11 shear walls

47
1 Shear walls column stiffening element stiffness: EI wall >> EI column

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1

Shear walls

• column – stiffening element

• stiffness: EIwall >> EIcolumn

2

Rigidity of stiffening elements

H2 >> H1Rigid walls – decrease of deflections δ and more simple foundations

Stiffening elements

Column systems without stiffening elements –deflections and big bending moments

Almost all lateral loads –carried by shear walls and stiffening elements

3

Shear walls (stiffening elements)

• contribute to overall stiffness of the building + provide stability

• correct design of shear walls = condition of reliability and efficiency of the structure

• transfer lateral loads to basement• shall be designed so that all lateral load is

carried by shear walls (stiffening elements)

Principles of arrangement of stiffening elements

• use of existing structural members →• stiffening elements shall carry all lateral load• stiffening elements shouldn't restrain

deformations due to shrinkage, creep and temperature changes

• avoid torsion of the building• stiffening elements shall carry a big part of

vertical loads• stiffening elements should have big moments

of inertia• of stiffening elements´ axis shouldn't intersect

in one point

4

no point of intersection of stiffening element axis

Behaviour and design of shear walls

• according to governing deformation

Deformation• predominantly bending• predominantly shear

5

• If walls with different type of deformation are combined in one structure, bending and shear deformation (stiffness) must be considered in analysis.

• Deflection due to – Bending– Shear (usually neglected)

6

Stiffness of walls

Stiffness in bending

4% (16%) proportional share of shear in total stiffness – shear effect neglected

Combined stiffness Stiffness in shear

32

≥HB

51

81

HB

32

51

≥≥HB

Stiffness of walls

(Mohr principle)• Load the dual beam with moment

factored by stiffness

• Deflection ≈ moment on dual beam due to factored load M

EIWHM =

EIWHHH

EIWHy

332

21 3

==

7

Stiffness of walls

Stiffness = load that cause deflection equal to one

yHEIW 3

3= 3

3HEIK =

Stiffness in bending

3

3HEIK =

8

Shear stiffness

deflection due to shear

shear stiffness

9

Shear stiffness

HAGK S κ

=

Combined (complex) stiffness

shearbendingcombined KKK111

+=

10

Procedure of analysis of a building with shear walls

1st step • Divide lateral load to particular stiffening

elements (shear walls)

2nd step• Design of the wall• Check the wall stability (or max. deflection)

3rd step• Design of reinforcement

Dividing of load to walls –according to stiffness of walls

11

• stiffening elements + other loadbearingelements = statically indeterminate system

• simplified idealisation – preliminary design– interpretation and check of detailed analysis

Proportioning of lateral load to stiffening elements

According to layout• statically determinate

- proportioning from equilibrium of forces- (the most simple case: shear walls)

• statically indeterminate- further simplifying assumptions

12

Simplified method – idealisation

• flooring plates rigid for transferring lateral load

• pinned connection of floor slab and stiffening element, rigidity of the floor slab in its plane

Idealisation• stiffness of shear walls is

negligible perpendicular to the plane of the wall and negligible stiffness in torsion; stiffness of other vertical elements is also negligible

• stiffening elements don't change section along height

• buildings with rectangle shape may be analyzed separately with two directions of lateral load only

EIy>>EIx

EIx≈ 0, stiffness in torsion ≈ 0

wx

wy

13

• According to layout of stiffening elements the analysis is performed as:– 2D or – 3D.

• Stiffening systems: – statically determinate– statically indeterminate

For statically determinate systems the load carried by one wall may be determined from equilibrium conditions.

– graphically– by numerical calculations

14

Graphical determination of load carried by shear walls in statically determinate system

Calculation of load carried by one wall (statically determinate system)

Here: stiffnesses of all walls are equal

15

Dividing of wind load to particular walls (statically determinate system) – example

Wind load could be solve for 2 directions (x and y) and separately

Equilibrium conditions (in a plan): forces, momentsHere: stiffnesses of all walls are equal

12 12 6

lx = 30

B

C

A

4 x

A =

2

l y =

12

6 Wx

Wy y

x

cmcmcm EEIE ⋅=⋅⋅= 6,362,0121 3

xA WW =

152302

22xyxy

x

yC

WWWWlTW

W +=⋅

+=+=1522

xy

x

yB

WWlTW

W −=−=

12 12 6

lx = 30

B

C

A

4 x

A =

2

l y =

12

6 Wx

Wy y

x

Height of the building 33m. Three walls of same dimensions (B=6m, H=33m, t=0,2m) and same stiffness.

Load Wx: carried by wall A

Wall A is in nonsymetric position - Wx applies a torsional moment T= Wx .xA

Load Wy is carried by walls B and C. WB =WC= Wy /2Torsion: WB = -T/l, WC= T/l

Total lateral load for walls B,C:

16

Dividing of wind load to particular walls –general procedure

Due to lateral load the floor slab moves and rotates. 3 unknowns: translations Δx, Δy, rotation φ→ 3 equilibrium equations

∑=

=n

iixx WW

1∑

=

=m

ijyy WW

1

∑ ∑−=− ixijyjyxixyj yWxWeWeW

Origin – in centre of gravity

n

x i ii

K y1

0=

=∑m

yj jj

K x1

0=

=∑

stiffness of wall j in direction y stiffness of wall i in direction x

17

translation of point

forces in walls due to translation and rotation

i ix x yϕ∆ = ∆ −

i iy y xϕ∆ = ∆ +

( )xi i xiW x y Kϕ= ∆ −

( ) yjjyj KxyW ϕ+∆=

From equilibrium conditions we obtain Δx, Δy, φ

Consequently load of particular walls may be determined

n

x i ii

K y1

0=

=∑m

yj jj

K x1

0=

=∑

i ix x yϕ∆ = ∆ − i iy y xϕ∆ = ∆ +

( )xi i xiW x y Kϕ= ∆ −( ) yjjyj KxyW ϕ+∆=

new system of coordinatesorigin of coordinates – centroid of stiffness (centroid of bending)

translations of point

forces in walls due to translation and rotation

18

equilibrium of forces and moments (load and forces due to translations and rotation)

∑=

=n

iixx WW

1∑

=

=m

ijyy WW

1∑ ∑−=− ixijyjyxixyj yWxWeWeW

substitution (unknown Δx, Δy, φ) ;

∑=

=∆ n

ixi

x

K

Wx

1∑

=

=∆ m

jjy

y

K

Wy

1∑ ∑

= =

+

−= n

i

m

jjjyixi

yxxy

xKyK

eWeW

1 1

22ϕ

( ) yjjyj KxyW ϕ+∆=

calculation of forces in particular walls:

effect of bending effect of torsion

After the total load is distributed to particular shear walls→

design and check of each wall

19

Design of shear walls

• common requirement: no tensile stresses at bottom of the wall for operational load(repeated changing of compression and tensile is unsuitable)

Tensile at bottom section of the wall

20

Shear wall - stability check

• Check of stability – EQU • possible demand: check of stress at the

bottom-section of the wall (service load; characteristic value) : no tension

• Check of max deflectionfmax ≤ flim

Stability of a wall

Mdst ≤Mstb

Stability condition – EQU

Destabilizing moment

Stabilizing moment

γG = 1,1 γQ = 1,5 γG = 0,9 γQ = 0

21

Note:

event. check of the stability of the whole system (building)

Elastic behaviour of wall

M N V

principle tensile ~ fct → reinforcement according to detailing

22

Walls with predominantly shear deformation

yI

MAN

x +=σ 0=yσItSV

⋅⋅

22

2,1 22τ

σσσσσ +

+= yxyx

ctdf≤1σ ctdf⟩1σ

M, N, V

Linear analysis

23

• Heavily loaded walls with shear straining could be analysed by means of strut-and-tie method.

Tall walls (predominantly bending deformation)

• simplified analysis method: reinforcement design from normal stresses at bottom section of the wall

• Method is suitable for walls ca 5 m long

24

Simplified analysis of a tall wall

1. check of the stress at bottom

2. stability check

3. reinforcement design

γG = 1 γQ = 1

γG = 1,35 γQ = 1,5

γG = 1,1 γQ = 1,5

γG = 0,9 γQ = 0

γG = 1 γQ = 1,5 γQ =0

≤51

81

HB

Wall = cantilever, stress at bottom section 1-1´

25

recapitulation: Simplified analysis of tall wall (dominating bending stiffness) –procedure:

1. total horizontal load2. dividing of load to particular walls (according to

stiffness of walls)3. design of a wall

i. check of the stress at bottoma. wind load (γQ = 1) and vertical load (γG = 1 γQ = 1)

ii. stability check iii. reinforcement design :

a. max. wind load (γQ = 1,5) and max. vertical load (γG = 1,35γQ = 1,5)

b. max. wind load (γF = 1,5) a min. vertical load (γG = 1) → if tensile occurs – design tensile reinforcement

ad 3) wall design

charakteristické γF návrhové γF návrhové γF návrhovéStálé

0,9 1,1 1,35

Celkem stálé g k = g d = g d = g d = Nahodilé

1,5 1,5 1,5

q k = q d = q d = q d = Celkem (g+q) k = (g+q) d = (g+q) d = (g+q) d =

26

ad 3) wall design

ad 3) wall design– stability

γG = 1,1 γQ = 1,5

γG = 0,9 γQ = 0

27

ad 3) wall design – reinforcement design

• combination M + N

– moment from the central plane– slenderness– imperfections

wind load

ad 3) wall design – reinforcement designwind load + vertical loads

1 m

4,4 m

γG = 1,35 γQ = 1,5 vertical loads γG = 1, γQ = 0

horizontal loads γG = 1,35 γQ = 1,5

tensile reinforcement

28

ad 3) wall design – reinforcement design

N [kN]

-6000,0

-5000,0

-4000,0

-3000,0

-2000,0

-1000,0

0,0

1000,0

2000,0

-200,0 -150,0 -100,0 -50,0 0,0 50,0 100,0 150,0 200,0

M [kNm]

[M Ed,N Ed] tlak

[M Ed,N Ed] tah

ad 3) wall design – reinforcement design

Compression in footing section: (2nd order effect - slender, short column)

Tensile in footing section:

Main (vertical) reinforcementAs,min =0,002Ac (each surface 1/2 As,min )spacing ≤3t, ≤400mm

ydSt fAN ⋅=

sScdCEd AfAN σ⋅+=

29

Procedure of analysis of a building with shear walls

1st step • Divide lateral load to particular stiffening

elements (shear walls)

2nd step• Design of the wall• Check the wall stability (or max. deflection)

3rd step• Design of reinforcement

30

Simplified method of analysis of high wall with bending stiffness (shear effect neglected)

No tensile stress for service (operational) load in footing bottom

Design of reinforcement (compression force) -load situations

• max. lateral load + min. vertical load • max.lateral load + max. vertical load

51

81

HB

ad 3) wall design– stability

γG = 1,1 γQ = 1,5

γG = 0,9 γQ = 0

31

max. wind load (γQ = 1,5) and max. vertical load (γG = 1,35γQ = 1,5)

max. wind load (γF = 1,5) a min. vertical load (γG = 1) → if tensile occurs – design tensile reinforcement

Reinforcement design – normal forces

NRd = 0,8 Ac fcd + As σs

NRd = As fyd

Stiffening walls with openingsa) flexible cross beams - coupled wallsb) stiff cross beams - one wallc) medium cross beams

32

Flexible coupling beam

M M lI I1

1

1 2=

+

M M lI I2

2

1 2=

+

two cantilevers - connected - same deformation- proportion of moment - due to stiffness

Stiff cross beams

design of coupling (cross) beam

One wall in calculation of moment of inertia Ired use section

33

Z t hi i= τ

τiSid

red

V St I

=

ZV Sh

IiSid

red=

M Z lSd i=12

V ZSd i=

Design of coupling (cross) beam

shear force in i-storey:

substitution:

Reinforcement in coupling beam:

Medium cross beams

Stiffness of cross beam {0, ∞}2

21 HwM ⋅=

lZMMM ⋅++= 21

ZZ

M2

M1

l

34

Medium cross beams

Structure could be analysed as a frame:

part of the tie-beam has infinite stiffness

part at the opening is flexible

Reinforcement of coupling beam

• cantilever• ties• edge reinforcement• diagonal reinforcement• edge of the wall – U-shaped reinforcement

35

Reinforcement - detailing

Main - longitudinal reinforcement

As,min =0,002Ac (at least 1/2 As,min each edge of wall)spacing ≤3t, ≤400mm

As,max =0,04Ac ......... 0,08 Ac (in lapping area)

Secondary (horizontal) reinforcement

As ≥ 25% of main As

≥ 0,001Ac

spacing ≤ 400mm

Purpose:

• keep position of main reinforcement during grouting

• shrinkage cracking, cracks due to temperature changes

36

Ties: if As of main reinf. ≥ 2% Ac

spacing and ∅ − same as for columns

∅ ≥1/4 ∅ of main reinf.

≥ 6mm

spacing ≤ 12 ∅ of main reinf.

≤ t

≤ 300mm

reduced spacing of ties below and above ceiling – distance 4t

at least 4 ties per 1m2 of wall

If reinf. is provided as a weld mesh : φ ≤ 16 mm, cover > 2 φ.

buď

nebo

either

or

ties

lappingties

37

38

39

40

41

example

Wx

Wy

lx = 36

l y =

24

A

B

E D

C

x x0

y0 y

x0 = 4 xD = 2

yB

= -9

y

0 = 3

yA

= 9

xC = -16 xE = 14

6 6 6 6 6 6

6

6

6

6

Wx = 800 kN Wy = 1200 kN.

43

)18612(=

++−=

++⋅+⋅+⋅

=I

IIII

xIxIxIxEDC

EEDDCCo

32

)612(=

−=

+⋅+⋅

=I

III

xIxIyBA

BBAAo

centre of bending :

translations, rotation

resultant forces in walls

IEIEIEIEW

K

WxcmcmBcmAcm

xn

ixi

x 4002800

==+

==∆

∑ IEIEIEIEIEw

K

Wy

cmcmEcmDcmCcm

xn

jyj

y 40031200

==++

==∆

( )( ) IEIIIIIExKyK

eWeW

cmEDCBAcmn

i

m

ijjyixi

yxxy 88,31421699

38004120022222

1 1

22−=

⋅+⋅+⋅+⋅+⋅⋅+−⋅

=+

−=

∑ ∑= =

ϕ

( ) ( ) 365988,3400 =⋅−=⋅−∆= AcmAA IEyxW ϕ

( ) ( ) 435988,3400 =⋅+=⋅−∆= BcmBB IEyxW ϕ

( ) ( ) 9,3371688,3400 =⋅−=⋅+∆= CcmCC IExyW ϕ

( ) ( ) 8,407288,3400 =⋅+=⋅+∆= DcmDD IExyW ϕ

( ) ( ) 4,4541488,3400 =⋅+=⋅+∆= EcmEE IExyW ϕ

42

Proportioning of lateral load –statically indeterminate system

• different stiffness of different elements (different character of deformation)

simplifying assumptions for deformation - composite element

Design and analysis of combinedsystems

Effect of loads are usually solved separately for parts of the building or particular elements (assumed elastic

behaviour).

43

Analysis model – stability

• stiffening elements + other load-bearing elements = statically indefinite system

• numerical model – computer analysis• laboratory tests (tunnel)• simplified idealisation • deformation properties of the idealised structure

shall not differ from real behaviour

Analysis model

• In linear analysis effect of vertical and lateral loads may be analysed separately and superimposed

• The whole structure may be divided to structural parts or elements and these are solved separately

• Non-linear solution

44

Different behaviour of vertical elements

• different properties – stiffness –simplification – idealisation

Char. deformation of column system, wall system and combination

frame wall combined structure

45

linear solution – simplification • separate effects of loads → and other• divide to particular elements• analyse particular elements

• (nonlinear solution – all loads analysed together)

Structural model

• column – stiffening element

• stiffness: EIwall >> EIcolumn

46

Combination of different types of elements in one system

Interaction of different types of vertical elements – idealisation

• composite prism• substitute stiffness (continuous continuum

along the building height

RIGID slabs

47

Dividing of effects of loads to particular elements

• assumptions:– slabs rigid in their central plane– pinned connection of slabs and vertical stiffening

elements– slabs transfer load to vertical stiffening elements – stiffening element = cantilever fixed to basement

• According to arrangement of stiffening elements in a plan – plane or spatial (3D) system is assumed