PoundingbetweenAdjacentFrameStructuresunderEarthquake ... · 2019. 7. 30. · advantages as follows: without the system global dynamics equations, high programming, low order of system

Research ArticlePounding between Adjacent Frame Structures under EarthquakeExcitation Based on Transfer Matrix Method of Multibody Systems

Yin Zhang, Jianguo Ding , Hui Zhuang, Yu Chang, Peng Chen, Xiangxiang Zhang,Wenhao Xie, and Jin Fan

School of Science, Nanjing University of Science and Technology, Nanjing 210094, China

Correspondence should be addressed to Jianguo Ding; [email protected]

Received 17 December 2018; Accepted 6 February 2019; Published 18 March 2019

Academic Editor: Flavio Stochino

Copyright © 2019 Yin Zhang et al. ,is is an open access article distributed under the Creative Commons Attribution License,which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited.

In this paper, the case of two adjacent frame structures is studied by establishing amechanical model based on the transfermatrixmethodof multibody system (MS-TMM).,e transfer matrices of the related elements and total transfer equation are deduced, combining withtheHertz-dampmode.,e pounding process of two adjacent frame structures is calculated by compiling the relevantMATLAB programduring severe ground motions. ,e results of the study indicate that the maximum error of the peak pounding forces and the peakdisplacements at the top of the frame structure obtained by theMS-TMMandANSYS are 6.22% and 9.86%, respectively. Comparing thecalculation time by ANSYS and MS-TMM, it shows that the computation efficiency increases obviously by using the MS-TMM. ,epoundingmainly occurs at the top of the short structure; meanwhile, multiple pounding at the same timemay occur when the separationgap is small. ,e parametric investigation has led to the conclusion that the pounding force, the number of poundings, the moment ofpounding, and the structural displacement are sensitive to the change of the seismic peak acceleration and the separation gap size.

1. Introduction

Pounding will occur when the relative displacement of ad-jacent buildings is greater than the width of their separationgaps under the excitation of earthquake; it will directly affectthe failure mode and the degree of damage of the structure.During the 1985 Mexico City earthquake, about 40% ofdamaged structures were subjected to pounding and about15% of buildings collapsed due to collision [1]. In the 1989Loma Prima earthquake, there were about 200 impact eventsin San Francisco, Oakland, Santa Cruz, and Watsonvilleinvolving more than 500 buildings [2]. In the 1995 Hanshinearthquake, the 2008 Wenchuan earthquake, and the 2010Yushu earthquake, a considerable part of the damage causedby structural pounding was discovered [3–5]. In the past,many countries did not specify the setting of the separationgap that results in the distance between adjacent buildingsbeing very close to or even zero inmany buildings, whichmaylead buildings collide with each other under the excitation ofearthquakes. At present, most countries have establishedregulations on separation gaps, but the pounding may occurin the event of a rare earthquake [6]. ,erefore, the pounding

of structures during the earthquake has attracted more andmore attention from earthquake-resistant workers, and theyhave carried out many related studies [7].

Nowadays, there are two main methods for structuralpounding research: the classical contact method and thecontact force-based method. ,e classical method is basedon the momentum conservation of the system and theNewtonian velocity recovery coefficient [8]. Papadrakakiset al. [9], Desroches and Muthukumar [10], and Mahmoudet al. [11] analyzed the structural pounding problem basedon the classical method; the application of this method isgreatly limited because it cannot reflect the impact factorssuch as impact force and impact deformation and is not easyto combine with the existing software. An analytical tech-nique based on the contact force-based method is developed,where the contact element is activated when the structurescome into contact. ,e contact element uses an equivalentspring element and an equivalent damper element to sim-ulate the interaction and energy dissipation during thecollision which is placed in the event of a pounding and iswithdrawn when disengaged. Scholars have conducted ex-tensive researches on the contact force-based method, such

HindawiAdvances in Civil EngineeringVolume 2019, Article ID 5706015, 31 pageshttps://doi.org/10.1155/2019/5706015

mailto:[email protected]://orcid.org/0000-0002-2569-4665https://creativecommons.org/licenses/by/4.0/https://creativecommons.org/licenses/by/4.0/https://doi.org/10.1155/2019/5706015

as the linear spring pounding model which uses only onespring element to simulate structural pounding [12]. Inorder to reflect the nonlinear process of pounding, the Hertzmodel was used to simulate structural pounding responses[13, 14]. In order to further accurately simulate the structuralcollision response, the Kelvin model [15] and the Hertz-damp model [16, 17], which can consider the energy con-sumption of pounding, have been proposed and applied.

A number of researchers have studied the problem ofpounding for adjacent structural under earthquake: Efrai-miadou et al. [18, 19] performed seismic pounding responseanalysis of the different layer height structures. Naderpouret al. [20] studied the case of pounding between two adjacentbuildings by the application of single degree-of-freedomstructural model. Ghandil and Aldaikh [21] developed aseries of SSSI models to accurately study the problem ofSSSI-included pounding of two adjacent buildings. Kar-ayannis and Naoum [22] investigated the influence of twoadjacent structures with different stories, different layouts,and initial distance on torsion under earthquake. Further-more, some more recent numerical analyses have beencarried out to study the influence of different parameters inpounding of buildings [23, 24].

,e multibody system is a system in which a number ofrigid and flexible bodies are connected in some way. ,ecurrent variousmultibody system dynamicsmethods have thefollowing common features: it is necessary to establish theglobal dynamics equations of the system; the global dynamicsequation of the complex system involves high-order matricesand makes the computational workload large. Rui et al.established the transfer matrix method of multibody system(MS-TMM) by combining transfer matrix method withmodern calculation method in 1993 [25], which has theadvantages as follows: without the system global dynamicsequations, high programming, low order of system matrix,and high computational efficiency [26]. ,e MS-TMM wasmainly divided into the transfer matrix method for linearmultibody systems [27] and discrete time transfer matrixmethod [28]. ,e discrete time transfer matrix method issuitable for linear time-varying, nonlinear, large-motion, andgeneral multibody systems. In the field of civil engineering,some applications have been carried out on the application oftheMS-TMM. For example, Ding et al. applied this method tothe vibration analysis of building structures and the dynamicresponse analysis of structures under earthquake action. ,eyhave studied new single-story frame bent structures [29],portal frame structures [30], frame structures [31], reinforcedconcrete shear wall structures [32], etc. ,e results show thatthe MS-TMM calculation results are similar to the finiteelement calculation results, and the calculation efficiency issignificantly improved.

Many scholars have studied the pounding problem ofadjacent structures under earthquakes. However, most ofthem use finite element software for simulation calculation,such as ANSYS, ABAQUS, LUSAS, and DRAIN-2DX, whichhas a large computational workload and is time-consuming.,erefore, it is an important research direction to seek ahighly efficient calculation method. In this paper, the MS-TMM is introduced into the study of pounding between two

adjacent frame structures under the excitation of earth-quake. First, the appropriate pounding model is selected.Next, the mechanical model is established based on MS-TMM. ,en, the transfer matrices of elements and the totaltransfer equation of the structure are derived. Finally, thecorresponding MATLAB program is compiled to analyzethe influence of the separation gap size and the peak seismicacceleration on pounding process of two unequal adjacentframe structures during severe ground motions. Meanwhile,the results calculated by MS-TMM and ANSYS are com-pared further.

2. Pounding Model Selection

Based on the contact force-based method, scholars haveconducted extensive research on structural poundingproblems; the structural pounding analysis model is shownin Figure 1, where m1 and m2 represent the masses of the twocolliding bodies, u1 and u2 are the corresponding dis-placements, v10 and v20 are the speeds of the initial contactmoments of the two colliding bodies, respectively, v1 and v2are the corresponding speeds of the colliding bodies atseparation moment, k represents the stiffness coefficient ofthe contact unit, c is the damping coefficient, and gp is theinitial gap.

When different pounding models are used to simulatethe structural pounding process, the force-deformationrelationship expressions of the contact elements are asfollows.

(1) Linear elastic model [12]:

Fc �kδ δ > 0

0 δ ≤ 0 , δ � u1 − u2 −gp, (1)

where Fc is the pounding impact force and δ is therelative deformation of two colliding bodies duringcontact.

(2) Hertz model [13, 14]:

Fc �kδ1.5 δ > 0

0 δ ≤ 0

⎧⎨

⎩

⎫⎬

⎭, δ � u1 − u2 −gp. (2)

(3) Kelvin model [15]:

Fc �kδ + c _δ δ > 0

0 δ ≤ 0

⎧⎪⎨

⎪⎩

⎫⎪⎬

⎪⎭,

_δ � _u1 − _u2,

c � 2ζ

��

km1m2

m1 + m2 ,

ζ � −ln e

��

π2 +(ln e)2 ,

e � −v1 − v2

v10 − v20,

(3)

2 Advances in Civil Engineering

where _δ is the relative deformation speed of thecollision body during the pounding, ζ is the corre-sponding damping ratio, _u1 and _u2 are the de-rivatives of m1 and m2 for time t, and e is theNewtonian speed recovery coefficient before andafter pounding.

(4) Hertz-damp model [16, 17]:

Fc �kδ1.5 + c _δ δ > 0

0 δ ≤ 0

⎧⎪⎨

⎪⎩

⎫⎪⎬

⎪⎭,

c � λδ1.5,

λ �3k 1− e2( 4 v10 − v20(

,

(4)

where λ is the hysteresis damping coefficient, k is thecontact stiffness, reinforced concrete is usually takenas 2.0 × 106 kN/m3/2, and e represents the energyrecovery coefficient, which is usually taken as 0.65 inthe typical concrete structure.

,e linear spring model cannot simulate the energydissipation and the changes in local stiffness with com-pression. ,e Kelvin model also cannot representchanges in the compression stiffness of the pounding, butit can represent the energy dissipation; however, whenpounding is from the maximum compression to theuncompressed regression, the Kelvin model’s simulatedpounding force appears as a tensile force for a period oftime before the movement is about to come out of contact,which is inconsistent with the facts. ,e Hertz model cansimulate changes in compression stiffness, but does notrepresent energy dissipation during pounding. ,e Hertz-damp model improves this shortcoming of the Hertzmodel; therefore, this paper selects the Hertz-dampmodel.

3. Mechanical Model of a Frame Structure

In this paper, a frame structure is used as a multibodysystem. Since the planar arrangement of the frame structureis generally regular and symmetrical, it is simplified into aplanar structure for calculation and analysis. Model sim-plification concerns three parts: the first part is the sim-plification of the column, based on the discretization idea,the column is equivalent to several concentrated massesconnected by vertical elastic beams; the second part is the

beam-column joint, which is equivalent to a plane hingeunit; and the third part is the connection of the plate and thebeam, which is simplified as a transverse elastic beamconnected to the concentrated mass. ,e self-weight andexternal load of the structure are applied to the concentratedmass, ignoring the axial deformation of the elastic beam,only considering the lateral deformation. In summary, themechanical model of the frame structure is shown inFigure 2.

4. Derivation of Element Transfer Matrix

In this paper, the discrete time transfer matrix method isused to analyze the pounding response of two structuresunder the excitation of earthquake. For time-varyingsystems, there is no linear relationship between statevectors; so, we need to introduce a linearization idea tolinearize the state vector in order to transfer the force anddisplacement between points. When the time step is smallto a certain extent, the relationship between physicalparameters can be approximated, seen as linear in thephysical process corresponding to each time step. In thispaper, the linearization of acceleration and velocity isperformed by the Newmark-β method [33]; the relation-ship between velocity, acceleration, and displacement is asfollows:

€xt+Δt � Axt+Δt + B,

_xt+Δt � Cxt+Δt + D, (5)

where

A �1

βΔt2,

B � A −x ti−1( −x′ ti−1( Δt−(0.5− β)x″ ti−1( Δt2

,

C �c

βΔt2,

D � x′ ti−1( −(1− c)x″ ti−1( Δt + cBΔt.

(6)

When calculating the pounding response of two framestructures under earthquake, it is necessary to consider theseismic excitation and pounding force; therefore, theconcept of extending the transfer matrix is introduced hereto distinguish the transfer matrix based on the linearmultibody system transfer matrix method and the

m1 m2gp

u2u1 k

c

Figure 1: Equivalent model of pounding between two colliding bodies.

Advances in Civil Engineering 3

multibody system discrete time transfer matrix method.,e state vector of elements in physical coordinates isdefined as

z � x, y, θz, mz, qx, qy, 1 T, (7)

where x is the displacement in x direction, y is the dis-placement in y direction, θz is the angular displacementaround z, mz is the internal moment, qx is the internalforce in x direction, and qy is the internal force in y di-rection. ,e external force is fx,c � −m €xg + FC, where FCrepresents the pounding force, and it is assumed in thispaper that the pounding only occurs at the concentratedmass point.

4.1. Concentrated Mass at One Input End and One OutputEnd. ,e concentrated mass of mass m is shown inFigure 3(a), the input end is I and the output end is O.According to its stress balance and deformation relationship,we have

xO � xI,

yO � yI,

θz,O � θz,I,

mz,O � mz,I,

⎧⎪⎪⎪⎪⎪⎨

⎪⎪⎪⎪⎪⎩

qy,O � qy,I,

qx,O � qx,I + fx,c −m €xI − c _xI − kxI.

(8)

,e transfer equation is zO � UzI; combining equations(8) and (5), we can get the transfer matrix as follows:

U �

1 0 0 0 0 0 0

0 1 0 0 0 0 0

0 0 1 0 0 0 0

0 0 0 1 0 0 0

−(mA + cC + k) 0 0 0 1 0 −mBxI − cDxI + fx,c0 0 0 0 0 1 0

0 0 0 0 0 0 1

⎡⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎣

⎤⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎦

.

(9)

4.2. Concentrated Mass at One Input End and Two OutputEnds. ,e concentrated mass of mass m is shown inFigure 3(b), the input end is I and the output ends areO1 andO2. According to its stress balance and deformation re-lationship, we have

xO1 � xO2 � xI,

yO1 � yO2 � yI,

θz,O1 � θz,O2 � θz,I,

⎧⎪⎪⎪⎨

⎪⎪⎪⎩

mz,O1 + mz,O2 � mz,I,

qy,O1 + qy,O2 � qy,I,

qx,O1 + qx,O2 � qx,I + fx,c −m €xI − cζ _xI − kxI.

⎧⎪⎪⎪⎨

⎪⎪⎪⎩

(10)

,e transfer equation is UOzO � UIzI � UOzO1zO2

�

UIzIz0

; combining equations (10) and (5), we can get the

transfer matrix as follows:

Figure 2: Mechanical model of frame structure.


UO �

I3 03×11

U1 U1

01×6 I1 01×7

U3 −U3

03×7 U3

01×6 I1 01×7

⎡⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎣

⎤⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎦

,

UI �

I3 03×11

U2 U4 U1

01×6 I1 01×7

03×7 03×7

U3 03×7

01×6 I1 01×7

⎡⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎣

⎤⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎦

,

(11)

whereU1 � O3×3 I3 O3×1 ,

U2 �

0 0 0

−(mA + cC + k) 0 0

0 0 0

⎡⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎣

⎤⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎦,

U3 � I3 O3×4 ,

U4 �

1 0 0 0

0 1 0 −mBxI + cζDxI + fx,c0 0 1 0

⎡⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎣

⎤⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎦,

z0 � [0, 0, 0, 0, 0, 0, 0]T.

(12)

According to the transfer equation, the state vector of theoutput can be obtained as follows:

zO1 � E1U−1O UI E3zI1 + E4zI2 ,

� E1U−1O UIE3zI1 + E1U

−1O UIE4zI2,

� U11zI1 + U12zI2,



−1O UIE4zI2,

� U21zI1 + U22zI2,

(13)

whereE1 � I7 O7×7 ,

E2 � O7×7 I7 ,

E3 �I7×7O7×7

,

E4 �O7×7I7×7

.

(14)

4.3. Concentrated Mass at Two Input Ends and Two OutputEnds. ,e concentrated mass of mass m is shown inFigure 3(c), the input ends are I1 and I2 and the output endsare O1 and O2. According to its stress balance and de-formation relationship, we have

xO1 � xO2 � xI1 � xI2,

yO1 � yO2 � yI1 � yI2,

θz,O1 � θz,O2 � θz,I1 � θz,I2,

⎧⎪⎪⎪⎨

⎪⎪⎪⎩

mz,O1 + mz,O2 � mz,I1 + mz,I2,

qy,O1 + qy,O2 � qy,I1 + qy,I2,

qx,O1 + qx,O2 � qx,I1 + qx,I2 + fx,c −m €xI − cζ _xI − kxI.

⎧⎪⎪⎪⎨

⎪⎪⎪⎩

(15)

,e transfer equation is UOzO � UIzI � UOzO1zO2

�

UIzI1zI2

; combining equations (15) and (5), we can get the

transfer matrix as follows:

yqx,O

qx,I

fx,C

y qy,O2

qy,O1qy,I

qx,O2

qx,O1

mz,O2

mz,I

mz,O1

qx,I fx,c(a) (b)

yqy,O2

qy,O1

qx,O2mz,O2

qx,O1mz,O1

qy,I2mz,I2

mz,I1

qx,I2

qx,I1

qy,I1

fx,c

yqy,O

qx,Omz,Omz,I1

qy,I2mz,I2

qx,I2

qx,I1

qy,I1

fx,c

(c) (d)y

x

θz,O

θz,Ijqy,O

qx,I

qy,IEI

l

mz,O

mz,I

qx,O

y

x I O

kyk

kx

(e) (f )

Figure 3: Elements schematic. (a) Concentrated mass at one inputend and one output end, (b) concentrated mass at one input endand two output ends, (c) concentrated mass at two input ends andtwo output ends, (d) concentrated mass at two input ends and oneoutput end, (e) the longitudinal elastic beam, and (f) planar elastichinge.


U0 �

I3 O3×11U1 U1O1×6 I1 O1×7U3 −U3O3×7 U3O1×6 I1 O1×7 O

⎡⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎣

⎤⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎦

,

UI �

I3 O3×11U2 U4 U1O1×6 I1 O1×7O3×7 O3×7U3 O3×7O1×6 I1 O1×7

⎡⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎣

⎤⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎦

.

(16)

According to the transfer equation, the state vector of theoutput can be obtained as follows:



−1O UIE4zI2,

� U11zI1 + U12zI2,



−1O UIE4zI2,

� U21zI1 + U22zI2.

(17)

,e U1, U2, U3, U4, E1, E2, E3, and E4 are the same as inequations (12) and (14).

4.4. Concentrated Mass at Two Input Ends and One OutputEnd. ,e concentrated mass of mass m is shown inFigure 3(d), the input ends are I1 and I2 and the output end isO. According to its stress balance and deformation re-lationship, we have

xO � xI1 � xI2,

yO � yI1 � yI2,

θz,O � θz,I1 � θz,I2,

⎧⎪⎪⎪⎨

⎪⎪⎪⎩

mz,O � mz,I1 + mz,I2,

qy,O � qy,I1 + qy,I2,

qx,O � qx,I1 + qx,I2 + fx,c −m €xI − cζ _xI − kxI.

⎧⎪⎪⎪⎨

⎪⎪⎪⎩

(18)

,e transfer equation is zO � UzI � UzI1zI2

� U(E3zI1+

E4zI2); combining equations (18) and (5), we can get thetransfer matrix as follows:

U �

U3 O3×7U2 U4 U1O1×6 I1 O1×7

⎡⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎣

⎤⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎦. (19)

,e U1, U2, U3, U4, E3, and E4 are the same as inequations (12) and (14).

4.5. Massless Elastic Beam. ,e longitudinal elastic beamwith a length of l and a flexural rigidity of k is shown inFigure 3(e), the input end is I and the output end is O.According to its stress balance and deformation relationship,we have

qx,O � qx,I,

qy,O � qy,I,

−mz,I + mz,O − qx,Ol � 0,

⎧⎪⎪⎪⎪⎪⎨

⎪⎪⎪⎪⎪⎩

xO � xI + θz,Il +mz,Il

2

2EI+

qx,Il3

6EI,

yO � yI,

θz,O � θz,I +mz,Il

EI+

qx,Il2

2EI.

⎧⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎨

⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎩

(20)

,e transfer equation is zO � UzI; combining equation(20), we can get the transfer matrix as follows:

U �

1 0 ll2

2EIl3

6EI0 0

0 1 0 0 0 l

0 0 1l

EI

l2

2EI0 0

0 0 0 1 l 0 0

0 0 0 0 1 0 0

0

0

0

0

0

0

0

0

0

0

1

0

0

1

⎡⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎣

⎤⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎦

. (21)

Similarly, the transfer matrix of transverse masslessbeams is

U �

1 0 0 0 0 0 l

0 1 ll2

2EIl3

6EI0 0

0 0 1l

EI

l2

2EI0 0

0 0 0 1 0 l 0

0 0 0 0 1 0 0

0

0

0

0

0

0

0

0

0

0

1

0

0

1

⎡⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎣

⎤⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎦

. (22)


4.6. Planar Elastic Hinge. A plane elastic joint with angularstiffness of k, lateral spring stiffness of kx, and longitudinalspring stiffness of ky is shown in Figure 3(f), and the transferequation is

zO � UzI, (23)

where

U �

1 0 0 0 −1kx

0 0

0 1 0 0 0 −1ky

0

0 0 11k

0 0 0

O4×3 I4

⎡⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎣

⎤⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎦

. (24)

5. Derivation of the Total Transfer Equation ofFrame Structure

,e transmission direction in this paper is shown in Figure 4.,e bottom ends of the frame column a01, a02, . . . , a0(n−1) arethe input ends, and a0n is the output end. ,e mechanicalelement model number description is shown in the figure: thebottom ends of the frame columns are numbered asa01, a02, . . . , a0n from left to right; the mass points from the leftto the right of the first beam-column joint are numbered asa11, a21, . . . , an1; themass points from the left to the right of thesecond beam-column joint are numbered as a12, a22, . . . , an2,and so on; the mass point of the beam-column joint of thenth row of the mth layer is numbered as Anm; the first layer

of beams is numbered as L11, L21, . . . , L(n−1)1 from left toright, and so on; the number of the (n− 1)th beam of themth layer is L(n−1)m. Take the number L(n−1)m beam andnumber Anm column as examples to illustrate the internalnumbering of the beam and the inside of the column, as shownin Figure 5.

,e state vector that defines the elements in physicalcoordinates is

z � x, y, θz, mz, qx, qy, 1 T

. (25)

Taking frame column Ani and beam L(n−1)i as examples,the transfer matrices are derived as follows:

UAni � Uani,tUani,t−1 · · ·Uani,2Uani,1,

UL(n−1)i � Ul(n−1)i,kUl(n−1)i,k−1 · · ·Ul(n−1)i,2Ul(n−1)i,1.(26)

For the side cross-frame column A11, input from the a01and passed up, the transfer equation can be obtained asfollows:

Za11A11 � UA11Za01A11 � Ua11,tUa11,t−1 . . .Ua11,2Ua11,1Za01A11.

(27)

For a11, the input is Za11A11 and the outputs areZa11A12 and Za11L11, and the transfer matrices of the con-centrated mass output according to one end input are

Za11L11 � U1a11Za11A11 � E1U

−1Oa11

UIa11E3 � U1a11UA11Za01A11,

Za11A12 � U2a11Za11A11 � E2U

−1Oa11

UIa11E3 � U2a11UA11Za01A11.

(28)

In the same way, we continue to pass up and sort out thefollowing equation:

Za12L12 � U1a12UA12U

2a11UA11Za01A11

Za13L13 � U1a13UA13U

2a12UA12U

2a11UA11Za01A11

⋮

Za1(m−1)L1(m−1) � U1a1(m−1)

UA1(m−1)U2a1(m−2)

UA1(m−2) · · ·U2a12UA12U

2a11UA11Za01A11

Za1mL1m � Ua1mUA1mU2a1(m−1)

UA1(m−1)U2a1(m−2)

UA1(m−2) · · ·U2a12UA12U

2a11UA11Za01A11.

(29)

Equation (29) can be organized into a matrix form:

Za11L11Za12L12⋮

Za1(m−1)L1(m−1)Za1mL1m

⎡⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎣

⎤⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎦

�

U1a11UA11U1a12UA12U

2a11UA11

⋮

U1a1(m−1)UA1(m−1)U2a1(m−2)

UA1(m−2) · · ·U2a12UA12U

2a11UA11

Ua1mUA1mU2a1(m−1)

UA1(m−1)U2a1(m−2)

UA1(m−2) · · ·U2a12UA12U

2a11UA11

⎡⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎣

⎤⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎦

Za01A11 � UA1Za01A11, (30)

where Za11L11,Za12L12, · · · ,Za1(m−1)L1(m−1),Za1mL1m are state vec-tors at the left end of beams L11, L12, · · · , L1(i−1), L1i and thetransfer matrix of each beam of the first span is represented

by UL11,UL12, · · · ,UL1(m−1),UL1m; according to its transfermatrix and transfer equation, the state vector at the right endof each beam is


Za21L11Za22L12⋮


⎡⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎣

⎤⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎦

�

UL11UL12⋱

UL1(m−1)UL1m

⎡⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎣

⎤⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎦

U1a11UA11U1a12UA12U

2a11UA11

⋮

U1a1(m−1)UA1(m−1)U2a1(m−2)

· · ·U2a12UA12U2a11UA11

Ua1mUA1mU2a1(m−1)

· · ·U2a12UA12U2a11UA11

⎡⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎣

⎤⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎦

Za01A11. (31)

ani,t–1

ani,t

ani,t–2

ani,2

ani,3

ani,1

(a)

l(n–1)i,1 l(n–1)i,3

l(n–1)i,2 l(n–1)i,4

l(n–1)i,m–2 l(n–1)i,m–1

(b)

Figure 5: Mechanical model of beam and column. Calculation diagram of (a) frame column Ani and (b) beam L(n−1)i.

a1ma2m a(n–1)m

a2(m–1)

L2(m–1)L1(m–1)

L11

A11

a11L21a21

A21

a1(m–1)

A1(m–1) A2(m–1)

a(n–1)(m–1)

L(n–1)(m–1)

A(n–1)(m–1)

a(n–1)1

A(n–1)1

an1

An1

a01 a02 a0(n–1) a0n

L(n–1)1

an(m–1)

An(m–1)

anm

Anm

L(n–1)m

A(n–1)mA2mA1m

L1m L2m

Figure 4: Frame structure mechanics model.


UL1 can be defined as follows:

UL1 �

UL11UL12⋱

UL1(m−1)UL1m

⎡⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎣

⎤⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎦

. (32)

For the intermediate frame column a2, pass up frominput a02, the transfer equation of a21 is

Za21A21 � UA21Za02A21. (33)

Element a21 is a concentrated mass at one input end andtwo output ends, the input ends are Za21A21 and Za21L11, theoutput ends are Za21L21 and Za21A22, so we have

Za21L21 � U11a21Za21L11 + U

12a21Za21A21

� U11a21Za21L11 + U12a21UA21Za02A21,

Ζa21A22 � U21a21Za21L11 + U

22a21Za21A21

� U21a21Za21L11 + U22a21UA21Za02A21.

(34)

In the same way, we continue to pass up and sort out

Za22L22 � U11a22Za22L12 + U

12a22UA22U

21a21Za21L11 + U

12a22UA22U

22a21UA21Za02A21,

Za23L23 � U11a23Za23L13 + U

12a23UA23U

21a22Za22L12 + U

12a23UA23U

21a22UA22U

21a21Za21L11

+ U12a23UA23U22a22UA22U

22a21UA21Za02A21,

⋮

Za2mL2m � U11a2m

Za2mL1m + U12a2m

UA2mU21a2(m−1)

Za2(m−1)L1(m−1) + · · · + U12a2m

UA2mU21a2(m−1)

UA2(m−1) · · ·UA22U21a21Za21L11

+ U12a2mUA2mU22a2(m−1)

UA2(m−1) · · ·U22a21UA21Za02A21,

(35)

which is organized into a matrix form

Za21L21

Za22L22

⋮

Za2(m−1)L2(m−1)

Za2mL2m

⎡⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎣

⎤⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎦

�

U11a21 0 · · · 0 0

U12a22UA22U21a21

U11a22 · · · 0 0

⋮ ⋮ ⋱ ⋮ ⋮

U12a2(m−1)UA2(m−1)U21a2(m−2)

· · ·UA22U21a21

U12a2(m−1)UA2(m−1)U21a2(m−2)

· · ·UA23U21a22

· · · U11a2(m−1) 0

U12a2mUA2mU21a2(m−1)

UA2(m−1)

· · ·UA22U21a21


UA2(m−1)

· · ·UA23U21a22

· · · U12a2mUA2mU21a2(m−1)

U11a2m

⎡⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎣

⎤⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎦

,

Za21L11

Za22L12

⋮

Za2(m−1)L1(m−1)

Za2mL1m

⎡⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎣

⎤⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎦

+

U12a21UA21

U12a22UA22U22a21UA21

⋮

U12a2(m−1)UA2(m−1)U22a2(m−2)

UA2(m−2) · · ·U22a21UA21


UA2(m−1) · · ·U22a21UA21

⎡⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎣

⎤⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎦

Za02A21,

(36)


U1A2 and U1A2 can be defined as follows:

U1A2 �

U11a21 0 · · · 0 0

U12a22UA22U21a21

U11a22 · · · 0 0

⋮ ⋮ · · · ⋮ ⋮U12a2(m−1)UA2(m−1)U

21a2(m−2)

· · ·UA22U21a21

U12a2(m−1)UA2(m−1)U21a2(m−2)

· · ·UA23U21a22

· · · U11a2(m−1) 0


UA2(m−1)· · ·UA22U

21a21


UA2(m−1)· · ·UA23U

21a22

· · · U12a2mUA2mU21a2(m−1)

U11a2m

⎡⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎣

⎤⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎦

,

U2A2 �

U12a21UA21U12a22UA22U

22a21UA21

⋮U12a2(m−1)UA2(m−1)U

22a2(m−2)

UA2(m−2) · · ·U22a21UA21


UA2(m−1) · · ·U22a21UA21

⎡⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎣

⎤⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎦

,

(37)

where Za21L21,Za22L22, · · · ,Za2(m−1)L2(m−1),Za2mL2m are state vectorsat the left end of the beam L21, L22, · · · , L2(m−1), L2m and thetransfer matrices of the second span beam are represented by

UL21,UL22, · · · ,UL2(m−1),UL2m; according to the transfer matrixand the transfer equation of a span beam, the state vector ofthe right end of each beam can be obtained as follows:

Za31L21Za32L22⋮


⎡⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎣

⎤⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎦

�

UL21UL22⋱

UL2(m−1)UL2m

⎡⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎣

⎤⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎦

Za21L21Za22L22⋮


⎡⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎣

⎤⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎦

� UL2

Za21L21Za22L22⋮


⎡⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎣

⎤⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎦

. (38)

,e transfer of the state vector of the intermediate frameis the same as that of the frame column a2, so it can be

derived in the same way. ,e transfer equation of the framecolumn aj is

Zaj1Lj1Zaj2Lj2⋮

Zaj(m−1)Lj(m−1)ZajmLjm

⎡⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎣

⎤⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎦

�

U11aj1 0 · · · 0 0

U12aj2UAj2U21aj1

U11aj2 · · · 0 0

⋮ ⋮ ⋱ ⋮ ⋮U12aj(m−1)UAj(m−1)U

21aj(m−2)

· · ·UAj2U21aj1

U12aj(m−1)UAj(m−1)U21aj(m−2)

· · ·UAj3U21aj2

· · · U11aj(m−1) 0

U12ajmUAjmU21aj(m−1)

UAj(m−1)· · ·UAj2U

21aj1



21aj2

· · · U12ajmUAjmU21aj(m−1)

U11ajm

⎡⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎣

⎤⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎦

,

Zaj1L(j−1)1Zaj2L(j−1)2⋮

Zaj(m−1)L(j−1)(m−1)ZajmL(j−1)m

⎡⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎣

⎤⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎦

+

U12aj1UAj1U12aj2UAj2U

22aj1UAj1

⋮


UAj(m−2) · · ·U22aj1UAj1



⎡⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎣

⎤⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎦

Za0jAj1,

(39)


U1Aj and U2Aj can be defined as follows:

U1Aj �

U11aj1 0 · · · 0 0

U12aj2UAj2U21aj1

U11aj2 · · · 0 0

⋮ ⋮ ⋱ ⋮ ⋮U12aj(m−1)UAj(m−1)U

21aj(m−2)

· · ·UAj2U21aj1


· · ·UAj3U21aj2

· · · U11aj(m−1) 0



21aj1



21aj2

· · · U12ajmUAjmU21aj(m−1)

U11ajm

⎡⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎣

⎤⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎦

,

U2Aj �

U12aj1UAj1U12aj2UAj2U

22aj1UAj1

⋮





⎡⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎣

⎤⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎦

,

(40)

where Zaj1Lj1,Zaj2Lj2, · · · ,Zaj(m−1)Lj(m−1),ZajmLjm are state vectorsat the left end of beams Lj1, Lj2, · · · , Lj(m−1), Ljm, and usingULj1,ULj2, · · · ,ULj(m−1),ULjm to represent the transfer matrix of

the jth span of each beam, the state vector at the right end ofeach beam can be obtained as follows:

Za(j+1)1Lj1Za(j+1)2Lj2⋮

Za(j+1)(m−1)Lj(m−1)Za(j+1)iLji

⎡⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎣

⎤⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎦

�

ULj1ULj2⋱

ULj(m−1)ULjm

⎡⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎣

⎤⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎦

Zaj1Lj1Zaj2Lj2⋮


⎡⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎣

⎤⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎦

� ULj

Zaj1Lj1Zaj2Lj2⋮


⎡⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎣

⎤⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎦

. (41)

For the rightmost column an, Zan1L(n−1)1,Zan2L(n−1)2, · · · ,Zan(m−1)L(n−1)(m−1),ZanmL(n−1)m are the state vectors on the rightside of the (n− 1) span beam. Passing down from anm, thetransfer equation is

ZanmAnm � UanmZanmL(n−1)m. (42)

From the transfer relationship, the following equationcan be obtained in turn:

Zan(m−1)An(m−1) � U11an(m−1)

Zan(m−1)L(n−1)(m−1) + U12an(m−1)

UAnmUanmZanmL(n−1)m,

Zan(m−2)An(m−2) � U11an(m−2)

Zan(m−2)L(n−1)(m−2) + U12an(m−2)

UAn(m−1)U11an(m−1)

Zan(m−1)L(n−1)(m−1) + U12an(m−2)

UAn(m−1)U12an(m−1)

UAnmUanmZanmL(n−1)m,

⋮

Zan1An1 � U11an1Zan1L(n−1)1 + U

12an1UAn2U

11an2Zan2L(n−1)2 + · · · + U

12an1UAn2U

12an2UAn3 · · ·UAniUanmZanmL(n−1)m.

(43)

For a0n, the input end is Zan1An1 and the output end isZa0nAn1; so the transfer equation is expressed as follows:

Za0nAn1 � UAn1Zan1An1. (44)

Substituting equation (43) into equation (44),

Za0nAn1 � UAn1U11an1Zan1L(n−1)1 + UAn1U

12an1UAn2U

11an2Zan2L(n−1)2

+ · · · + UAn1U12an1UAn2U

12an2UAn3 · · ·UAnmUanmZanmL(n−1)m.

(45)

Equation (45) can be organized into a matrix form asfollows:


Za0nAn1 � UAn1U11an1

UAn1U12an1UAn2U

11an2

· · · UAn1U12an1UAn2 · · ·UAnmUanm

Zan1L(n−1)1Zan2L(n−1)2⋮

ZanmL(n−1)m

⎡⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎣

⎤⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎦

� UAn

Zan1L(n−1)1Zan2L(n−1)2⋮

ZanmL(n−1)m

⎡⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎣

⎤⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎦

. (46)

Substituting the above derivation formula into equation(46), the relationship between the output end and the inputend is

Za0nAn1 � UAnUL(n−1)U2A(n−1)Za0(n−1)A(n−1)1 + UAnUL(n−1)U

1A(n−1)UL(n−2)U

2A(n−2)Za0(n−2)A(n−2)1+

⋮

UAnUL(n−1)U1A(n−1)UL(n−2) · · ·U

1A2UL1UA1Za01A11.

(47)

,en, the total transfer equation of the frame structure is

UallZall � 0, (48)where

Uall �UAnUL(n−1)U1A(n−1)UL(n−2)

· · ·U1A2UL1UA1· · ·

UAnUL(n−1)U1A(n−1)

UL(n−2)U2A(n−2)UAnUL(n−1)U2A(n−1) −I

⎡⎢⎢⎢⎢⎣ ⎤⎥⎥⎥⎥⎦, (49)

Zall �

Za01A11Za02A21⋮

Za0(n−1)A(n−1)1Za0nAn1

⎡⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎣

⎤⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎦

. (50)

6. Example Analysis

Two adjacent frame structures are selected as models, one ofwhich is a 5-layer structure (J1) and the other is a 10-layerstructure (J2); the height of each layer of the two structures is3.2m, the concrete grade is C30, the strength grade of thelongitudinal reinforcement of the column and the beam isHRB400, the strength grade of the stirrup is HPB300, and thecross-sectional dimensions of the J1 and J2 structural col-umns are 500mm∗ 500mm and 600mm∗ 600mm, re-spectively, and the beam cross-section dimensions are600mm∗ 300mm. ,e structural schematic is shown inFigure 6.

A simplified model of the structure based on the MS-TMM is shown in Figure 7. J1 simplifies each frame columninto four concentrated masses and four sections of masslessbeams; the first span of each beam is simplified into twoelastic hinges, three concentrated masses, and three sectionsof massless elastic beams; the second span of each beam issimplified into two planar elastic hinges, seven concentratedmasses, and seven sections of massless elastic beams. J2simplifies each frame column into four concentrated masses

and four sections of massless beams; simplifies each of thefirst and third span of each beam into two planar elastichinges, seven concentrated masses, and seven sections ofmassless elastic beams; simplifies each beam of the secondspan into two elastic hinges, three concentrated masses, andthree sections of massless elastic beams. ,e specific pa-rameters are shown in Table 1.

When the peak acceleration of seismic is large, thestructure may exhibit elastoplastic deformation. ,erefore,the stiffness and damping of the structural material shouldbe considered as a function of time, that is, determines thestructural restoring force model. Currently used resiliencemodels include bilinear restoring model and trilinear re-storing model. ,e bilinear restoring model is too simpleand rough; so, this paper uses the trilinear restoring model,as shown in Figure 8.

,e steps for calculating the pounding response of twostructures under earthquake based on MS-TMM are shownin Figure 9. In this paper, the calculation is achieved bycompiling the relevant MATLAB program.

In this paper, two natural earthquake waves (the ElCentro earthquake wave and the Taft earthquake wave)


Ten-layer structure (J2) Five-layer structure (J1)

2400

D

C

B

A

6900

6900

7200

6900

2100

0

1 32

D

C

B

A

6000

14400

2400 6000

6900

7200

6900

2100

0

1 32 4

(a)

1

2400 2400

14400

6900 6000 6000

9300

13 32 2 4

10∗

3200

5∗32

00

(b)

Figure 6: Structure diagram. (a) Structural plan. (b) Structural elevation.


and one artificial earthquake wave (the Nanjing earth-quake wave) are chosen. We adjust the peak acceleration to35 cm/s2, 70 cm/s2, and 220 cm/s2, respectively, and the

pounding response of the structure under different sep-aration gap sizes is calculated, as shown in Table 2. In orderto compare and analyze the pounding response of adjacentstructures under the excitation of earthquake based onMS-TMM, this paper also uses the element softwareANSYS to perform modeling and calculation. In theanalysis of ANSYS, because the structure is regular,BEAM161 is selected as the model for both the beam andthe column elements, and SHELL63 is selected as themodel for the floor elements, the contact type is defined asASSC, as shown in Figure 10.

7. Pounding Force Analysis

,e time history of the pounding force based on MS-TMMand ANSYS is shown in Figures 11 to 14; it can be seen thatthe results obtained by the two methods are similar. It isfound that changing the peak value of seismic accelerationand the width of separation gap will have a great influenceon pounding force, the number of pounding, and themoment of pounding. Under the earthquake with samepeak acceleration, as the width of the separation gap de-creases, the pounding force and the number of poundingsincrease. When the separation gap width is the same, as thepeak acceleration of the earthquake increases, thepounding force and the number of poundings increase. It

L110a110

a19 L19 a29 a39L29 a14 a24 a34L14 L34 a44L24

a110,3

a110,2

a110,1

L11a11a11,3

a11,2

a11,1

a01 a02 a03 a01 a02 a03 a04

L21 L11 L21 L31a12a21,3

a21,2

a21,1

a13a31,3

a31,2

a31,1

a12 a13a21,3

a21,2

a21,1

a14a41,3

a41,2

a41,1

a31,2

a31,2

a31,1

a11a11,3

a11,2

a11,1

a210a210,3

a210,2

a210,1

a310a310,3

a310,2

a310,1

a15a15,3

a15,2

a15,1

a25 a35a25,3

a25,2

a25,1

a45a45,3

a45,2

a45,1

a35,3

a35,2

a35,1

L210 L15 L25 L35

Figure 7: Simplified model diagram.

Table 1: Structure parameter.

Structure

Column Planar elastic hinge First span beam Second span beam ,ird span beamElasticbeamlength(m)

Concentratedmass (kg)

Kx(N·m/rad)

Ky(N·m/rad)

K (N·m/rad)

Elasticbeamlength(m)






J1 0.8 1723.8 1.6∗109 1.6∗109 6.3∗109 0.75 2752.8 0.6 2592.6 0.75 2752.8J2 0.8 2337.3 1.6∗109 1.6∗109 6.3∗109 0.6 2681.28 0.8625 3303.72 — —

F

Fy

1

2 3(8)

6(11)

9

7

5

0 4 10

Fc

xcx

xy

k4

k1

k2

k3

Figure 8: Trilinear restoring model.


can be seen from Figure 11 that when the acceleration peakis 35 cm/s2 and the separation gap width is 30mm, J1, a15and J2, a35 will not pound. As shown in Figure 12, when thepeak acceleration is 70 cm/s2 and the width of the gap is60mm, there will be no pounding between J1, a15 and J2,a35. As shown in Figure 13, when the peak acceleration is220 cm/s2 and the width of the gap is 150mm, there will beno pounding between J1, a15 and J2, a35. It can be seen fromFigure 14 that when the acceleration peak is 400 cm/s2 andwhen the gap size is 220mm, J1, a15 and J2, a35 will notpound.

According to Table 3, we can see, under the excitationof the El Centro wave with a peak acceleration of 35 cm/m2, when the widths of the separation gaps are 10mm and20mm, respectively, the errors of the peak value of thepounding force calculated by the two methods are 5.59%and 3.88%, respectively. Under the excitation of the ElCentro wave with a peak acceleration of 70 cm/m2, whenthe widths of the separation gaps are 10mm and 40mm,

respectively, the errors of the peak value of the poundingforce calculated by the two methods are 6.22% and 4.62%,respectively. Under the excitation of the El Centro wavewith a peak acceleration of 220 cm/m2, when the widths ofthe separation gaps are 100mm and 130mm, respectively,the errors of the peak value of the pounding force cal-culated by the two methods are 5.37% and 4.39%, re-spectively. Under the excitation of the El Centro wave witha peak acceleration of 400 cm/m2, when the widths of theseparation gap are 200mm and 210mm, respectively, theerrors of the peak value of the pounding force calculated bythe two methods are 6.25% and 5.32%, respectively. Evi-dently, the results of the two methods are almost the same;however, using ANSYS to calculate the pounding responseof two adjacent frame structures under the excitation ofearthquake takes about 8 hours, while the calculation byMS-TMM only takes about 20minutes; its time con-sumption is only 1/24 of ANSYS, and the calculation speedadvantage is very obvious.

Start

Establish two structural mechanics models, break up the wholeinto parts, determine the state vector

Establish the extended transfer matrix of each element and thesystem total transfer matrix, Uall

Determine the initial conditions and boundary conditions of thetwo structures at time = 0

Determine the linearization parameter value at time ti and determinethe transfer matrix and total transfer matrix of each element

Determine the unknowns in the boundary state vector based on thetotal transfer equation and boundary conditionsDetermine the

stiffness of eachcomponent based

on the trilinearrestoring model

Calculate thepounding force

based on theextracted data

and formula (4)Determine the state vector of the element at time = ti accordingto the transfer equation

Extract the required response data from the state vector of the corresponding component

ti < T ti < T

ti ≥ T

End

i = 1i = i + 1i = i + 1

Figure 9: ,e process for calculation of pounding response based on MS-TMM.

Table 2: Seismic wave peak acceleration peak and separation gap size description table.Seismic acceleration peak 35 cm/s2 70 cm/s2 220 cm/s2 400 cm/s2

Separation gap (mm) 10 20 30 10 40 60 100 130 150 100 200 210


Figure 10: Analyzed ANSYS model of two adjacent frame structures.

El Centro-ANSYSEl Centro-MSTMM

Nanjing-MSTMMTaft-MSTMM

1.8E + 05

Poun

ding

forc

e (N

)

1.6E + 051.4E + 051.2E + 051.0E + 058.0E + 046.0E + 044.0E + 042.0E + 040.0E + 00

0 2 4Time (s)

6 8 10

(a)



Poun

ding

forc

e (N

)

1.6E + 05

1.4E + 05

1.2E + 05

1.0E + 05

8.0E + 04

6.0E + 04

4.0E + 04

2.0E + 04

0.0E + 000 2 4

Time (s)6 81 3 5 7 9 10

(b)



Poun

ding

forc

e (N

)

8.80E + 066.60E + 064.40E + 062.20E + 060.00E + 00

–2.20E + 06–4.40E + 06–6.60E + 06–8.80E + 06

0 2 4Time (s)

6 81 3 5 7 9 10

(c)

Figure 11: Time-history diagram of J2, a35-J1, a15 pounding force under seismic wave (peak acceleration is 35 cm/s2). (a) 10mm, (b) 20mm,

and (c) 30mm.


8. Displacement Analysis

Select the point of J1, a15 and the point J2, a35 for study.,edisplacement time-history diagrams are shown in Fig-ures 15 to 22. It can be seen from the comparison that thetrends of the displacement time-history diagram obtainedby the two methods are almost the same. As illustrated inthe figure, the acceleration peak of seismic wave and theseparation gap both have an effect on the structural dis-placement response. ,e main feature is that the dis-placement on the pounding side is limited, which makesthe displacement reaction produce obvious directionaldifference, and the difference becomes more significantwith the increase of the acceleration peak and the decreaseof the separation gap. Statistical comparisons are madebetween the displacement peaks obtained by the twomethods, as shown in Tables 4 and 5.

As shown in Tables 4 and 5, under the excitation ofseismic wave with peak acceleration of 35 cm/s2, 70 cm/s2,and 220 cm/s2, the peak displacement of J1, a15 graduallyincreases with the increase of the separation gap widthwhile the peak displacement of J2, a310 gradually de-creases as the width of the separation gap increases; thatis, poundings suppress the peak displacement at the top

of the lower structure and magnify the peak displacementat the top of the higher structure. However, under theexcitation of the earthquake with a peak acceleration of400 cm/s2, the J1, a15 peak displacement will decreaseslightly with the increase of the separation gap width; forJ2, a310, it will decrease slightly with the increase of theseparation gap width.

Under the excitation of the El Centro wave with a peakacceleration of 35 cm/s2, when the separation gaps are10mm, 20mm, and 30mm, respectively, the peak dis-placement errors of J1, a15 and J2, a310 calculated by MS-TMM and ANSYS are 6.51%, 4.87%, and 4.50%, and5.16%, 7.42%, and 5.70%, respectively. Under the exci-tation of the El Centro wave with an acceleration peak of70 cm/s2, when the separation gaps are 10mm, 40mm,and 50mm, respectively, the peak displacement errors ofJ1, a15 and J2, a310 calculated by MS-TMM and ANSYS are9.86%, 5.01%, and 5.71% and 8.22%, 5.57%, and 6.20%,respectively. Under the excitation of the El Centro wavewith an acceleration peak of 220 cm/s2, when the sepa-ration gaps are 100mm, 130mm, and 150mm, re-spectively, the peak displacement errors of J1, a15 and J2,a310 calculated by MS-TMM and ANSYS are 5.17%, 7.83%,and 4.47% and 5.28%, and 7.20%, 5.75%, respectively.

2.4E + 05

2.1E + 05

1.8E + 05

1.5E + 05

1.2E + 05

9.0E + 04

6.0E + 04

3.0E + 04

0.0E + 00

Poun

ding

forc

e (N

)

0 2 4Time (s)

6 81 3 5 7 9 10



(a)

2.07E + 05

1.84E + 05

1.61E + 05

1.38E + 05

1.15E + 05

9.20E + 04

6.90E + 04

4.60E + 04

2.30E + 04

0.00E + 00

Poun

ding

forc

e (N

)

0 2 4Time (s)

6 81 3 5 7 9 10



(b)

Poun

ding

forc

e (N

)

8.80E + 066.60E + 064.40E + 062.20E + 060.00E + 00

–2.20E + 06–4.40E + 06–6.60E + 06–8.80E + 06

0 2 4Time (s)

6 81 3 5 7 9 10



(c)

Figure 12: Time-history diagram of J2, a35-J1, a15 pounding force under seismic wave (peak acceleration is 70 cm/s2). (a) 10mm, (b) 40mm,

and (c) 60mm.


Under the excitation of the El Centro wave with an ac-celeration peak of 400 cm/s2, when the separation gaps are100mm, 200mm, and 210mm, respectively, the peakdisplacement errors of J1, a15 and J2, a310 calculated byMS-TMM and ANSYS are 5.83%, 6.76%, and 5.16% and6.24%, 4.15%, and 5.47%, respectively. However, usingANSYS to calculate the pounding response of two adja-cent frame structures under the excitation of earthquaketakes about 8 hours, while the calculation by MS-TMMonly takes about 20minutes; its time consumption is only1/24 of ANSYS, and the calculation speed advantage isvery obvious.

9. Analysis of Pounding ProcessBased on MS-TMM

It is found that the influence of the three seismic waves onthe pounding response of adjacent buildings is basically thesame in the above researches; so, we only study the poundingresponse of adjacent buildings under the excitation of the ElCentro wave in next research. ,e MS-TMM is used tocalculate the pounding response of J1 and J2 under theexcitation of the El Centro wave, and the deformation of thestructure is drawn at the moment when the pounding forceis greater than 400KN, as shown in Figures 23 to 26. It can beseen from the figure that the pounding mainly occurs at the

apex of the shorter structure. Meanwhile, multiple poundingat the same timemay occur when the separation gap is small,which will have a more adverse effect on the structure.Seismic peak acceleration and separation gap size will affectthemoment of pounding.,e pounding force is small exceptfor a few moments, and the rest is small when the separationgap is small.

10. Analysis of Shear Force Based on MS-TMM

As illustrated in Figures 27 and 28, under the same peakseismic excitation, although the separation gap is different, theshear force of each column of the adjacent two structures isconsistent before the first pounding. ,e shear force historiesof J1-A11 and J2-A31 at the peak seismic acceleration of220 cm/s2 and 400 cm/s2 are quite different from these whosepeak seismic acceleration are 35 cm/s2 and 70 cm/s2, since thestructural stiffness is reduced and the period is increasedunder the seismic excitation of higher peak acceleration; as aresult, the smaller frequency in the seismic wave is amplified.In addition, the influence of pounding on the base shear forceof the shorter structure is greater than that of the base shearforce of the higher structure, this is because the poundingforce of the two structures is the same, but the base shear forceof the shorter structure is smaller than the base shear of thehigher structure.



Poun

ding

forc

e (N

)4.68E + 054.16E + 053.64E + 053.12E + 052.60E + 052.08E + 051.56E + 051.04E + 055.20E + 040.00E + 00

0 2 4Time (s)

6 81 3 5 7 9 10

(a)



Poun

ding

forc

e (N

)

9.60E + 048.40E + 047.20E + 046.00E + 044.80E + 043.60E + 042.40E + 041.20E + 040.00E + 00

0 2 4Time (s)

6 81 3 5 7 9 10

(b)



Poun

ding

forc

e (N

)

8.80E + 066.60E + 064.40E + 062.20E + 060.00E + 00

–2.20E + 06–4.40E + 06–6.60E + 06–8.80E + 06

0 2 4Time (s)

6 81 3 5 7 9 10

(c)

Figure 13: Time-history diagram of J2, a35-J1, a15 pounding force under seismic wave (peak acceleration is 220 cm/s2). (a) 100mm,

(b) 130mm, and (c) 150mm.


As illustrated in Figure 29, when the peak value of theseismic acceleration is 35 cm/s2, the maximum shear forcesat the bottom of J1-A11 and J2-A31 are 95KN and 138KN,respectively; when the peak value of the seismic accelerationis 70 cm/s2, the maximum shear forces at the bottom of J1-A11 and J2-A31 are 181KN and 298KN, respectively; whenthe peak value of the seismic acceleration is 220 cm/s2, the

maximum shear forces at the bottom of J1-A11 and J2-A31are 347KN and 587KN, respectively; and when the peakvalue of the seismic acceleration is 400 cm/s2, the maximumshear forces at the bottom of J1-A11 and J2-A31 are 485KNand 845KN, respectively. ,ese indicate that the maximumshear force of the two structural side columns increasessignificantly by increasing the peak seismic acceleration.



Poun

ding

forc

e (N

)6.24E + 055.46E + 054.68E + 053.90E + 053.12E + 052.34E + 051.56E + 057.80E + 040.00E + 00

0 2 4Time (s)

6 81 3 5 7 9 10

(a)



Poun

ding

forc

e (N

)

9.90E + 038.80E + 037.70E + 036.60E + 035.50E + 034.40E + 033.30E + 032.20E + 031.10E + 030.00E + 00

0 2 4Time (s)

6 81 3 5 7 9 10

(b)



Poun

ding

forc

e (N

)

8.80E + 066.60E + 064.40E + 062.20E + 060.00E + 00

–2.20E + 06–4.40E + 06–6.60E + 06–8.80E + 06

0 2 4Time (s)

6 81 3 5 7 9 10

(c)

Figure 14: Time-history diagram of J2, a35-J1, a15 pounding force under seismic wave (peak acceleration is 400 cm/s2). (a) 100mm,

(b) 200mm, and (c) 220mm.

Table 3: Peak pounding force of J2, a35-J1, a15 under seismic wave.

Seismic peakacceleration

Separationgap El Centro-MS-TMM (KN)

El Centro-ANSYS(KN) Error (%) Nanjing-MS-TMM (KN) TAFT-MS-TMM (KN)

35 cm/s210mm 161 143 5.59 147 16920mm 134 129 3.88 128 14030mm 0 0 0 0 0

70 cm/s210mm 239 225 6.22 221 23540mm 181 173 4.62 163 18960mm 0 0 0 0 0

220 cm/s2100mm 412 391 5.37 389 424130mm 95 91 4.39 56 81150mm 0 0 0 0 0

400 cm/s2100mm 697 656 6.25 637 683200mm 9.4 8.9 5.32 0 0210mm 0 0 0 0 0


1.68E – 021.26E – 028.40E – 034.20E – 030.00E + 00

–4.20E – 03–8.40E – 03–1.26E – 02–1.68E – 02

0 1 2 3 4 5 6Time (s)

7 8 9 10

Disp

lace

men

t (m

)



(a)

1.88E – 021.41E – 029.40E – 034.70E – 030.00E + 00

–4.70E – 03–9.40E – 03–1.41E – 02–1.88E – 02

0 1 2 3 4 5 6Time (s)

7 8 9 10

Disp

lace

men

t (m

)



(b)2.65E – 022.12E – 021.59E – 021.06E – 025.30E – 030.00E + 00

–5.30E – 03–1.06E – 02–1.59E – 02

0 1 2 3 4 5 6Time (s)

7 8 9 10

Disp

lace

men

t (m

)



(c)

Figure 15: Time-history diagram of J1, a15 displacement under seismic wave (peak acceleration is 35 cm/s2). (a) 10mm, (b) 20mm, and (c)

30mm.

Disp

lace

men

t (m

)

3.95E – 02

0 1 2 3 4 5 6 7 8 9 10Time (s)

3.16E – 022.37E – 021.58E – 027.90E – 030.00E + 00

–7.90E – 03–1.58E – 02–2.37E – 02



(a)

Disp

lace

men

t (m

)

3.56E – 022.67E – 021.78E – 028.90E – 03

0 1 2 3 4 5 6 7 8 9 10Time (s)

0.00E + 00–8.90E – 03–1.78E – 02–2.67E – 02–3.56E – 02



(b)

Figure 16: Continued.


Figure 29 also shows that pounding suppresses themaximum base shear force of the shorter structural sidecolumn and enlarges the maximum base shear force of thehigher structural side column, since the pounding mainlyoccurs in the negative movement of the two structures

under the excitation of the El Centro wave; meanwhile, thebase shear force of the two structural side columns ispositive, and the pounding produces a positive force onthe shorter structure and a negative force on the higherstructure.

Disp

lace

men

t (m

)

5.50E – 024.40E – 023.30E – 022.20E – 021.10E – 020.00E + 00

–1.10E – 02–2.20E – 02–3.30E – 02

0 1 2 3 4 5 6 7 8 9 10Time (s)



(c)

Figure 16: Time-history diagram of J1, a15 displacement under seismic wave (peak acceleration is 70 cm/s2). (a) 10mm, (b) 40mm, and

(c) 60mm.

Disp

lace

men

t (m

)

1.45E – 011.16E – 018.70E – 025.80E – 022.90E – 020.00E + 00

–2.90E – 02–5.80E – 02–8.70E – 02

0 1 2 3 4 5 6 7 8 9 10Time (s)



(a)

Disp

lace

men

t (m

)

1.55E – 011.24E – 019.30E – 026.20E – 023.10E – 020.00E + 00

–3.10E – 02–6.20E – 02–9.30E – 02

0 1 2 3 4 5 6 7 8 9 10Time (s)



(b)

Disp

lace

men

t (m

)

1.70E – 011.36E – 011.02E – 016.80E – 023.40E – 020.00E + 00

–3.40E – 02–6.80E – 02–1.02E – 01

0 1 2 3 4 5 6 7 8 9 10Time (s)



(c)


(c) 150mm.


3.0E – 012.5E – 012.0E – 011.5E – 011.0E – 015.0E – 020.0E + 00

–5.0E – 02–1.0E – 01–1.5E – 01

Disp

lace

men

t (m

)

0 1 2 3 4 5 6 7 8 9 10Time (s)



(a)

2.5E – 012.0E – 011.5E – 011.0E – 015.0E – 020.0E + 00

–5.0E – 02–1.0E – 01–1.5E – 01

Disp

lace

men

t (m

)

0 1 2 3 4 5 6 7 8 9 10Time (s)



(b)

2.60E – 012.08E – 011.56E – 011.04E – 015.20E – 020.00E + 00

–5.20E – 02–1.04E – 01–1.56E – 01

Disp

lace

men

t (m

)

0 1 2 3 4 5 6 7 8 9 10Time (s)



(c)


(c) 210mm.

Disp

lace

men

t (m

)

5.0E – 024.0E – 023.0E – 022.0E – 021.0E – 02

–1.0E – 02–2.0E – 02–3.0E – 02–4.0E – 02

0.0E + 00

0 1 2 3 4 5 6 7 8 9 10Time (s)



(a)

Disp

lace

men

t (m

)

5.0E – 024.0E – 023.0E – 022.0E – 021.0E – 02

–1.0E – 02–2.0E – 02–3.0E – 02–4.0E – 02–5.0E – 02

0.0E + 00

0 1 2 3 4 5 6 7 8 9 10Time (s)



(b)



11. Conclusion

In this paper, a mechanical model based on the transfermatrix method of multibody systems (MS-TMM) is

established, the transfer matrix of the related elements andoverall transfer equation are deduced, combining theHertz-damp model, and the corresponding MATLABprogram is compiled to determine the pounding process

Disp

lace

men

t (m

)

4.80E – 02

3.60E – 02

2.40E – 02

1.20E – 02

0.00E + 00

–1.20E – 02

–2.40E – 02

–3.60E – 02

–4.80E – 020 1 2 3 4 5 6 7 8 9 10

Time (s)



(c)


30mm.

Disp

lace

men

t (m

)

1.0E – 01

8.0E – 02

6.0E – 02

4.0E – 02

2.0E – 02

0.0E + 00

–2.0E – 02

–4.0E – 02

–6.0E – 02

–8.0E – 020 1 2 3 4 5 6 7 8 9 10

Time (s)



(a)

Disp

lace

men

t (m

)1.20E – 019.60E – 027.20E – 024.80E – 022.40E – 020.00E + 00

–2.40E – 02–4.80E – 02–7.20E – 02–9.60E – 02

0 1 2 3 4 5 6 7 8 9 10Time (s)



(b)

Disp

lace

men

t (m

)

9.60E – 027.20E – 024.80E – 022.40E – 020.00E + 00

–2.40E – 02–4.80E – 02–7.20E – 02–9.60E – 02

0 1 2 3 4 5 6 7 8 9 10Time (s)



(c)


60mm.


2.40E – 011.92E – 011.44E – 019.60E – 024.80E – 020.00E + 00

–4.80E – 02–9.60E – 02–1.44E – 01–1.92E – 01

Disp

lace

men

t (m

)

0 1 2 3 4 5 6 7 8 9 10Time (s)



(a)

2.35E – 011.88E – 011.41E – 019.40E – 024.70E – 020.00E + 00

–4.70E – 02–9.40E – 02–1.41E – 01

Disp

lace

men

t (m

)

0 1 2 3 4 5 6 7 8 9 10Time (s)



(b)2.5E – 012.0E – 011.5E – 011.0E – 015.0E – 020.0E + 00

–5.0E – 02–1.0E – 01–1.5E – 01

Disp

lace

men

t (m

)

0 1 2 3 4 5 6 7 8 9 10Time (s)



(c)


(c) 150mm.


Nanjing-MSTMMTa�-MSTMM

–3.2E – 01–2.4E – 01–1.6E – 01–8.0E – 02

0.0E + 008.0E – 021.6E – 012.4E – 013.2E – 014.0E – 01

Disp

lace

men

t (m

)

1 2 3 4 5 6 7 8 9 100Time (s)

(a)


Nanjing-MSTMMTa�-MSTMM

–2.28E – 01–1.52E – 01–7.60E – 02

0.00E + 007.60E – 021.52E – 012.28E – 013.04E – 013.80E – 01

Disp

lace

men

t (m

)

1 2 3 4 5 6 7 8 9 100Time (s)

(b)



of two adjacent frame structures during severe groundmotions. e following conclusions are drawn from theresults:

(1) e pounding response trends of two adjacent framestructures that under the El Centro wave obtained bythe MS-TMM are similar to the responses obtained

Table 5: Peak displacement of J2, a310 under the excitation of seismic wave.


Separationgap

El Centro-MS-TMM(mm)

El Centro-ANSYS(mm)

Error(%)

Nanjing-MS-TMM(mm)

TAFT-MS-TMM(mm)

35 cm/s210mm 37.91 36.05 5.16 36.92 38.3420mm 40.37 37.58 7.42 40.67 41.3030mm 40.45 43.00 5.70 41.81 42.02

70 cm/s210mm 70.94 65.55 8.22 69.84 73.4940mm 84.72 80.25 5.57 84.26 87.3560mm 84.89 80.07 6.20 85.38 88.17

220 cm/s2100mm 172.22 163.58 5.28 181.35 184.26130mm 166.77 155.57 7.20 170.51 171.28150mm 160.17 151.46 5.75 164.22 168.93

400 cm/s2100mm 286.38 269.56 6.24 293.32 299.08200mm 283.39 272.10 4.15 279.57 285.19210mm 281.14 266.57 5.47 279.57 285.19

Table 4: Peak displacement of J1, a15 under the excitation of seismic wave.


Separationgap

El Centro-MS-TMM(mm)

El Centro-ANSYS(mm)

Error(%)

Nanjing-MS-TMM(mm)

TAFT-MS-TMM(mm)

35 cm/s210mm 13.25 12.44 6.51 10.76 15.4120mm 15.71 14.98 4.87 12.51 17.5230mm 17.17 16.43 4.5 14.17 18.39

70 cm/s210mm 26.62 24.23 9.86 24.47 29.3640mm 32.29 30.75 5.01 29.18 33.2960mm 36.48 34.51 5.71 33.79 37.15

220 cm/s2100mm 94.35 89.71 5.17 96.61 100.77130mm 108.74 100.84 7.83 113.54 109.13150mm 126.73 121.31 4.47 124.49 125.57

400 cm/s2100mm 179.27 169.39 5.83 189.03 170.16200mm 162.60 152.3 6.76 167.81 163.25210mm 163.71 155.67 5.16 167.81 163.25

E1 Centro-ANSYSE1 Centro-MSTMM


–2.96E – 01–2.22E – 01–1.48E – 01–7.40E – 02

0.00E + 007.40E – 021.48E – 012.22E – 012.96E – 013.70E – 01

Disp

lace

men

t (m

)1 2 3 4 5 6 7 8 9 100

Time (s)

(c)


(c) 210mm.


J1-a1 (3.76s)J2-a3 (3.76s)J1-a1 (4.68s)J2-a3 (4.68s)

J1-a1 (6.44s)J2-a3 (6.44s)J1-a1 (7.04s)J2-a3 (7.04s)

0

5

10

15

20

25

30

35

40

Num

ber o

f par

ticle

s

–0.01 0.00 0.01 0.02 0.03–0.02Displacement (m)

(a)

J1-a1 (7.12s)J2-a3 (7.12s)J1-a1 (7.96s)J2-a3 (7.96s)

J1-a1 (8.84s)J2-a3 (8.84s)J1-a1 (9.7s)J2-a3 (9.7s)

–0.03 –0.02 –0.01 0.00 0.01 0.02–0.04Displacement (m)

0

5

10

15

20

25

30

35

40

Num

ber o

f par

ticle

s

(b)

Figure 24: J1-a1, J2-a3 deformation diagram of the structure pounding moment under the excitation of El Centro wave (70m/s2). (a) 10mmand (b) 40mm.

J1-a1 (6.12s)J2-a3 (6.12s)J1-a1 (7.34s)J2-a3 (7.34s)J1-a1 (8.14s)

J2-a3 (8.14s)J1-a1 (8.8s)J2-a3 (8.8s)J1-a1 (9.84 s)J2-a3 (9.84 s)

0

5

10

15

20

25

30

35

40

Num

ber o

f par

ticle

s

0.00 0.01 0.02–0.01Displacement (m)

(a)

0

5

10

15

20

25

30

35

40

Num

ber o

f par

ticle

s

J1-a1 (4.74s)J2-a3 (4.74s)J1-a1 (7.12s)

J2-a3 (7.12s)J1-a1 (8.84s)J2-a3 (8.84s)

–0.01 0.00 0.01–0.02Displacement (m)

(b)

Figure 23: J1-a1, J2-a3 deformation diagram of the structure pounding moment under the excitation of El Centro wave (35m/s2). (a) 10mmand (b) 20mm.


by ANSYS, the biggest error of the peak poundingforce is 6.22%, the biggest error of the maximumdisplacement for the top of the structure with fewerstories and the structure with more stories is 9.86%

and 8.22%, respectively, while the calculation timebased on the MS-TMM method was approximately1/24 of that based on the ANSYS method. erefore,analysis of the pounding between adjacent frame

J1-a1 (2.42s)J2-a3 (2.42s)

J1-a1 (3.28s)J2-a3 (3.28s)

0

5

10

15

20

25

30

35

40

Num

ber o

f par

ticle

s

–0.06 –0.03 0.00 0.03–0.09Displacement (m)

(a)

J1-a1 (3.18s)J2-a3 (3.18s)

0

5

10

15

20

25

30

35

40

Num

ber o

f par

ticle

s

–0.09 –0.06 –0.03 0.00–0.12Displacement (m)

(b)

Figure 25: J1-a1, J2-a3 deformation diagram of the structure pounding moment under the excitation of El Centro wave (220m/s2).(a) 100mm and (b) 130mm.

J1-a1 (2.48s)J2-a3 (2.48s)

J1-a1 (2.84s)J2-a3 (2.84s)

0

5

10

15

20

25

30

35

40

Num

ber o

f par

ticle

s

–0.05 0.00 0.05 0.10–0.10Displacement (m)

Figure 26: J1-a1, J2-a3 deformation diagram of the structure pounding moment under the excitation of El Centro wave (400m/s2, 100mm).


30mm20mm10mm

–1.36E + 05–1.02E + 05–6.80E + 04–3.40E + 04

0.00E + 003.40E + 046.80E + 041.02E + 051.36E + 05

Shea

r for

ce (N

)

1 2 3 4 5 6 7 8 9 100

Time (s)

(a)

60mm40mm10mm

Shea

r for

ce (N

)

–3.2E + 05

–2.4E + 05

–1.6E + 05

–8.0E + 04

0.0E + 00

8.0E + 04

1.6E + 05

2.4E + 05

3.2E + 05

1 2 3 4 5 6 7 8 9 100

Time (s)

(b)


30mm20mm10mm

–1.24E + 05

–9.30E + 04

–6.20E + 04

–3.10E + 04

0.00E + 00

3.10E + 04

6.20E + 04

9.30E + 04

1.24E + 05

Shea

r for

ce (N

)

1 2 3 4 5 6 7 8 9 100

Time (s)

(a)

60mm40mm10mm

–2.0E + 05–1.5E + 05–1.0E + 05–5.0E + 04

0.0E + 005.0E + 041.0E + 051.5E + 052.0E + 05

Shea

r for

ce (N

)

1 2 3 4 5 6 7 8 9 100

Time (s)

(b)

150mm120mm100mm

–4E + 05

–3E + 05

–2E + 05

–1E + 05

0E + 00

1E + 05

2E + 05

3E + 05

4E + 05

Shea

r for

ce (N

)

1 2 3 4 5 6 7 8 9 100

Time (s)

(c)

210mm200mm150mm

–6.00E + 05

–4.50E + 05

–3.00E + 05

–1.50E + 05

0.00E + 00

1.50E + 05

3.00E + 05

4.50E + 05

6.00E + 05

Shea

r for

ce (N

)

1 2 3 4 5 6 7 8 9 100

Time (s)

(d)

Figure 27: Base shear force history for J1-A11. (a) 35 cm/s2, (b) 70 cm/s2, (c) 220 cm/s2, and (d) 400 cm/s2.


structures under earthquake excitation using theMS-TMM not only guarantees calculation accuracybut also has high computational eciency.

(2) e pounding force and the number of poundingsincrease with the decrease of the separation gap sizewhen the two adjacent structures under same peakseismic excitation; the pounding force and thenumber of poundings increase with the increase ofpeak seismic excitation when the two adjacentstructures have same separation gap size.

(3) Pounding of two adjacent structures will occur whenthe acceleration peak is 35 cm/s2 and the separationgap is not more than 20mm, the acceleration peak is70 cm/s2 and the separation gap is not more than40mm, the acceleration peak is 220 cm/s2 and theseparation gap is not more than 130mm, the ac-celeration peak is 400 cm/s2 and the antiseismic jointwidth is not more than 200mm. e displacement

response of the two structures shows obvious di-rectional dierences due to the pounding. e dif-ference becomes more signicant with the increaseof the acceleration peak and the decrease of theseparation gap.

(4) e pounding mainly occurs at the top of thestructure with fewer stories, while multiple poundingat the same timemay occur when the separation gap issmall. Seismic peak ground acceleration and separa-tion gap size will aect the moment of pounding. epounding force is small except for a fewmoments, andthe rest is small when the separation gap is small. einuence of pounding on the base shear force of theshorter structure is greater than the base shear force ofthe higher structure. Pounding suppresses the max-imum base shear force of the shorter structural sidecolumn and enlarges the maximum base shear forceof the higher structural side column.

150mm130mm100mm

–5.40E + 05

–3.60E + 05

–1.80E + 05

0.00E + 00

1.80E + 05

3.60E + 05

5.40E + 05

7.20E + 05Sh

ear f

orce

(N)

1 2 3 4 5 6 7 8 9 100

Time (s)

(c)

210mm200mm100mm

–8.40E + 05–6.30E + 05–4.20E + 05–2.10E + 05

0.00E + 002.10E + 054.20E + 056.30E + 058.40E + 05

Shea

r for

ce (N

)

2 4 6 8 100

Time (s)

(d)

Figure 28: Base shear force history for J2-A31. (a) 35 cm/s2, (b) 70 cm/s2, (c) 220 cm/s2, and (d) 400 cm/s2.

10 20 30 10 60

Separation gap (mm)

100 100 200 21015013040

400cm/s2220cm/s235

Documents

PoundingbetweenAdjacentFrameStructuresunderEarthquake ... · 2019. 7. 30. · advantages as follows: without the system global dynamics equations, high programming, low order of system