THE EFFECT OF FLEXIBLE HORIZONTAL · PDF fileMasahide MURAKAMI1, Peter J MOSS2, Athol J CARR3...

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1 Dept of Architecture, School of Science & Engineering, Kinki University,Osaka, Japan. Email:[email protected] Dept of Civil Engineering, University of Canterbury, Christchurch, New Zealand.3 Dept of Civil Engineering, University of Canterbury, Christchurch, New Zealand.4 Inayama Architect, Musahino, Tokyo, Japan.

THE EFFECT OF FLEXIBLE HORIZONTAL DIAPHRAGMS ON THE SEISMICTORSONAL RESISTANCE OF SYSTEMS WITH DUCTILE WALLS

Masahide MURAKAMI1, Peter J MOSS2, Athol J CARR3 And Mashiro INAYAMA4

SUMMARY

The roofs and floors of timber houses are often forced to behave as horizontal diaphragms in orderto transfer the shear force from the upper storey to the walls of the first storey, particularly whenthe second storey is set back from the first storey. There has been no way to predict that kinds ofdamage for seismic assessments of existing buildings up to now. A seismic design procedure fortorsional effects on ductile wall systems connected to a rigid horizontal diaphragm has beenproposed by T. Paulay. A new design or seismic assessment procedure in which Paulay’s torsionalmodel is extended to ductile wall systems connected to flexible diaphragms is presented in thispaper. The shear stresses acting on each diaphragm between the walls are estimated by using thedifference between the distinct earthquake induced lateral force and the ultimate shear capacity ofeach wall. The storey drift of each wall is computed by superimposing the local diaphragm shearangles on the overall twist angle of the floor computed assuming a rigid diaphragm. Nonlineardynamic time history analyses were carried out to verify this design or assessment procedure.

INTRODUCTION

The roofs and floors of timber houses are often forced to behave as horizontal diaphragms in order to transfer theshear force from the upper storey to the walls of the first storey, particularly when the second storey is set backfrom first storey. Some of the timber houses collapsed in the Hyogoken-Nanbu earthquake in Japan in 1995because of the lack of shear capacity of their roofs as shown in Photo.1. There is no way to predict that kind ofdamage for seismic assessments of existing buildings up to now. The objective of this paper is to propose an ideato predict the damage of timber houses whose second storey is set back from the first storey via weak roofs orfloors.

The seismic design procedure for the torsional effects on ductile wall systems connected to a rigid horizontaldiaphragm has been proposed by T. Paulay[1997]. A new design or seismic assessment procedure in whichPaulay’s torsional model is extended to ductile wall systems connected to a flexible diaphragm is presented inthis paper.

PROPOSED DESIGN AND /OR ASSESSMENT PROCEDURE

2.1. Torsional Behaviours of Ductile Wall Systems with Rigid Diaphragms

The seismic design procedure for the torsional effects on ductile wall systems connected to a rigid horizontaldiaphragm has been proposed by T. Paulay as shown in Fig.1, where the shear force - storey drift relationship ofa wall system is assumed to be bi-linear in this theory. The earthquake induced lateral force, which is balancedwith the total ultimate shear capacity of a wall system in the Y direction, acts at the centre of mass, CM, in thisrigid diaphragm theory. The centre of resistance, CR, means the position of the force line of the total ultimateshear capacity of a wall system in the Y direction. The twist due to the torque occurred by the differencebetween the centre of mass and the centre of resistance can be restrained only by the elastic rotational stiffness“kt” carried by a wall system in the X direction when all walls in the Y direction deform beyond the yield

displacement because the rotational stiffness of a wall system in the Y direction is zero after yield. The walldisplacements are distributed linearly as shown in Fig.2, where “θt” is the twist angle calculated by (ΣQui)e/kt,and “e” is the distance between CM and CR. The system displacement in a wall system with a rigid diaphragm isthe displacement at the centre of mass. The system displacement can be determined by satisfying thedisplacement ductility capacity of a critical wall in this wall system.

2.2. Torsional Behaviours of Ductile Wall systems with Flexible Diaphragms

Inertia forces caused by horizontal acceleration are developed throughout a structure as earthquake inducedlateral forces. The concept of the distinct earthquake induced lateral force that belongs to each wall is introducedin this calculation procedure. The distinct earthquake induced lateral force can be calculated from the masssupported by each wall as shown in Fig.3 by assuming that uniform horizontal acceleration “a” is exerted to eachmass. The total ultimate shear capacity of the wall system in the Y direction is balanced with the totalearthquake induced lateral force of the structure. The shear force acting on each local diaphragm between wallscan be calculated from the difference between the ultimate shear capacity and the earthquake induced lateralforce of each wall, and the distribution of shear forces acting on a diaphragm can be drawn as shown in Fig.4.The equilibrium of moments can be expressed as Eq.(1) by assuming that all walls in the Y direction yieldcompletely and that all walls in the X direction stay in the elastic range.

(1)

The twist angle “θt” can be calculated without the calculation of the distance between CM and CR, “e”, by usingEq.(1). Eq.(1) can be derived by making a linear elastic analysis of a statically determinate structure as shown inFig.5. Flexural deformation is neglected in this procedure because shear deformation is larger than flexuraldeformation when a flexible diaphragm is deformed. The local diaphragm shear stress “τij” can be calculated asshown in Fig.5. The local diaphragm shear angle is given by γ=τ/G (γ: engineering shear strain, G: shearstiffness). The storey drift of each wall is computed by superimposing the local diaphragm shear angles on theoverall twist angle of the diaphragm computed assuming a rigid diaphragm as shown in Fig.6.

The system displacement can be represented as the average displacement of the wall system in the Y directionbecause the displacement at the centre of mass may depend on the shear deformation of a local diaphragmadjacent to the centre of mass.In actual buildings, the number of walls is larger than the example shown in Fig.5 and the shape of a diaphragmis more complicated. The procedure of modelling a statically determinate structure is as follows in these cases;

(1) Walls in the X direction are modelled into elastic springs supported with rollers.(2)A diaphragm is modelled into truss elastic elements or quadrilateral elastic finite elements supported at onenode, i.e. one wall in the Y direction, with a roller in the Y direction.(3) Induced forces “Pi-Qui” in the Y direction are applied to the other walls. Then the shear stress acting on eachdiaphragm, the shear force acting of each wall in the X direction, and relative displacement of each wall in the Ydirection relative to the supported node can be determined by a linear elastic analysis.

3.VERIFICATION WITH TIME HISTORY ANALYSES

3.1 Parameters of Numerical Studies

Time history analyses with two different kinds of structures were carried out to verify this simple design and/orassessment procedure. The parameters for the numerical studies are shown in Table 1. The model code“AXXX”, where the symbol “X” indicates any symbol in Table (a), are the structures in which the shear stiffnessand ultimate shear capacity of Frame 3 are the largest and those of Frame 2 are the smallest. “ARXX” are thestructures with a rigid diaphragm. “ASXX” are the structures with a flexible diaphragm. The displacement ofFrame 2 of “ASXX” is larger than that of “ARXX”. There is no wall at the position of Frame 1 in “BXXX”,therefore the displacements at Frame 1 are the largest due to twist. The Stewart degrading stiffness hysteresisrule[1987] is used for all springs except “ky3” in “XXXB” as illustrated in the inset figure in Table 1. The yielddisplacement “dy” and the ultimate displacement “du” of all springs have the same values in the structures“XXXU”. The values of “dy” and “du” of “ky3” are larger than those of “ky1” and “ky2” in the structures“XXXS”. The Stewart hysteresis rule is used for “ky1” and “ky2” and the bi-linear hysteresis rule is used for“ky3” in the structures “XXXB”. The parameters for the base shear coefficients and excitations are given in Table1. 5% initial stiffness Rayleigh damping is used throughout these numerical studies.

0)(∑ =⋅−+⋅ iuiitt lQPk θ

The response of a single degree of freedom (SDOF) system concerned to each model was computed as follows:The tri-linear envelope curves of the Stewart hysteresis rule in the SDOF system “XXXU” and “XXXS” aredetermined by making their dissipated energies equal to those of the total envelope curves of “ky1”,”ky2”, and“ky3”. The co-ordinates of the envelope curve of “XXXU” are (dy=2.5cm, Fy=2000kgf) and (du=5cm,Fu=3000kgf), and those of “XXXS” are (dy=2.55cm,Fy=2000kgf) and (du=5.1cm,Fu=3000kgf). The springs“ky1”,”ky2”, and “ky3” are connected in parallel at the centre of mass in the SDOF system “XXXB”. The totalyield shear capacity and the total ultimate shear capacity in the Y direction are the same values throughout thesenumerical studies. The total shear capacities of “XX3X” in the X direction are 75% of those in the Y direction,and those of “XX6X” in the X direction are 150% of those in the Y direction.The characteristics of twist calculated by linear static analyses mentioned above are shown in Fig.7 whererelative displacements against Frame 2 are plotted. Frame 2 is supported with a roller in the Y direction andelastic stiffnesses “ko” are used for “kx”. The order of local diaphragm shear angles “γ” is similar to that of thetwist angle of a rigid diaphragm “θt” when a diaphragm is flexible.

3.2 Prediction of System Displacements from Responses of SDOF Systems

The typical time histories of the averaged displacement of Frames 1, 2, and 3 compared with the displacementresponse of the SDOF system are shown in Figs.8 where the model codes are “XXXU” and the ultimate shearcoefficients are 0.3. The response waveforms are generally similar to those of SDOF systems. The maximumdisplacement responses of the SDOF system versus the maximum values of the averaged displacement of Frame1, 2, and 3 are plotted in Fig.9. Most of the models except a few whose Cbu is 0.4 are deformed beyond “du” atthe maximum displacement responses. The maximum values of the averaged displacement of Frame 1, 2, and 3are generally smaller than the maximum displacement responses of the SDOF system as shown in Fig.9.

3.3 Prediction of the Maximum Displacement of Each Frame from the System Displacement

The system displacement “δave” can be defined by Eq.(2), where “δn” is the maximum displacement response of“Frame n”, and ∆1…∆n (∆n=0)…∆i are the relative displacements of frames against “Frame n” which issupported by a roller in linear static analysis as shown in Fig.5

(2a)

(2b)

The value of “δn“ was determined by giving the system displacement “δave” which is equal to the maximum ofaverage displacement of Frame 1, 2, and 3 computed with a time history analysis to verify Eq.(2). Then themaximum displacement response of each frame “∆j+δn” was found. The maximum displacements of each frameby time history analyses versus the predicted displacement of each frame by the procedure mentioned above areplotted in Fig.10. The main reason for differences between them is due to overestimation of the predicted twistangle of “BSXX” in linear static analyses.

3.4 Prediction of the Maximum Displacement of Each Frame from the Maximum Displacement of SDOFSystem

The maximum displacement response of each frame “∆j+δn” can be calculated from Eq.(2) by giving “δave”which is equal to the maximum displacement response of the SDOF system. The maximum displacement ofeach frame by time history analyses versus the displacement of each frame predicted from the maximumdisplacement response of the SDOF system by using Eq.(2) are plotted in Fig.11. The errors of Fig.11 are largerthan those of Fig.10 because the errors of Fig.11 contain those of Figs.9 and 10. The displacement of each framepredicted from the maximum displacement response of the SDOF system is 1.5 times greater than the maximumdisplacement response of each frame by time history analysis on the average, and approximately 2.5 timesgreater on the maximum. It is generally recognised that wide scattering of plots occurs when diaphragms areflexible because the number of degree of freedom of the structures with flexible diaphragms is greater than thatof structures with rigid diaphragms.

3.5 Prediction of Shear Forces Acting on Perpendicular Walls and Diaphragms

The maximum shear forces acting on perpendicular walls and diaphragms computed by time history analyses areplotted in Figs.12 and 13 respectively, in which the shear forces predicted based on linear static analyses are

∑=

+∆=i

jnjave i 1

)(1 δδ

∑ ∆−==

i

jjaven i 1

1δδ

indicated as the lines “Calc.”. All plots are shifted from left to right when the ultimate shear coefficient “Cub“goes from 0.2 to 0.4 in each group. The mean values and standard deviations of the ratios of shear forces bytime history analyses against those by linear static analyses throughout the group “XRXX” or “XSXX” are alsoshown. It is generally recognised that shear forces acting on perpendicular walls in structures with rigiddiaphragms are larger than those with flexible diaphragms. Some plots of “BX3X” reach the ultimate shearcapacity “Fu” because the predicted value is larger than “Fu”. The predicted shear forces acting on perpendicularwalls are close to the mean value of shear forces computed by time history analyses. The predicted shear forcesacting on diaphragms are close to the mean values of computed shear forces acting on diaphragms except thediaphragm between Frame1 and 2 of “AXXX”. It is generally recognised that the plots of structures with flexiblediaphragms are more scattered than those of structures with rigid diaphragms because the number of degree offreedom of structures with flexible diaphragms becomes greater than that of structures with rigid diaphragms.

4.CONCLUSION

A seismic assessment procedure to predict torsional behaviour of ductile wall systems with flexible diaphragmshas been proposed in this paper.

The displacement of each frame predicted from the maximum displacement response of the SDOF system is 1.5times greater than the maximum displacement response of each frame by time history analyses on the average,and approximately 2.5 times greater on the maximum.

It is generally recognised that the predicted shear forces acting on perpendicular walls and diaphragms ofstructures with flexible diaphragms are more scattered than those of structures with rigid diaphragms because thenumber of degree of freedom of structures with flexible diaphragms becomes greater than that of structures withrigid diaphragms.

The accuracy of this simple procedure is not good enough but these predicted values may be utilized as atentative criterion by introducing a safety factor after further numerical studies.

REFERENCE

T.Paulay (1997), “Seismic Torsional Effects on Ductile Structural Wall System”, Journal of EarthquakeEngineering, Vol.1, No.4, Imperial College Press, pp.721-745W.Stewart (1987), The Seismic Design of Plywood Sheathed Shear Wall, Phd. Thesis, University of Canterbury,New Zealand

ACKNOWLEDGEMENT

The authors would like to thank T.Paulay, Emeritus Professor of University of Canterbury, for his advice.

Original shape of Deformation of collapsed house flexible diaphragm

A collapsed house due to weak roof at Kobe in 1995 This type of NZ house may have a similar problem.Photo. 1: Problems of flexible diaphragms

eC M

C RM y= ΣQ u i

e

k x1

k x2

B

l 1 l 2

Q u 1 Q u2 Q u3

k t=k x1 k x2 B

2

k x1 + k x2

M t=ΣQ u i e

θt=M t/k t

Y

X

δ1δ2

δ3δs

θt

C M

M 1 M 2 M 3

Q u 1 Q u 2 Q u 3

P 1=a M 1 P 2=a M 2 P 3=a M 3

ΣP i= a ΣM i= ΣQ u i

Figure 1: Rigid diaphragm theory Figure 2: Deformation pattern Figure 3: Distinct earthquake of rigid diaphragm theory induced lateral forces

P1 -

Qu1

P2-Q

u2

P3-Q

u3

V 23

V 12

δ1

δ2δ3

θ t

γ1 2= τ12/G 1 2

γ 23= τ 2 3 /G 23

H w

H w

H 12 =τ 1 2l 1H 2 3 = τ2 3 l 2

H 1 2 H 2 3

V 1 2

V 2 3V 1 2

V 2 3

B

l 1 l 2

P 1- Q u1 P 2- Q u2 P 3-Q u3

τ12 = V 1 2 /B τ2 3 = V23 /B

Figure 4: Shear force Figure 5: Analytical model Figure 6: Modified displacement distribution pattern

F

F u

d u

ko

ko

k u = k o /1 0 0 00

d

B i-l in ear m o d e l

m

2 m

m

m

m

2 m

k y2

k y1 k y3

k x

G G

kx

F ra m e N o . 1 2 3

T yp e A

3.5

m

3 .5 m 3 .5 m

m

m

m

m

m

m

k y2

k y3 k x

G G

k x

T y p e B

3.5

m

F ra m e N o . 1 2 3

3 .5 m 3 .5 m

Y

X

0 .2 5 F y

Fy

F u = 1 .5 F y

ko

ku=1.45ko

0 .0 9 d m a x

d m a xd

0 .5kok u = 0

dy

k p=k o(dy/dma x) 0 .3 8

kp

F

d u = 2 d y

W .S tew art m o d e l

Base shear coefficient at ultimate state=0.2, 0.3, 0.4

Excitations in Y direction (Vmax=50cm/s (50kine))El.Centro 1940 NS, Taft EW, Kobe NS,Hachinohe NS

(a) Definition of model codeSymbol A

BRS

36

USB

Ref. Table No. (b) (c) (d) (e)

(b) Strengths of springs in Y directionky1(kgf) ky2(kgf) ky3(kgf)Sym

-bolSpring

arrangement Fy Fu Fy Fu Fy FuA Type A 700 1050 200 300 1100 1650B Type B 0 0 1000 1500 1000 1500

(c)Elastic stiffnesses of diaphragm springsSymbol G(kgf/cm)

R 1000000 (rigid)S 100 (flexible)

(d) Characteristics of springs in X directionSymbol ko Fy dy Fu du

3 300 750 2.5 1125 56 600 1500 2.5 2250 5

Hysteresis =W.Stewart model unit:kgf, cm

(e) Displacement and hysteresis rulesky1 ky2 ky3Symb

ol Model dy du Model dy du Model dy duU S 2.5 5 S 2.5 5 S 2.5 5S S 2 4 S 2 4 S 3 6B S 2 4 S 2 4 B 3 4.5

S=W.Stewart model, B=Bi-linear model unit:cm

Table 1: Parameters of numerical studies

(a) XXXU (b) XXXS (c) XXXBFigure9: Prediction of system displacements

0

2

4

6

8

1 0

1 2

1 4

1 6

1 8

2 0

0 2 4 6 8 1 0 1 2 1 4 1 6 1 8 2 0

M e a n o f Y / X = 1 .2 0

S t de v . o f Y / X = 0 .2 7

d u

d u

Y = X

( Y )

( X )

B la c k m a r k s in d i c a te A R 3 U , A R 6U , B R 3 U , B R 6 U .

W h i te m a r k s i n d i c a te A S 3 U , A S 6 U , B S 3 U , B S 6 U .

E l C e n tro

K o b e

T a ft

H a c h in o h e

(5 0 k in e )

M a x im u m o f a v e r a g e d d is p la c e m e n t

o f to rs io n a l re s p o n s e ( c m )

Max

imum

dis

pla

cem

ent

of S

.D.O

.F.

syst

em (

cm)

0

2

4

6

8

1 0

1 2

1 4

1 6

1 8

2 0

0 2 4 6 8 1 0 1 2 1 4 1 6 1 8 2 0

M e a n o f Y / X = 1 .3 6

S t de v . o f Y / X = 0 .3 2

d u

d u

Y = X

( X )

( Y )

B la c k m a r k s i n d i c a te A R 3 S , A R 6S , B R 3S , B R 6 S .

W h i t e m a r k s i nd i c a t e A S 3 S , A S 6 S , B S 3 S , B S 6 S .

E l C e n tro

K o b e

T a ft

H a c h in o h e

(5 0 k in e )


o f to rs io n a l re s p o n s e ( c m )

Max

imum

dis

pla

cem

ent

of S

.D.O

.F.

syst

em (

cm)

0

2

4

6

8

1 0

1 2

1 4

1 6

1 8

2 0

2 2

2 4

2 6

0 2 4 6 8 1 0 1 2 1 4 1 6 1 8 2 0 2 2 2 4 2 6

M e a n o f Y / X = 1 .3 9

S t de v . o f Y / X = 0 .3 2

d u

d u

Y = X

( X )

(Y )

B l a c k m a r k s i n d ic a te A R 3 B , A R 6B , B R 3 B , B R 6 B .

W h it e m a rk s i nd ic a t e A S 3 B , A S 6 B , B S 3 B , B S 6 B .

E l C e n tro

K o b e

T a ft

H a c h in o h e

(5 0 k in e )


o f to r s io n a l re s p o n s e (c m )

Max

imum

dis

pla

cem

ent

of S

.D.O

.F.

syst

em (

cm)

1 2 3- 2 0

- 1 5

- 1 0

- 5

0

5

1 0

1 5

2 0

2 5

A R 3 X

A R 6 X

B R 3 X

B R 6 X

A S 3 X

A S 6 X

B S 3 X

B S 6 X

Re

lativ

e di

spla

cem

ent (

cm)

F ra m e N o .

0 2 4 6 8 1 0- 1 0

- 5

0

5

1 0

E l.C e n t ro ( 5 0 k in e ) U lt im a te B a s e S h e a r C o e ff ic ie n t =0 . 3

S .D .O .F .

A R 3 U ,A R 6 U ,A S 3 U ,A S 6 U

B R 3 U ,B R 6 U ,B S 3 U ,B S 6 U

Dis

pla

cem

ent(

cm)

T im e ( S e c . )

0 2 4 6 8 1 0

- 1 5

- 1 0

- 5

0

5

1 0

1 5 S . D .O .F .

A R 3 U ,A R 6 U ,A S 3 U ,A S 6 U

B R 3 U ,B R 6 U ,B S 3 U ,B S 6 U

K o b e ( 5 0 k in e ) U lt im a t e B a s e S h e a r C o e ff ic i e n t = 0 .3

Dis

pla

cem

en

t(cm

)

T im e ( S e c . )

Figure7: Static characteristics Figure8: Typical displacement time histories

Figure10: Prediction of maximum displacement of each Frame from system displacement

0 5 1 0 1 5 2 0 2 5 3 0 3 5 4 00

5

1 0

1 5

2 0

2 5

3 0

3 5

4 0

M e a n o f Y /X = 1.3 1

S td e v . o f Y /X = 0 .4 0

( X )

( Y )

A R X X

A S X X

B R X X

B S X X

F ra m e 1

Pre

dit

ed

ma

xim

um

dis

pla

ce

me

nt

(cm

)

M a x im u m d i s p la c e m e n t b y t im e h i s to ry a n a ly s is ( c m )

0 5 1 0 1 5 2 0 2 5 3 0 3 5 4 00

5

1 0

1 5

2 0

2 5

3 0

3 5

4 0

M e a n o f Y /X = 1.0 4

S td e v . o f Y /X = 0 .1 4

( X )

( Y )

F ra m e 2

A R X X

A S X X

B R X X

B S X X

Pre

dit

ed

ma

xim

um

dis

pla

ce

me

nt

(cm

)


0 5 1 0 1 5 2 0 2 5 3 0 3 5 4 00

5

1 0

1 5

2 0

2 5

3 0

3 5

4 0

M e a n o f Y /X = 0.7 4

S td e v . o f Y /X = 0 .8 7

( X )

( Y )

F ra m e 3

A R X X

A S X X

B R X X

B S X X

Pre

dit

ed

ma

xim

um

dis

pla

ce

me

nt

(cm

)


Figure11: Prediction of maximum displacement of each Frame from response of SDOFsystem

0 5 10 15 20 25 30 35 400

5

10

15

20

25

30

35

40

(X)

(Y)

Mean of Y /X =1.50

S tdev. of Y /X =0.45

Y=X

Y=2.5X

Frame 1

ARXX

ASXX

BRXX

BSXX

Pre

dite

d m

axi

mu

m d

isp

lac

em

en

t (c

m)

Maximum displacement by time histo ry ana lys is (cm)

0 5 10 15 20 25 30 35 400

5

10

15

20

25

30

35

40

(X)

(Y)

M ean o f Y/X=1.39

Stdev. o f Y/X=0.43

Y=2 .5X

Y=XFrame 2

ARXX

ASXX

BRXX

BSXX

Pre

dite

d m

axi

mu

m d

isp

lace

me

nt

(cm

)

Maximum displacement by time history ana lysis (cm)

0 5 10 15 20 25 30 35 400

5

10

15

20

25

30

35

40

(X)

(Y)

Y=X

M ean of Y/X=0.83

Stdev. o f Y/X=0.67

Frame 3

ARXX

ASXX

BRXX

BSXX

Pre

dite

d m

axi

mu

m d

isp

lace

me

nt

(cm

)

Maximum displacement by time history ana lysis (cm)

0 1 2 3 4 5 6 70

200

400

600

800

1000

1200 Me an= 0.8 5 S td ev.=0.36Me an= 1.0 7 S td ev.=0.34

Fu

Fy

Calc.

AS3BAS3S

AS3UAR3BAR3S

AR3U

Ind

uc

ed

sh

ea

r fo

rce

in X

dir

ec

tion

(kg

f)

0 1 2 3 4 5 6 7

0

500

1000

1500

2000

M ean =1.11 Std ev.=0.51M ean =1.87 Std ev.=0.64

Fu

Fy

Calc.

AS6BAS6S

AS6UAR6BAR6S

AR6U

Ind

uc

ed

sh

ea

r fo

rce

in X

dir

ec

tion

(kg

f)

0 1 2 3 4 5 6 70

200

400

600

800

1000

1200

1400

M ea n=0.61 Std ev.=0.11M ea n=0.60 Std ev.=0.09

Cb u=0 .4

Cb u=0 .3

Cb u=0 .2

Fu

Fy

Calc.

BS3BBS3S

BS3UBR3BBR3S

BR3U

Ind

uc

ed

sh

ea

r fo

rce

in X

dir

ec

tion

(kg

f)

0 1 2 3 4 5 6 70

500

1000

1500

2000

M ean= 0.89 Std ev.=0.21M ean= 1.18 Std ev.=0.17

Fu

Fy

Calc.

BS6BBS6S

BS6BR6BBR6S

BR6

Ind

uce

d s

he

ar

forc

e in

X d

ire

ctio

n (

kgf)

　　　　Figure12: Shear forces of perpendicular walls

E l. C en t ro K o b e T a ft H a c h in o h eK e y

0 1 2 3 4 5 6 7

0

2 0 0

4 0 0

6 0 0

8 0 0

1 0 0 0

1 2 0 0

1 4 0 0

M e a n = 2 .6 4 S td e v .= 0 .6 4M e a n = 1 .2 1 S td e v .= 0 .3 0

C a lc .

A S 3 BA S 3 S

A S 3 UA R 3 BA R 3 S

A R 3 U

Ind

uc

ed

sh

ea

r fo

rce

in

Sla

b 1

-2 (

kg

f)

0 1 2 3 4 5 6 7

0

2 0 0

4 0 0

6 0 0

8 0 0

1 0 0 0

1 2 0 0

1 4 0 0

M e a n = 2 . 5 6 S td e v . = 0 . 6 0M e a n = 1 . 4 2 S td e v . = 0 . 3 5

C a lc .

A S 6 BA S 6 S


A R 6 U

Ind

uc

ed

sh

ea

r fo

rce

in

Sla

b 1

-2 (

kg

f

0 1 2 3 4 5 6 7

0

2 0 0

4 0 0

6 0 0

8 0 0

1 0 0 0

1 2 0 0

1 4 0 0


C a lc .

A S 3 BA S 3 S


A R 3 U

Ind

uc

ed

sh

ea

r fo

rce

in

Sla

b 2

-3 (

kg

f

0 1 2 3 4 5 6 7

0

2 0 0

4 0 0

6 0 0

8 0 0

1 0 0 0

1 2 0 0

1 4 0 0


C a lc .

A S 6 BA S 6 S


A R 6 U

Ind

uc

ed

sh

ea

r fo

rce

in

Sla

b 2

-3 (

kg

f

0 1 2 3 4 5 6 7

0

2 0 0

4 0 0

6 0 0

8 0 0

1 0 0 0

1 2 0 0

1 4 0 0


C a lc .

B S 3 BB S 3 S

B S 3 UB R 3 BB R 3 S

B R 3 U

Ind

uc

ed

sh

ea

r fo

rce

in

Sla

b 1

-2 (

kg

f

0 1 2 3 4 5 6 7

0

2 0 0

4 0 0

6 0 0

8 0 0

1 0 0 0

1 2 0 0

1 4 0 0

1 6 0 0

1 8 0 0 M e a n = 1 .0 5 S td e v . = 0 . 3 1M e a n = 1 .0 4 S td e v . = 0 . 1 1

C a lc .

B S 6 BB S 6 S


B R 6 U

Ind

uc

ed

sh

ea

r fo

rce

in

Sla

b 1

-2 (

kg

f

0 1 2 3 4 5 6 7

0

2 0 0

4 0 0

6 0 0

8 0 0

1 0 0 0


C a lc .

B S 3 BB S 3 S


B R 3 U

Ind

uc

ed

sh

ea

r fo

rce

in

Sla

b 2

-3 (

kg

f

0 1 2 3 4 5 6 7

0

2 0 0

4 0 0

6 0 0

8 0 0

1 0 0 0

M e a n = 0 . 7 9 S td e v .= 0 . 4 3M e a n = 0 . 8 8 S td e v .= 0 . 3 3

C a lc .

B S 6 BB S 6 S


B R 6 U

Ind

uc

ed

sh

ea

r fo

rce

in

Sla

b 2

-3 (

kg

f

Figure13: Shear forces acting on diaphragms

THE EFFECT OF FLEXIBLE HORIZONTAL · PDF fileMasahide MURAKAMI1, Peter J MOSS2, Athol J CARR3...

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Transcript of THE EFFECT OF FLEXIBLE HORIZONTAL · PDF fileMasahide MURAKAMI1, Peter J MOSS2, Athol J CARR3...