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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
At least two curtains of reinforcement are needed
in the wall if the in-plane factored shear exceed a .alue of
( )kN V kN A
%
""&6603.30
6
30)4000(200
6
=>==
/inimum reinforcement
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
"
"
=
===≥
=+−==+−=
=
Check long direction
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
At least two curtains of reinforcement are needed
in the wall if the in-plane factored shear exceed a .alue of
( )kN V kN A
%
""&3826.821
6
30)4500(200
6
=>==
/inimum reinforcement
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
Check long direction
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Check long direction
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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