Cantilever interaction of shear walls and frames

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Authors Patel, Chimanbhai N., 1937-

Publisher The University of Arizona.

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CANTILEVER INTERACTION OF SHEAR WALLS AND FRAMES

V .Chimanbhai N . Patel

A Thesis Submitted to the Faculty of the

DEPARTMENT OF CIVIL ENGINEERINGIn Partial Fulfillment of the Requirements

For the Degree ofMASTER OF SCIENCE

In the Graduate CollegeTHE UNIVERSITY OF ARIZONA

1 9 6 8

STATEMENT BY AUTHOR

This thesis has been submitted in partial fulfillment of requirements for an advanced degree at The University of Arizona and is deposited in the University Library to be made available to borrowers under rules of the Library.

Brief quotations from this thesis are allowable without special permission, provided that accurate acknowledg­ment of source is made. Requests for permission for extended quotation from or reproduction of this manuscript in whole or in part may be granted by the head of the major department or the Dean of the Graduate College when in his judgment the proposed use of the material is in the interests of scholar­ship, In all other instances, however, permission must be obtained from the author.

SIGNED:

APPROVAL BY THESIS DIRECTOR This thesis has been approved on the date shown below:

Andrew W.^RossProfessor of Civil Engineering

h/t ct* / O

ACKNOWLEDGMENT

The author wishes to express his deep gratitude to his thesis director, Professor Andrew W. Ross, for his active interest, helpful guidance and constructive suggestions„

He would also like to express his appreciation to Rod Gomez and Assoc., Consulting Engineers, Tucson, Arizona, for the practical training he received.

TABLE OF CONTENTS

LIST OF ILLUSTRATIONS.......LIST OF TABLES..ABSTRACT,CHAPTER

12

INTRODUCTION.................THE EQUIVALENT COLUMN METHOD.

A. Physical Analysis..............B . Twin Cantilevers ...............

Initial Assumptions ............Method of Analysis.............Derivation of General Equations

INTERACTION OF SHEAR WALLS AND FRAMESA. Concept and Method of Analysis ..........

Sys tern W ................................System F................................

B . Iterative Procedure to be Used.........Equilibrium conditions ..................Step 1: Free Deflection of Shear Wall

(conjugate beam method)...............Step 2: Initial Deflection and Rotation. Step 3: Induced Fixed End Moments.......Step 4: Story Shears in Frame...........Step 5: Concentrated Moments and Net

De if 1 c 11 on s. . .. .. . .. ... . ... ... . .. .. ... Step 6: The Forced Convergence Correction De s i gn E 3c amp 1 e .........................

B \S E RO T AT I ................................A. Base Rotation (General)................

Overturning Moment.....................Base Rotation Formula..................

S 1 1 I. 2C ajlip 1 5 oooe.o. ..................

Page

> • vi . . vii , „ vii

. 5

. 7

. 7

. 8

. 9

.15

.15

.15

.16

.16

.16

.19

.21

.22,23.24.26.27.39.39.40.41.42

Table of Contents (Continued)

B . Omission of Khan’s Charts........ .Derivation of Algebraic Equation .

5 C U 5 I ON eeooeeeoeooo»»e»6eoo6eeeaeeseoeeAOfl»»APPENDIX

I A. Free Delections and Rotations: Shear Wall...B . Moment Distributions ........................

1. Proposed Method 1st Trial...............2. Proposed Method Final Trial.............3. Basic Method Final Trial................

C. Tabulated results of the deformations of thewall from 1 through 6 trials..............

11 f EK 2.012 S . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

REFERENCE S..............................................

v

Page.45.54.57

,60.65.66.67.68

.69

. 76

• 79

LIST OF ILLUSTRATIONS Figure Page

1 Typical Deflected Shapes....... 62 Substitute Cantilevers ...... 103 Idealized Structure and Free Deflection of the

a l l a a a a « i a a « a Q Q » t t a a a o a o a o a o 6 B O O o a n a n o a O & a a s o g p > T 7

4 Fixed End Moments From Deflected Shape of theF r ame a o @ * @ * @ B @ a @ @ a a a a * * a e o @ a o a a a a a a @ a o a a a c a B B a l 8

5 Shear Wall Flexural Equivalency..................206 Interacting Forces and Moments of Combined

S y* S t e Xfi « a e e e o a O Q a e a 6 6 o a e e a e e e o e o o e e o 6 B a o a a « e e o e 2 5

7 Shear Walled Frame Elevation for IllustratedE 2C amp 1 e aooaeoooeooooooooaaoodaaooooaooaoaaooaa28

8 Substitute Cantilevers for Illustrated Example...299 Forces and Moments After Moment Distribution...... 3510 Rate of Convergence of Deflections at Second

F 1 O O r o a e e o e o o o o o e d e o o o o e o o o e o o o o a o o o o e o g a a o B o e 4911 Rate of Convergence of Deflections at 5th Floor.,5012 Rate of Convergence of Deflections at 9th Floor..5013- Rate of Convergence of Rotations at 2nd Floor.... 5114 Rate of Convergence of Rotations at 5th Floor.... 5115 Rate of Convergence of Rotations at 9th Floor.... 5216 Deflection vs. Height (with base rotation)....... 53

vi

LIST OF TABLES

Table1

34

Horizontal Deflection and Rotation for First Iterative Trial...

Final Bending Moment on the Wall at the End of 1st Trial.,

I* !3L XI £11 tiZ S XX1 S . . ... . . . . . . . . . . . . . . . . . . . . . . . . . o e o .

Total Free Horizontal Deflections and Rotations at Each Story.

Page

, . 31

.36 , .38

475 Final Results With and Without Base Rotation, . e o e .486 Deflection and Rotation for First Cycle ..... 70

7 Deflection and Rotation for Second Cycle..... . .... 718 Deflection and Rotation for Third Cycle...... ..... 72

9 Deflection and Rotation for Fourth Cycle..... ..... 7310 Deflection and Rotation for Fifth Cycle......

11 Deflection and Rotation for Sixth Cycle...... .....75

vii

ABSTRACT

If a building consisting of frames and shear walls is replaced by two representative cantilevers; then one cantilever represents the frame and the other the shear wall or assembly of shear walls„ In the analysis of the interaction of the shear wall and frames, use of this concept of twin cantilevers is made in order to save a considerable amount of manual work in moment distribution„

For an approximate deflected shape of the frame, the developed forces and moments are calculated at the joint between shear wall and frame at each story. The balance of forces and moments (loading and frame) are applied to the shear wall to find the deflection of the wall at each story. This iterative procedure is carried on until the following conditions are satisfied.

That; the horizontal deflection must be the samein both cantilevers at corresponding levels.

That; the summation of shears developed in both cantilevers must be equal to the total external shear (due to loading) at every story.

Effects of base rotation of the shear wall are included also in the analysis.

ix

This thesis is an attempt to both simplify and abbreviate existing techniques for accomplishing the above by combining methods suggested in recent literature.

CHAPTER 1 INTRODUCTION

As multistory building construction has evolved from bearing wall types of the late 19th century and the rigid steel frames of the 1930’s to the delicate curtain wall buildings of the 1950*5, the interior compositions have changed along with the exterior skins (3)„, The trends toward column-free interiors, long spans, minimum floor to floor heights and maximum rentable area together with the increased popularity of reinforced concrete as a construction material have resulted in the use of the shear wall as the principal lateral load resisting member in many multistory buildings. Frequently the service core of the building has provided an excellent location, for the shear wall since enclosure walls are required there anyhow.

Since a large number of tall buildings are now being constructed, and present practice in this type of construc­tion is to provide shear walls along with the frames to resist lateral loads due to wind or earthquake, the design of shear walls has been the subject of considerable discussion in the past few years,

The simplest approach to the design problem is to consider the shear wall as an independent member and design it as a vertical cantilever. But since the steel framing

1

actually does resist some of the lateral loading, such design is quite conservative with respect to shear walls; whereas the frames are underdesigned.

A number of excellent articles have been published on the subject of interaction of shear walls with frames in multistory structures (2,3,4,5). The analyses presented in the aforementioned papers are, of course, interesting and applicable to the design, but require use of a computer that may not always be available to office practice.

As has been mentioned above, the distribution of lateral forces between the frames and the walls should result in more economical structures because, in practical cases,' the results of an exact analysis will indicate a reduction of reinforcement in the shear wall. On the other hand, the building code requirement (6) of the one-third increase in allowable wind or earthquake stresses will generally permit accommodation of the additional stresses in the frame with no need for additional reinforcement over the major part of most tall structures =

Recent building regulations are influenced by the concept that structures designed for earthquake regions must serve two functions; (1) for frequent small shocks, they must be capable of controlling damage to nonstructurai elements in a building (partitions, skin, ducts, water and soil lines*etc., which, incidentally, may amount to more than 701 of the cost of the building), and (2) for several

earthquakes, the structure must have adequate ductility to accommodate large lateral deflections, with little, if any, loss in capacity. The design procedure presented in this thesis with consideration of code requirements relating to lateral loading will result in the shear wall braced structures accomplishing both functions (1) and (2) outlined above. The design information presented in Khan’s article(1) is helpful to engineers to establish more precise and economical reinforcing requirements. But the moment distribution is very time consuming for multibayed multistory buildings. Moreover, the design procedure requires successive adjustments to story deflections, so that it too is iterative.

An attempt has been made to both simplify and abbreviate existing techniques by combining methods suggested in recent literature, with the care not to dissatisfy the equilibrium conditions (Chapter 3).

Base rotation is considered in Chapter 4 in order to determine its effect on the moments and shears in the final solution. An investigation is made also into the rate of closure of the iteration when Khan’s charts (1) are not used to get the deflected shape close to the final correct deflected shape.

Although the analytical method considered is applicable for free-standing walls, or the enclosure around elevator shaft or stairs, it is not applicable to walls that are filled-in panels bounded by steel framing. The constraints imposed by the boundary connections by the frame would prevent, or modify, deflection as a contilever. Similarly the presence of such filled-in walls would modify the deflection of the frame members.

Note: The symbols adopted for use in this thesis are definedwhere they first appear and are listed alphabetically in Appendix II.

CHAPTER 2 THE EQUIVALENT COLUMN METHOD

A, Physical Analysis.The interaction of a shear wall and a frame is a

special case of indeterminancy in which two basically different components are tied together to produce one structure„ If the frame alone is considered to take the full lateral load, it would develop moments in columns and beams to resist the total shear at each story while the effects of overturning would normally be considered secondary and, in most cases, negligible„ In resisting all lateral loads, a frame would deflect as in Fig. 1(a). The floors would remain essentially level even though the joints would rotate. If a shear wall, on the other hand, is considered to resist all the lateral loads, it would develop moments at each floor equal to the overturning moment at that level and the deflected shape, Fig. 1(b) would be that of a canti­lever. .

If a shear wall and a frame exist together in a building, each one will try to obstruct the other from taking its natural free deflected shape, and as a result a distribution of forces between the two results. As shown in Fig. 1(c), the frame will restrain or pull the shear wall

b Free Wall/77T77777/777Z77-

Combined Frame S WallcFree FrameOx

Figure 1. Typical Deflected Shapes.

back in upper stories, while in the lower regions the opposite will occur.

The conflicting deflection characteristics of the frame and the shear Wall can be considered if the structure is first divided into two separated systems (1) frame, and(2) shear wall. Then for an approximate deflected shape of the frame, the developed forces and moments are calculated at the joint between shear wall and frame at each story. The difference of forces and moments between the external loading and frame are applied to the shear wall to find the deflection of the wall at each story, and these deflections are compared with those previously assumed. The procedure is repeated until (1) horizontal deflection must be same in both systems at corresponding levels, and (2) the summation

of shears developed in both systems must be equal to the total external shear (due to loading) at every story. Since this is essentially a successive approximation procedure., equality means an acceptably low level of inequality.

In this chapter, an attempt has been made to mathematically represent the frame as one cantilever and the shear wall as a second.

B. Twin Cantilevers.Initial Assumptions

The assumptions which are .usually made in the analysis of shear walls are as follows.

1* Shear walls, have moment and shear resisting connections with the adjacent framework.

2, Shear walls act mainly as vertical cantilevers fixed at the footing level.

3. The entire structure is tied and braced firmly so • that the building tends to act as a single unit.

4= The floor'slabs are infinitely rigid in their ownplane. Since the rotation is inversely proportional ■to the flexural rigidity, the slabs are considered not to undergo any rotation or distortion in the horizontal plane. But in case of flat slab design, the slab does undergo flexure in the vertical plane. Thus, the method is still valid for flat slab design.

5. Initially, it is assumed that the slopes in the frame at any particular floor are the same; also that the slopes in the walls at any floor level are the same though different from that in the frame.

Method of Analysis:The building consisting of frames and walls is

replaced by two representative cantilevers (Fig. 2). The Substitute cantilever for the frame includes the stiffnesses - of the columns and beams. The other cantilever represents the shear wall or assembly of the shear walls. These

two cantilevers are tied together at each floor level so that1 the sum of the shears at any story developed by two

cantilevers is equal to the total shear acting on the building,

2. the slope at one end of the link members joining thewall and frame represents the slope in the frame work,whereas the slope at the other end represents the slope of the shear wall,

3. the lateral displacement of the two cantilevers at any floor level is the same.

Derivation of General Equations:The basic slope deflection equations for a beam are

MABskBA^2dA+dB+UAB^ “m FAB° 0 ° ° ° ° (1)

MBAok B A 2 dB+ dA+ UB A-* ^ B A °»°(2)

where resultant end moment at point A of member AB,positive if clockwise and vice versa.

pM fixed end moment at point A of member AB,positive if clockwise and vice versa.

d^ ® 2E6.* deformation at AA

Referring to Fig. 2(a), for joint n, for equilibrium to

exist. . :

7TTJT7T777772V±_J

Shear-Walled Frame

Substitute Frame Canti­lever

Substitute ■Shear Wall

JointStiffness

n-1

b .. Twin CantileversFigure 2. Substitute Cantilevers

Expressing the above relations in terms of the slope deflection equations (1) and (2).

*kn(2d^ dn"'>°-

Where, stiffness of columns between floor n and n-1

f 6EDnUn« ^— Dnas relative displacement of thetwo ends of member n

E ° modulus of elasticityhn® height of nth story

kE” stiffness of beam at nth floor, n

But as per assumption, 5,f f f f f f en'"en'"“6n " dn'“dn"'"dn'

Where joint rotation of frame work at nth floor.So,substituting this deformation condition in previousequation

< ( 2kn»2kn»lt3kn*3n ' ^ kn < - V kn*ldL v kn <

*kn*lUn-H*0-' ’ ••Similarly for equilibrium to exist at joint n

12

For equilibrium to exist at joint n 19

Hi1,(n+1)1*Mn •,(n-1),+Mn',n"+Mn",n“° ”s 0 — •(5)

There will be vertical movements at the connecting points of the "link" beams with the shear wall. The vertical movement at any connecting point will be equal to the slope at that point multiplied by half the width of the shear

wall. Vertical movement at point n"""0n"^n ’ 1

where e^n° rotation of shear wall at nth floor

&Y 58 half the width of an actual shear wall at nthTifloor.

s°< un',n"‘^ S ^^n

Now expressing equation (5) in terms of the slope deflection equations (1) and (2)

k Cn*l) ' '*Un.l)"kn'(2dn'ld(n-D.WjX

n

Where. stiffness of link beamn

d^„s deformation at joint between link beam and wall at nth floor.

Therefore,

dn^2k(n*l) ,'!'2kn ,'!‘3kn 9's'2klb^+k(n+l) ,dn+l

*kn'<-l,k(n»l) 'U( „ » l ) < ' UX b (1<- i K -°n

Adding equations (4)» (4a)» and (6),

dn (2kn'>'2kn"''>2kn ,'>2kn*l'>2k (n-H),,,<'2k Cn+l) '

<6kn»6kn-42knb^ dn»l(kn.l*k(n»l)"'

»k (n+l)')+dn-l(kn+kn"'+kn')+knb ( dnn

^Un (kn^kn"' *kne *dn+1 (kn+ l*k (n+l)"' *k (n+l) ' =0 ’ ’ “

Now, making the following substitutions in equation (7)

Kn-kn"kn"'»kn'=Ek^

kn+l*kn+l*k (n+l)"'*k (n+l)'*^kn+l(These are the sums of all the column stiffnesses at the nth and the n^lth stories, respectively«)

14

Kn'lcn ^ n ' ' !;kn

(This is the sum of the stiffnesses of all the beams except the link beams at the nth floor.)

Klb=rklb n n3£w

n

(These are the sums of stiffnesses of all the link beams.)

dn(2K^ 2Kntl*6Kn*2Knb>*Kn » l 4 l ,

»Kndn . l ^ < ,KX < » l UnM-°

Say, Kn- 2 K ^ 2 K ^ 1<-6Kb *2K^b ... ...(8a)

Substituting equation (8a) into equation (8),

KndntKn»ldn»l"Kndn-llkIFdn4KnUn"Kn»lUn»l=0‘'' ••• ™

Equation (9) shows that the structure shown in Fig. 2(a) can be represented as shown in Fig. 1(b). Thus the frame system can be represented as one cantilever and the assembly of shear walls as the other. These twin cantilevers are tied together by link beams so that the entire structure is looked at as a single unit. Kn (Eqn. 8a) is the stiffness of the joint between frame column and link beam at nth story,

CHAPTER 3INTERACTION OF SHEAR WALLS AND FRAMES

A. Concept and Method of Analysis„As has been seen in Chapter 2, the concept of twin

cantilevers reduces the multibay$ multistory shear^walled structure into the single-bay multistory shear-walled structure. Then the analysis is performed in two stages.In the first stage of analysis of the structure, it is necessary to determine the deflected shape and the amount of lateral load distributed to the walls and frame, respectively,at each story. For this purpose, the structure is separatedinto two distinct systems; system W and system F.

1. System W : Shear wall or assembly of shear walls.This system can have any configuration. Walls are extended over the entire height of the structure.The stiffness of this shear wall system at anystory equals the sum of the stiffnesses of allshear walls regardless of their shape and size.Shape and size should be considered in computing an average the distance from the neutral axis ofthe system W to its extreme fiber. Thus, thecoupled shear walls can be represented in high multistory buildings as a single wall with an

16

equivalent stiffness equal to the sum of the stiff­nesses of both shear walls.

2. System F: This system includes all columns andbeams contributing to the lateral stiffness. The link beams, members linking the frames with the shear walls, are also included in this system. In twin cantilevers, the stiffnesses of columns, beams and link beams are simply the sum of the stiffnesses of all such members in the structure. The link beam span is an average of the link beam spans of the structure when this spans are within the same range of magnitude.

Then the analysis of a single-bay shear-walled multistory frame system is performed by an iterative solution presented subsequently.

B. Iterative Procedure to be Used.Equilibrium conditions:

1. The horizontal deflection must be the same in both cantilevers at corresponding levels.

2. The summation of shears developed in both cantilevers must be equal to the total external shear (due to loading) at every story.

3. Link members connecting two cantilevers must undergo the same rotations and vertical translations as those of system W and system F at their point of connection.

17

Link Beams

LO

Tf"

torC

(SJrC

P

P

P

P

P

777/77777777- H ^ q

7777 LB

5

4

3

2

1

TTtTJTTTTHT . 2 C ,

(a) Idealized Structure (b) Free Deflection of System W.

Figure 3. Idealized Structure and Free Deflection of the Wall.

18

A r -A

3

1

"Force-Fitting" System F Fixed End Moment From "Force Fitting"

Figure 4. Fixed End Moments From Deflected Shape of the Frame.

19The foregoing three requirements of compatibility and equil­ibrium can be achieved by the following six steps of analysis.

Step 1: Free Deflection of Shear Wall (conjugate beam method)Total computed external loads (wind or seismic) on

the idealized structure (Fig. 3a) are directly applied to shear wall at each floor level. The slopes and deflections of the shear wall at each floor level are determined by the conjugate beam method shown hereinafter.

Conjugate Beam Method.In the conjugate beam method (8), the relationship

between the real beam and the conjugate beam are as follows;(a). The span of the conjugate beam is equal to the

span of the real beam.(b) The load of the conjugate beam is the M/EI diagram

of the real beam.(c) The shear at any section of the conjugate beam

is equal to the slope of the corresponding section of the real beam.

(d) The moment at any section of the conjugate beam is equal to the deflection at the corresponding section of the real beam.

A A(a) Real Beam (b) Conjugate Beam

where Iw=moment of intertia of shear wall.

E “Modulus of elasticity of concrete (shear wall)

Figure 5. Shear Wall Flexural Equivalency.

The real beam is fixed at A (Fig. 5a)

Deflection =0 . .In conjugate beam; at end A shear =0

and moment“0So end A is free in the conjugate beam as shown in Fig. 5. Similarly, in the real beam, the slope and deflection are not zero at free end B, hence the end B is fixed in the conjugate beam as shown in Fig. 5b.For example: The slope at pt. C in the real beam;

. .At end A; Rotation 6=0

8^=shear at pt. C in the conjugate beam

and the deflection at pt. C in the real beam;A moment at pt. C in the conjugate beam

Thus, the free horizontal deflection and rotation can be calculated by the conjugate beam method at any point on the shear wall. The free horizontal deflection, rotation and vertical deflection at any point, i, are denoted as A ^ , efi and respectively. The deflections one floor aboveand below are + and respectively.

Step 2: Initial Deflection and Rotation.For quick convergence, initial deflection and

rotations are assumed or approximated from Figs. 32 through 38 given in Khan's article (1). In the absence of a good estimate, however, the deflected shape is assumed as the free deflected shape of the shear wall, which would mean that, in the first cycle, initial deflection and rotation at the ith floor would be

andAii(l)”Afi

6ii(l)”8fi

System F is forced to undergo the assumed deflections at each floor (Fig. 4a). This also requires that the connecting members at each floor must have the same rotations and

22

vertical translation as system W at their points of connec­tion with system W.

Step 3; Induced Fixed End Moments.The moments induced by "force-fitting" can be

determined directly by using moment distribution„ The foreed-fitted frame shown in Fig„ 4a has known story deflec­tion and rotations at the connecting points; hence, for uniform columns and beam sections» the fixed end moment at the beginning of moment distribution (Fig. 4b) would be for columns at ith story

6EI .FMci=(—

i

At the ith floor for "link" beams at their shear wall end,

4EI. . 6EI, .FMbiw” (1 -- V (— y - ) A vi

b Lb

but Avi=Zw8i

2EI, • I• - FMbiw35C - ™ ) [ 2 ^ 3 ( I )]9i

Lb b

and at the ith floor for "link" beams at their frame end2EIb . 6EL.

FMbif“ C~f— 0i+ C~7T“ ) Avi *Lb ' bb

But again

Distribution Factors:As has been seen in Chapter 2, the joint stiffness,

in the twin cantilever, at any nth joint is as follows:

S r 2Kn*2KS*l+6Kn*2Knb -

Therefore,Distribution Factors:

2KC 2K<-Lower column^ n

Kn 2KC-o-2Kc. 1*6Kb-$-2K b n n+1 n n

2Kn+l 2Kn^lUpper c o l u m n -=— ------ 5-rx" 2Kn*2K^ V 6Kn"2Kn

2Klb 2KlbLink beam - =— -- r-2----5----rg •

n 2K5+2Kn,l+6Kn+2Kn

Knowing the fixed end moments and the distribution factors, the moment distribution can be run to get final moments in the members. When a known, fixed si desway is imposed on a structure, as in this case, the cumbersome sideway corrections to the moment distribution are not required, and the solution will converge rapidly.

Step 4: Story Shears in Frame.After force-fitting system F to system W, the total

shears in each story of system F as well as moments and

24reactions applied on system W by the connecting links are computed (Fig. 6a). The shears generated by force-fitting can be used directly in the next step. The resulting horizontal forces P ' are shown in Fig. 6 only for illustra­tion purposes. These interaction forces may be either positive or negative at different floors.

Step 5: Concentrated Moments and Net Deflections.All shear forces and moments generated by force-

fitting system F are applied to the isolated free system W (Fig. 6b). At any story i, 1VL and should be replaced by moment (Fig. 6 c)

M ' - W i V

The moments and forces shown in Fig. 6b will cause negativedeflections and rotations of system % A . and 8 . respectively.' ai axThe balance of forces and moments (loading and frame) are applied to the shear wall in order to find the net deflections

and rotations, and @e£ respectively, at each story.So at the end of first cycle, the deflections and rotations equations can be written as follows:

Aei(l)=Afi~Aai(l)

9ei(l)=6fi"6ai(l)

or in general at the end of the nth cycle

Aei (n)~Aii (n) "Aai (n)

2.5

VS

V4

V3

V2

VI- V

4-V4 -V5

3 3 "2

(a) Forces’ and Moments in System F after Moment Distribution.

%p-i

crIIcrM

CTMcrM

/777T//7/7777

(b) Forces and MomentsFrom System F Applied to System W.

N.A.

— yf—

Vi

M'=Mi+Rvi(^w)

(c) Total Concentrated Bending Moments.

Figure 6 . Interacting Forces and Moments of Combined System.

26

and

9ei(n) 6ii(n) 9ai(n)°

This is the end of one cycle of iteration. For the stable condition to occur, the initial deflections at any floor "i" at the beginning of nth cycle, A^(n) must be the same as the end (net) deflections Agi^) the completion of the nth cycle. However in many cases in the first cycle, A ^ is negative indicating that the iteration is divergent. The generalization of this method of solution therefore depends on the use of a proper "forced - convergence - correction" to be applied to the initial deformations of the nth cycle

Aii(n) an^ 0ii(n)> t0 obtain the initial trial deformations

of the (n-s-l)th cycle, Aii(-n^1) and 6ii(n-i-l) •

In Fig. 6b it will be noted that the axial effects of the link beam reactions (R^) on the shear wall are neglected. This is compatible with the neglect of axial forces in most frame analysis methods. The two major contributors to shear wall behavior (concentrated moment , and horizontal loading, P^) are considered.

Step 6 ; The Forced Convergence Corrections.The convergence corrections are derived from the

hypothesis that in each cycle the movement of system W

at each floor with respect to its free deflected shape is

lineally proportional to the movement of system F with

27respect to the vertical line. Therefore, it can be shown that if at the nth cycle the initial trial values at the ith floor were and 6^ and the end values were

Aei(n) and 8ei(n)' the initial trial values at the (n^l)th cycle should be as follows:

Aii(n)

and0 = 6 JL9ei(n) ~9ii (n)ii(n-M) iiCn)^ e f i - e e i ( n )

eii(n)

The values A and 0 obtained by above two equations are used as initial values for the next cycle, and the procedure ,is repeated beginning with the second step outlined previously.This iterative procedure is carried on until the netdeflection is equal to the initial deflection A ^ .At the end of each cycle, A ^ and A ^ should be checkeduntil the convergence is within a specific acceptable tolerance,

for example 5% to 10%, based on the designer's judgment.

Design Example:Data: The dimensions and lateral forces are shown in theFig. 7; whereas the elastic properties are listed as follows:

Column Stiffness kG=800 in^Beam Stiffness k^=200 in^

28

Shear Wallk lh-200 in3 I -200 ft4

Area A =30 ft3

E =3420K/in

As has been seen in Chapter 2, the structure shown in Fig. 7 can be represented as the single bay shear-walled frame structure (Fig. 8).

Story10th

9th8th

7th

6 th

5 th

4th

3rd

2nd

1st

Ground

12K

22K

26K

24

24

24

24

24

24

24

7 7

H-25 25 9'

sr-4iiH<SUo ,

HFiguie 7. Shear Walled Frame Elevation for Illustrated

Example.

29Where Kc=j:kc -800 + 800

=1600 in3 Klb = 2:klb = 200 in3

=200 in3Joint stiffness at any joint "n"

Kn"2Kn+2K^ i +6Kn + 2K12

22

2624

K

KK

2 4

24K

2 4

24

24

K

K

24

12

K

K

K'

Klb

Figure 8 . Substitute Cantilevers for Illustrated Example

. . For top joint K10 = 2 (1600)+2 (0)+6 (200)+ 2 (200)=3200+0+1200+400 -4800 in3

For any intermediate jointKn®2(1600)+2(1600)+6(200)+2(200)

*3200+3200+1200+400 *8000 in^

Distribution Factors? ? % ? nnTop joint: link beam = * *0,08Y 4 OV UK10

Column = - i ” 1600K10 4800

= 0.33

2KnbIntermediate joint: link beam =----Kn

.2x2008000

= 0.05

2Kn%n.2(1600)8000

= 0.4

Upper or lower col,

Free Deformations of Shear Wall.Neglecting the frame, at the first stage, the

external lateral loads are applied to the shear wall and thus free deflections and rotations are calculated in

31Appendix 1(A) at each floor level0 These deflections and rotations are tabulated in Table 1. The third column of Table 1 shows the factors from Khan’s (1) chart in order to approximate the deflected shape close to the finaldeflected shape of the structure. The deflected shape ofthe force-fitting system is shown in Fig. 4a,

Table 1. Horizontal Deflection Iterative Trial.

and Rotation for First

StoryFreeDeflection

Aif

FreeRotations

6if

Reductionfactor

F Aii(l) 9i i (1)

1 0.13 .00158 .006 0.04 .000034

2 0.46 .00 2 85 .022 0.14 .000125

3 0.96 .00364 .045 0.287 .000255

4 1.57 .00440 .077 0.49 .000436

5 2.28 .00493 .110 0.70 .00063

6 3.05 .00527 . 150 0.955 .00087

7 3.86 .0055 .193 1.23 .00109

8 4.69 .0056 .2 34 1.49 .00132

9 5.53 .00565 .280 1.782 .00158

10 6.37 .00566 . 325 2.07 .00184

The force-fitting system is shown in Fig. 4a 6

The induced moment in

FMbiw-CZEKbi)(2*3^ ) e ib

link beam at shear wall end

2x3420x200 ,lA_4-SVa' - n - 23~~ i

=114x10 (2.54)Bi

=289„56x10 9i

32

At 10th story

. . FMbl0w=2 89.56xl03x.00184 =533 K Ft.

Similarly

FMbq*r457 K Ft-FMb8w=384

FMb 7w=315

FMb6w=252

FMbSw=182 .

FMb4w=126

FMb3w= 7 3 "8

FMb2w= 3 6 -2

FMblw" 9 °83

Now induced fixed end moment in link beam on frame end,

LsFMbif-2EKbi(1+3r > eib

,2x5420x200^1+ 34 .5)Q12 25 1

=114x1.54xlO30i

53

At the 10th story

FMbl0 f“ 1 1 4 x l o 5 4 x l 0 3 * ’00184=323 KFt

Similarly

FMb9f“ 2 7 7 K «Ft

™ b 8 f =232

FMb7f«191

™b6f=155

FMbsf»110 KFt

FMb4f=76.3

FMb3f='44 -7

FMb2f=22.0

FMblf=5.95

Induced P»E«M in ColumnsAt any story i,

6EI

i

6EK=~h (‘i"4 ! - ! 1i

6x3420x1600,, . .=----------- (A.-A, i)12x12x12 1 1 i

= 19xl03(Ai-Ai_1)

At 1 0th Story

FMclO=19x 1 0 3 (2 „=5470 K.Ft

Similarly ;

FMc9=5540 K ’

FMcg=4940 K '

FM .=5220 C 7 K'

FMc6=4840 K !

FMc5=3990 K'

FMc4=3860 K*

FMc3=2790 K'

FMc2=1900 K*

FMc1=760 K ’

After running the moment distribution (see Appendix 1(B) „ the final forces and moments are shown in Fig„ 9„ The story shears in the frame and the shear wall, concentrated bending moment and then final bending moments are tabulated in the Table 2,

35

=, -217

. 153

113K

143

126

81

122

58

52

54.36

695K'i V - 217%%rt / 8564

K 1* 54.8 ' 561h 40

41.28d4 6 ' 2 4

1B 4°71734.96 v_ 335

45r30.04 rx 265 -lt> - 4124. 88 f

124.9 ■p6416.44 , 104.2

^ 6 #11.65 52.63

27.1 6.32if: 24,9

939K'

I

V;

V:

v

I

932

769

677

564

470

377

199

156.6

81.10

nrrrrrTT

Figure 9. Forces and Moments After Moment Distribution.

36

Table 2. Final Bending Moment on the Wall at the End of 1st Trial.

Story

10th 9th 8th 7th 6th 5 th 4th 3rd 2nd 1st

Ground

ExternalShearsKips

123460

84108132156180204228240

Frame Story Shear Cone. FinalStory Shears in Wall Bending Bending

Kips Kips Moments MomentsK* K'

21715311314312681

122585227.10

■205 ■119 • 53■ 59■ 18 51

34122

152 200 .9 212.9

939932769677564470

377199156.681.100

939 4331 6528 7841 9113 9 893 9658 9449 8141 6398.7 3988

The final moments will give the net deformations of the shear wall at the end of the first trial, ^ and9ei(i). The calculations for these net deformations are done by conjugate beam method and tabulated in Table 6 in

Appendix 1(C) .

37

Deformations for Second Cycle,The initial trial values for the next cycle is

obtained as follows:

1.., * 'iciCr.)~Aii(n)

4ii(n)

Aii(l)

For example at 10th story

4 iioc2)"^*°7+2.07

=2.07-1.22 =0.85 in.

Similarly for all the stories the deflections and rotations are calculated, and the results are shown in Table 6 (Appendix

1(C) .This iterative procedure is carried on until the

initial deflections are equal to the final deflectionsAei(n)* [The results of intermediate iterations are tabulated in Appendix C, Table 7 through Table 11.]

The final results are tabulated and compared with those obtained by Khan's Method in Table 3.

38Table 3. Final Results

Shear Wall only; Cantilever Proposed Method Khan's MethodAnalysis

Floor ShearKips

BendingMomentsK'

Deflec­tion in.

ShearKips

BendingMomentK'

Concen­trated Bend Mom K'

Deflec­tions . in.

ShearKips

Bend ' Mom K'

Concen­tratedB.M.K'

Defletions

■ in.

10 . 12 0 6.37 -27.1 173 173 0.89 -27.1 169.8 169.8 0.929 34 -144 5.53 - 9.0 704.2 206.5 0.83 - 7 699.0 204.0 0.86

8 60 -552 4.69 39.2 1055.2 . 242.5 0.76 40.2 1012.0 229.0 0.797 84 -1272 3.86 13.0 877.3 292.5 0.69 13 815.6 286.0 0.716 108 -2 280 3.05 68.6 1063.8 342 .5 0.59 69.7 996.3 336.7 0.615 132 -3576 2.28 56.5 624.6 384 0.49 5 7.7 535.3 375.3 0.50

4 156 -5160 1.57 97.5 359.1 412.5 0.37 98 248.8 406.0 0.38.3 180 -7032 0.96 104.0 -406.4 404.5 0.25 105 -529.2 398.0 0.252 204 -9192 0.46 '153.5 -1291.9 362.5 0.13 154 -1432.7 356.5 0.13

■ 1 228 -11640 0.13 192.6 -2898 235.2 . 0.04 192.6- 3056.2 224.5 0.04

Ground 240 -14376 0 204.6 -5209.2 .0 0 204.6- 5 368.2 0 0

CHAPTER 4 BASE ROTATION

A . Base Rotation (General).The engineer is sometimes confronted with the

question of whether the shear wall bases should be fixed or free to rotate. At the other times he is compelled to design the footings for a central load and a moment, and for a limited amount of rotation„ Therefore an understanding of the rotation characteristic of the shear wall base and footing is essential„

When the lower end of a shear wall is subjected to a bending moment, the joint between the shear wall and the footing must be strong enough to transfer the stresses„ But this can be overcome by embedding dowels in the footing, and then the shear wall can be considered fully fixed to the footing. Once the shear wall is fixed to the footing, the base rotation of the shear wall is caused only by the elastic deformation due to the greater soil compression at the toe of the base, which is generally small and insignificant.

Regardless of degree of fixity between the shear wall base and the footing, the moment from the shear wall will cause unsymmetrical soil pressure. The soil pressure is assumed to have straight line or planar distribution. Unfortunately the pressure distribution is not likely to be

39

40planar and cannot be determined quantitatively (10). There­fore, the rotation of a footing acted ©n by a moment or an eccentric loading can only be estimated on the basis of some simple calculations guided by good engineering judgment.For example, small and shallow footings on sand are prone to rotation because the sand readily runs out from under the toe of the footing. If the footing is located at greater depth, the sand is subjected to a confining pressure due to the weight of the overlying soil. The relative effect of the edge condition diminishes as the size of footing increases. It becomes apparent that small and shallow footings on granular soils should not be relied upon for providing fixity to the shear wall base.

Contrary to the sand, clay and clayey soils resemble elastic material and are capable of resisting a concentrated stress at the edge. Furthermore since a large portion of the settlement of footings on clay is due to consolidation resulting from bending moment acting for a long period of time, so the bending moment due to the wind or seismic load acting in short duration would not cause significant rotation.

Overturning Moment.The axial load from earthquake force on vertical

elements and footings in every building or structure may be modified and the overturning moment at the base of the building or structure shall be determined in accordance with

the following formula.

41

M^”JEF hj, ,„o . . .Uniform Building Code Sec. 2314,p. 115

=JxCantilever Moment

, , 0.5where, 3 M

and T* *05Hv r

Fi=Lateral force applied to a level designated as "i" h^=Height in feet above the base to the level designated

as "i".D“ Dimension of the building in feet in a direction parallel

to the applied forces.H= The height of the main portion of the building in feet

above the base.

The required value of "J" shall not be less than 0.33 nor more than 1.00. "J" should be 1.00 for elevated tank supported with four or more cross braced legs.

Base Rotation Formula.The joint between shear wall and footing is assumed

to be strong enough so that the stresses due to bending moment at the lower end of the shear wall would be transferred to the soil.

MbcMax stress in soil = — —

2

where c* half length of the footing mom, of inertia of footing.

Now,MvC

max, vertical deformation in soil ■ ——r-I

where k- subgrade soil modulus in lbs, per sq, in, per inch of deformation.

Base rotation: eR-settlellient& c

Mbc- ?

c

ifk' !

The translation of the shear wall at the ground floor level is prevented in every step of the computations. Thus, only the appropriate rotation at the base of the shear wall will be permitted. The design loads can be applied directly to the shear wall, and the iterative analysis should give the desired results.

Design Example.So far it has been assumed that full fixity exists

at the base of the shear wall as in the numerical example presented in the previous chapter. However, the theory can be expanded to include the case where the base can rotate due

to elastic support. This may occur if the supporting soil is elastic or if the wall is resting on columns which may have uneven settlements„ In either case, a horizontal move­ment at the base is considered prevented.

The base rotation can easily be included in the . aforementioned iterative solution by increasing the rotation at all stories by the amount of base rotation and increasing horizontal deflection at all stories by the product of base rotation and the distance from the base to each story.

In the previous example,Cantilever mom.=-14376 K.Ft.

overturning mom. =J£F^h^=Jxcantilever moment.

But J= 9-^L 3\/T2

and T-’— iV F i

_.05(120) _ 6 V(59) 7.67

= 0.78

j„ 0.5 _ 0.53 \/(0 . 78) 2 0 . 85

= 0.59

overturning moment® 0.59(-14376)

M, 44Now, base rotation 9^-

^K.

Take k average =200 lbs/in /in (k varies between 50 and 500) use footing size 6 'x25’

=\I-=~x6*253=7812.5 Ft4 f 12

9 . 8480xl03 B 7812.5(200x1728)

=.0031 radian Horizontal deflectionat any floor due to base =Base Rotation^ Distance inrotation inches from

base to that floor.

..Hor. Defl-’at 10th floor ^0.0031*(120x12)due to base rotation r4 o x n o

In the same way, the horizontal deflections due to base rotation are found out for all stories and are tabulated in Table 4.

Once the total deflections and rotations are known, the rest of the iterative procedure is followed exactly the same way as presented in Chapter 3. The final results are tabulated in Table 5 along with the previous results obtained without inclusion of base rotation for the comparison purpose.

45

The overall results are very close in both cases, with and without inclusion of base rotation.

Bo Omission of Khan's Charts»In this chapter, the numerical example includes the

effect of base rotation of the shear wall, as well as the effect on rate of convergence if Khan's charts (1) are not used to obtain approximate deflections for each trial.

Starting with total free deflected shape, seven iterative trials have been made. The initial and net deflec­tion versus numbers of trials graphs are plotted for each story, and these seemingly converging graphs are extrapolated to a point of intersection» These intersecting values of deflections are very close to the correct values. In the same way, the approximate values of rotations are also obtained. For the ninth, fifth and second stories these deflection and rotation graphs (Fig. 10 to Fig. 15) are presented for illustration purposes. Beginning with these close values of deflections and rotations, two more trials were needed and the subsequent third tfial is the final solution. The example in the previous chapter with the exclusion of the base rotation also needed seven trials to come to the final solution even when using Khan's charts values.

This demonstrates that the technique of using the

deflected shape obtained from the developed graphs from the results of the example itself converges as fast as the use. of Khan's charts.

46Some A^ and Ag curves cross rather than truly

converging, as shown in Fig. 10 to 15; but this is to be expected since the deflection at each story is affected by the deflections of all other stories. The initial deflec­tion values, as found from the relationship;

A = A AAei(n)~Aii(n)■ii(n*l) ii(n) ^ Afi-Aei(n)

Aii(n)

do not reflect any effects from the other stories of the structure. By using the curve intersection points, known or extrapolated, as final deflection values, results very close to the correct values are obtained with a minimum of successive approximations.

Table 4. Total Free Horizontal Deflections and Rotations at Each Story,

CD

Hori.FreeAfi (2).

Deflections hDue to base rotation ^

Total Free (4)=(2) + (3)

Free8fi (5)

Rotations 8

Base Rotation 9B (6) -

Total Fr< (7) = (5) +

10 6,37 4.50 5s-OOo .00566 .0031 ,00876

9 5,53 4.05 9.58 .00565 ,0031 .008758 4.69 3.60 8.29 .00560 .0031 .00877 3.86 3.15 7.01 .00550 .0031 .00866 3.05 2.70 5.75 .00527 .0031 .008375 2.28 2.25 4.53 .00493 .0031 .008034 1.57 1.80 3.37 .00440 .0031 .00753 0.96 1.35 2.31 .00364 .0031 .006742 0.46 0.90 1.36 .00284 .0031 .00594

1 0.13 0.45 O OO .00158 .0031 .00468

Ground 0 0 0 0 .0031 .0031

Table 5. Final Results With and Without Base Rotation.

With Base Rotation Without Base Rotation

Story ShearK,

Concen. B’MK.Ft

Bend’g Mom. ^ ,

De£ln ° inch.

ShearK.

Concen. B . M „ j£,

Bend’g Mom. ^ ,

Deflectinch

10 -39.7 193.2 193.2 0.88 -27.1 173 173 0.899 - 0.3 231.2 900.8 0.81 - 9.0 206.5 704.2 0.838 18.5 253.5 1157.9 0.74 39.2 242.5 1055.2 0.76

7 57.2 283.5 1219.4 0.67 13.0 292.5 877.3 0.696 47.4 324.0 857.0 0.60 68.6 342.5 1063.8 0.59S 61.3 382.2 670.4 0.50 56.5 384.0 624.6 0.494 95.9 409.0 343.8 0.38 97.5 412.5 359.1 0.373 110.7 380.5 -426.5 0.26 104.0 404.5 -406.4 0.252 143.3 399.0 -1357.9 0.13 153.5 362 .5 -1291.9 0.131 199.4 279.0 -2798.5 0.04 192.6 235.2 -2898.0 0 .04

Ground 211.4 0 -5191.3 0 204.6 0 -5209.2 0

Deflection

in Inches

A

0.8

0.6

0.4

0.2

*e2 (n)I________1 I

A ?r >= Net deflection at 2nd floor at end of nth cycle of iteration.

2 fnV Deflection at 2nd floor at 1 beginning of nth cycle of

iteration.

3 4 5Number of Trials

10

Figure 10. Rate of Convergence of Deflections at Second Floor.

Deflection

in inches

Deflection

in inches

50iv: .y--

e5 (n)

Deflection at 5th floor at end

of nth cycle of iteration.

A.rV \=Deflection at 5th 1 floor at beginning

of nth cycle of iteration.

=#=-3 8

Figure 11. Rate of Convergence of Deflections at 5th Floor.

1,0

Ai9(n)

e9 (n)

A q/- ^“Net Deflection at 9th . } floor at end nth cycle

of iteration.A.Q , ^Deflection at 5th floor 1 \ ' at beginning of nth

cycle of iteration.

4 5 6Number of Trials

Figure 12. Rate of Convergence of Deflections at 9th Floor. . .

Rotation

in Radians

, .

Rotation

in Ra

dians

000 8

0006

,0004

.0002

51

e2 (n)8f,?fnY~Rotation in shear wall

- at 2nd floor at the end of nth cycle.

' 0* o/tl'i“■Rotation in shear wall _ J at 2nd floor at the

beginning of nth cycle.

Number of Trials

Figure 13. Rate of Convergence of Rotations at 2nd Floor.

,0010

,0008

0006

.0004

0002

e5 (n)

2 .

6 rrr^^RDtatiori in shear wall ■ at 5th floor at the end

of nth cycle.GiSfnl'Rotation in shear wall

: J at 5th floor at beginning of nth cycle.

4 5 6Numbers of Trials

Figure 14. Rate of Convergence of Rotations at . 5th Floor.

Rotation

in Radians

52

.0008

0006i9(n)

.0004

0002

Number of Trials

- .0002

e Qr ^Rotation in shear wall "I at 9th floor at the cnd=

of nth cycle.0 .Q r N=Rotation in shear wall 1* ' at 9th floor at the

beginning of nth cycle.

ei9 (n)-.0004

-.0006

Figure 15. Rate of Convergence of Rotations at 9th Floor.

Heig

ht53

Wall-column stiffness ratio, — -=96Sc

Column-Be am stiffness ratio, — ^=4SB

i

Deflections in inches

Figure 16. Deflection vs. Height (with base rotation).

54

Derivation of Algebraic Equation.The deflect ions versus story height graph (Fig. 16)

is plotted and the algebraic equation for the graph is derived as follows:The general equation for curve is

y«Axn+Bx+CBut x=0

C= 0y=0

yssAxn+Bx conditions from graphxB0 .6 y*0 .6

x=0.36 y=0.33x=0.216 y=0.16

.•.0.6 -A(0.6)n + 0.6B 0.33=A(0.36)n+0.36B 0.33-A(0.6)2n+0.36B0.36=6A(0.6)n+0.36(B)

2n , ,,n 0.6-A(0.6)n*0.6Band 0.16=A(.216)n*216B

-.03 =A(0.6)-6A(0.6)

. . .03 0.16-A(0.6)3n+.216B0.6(0.6)n-(0.6)2n .216=.36A(.6)n+.216B

-.056=A[.63n-.36(.6)n ]

.'.A- •°56. 36(.6)n -(.6)3n

.£3________ , .056,6 (.6 )n -(.6 )2n .36(.6)n-(.6)3n

5536(.6)n-C.6)3n6(.6)n -(.6)2n .056

.03 1.865

.•.0.36(.6)n-(.6)3n=1.12(.6)n-1 . 86 5(.6) 2n

.'.0.36-0.62n=l.12-1.865(0.6)n

.•.0.62n-l. 865(.6)n- .76-0

n ,n +1.865+ 3.48-3.04U . o c ______ —__________

1.865*.632

=1.248 or 0.616

.*.n log.6= log 1 . 248 or log 0.616

n - M £6 or L 1 9 T. 778 T.778

_+0.096 __ -0.21" ' L L L - 0.222

*-0.432 or 0.95

Now .030 .6 (0 .6 )n -(0 .6 )2n

substituting values of "n" in above equation ne-0.432 A«-0.0 37

and

5bn = 0 .95 A®-3.00

fl-~A(0.6)n+0 .6 0.6

n = -0.432 and A=-.037 B=1.076n = 0.95 and A=- 3.0 B=4.08

y=-. 0 3 7 x " + 0 76x-^-^-x0 "432

ory=-3x0'^^>4.08x=4.08x-3x°'9^

From Khan’s charts (1) (Fig. 32 through Fig. 38), it is clear that the variation of column-beam stiffnessscratio, — does not radically affect the graph. But wall-

s«column stiffness ratio, — has seemingly important influence

cSs

So for constant value of — (i.e. 96), our derived algebraicSc

equations give the deflected shape close to the finalSs

deflected shape regardless of the value of — ratio.SB

CHAPTER 5 CONCLUSIONS

A practical method of analysis for the multistory framed structure with moment- resisting joints that is also braced with shear walls has been proposed in this thesis.The method does not require simplifying assumptions that are not ultimately checked and yet gives the designer some freedoms as follows:

1. Variable story heights are permitted.2. Constant shear walls are not needed and the

openings in shear walls are allowable.3. Variable sections of columns and girders are allowed.

The common design assumptions that all horizontal loads are carried by the shear walls are not correct over the entire height of the structure. Furthermore the distribution ofthe lateral shear between the frame and the shear wall depends not only on their relative stiffnesses but on the number of stories as well.

The iterative procedure adapted in this thesis is laborious and requires time and patience. In spite of that, the procedure is simple and a problem can be solved on a sliderule or a small desk calculator in the design offices.

57

58The concept of the substitute cantilever for the

frame reduces the multibayed structure into the single bay structure. A numerical example is presented to illustrate the design procedure and simplicity of the calculations.For the final trial of the example, the moment distributions involved in both the proposed and basic methods are presented in Appendix 1(B) in order to show that a considerable amount of work is saved by the proposed moment distribution.

The use of Khan's Charts (1) to get the approximated final deflected shape of the frame is recommended in order to obtain faster convergence of the iteration.

The method presented is easily applicable to the design of the shear wall buildings of any height for maximum economy with adequate control over the required strength and ductility of all structural elements.

Base rotation is considered in Chapter 4 and results both, with and without base rotations, are tabulated in Table 5 for comparison purposes. The overall difference for horizontal shear and bending moment is within four percent for this example building in both cases with and without base rotation. The maximum difference in the deflection is also four percent. Thus, it seems that the base rotation has insignificant influence in the iterative analysis for at least some building proportions.

59The same numerical example is solved again in

Chapter 4 with inclusion of the base rotation and without use of Khan’s charts to obtain an approximate close deflected shape to start with. It is seen that the example without use of charts requires three more trials than that with use of charts. Therefore, if the charts are not available, the graphs can be plotted for first six or seven iterative trials starting from free deflected shape with or without base rotation. Then the intersection values which are correct or quite close to correct values can be read directly if the graphs intersect otherwise extend to let them intersect.

The algebraic equation that is derived from deflection versus height graph is restricted to concentrated loading and to wall-column stiffness ratio of about 1 0 0 .

APPENDIX 1(A)

Free Deflections and Rotations; Shear Wall.

Story External Load Story Shear Bend. Mom.K K K

10 12 12 0

9 22 34 -1448 26 60 -5527 24 84 -12726 24 108 -2280

5 24 132 -35764 24 156 -51603 24 180 -70322 24 204 -9192

1 24 228 -11640

Ground 12 240 -14376

Real Beam

Conj ugate Beam

12010'=120'

60

24 24 24 24 24 24 24 26 22 12i i i i i i i i i i3 4 5 6 8 9

►—< ►—iU3 w\(N1inin rH

10

61Area A , - 11376.11640

2 EIw

/.EI^-156,096 K.Ft

S i m i l a r l y ,

EIA2 = 124992El 2 > w n 97, 344El

v A 4 = 73, 152El > in it 52, 416Elw A 6 ” 35, 136

Elw A7 = 2 1 ,312Elw A 8' 1 0,944Elw A9* 4, 176ElwA 10= 864

x ,11640+2(14376) x12 1 11640+14376 3

= H ! 2i ,4 26016

=6.2 Ft.

S i m i l a r l y ,

X 2 = 6.23 Ft

X3 = 6.27 Ft

X4 = 6.31

x5 = 6.37

*12

62

X6 = 6 ,,32

x7 = 6 «.57

x 8 = 6 ., 78

x9 = 7., 15

x 10 = 8 .,00

156 ,096

= 156,096______(3420x144)200

= 156,096 9.85x10?

* 0.00158 Radians

6f2‘Al* A 2

156,096+124,992 = =----9.85x10

=0.00285 Rad.

=0.0036

Rotations.

63

Similarly,

0 “0.00 44 Rad.

0 , -0.004935

0, =0.005276

0 r “0.0055 7

0 r “0.0056 1 8

0r =0.00565 9

6r “0.00566

Deflections.Afl=Aixi

,156,096x6.2 xl2 (3420x144) 200

,156,096x6.28.208x106

=0.128 -0.13 in.

Af C12**)*A2x2,156,096(18.2)♦124,992(6.23)

8. 208x106=0.46 in.

6 4

Af «A1(24+x^)C12>x2)

=0.96 in.

Similarly,

A r =1.57 4

A r =2.285

A , =3.056

A r =3.867

A * =4.698

A r =5.53 9

A r =6.37 t 10

APPENDIX 1(B)

Moment Distributions.

1. Proposed method - First Trial2. Proposed Method - Final Trial3. Basic Method - Final Trial

.25

~TJ~-5470 ♦4293 ♦1724 -1509 - 749

irir♦1610

-1064 ♦1221

0.4 0.15-5540♦ 4094 1537♦ 2146-1636

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Ground

Final Trial Proposed Method

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10th

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10th

9 th

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8th

15976-35♦16-7.5♦ 3.5 7T2

6 th

89.5-40

- 8.5 ♦ 4.0

5th

4th

3rd

Ground

193

O'

APPENDIX 1(C)

Tabulated results of the deformations of the wall from 1 through 6 trials.

Table 6. Deflection and

Deflection

Story Free Initial Net Initial forAfi 4 ii(l) Aei(l) "ext ere!4. in. in. ii(2)ln- in.

10 6.37 2.07 -7.14 0.85

9 5.53 1.782 -5.61 0.76

8 4.69 1.49 -5.21 0.61

7 3.86 1.23 -3.32 0.564

6 3.05 0.955 -2.36 0.46

5 2.28 0.70 -1.57 0.354 1.57 0.49 -0.94 0.256

3 0.96 0.287 -0.49 0.16

2 0.46 0.14 -0.195 0.081

1 0.13 0 .04 -0.0425 0.0245

Rotation for First Cycle.

RotationsFree Initial Net Initial for

:le

Rad.6fi 6ii(l) eei(l) cycle

Rad. Rad Rad ii(2)

00566 .00184 - .009 0.00064

00565 .00158 - .00866 0.000550056 .001325 - .00802 0.0004950055 .00109 -.00714 0.0004300527 .00087 -.0061 0.00037

00493 .00063 - .00495 0.00030044 .000436 -.00375 0.000224

00364 .000255 -.00259 0.00013500285 .000125 - .00152 0.00007800158 .000034 -.00063 0.000028

109

8

765

4321

Table 7. Deflection and Rotation for Second Cycle.

Deflections RotationsFreeAfi in.

InitialAii(2) in.

NetAei(2) in.

Initial for next cycle6 ii(3)

in.

FreeefiRad.

Initial8ii(2) Rad.

Net*ei(2) Rad.

6.37 0.85 1.31 0.92 .00566 .00064 .000414

5.53 0.76 1.26 0.835 .00565 .00055 .000475

4.69 0.61 1.175 0.6937 .0056 .000495 .000675

3.86 0.564 1.06 0.647 .0055 .00043 .0009053.05 0.46 0.92 0.542 .00527 .00037 .00108

2.28 0.35 0.75 0.4245 .00493 .0003 .00126

1.57 0.256 0.56 0.3175 .0044 .000224 .00135

0.96 0.16 0.367 0.204 .00364 .000135 .00131

0.46 0.081 0.19 0.106 .00284 .000078 .00112

0.13 0.0245 0.054 0.0317 .00158 .000028 .0007

Table 8. Deflection and

DeflectionsInitial for next cycleAii(4) in.

10 6.37 0.92 0.99 0.9302

9 5.53 0.835 0.97 0.856

8 4.69 0.6937 0.92 0.729

7 3.86 0.647 0 . 86 0 .685

6 3.05 0.542 0.753 0.5815 2.28 0.4245 0.625 0.46554 1.57 0.3175 0.471 0.35173 0.96 0.204 0.314 0.2306

2 0.46 0.106 0.1625 0.1223

1 0.13 0.0317 0.0317 0.0361

Story Free Initial NetAfi 4ii (3) 4ei (3)in. in. in.

Rotation for Third Cycle.

Rotations

Initial for next cycle6 ii(4)Rad.

00566 .00064 .0001192 .000656

00565 .00055 .000185 .0005110056 .000495 .00039 .00050060055 .00043 .000621 .000484

00527 .00037 .00081 .000461300493 .0003 .00099 .00042670044 .000224 .00109 .000366500364 .000135 .0011 0.0002653

00284 .000078 .000955 0.0001737

00158 .000028 .00061 0.0000568

Free6fi Rad.

Initial6ii(3)Rad.

6Netei(2)Rad.

Table 9. Deflections and

DeflectionsStory Free Initial Net Initial for

sfi 6 ii(4) &ei(4) n=xt ^clein. in. in. ii(5)

in.

10 6.37 0.9302 0.905 0.929 5.53 0.856 0.884 0.878 4.69 0.729 0.846 0.797 3.86 0.685 0.775 0.736 3.05 0.581 0.678 0.635 2.28 0.4655 0.57 0.524 1.57 0.3517 0.428 0.393 0.96 0.2306 0.297 0.26

2 0.46 0.1223 0.154 0.14

1 0.13 0.0361 0.045 0.04

Rotations for Fourth Cycle.

RotationsFree Initial Net Initial for6fi eii(4) eei(4) n=xt c>rcleRad. Rad. Rad. ii(S)

Rad.

00566 .000565 .00058 .000318100565 .000511 .000845 .0005040056 .0005006 .000977 .00062110055 .000484 .00098 .00067300527 .0004613 .000887 .00065700493 .0004267 .00073 .0005960044 .0003665 .000576 .000480

00364 .0002653 .000375 .00043800284 .0001737 .000197 .00035400158 .0000568 .000139 .000352

Table 10. Deflection and Rotation for Fifth Cycle.

Deflections RotationsStory Free Initial Net Initial for Free Initial Net Initial for

afi \i(S) 6ei(S) nef c>rcle efi eii(5) 6ei(S) n=xt cyclein. in. in. ii(6 ) Rad. Rad. Rad. ii(6 )

in. Rad.

10 6.37 0.92 1.03 0.939 0.00566 .0003181 .00059 .0003679 5.53 0.87 0.956 0.884 0.00565 .000509 .000636 .0003648 4.69 0179 0.874 0.804 0.0056 .0006211 .000755 .000464

7 3.86 0.73 0.78 0.74 0.0055 .000673 .000886 .0005186 3.05 0.63 0.665 0.637 0.00527 .000657 .00083 .0006175 2.28 0.52 0.538 0.524 0.00493 .000596 .000912 .0006934 1.57 0.39 0.403 0.393 0.0044 .000480 .000952 .0007183 0.96 0.26 0.267 0.262 0.00364 .000438 .000938 .0006802 0.46 0.14 0.14 0.14 0.00284 .000354 .00081 .0006291 0.13 0.04 0.04 0.04 0.00158 .000352 .000524 .000366

Table 11. Deflection and Rotation For Sixth Cycle.

Free Initial Net Initial for Free Initial Net Initial forStory Afi iii(6) Aei(6) next cycle 8fi e.i(6) 8ei(6) next cycle

in. in. in. Aii(7) Rad. Rad. Rad. 9ii(7)in. Rad.

10 6.37 0.939 0.87 0.89 .00566 .000367 .000484 .0004259 5.53 0.884 0.79 0.83 .00565 .000364 .000732 .000680

8 4.69 0.804 0.75 0.76 .0056 .000464 .00083 .0007557 3.86 0.74 0.65 0.69 .0055 .000518 .00083 .0007746 3.05 0.637 0.55 0.59 .00527 .000617 .000765 .0007295 2.28 0.52 0.47 0.49 .00493 .000693 .00067 .0006434 1.57 0.393 0.35 0.37 <0044 .000718 .000582 .0005503 0.96 0.26 0.25 0.25 .00364 .000680 .000473 .0004682 0.46 0.14 0.13 0.13 .00284 .000629 .000373 .000368

1 0.13 0.04 0.04 0.04 .00158 .000366 .000367 .000347

cn

APPENDIX II NOTATIONS

D = The dimension of the building in feet in a directionparallel to the applied forces.

D = Relative displacement of the two ends of member n.n *d*7 = 2Eef= Deformation at joint n .n n

E = Modulus of elasticityFMci = Fixed end moment of column at ith story.FMbiwa Fixed end moment of linkbeam at ith story at its wall

end.FMbif® Fixed end moment of linkbeam at ith story at its

frame end.F « Lateral forces applied to a level designated as "i".

hn = Height of nth story.

= Height in feet above the base to the level designatedas "i".

IP = Moment of inertia of wall footing.I » Moment of inertia of shear wall,wJ = Numerical coefficient for base moment as specified in

Chapter 4.k = Subgrade soil modulus in pounds per square inch per

inch of deformation.

76

77

= zk^= The sum of all column stiffnesses at nth story.

K" ' = I k ^ = The sum of all linkbeam stiffnesses at nth storv. n n y

kc * Stiffness of column between floor n and n-1. nk^ = Stiffness of beam at nth floor, nk ^ = Stiffness of linkbeam at nth floor, n

Zw = Half the width of shear wall at nth floor, nMas “ Resultant end moment at point A of member AB

= Fixed end moment at point A of member AB

= Overturning moment at the base of the shear wall.i

NT » Concentrated moment on shear wall at floor i.

- Applied moment on shear wall by linkbeam at floor i.= External load at floor i.

Rvi = Vertical reaction of the link beam at the shear wall at floor i.

T = Fundamental period of vibration of the structure in seconds.

8^ = Joint rotation of framework at nth floor.

8W = Rotation of shear wall at nth floor.n

6 f 53 Free rotation of wall at floor i.8ii(n)^Rotation in shear wall at ith floor at the beginning

of nth cycle.

78

6ei(n)*Rotati°n shear wall at ith floor at the end of nth cycle.

A = F r e e deflection of wall at floor i.

A i i ^ )=Deflection at ith floor at beginning of nth cycle of iteration.

A . , x=Net deflection at ith floor at end of nth cycle of ei (nJiteration.

REFERENCES

Khan, Fazlur R., John A. Sbarounis, "Interaction of Shear Walls and Frames." Journal of A.S.C.E.Vol. 90 STS, p. 285.

Tamhankar, M. G., J. P. Jain, G. S. Ramaswamy, "The Concept of Twin Cantilevers in the Analysis of Shear-Walled Multistorey Buildings." Indian Cone. Journal p. 488 Dec. 1966.

Gould, Phillip L., "Interaction of Shear Wall-Frame Systems in Multistory Buildings." Journal of Am. Cone. Inst. p. 45, Jan. 1965.

Cardan, Bernhard, "Concrete Shear Walls Combined With Rigid Frames in Multistory Buildings Subject to Lateral Loads." Journal of Am. Cone. Inst. p. 299 Sept. 1961.

Thadani, B. N., "Analysis of Shear Wall Structures." Indian Cone. Journ. p. 97, March 1966.

"Building Code Requirements for Reinforced Concrete," Amer. Concrete Inst. Detroit, Mich., ACI 318-63, 1963.

Blume, John A., Nathan M. Newmark and Leo H. Corning, "Design of Multistory Reinforced Concrete Buildings for Earthquake Motions," Portland Cement Ass’n., Chicago, 111., 1961.

Kinney, J. S., "Indeterminate Structural Analysis," Addison - Wesley Publishing Co., Inc. Reading, Massachusetts, 1957.

Sexton, H. J., Discussion of "Concrete Shear WallsCombined with Rigid Frames in Multistory Buildings Subject to Lateral Loads," by Bernard Cardan,Journ. of the Amer. Cone. Inst., part 2, Vol. 58, March, 1962, pp. 825-827.

Teng, Wayne C., "Foundation Design", Prentice-Hall, Inc New jersey, 1962, pp. 137-145.