shear wall design 2

8/9/2019 Shear Wall Design 2

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Seismic Design of

ConcreteStructure


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Seismic Design of ConcreteStructure

• Earthquakes occur in many regions of the world. In certain locations where the intensityof the ground shaking is small, the designerdoes not have to consider seismic eects.

• In other locations-articularly in regions nearan active geological fault !a fracture line inthe rock structure", such as the San #ndreasfault that runs along the western coast ofCalifornia-large ground motions frequently

occur that can damage or destroy $uildingsand $ridges in large areas of cities

• #ssuming the $uilding is %&ed at its $ase, thedislacement of 'oors will vary from (ero at

the $ase to a ma&imum at the roof


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Earthquake focus and

eicenter


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Distri$ution of magnitude) or greater earthquakes,

*+ - *++


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here are several analytical rocedures todetermine the magnitude of the $ase shearfor which $uildings must $e designed, we will

only consider the equivalent lateral force procedure, descri$ed in the #/SI0#SCE and 12C standard. 1sing this

rocedure, we comute the magnitude of the$ase shear as


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a$le 34Design Coeicients 5 6actors for 2asic Seismic-6orce-7esistin


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Story Drift 5 he 8-Delta Eect


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Allowable Story Drift, ∆


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Important Factor


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9verturning


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E&amle *

• Determine the design seismic forces acting at eachoor of the six-story oice building in Figure below

• D.L=8kN/m2

• L.L=2.5kN/m2

• Shear wall sstem

!"m

#$m

N


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( )

( )

( )

( )

kN

I R

W S V

kN W IS V

kN V

I

Wall Shear R

T

kN W S

I RT

W S V

DS

DS

D

D

1134

14

)21600(21.0

20021600)21.0)(1(0441.0)(0441.0

1169)(462.0

2160010.0

1

4

462.0200488.0

21600)6)(18)(25(810.0

max

min

14

1

1

43

===

===

==

=

===

===

=


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T=0.462 < 0.5 then k=1


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Note:

Column 4 = column 3a * Column 2

Fx = column 5 * (V = 1134)))

Vx = cumulative column 6

Mx = moment !om Fx can calculate" !om #$ea! %!ea


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Shear :all Design


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Shear walls &!ovi"e a$i'$ in&lane #tine## an"

#t!en't$ o! ot$ late!al an"

'!avit loa"#+ an" a!e i"eall

#uitale o! tall uil"in'#+

e#&eciall t$o#e conceive" in

!eino!ce" conc!ete. ,all

uil"in'# "e#i'ne" to ca!! t$e

enti!e late!al loa"in' t$!ou'$

#$ea! -all# can e economical to$ei'$t# o a!oun" 40 #to!ie#.

,alle! #t!uctu!e# #$oul" comine

#$ea! -all# -it$ ot$e! #t!uctu!al

##tem#.

Introduction


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Shear :all design


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Shear :all Design Stes

I

MC

A

P

l !!h I

"

## $

±=

==1212

33

2.0 "" <

!- %alculate &xternal 'oad

%

%

%

P

V

M

#- boundary element chec(

)hen boundary elements are not re*uiredif


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#! l l )25.02.0( −≈

#l

#!

#! l l )25.02.0( −≈

!! #!

#l

)hen choose the one of+


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,- 'ongitudinal einforcement

At least two curtains of reinforcement are needed

in the wall if the in-plane factored shear exceed a .alue of

6

""& A

##"& !l A =

+==⇒

=

+==⇒

>

'n

"

"&n%#

#

'n

"

"&n%

#

#

AV V l

hi

AV V l

hi

ρ φ φ

ρ φ φ

4 5.1

6 2

( ) 6

4

6.0

max

""&

n

A

V =

=φ


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/inimum reinforcement

6

0025.0

0025.0

6

""&

%

h

&

""&

%

AV i

A

V i

<

==

>

ρ

ρ

0025.0

0015.0

16 a!o!

0020.0 0012.0

16 a!o!

=

=>

==

≤

h

&

h

&

ρ

ρ

φ

ρ ρ

φ


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0- 1erify ade*uacy of shear wall section at itsbase under combined axial load and bendingmoment

)2(inent!eino!cemo a!ea

element# oun"a!*inent!eino!cemo a!ea

"ia'!amninte!actio+

!# s&

s!

s& s s!

$ s

$

%

# $

%

%

%

l l A

A

A A A

A A

A

P

l A

M

P M

−=

=−=

=

→

ρ

ρ

#! l l )25.02.0( −≈

!! #!

#l


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$- 2oundary element trans.erse reinforcement

'h

"

"h

$

" sh

'h

"" sh

A

AhS A

hS A

13.0

09.0

−≥

≥

"h

h

! "!

You must check it in two dimension

E l *


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E&amle *Design the following shear wall

!- %alculate &xternal 'oad

Shear Force

Moment

(l # 4=

(!# 2.0=


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( )

!e/ui!e"elementoun"a!

6)30(2.04.1810*06.1

28800

42.0

1500

06.112

42.0

12

26

33

M)a( MN

l !

I

"

## $

=>=

×+

×=

===

−

#- boundary element chec(

4)25.02.0(8.0 −≈

3.0=!!2.0=#!

4=#l


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,- 'ongitudinal einforcement



( )kN V kN A

%

""&6603.30

6

30)4000(200

6

=>==


6

""&

%

AV


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0- 1erify ade*uacy of shear wall section at itsbase under combined axial load and bendingmoment

{ } ( )

( )

1816element oun"a!*eac$o!25.38

5.6)13.12(100

10010000)101(01.0

01.0

9.0

21.05.1101

101500

31.02.2

4000101

108800

10113.08.022.08.024

101500

.108800

2

2

25.06.2

226

6

3

6

6

262

3

6

φ

ρ

ρ

γ

%se"(

"( A A A

"((( A A

ksi MPa A

P

ksi MPa

l A

M

((( A

N P

( N M

s& s s!

$ s

$

%

# $

%

$

%

%

=××−=−=

==×===

=

==××

=

==

×

×=

×==×+××−=

×=

×=

#! l l )25.02.0( −≈

!! #!

#l


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$- 2oundary element trans.erse reinforcement

'h

"

"h

$

" sh

'h

"" sh

A

AhS A

hS A

13.0

09.0

−≥

≥

"h

h

! "!

22

4.52.54342030)04)(120(09.009.0

4.20)8.13(230

4.0)8.13(280

120

"((( hS A

"(!

"(h

((S

'h

"" sh

"

"

===≥

=+−==+−=

=

Check short directionCheck short direction

Check long direction


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2

22

6.1

6.1160420

30)204)(120(09.009.0

4.20)8.13(230

4.0)8.13(280

120

"( A

"(((

!S A

"(!

"(h

((S

sh

'h

"" sh

"

"

=

===≥

=+−==+−=

=


in #$o!t "i!ection u#e 2 le'# clo#e $oo& 2 inte!nal le'#

in #$o!t "i!ection u#e 2 le'# clo#e $oo&

10φ

26.5)01.2(2)9.0(2 "(=+=

10φ

16φ

26.1)9.0(2 "(==


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Design E&amle

5.0

5.0

5.0

5.0

5.0 D.L=8kN/m2 L.L=2.5kN/m2

20.0

20.0

3.0

5.0

L

3.0

3.0 3.01

2

7.5

10

Design Shear wall ! 3 # in the 4gu

5a6a %ity


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20.0

20.0

3.0

5.0

L

3.0

3.0 3.01

2

7.5

10

!- calculate the length of shear wall !

∑∑

=

=n

i i

ii

I

' I '

1

( )

( )

10

+0166.012

2.0

5.

+08.212

2.05

1

33

1

2

3

2

−=

==

=

==

'

* *

I

'

I

( *

*

I I *

I I ' I ' I

I ' I '

n

i i

ii

5.4

0)10(0166.0)5.(08.2

0)10(0166.0)5.(08.2

1

3

21

3

21

2211

1

=

=−

=+

−+=

+

+==∑∑=

x

y


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( )

( )

( )

( )

kN

I R

W S V

kN W IS V

kN V

I Wall Shear R

T

kN W

S

I RT

W S V

DS

DS

D

D

1050

25.14

)16000(21.02.18516000)21.0)(25.1(0441.0)(0441.0

909)(55.0

1600010.0

25.1 4

55.0250488.0

16000)5)(20)(20(8

10.0

max

min

25.14

1

1

43

======

==

==

==

===

=

#- Seismic Shear for the total building

6or the total 2uildingT=0 55+ 0 5 02511 50+ −Tk


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T=0.55+ 0.5 025.112

5.0 =+= T k


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1 7 898 (:/ 7!;


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20.0

20.0

3.0

5.0

L

3.0

3.0 3.01

2

7.5

10

,- In this example we will neglect the eect of twist

For shear wall :o !

42.012

52.012

5.42.0

125.42.0

1

33

3

=+

==××

×

=∑

n

i

i

i

I

I a

(kN M a M kN V aV

i

i

.025162642.0)(38290942.0)(=×=×=

=×=×=

For shear wall :o #

58.0=a

(kN M a M

kN V aV

i

i

.902162658.0)(

52890958.0)(

=×=×==×=×=

1

2

0 Shear wall design


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( )

!e/ui!e"elementoun"a!

6)30(2.002.1210*519.1

25.2025

5.42.0

1500

519.112

5.42.0

12

23

33

M)a( MN

l !

I

"

##

$

=>=

×+

×=

===−

boundary element chec(

5.4)2.0(9.0 ×≈

(l #

5.4=

(!#

2.0=

0- Shear wall design


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'ongitudinal einforcement



( )kN V kN A

%

""&3826.821

6

30)4500(200

6

=>==


6

""&

%

AV


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1erify ade*uacy of shear wall section at itsbase under combined axial load and bendingmoment

( )

1616element oun"a!*eac$o!8.32

6.65)13.12(90

909000)109.0(01.001.0

9.0

24.066.1109.0

101500

25.03.14500109.0

10025

109.09.02.05.4

101500

.108800

2

2

25.0

)9.0(25.4

226

6

3

6

6

262

3

6

φ

ρ

ρ

γ

%se"(

"( A A A

"((( A A

ksi MPa A

P

ksi MPal A

M

((( A

N P

( N M

s& s s!

$ s

$

%

# $

%

$

%

%

=××−=−=

==×===

=

==××

=

==

×

×=

×==×=×=

×=

−

5.4)2.0(9.0 ×≈

(l #

5.4=

(!#

2.0=


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2oundary element trans.erse reinforcement

'h

"

"h

$

" sh

'h

"" sh

A

AhS A

hS A

13.0

09.0

−≥

≥

"h

h

! "!

22

2.59.51642030)804)(100(09.009.0

4.10)8.13(220

4.80)8.13(290

100

"((( hS A

"(!

"(h

((S

'h

"" sh

"

"

===≥

=+−==+−=

=

Check short directionCheck short direction



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in #$o!t "i!ection u#e 2 le'# clo#e $oo& 2 inte!nal le'#

in #$o!t "i!ection u#e 2 le'# clo#e $oo&

10φ

26.5)01.2(2)9.0(2 "(=+=

10φ

16φ

26.1)9.0(2 "(==

22

.08.66420

30)104)(100(09.009.0

4.10)8.13(220

4.80)8.13(290

100

"(((

hS A

"(!

"(h

((S

'h

"" sh

"

"

===≥

=+−==+−=

=

epeat the last procedure for Shear wall #


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shear wall design 2

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