0.10
0.15
-1.0
-1.0
0.20
-1.0
0.25
2.0
1.0
2.0
1.5
1.5
1.0
1.0
1.0
0.35
0.40
0.45
1.0
1.5
1.0
1.0
1.5
1.0
0.30
1.0
1.0
1.5
1.0
2.0
1.0
1.5
1.0
1.0
2.0
2.0
1.5
2.0
1.5
1.0
(a) Stress / σ 0
(b) Particle Velocity / V0
Fig. 11 Propagation of Stress and Particle Velocity during a Collision of Two Rods with Different
Lengths
reflected at the other end reaches the contact surface in rod 1. The compression stress once generated is
released until this unit time in rod 1. However, since rod 2 is still in compression at this time, the contact
continues. At 0.4 unit time, a tension wave generated by reflection at the other end reaches the contact
surface by releasing compression in rod 2. At this time, the velocity has changed from 2V0 to V0 in rod 1
and from V0 to 1.5V0 in rod 2. As a consequence, the contact completes and a separation starts to occur
at o.4 unit time. It should be noted that an oscillation remains after the separation in rod 2.
Figs 12, 13 and 14 show a numerical simulation. The impact spring stiffness k I was set so that γ = 1.0 by
Eq. (10). Rods 1 and 2 were idealized by 10 and 20 linear beam elements, respectively. The numerical
time increment ∆t was set as 10 2000 (=0.00158) times TI by Eq. (15). Similar to the analysis of a
pounding between two rods with the same length, although wave fronts are not sharp enough and there are
some numerical oscillations in analysis, the overall responses during a pounding are well represented. It is
interesting to note that the contact completes at 0.2 unit time in the analysis. As described above, rod 1 is
ready to separate at 0.2 unit time. However since rod 2 is still in compression, the contact continues until
3
Numerical Solution
Exact Solution
2
1
0
0
0.2
0.4
0.6
Unit Time
0.8
Paricle Velocity / V 0
Paricle Velocity / V 0
3
2
1
0
1
0
0.2
(a) Rod 1
0.4
0.6
Unit Time
0.8
1
0.8
1
(b) Rod 2
(1) Contact Surface
3
Exact Solution
Numerical Solution
2
1
0
0
0.2
0.4
0.6
Unit Time
0.8
Paricle Velocity / V 0
Paricle Velocity / V 0
3
2
1
0
1
0
0.2
(a) Rod 1
0.4
0.6
Unit Time
(b) Rod 2
(2) Free Surface
Fig. 12 Variation of Particle Velocity
0.4 unit time in the exact solution. In analysis, two rods are not in contact after 0.2 unit time, but stay with
a very small gap until 0.4 unit time. The separation starts to increase at 0.4 unit time.
Difference with Pounding of Two Rigid Bodies
If we compute the velocity of two rods after a collision assuming that they can be idealize as a perfect
elastic collision of two rigid bodies with a mass of ρAL and 2 ρAL , the preservation of momentum and
the condition of perfect elastic collision are written as
ρAL ⋅ 2V0 + 2 ρAL ⋅ V0 = ρAL ⋅ V1′ + 2 ρAL ⋅ V2′
e=−
V2′ − V1′
=1
V0 − 2V0
(16)
(17)
in which V1′ and V2′ represent the velocity of rods 1 and 2 after a collision, respectively, Therefore we
obtain V ′ = 2 3 ⋅ V and V ′ = 5 3 ⋅ V . As a consequence, the kinematic energy and momentum after a
1
0
2
o
pounding are
2
E=
2
1 2 
1 5 
m V0  + 2m V0  = 3mV0 2
2 3 
2 3 
5
2
P = m V0 + 2m V0 = 4mV0
3
3
(18)
(19)
1
Stress / σ0
Stress / σ0
1
0
-1
0
-1
0
0.2
0.4
0.6
Unit Time
0.8
1
0
0.2
(a) Rod 1
0.4
0.6
Unit Time
0.8
1
0.8
1
(b) Rod 2
(1) Contact Surface
1
Stress / σ0
Stress / σ0
1
0
-1
0
-1
0
0.2
0.4
0.6
Unit Time
0.8
1
0
0.2
(a) Rod 1
0.4
0.6
Unit Time
(b) Rod 2
(2) Free Surface
Fig. 13 Variation of Stress
Relative Displacement
0.02
0
-0.02
0
0.2
Unit Time
0.4
Fig. 14 Relative Displacement between Two Rods during a Collision
Because the kinematic energy and momentum before the collision is 3mV0 2 and 4mV0 , respectively,
both kinematic energy and momentum are preserved.
On the other hand, because the velocity after a collision is V1′ = V0 and V2′ = 3 2 ⋅ V0 based on the wave
propagation theory, the kinematic energy and momentum are
2
E=
1
1 3 
11
mV0 2 + 2m V0  = mV0 2
2
2 2 
4
3
P = mV0 + 2m V0 = 4mV0
2
(20)
(21)
1
1
2
Unit Time
3
2
1
0
1
2
Unit Time
3
1
1
2
Unit Time
0
1
2
Unit Time
3
4
0
1
2
Unit Time
3
4
0
1
2
Unit Time
3
4
3
2
1
(2) ∆t TI =1
2
0
1
0
-1
4
3
0
-1
2
(1) ∆t TI =0.1
3
0
-1
3
0
-1
4
Velocity/ V0
Velocity/ V0
0
Velocity/ V0
2
0
-1
Velocity/ V0
Exact Solution
Numerical Solution
Velocity/ V0
Velocity/ V0
3
3
4
3
2
1
0
-1
(3) ∆t TI =5
Fig. 15 Effect of Time Interval on the Responses at Contact Surface
Therefore, the momentum is preserved but the kinematic energy is not preserved. This is because an
oscillation remains in rod 2 after the separation. It is important to note that the idealization of flexible rods
by rigid bodies induces an error in the estimate of the velocities after a collision.
Effect of Time Interval of Numerical Integration
Fig. 15 shows the variation of the particle velocity of two rods at the contact surface. Although there
remains an oscillation in the computed particle velocities of two rods, the velocity after a pounding
changes to V0 and 1.5V0 in the rods 1 and 2, respectively, when ∆t TI is equal to 0.1 and 1.0, while it
changes to 2 3 ⋅ V0 and 5 3 ⋅ V0 in the rods 1 and 2, respectively, when ∆t TI is equal to 5.0. It is noted
that the velocities after a pounding which are computed using ∆t TI =5.0 correspond to those estimated
by assuming that the rods are two rigid bodies. Consequently it is suggested to set the time interval of
numerical integration as
∆t < TI
(22)
CONCLUSIONS
An application of the impact spring was clarified based on two numerical examples of an entire process of
pounding of two elastic rods in their longitudinal direction. Based on the results presented herein, the
following conclusions may be deduced:
1) The magnitude of impact spring stiffness is less sensitive to the magnitude of axial force and velocity
after a pounding.
2) Variation of the particle velocity and impact force depends on a parameter γ defined by Eq. (10). The
impact force is less sensitive to the magnitude of impact spring stiffness if the parameter γ is kept
constant. However the impact force sharply increases as the parameter γ increases. Small separations
and contacts repeat at the contact surface when the parameter γ is excessively larger than 1.0. On the
other hand, the contact duration TI becomes excessively longer than the exact duration under the
parameter γ smaller than 1.0. As a consequence, the overall behavior of a pounding can be
represented by choosing the parameter γ nearly 1.0.
3) Even if a rod is idealized by more number of elements, the relative velocity between two rods, particle
velocities, stress induced in two rods and impact force change less significantly. However the
response acceleration increases as the number of element increases.
4) Although the kinematic energy is preserved during a collision of two elastic rods with the same
length, it is not preserved during a pounding of two elastic rods with different length because an
oscillation remains in the longer rods after separation. As a consequence, the theory of rigid body
collision fails to predict the velocity after a collision.
5) The time interval of numerical integration has to be short enough to satisfy Eq. (22). If the time
interval is excessively longer than Eq. (22), the velocities of two elastic rods after a collision
approaches the values predicted by the theory of rigid body collision.
REFERENCES
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decks subjected to a strong ground motion.” 12th World Conference on Earthquake Engineering,
Paper No. 1435 (CD-ROM), Auckland, New Zealand.
3. Tseng, W.S. and Penzien, J. (1973) “Analytical investigations of the seismic response of long multispan highway bridges.” Report No. EERC 73-12, Earthquake Engineering Research Center,
University of California, Berkeley, USA.
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