THE STRUCTURAL DESIGN OF TALL AND SPECIAL BUILDINGS

Struct. Design Tall Spec. Build. 13, 73–87 (2004)

Published online in Wiley Interscience (www.interscience.wiley.com). DOI:10.1002/tal.235

SHEAR WALL WITH OUTRIGGER TRUSSES ON WALL AND

COLUMN FOUNDATIONS

J. C. D. HOENDERKAMP

Department of Architecture and Building, Eindhoven University of Technology, Eindhoven, The Netherlands

SUMMARY

A graphical method of analysis is presented for preliminary design of outrigger truss-braced high-rise shear wall

structures with non-fixed foundation conditions subject to horizontal loading. The method requires the calculation of six structural parameters: bending stiffness for the shear wall, bending and racking shear stiffnesses for

the outrigger, an overall bending stiffness contribution from the exterior columns, and rotational stiffnesses for

the shear wall and column foundations.

The method of analysis employs a simple procedure for obtaining the optimum location of the outrigger up

the height of the structure and a rapid assessment of the influence of the individual structural elements on the

lateral deflections and bending moments of the high-rise structure. It is concluded that all six stiffnesses should

be included in the preliminary analysis of a proposed tall building structure as the optimum location of the outrigger as well as the reductions in horizontal deformations and internal forces in the structure can be significantly

influenced by all the structural components. Copyright © 2004 John Wiley & Sons, Ltd.

1.

INTRODUCTION

The outrigger-braced high-rise shear wall shown in Figure 1 consists of a centrally located wall with

two equal-length trusses positioned at a distance x from the top of the structure. The outriggers connect

the shear wall to columns in the façade of the structure. This way the horizontally applied load will

force the wall to behave compositely with the exterior structure by introducing axial forces in the

columns. These forces form a restraining moment which is in the oposite direction to the bending

moment from the horizontal load. This effect will decrease the bending moments in the wall from outrigger level down to the base and it will reduce the horizontal deflections of the structure. The bending

moments in the shear wall and axial forces in the columns are resisted by foundation structures on

piles which are indicated by translational springs. Combined with the laterally applied loading the

restraining moment will force a triple curvature in the wall up the height and double curvature in the

trusses as shown in Figure 1.

In the horizontal deflection analysis of outrigger-braced shear walls on fixed foundations it has been

shown (Stafford Smith and Coull, 1991; Stafford Smith and Salim, 1981) that the wall can be represented by a single flexural stiffness parameter. The outriggers were assumed to be prismatic members

rigidly connected to the wall and hinge connected to the exterior columns. The resulting single curvature behaviour of these flexural members was represented by a single bending stiffness parameter.

* Correspondence to: J. C. D. Hoenderkamp, Department of Architecture and Building, Structural Design Group, Eindhoven

University of Technology, P.O. Box 513, Route 7 5600 MB Eindhoven, The Netherlands.

E-mail: j.c.d.hoenderkamp@bwk.tue.nl

Copyright © 2004 John Wiley & Sons, Ltd.

Received January 2003

Accepted February 2003

74

J. C. D. HOENDERKAMP

outrigger

x

outrigger

h

H

facade

column

shear

wall

facade

column

piles

b

c

c

b

ground level

deflected shape

Figure 1. Shear wall with outrigger trusses

It was further taken that the columns are pin connected to a fixed foundation and could thus be represented by a parameter which represents the axial stiffnesses of the columns only. With three stiffness parameters representing the wall, outriggers and exterior columns it was possible to combine

them in a single dimensionless parameter which allowed a graphical procedure to obtain the optimum

location of the outriggers such that they would cause the largest reduction in horizontal deflection at

the top of the structure.

Figure 1 also shows simplified models for the foundations of the central shear wall and exterior

columns. In the structural analysis the individual elements are only subjected to forces resulting from

horizontal loading on the structure. The foundations under the pin-connected façade columns are taken

to be piled foundations which act in a vertical direction only. They are modelled as linear springs. The

shear wall foundation is only subjected to a bending moment. The net axial load on this foundation

as a result of lateral loading is zero as it is positioned in the centre of a symmetric structure. The foundation of the wall can thus be modelled by a rotational spring with a rotational spring constant. The

outriggers consist of trusses of which the deformations due to shear cannot be ignored. The shear stiffness of the trusses in addition to their flexural stiffness must be included in the analysis. In order to

allow the truss to be subjected to flexural deformations it will be necessary to assume the structural

floors of the building to be connected to the trusses at the exterior columns and shear wall locations

only; i.e. the floor structure is not part of the outrigger structure but it will cause the shear wall and

exterior columns to have identical rotations at all floor levels. Further simplifying assumptions are:

the structure behaves linear elastically; the sectional properties of the shear wall, exterior columns and

outriggers are uniform througout their height or length; and the distribution of the lateral loading is

uniformly distributed along the height of the structure.

A compatibility equation is to be developed for the rotations in the wall and outrigger at the intersection of the neutral line of the outriggers and the face of the shear wall. This leads to an expression

Copyright © 2004 John Wiley & Sons, Ltd.

Struct. Design Tall Spec. Build. 13, 73–87 (2004)

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SHEAR WALL WITH OUTRIGGER TRUSSES

for the restraining moment allowing the reduction in horizontal deflection at the top to be determined.

Maximizing this reduction will yield the optimum location of the outrigger.

2.

SHEAR WALL

The rotations in the shear wall are the result of a uniformly distributed horizontal load w, and a restraining moment M caused by a reverse action of the outriggers. They will cause rotations at outrigger

level due to bending in the wall and rotation of the wall foundation as presented in Figure 2. For the

shear walls it will be assumed that plane sections remain plane in bending so that the rotations at the

shear wall centre line are identical to those at the face of the wall.

2.1

Rotations due to lateral loading w

The rotation in the reinforced concrete shear wall at level x can quite simply be expressed as follows:

q s ; b ;w =

w( H 3 - x 3 )

6 EIs

(1)

where H is the total height of the structure, x is the distance measured from the top of the stucture and

EIs is the flexural stiffness of the shear wall. The rotation of the wall due to the rotation of its foundation is

wH 2

2Cs

q s ;Cs ;w =

(2)

in which Cs is the rotational stiffness of the shear wall foundation.

2.2

Rotations due to restraining moment M

The rotation in the shear wall at level x as a result of bending in the wall is given by

q s ;b ;M = -

qs;b;w

qs;b;M

M ( H - x)

EIs

(3)

qs;Cs;w

qs;Cs;M

x

H

EIs

EIs

Cs

Cs

Figure 2. Rotations in shear wall

Copyright © 2004 John Wiley & Sons, Ltd.

Struct. Design Tall Spec. Build. 13, 73–87 (2004)

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J. C. D. HOENDERKAMP

The rotation of the shear wall and its foundation are

M

Cs

q s ;Cs ;M = -

(4)

It is noted here that the rotations caused by restraining moment M are in the opposite direction to those

by horizontal load w.

3.

OUTRIGGER

The rotations in the outrigger due to the restraining moment are obtained by splitting up this action

as shown in Figure 3, where the outriggers have been separated from the braced frame for clarity. The

restraining moment in the structure due to outrigger action is the product of the axial force in the exterior columns and the distance between them:

M = Fa (2 b + 2c) = Fa (2l)

(5)

in which Fa is the restraining force in the exterior columns, l is the distance from the exterior column

to the centre line of the shear wall and b is the length of the flexible outrigger measured from the

façade column to the outrigger/shear wall interface. Considering the free body diagram of a single

outrigger, then

Fa ¥ b = Fr ¥ h

(6)

where Fr represents the complementary shear forces in the outriggers and h is the vertical dimension

of the outrigger. The free body diagram in the centre of Figure 3 allows an expression for the restraining moment on the shear wall to be written as

M = 2 Fa c + 2 Fr h

Fr

Fa

Fr

Fr

outrigger

Fa

Fa

Fr

x

Fr

Fa

Fa

outrigger

(7)

Fr

Fr

h

Fa

H

Fr

shear

wall

b

c

c

b

Figure 3. Free body diagram restraining forces

Copyright © 2004 John Wiley & Sons, Ltd.

Struct. Design Tall Spec. Build. 13, 73–87 (2004)

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SHEAR WALL WITH OUTRIGGER TRUSSES

qr;b;Fr

Fr/2

Fr/2

Fr/2

Fr/2

(a)

qr;s;Fr

Fr

h

Fr

b

a

(b)

Figure 4. Bending and shear deformations in truss

Substituting Equations (6) and (7) into Equation (5) will lead to an expression for the effective restraining moment causing bending and shear deformations in the outriggers as shown in Figure 4.

2Fr h =

M

a

(8)

where a dimensionless parameter

a=

3.1

l

b

(9)

Rotations due to restraining force Fr

The double curvature bending in the outrigger as shown in Figure 4(a) is caused by axial strain in the

top and bottom cords. The rotations in the truss at both ends are identical because the structural floors

are assumed to be connected to the outrigger at the columns and shear wall only. The rotation due to

bending at the outrigger/wall interface is given by

q r ;b ;Fr =

2 Fr h

Mb

=

Â (12 EIr b) 24aEIr

(10)

where EIr is the bending stiffness of the outrigger, which can be obtained as follows:

EIr = EAr h 2 2

(11)

where Ar is the cross-sectional area of the top and bottom chords of the rigger.

The rotation due to racking shear results from strain in the diagonals as is indicated in Figure 4(b)

and can be expressed as

Copyright © 2004 John Wiley & Sons, Ltd.

Struct. Design Tall Spec. Build. 13, 73–87 (2004)

78

J. C. D. HOENDERKAMP

q r ;s ;Fr =

2 Fr h

M

=

hÂ GAi haGAr

(12)

It has been shown earlier (Hoenderkamp and Snijder, 2000) that the racking shear stiffness of a rigger,

GAr, is the sum of the individual racking shear stiffnesses of all the bracing segments in that rigger.

So for the outrigger structure

n

GAr = Â GAi

(13)

i =1

where n represents the total number of segments in the two outriggers and GAi is the racking shear

stiffness of a single segment of width a. For several types of bracing systems the racking shear stiffness is given Appendix A.

3.2

Rigid body rotations of outrigger

The wide-column behaviour of the shear wall due to bending caused by horizontal load w will result

in rigid body rotations of the outriggers as shown in Figure 5. This reverse rotation is given by

d s ;b ;w ˆ

c

q r ;b ;w = - Ê

= - (q r ;b ;w )

Ë b ¯

b

(14)

Substituting Equation (1) into Equation (14) yields

3

3

Ï w( H - x ) ¸ c

q r ;b ;w = - Ì

˝

Ó 6 EIs

˛b

(15)

An additional reverse rotation in the trusses occurs due to the rotation of the wall foundation when

subjected to horizontal loading and is shown in Figure 6. This rigid body rotation for the outrigger

can be expressed as follows:

q r ;Cs ;w = - Ê

Ë

d s ;Cs ;w ˆ

c

= - (q s ;Cs ;w )

¯

b

b

(16)

qs;b;w

qr;b;w

qr;b;w

d s;b;w

b

d s;b;w

c

c

b

Figure 5. Rigid body rotation of outriggers due to bending in shear wall

Copyright © 2004 John Wiley & Sons, Ltd.

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SHEAR WALL WITH OUTRIGGER TRUSSES

qs;Cs;w

qr;Cs;w

qr;Cs;w

d s;Cs;w

d s;Cs;w

c

b

c

b

Figure 6. Rigid body rotation of outriggers due to rotation of wall foundation

substituting Equation (2) into Equation (16) yields

2

Ï wH ¸ c

q r ;Cs ;w = - Ì

˝

Ó 2Cs ˛ b

(17)

Restraining moment M will also cause rigid body rotations of the outriggers. The shear wall will deflect

in the opposite direction as shown in Figures 5 and 6. The rigid body rotation of the outriggers will

now be in the clockwise direction. For bending in the wall:

q r ;b ;M = - Ê

Ë

c

d s ;b ;M ˆ

= - (q s ; b ;M )

b ¯

b

(18)

Substituting Equation (3) into Equation (18) yields

M ( H - x) ¸ c

q r ;b ;M = ÏÌ

˝

Ó EIs ˛ b

(19)

For rotation of the shear wall foundation:

q r ;Cs ;M = - Ê

Ë

d s ;Cs ;M ˆ

c

= - (q r ;Cs ;M )

b ¯

b

(20)

Substituting Equation (4) into Equation (20) yields

M c

q r ;Cs ;M = ÏÌ ¸˝

Ó Cs ˛ b

3.3

(21)

Rotations due to restraining force Fa

The restraining forces in the exterior columns will cause two more rigid body rotations of the outriggers: one resulting from shortening and lengthening of the exterior columns and another due to vertical displacements in the column foundations.

Copyright © 2004 John Wiley & Sons, Ltd.

Struct. Design Tall Spec. Build. 13, 73–87 (2004)

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J. C. D. HOENDERKAMP

The first outrigger rotation can be defined as the column change in length divided by the length of

the outrigger. This leads to

q r ;a ;Fa =

Fa ( H - x )

bEAc

(22)

where Ac is the cross-sectional area of the exterior column.

Defining an overall stiffness parameter for the exterior column structure to be

EIc = 2 EAc l 2

(23)

and substituting Equations (5) and (23) into Equation (22) will yield the following expression:

q r ;a ;Fa =

M ( H - x )a

EIc

(24)

The second outrigger rotation can be defined as the foundation displacement divided by the length of

the outrigger, which yields

q r ;Cc ;Fa =

Fa k aM

= 2

b

2l k

(25)

where k represents the translational stiffness of the column foundation. Defining an overall rotational

stiffness parameter for the combined column foundations as

Cc = 2 l 2 k

(26)

and substituting Equation (26) into Equation (25) leads to the following simplified expression:

q r ;Cc ;Fa =

4.

aM

Cc

(27)

HORIZONTAL DEFLECTION

Compatibility in rotation at level x requires that the rotations in the shear wall, qs, and outrigger,

qr, at their interface are identical. Substituting for the individual rotations yields the following

expression:

3

3

2

w( H 3 - x 3 ) wH 2 M ( H - x ) M

Ï w( H - x ) ¸ c Ï wH ¸ c

+

= -Ì

˝ -Ì

˝

6 EIs

2Cs

EIs

Cs

Ó 6 EIs

˛ b Ó 2Cs ˛ b

Mb

M

M ( H - x ) ¸ c Ï M ¸ c aM ( H - x ) a M

+

+

+ ÏÌ

+

˝ +Ì ˝ +

24aEIr haGAr Ó EIs ˛ b Ó Cs ˛ b

EIc

Cc

(28)

Simplifying leads to

w( H 3 - x 3 ) wH 2

H-x H-x

b

1

1

1

+

= M ÏÌ

+

+

+

+

+ ¸˝

2

2

6 EIs

2Cs

EIc

Ó EIs

24a EIr ha GAr Cs Cc ˛

Copyright © 2004 John Wiley & Sons, Ltd.

(29)

Struct. Design Tall Spec. Build. 13, 73–87 (2004)

SHEAR WALL WITH OUTRIGGER TRUSSES

81

Setting two characteristic flexibility parameters as

H

H

+

EIs EIc

(30)

b

1

1

1

+

+

+

24a 2 EIr ha 2 GAr Cs Cc

(31)

S1 =

and

S2 =

leads to the following expression for the restraining moment:

3

3

2

H

Ï w( H - x ) wH ¸Ï

¸

M=Ì

+

˝Ì

˝

(

)

2Cs ˛Ó H - x S1 + HS2 ˛

Ó 6 EIs

(32)

The horizontal deflection at the top of the building can be now obtained as follows:

ytop =

wH 4 wH 3 M ( H 2 - x 2 ) MH

+

8 EIs

2Cs

2 EIs

Cs

(33)

where the first two terms on the right-hand side represent the ‘free’ horizontal deflections of the shear

wall at the top due to bending of the wall and rotation of its foundation as a direct result of lateral

loading. The third term is due to the restraining moment M which causes a reverse deflection and rotation at outrigger level, resulting in a deflection reduction at the top. The last term represents a horizontal deflection at the top of the wall due to reverse foundation rotation.

5.

OPTIMUM LOCATION OF OUTRIGGER

The reduction in horizontal deflection at the top of the structure due to restraining moment M is

represented by the last two terms on the right-hand side of Equation (33) and can be expressed as

2

2

H¸

ÏH - x

yred = M Ì

+ ˝

Cs ˛

Ó 2 EIs

(34)

This reduction is maximized by differentiating it w.r.t. x, setting it equal to zero and solving for x.

After simplifying this leads to

d ÈÏ

5 - 3x 2 - 2 x 3

6 ¸Ï

1

¸˘ = 0

2

3

5

+

˝˙

ÍÌ1 - x - x + x +

2 ˝Ì

(g H )

dx ÎÓ

(g H ) ˛Ó1 - x + w ˛˚

(35)

in which two characteristic non-dimensional parameters for the outrigger-braced shear wall structure

have been set as follows:

gH =

Copyright © 2004 John Wiley & Sons, Ltd.

HCs

EIs

(36)

Struct. Design Tall Spec. Build. 13, 73–87 (2004)

82

J. C. D. HOENDERKAMP

0.2

Rigger Location, x/H

0.3

0.4

Values of g H

> 100

0.5

10

6

0.6

4

3

2

0.7

0.0

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9

1.0

Values of w

Figure 7. Optimum location of outrigger

and

w=

S2

S1

(37)

A dimensionless location parameter for the optimum position of the outrigger has been set as

x=

x

H

(38)

Figure 7 shows a graphical presentation of optimum outrigger locations up the height of a shear wall

braced structure on wall and column foundations as a function of two non-dimensional characteristic

parameters, gH and w.

6.

EXAMPLE

The structural floor plan of a 29-storey, 87 m high building in Figure 8 shows the arrangement of four

identical shear walls with horizontal steel trusses on both sides. Each truss is 9 m long, has a single

storey height of 3 m and comprises five X-braced segments. The building is subjected to a uniformly

distributed lateral load of 1·6 kN/m2. The flexural stiffness of the reinforced concrete shear wall EIs =

1·5 ¥ 109 kNm2. For the horizontal trusses: top and bottom chords Ar = 1·78 ¥ 10-2 m2, diagonals Ad =

9·726 ¥ 10-3 m2 and for the exterior columns Ac = 3·12 ¥ 10-2 m2. The elastic modulus of steel E = 2·1

¥ 108 kN/m2. The rotational stiffness of the shear wall foundation Cs = 2·0 ¥ 108 kNm and the translational stiffness of the column foundations k = 4·0 ¥ 105 kN/m.

The detailed calculations are shown here for a single shear wall with one outrigger on both sides.

The flexural stiffness of the outrigger structure is given by Equation (11):

Copyright © 2004 John Wiley & Sons, Ltd.

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SHEAR WALL WITH OUTRIGGER TRUSSES

Shear walls with outriggers

9m

9m

9m

9m

9m

9m

9m

9m

w

Figure 8. Structural floor plan

EIr = EAr h 2 2 = 2◊1 ¥ 10 8 ¥ 1◊78 ¥ 10 -2 ¥ 32 2 = 1◊682 ¥ 10 7 kNm 2

The racking shear stiffness is the sum for 10 segments. Equation (A.1) yields

GAr = 10

2a 2 h

2 ¥ 18

◊ 2¥3

EA

=

¥ 2◊1 ¥ 10 8 ¥ 9◊726 ¥ 10 -3 = 9◊272 ¥ 10 6 kN

d

1◊ 5

d3

(18

◊ 2 + 32 )

The global second moment of area of the exterior columns can be obtained from Equation (23):

2

9

EIc = 2 EAc l 2 = 2 ¥ 2◊1 ¥ 10 8 ¥ 312

◊ ¥ 10 -2 ¥ Ê 9 + ˆ = 2◊388 ¥ 10 9 kNm 2

Ë

2¯

An overall rotational stiffness parameter for the foundations under the exterior columns is given by

Equation (26):

Cc = 2 l 2 k = 2 ¥ 13◊52 ¥ 4 ¥ 10 5 = 1◊458 ¥ 10 8 kNm

The dimensionless parameter a in Equation (9) yields

a=

l 13◊5

=

= 1◊5

b

9

The characteristic parameters S1 and S2 are given by Equations (30) and (31):

S1 =

H

H

87

87

-1

+

=

+

= 9◊443 ¥ 10 -8 (kNm)

EIs EIc 1◊5 ¥ 10 9 2◊388 ¥ 10 9

Copyright © 2004 John Wiley & Sons, Ltd.

Struct. Design Tall Spec. Build. 13, 73–87 (2004)

84

J. C. D. HOENDERKAMP

b

1

1

1

+

+

+

2

2

C

C

24a EIr ha GAr

s

c

9

1

1

1

-1

=

+

+

+

= 3◊775 ¥ 10 -8 (kNm)

2

7

2

6

8

8

24 ¥ 1◊5 ¥ 1◊682 ¥ 10

3 ¥ 1◊5 ¥ 9◊272 ¥ 10

2◊0 ¥ 10

1◊458 ¥ 10

S2 =

The characteristic non-dimensional parameters for the structure can now be obtained from Equations

(36) and (37):

gH =

HCs 87 ¥ 2◊0 ¥ 10 8

=

= 11◊60

EIs

1◊5 ¥ 10 9

w=

S2 3◊775 ¥ 10 -8

=

= 0◊400

S1 9◊443 ¥ 10 -8

From the diagram in Figure 7 it can now quite easily be determined that the optimum location of the

outrigger will be at x/H = 0·34. Locating the outrigger at a mid-storey level nearest to the theoretical

optimum location, i.e. 28·5 m from the top, and using Equation (32) yields the restraining moment:

3

3

H

Ï w(H - x ) wH ¸ Ï

¸

M=Ì

+

˝Ì

˝

2Cs ˛ Ó (H - x )S1 + HS2 ˛

Ó 6 EIs

3

3

18 ¥ 87 ¸ Ï

87

Ï18(87 - 28◊5 )

¸

=Ì

+

˝

9

8 ˝Ì

(

)

2 ¥ 2◊0 ¥ 10 ˛ Ó 87 - 28◊5 ¥ 9◊443 ¥ 10 -8 + 87 ¥ 3◊775 ¥ 10 -8 ˛

Ó 6 ¥ 1◊5 ¥ 10

= 1◊592 ¥ 10 4 kNm

This is a 23·4% reduction in the bending moment at the base of the shear wall. It is not the maximum

possible moment reduction for the frame; this will occur with the outrigger at a lower position. The

reduction occurs between outrigger level and the base of the structure.

The horizontal deflections and reductions at the top of the structure are given by Equation (33):

wH 4 wH 3 M (H 2 - x 2 ) MH

+

8 EIs

2Cs

2 EIs

Cs

4

3

18 ¥ 87

18 ¥ 87

1◊592 ¥ 10 4 (87 2 - 28◊52 ) 1◊592 ¥ 10 4 ¥ 87

=

+

9

8

8 ¥ 1◊5 ¥ 10

2 ¥ 2◊0 ¥ 10

2 ¥ 1◊5 ¥ 10 9

2◊0 ¥ 10 8

m

= 0◊0859 + 0◊0296 - 0◊0358 - 0◊0069 = 0◊0728m

Ytop =

The total reduction in horizontal deflection is 37·0%. The values in Table 1 give percentage reductions of lateral deflections at the top of the structure and of bending moments in the shear wall for the

optimum outrigger location. This location, xopt, is defined as the position for which the deflection reduction has a maximum value. The bottom row gives the horizontal deflections at the top of the structure. The percentages given in parentheses indicate values obtained from computer stiffness matrix

analyses.

In the column marked ‘Flexible’ all stiffness parameters have been given finite values. In the third

column the rotational stiffness of the shear wall foundation has been given an infinite value, the other

stiffnesses remaining unchanged. Similar reasoning has been applied to columns 4, 5 and 6.

Copyright © 2004 John Wiley & Sons, Ltd.

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SHEAR WALL WITH OUTRIGGER TRUSSES

Table 1. Structural performance

w

gH

xopt. (m)

% Mred

% yred

ytop (mm)

Flexible

Cs = •

Cc = •

GAr = •

Cs = Cc = GAr = •

0·400

11·6

28·5

23·4 (23·5)

37·0 (37·1)

72·8

0·347

•

25·5

18·9 (19·0)

34·6 (34·7)

56·2

0·327

11·6

31·5

25·7 (25·8)

39·8 (39·9)

69·6

0·231

11·6

34·5

29·4 (29·5)

44·3 (44·5)

64·3

0·105

•

34·5

27·1 (27·2)

45·7 (45·9)

46·7

The results show that increasing the stiffness of the exterior column foundations and the racking

shear of the outrigger yields larger reductions for the horizontal deflection and bending moments in

the shear wall. Increasing the stiffness of the shear wall foundation, however, will cause the outrigger to be less effective. The last column assumes infinite stiffnesses for the foundations and the racking

shear of the outrigger. This can lead to seriously underestimating the deflections.

7.

CONCLUSIONS

The analysis of shear wall structures with outriggers on non-fixed foundations subjected to lateral

loading is similar to that of outrigger structures on fixed foundations. The method requires six structural parameters: bending stiffness for the wall (EIs), bending and racking shear stiffnesses for the outrigger (EIr and GAr), an overall bending stiffness contribution from the exterior columns (EIc) and

rotational stiffnesses from the shear wall and column foundations (Cs and Cc).

In the analysis of a shear wall with outrigger trusses on non-fixed foundations, the structural properties can be combined into two characteristic flexibility parameters: S1 for the vertical structural elements, and S2 representing the outrigger and foundation structures.

The simplified model for analysis yields two characteristic non-dimensional parameters: w and gH,

which can be used in a diagram to determine the optimum level of the outrigger structure.

Increasing the rotational stiffness of the shear wall foundation will move the optimum location of

the outrigger higher up the structure but will not improve its efficiency.

Increasing the stiffness of the foundations of the exterior columns will lower the optimum location

of the outrigger structure and improve its performance w.r.t. horizontal deflection and moment

reduction.

Increasing the racking shear stiffness of the outriggers will also lower the optimum location of the

outriggers and improve its performance.

The reductions in horizontal deflections and bending moments of the shear wall are influenced by

all structural parameters. It is therefore suggested that all stiffnesses be included in the preliminary

design of a proposed tall building structure.

APPENDIX A:

RACKING SHEAR STIFFNESS OF BRACING SYSTEMS

Figure 9 shows several types of bracing that can be used in the vertical and horizontal trusses of the

structure. The racking shear stiffnesses of these bracing systems are given for a standard segment with

a length a and a height h. All connections are taken to be pinned with exceptions for K- and kneebracings where the braces are pin connected at the top to continuous beams of length a. It should be

noted that the vertical members do not have any influence on the racking shear stiffness of the segment.

Copyright © 2004 John Wiley & Sons, Ltd.

Struct. Design Tall Spec. Build. 13, 73–87 (2004)

86

J. C. D. HOENDERKAMP

Ar

Ad

d

Ad

d

h

Ad

Ad

Ar

(b)

(a)

a

a

m

e

m

Ar, Ir

Ar

d

h

Ad

Ad

Ar, Ir

Ar

a

Ad

d

(c)

(d)

a

Figure 9. Typical truss bracings

The racking shear stiffness for an X-braced segment as shown in Figure 9(a) is given by

GAi ;X =

2a 2 h

EAd

d3

(A.1)

where Ad is the cross-sectional area of a diagonal and d represents its length.

The racking shear stiffness of the K-braced segment in Figure 9(b) is

GAi ;K =

a 2 hE

2d 3

a3

+

4 Ar

Ad

(A.2)

in which Ar is the cross-sectional area of the horizontal member. Bracing systems with a single diagonal, N-bracing, as shown in Figure 9(c) have the following racking shear stiffness:

GAi ;N =

a 2 hE

d 3 a3

+

Ad Ar

(A.3)

The racking shear stiffness of a full-height knee-braced segment as indicated in Figure 9(d) can be

expressed as

GAi ;KB =

Copyright © 2004 John Wiley & Sons, Ltd.

2 m 2 hE

d 3 m3 m2e2h2

+

+

Ad Ar

6 aIr

(A.4)

Struct. Design Tall Spec. Build. 13, 73–87 (2004)

SHEAR WALL WITH OUTRIGGER TRUSSES

87

in which m is the horizontal distance between column and cord connections to bracing, e is the

horizontal distance between tops of knee-braces and Ir indicates the second moment of area for the

horizontal member.

ACKNOWLEDGEMENTS

The author wishes to record his appreciation of H. H. Snijder, M. C. M. Bakker and M. R. Trouw for

their contributions to this project.

REFERENCES

Hoenderkamp JCD, Snijder HH. 2000. Simplified analysis of façade rigger braced high-rise structures. Structural

Design of Tall Buildings 9: 309–319.

Stafford Smith B, Coull A. 1991. Tall Building Structures. Wiley: New York.

Stafford Smith B, Salim I. 1981. Parameter study of outrigger-braced tall building structures. Journal of the Structural Division, ASCE 107(ST10): 2001–2013.

Copyright © 2004 John Wiley & Sons, Ltd.

Struct. Design Tall Spec. Build. 13, 73–87 (2004)

Struct. Design Tall Spec. Build. 13, 73–87 (2004)

Published online in Wiley Interscience (www.interscience.wiley.com). DOI:10.1002/tal.235

SHEAR WALL WITH OUTRIGGER TRUSSES ON WALL AND

COLUMN FOUNDATIONS

J. C. D. HOENDERKAMP

Department of Architecture and Building, Eindhoven University of Technology, Eindhoven, The Netherlands

SUMMARY

A graphical method of analysis is presented for preliminary design of outrigger truss-braced high-rise shear wall

structures with non-fixed foundation conditions subject to horizontal loading. The method requires the calculation of six structural parameters: bending stiffness for the shear wall, bending and racking shear stiffnesses for

the outrigger, an overall bending stiffness contribution from the exterior columns, and rotational stiffnesses for

the shear wall and column foundations.

The method of analysis employs a simple procedure for obtaining the optimum location of the outrigger up

the height of the structure and a rapid assessment of the influence of the individual structural elements on the

lateral deflections and bending moments of the high-rise structure. It is concluded that all six stiffnesses should

be included in the preliminary analysis of a proposed tall building structure as the optimum location of the outrigger as well as the reductions in horizontal deformations and internal forces in the structure can be significantly

influenced by all the structural components. Copyright © 2004 John Wiley & Sons, Ltd.

1.

INTRODUCTION

The outrigger-braced high-rise shear wall shown in Figure 1 consists of a centrally located wall with

two equal-length trusses positioned at a distance x from the top of the structure. The outriggers connect

the shear wall to columns in the façade of the structure. This way the horizontally applied load will

force the wall to behave compositely with the exterior structure by introducing axial forces in the

columns. These forces form a restraining moment which is in the oposite direction to the bending

moment from the horizontal load. This effect will decrease the bending moments in the wall from outrigger level down to the base and it will reduce the horizontal deflections of the structure. The bending

moments in the shear wall and axial forces in the columns are resisted by foundation structures on

piles which are indicated by translational springs. Combined with the laterally applied loading the

restraining moment will force a triple curvature in the wall up the height and double curvature in the

trusses as shown in Figure 1.

In the horizontal deflection analysis of outrigger-braced shear walls on fixed foundations it has been

shown (Stafford Smith and Coull, 1991; Stafford Smith and Salim, 1981) that the wall can be represented by a single flexural stiffness parameter. The outriggers were assumed to be prismatic members

rigidly connected to the wall and hinge connected to the exterior columns. The resulting single curvature behaviour of these flexural members was represented by a single bending stiffness parameter.

* Correspondence to: J. C. D. Hoenderkamp, Department of Architecture and Building, Structural Design Group, Eindhoven

University of Technology, P.O. Box 513, Route 7 5600 MB Eindhoven, The Netherlands.

E-mail: j.c.d.hoenderkamp@bwk.tue.nl

Copyright © 2004 John Wiley & Sons, Ltd.

Received January 2003

Accepted February 2003

74

J. C. D. HOENDERKAMP

outrigger

x

outrigger

h

H

facade

column

shear

wall

facade

column

piles

b

c

c

b

ground level

deflected shape

Figure 1. Shear wall with outrigger trusses

It was further taken that the columns are pin connected to a fixed foundation and could thus be represented by a parameter which represents the axial stiffnesses of the columns only. With three stiffness parameters representing the wall, outriggers and exterior columns it was possible to combine

them in a single dimensionless parameter which allowed a graphical procedure to obtain the optimum

location of the outriggers such that they would cause the largest reduction in horizontal deflection at

the top of the structure.

Figure 1 also shows simplified models for the foundations of the central shear wall and exterior

columns. In the structural analysis the individual elements are only subjected to forces resulting from

horizontal loading on the structure. The foundations under the pin-connected façade columns are taken

to be piled foundations which act in a vertical direction only. They are modelled as linear springs. The

shear wall foundation is only subjected to a bending moment. The net axial load on this foundation

as a result of lateral loading is zero as it is positioned in the centre of a symmetric structure. The foundation of the wall can thus be modelled by a rotational spring with a rotational spring constant. The

outriggers consist of trusses of which the deformations due to shear cannot be ignored. The shear stiffness of the trusses in addition to their flexural stiffness must be included in the analysis. In order to

allow the truss to be subjected to flexural deformations it will be necessary to assume the structural

floors of the building to be connected to the trusses at the exterior columns and shear wall locations

only; i.e. the floor structure is not part of the outrigger structure but it will cause the shear wall and

exterior columns to have identical rotations at all floor levels. Further simplifying assumptions are:

the structure behaves linear elastically; the sectional properties of the shear wall, exterior columns and

outriggers are uniform througout their height or length; and the distribution of the lateral loading is

uniformly distributed along the height of the structure.

A compatibility equation is to be developed for the rotations in the wall and outrigger at the intersection of the neutral line of the outriggers and the face of the shear wall. This leads to an expression

Copyright © 2004 John Wiley & Sons, Ltd.

Struct. Design Tall Spec. Build. 13, 73–87 (2004)

75

SHEAR WALL WITH OUTRIGGER TRUSSES

for the restraining moment allowing the reduction in horizontal deflection at the top to be determined.

Maximizing this reduction will yield the optimum location of the outrigger.

2.

SHEAR WALL

The rotations in the shear wall are the result of a uniformly distributed horizontal load w, and a restraining moment M caused by a reverse action of the outriggers. They will cause rotations at outrigger

level due to bending in the wall and rotation of the wall foundation as presented in Figure 2. For the

shear walls it will be assumed that plane sections remain plane in bending so that the rotations at the

shear wall centre line are identical to those at the face of the wall.

2.1

Rotations due to lateral loading w

The rotation in the reinforced concrete shear wall at level x can quite simply be expressed as follows:

q s ; b ;w =

w( H 3 - x 3 )

6 EIs

(1)

where H is the total height of the structure, x is the distance measured from the top of the stucture and

EIs is the flexural stiffness of the shear wall. The rotation of the wall due to the rotation of its foundation is

wH 2

2Cs

q s ;Cs ;w =

(2)

in which Cs is the rotational stiffness of the shear wall foundation.

2.2

Rotations due to restraining moment M

The rotation in the shear wall at level x as a result of bending in the wall is given by

q s ;b ;M = -

qs;b;w

qs;b;M

M ( H - x)

EIs

(3)

qs;Cs;w

qs;Cs;M

x

H

EIs

EIs

Cs

Cs

Figure 2. Rotations in shear wall

Copyright © 2004 John Wiley & Sons, Ltd.

Struct. Design Tall Spec. Build. 13, 73–87 (2004)

76

J. C. D. HOENDERKAMP

The rotation of the shear wall and its foundation are

M

Cs

q s ;Cs ;M = -

(4)

It is noted here that the rotations caused by restraining moment M are in the opposite direction to those

by horizontal load w.

3.

OUTRIGGER

The rotations in the outrigger due to the restraining moment are obtained by splitting up this action

as shown in Figure 3, where the outriggers have been separated from the braced frame for clarity. The

restraining moment in the structure due to outrigger action is the product of the axial force in the exterior columns and the distance between them:

M = Fa (2 b + 2c) = Fa (2l)

(5)

in which Fa is the restraining force in the exterior columns, l is the distance from the exterior column

to the centre line of the shear wall and b is the length of the flexible outrigger measured from the

façade column to the outrigger/shear wall interface. Considering the free body diagram of a single

outrigger, then

Fa ¥ b = Fr ¥ h

(6)

where Fr represents the complementary shear forces in the outriggers and h is the vertical dimension

of the outrigger. The free body diagram in the centre of Figure 3 allows an expression for the restraining moment on the shear wall to be written as

M = 2 Fa c + 2 Fr h

Fr

Fa

Fr

Fr

outrigger

Fa

Fa

Fr

x

Fr

Fa

Fa

outrigger

(7)

Fr

Fr

h

Fa

H

Fr

shear

wall

b

c

c

b

Figure 3. Free body diagram restraining forces

Copyright © 2004 John Wiley & Sons, Ltd.

Struct. Design Tall Spec. Build. 13, 73–87 (2004)

77

SHEAR WALL WITH OUTRIGGER TRUSSES

qr;b;Fr

Fr/2

Fr/2

Fr/2

Fr/2

(a)

qr;s;Fr

Fr

h

Fr

b

a

(b)

Figure 4. Bending and shear deformations in truss

Substituting Equations (6) and (7) into Equation (5) will lead to an expression for the effective restraining moment causing bending and shear deformations in the outriggers as shown in Figure 4.

2Fr h =

M

a

(8)

where a dimensionless parameter

a=

3.1

l

b

(9)

Rotations due to restraining force Fr

The double curvature bending in the outrigger as shown in Figure 4(a) is caused by axial strain in the

top and bottom cords. The rotations in the truss at both ends are identical because the structural floors

are assumed to be connected to the outrigger at the columns and shear wall only. The rotation due to

bending at the outrigger/wall interface is given by

q r ;b ;Fr =

2 Fr h

Mb

=

Â (12 EIr b) 24aEIr

(10)

where EIr is the bending stiffness of the outrigger, which can be obtained as follows:

EIr = EAr h 2 2

(11)

where Ar is the cross-sectional area of the top and bottom chords of the rigger.

The rotation due to racking shear results from strain in the diagonals as is indicated in Figure 4(b)

and can be expressed as

Copyright © 2004 John Wiley & Sons, Ltd.

Struct. Design Tall Spec. Build. 13, 73–87 (2004)

78

J. C. D. HOENDERKAMP

q r ;s ;Fr =

2 Fr h

M

=

hÂ GAi haGAr

(12)

It has been shown earlier (Hoenderkamp and Snijder, 2000) that the racking shear stiffness of a rigger,

GAr, is the sum of the individual racking shear stiffnesses of all the bracing segments in that rigger.

So for the outrigger structure

n

GAr = Â GAi

(13)

i =1

where n represents the total number of segments in the two outriggers and GAi is the racking shear

stiffness of a single segment of width a. For several types of bracing systems the racking shear stiffness is given Appendix A.

3.2

Rigid body rotations of outrigger

The wide-column behaviour of the shear wall due to bending caused by horizontal load w will result

in rigid body rotations of the outriggers as shown in Figure 5. This reverse rotation is given by

d s ;b ;w ˆ

c

q r ;b ;w = - Ê

= - (q r ;b ;w )

Ë b ¯

b

(14)

Substituting Equation (1) into Equation (14) yields

3

3

Ï w( H - x ) ¸ c

q r ;b ;w = - Ì

˝

Ó 6 EIs

˛b

(15)

An additional reverse rotation in the trusses occurs due to the rotation of the wall foundation when

subjected to horizontal loading and is shown in Figure 6. This rigid body rotation for the outrigger

can be expressed as follows:

q r ;Cs ;w = - Ê

Ë

d s ;Cs ;w ˆ

c

= - (q s ;Cs ;w )

¯

b

b

(16)

qs;b;w

qr;b;w

qr;b;w

d s;b;w

b

d s;b;w

c

c

b

Figure 5. Rigid body rotation of outriggers due to bending in shear wall

Copyright © 2004 John Wiley & Sons, Ltd.

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79

SHEAR WALL WITH OUTRIGGER TRUSSES

qs;Cs;w

qr;Cs;w

qr;Cs;w

d s;Cs;w

d s;Cs;w

c

b

c

b

Figure 6. Rigid body rotation of outriggers due to rotation of wall foundation

substituting Equation (2) into Equation (16) yields

2

Ï wH ¸ c

q r ;Cs ;w = - Ì

˝

Ó 2Cs ˛ b

(17)

Restraining moment M will also cause rigid body rotations of the outriggers. The shear wall will deflect

in the opposite direction as shown in Figures 5 and 6. The rigid body rotation of the outriggers will

now be in the clockwise direction. For bending in the wall:

q r ;b ;M = - Ê

Ë

c

d s ;b ;M ˆ

= - (q s ; b ;M )

b ¯

b

(18)

Substituting Equation (3) into Equation (18) yields

M ( H - x) ¸ c

q r ;b ;M = ÏÌ

˝

Ó EIs ˛ b

(19)

For rotation of the shear wall foundation:

q r ;Cs ;M = - Ê

Ë

d s ;Cs ;M ˆ

c

= - (q r ;Cs ;M )

b ¯

b

(20)

Substituting Equation (4) into Equation (20) yields

M c

q r ;Cs ;M = ÏÌ ¸˝

Ó Cs ˛ b

3.3

(21)

Rotations due to restraining force Fa

The restraining forces in the exterior columns will cause two more rigid body rotations of the outriggers: one resulting from shortening and lengthening of the exterior columns and another due to vertical displacements in the column foundations.

Copyright © 2004 John Wiley & Sons, Ltd.

Struct. Design Tall Spec. Build. 13, 73–87 (2004)

80

J. C. D. HOENDERKAMP

The first outrigger rotation can be defined as the column change in length divided by the length of

the outrigger. This leads to

q r ;a ;Fa =

Fa ( H - x )

bEAc

(22)

where Ac is the cross-sectional area of the exterior column.

Defining an overall stiffness parameter for the exterior column structure to be

EIc = 2 EAc l 2

(23)

and substituting Equations (5) and (23) into Equation (22) will yield the following expression:

q r ;a ;Fa =

M ( H - x )a

EIc

(24)

The second outrigger rotation can be defined as the foundation displacement divided by the length of

the outrigger, which yields

q r ;Cc ;Fa =

Fa k aM

= 2

b

2l k

(25)

where k represents the translational stiffness of the column foundation. Defining an overall rotational

stiffness parameter for the combined column foundations as

Cc = 2 l 2 k

(26)

and substituting Equation (26) into Equation (25) leads to the following simplified expression:

q r ;Cc ;Fa =

4.

aM

Cc

(27)

HORIZONTAL DEFLECTION

Compatibility in rotation at level x requires that the rotations in the shear wall, qs, and outrigger,

qr, at their interface are identical. Substituting for the individual rotations yields the following

expression:

3

3

2

w( H 3 - x 3 ) wH 2 M ( H - x ) M

Ï w( H - x ) ¸ c Ï wH ¸ c

+

= -Ì

˝ -Ì

˝

6 EIs

2Cs

EIs

Cs

Ó 6 EIs

˛ b Ó 2Cs ˛ b

Mb

M

M ( H - x ) ¸ c Ï M ¸ c aM ( H - x ) a M

+

+

+ ÏÌ

+

˝ +Ì ˝ +

24aEIr haGAr Ó EIs ˛ b Ó Cs ˛ b

EIc

Cc

(28)

Simplifying leads to

w( H 3 - x 3 ) wH 2

H-x H-x

b

1

1

1

+

= M ÏÌ

+

+

+

+

+ ¸˝

2

2

6 EIs

2Cs

EIc

Ó EIs

24a EIr ha GAr Cs Cc ˛

Copyright © 2004 John Wiley & Sons, Ltd.

(29)

Struct. Design Tall Spec. Build. 13, 73–87 (2004)

SHEAR WALL WITH OUTRIGGER TRUSSES

81

Setting two characteristic flexibility parameters as

H

H

+

EIs EIc

(30)

b

1

1

1

+

+

+

24a 2 EIr ha 2 GAr Cs Cc

(31)

S1 =

and

S2 =

leads to the following expression for the restraining moment:

3

3

2

H

Ï w( H - x ) wH ¸Ï

¸

M=Ì

+

˝Ì

˝

(

)

2Cs ˛Ó H - x S1 + HS2 ˛

Ó 6 EIs

(32)

The horizontal deflection at the top of the building can be now obtained as follows:

ytop =

wH 4 wH 3 M ( H 2 - x 2 ) MH

+

8 EIs

2Cs

2 EIs

Cs

(33)

where the first two terms on the right-hand side represent the ‘free’ horizontal deflections of the shear

wall at the top due to bending of the wall and rotation of its foundation as a direct result of lateral

loading. The third term is due to the restraining moment M which causes a reverse deflection and rotation at outrigger level, resulting in a deflection reduction at the top. The last term represents a horizontal deflection at the top of the wall due to reverse foundation rotation.

5.

OPTIMUM LOCATION OF OUTRIGGER

The reduction in horizontal deflection at the top of the structure due to restraining moment M is

represented by the last two terms on the right-hand side of Equation (33) and can be expressed as

2

2

H¸

ÏH - x

yred = M Ì

+ ˝

Cs ˛

Ó 2 EIs

(34)

This reduction is maximized by differentiating it w.r.t. x, setting it equal to zero and solving for x.

After simplifying this leads to

d ÈÏ

5 - 3x 2 - 2 x 3

6 ¸Ï

1

¸˘ = 0

2

3

5

+

˝˙

ÍÌ1 - x - x + x +

2 ˝Ì

(g H )

dx ÎÓ

(g H ) ˛Ó1 - x + w ˛˚

(35)

in which two characteristic non-dimensional parameters for the outrigger-braced shear wall structure

have been set as follows:

gH =

Copyright © 2004 John Wiley & Sons, Ltd.

HCs

EIs

(36)

Struct. Design Tall Spec. Build. 13, 73–87 (2004)

82

J. C. D. HOENDERKAMP

0.2

Rigger Location, x/H

0.3

0.4

Values of g H

> 100

0.5

10

6

0.6

4

3

2

0.7

0.0

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9

1.0

Values of w

Figure 7. Optimum location of outrigger

and

w=

S2

S1

(37)

A dimensionless location parameter for the optimum position of the outrigger has been set as

x=

x

H

(38)

Figure 7 shows a graphical presentation of optimum outrigger locations up the height of a shear wall

braced structure on wall and column foundations as a function of two non-dimensional characteristic

parameters, gH and w.

6.

EXAMPLE

The structural floor plan of a 29-storey, 87 m high building in Figure 8 shows the arrangement of four

identical shear walls with horizontal steel trusses on both sides. Each truss is 9 m long, has a single

storey height of 3 m and comprises five X-braced segments. The building is subjected to a uniformly

distributed lateral load of 1·6 kN/m2. The flexural stiffness of the reinforced concrete shear wall EIs =

1·5 ¥ 109 kNm2. For the horizontal trusses: top and bottom chords Ar = 1·78 ¥ 10-2 m2, diagonals Ad =

9·726 ¥ 10-3 m2 and for the exterior columns Ac = 3·12 ¥ 10-2 m2. The elastic modulus of steel E = 2·1

¥ 108 kN/m2. The rotational stiffness of the shear wall foundation Cs = 2·0 ¥ 108 kNm and the translational stiffness of the column foundations k = 4·0 ¥ 105 kN/m.

The detailed calculations are shown here for a single shear wall with one outrigger on both sides.

The flexural stiffness of the outrigger structure is given by Equation (11):

Copyright © 2004 John Wiley & Sons, Ltd.

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83

SHEAR WALL WITH OUTRIGGER TRUSSES

Shear walls with outriggers

9m

9m

9m

9m

9m

9m

9m

9m

w

Figure 8. Structural floor plan

EIr = EAr h 2 2 = 2◊1 ¥ 10 8 ¥ 1◊78 ¥ 10 -2 ¥ 32 2 = 1◊682 ¥ 10 7 kNm 2

The racking shear stiffness is the sum for 10 segments. Equation (A.1) yields

GAr = 10

2a 2 h

2 ¥ 18

◊ 2¥3

EA

=

¥ 2◊1 ¥ 10 8 ¥ 9◊726 ¥ 10 -3 = 9◊272 ¥ 10 6 kN

d

1◊ 5

d3

(18

◊ 2 + 32 )

The global second moment of area of the exterior columns can be obtained from Equation (23):

2

9

EIc = 2 EAc l 2 = 2 ¥ 2◊1 ¥ 10 8 ¥ 312

◊ ¥ 10 -2 ¥ Ê 9 + ˆ = 2◊388 ¥ 10 9 kNm 2

Ë

2¯

An overall rotational stiffness parameter for the foundations under the exterior columns is given by

Equation (26):

Cc = 2 l 2 k = 2 ¥ 13◊52 ¥ 4 ¥ 10 5 = 1◊458 ¥ 10 8 kNm

The dimensionless parameter a in Equation (9) yields

a=

l 13◊5

=

= 1◊5

b

9

The characteristic parameters S1 and S2 are given by Equations (30) and (31):

S1 =

H

H

87

87

-1

+

=

+

= 9◊443 ¥ 10 -8 (kNm)

EIs EIc 1◊5 ¥ 10 9 2◊388 ¥ 10 9

Copyright © 2004 John Wiley & Sons, Ltd.

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84

J. C. D. HOENDERKAMP

b

1

1

1

+

+

+

2

2

C

C

24a EIr ha GAr

s

c

9

1

1

1

-1

=

+

+

+

= 3◊775 ¥ 10 -8 (kNm)

2

7

2

6

8

8

24 ¥ 1◊5 ¥ 1◊682 ¥ 10

3 ¥ 1◊5 ¥ 9◊272 ¥ 10

2◊0 ¥ 10

1◊458 ¥ 10

S2 =

The characteristic non-dimensional parameters for the structure can now be obtained from Equations

(36) and (37):

gH =

HCs 87 ¥ 2◊0 ¥ 10 8

=

= 11◊60

EIs

1◊5 ¥ 10 9

w=

S2 3◊775 ¥ 10 -8

=

= 0◊400

S1 9◊443 ¥ 10 -8

From the diagram in Figure 7 it can now quite easily be determined that the optimum location of the

outrigger will be at x/H = 0·34. Locating the outrigger at a mid-storey level nearest to the theoretical

optimum location, i.e. 28·5 m from the top, and using Equation (32) yields the restraining moment:

3

3

H

Ï w(H - x ) wH ¸ Ï

¸

M=Ì

+

˝Ì

˝

2Cs ˛ Ó (H - x )S1 + HS2 ˛

Ó 6 EIs

3

3

18 ¥ 87 ¸ Ï

87

Ï18(87 - 28◊5 )

¸

=Ì

+

˝

9

8 ˝Ì

(

)

2 ¥ 2◊0 ¥ 10 ˛ Ó 87 - 28◊5 ¥ 9◊443 ¥ 10 -8 + 87 ¥ 3◊775 ¥ 10 -8 ˛

Ó 6 ¥ 1◊5 ¥ 10

= 1◊592 ¥ 10 4 kNm

This is a 23·4% reduction in the bending moment at the base of the shear wall. It is not the maximum

possible moment reduction for the frame; this will occur with the outrigger at a lower position. The

reduction occurs between outrigger level and the base of the structure.

The horizontal deflections and reductions at the top of the structure are given by Equation (33):

wH 4 wH 3 M (H 2 - x 2 ) MH

+

8 EIs

2Cs

2 EIs

Cs

4

3

18 ¥ 87

18 ¥ 87

1◊592 ¥ 10 4 (87 2 - 28◊52 ) 1◊592 ¥ 10 4 ¥ 87

=

+

9

8

8 ¥ 1◊5 ¥ 10

2 ¥ 2◊0 ¥ 10

2 ¥ 1◊5 ¥ 10 9

2◊0 ¥ 10 8

m

= 0◊0859 + 0◊0296 - 0◊0358 - 0◊0069 = 0◊0728m

Ytop =

The total reduction in horizontal deflection is 37·0%. The values in Table 1 give percentage reductions of lateral deflections at the top of the structure and of bending moments in the shear wall for the

optimum outrigger location. This location, xopt, is defined as the position for which the deflection reduction has a maximum value. The bottom row gives the horizontal deflections at the top of the structure. The percentages given in parentheses indicate values obtained from computer stiffness matrix

analyses.

In the column marked ‘Flexible’ all stiffness parameters have been given finite values. In the third

column the rotational stiffness of the shear wall foundation has been given an infinite value, the other

stiffnesses remaining unchanged. Similar reasoning has been applied to columns 4, 5 and 6.

Copyright © 2004 John Wiley & Sons, Ltd.

Struct. Design Tall Spec. Build. 13, 73–87 (2004)

85

SHEAR WALL WITH OUTRIGGER TRUSSES

Table 1. Structural performance

w

gH

xopt. (m)

% Mred

% yred

ytop (mm)

Flexible

Cs = •

Cc = •

GAr = •

Cs = Cc = GAr = •

0·400

11·6

28·5

23·4 (23·5)

37·0 (37·1)

72·8

0·347

•

25·5

18·9 (19·0)

34·6 (34·7)

56·2

0·327

11·6

31·5

25·7 (25·8)

39·8 (39·9)

69·6

0·231

11·6

34·5

29·4 (29·5)

44·3 (44·5)

64·3

0·105

•

34·5

27·1 (27·2)

45·7 (45·9)

46·7

The results show that increasing the stiffness of the exterior column foundations and the racking

shear of the outrigger yields larger reductions for the horizontal deflection and bending moments in

the shear wall. Increasing the stiffness of the shear wall foundation, however, will cause the outrigger to be less effective. The last column assumes infinite stiffnesses for the foundations and the racking

shear of the outrigger. This can lead to seriously underestimating the deflections.

7.

CONCLUSIONS

The analysis of shear wall structures with outriggers on non-fixed foundations subjected to lateral

loading is similar to that of outrigger structures on fixed foundations. The method requires six structural parameters: bending stiffness for the wall (EIs), bending and racking shear stiffnesses for the outrigger (EIr and GAr), an overall bending stiffness contribution from the exterior columns (EIc) and

rotational stiffnesses from the shear wall and column foundations (Cs and Cc).

In the analysis of a shear wall with outrigger trusses on non-fixed foundations, the structural properties can be combined into two characteristic flexibility parameters: S1 for the vertical structural elements, and S2 representing the outrigger and foundation structures.

The simplified model for analysis yields two characteristic non-dimensional parameters: w and gH,

which can be used in a diagram to determine the optimum level of the outrigger structure.

Increasing the rotational stiffness of the shear wall foundation will move the optimum location of

the outrigger higher up the structure but will not improve its efficiency.

Increasing the stiffness of the foundations of the exterior columns will lower the optimum location

of the outrigger structure and improve its performance w.r.t. horizontal deflection and moment

reduction.

Increasing the racking shear stiffness of the outriggers will also lower the optimum location of the

outriggers and improve its performance.

The reductions in horizontal deflections and bending moments of the shear wall are influenced by

all structural parameters. It is therefore suggested that all stiffnesses be included in the preliminary

design of a proposed tall building structure.

APPENDIX A:

RACKING SHEAR STIFFNESS OF BRACING SYSTEMS

Figure 9 shows several types of bracing that can be used in the vertical and horizontal trusses of the

structure. The racking shear stiffnesses of these bracing systems are given for a standard segment with

a length a and a height h. All connections are taken to be pinned with exceptions for K- and kneebracings where the braces are pin connected at the top to continuous beams of length a. It should be

noted that the vertical members do not have any influence on the racking shear stiffness of the segment.

Copyright © 2004 John Wiley & Sons, Ltd.

Struct. Design Tall Spec. Build. 13, 73–87 (2004)

86

J. C. D. HOENDERKAMP

Ar

Ad

d

Ad

d

h

Ad

Ad

Ar

(b)

(a)

a

a

m

e

m

Ar, Ir

Ar

d

h

Ad

Ad

Ar, Ir

Ar

a

Ad

d

(c)

(d)

a

Figure 9. Typical truss bracings

The racking shear stiffness for an X-braced segment as shown in Figure 9(a) is given by

GAi ;X =

2a 2 h

EAd

d3

(A.1)

where Ad is the cross-sectional area of a diagonal and d represents its length.

The racking shear stiffness of the K-braced segment in Figure 9(b) is

GAi ;K =

a 2 hE

2d 3

a3

+

4 Ar

Ad

(A.2)

in which Ar is the cross-sectional area of the horizontal member. Bracing systems with a single diagonal, N-bracing, as shown in Figure 9(c) have the following racking shear stiffness:

GAi ;N =

a 2 hE

d 3 a3

+

Ad Ar

(A.3)

The racking shear stiffness of a full-height knee-braced segment as indicated in Figure 9(d) can be

expressed as

GAi ;KB =

Copyright © 2004 John Wiley & Sons, Ltd.

2 m 2 hE

d 3 m3 m2e2h2

+

+

Ad Ar

6 aIr

(A.4)

Struct. Design Tall Spec. Build. 13, 73–87 (2004)

SHEAR WALL WITH OUTRIGGER TRUSSES

87

in which m is the horizontal distance between column and cord connections to bracing, e is the

horizontal distance between tops of knee-braces and Ir indicates the second moment of area for the

horizontal member.

ACKNOWLEDGEMENTS

The author wishes to record his appreciation of H. H. Snijder, M. C. M. Bakker and M. R. Trouw for

their contributions to this project.

REFERENCES

Hoenderkamp JCD, Snijder HH. 2000. Simplified analysis of façade rigger braced high-rise structures. Structural

Design of Tall Buildings 9: 309–319.

Stafford Smith B, Coull A. 1991. Tall Building Structures. Wiley: New York.

Stafford Smith B, Salim I. 1981. Parameter study of outrigger-braced tall building structures. Journal of the Structural Division, ASCE 107(ST10): 2001–2013.

Copyright © 2004 John Wiley & Sons, Ltd.

Struct. Design Tall Spec. Build. 13, 73–87 (2004)

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