seismic design of concrete structure. seismic design of concrete structure earthquakes occur in many...
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Seismic Design of Concrete Structure
Seismic Design of Concrete Structure
• Earthquakes occur in many regions of the world. In certain locations where the intensity of the ground shaking is small, the designer does not have to consider seismic effects.
• In other locations-particularly in regions near an active geological fault (a fracture line in the rock structure), such as the San Andreas fault that runs along the western coast of California-large ground motions frequently occur that can damage or destroy buildings and bridges in large areas of cities
• Assuming the building is fixed at its base, the displacement of floors will vary from zero at the base to a maximum at the roof
Earthquake focus and epicenter
Distribution of magnitude 5 or greater earthquakes,
1980 - 1990
There are several analytical procedures to determine the magnitude of the base shear for which buildings must be designed, we will
only consider the equivalent lateral force procedure, described in the ANSI/ASCE and UBC standard. Using this procedure, we compute the magnitude of the base shear as
21.0
10.01
DS
D
S
S
Palestine
Table 2:Design Coefficients & Factors for Basic Seismic-Force-Resisting Systems
Story Drift & The P-Delta Effect
Allowable Story Drift,
Important Factor
Overturning
Example 1• Determine the design seismic forces acting at each
floor of the six-story office building in Figure below.• D.L=8kN/m2• L.L=2.5kN/m2• Shear wall system
18m
25m
N
kN
IR
WSV
kNWISV
kNV
I
WallShearR
T
kNW
S
IRT
WSV
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(8
10.0
max
min
14
1
1
43
T=0.462 < 0.5 then k=1
Note:Column 4 = column 3a * Column 2
Fx = column 5 * (V = 1134) ((
Vx = cumulative column 6Mx = moment from Fx can calculated from shear Area
Shear Wall Design
Shear walls provide a high in-plane stiffness and strength for both lateral and gravity loads, and are ideally suitable for tall buildings, especially those conceived in reinforced concrete. Tall buildings designed to carry the entire lateral loading through shear walls can be economical to heights of around 40 stories. Taller structures should combine shear walls with other structural systems.
Introduction
Shear Wall design
Shear Wall Design Steps
I
MC
A
Pf
lbbhI
c
wwg
1212
33
'2.0 cc ff
1- Calculate External Load
u
u
u
P
V
M
2- boundary element check
Then boundary elements are not requiredif
wb ll )25.02.0(
wl
wb
wb ll )25.02.0(
bbwb
wl
Then choose the one of:
3- Longitudinal Reinforcement
At least two curtains of reinforcement are needed in the wall if the in-plane factored shear exceed a value of
6
'ccv fA
wwcv blA
ync
cvnuw
w
ync
cvnuw
w
ff
AVVl
hif
ff
AVVl
hif
4 5.1
6 2
'
'
6
4
6.0
'
maxccv
n
fAV
Minimum reinforcement
6
0025.0
0025.0 6
'
'
ccvu
h
v
ccvu
fAVif
fAVif
0025.0
0015.0
16bar for
0020.0
0012.0
16bar for
h
v
h
v
4- Verify adequacy of shear wall section at its base under combined axial load and bending moment
)2(in ent reinforcem of area
elementsboundary in ent reinforcem of area
diagramn interactio,
bwsv
sb
svssb
gs
g
u
wg
u
u
u
llA
A
AAA
AA
A
P
lA
M
P
M
wb ll )25.02.0(
bbwb
wl
5- Boundary element transverse reinforcement
yh
c
ch
gcsh
yh
ccsh
f
f
A
AhSA
f
fhSA
'
'
1 3.0
09.0
ch
h
b cb
You must check it in two dimension
Example 1 Design the following shear wall
1- Calculate External Load
Shear Force
Moment
mlw 4
mbw 2.0
requiredelement Boundary
6)30(2.0/4.1810*067.1
28800
42.0
1500
067.112
42.0
12
26
33
MpamMNf
lbI
c
wwg
2- boundary element check
4)25.02.0(8.0
3.0bb2.0wb
4wl
3- Longitudinal Reinforcement
At least two curtains of reinforcement are needed in the wall if the in-plane factored shear exceed a value of
kNVkNfA
uccv 6603.730
6
30)4000(200
6
'
Minimum reinforcement
6
'ccv
u
fAV
directioneach 25@12
45
44002001000002.00020.0
directioneach 25@12
45
4.224020010000012.00012.0
16bar for
max
22
max
22
cmuse
cmS
cmmmA
cmuse
cmS
cmmmA
svh
svv
4- Verify adequacy of shear wall section at its base under combined axial load and bending moment
1816 element boundary each for 25.38
5.76)13.12(100
10010000)101(01.0
01.0
9.0
21.05.1101
101500
31.02.24000101
108800
10113.08.022.08.024
101500
.108800
2
225.06.2
226
6
3
6
6
262
3
6
usecm
cmAAA
cmmmAA
ksiMPaA
P
ksiMPalA
M
mmmA
NP
mNM
svssb
gs
g
u
wg
u
g
u
u
wb ll )25.02.0(
bbwb
wl
5- Boundary element transverse reinforcement
yh
c
ch
gcsh
yh
ccsh
f
f
A
AhSA
f
fhSA
'
'
1 3.0
09.0
ch
h
b cb
22'
4.52.543420
30)704)(120(09.0 09.0
4.20)8.13(230
4.70)8.13(280
120
cmmmf
fhSA
cmb
cmh
mmS
yh
ccsh
c
c
Check short directionCheck short direction
Check long direction
2
22'
6.1
6.1160420
30)204)(120(09.0 09.0
4.20)8.13(230
4.70)8.13(280
120
cmA
cmmmf
fbSA
cmb
cmh
mmS
sh
yh
ccsh
c
c
Check long direction
in short direction use 2 legs close hoop & 2 internal legs
in short direction use 2 legs close hoop
1026.5)01.2(2)79.0(2 cm
10
16
26.1)79.0(2 cm
Design Example
5.0
5.0
5.0
5.0
5.0D.L=8kN/m2L.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 1 & 2 in the figure belowGaza City
20.0
20.0
3.0
5.0
L
3.0
3.0 3.01
2
7.5
10
1- calculate the length of shear wall 1
n
i i
ii
I
yIy
1
10
,0166.012
2.0
5.7
,08.212
2.05
1
33
1
2
3
2
y
LL
I
y
I
mL
L
II
L
II
yIyI
I
yIy
n
i i
ii
5.4
0)10(0166.0)5.7(08.2
0)10(0166.0)5.7(08.2
1
3
21
3
21
2211
1
x
y
kN
IR
WSV
kNWISV
kNV
I
WallShearR
T
kNW
S
IRT
WSV
DS
DS
D
D
1050
25.14
)16000(21.0
2.18516000)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
2- Seismic Shear for the total building
For the total BuildingT=0.55> 0.5 025.11 2
5.0 Tk
V = 909 kNM =16726 kN.m
For shear walls
)(05.0.
)(
)(
22
1
LRCxIyI
yIb
I
Ia
MbaM
VbaV
iixiiy
ii
n
ii
i
i
i
x
y
20.0
20.0
3.0
5.0
L
3.0
3.0 3.01
2
7.5
10
In this example without neglect the effect of twist force
For shear wall No. 1
42.012
52.012
5.42.0
125.42.0
1
33
3
n
ii
i
I
Ia
1
2
024.0
)2005.0()10(4)5.7()10(
102
1232.02
1252.02
125.42.0
125.42.0
333
3
b
b
20.0
20.0
3.0
5.0
L
3.0
3.0 3.01
2
7.5
10
3- In this example we will neglect the effect of twist force
For shear wall No. 1
42.012
52.012
5.42.0
125.42.0
1
33
3
n
ii
i
I
Ia
mkNMaM
kNVaV
i
i
.70251672642.0)(
38290942.0)(
For shear wall No. 2
58.0a
mkNMaM
kNVaV
i
i
.97021672658.0)(
52890958.0)(
1
2
requiredelement Boundary
6)30(2.0/072.1210*519.1
25.27025
5.42.0
1500
519.112
5.42.0
12
23
33
MpamMNf
lbI
c
wwg
boundary element check
5.4)2.0(9.0
mlw 5.4
mbw 2.0
4- Shear wall design
Longitudinal Reinforcement
At least two curtains of reinforcement are needed in the wall if the in-plane factored shear exceed a value of
kNVkNfA
uccv 3826.821
6
30)4500(200
6
'
Minimum reinforcement
6
'ccv
u
fAV
directioneach 25@12
45
44002001000002.00020.0
directioneach 25@12
45
4.224020010000012.00012.0
16bar for
max
22
max
22
cmuse
cmS
cmmmA
cmuse
cmS
cmmmA
svh
svv
Verify adequacy of shear wall section at its base under combined axial load and bending moment
1616 element boundary each for 8.32
6.65)13.12(90
909000)109.0(01.0
01.0
9.0
24.066.1109.0
101500
25.073.14500109.0
107025
109.09.02.05.4
101500
.108800
2
225.0
)9.0(25.4
226
6
3
6
6
262
3
6
usecm
cmAAA
cmmmAA
ksiMPaA
P
ksiMPalA
M
mmmA
NP
mNM
svssb
gs
g
u
wg
u
g
u
u
5.4)2.0(9.0
mlw 5.4
mbw 2.0
Boundary element transverse reinforcement
yh
c
ch
gcsh
yh
ccsh
f
f
A
AhSA
f
fhSA
'
'
1 3.0
09.0
ch
h
b cb
22'
2.59.516420
30)804)(100(09.0 09.0
4.10)8.13(220
4.80)8.13(290
100
cmmmf
fhSA
cmb
cmh
mmS
yh
ccsh
c
c
Check short directionCheck short direction
Check long direction
Check long direction
in short direction use 2 legs close hoop & 2 internal legs
in short direction use 2 legs close hoop
1026.5)01.2(2)79.0(2 cm
10
16
26.1)79.0(2 cm
22'
7.08.66420
30)104)(100(09.0 09.0
4.10)8.13(220
4.80)8.13(290
100
cmmmf
fhSA
cmb
cmh
mmS
yh
ccsh
c
c
Repeat the last procedure for Shear wall 2
