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Tomohiro Nakano

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The SIJ Transactions on Computer Networks & Communication Engineering (CNCE), Vol. 2, No. 3, May 2014
ISSN: 2321-2403 © 2014 | Published by The Standard International Journals (The SIJ) 36
Abstract
—
An equation of motion for structure under both inertial force and the relative displacement between its support points was formulated, and a numerical method for applying the equation to non-linear dynamic response analysis was developed. Using the formulation and the method, the author simulated the dynamic responses of 2 elastic structures and that of an inelastic structure subjected to both the inertial force and the relative displacement between support points, to prove the method can evaluate the effect of the inertial force. In addition, the non-linear responses of the inelastic structure under some pairs of relative displacement waves, with/without the consideration of inertial force, were carried out to evaluate the influence of the inertial force. As a result of the analyses, the method could represent both the effect of inertial force and that of the relative displacement. It was also shown that the effect of inertial force is of great importance for evaluating the damage of structure caused by dynamic relative displacement of its support points.
Keywords
—
Equation of Motion; Inertial Force; Non-Linear Dynamic Response Analysis; Relative Displacement; Time History Response of Curvature.
Abbreviations
—
Equation of Motion (EOM).
I.
I
NTRODUCTION
HE development of computer science over the last few decades has contributed extremely to various engineering fields. It is difficult to carry out experiments using real civil structures, since they are generally pretty huge; therefore we have predicted the behaviour of civil structures by the use of mathematical and dynamical methods. To apply the methods, computational mechanics is essential to achieve the prediction. In solving the time-dependent problems, the Newmark method have been applied to solve a system of linear ordinary differential equations of second order in time.
The progress in computer science has made it possible to apply the method to various problems, such as moving load problem [Eftekhari & Khani, 2010] and dynamical contact problem [Deuflhard et al., 2008]. This fact indicates that the Newmark method has the potential to solve a variety of vibration problems. In fact, several computer algorithms have elucidated structural response under seismic loads. Furthermore, the explication of the earthquake ground motions in the epicentral region, including fault movements are elucidated, also [Hori, 2007; Campbell et al., 2009]. Nevertheless, the way to predict the interaction between the structural behaviour, which is supported by plural points (e.g., bridges) and the fault movements, has not been clarified so far. In other words, at present, there is no way to predict the dynamic behaviour of the structure supported by plural points, which move with relative displacement, because existent computer algorithms can deal only with the effect of the inertial force, or only with the statically relative displacement. To the author
’
s best knowledge, the only attempt to represent the interaction between the inertial force and the relative displacement was made by Otsuka et al., (2007). Their method was to introduce stiff springs between fixed points and support points, that is, an indirect method using existing algorithm. Therefore, the interaction will be able to be represented by the development of the new computer algorithm, which can consider both the effect of inertial force and that of relative displacement. The motivation of this research are that: a) the behaviour of structures, considering the inertial force and the relative displacement between support points, should be represented by a direct method, and b) it is important to investigate the effect of the interaction between the inertial force and the relative displacement between support points.
T
*Associate Professor, Department of Civil Engineering, Tokai University, Hiratsuka, Kanagawa, JAPAN. E-Mail: tom_nakano{at}tokai-u{dot}jp
Tomohiro Nakano*
A Numerical Method for Dynamic Response Analysis of Structure Subjected to Relative Displacement between Support Points
The SIJ Transactions on Computer Networks & Communication Engineering (CNCE), Vol. 2, No. 3, May 2014
ISSN: 2321-2403 © 2014 | Published by The Standard International Journals (The SIJ) 37
For that reason, the objectives of this research are to: a) formulate the Equation Of Motion (EOM) considering the inertial force and the relative displacement between support points, b) provide a computer algorithm for non-linear dynamic response analysis using the EOM, and c) show the applicability of the direct method to linear and non-linear dynamic response analyses. The author developed an algorithm, which is composed of a static analysis part and a dynamic one. The algorithm also makes it possible to calculate the structural behaviour even in the non-linear range. Though the application of the method to some real structures had already been showed in some papers [Nakano & Ohta, 2008; Nakano, 2013], the analyses focused only on the non-linear phenomena of bridge structures. However, this method can be applied not only to non-linear structures but also to the fields concerning elastic vibration problems of parts or components to evaluate the effect of the inertial force (e.g., mechanical or electrical engineering). The contributions of this paper are: a) establishment of a direct method to solve a vibration problem considering the inertial force and the relative displacement between support points, b) achievement of the application of the method to some problems, and c) presentation of non-linear time-history phenomena for the purpose to show the interaction between the inertial force and the dynamic relative displacement.
II.
T
HE
E
QUATION OF
M
OTION WITH
R
ELATIVE
D
ISPLACEMENT
2.1.
The Equation of Motion for Elastic Structure with Relative Displacement
Let us consider a finite element model of a structure composed of some elements (black solid line in figure 1) [Nakano & Ohta, 2008; Nakano, 2013]. In the absolute coordinates, the Equation Of Motion (EOM) for the structure without the damping effect can be expressed as follows.
+
=
(1)
Figure 1: Division of Displacement Vector
Here,
is the mass matrix,
the stiffness matrix,
u
the absolute displacement vector and
the external force vector. We now divide these matrices and vectors as
=
000
000
=
(2)
=
=
Here, the subscripts
,
of the components in the matrices or the vectors denote the degrees of freedoms corresponding to the support points
,
(displacement controlled points) and
the degree of freedoms of other points (load controlled points). Next, we consider the condition when the support points
,
displace
and
statically. The stiffness equation under the condition can be written as;
0
=
0
(3) In Eq.(3),
0
denotes the displacement vector of the load controlled points,
0
the static load vector such as dead loads,
and
are the reaction force vectors caused by the displacements
and
. From the first row of Eq.(3), we have
0
+
+
=
0
(4) Then, we introduce the
1
, which is the displacement vector caused by the inertial force (i.e., dynamic effects) at the load controlled points (Figure 1). The relation among
0
,
1
and the total displacement vector
should be
=
0
+
1
(5) The load vector
in Eq.(1) is just the same as
0
, that is,
=
0
. Considering these relations, Eq.(1) yields
0
+
1
+
0
+
1
+
+
=
0
(6) Substituting Eq.(4) into Eq.(6), we have
1
+
1
=
−
0
(7) In the case that the structure is elastic, there exists the inverse of
. Therefore, Eq.(4) can be rewritten as follows.
0
=
−
1
0
−
−
(8) Using Eq.(8), we express the Eq.(7) as
1
+
1
=
−
1
+
(9) Eq.(9) indicates that when we have the acceleration record of the support points
A
and
B
in advance, we can predict the dynamic response considering both the inertial force and the relative displacement of the support points. One of the important points when applying this equation to dynamic analysis is that
1
and
1
are the “relative”
displacement and the
“relative” acceleration. Therefore, when
we evaluate the absolute displacement and the absolute acceleration, we have to use the relations
=
0
+
1
(10)
=
0
+
1
(11)
2.2.
Incremental Form of the Equation of Motion
To simulate the structure in non-linear range, we have to use the incremental form of the EOM. The incremental form of Eq.(9) can be expressed as
∆
1
+
∆
1
=
−
1
∆
+
∆
(12) Here, the notation of
means the tangent stiffness matrix.
u
S0
u
S1
u
A
u
B
The SIJ Transactions on Computer Networks & Communication Engineering (CNCE), Vol. 2, No. 3, May 2014
ISSN: 2321-2403 © 2014 | Published by The Standard International Journals (The SIJ) 38
However, the right hand side in Eq.(12), which is the external force vector, cannot be determined as a constant vector, because the tangent stiffness matrix varies during the non-linear analysis. This fact will make it difficult to solve the equation. Accordingly, we use the incremental form of Eq.(7) instead of Eq.(12) to carry out the non-linear analysis. The incremental form of Eq.(7) is
∆
1
+
∆
1
=
−
∆
0
(13) In this equation,
∆
0
is the increment of the second order differential of
0
, which can be obtained by solving Eq.(3). In the time history response analysis, we can determine the
∆
0
using the
0
in previous step and the
0
before the previous step. Using Eq.(13), we can calculate the response of the inertial force and relative displacements, even though the structure is in the non-linear range. To apply Eq.(13), we have to use a special algorithm, which is composed of a static analysis part and a dynamic one. The algorithm of the time history response analysis can be found in figure 2.
Figure 2: Integration Procedure of Dynamic Analysis
TIMEINCREMENT LOOP :
n
,
STATEDETERMINATION BYSTATIC ANALYSIS
EnterNewton-Raphsoniteration loop : k=1,2,3, ... untilconvergenceSolveAssemble structure resisting force vector Computeunbalancedforcevector GotonextNewton-Raphsoniteration
CALCULATE , ANDSTATEDETERMINATION BYDYNAMIC ANALYSIS
EnterNewton-Raphsoniteration loop : k=1,2,3, ... untilconvergenceSolve and by Newmark methodandupdate andAssemble structure resisting forceComputeunbalancedforcevector GotonextNewton-Raphsoniteration
GOTONEXTTIME STEP
tu141u21Cu21ut1MuMP
1n0S1n0SSS1n0S1n0SS0SS
1nAnAA
uuu
1nBnBB
uuu
BA0St0B0A0S
uuuK FFF
0S0S0S
uuu
0R
F
0R 0U
FFF
0S
u
0S
u
0SS
uMP
S2SStSS
Mt1Ct21K K
PK u
11S
1S1S1S
uuu
1S
u
1S
u
1S
u
1S
u
R
P
R 1SS1SS
DR
PuCuMP
DR U
PPP
The SIJ Transactions on Computer Networks & Communication Engineering (CNCE), Vol. 2, No. 3, May 2014
ISSN: 2321-2403 © 2014 | Published by The Standard International Journals (The SIJ) 39
III.
D
YNAMIC
R
ESPONSE
A
NALYSIS
O
F
E
LASTIC
B
EAMS
U
NDER
I
NERTIAL
F
ORCE AND
R
ELATIVE
D
ISPLACEMENT
3.1.
Target Structure and Applied Displacement Waves
This study is concerned with the EOM for elastic structure under relative displacement and its application to the non-linear analysis. This chapter presents some examples of the elastic beams under dynamic relative displacement at their support points, obtained by the EOM suggested in this study. Now, we consider two simple beams with length of 10m as shown in Figure 3. Both beams have the elastic modulus of E=200[GPa], the sectional area of A=0.01[m
2
], the moment of inertia of I=3.0×10
-6
[m
4
] and the mass of unit length of 0.05[ton/m]. One has a mass of 1000[ton] at the centre (Beam-A), the other two masses of 300ton at 3m from both ends (Beam-B).
Figure 3: Example Simple Beams
Then, we input the vertical displacement waves at both support points. The vertical displacement waves are as follows.
0
=
−
2
+
2 (0
≤
<1.0)
−
1.0
1.0
≤≤
21.0
(14) Here, the
0
(
)
has the period of T=2.0[s]. Now, 4 displacement waves are prepared. Figure 4 shows the amplitude and the wave form of each wave.
Figure 4: Prepared Displacement Waves
To investigate the dynamic responses of the beams, we apply some pairs of these waves at the support points. Incidentally, the reason why the beams hold the unrealistic mass is to make it possible to visualize the deformation of the beams. In addition to this, for simplification, the effect of gravity is neglected purposely.
3.2.
Dynamic Response of the Beam under the Couple of Waves of the Same Phase
Figure 5 (Left: a, Right: b): The Deformation of Beam-A under the Couple of Waves of Same Phase at the time of
t
=10.5, 11.0, 11.5 and 12.0[s]
According to Figure 5-a, though both ends move between -5cm and +5cm, the centre of the beam is displaced just a little (0.5cm). The reason is that the stationary large mass has the tendency to stay in its srcinal location, and that the large mass in motion tends to keep moving in the same direction. Figure 5-b shows the deformation of Beam-A under the pair of waves of the coordinate phase, however the waves have different amplitudes to each other. Though the figure 5-a can be obtained by the general EOM, the general EOM cannot deal with the problem as figure 5-b. However, the EOM suggested in this study makes it possible to represent the result considering both the inertial force and the relative displacement. In fact, figure 5-b indicates similar deformation forms as figure 5-a, which include the effect of inertial forces.
3.3.
Dynamic Response of the Beam under the Couple of Waves of Opposite Phase
Figure 6 shows the deformation of Beam-B under the couple of waves of the opposite phase. Because the static analysis shows that the beam centre (i.e., at the 1000ton mass of Beam-A) keeps the same location in this case, the Beam-B is adapted to investigate the effect of the inertial force.
1000 [ton]5.0 [m]5.0 [m]300 [ton]3.0 [m]3.0 [m]4.0 [m]300 [ton]
Beam -ABeam -B
0 5 10 15 20-0.100.1
D i s p l a c e m e n t [ m ]
Time [sec]0 5 10 15 20-0.100.1
D i s p l a c e m e n t [ m ]
Time [sec]0 5 10 15 20-0.100.1
D i s p l a c e m e n t [ m ]
Time [sec]0 5 10 15 20-0.100.1
D i s p l a c e m e n t [ m ]
Time [sec]
Wave 1 (A=0.05m)Wave 2 (A=0.075m)Wave 3 (A= -0.05m)Wave 4 (A= -0.075m)
1000 [ton]Wave 1Wave 2
0 5 10-0.100.1
D i s p l a c e m e n t [ m ]
t = 10.5[s]0 5 10-0.100.1
D i s p l a c e m e n t [ m ]
t = 11.0[s]0 5 10-0.100.1
D i s p l a c e m e n t [ m ]
t = 11.5[s]0 5 10-0.100.1
D i s p l a c e m e n t [ m ]
t = 12.0[s]
1000 [ton]Wave 1Wave 1
0 5 10-0.100.1
D i s p l a c e m e n t [ m ]
t = 10.5[s]0 5 10-0.100.1
D i s p l a c e m e n t [ m ]
t = 11.0[s]0 5 10-0.100.1
D i s p l a c e m e n t [ m ]
t = 11.5[s]0 5 10-0.100.1
D i s p l a c e m e n t [ m ]
t = 12.0[s]

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